The real treasure was the numbers we made along the way
22
$begingroup$
Your task is to write a program, function or snippet (yes, snippets are allowed) that simply outputs an integer. However, you must be able to separate your submission into prefixes that also produce distinct integers. You cannot use any bytes that have appeared in previous prefixes. For example, we can have the prefixes:
1 # 1 (Now we can't use 1)
1-6 # -5 (Now we can't use - or 6)
1-6/3 # -1 (Now we can't use / or 3)
1-6/3+0xA # 9 Final submission
Rules
- Your goal is to create to try and create the most unique integers, while keeping them close to zero.
- The scoring system is
((number of unique integers)**3)/(sum of absolute values), where the higher your score, the better. The above example scores $(4^3)/(1+lvert-5rvert+lvert-1rvert+9) = 64/16 = 4$.
- The scoring system is
- There should be at least two unique integers (no dividing by zero!)
- Please format your answer similar to:
# Language, $(4^{3})/16 = 4$
1-6/3+0xA (the full program)
- Mention if your submission is composed of snippets that evaluate to a value, functions or full programs.
- List each of the prefixes and (optionally) an explanation for how they work.
number code-challenge restricted-source
asked Feb 11 at 0:49
Jo KingJo King
23.1k253121
$endgroup$
|
show 10 more comments
22
$begingroup$
Your task is to write a program, function or snippet (yes, snippets are allowed) that simply outputs an integer. However, you must be able to separate your submission into prefixes that also produce distinct integers. You cannot use any bytes that have appeared in previous prefixes. For example, we can have the prefixes:
1 # 1 (Now we can't use 1)
1-6 # -5 (Now we can't use - or 6)
1-6/3 # -1 (Now we can't use / or 3)
1-6/3+0xA # 9 Final submission
Rules
- Your goal is to create to try and create the most unique integers, while keeping them close to zero.
- The scoring system is
((number of unique integers)**3)/(sum of absolute values), where the higher your score, the better. The above example scores $(4^3)/(1+lvert-5rvert+lvert-1rvert+9) = 64/16 = 4$.
- The scoring system is
- There should be at least two unique integers (no dividing by zero!)
- Please format your answer similar to:
# Language, $(4^{3})/16 = 4$
1-6/3+0xA (the full program)
- Mention if your submission is composed of snippets that evaluate to a value, functions or full programs.
- List each of the prefixes and (optionally) an explanation for how they work.
number code-challenge restricted-source
asked Feb 11 at 0:49
Jo KingJo King
23.1k253121
$endgroup$
2
$begingroup$
The Final submission line appears to have swapped the 3 for a 2 (I haven't edited to correct this in case I've misunderstood)
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– trichoplax
Feb 11 at 1:23
3
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It looks to me like you can get a near-optimal score in Python by writingprint 123-len(open(__file__).read())followed by#then one of each byte not used so far (except some control characters), with prefixes chopping those bytes. Adjust 123 to whatever centers the range of outputs around 0. For a better score, the non-comment part of the program can be rewritten using only the 7 distinct characters that are universal for Python. Is this all valid?
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– xnor
Feb 11 at 1:37
1
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@JoKing I do include negatives, when the difference is negative.
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– xnor
Feb 11 at 1:41
1
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@JoKing This method also be done in languages where a string can be written without a closing quote.
$endgroup$
– xnor
Feb 11 at 1:47
2
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I can't believe so few other people noticed it and upvoted the earlier comment. :P I just noticed that @trichoplax spotted the typo, too.
$endgroup$
– Peter Cordes
Feb 12 at 11:07
|
show 10 more comments
22
22
22
2
$begingroup$
Your task is to write a program, function or snippet (yes, snippets are allowed) that simply outputs an integer. However, you must be able to separate your submission into prefixes that also produce distinct integers. You cannot use any bytes that have appeared in previous prefixes. For example, we can have the prefixes:
1 # 1 (Now we can't use 1)
1-6 # -5 (Now we can't use - or 6)
1-6/3 # -1 (Now we can't use / or 3)
1-6/3+0xA # 9 Final submission
Rules
- Your goal is to create to try and create the most unique integers, while keeping them close to zero.
- The scoring system is
((number of unique integers)**3)/(sum of absolute values), where the higher your score, the better. The above example scores $(4^3)/(1+lvert-5rvert+lvert-1rvert+9) = 64/16 = 4$.
- The scoring system is
- There should be at least two unique integers (no dividing by zero!)
- Please format your answer similar to:
# Language, $(4^{3})/16 = 4$
1-6/3+0xA (the full program)
- Mention if your submission is composed of snippets that evaluate to a value, functions or full programs.
- List each of the prefixes and (optionally) an explanation for how they work.
number code-challenge restricted-source
asked Feb 11 at 0:49
Jo KingJo King
23.1k253121
$endgroup$
Your task is to write a program, function or snippet (yes, snippets are allowed) that simply outputs an integer. However, you must be able to separate your submission into prefixes that also produce distinct integers. You cannot use any bytes that have appeared in previous prefixes. For example, we can have the prefixes:
1 # 1 (Now we can't use 1)
1-6 # -5 (Now we can't use - or 6)
1-6/3 # -1 (Now we can't use / or 3)
1-6/3+0xA # 9 Final submission
Rules
- Your goal is to create to try and create the most unique integers, while keeping them close to zero.
- The scoring system is
((number of unique integers)**3)/(sum of absolute values), where the higher your score, the better. The above example scores $(4^3)/(1+lvert-5rvert+lvert-1rvert+9) = 64/16 = 4$.
- The scoring system is
- There should be at least two unique integers (no dividing by zero!)
- Please format your answer similar to:
# Language, $(4^{3})/16 = 4$
1-6/3+0xA (the full program)
- Mention if your submission is composed of snippets that evaluate to a value, functions or full programs.
- List each of the prefixes and (optionally) an explanation for how they work.
number code-challenge restricted-source
number code-challenge restricted-source
asked Feb 11 at 0:49
Jo KingJo King
23.1k253121
asked Feb 11 at 0:49
Jo KingJo King
23.1k253121
edited Feb 12 at 11:11
Jo King
asked Feb 11 at 0:49
Jo KingJo King
23.1k253121
asked Feb 11 at 0:49
Jo KingJo King
23.1k253121
asked Feb 11 at 0:49
Jo KingJo King
23.1k253121
23.1k253121
2
$begingroup$
The Final submission line appears to have swapped the 3 for a 2 (I haven't edited to correct this in case I've misunderstood)
$endgroup$
– trichoplax
Feb 11 at 1:23
3
$begingroup$
It looks to me like you can get a near-optimal score in Python by writingprint 123-len(open(__file__).read())followed by#then one of each byte not used so far (except some control characters), with prefixes chopping those bytes. Adjust 123 to whatever centers the range of outputs around 0. For a better score, the non-comment part of the program can be rewritten using only the 7 distinct characters that are universal for Python. Is this all valid?
$endgroup$
– xnor
Feb 11 at 1:37
1
$begingroup$
@JoKing I do include negatives, when the difference is negative.
$endgroup$
– xnor
Feb 11 at 1:41
1
$begingroup$
@JoKing This method also be done in languages where a string can be written without a closing quote.
$endgroup$
– xnor
Feb 11 at 1:47
2
$begingroup$
I can't believe so few other people noticed it and upvoted the earlier comment. :P I just noticed that @trichoplax spotted the typo, too.
$endgroup$
– Peter Cordes
Feb 12 at 11:07
|
show 10 more comments
2
$begingroup$
The Final submission line appears to have swapped the 3 for a 2 (I haven't edited to correct this in case I've misunderstood)
$endgroup$
– trichoplax
Feb 11 at 1:23
3
$begingroup$
It looks to me like you can get a near-optimal score in Python by writingprint 123-len(open(__file__).read())followed by#then one of each byte not used so far (except some control characters), with prefixes chopping those bytes. Adjust 123 to whatever centers the range of outputs around 0. For a better score, the non-comment part of the program can be rewritten using only the 7 distinct characters that are universal for Python. Is this all valid?
$endgroup$
– xnor
Feb 11 at 1:37
1
$begingroup$
@JoKing I do include negatives, when the difference is negative.
$endgroup$
– xnor
Feb 11 at 1:41
1
$begingroup$
@JoKing This method also be done in languages where a string can be written without a closing quote.
$endgroup$
– xnor
Feb 11 at 1:47
2
$begingroup$
I can't believe so few other people noticed it and upvoted the earlier comment. :P I just noticed that @trichoplax spotted the typo, too.
$endgroup$
– Peter Cordes
Feb 12 at 11:07
2
2
$begingroup$
The Final submission line appears to have swapped the 3 for a 2 (I haven't edited to correct this in case I've misunderstood)
$endgroup$
– trichoplax
Feb 11 at 1:23
$begingroup$
The Final submission line appears to have swapped the 3 for a 2 (I haven't edited to correct this in case I've misunderstood)
$endgroup$
– trichoplax
Feb 11 at 1:23
3
3
$begingroup$
It looks to me like you can get a near-optimal score in Python by writing
print 123-len(open(__file__).read()) followed by # then one of each byte not used so far (except some control characters), with prefixes chopping those bytes. Adjust 123 to whatever centers the range of outputs around 0. For a better score, the non-comment part of the program can be rewritten using only the 7 distinct characters that are universal for Python. Is this all valid?$endgroup$
– xnor
Feb 11 at 1:37
$begingroup$
It looks to me like you can get a near-optimal score in Python by writing
print 123-len(open(__file__).read()) followed by # then one of each byte not used so far (except some control characters), with prefixes chopping those bytes. Adjust 123 to whatever centers the range of outputs around 0. For a better score, the non-comment part of the program can be rewritten using only the 7 distinct characters that are universal for Python. Is this all valid?$endgroup$
– xnor
Feb 11 at 1:37
1
1
$begingroup$
@JoKing I do include negatives, when the difference is negative.
$endgroup$
– xnor
Feb 11 at 1:41
$begingroup$
@JoKing I do include negatives, when the difference is negative.
$endgroup$
– xnor
Feb 11 at 1:41
1
1
$begingroup$
@JoKing This method also be done in languages where a string can be written without a closing quote.
$endgroup$
– xnor
Feb 11 at 1:47
$begingroup$
@JoKing This method also be done in languages where a string can be written without a closing quote.
$endgroup$
– xnor
Feb 11 at 1:47
2
2
$begingroup$
I can't believe so few other people noticed it and upvoted the earlier comment. :P I just noticed that @trichoplax spotted the typo, too.
$endgroup$
– Peter Cordes
Feb 12 at 11:07
$begingroup$
I can't believe so few other people noticed it and upvoted the earlier comment. :P I just noticed that @trichoplax spotted the typo, too.
$endgroup$
– Peter Cordes
Feb 12 at 11:07
|
show 10 more comments
16 Answers
16
active
oldest
votes
18
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TI-BASIC (TI-84+ CE), $238^3/14161 = 952.0$
118-length("+×/ABCDEFGHIJK...
The 256 bytes in TI-BASIC break down thusly:
- 240 bytes are one-byte tokens that can be quoted
- 11 are the start of two-byte tokens
- 2 are unused
- The remaining 3 are the
",→, and newline characters which break strings.
The string contains all 237 allowable bytes in the first category not already in 118-length(.
The program calculates 238 distinct integers, ranging from 118 down to -119. The score is therefore $$ frac{238^3}{sum_{n = -119}^{118} |n|} = 952.$$
answered Feb 11 at 2:58
lirtosiastlirtosiast
18.4k437109
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1
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Wouldn't the string have to end with " which is a byte that was already used? Or why not? I am a bit confused by this also in other answers this seems to happen so probably I miss something.
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– findusl
Feb 13 at 8:42
2
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@findusl TI-BASIC has a tendency to automatically complete parentheses and strings for you. so you could writemean({1,2,3, which would complete tomean({1,2,3}). strings are also completed:Disp "Hi!completes toDisp "Hi!". Therefore the final closing quote is unnecessary
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– Conor O'Brien
Feb 13 at 12:23
add a comment |
13
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vim, 14³ / 53 = 51.77
i10<esc>X<c-x>r2hx<c-a>Pa<del>4<c-c>d^s5<c-o>I-<c-o>$<c-o>C6<c-h>7
Try it online! (all prefixes)
Breakdown:
i1 1
0 10
<esc>X 0
<c-x> -1 <c-x> decrements
r2 -2
hx 2 hx is X, but we've already used that
<c-a> 3 <c-a> increments
P -3 paste the - from two steps ago
a<del>4 -4
<c-c>d^ 4 <c-c> is an <esc> alternative
s5 5
<c-o>I-<c-o>$<c-o> -5 <c-o> allows a single normal mode input
C6 -6 trailing <c-o> from previous line used here
<c-h>7 -7 <c-h> = backspace
answered Feb 11 at 2:20
Doorknob♦Doorknob
54.8k17114350
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are you not reusing the < byte? Or is that a placeholder for a different byte?
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– findusl
Feb 13 at 13:20
1
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@findusl no -<esc>represents the escape key and<c-char>represents ctrl+char
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– Doorknob♦
Feb 13 at 14:20
add a comment |
10
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Lenguage, 256³ / 16384 = 1024
The source code of the solution can be described as follows:
183662808731984092 0x00 bytes
4775233027031586288 0x01 bytes
27 0x02 bytes
81 0x03 bytes
128931291729852829668 0x04 bytes
108 0x05 bytes
648 0x06 bytes
81 0x07 bytes
2079 0x08 bytes
27 0x09 bytes
3481144876706026401009 0x0A bytes
108 0x0B bytes
2808 0x0C bytes
27 0x0D bytes
17496 0x0E bytes
81 0x0F bytes
2079 0x10 bytes
27 0x11 bytes
56106 0x12 bytes
756 0x13 bytes
93990911671062712827216 0x14 bytes
108 0x15 bytes
2808 0x16 bytes
27 0x17 bytes
75789 0x18 bytes
756 0x19 bytes
474552 0x1A bytes
27 0x1B bytes
81 0x1C bytes
56025 0x1D bytes
756 0x1E bytes
1514835 0x1F bytes
20439 0x20 bytes
2537754615118693246334805 0x21 bytes
108 0x22 bytes
2808 0x23 bytes
27 0x24 bytes
75789 0x25 bytes
756 0x26 bytes
2046276 0x27 bytes
20439 0x28 bytes
12812877 0x29 bytes
108 0x2A bytes
729 0x2B bytes
2079 0x2C bytes
27 0x2D bytes
1512648 0x2E bytes
40920957 0x2F bytes
1689866867948183303767857 0x30 bytes
66829507740256534347823839 0x31 bytes
2808 0x32 bytes
27 0x33 bytes
75789 0x34 bytes
756 0x35 bytes
2046276 0x36 bytes
20439 0x37 bytes
55249425 0x38 bytes
551880 0x39 bytes
345947652 0x3A bytes
2916 0x3B bytes
27 0x3C bytes
19656 0x3D bytes
56133 0x3E bytes
40842225 0x3F bytes
1104865839 0x40 bytes
45626405434600949201732247 0x41 bytes
1804396708986926427391243545 0x42 bytes
27 0x43 bytes
75789 0x44 bytes
756 0x45 bytes
2046276 0x46 bytes
20439 0x47 bytes
55249425 0x48 bytes
551880 0x49 bytes
1491734448 0x4A bytes
9355487364 0x4B bytes
78732 0x4C bytes
756 0x4D bytes
530685 0x4E bytes
1515591 0x4F bytes
1102740075 0x50 bytes
29831377653 0x51 bytes
1231912946734225628446770696 0x52 bytes
48718711142647013539563576444 0x53 bytes
2066715 0x54 bytes
55801305 0x55 bytes
1506635235 0x56 bytes
292874988897 0x57 bytes
2125764 0x58 bytes
14348907 0x59 bytes
40920957 0x5A bytes
29773982025 0x5B bytes
33261649561824092773510005450 0x5C bytes
1315405200851469365568216563988 0x5D bytes
55801305 0x5E bytes
1506635235 0x5F bytes
7948303851537 0x60 bytes
57395628 0x61 bytes
387420489 0x62 bytes
1104865839 0x63 bytes
803897514675 0x64 bytes
898064538169250504884770147177 0x65 bytes
35515940422989672870341847227676 0x66 bytes
1506635235 0x67 bytes
40679151345 0x68 bytes
214604203991472 0x69 bytes
1549681956 0x6A bytes
10460353203 0x6B bytes
29831377653 0x6C bytes
24001245795414021635055080475036 0x6D bytes
246496735155742018538946394995 0x6E bytes
958930391420721167499229875147252 0x6F bytes
40679151345 0x70 bytes
5795411844856032 0x71 bytes
41841412812 0x72 bytes
282429536481 0x73 bytes
805447196631 0x74 bytes
648033636476178584146487172825972 0x75 bytes
108 0x76 bytes
6655411849205034500551552664784 0x77 bytes
25891120568359471522479206628975804 0x78 bytes
1098337086315 0x79 bytes
156484875126743748 0x7A bytes
21747074309037 0x7B bytes
17496908184856821771955153666301244 0x7C bytes
108 0x7D bytes
2808 0x7E bytes
27 0x7F bytes
81 0x80 bytes
179696119928535931514891921949087 0x81 bytes
699060255345705731106938578982346708 0x82 bytes
4225121283523411674 0x83 bytes
472416520991134187843376319996477587 0x84 bytes
108 0x85 bytes
2808 0x86 bytes
27 0x87 bytes
75789 0x88 bytes
108 0x89 bytes
729 0x8A bytes
2079 0x8B bytes
27 0x8C bytes
4851795238070470150902081892625349 0x8D bytes
18874626894334054739887341632523361116 0x8E bytes
114078274655132115171 0x8F bytes
12755246066760623071771160639904894849 0x90 bytes
108 0x91 bytes
2808 0x92 bytes
27 0x93 bytes
75789 0x94 bytes
756 0x95 bytes
2046276 0x96 bytes
2916 0x97 bytes
27 0x98 bytes
19656 0x99 bytes
56133 0x9A bytes
130998471427902694074356211100885179 0x9B bytes
854006569949556303994892977044130020645 0x9C bytes
108 0x9D bytes
2808 0x9E bytes
27 0x9F bytes
75789 0xA0 bytes
756 0xA1 bytes
2046276 0xA2 bytes
55269864 0xA3 bytes
78732 0xA4 bytes
756 0xA5 bytes
530685 0xA6 bytes
1515591 0xA7 bytes
3536958728553372740007617699723899860 0xA8 bytes
23058177388638020207862110380191510557388 0xA9 bytes
108 0xAA bytes
2808 0xAB bytes
27 0xAC bytes
75789 0xAD bytes
756 0xAE bytes
2046276 0xAF bytes
20439 0xB0 bytes
55249425 0xB1 bytes
1492286328 0xB2 bytes
2125764 0xB3 bytes
14348907 0xB4 bytes
40920957 0xB5 bytes
622666287378897486676257185943063330345696 0xB6 bytes
108 0xB7 bytes
2808 0xB8 bytes
27 0xB9 bytes
75789 0xBA bytes
756 0xBB bytes
2046276 0xBC bytes
20439 0xBD bytes
55249425 0xBE bytes
1492286328 0xBF bytes
40291730856 0xC0 bytes
57395628 0xC1 bytes
387420489 0xC2 bytes
1104865839 0xC3 bytes
16811989759230232140258944020462709919333792 0xC4 bytes
108 0xC5 bytes
2808 0xC6 bytes
27 0xC7 bytes
75789 0xC8 bytes
756 0xC9 bytes
2046276 0xCA bytes
20439 0xCB bytes
55249425 0xCC bytes
551880 0xCD bytes
1491734448 0xCE bytes
40291730856 0xCF bytes
1087876733112 0xD0 bytes
1549681956 0xD1 bytes
10460353203 0xD2 bytes
29831377653 0xD3 bytes
453923723499216267786991488552493167822012492 0xD4 bytes
2808 0xD5 bytes
27 0xD6 bytes
75789 0xD7 bytes
756 0xD8 bytes
2046276 0xD9 bytes
20439 0xDA bytes
55249425 0xDB bytes
551880 0xDC bytes
1491734448 0xDD bytes
40291730856 0xDE bytes
1087876733112 0xDF bytes
29372671794024 0xE0 bytes
324270949293 0xE1 bytes
805447196631 0xE2 bytes
12255940534478839230248770190917315531194337311 0xE3 bytes
76545 0xE4 bytes
2066715 0xE5 bytes
55801305 0xE6 bytes
1506635235 0xE7 bytes
832318279426764 0xE8 bytes
21747074309037 0xE9 bytes
330910394430928659216716795154767519342247107424 0xEA bytes
2066715 0xEB bytes
55801305 0xEC bytes
1506635235 0xED bytes
22472634223673946 0xEE bytes
8934580649635073798851353469178723609411678244474 0xEF bytes
55801305 0xF0 bytes
1506635235 0xF1 bytes
40679151345 0xF2 bytes
606761124039196515 0xF3 bytes
241233677540146992568986543667825537454115312600825 0xF4 bytes
1506635235 0xF5 bytes
40679151345 0xF6 bytes
6513309293583968799362636679031305893812560835614495 0xF7 bytes
40679151345 0xF8 bytes
1098337086315 0xF9 bytes
175859350926767157582791190333845259132939142561591365 0xFA bytes
1098337086315 0xFB bytes
4748202475022713254735362139013821996589386504264297360 0xFC bytes
29655101330505 0xFD bytes
128201466825613257877854777753373193907913435615136028720 0xFE bytes
3461439604291557962702078999341076235513663562296408699075 0xFF bytes
Each run of identical bytes combined with the previous runs constitutes a prefix, and the full program is all the strings described above concatenated.
This is an optimal score. (The programs print all numbers from -127 to 128 inclusive.)
To create this answer, first, I copied the Java answer here for generating brainfuck programs to print a string. I ran it with this main function:
public static void main(String args) {
for (int i = -127; i <= 128; ++i) generate(""+i);
}
Then, I ran this Ruby script to generate the human-readable text shown here:
#!/usr/bin/ruby
nums = File.open('f').map do |line|
line.chomp.sub(/^+/, '>+').chars.map{|ch|
'+-><.,'.index(ch).to_s(3).rjust(3,?0)
}.join.to_i 3
end
puts ([0]+nums.sort).each_cons(2).map.with_index{|x,i|"#{x[1]-x[0]} 0x#{i.to_s(16).rjust(2,?0).upcase} bytes"}*"n"
The largest program is 3594571896764310192036774345469579167648804468538578264427 bytes long (and prints -90).
answered Feb 11 at 3:17
Doorknob♦Doorknob
54.8k17114350
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1
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Props for the speed and practical problem-solving skill!
$endgroup$
– lirtosiast
Feb 11 at 3:30
add a comment |
6
$begingroup$
Javascript, $(111^{3})/2530 = 540.5656126482213$
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(Credit to this interesting utility for the conversion.) The above effectively evaluates into the following code:
(new Proxy({},{get:(a,p,b,n=p.length+3)=>n%2?(n-1)/2:-n/2})).$0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyzÀāłƃDŽȅɆʇˈ̉͊βόЍюҏӐԑՒ֓הٖؕڗۘܙݚޛߜరᡡⱱ㣣䤤奥榦秧訨驩ꪪ뫫쬬흝ﯯ&''|""**"",~{}
...which creates a proxy that converts the length of a property name into a number. The first prefix ends one character after the ., and uses only 6 unique characters, excluding the .$. The next 63 slices append 63 other single-byte characters which are valid in JS object property names. The next 44 slices append two-byte valid characters made up of unused bytes. The final slices append &'', |""**"", and ,~{}. The first slice returns -2, the second 2, then -3, then 3, -4, 4, -5, etc. up to the last three slices which are 0, 1, and -1.
answered Feb 11 at 7:58
Yair RandYair Rand
3415
$endgroup$
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3
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Brain-Flak, $4^3/4=16$
For this one we use brain-flak snippets so we are concerned with the value not what they do.
Complete program:
<>(())[]{}
Prefixes and output
0 : <>
1 : <>(())
-2 : <>(())[]
-1 : <>(())[]{}
Try it online!
Explanation
This is probably the best a brain-flak answer can get since we've used up all the character Brain-flak cares about, and we can't cluster the numbers closer to zero.
The first snippet is <>, <> changes the stack but otherwise evaluates to zero, so we start with zero. From here we add (()) this makes one. We could have just done (), because that is also one, but if we push a one now we will be able to recall it later with {} and . This is pretty much the only way {} and can create value so we do that. Since there is one thing on the stack counts as a 1 so we subtract 3 from the total with [] bringing us to -2. Lastly we use {} to pop the one we pushed earlier bringing us up to -1.
answered Feb 11 at 20:54
TRITICIMAGVSTRITICIMAGVS
35.1k10158369
$endgroup$
add a comment |
3
$begingroup$
Python 2, 32 34.7 40 $(11^{3})/32 = 41.59375$
After two hours of trial and error I finally got all integers from -4 to +5 (and 7).
0^cmp('','_')&3+1/~9*2-[-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]][==[]]|6<<7<<7>>8>>8and 5
This is a snippet with the following prefix snippets:
# -1
cmp(cmp,())
# +3 = (-1) & 3
cmp(cmp,())&3
# +7 = 3 + 4
cmp(cmp,())&3+4
# +2 = 3 + (-1) = 3 + floor(4/-10)
cmp(cmp,())&3+4/~9
# +1 = 3 + (-2) = 3 + 2*floor(4/-10)
cmp(cmp,())&3+4/~9*2
# -4 = 1 - 5
cmp(cmp,())&3+4/~9*2-1-1-1-1-1
# -2 = (-4) | 6
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6
# -3 = (-4) | 1 = (-4) | floor(6/2) = (-4) | (6>>2)
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8
# 0 = -3 and 0
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 0
# +5 = -3 and 05
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 05
# +4 = 5 ^ 1
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 05^True
Try it online!
answered Feb 12 at 11:32
ovsovs
19k21159
$endgroup$
1
$begingroup$
You can change the2*2to22&7(-3&6 instead of -3&4) to fix that repetition of*.
$endgroup$
– Kevin Cruijssen
Feb 12 at 11:47
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@KevinCruijssen thanks for the suggestion.
$endgroup$
– ovs
Feb 12 at 11:54
$begingroup$
You can free up'and_by usingcmp(cmp,())instead ofcmp('','_')If that is helpful.
$endgroup$
– TRITICIMAGVS
Feb 13 at 3:46
$begingroup$
You could also make the list section a little shorter and a lot easier to understand by doing--[-1-1-1-1-1], but that doesn't really improve your score.
$endgroup$
– TRITICIMAGVS
Feb 13 at 3:54
add a comment |
2
$begingroup$
Python 2, $11^3/53 = 25.1132075472$
015-8*2+6/3&9%7ifelse 4^ord('b')
Prefixes:
0 # 0
01 # 1
015 # 13 (octal)
015-8 # 5
015-8*2 # -3
015-8*2+6 # 3
015-8*2+6/3 # -1
015-8*2+6/3&9 # 9 (-1 bitwise-AND 9 = 9)
015-8*2+6/3&9%7 # 2
015-8*2+6/3&9%7ifelse 4 # 4
015-8*2+6/3&9%7ifelse 4^ord('b') # 12 (4 bitwise-XOR 8 -- OR would also work)
Not optimal, but the best I could personally find with this particular structure.
answered Feb 11 at 2:17
Aidan F. PierceAidan F. Pierce
833616
$endgroup$
add a comment |
2
$begingroup$
Python 2 $6^3/9 = 24$
False-False**0&1+[<[]][<]^~3>>2
Here are the prefixes and what they output
0 : False-False
-1 : False-False**0
1 : False-False**0&1
2 : False-False**0&1+[<[]][<]
-2 : False-False**0&1+[<[]][<]^~3
-3 : False-False**0&1+[<[]][<]^~3>>2
answered Feb 11 at 20:36
TRITICIMAGVSTRITICIMAGVS
35.1k10158369
$endgroup$
add a comment |
2
$begingroup$
Clean, 40$^3$/400 = 160
import StdEnv,Data.GenDefault,Gast.GenLibTest
derive genShow G
derive gDefault G
n = sum[(bitnot o bitnot) one \ n <-: s]-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one
s = (hd o show o d) undef
d :: G -> G
d g = gDefault{|*|} g
:: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`
Try it online!
Defines n which produces -20, -19, ..., 18, 19 depending on the length of the type G.
The smallest prefix of G looks like :: G = G, preceded by the assorted imports and definitions.
The largest prefix of G looks like :: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`, similarly preceded.
answered Feb 12 at 1:35
ΟurousΟurous
7,05211035
$endgroup$
$begingroup$
I think I'm missing something in this challenge, but… this answer (and a handful of others) is reusing "bytes that have appeared in previous prefixes", isn't it?
$endgroup$
– Blackhole
Feb 12 at 1:39
4
$begingroup$
@Blackhole As I (and I assume others) understood it, you can use bytes multiple times in the same prefix, but not in subsequent ones.
$endgroup$
– Οurous
Feb 12 at 1:40
add a comment |
2
$begingroup$
Batch, 176³/7744 = 704
@cmd/cset/a%~z0-110
: !"#$&'()*+,.23456789;<=>?ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`bfghijklnopqruvwxy{|} ¡¢£¤¥¦§¨©ª«¬®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ
The first prefix includes the colon and outputs the number -88. Subsequent prefixes output ascending numbers and the final program outputs 87.
answered Feb 12 at 19:51
NeilNeil
80.8k744178
$endgroup$
add a comment |
2
$begingroup$
Pyth, $(249^{3})/15500 = 996.0$
-C|l"x01x02x03x04x05x06x07x08tnx0bx0crx0ex0fx10x11x12x13x14x15x16x17x18x19x1ax1bx1cx1dx1ex1f !#$%&'()*+,./0123456789:;<=>?@ABDEFGHIJKLMNOPQRSTUVWXYZ^_`abcdefghijkmnopqrstuvwxyz{}~x80x81x82x83x84x85x86x87x88x89x8ax8bx8cx8dx8ex8fx90x91x92x93x94x95x96x97x98x99x9ax9bx9cx9dx9ex9fxa0¡¢£¤¥¦§¨©ª«¬xad®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ
All of the escaped characters are single bytes - I've presented the program in this form for ease of copying and pasting. This link is to the full program without escaped characters; as can be seen there are a lot of placeholders for the unprintable characters used.
The approach used is similar to lirtosiast's TI-BASIC answer, though I wrote my answer before browsing the other entries. The cord works as follows:
-C|l"...
| "|"
C Character code of the above, i.e. 124
- Subtract from the above...
l"... ... the length of the trailing string - the closing quote is inferred
The string is made of every single-byte character other than 0 (NULL), 127 (backspace), and the boilerplate characters (-C|l"). That way, each character of the string can be removed one at a time to produce strings of lengths 248 down to 0. These programs yield results in the range [-124 - 124], with sum of absolute values of 15500.
The code was generated using this program:
+J"-C\|l""sCM-U256+CMJ,Z127
J"-C\|l"" Assign the boilerplate code to J
CMJ Get the character code of each character in J
+ ,Z127 Append 0 and 127
-U256 Get values in range [0-256) which are not in the above
CM Get each character with code points in the above
s Concatenate into string
+J Prepend boilerplate
And the sum of the numbers generated from each of the truncated programs was found with this program:
sm.a.vd>._+J"-C\|l""sCM-U256+CMJ,Z127 5
+J"-C\|l""sCM-U256+CMJ,Z127 The full program string, as above
._ Get the prefixes of the above
> 5 Discard the first 5 (those prefixes which don't include all the boilerplate code)
m Map each of the above to:
.vd Evaluate as Pyth code
.a Take the absolute value of the above
s Take the sum
answered Feb 13 at 11:26
SokSok
4,007925
$endgroup$
add a comment |
2
$begingroup$
Jelly, 984
LNḂ¡:2µ}“¢¢¢£¤¥¦©¬®½¿€ÆÇÐÑ×ØŒÞßæçðıȷñ÷øœþ !"#$%&'()*+,-./013456789;<=>?@ABCDEFGHIJKMOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|~¶°¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾ƁƇƊƑƓƘⱮƝƤƬƲȤɓƈɗƒɠɦƙɱɲƥʠɼʂƭʋȥẠḄḌẸḤỊḲḶṂṆỌṚṢṬỤṾẈỴẒȦĊḊĖḞĠḢİĿṀṄȮṖṘṠṪẆẊẎŻạḅḍẹḥịḳḷṃṇọṛṣṭ§Äẉỵẓȧḃċḋėḟġḣŀṁṅȯṗṙṡṫẇẋẏż”‘
Try it online!
You can see the prefixes (separated by a double newline) here.
The numbers printed are, in order:
0 1 -1 -2 2 -3 3 -4 4 -5 5 -6 6 -7 7 -8 8 -9 9 -10 10 -11 11 -12 12 -13 13 -14 14 -15 15 -16 16 -17 17 -18 18 -19 19 -20 20 -21 21 -22 22 -23 23 -24 24 -25 25 -26 26 -27 27 -28 28 -29 29 -30 30 -31 31 -32 32 -33 33 -34 34 -35 35 -36 36 -37 37 -38 38 -39 39 -40 40 -41 41 -42 42 -43 43 -44 44 -45 45 -46 46 -47 47 -48 48 -49 49 -50 50 -51 51 -52 52 -53 53 -54 54 -55 55 -56 56 -57 57 -58 58 -59 59 -60 60 -61 61 -62 62 -63 63 -64 64 -65 65 -66 66 -67 67 -68 68 -69 69 -70 70 -71 71 -72 72 -73 73 -74 74 -75 75 -76 76 -77 77 -78 78 -79 79 -80 80 -81 81 -82 82 -83 83 -84 84 -85 85 -86 86 -87 87 -88 88 -89 89 -90 90 -91 91 -92 92 -93 93 -94 94 -95 95 -96 96 -97 97 -98 98 -99 99 -100 100 -101 101 -102 102 -103 103 -104 104 -105 105 -106 106 -107 107 -108 108 -109 109 -110 110 -111 111 -112 112 -113 113 -114 114 -115 115 -116 116 -117 117 -118 118 -119 119 -120 120 -121 121 -122 122 123
answered Feb 13 at 19:26
Erik the OutgolferErik the Outgolfer
31.9k429103
$endgroup$
add a comment |
1
$begingroup$
Python 3, $(6^{3})/12 = 18$
0x1-2+3|True*~4
This is a snippet.
The individual prefix snippets are:
0 # gives 0
0x1 # gives 1 (hex syntax)
0x1-2 # gives -1 (1 minus 2)
0x1-2+3 # gives 2 (-1 plus 3)
0x1-2+3|True # gives 3 (True == 1; 2 bitwise-OR 1 == 3)
0x1-2+3|True*~4 # gives -5 (~4 == -5; True*~4 == -5; 2|-5 == -5)
Python's bitwise operators treat negative integers as two's complement with unlimited 1s to the right, so -5 is treated as ...11111011, which when ORed with ...000010 gives ...11111011 unchanged.
answered Feb 11 at 1:32
pizzapants184pizzapants184
2,674716
$endgroup$
add a comment |
1
$begingroup$
Excel, $(6^{3})/12 = 18$
=0 0
=01 1
=01*2 2
=01*2-5 -3
=01*2-5+8 5
=01*2-5+8/4 -1
Prefix challenge does not gel well with Excel's formula structure
answered Feb 11 at 10:12
WernischWernisch
1,628317
$endgroup$
add a comment |
0
$begingroup$
Charcoal, 953/2256 = 380.04211
Ｉ⁻⁴⁸Ｌ !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|}~
The first prefix includes the space and outputs the number 47. Try it online! Subsequent prefixes output descending numbers and the final program outputs -47. Try it online!
answered Feb 11 at 10:53
NeilNeil
80.8k744178
$endgroup$
add a comment |
0
$begingroup$
Runic Enchantments, 1173/3422 = 468.034
>"ul'B-$; !#%&()*+,.:0123456789@<=?ACDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijkmnopqrstvwxyz{|}~€
Try it online!
And the prefix.
Prints [-58,58] inclusive.
Uses a similar tactic as other entries, using a progressively longer string, getting its length, subtracts B (decimal value of 66), and prints the result. Uses a non-printing character (shown in the code block as € because apparently that's what € evaluates to in HTML) in order to utilize byte value 0x80 (and consuming an otherwise unusable 0x00) and score one more value. Only byte values up to 127 that can't be represented in any fashion are 10 and 13, because they would cause the program to be two lines tall and not be read correctly by the instruction pointer.
answered Feb 14 at 3:38
Draco18sDraco18s
1,356619
$endgroup$
add a comment |
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18
$begingroup$
TI-BASIC (TI-84+ CE), $238^3/14161 = 952.0$
118-length("+×/ABCDEFGHIJK...
The 256 bytes in TI-BASIC break down thusly:
- 240 bytes are one-byte tokens that can be quoted
- 11 are the start of two-byte tokens
- 2 are unused
- The remaining 3 are the
",→, and newline characters which break strings.
The string contains all 237 allowable bytes in the first category not already in 118-length(.
The program calculates 238 distinct integers, ranging from 118 down to -119. The score is therefore $$ frac{238^3}{sum_{n = -119}^{118} |n|} = 952.$$
answered Feb 11 at 2:58
lirtosiastlirtosiast
18.4k437109
$endgroup$
1
$begingroup$
Wouldn't the string have to end with " which is a byte that was already used? Or why not? I am a bit confused by this also in other answers this seems to happen so probably I miss something.
$endgroup$
– findusl
Feb 13 at 8:42
2
$begingroup$
@findusl TI-BASIC has a tendency to automatically complete parentheses and strings for you. so you could writemean({1,2,3, which would complete tomean({1,2,3}). strings are also completed:Disp "Hi!completes toDisp "Hi!". Therefore the final closing quote is unnecessary
$endgroup$
– Conor O'Brien
Feb 13 at 12:23
add a comment |
18
$begingroup$
TI-BASIC (TI-84+ CE), $238^3/14161 = 952.0$
118-length("+×/ABCDEFGHIJK...
The 256 bytes in TI-BASIC break down thusly:
- 240 bytes are one-byte tokens that can be quoted
- 11 are the start of two-byte tokens
- 2 are unused
- The remaining 3 are the
",→, and newline characters which break strings.
The string contains all 237 allowable bytes in the first category not already in 118-length(.
The program calculates 238 distinct integers, ranging from 118 down to -119. The score is therefore $$ frac{238^3}{sum_{n = -119}^{118} |n|} = 952.$$
answered Feb 11 at 2:58
lirtosiastlirtosiast
18.4k437109
$endgroup$
1
$begingroup$
Wouldn't the string have to end with " which is a byte that was already used? Or why not? I am a bit confused by this also in other answers this seems to happen so probably I miss something.
$endgroup$
– findusl
Feb 13 at 8:42
2
$begingroup$
@findusl TI-BASIC has a tendency to automatically complete parentheses and strings for you. so you could writemean({1,2,3, which would complete tomean({1,2,3}). strings are also completed:Disp "Hi!completes toDisp "Hi!". Therefore the final closing quote is unnecessary
$endgroup$
– Conor O'Brien
Feb 13 at 12:23
add a comment |
18
18
18
$begingroup$
TI-BASIC (TI-84+ CE), $238^3/14161 = 952.0$
118-length("+×/ABCDEFGHIJK...
The 256 bytes in TI-BASIC break down thusly:
- 240 bytes are one-byte tokens that can be quoted
- 11 are the start of two-byte tokens
- 2 are unused
- The remaining 3 are the
",→, and newline characters which break strings.
The string contains all 237 allowable bytes in the first category not already in 118-length(.
The program calculates 238 distinct integers, ranging from 118 down to -119. The score is therefore $$ frac{238^3}{sum_{n = -119}^{118} |n|} = 952.$$
answered Feb 11 at 2:58
lirtosiastlirtosiast
18.4k437109
$endgroup$
TI-BASIC (TI-84+ CE), $238^3/14161 = 952.0$
118-length("+×/ABCDEFGHIJK...
The 256 bytes in TI-BASIC break down thusly:
- 240 bytes are one-byte tokens that can be quoted
- 11 are the start of two-byte tokens
- 2 are unused
- The remaining 3 are the
",→, and newline characters which break strings.
The string contains all 237 allowable bytes in the first category not already in 118-length(.
The program calculates 238 distinct integers, ranging from 118 down to -119. The score is therefore $$ frac{238^3}{sum_{n = -119}^{118} |n|} = 952.$$
answered Feb 11 at 2:58
lirtosiastlirtosiast
18.4k437109
edited Feb 11 at 3:13
answered Feb 11 at 2:58
lirtosiastlirtosiast
18.4k437109
answered Feb 11 at 2:58
lirtosiastlirtosiast
18.4k437109
answered Feb 11 at 2:58
lirtosiastlirtosiast
18.4k437109
18.4k437109
1
$begingroup$
Wouldn't the string have to end with " which is a byte that was already used? Or why not? I am a bit confused by this also in other answers this seems to happen so probably I miss something.
$endgroup$
– findusl
Feb 13 at 8:42
2
$begingroup$
@findusl TI-BASIC has a tendency to automatically complete parentheses and strings for you. so you could writemean({1,2,3, which would complete tomean({1,2,3}). strings are also completed:Disp "Hi!completes toDisp "Hi!". Therefore the final closing quote is unnecessary
$endgroup$
– Conor O'Brien
Feb 13 at 12:23
add a comment |
1
$begingroup$
Wouldn't the string have to end with " which is a byte that was already used? Or why not? I am a bit confused by this also in other answers this seems to happen so probably I miss something.
$endgroup$
– findusl
Feb 13 at 8:42
2
$begingroup$
@findusl TI-BASIC has a tendency to automatically complete parentheses and strings for you. so you could writemean({1,2,3, which would complete tomean({1,2,3}). strings are also completed:Disp "Hi!completes toDisp "Hi!". Therefore the final closing quote is unnecessary
$endgroup$
– Conor O'Brien
Feb 13 at 12:23
1
1
$begingroup$
Wouldn't the string have to end with " which is a byte that was already used? Or why not? I am a bit confused by this also in other answers this seems to happen so probably I miss something.
$endgroup$
– findusl
Feb 13 at 8:42
$begingroup$
Wouldn't the string have to end with " which is a byte that was already used? Or why not? I am a bit confused by this also in other answers this seems to happen so probably I miss something.
$endgroup$
– findusl
Feb 13 at 8:42
2
2
$begingroup$
@findusl TI-BASIC has a tendency to automatically complete parentheses and strings for you. so you could write
mean({1,2,3, which would complete to mean({1,2,3}). strings are also completed: Disp "Hi! completes to Disp "Hi!". Therefore the final closing quote is unnecessary$endgroup$
– Conor O'Brien
Feb 13 at 12:23
$begingroup$
@findusl TI-BASIC has a tendency to automatically complete parentheses and strings for you. so you could write
mean({1,2,3, which would complete to mean({1,2,3}). strings are also completed: Disp "Hi! completes to Disp "Hi!". Therefore the final closing quote is unnecessary$endgroup$
– Conor O'Brien
Feb 13 at 12:23
add a comment |
13
$begingroup$
vim, 14³ / 53 = 51.77
i10<esc>X<c-x>r2hx<c-a>Pa<del>4<c-c>d^s5<c-o>I-<c-o>$<c-o>C6<c-h>7
Try it online! (all prefixes)
Breakdown:
i1 1
0 10
<esc>X 0
<c-x> -1 <c-x> decrements
r2 -2
hx 2 hx is X, but we've already used that
<c-a> 3 <c-a> increments
P -3 paste the - from two steps ago
a<del>4 -4
<c-c>d^ 4 <c-c> is an <esc> alternative
s5 5
<c-o>I-<c-o>$<c-o> -5 <c-o> allows a single normal mode input
C6 -6 trailing <c-o> from previous line used here
<c-h>7 -7 <c-h> = backspace
answered Feb 11 at 2:20
Doorknob♦Doorknob
54.8k17114350
$endgroup$
$begingroup$
are you not reusing the < byte? Or is that a placeholder for a different byte?
$endgroup$
– findusl
Feb 13 at 13:20
1
$begingroup$
@findusl no -<esc>represents the escape key and<c-char>represents ctrl+char
$endgroup$
– Doorknob♦
Feb 13 at 14:20
add a comment |
13
$begingroup$
vim, 14³ / 53 = 51.77
i10<esc>X<c-x>r2hx<c-a>Pa<del>4<c-c>d^s5<c-o>I-<c-o>$<c-o>C6<c-h>7
Try it online! (all prefixes)
Breakdown:
i1 1
0 10
<esc>X 0
<c-x> -1 <c-x> decrements
r2 -2
hx 2 hx is X, but we've already used that
<c-a> 3 <c-a> increments
P -3 paste the - from two steps ago
a<del>4 -4
<c-c>d^ 4 <c-c> is an <esc> alternative
s5 5
<c-o>I-<c-o>$<c-o> -5 <c-o> allows a single normal mode input
C6 -6 trailing <c-o> from previous line used here
<c-h>7 -7 <c-h> = backspace
answered Feb 11 at 2:20
Doorknob♦Doorknob
54.8k17114350
$endgroup$
$begingroup$
are you not reusing the < byte? Or is that a placeholder for a different byte?
$endgroup$
– findusl
Feb 13 at 13:20
1
$begingroup$
@findusl no -<esc>represents the escape key and<c-char>represents ctrl+char
$endgroup$
– Doorknob♦
Feb 13 at 14:20
add a comment |
13
13
13
$begingroup$
vim, 14³ / 53 = 51.77
i10<esc>X<c-x>r2hx<c-a>Pa<del>4<c-c>d^s5<c-o>I-<c-o>$<c-o>C6<c-h>7
Try it online! (all prefixes)
Breakdown:
i1 1
0 10
<esc>X 0
<c-x> -1 <c-x> decrements
r2 -2
hx 2 hx is X, but we've already used that
<c-a> 3 <c-a> increments
P -3 paste the - from two steps ago
a<del>4 -4
<c-c>d^ 4 <c-c> is an <esc> alternative
s5 5
<c-o>I-<c-o>$<c-o> -5 <c-o> allows a single normal mode input
C6 -6 trailing <c-o> from previous line used here
<c-h>7 -7 <c-h> = backspace
answered Feb 11 at 2:20
Doorknob♦Doorknob
54.8k17114350
$endgroup$
vim, 14³ / 53 = 51.77
i10<esc>X<c-x>r2hx<c-a>Pa<del>4<c-c>d^s5<c-o>I-<c-o>$<c-o>C6<c-h>7
Try it online! (all prefixes)
Breakdown:
i1 1
0 10
<esc>X 0
<c-x> -1 <c-x> decrements
r2 -2
hx 2 hx is X, but we've already used that
<c-a> 3 <c-a> increments
P -3 paste the - from two steps ago
a<del>4 -4
<c-c>d^ 4 <c-c> is an <esc> alternative
s5 5
<c-o>I-<c-o>$<c-o> -5 <c-o> allows a single normal mode input
C6 -6 trailing <c-o> from previous line used here
<c-h>7 -7 <c-h> = backspace
answered Feb 11 at 2:20
Doorknob♦Doorknob
54.8k17114350
edited Feb 11 at 2:35
answered Feb 11 at 2:20
Doorknob♦Doorknob
54.8k17114350
answered Feb 11 at 2:20
Doorknob♦Doorknob
54.8k17114350
answered Feb 11 at 2:20
Doorknob♦Doorknob
54.8k17114350
54.8k17114350
$begingroup$
are you not reusing the < byte? Or is that a placeholder for a different byte?
$endgroup$
– findusl
Feb 13 at 13:20
1
$begingroup$
@findusl no -<esc>represents the escape key and<c-char>represents ctrl+char
$endgroup$
– Doorknob♦
Feb 13 at 14:20
add a comment |
$begingroup$
are you not reusing the < byte? Or is that a placeholder for a different byte?
$endgroup$
– findusl
Feb 13 at 13:20
1
$begingroup$
@findusl no -<esc>represents the escape key and<c-char>represents ctrl+char
$endgroup$
– Doorknob♦
Feb 13 at 14:20
$begingroup$
are you not reusing the < byte? Or is that a placeholder for a different byte?
$endgroup$
– findusl
Feb 13 at 13:20
$begingroup$
are you not reusing the < byte? Or is that a placeholder for a different byte?
$endgroup$
– findusl
Feb 13 at 13:20
1
1
$begingroup$
@findusl no -
<esc> represents the escape key and <c-char> represents ctrl+char$endgroup$
– Doorknob♦
Feb 13 at 14:20
$begingroup$
@findusl no -
<esc> represents the escape key and <c-char> represents ctrl+char$endgroup$
– Doorknob♦
Feb 13 at 14:20
add a comment |
10
$begingroup$
Lenguage, 256³ / 16384 = 1024
The source code of the solution can be described as follows:
183662808731984092 0x00 bytes
4775233027031586288 0x01 bytes
27 0x02 bytes
81 0x03 bytes
128931291729852829668 0x04 bytes
108 0x05 bytes
648 0x06 bytes
81 0x07 bytes
2079 0x08 bytes
27 0x09 bytes
3481144876706026401009 0x0A bytes
108 0x0B bytes
2808 0x0C bytes
27 0x0D bytes
17496 0x0E bytes
81 0x0F bytes
2079 0x10 bytes
27 0x11 bytes
56106 0x12 bytes
756 0x13 bytes
93990911671062712827216 0x14 bytes
108 0x15 bytes
2808 0x16 bytes
27 0x17 bytes
75789 0x18 bytes
756 0x19 bytes
474552 0x1A bytes
27 0x1B bytes
81 0x1C bytes
56025 0x1D bytes
756 0x1E bytes
1514835 0x1F bytes
20439 0x20 bytes
2537754615118693246334805 0x21 bytes
108 0x22 bytes
2808 0x23 bytes
27 0x24 bytes
75789 0x25 bytes
756 0x26 bytes
2046276 0x27 bytes
20439 0x28 bytes
12812877 0x29 bytes
108 0x2A bytes
729 0x2B bytes
2079 0x2C bytes
27 0x2D bytes
1512648 0x2E bytes
40920957 0x2F bytes
1689866867948183303767857 0x30 bytes
66829507740256534347823839 0x31 bytes
2808 0x32 bytes
27 0x33 bytes
75789 0x34 bytes
756 0x35 bytes
2046276 0x36 bytes
20439 0x37 bytes
55249425 0x38 bytes
551880 0x39 bytes
345947652 0x3A bytes
2916 0x3B bytes
27 0x3C bytes
19656 0x3D bytes
56133 0x3E bytes
40842225 0x3F bytes
1104865839 0x40 bytes
45626405434600949201732247 0x41 bytes
1804396708986926427391243545 0x42 bytes
27 0x43 bytes
75789 0x44 bytes
756 0x45 bytes
2046276 0x46 bytes
20439 0x47 bytes
55249425 0x48 bytes
551880 0x49 bytes
1491734448 0x4A bytes
9355487364 0x4B bytes
78732 0x4C bytes
756 0x4D bytes
530685 0x4E bytes
1515591 0x4F bytes
1102740075 0x50 bytes
29831377653 0x51 bytes
1231912946734225628446770696 0x52 bytes
48718711142647013539563576444 0x53 bytes
2066715 0x54 bytes
55801305 0x55 bytes
1506635235 0x56 bytes
292874988897 0x57 bytes
2125764 0x58 bytes
14348907 0x59 bytes
40920957 0x5A bytes
29773982025 0x5B bytes
33261649561824092773510005450 0x5C bytes
1315405200851469365568216563988 0x5D bytes
55801305 0x5E bytes
1506635235 0x5F bytes
7948303851537 0x60 bytes
57395628 0x61 bytes
387420489 0x62 bytes
1104865839 0x63 bytes
803897514675 0x64 bytes
898064538169250504884770147177 0x65 bytes
35515940422989672870341847227676 0x66 bytes
1506635235 0x67 bytes
40679151345 0x68 bytes
214604203991472 0x69 bytes
1549681956 0x6A bytes
10460353203 0x6B bytes
29831377653 0x6C bytes
24001245795414021635055080475036 0x6D bytes
246496735155742018538946394995 0x6E bytes
958930391420721167499229875147252 0x6F bytes
40679151345 0x70 bytes
5795411844856032 0x71 bytes
41841412812 0x72 bytes
282429536481 0x73 bytes
805447196631 0x74 bytes
648033636476178584146487172825972 0x75 bytes
108 0x76 bytes
6655411849205034500551552664784 0x77 bytes
25891120568359471522479206628975804 0x78 bytes
1098337086315 0x79 bytes
156484875126743748 0x7A bytes
21747074309037 0x7B bytes
17496908184856821771955153666301244 0x7C bytes
108 0x7D bytes
2808 0x7E bytes
27 0x7F bytes
81 0x80 bytes
179696119928535931514891921949087 0x81 bytes
699060255345705731106938578982346708 0x82 bytes
4225121283523411674 0x83 bytes
472416520991134187843376319996477587 0x84 bytes
108 0x85 bytes
2808 0x86 bytes
27 0x87 bytes
75789 0x88 bytes
108 0x89 bytes
729 0x8A bytes
2079 0x8B bytes
27 0x8C bytes
4851795238070470150902081892625349 0x8D bytes
18874626894334054739887341632523361116 0x8E bytes
114078274655132115171 0x8F bytes
12755246066760623071771160639904894849 0x90 bytes
108 0x91 bytes
2808 0x92 bytes
27 0x93 bytes
75789 0x94 bytes
756 0x95 bytes
2046276 0x96 bytes
2916 0x97 bytes
27 0x98 bytes
19656 0x99 bytes
56133 0x9A bytes
130998471427902694074356211100885179 0x9B bytes
854006569949556303994892977044130020645 0x9C bytes
108 0x9D bytes
2808 0x9E bytes
27 0x9F bytes
75789 0xA0 bytes
756 0xA1 bytes
2046276 0xA2 bytes
55269864 0xA3 bytes
78732 0xA4 bytes
756 0xA5 bytes
530685 0xA6 bytes
1515591 0xA7 bytes
3536958728553372740007617699723899860 0xA8 bytes
23058177388638020207862110380191510557388 0xA9 bytes
108 0xAA bytes
2808 0xAB bytes
27 0xAC bytes
75789 0xAD bytes
756 0xAE bytes
2046276 0xAF bytes
20439 0xB0 bytes
55249425 0xB1 bytes
1492286328 0xB2 bytes
2125764 0xB3 bytes
14348907 0xB4 bytes
40920957 0xB5 bytes
622666287378897486676257185943063330345696 0xB6 bytes
108 0xB7 bytes
2808 0xB8 bytes
27 0xB9 bytes
75789 0xBA bytes
756 0xBB bytes
2046276 0xBC bytes
20439 0xBD bytes
55249425 0xBE bytes
1492286328 0xBF bytes
40291730856 0xC0 bytes
57395628 0xC1 bytes
387420489 0xC2 bytes
1104865839 0xC3 bytes
16811989759230232140258944020462709919333792 0xC4 bytes
108 0xC5 bytes
2808 0xC6 bytes
27 0xC7 bytes
75789 0xC8 bytes
756 0xC9 bytes
2046276 0xCA bytes
20439 0xCB bytes
55249425 0xCC bytes
551880 0xCD bytes
1491734448 0xCE bytes
40291730856 0xCF bytes
1087876733112 0xD0 bytes
1549681956 0xD1 bytes
10460353203 0xD2 bytes
29831377653 0xD3 bytes
453923723499216267786991488552493167822012492 0xD4 bytes
2808 0xD5 bytes
27 0xD6 bytes
75789 0xD7 bytes
756 0xD8 bytes
2046276 0xD9 bytes
20439 0xDA bytes
55249425 0xDB bytes
551880 0xDC bytes
1491734448 0xDD bytes
40291730856 0xDE bytes
1087876733112 0xDF bytes
29372671794024 0xE0 bytes
324270949293 0xE1 bytes
805447196631 0xE2 bytes
12255940534478839230248770190917315531194337311 0xE3 bytes
76545 0xE4 bytes
2066715 0xE5 bytes
55801305 0xE6 bytes
1506635235 0xE7 bytes
832318279426764 0xE8 bytes
21747074309037 0xE9 bytes
330910394430928659216716795154767519342247107424 0xEA bytes
2066715 0xEB bytes
55801305 0xEC bytes
1506635235 0xED bytes
22472634223673946 0xEE bytes
8934580649635073798851353469178723609411678244474 0xEF bytes
55801305 0xF0 bytes
1506635235 0xF1 bytes
40679151345 0xF2 bytes
606761124039196515 0xF3 bytes
241233677540146992568986543667825537454115312600825 0xF4 bytes
1506635235 0xF5 bytes
40679151345 0xF6 bytes
6513309293583968799362636679031305893812560835614495 0xF7 bytes
40679151345 0xF8 bytes
1098337086315 0xF9 bytes
175859350926767157582791190333845259132939142561591365 0xFA bytes
1098337086315 0xFB bytes
4748202475022713254735362139013821996589386504264297360 0xFC bytes
29655101330505 0xFD bytes
128201466825613257877854777753373193907913435615136028720 0xFE bytes
3461439604291557962702078999341076235513663562296408699075 0xFF bytes
Each run of identical bytes combined with the previous runs constitutes a prefix, and the full program is all the strings described above concatenated.
This is an optimal score. (The programs print all numbers from -127 to 128 inclusive.)
To create this answer, first, I copied the Java answer here for generating brainfuck programs to print a string. I ran it with this main function:
public static void main(String args) {
for (int i = -127; i <= 128; ++i) generate(""+i);
}
Then, I ran this Ruby script to generate the human-readable text shown here:
#!/usr/bin/ruby
nums = File.open('f').map do |line|
line.chomp.sub(/^+/, '>+').chars.map{|ch|
'+-><.,'.index(ch).to_s(3).rjust(3,?0)
}.join.to_i 3
end
puts ([0]+nums.sort).each_cons(2).map.with_index{|x,i|"#{x[1]-x[0]} 0x#{i.to_s(16).rjust(2,?0).upcase} bytes"}*"n"
The largest program is 3594571896764310192036774345469579167648804468538578264427 bytes long (and prints -90).
answered Feb 11 at 3:17
Doorknob♦Doorknob
54.8k17114350
$endgroup$
1
$begingroup$
Props for the speed and practical problem-solving skill!
$endgroup$
– lirtosiast
Feb 11 at 3:30
add a comment |
10
$begingroup$
Lenguage, 256³ / 16384 = 1024
The source code of the solution can be described as follows:
183662808731984092 0x00 bytes
4775233027031586288 0x01 bytes
27 0x02 bytes
81 0x03 bytes
128931291729852829668 0x04 bytes
108 0x05 bytes
648 0x06 bytes
81 0x07 bytes
2079 0x08 bytes
27 0x09 bytes
3481144876706026401009 0x0A bytes
108 0x0B bytes
2808 0x0C bytes
27 0x0D bytes
17496 0x0E bytes
81 0x0F bytes
2079 0x10 bytes
27 0x11 bytes
56106 0x12 bytes
756 0x13 bytes
93990911671062712827216 0x14 bytes
108 0x15 bytes
2808 0x16 bytes
27 0x17 bytes
75789 0x18 bytes
756 0x19 bytes
474552 0x1A bytes
27 0x1B bytes
81 0x1C bytes
56025 0x1D bytes
756 0x1E bytes
1514835 0x1F bytes
20439 0x20 bytes
2537754615118693246334805 0x21 bytes
108 0x22 bytes
2808 0x23 bytes
27 0x24 bytes
75789 0x25 bytes
756 0x26 bytes
2046276 0x27 bytes
20439 0x28 bytes
12812877 0x29 bytes
108 0x2A bytes
729 0x2B bytes
2079 0x2C bytes
27 0x2D bytes
1512648 0x2E bytes
40920957 0x2F bytes
1689866867948183303767857 0x30 bytes
66829507740256534347823839 0x31 bytes
2808 0x32 bytes
27 0x33 bytes
75789 0x34 bytes
756 0x35 bytes
2046276 0x36 bytes
20439 0x37 bytes
55249425 0x38 bytes
551880 0x39 bytes
345947652 0x3A bytes
2916 0x3B bytes
27 0x3C bytes
19656 0x3D bytes
56133 0x3E bytes
40842225 0x3F bytes
1104865839 0x40 bytes
45626405434600949201732247 0x41 bytes
1804396708986926427391243545 0x42 bytes
27 0x43 bytes
75789 0x44 bytes
756 0x45 bytes
2046276 0x46 bytes
20439 0x47 bytes
55249425 0x48 bytes
551880 0x49 bytes
1491734448 0x4A bytes
9355487364 0x4B bytes
78732 0x4C bytes
756 0x4D bytes
530685 0x4E bytes
1515591 0x4F bytes
1102740075 0x50 bytes
29831377653 0x51 bytes
1231912946734225628446770696 0x52 bytes
48718711142647013539563576444 0x53 bytes
2066715 0x54 bytes
55801305 0x55 bytes
1506635235 0x56 bytes
292874988897 0x57 bytes
2125764 0x58 bytes
14348907 0x59 bytes
40920957 0x5A bytes
29773982025 0x5B bytes
33261649561824092773510005450 0x5C bytes
1315405200851469365568216563988 0x5D bytes
55801305 0x5E bytes
1506635235 0x5F bytes
7948303851537 0x60 bytes
57395628 0x61 bytes
387420489 0x62 bytes
1104865839 0x63 bytes
803897514675 0x64 bytes
898064538169250504884770147177 0x65 bytes
35515940422989672870341847227676 0x66 bytes
1506635235 0x67 bytes
40679151345 0x68 bytes
214604203991472 0x69 bytes
1549681956 0x6A bytes
10460353203 0x6B bytes
29831377653 0x6C bytes
24001245795414021635055080475036 0x6D bytes
246496735155742018538946394995 0x6E bytes
958930391420721167499229875147252 0x6F bytes
40679151345 0x70 bytes
5795411844856032 0x71 bytes
41841412812 0x72 bytes
282429536481 0x73 bytes
805447196631 0x74 bytes
648033636476178584146487172825972 0x75 bytes
108 0x76 bytes
6655411849205034500551552664784 0x77 bytes
25891120568359471522479206628975804 0x78 bytes
1098337086315 0x79 bytes
156484875126743748 0x7A bytes
21747074309037 0x7B bytes
17496908184856821771955153666301244 0x7C bytes
108 0x7D bytes
2808 0x7E bytes
27 0x7F bytes
81 0x80 bytes
179696119928535931514891921949087 0x81 bytes
699060255345705731106938578982346708 0x82 bytes
4225121283523411674 0x83 bytes
472416520991134187843376319996477587 0x84 bytes
108 0x85 bytes
2808 0x86 bytes
27 0x87 bytes
75789 0x88 bytes
108 0x89 bytes
729 0x8A bytes
2079 0x8B bytes
27 0x8C bytes
4851795238070470150902081892625349 0x8D bytes
18874626894334054739887341632523361116 0x8E bytes
114078274655132115171 0x8F bytes
12755246066760623071771160639904894849 0x90 bytes
108 0x91 bytes
2808 0x92 bytes
27 0x93 bytes
75789 0x94 bytes
756 0x95 bytes
2046276 0x96 bytes
2916 0x97 bytes
27 0x98 bytes
19656 0x99 bytes
56133 0x9A bytes
130998471427902694074356211100885179 0x9B bytes
854006569949556303994892977044130020645 0x9C bytes
108 0x9D bytes
2808 0x9E bytes
27 0x9F bytes
75789 0xA0 bytes
756 0xA1 bytes
2046276 0xA2 bytes
55269864 0xA3 bytes
78732 0xA4 bytes
756 0xA5 bytes
530685 0xA6 bytes
1515591 0xA7 bytes
3536958728553372740007617699723899860 0xA8 bytes
23058177388638020207862110380191510557388 0xA9 bytes
108 0xAA bytes
2808 0xAB bytes
27 0xAC bytes
75789 0xAD bytes
756 0xAE bytes
2046276 0xAF bytes
20439 0xB0 bytes
55249425 0xB1 bytes
1492286328 0xB2 bytes
2125764 0xB3 bytes
14348907 0xB4 bytes
40920957 0xB5 bytes
622666287378897486676257185943063330345696 0xB6 bytes
108 0xB7 bytes
2808 0xB8 bytes
27 0xB9 bytes
75789 0xBA bytes
756 0xBB bytes
2046276 0xBC bytes
20439 0xBD bytes
55249425 0xBE bytes
1492286328 0xBF bytes
40291730856 0xC0 bytes
57395628 0xC1 bytes
387420489 0xC2 bytes
1104865839 0xC3 bytes
16811989759230232140258944020462709919333792 0xC4 bytes
108 0xC5 bytes
2808 0xC6 bytes
27 0xC7 bytes
75789 0xC8 bytes
756 0xC9 bytes
2046276 0xCA bytes
20439 0xCB bytes
55249425 0xCC bytes
551880 0xCD bytes
1491734448 0xCE bytes
40291730856 0xCF bytes
1087876733112 0xD0 bytes
1549681956 0xD1 bytes
10460353203 0xD2 bytes
29831377653 0xD3 bytes
453923723499216267786991488552493167822012492 0xD4 bytes
2808 0xD5 bytes
27 0xD6 bytes
75789 0xD7 bytes
756 0xD8 bytes
2046276 0xD9 bytes
20439 0xDA bytes
55249425 0xDB bytes
551880 0xDC bytes
1491734448 0xDD bytes
40291730856 0xDE bytes
1087876733112 0xDF bytes
29372671794024 0xE0 bytes
324270949293 0xE1 bytes
805447196631 0xE2 bytes
12255940534478839230248770190917315531194337311 0xE3 bytes
76545 0xE4 bytes
2066715 0xE5 bytes
55801305 0xE6 bytes
1506635235 0xE7 bytes
832318279426764 0xE8 bytes
21747074309037 0xE9 bytes
330910394430928659216716795154767519342247107424 0xEA bytes
2066715 0xEB bytes
55801305 0xEC bytes
1506635235 0xED bytes
22472634223673946 0xEE bytes
8934580649635073798851353469178723609411678244474 0xEF bytes
55801305 0xF0 bytes
1506635235 0xF1 bytes
40679151345 0xF2 bytes
606761124039196515 0xF3 bytes
241233677540146992568986543667825537454115312600825 0xF4 bytes
1506635235 0xF5 bytes
40679151345 0xF6 bytes
6513309293583968799362636679031305893812560835614495 0xF7 bytes
40679151345 0xF8 bytes
1098337086315 0xF9 bytes
175859350926767157582791190333845259132939142561591365 0xFA bytes
1098337086315 0xFB bytes
4748202475022713254735362139013821996589386504264297360 0xFC bytes
29655101330505 0xFD bytes
128201466825613257877854777753373193907913435615136028720 0xFE bytes
3461439604291557962702078999341076235513663562296408699075 0xFF bytes
Each run of identical bytes combined with the previous runs constitutes a prefix, and the full program is all the strings described above concatenated.
This is an optimal score. (The programs print all numbers from -127 to 128 inclusive.)
To create this answer, first, I copied the Java answer here for generating brainfuck programs to print a string. I ran it with this main function:
public static void main(String args) {
for (int i = -127; i <= 128; ++i) generate(""+i);
}
Then, I ran this Ruby script to generate the human-readable text shown here:
#!/usr/bin/ruby
nums = File.open('f').map do |line|
line.chomp.sub(/^+/, '>+').chars.map{|ch|
'+-><.,'.index(ch).to_s(3).rjust(3,?0)
}.join.to_i 3
end
puts ([0]+nums.sort).each_cons(2).map.with_index{|x,i|"#{x[1]-x[0]} 0x#{i.to_s(16).rjust(2,?0).upcase} bytes"}*"n"
The largest program is 3594571896764310192036774345469579167648804468538578264427 bytes long (and prints -90).
answered Feb 11 at 3:17
Doorknob♦Doorknob
54.8k17114350
$endgroup$
1
$begingroup$
Props for the speed and practical problem-solving skill!
$endgroup$
– lirtosiast
Feb 11 at 3:30
add a comment |
10
10
10
$begingroup$
Lenguage, 256³ / 16384 = 1024
The source code of the solution can be described as follows:
183662808731984092 0x00 bytes
4775233027031586288 0x01 bytes
27 0x02 bytes
81 0x03 bytes
128931291729852829668 0x04 bytes
108 0x05 bytes
648 0x06 bytes
81 0x07 bytes
2079 0x08 bytes
27 0x09 bytes
3481144876706026401009 0x0A bytes
108 0x0B bytes
2808 0x0C bytes
27 0x0D bytes
17496 0x0E bytes
81 0x0F bytes
2079 0x10 bytes
27 0x11 bytes
56106 0x12 bytes
756 0x13 bytes
93990911671062712827216 0x14 bytes
108 0x15 bytes
2808 0x16 bytes
27 0x17 bytes
75789 0x18 bytes
756 0x19 bytes
474552 0x1A bytes
27 0x1B bytes
81 0x1C bytes
56025 0x1D bytes
756 0x1E bytes
1514835 0x1F bytes
20439 0x20 bytes
2537754615118693246334805 0x21 bytes
108 0x22 bytes
2808 0x23 bytes
27 0x24 bytes
75789 0x25 bytes
756 0x26 bytes
2046276 0x27 bytes
20439 0x28 bytes
12812877 0x29 bytes
108 0x2A bytes
729 0x2B bytes
2079 0x2C bytes
27 0x2D bytes
1512648 0x2E bytes
40920957 0x2F bytes
1689866867948183303767857 0x30 bytes
66829507740256534347823839 0x31 bytes
2808 0x32 bytes
27 0x33 bytes
75789 0x34 bytes
756 0x35 bytes
2046276 0x36 bytes
20439 0x37 bytes
55249425 0x38 bytes
551880 0x39 bytes
345947652 0x3A bytes
2916 0x3B bytes
27 0x3C bytes
19656 0x3D bytes
56133 0x3E bytes
40842225 0x3F bytes
1104865839 0x40 bytes
45626405434600949201732247 0x41 bytes
1804396708986926427391243545 0x42 bytes
27 0x43 bytes
75789 0x44 bytes
756 0x45 bytes
2046276 0x46 bytes
20439 0x47 bytes
55249425 0x48 bytes
551880 0x49 bytes
1491734448 0x4A bytes
9355487364 0x4B bytes
78732 0x4C bytes
756 0x4D bytes
530685 0x4E bytes
1515591 0x4F bytes
1102740075 0x50 bytes
29831377653 0x51 bytes
1231912946734225628446770696 0x52 bytes
48718711142647013539563576444 0x53 bytes
2066715 0x54 bytes
55801305 0x55 bytes
1506635235 0x56 bytes
292874988897 0x57 bytes
2125764 0x58 bytes
14348907 0x59 bytes
40920957 0x5A bytes
29773982025 0x5B bytes
33261649561824092773510005450 0x5C bytes
1315405200851469365568216563988 0x5D bytes
55801305 0x5E bytes
1506635235 0x5F bytes
7948303851537 0x60 bytes
57395628 0x61 bytes
387420489 0x62 bytes
1104865839 0x63 bytes
803897514675 0x64 bytes
898064538169250504884770147177 0x65 bytes
35515940422989672870341847227676 0x66 bytes
1506635235 0x67 bytes
40679151345 0x68 bytes
214604203991472 0x69 bytes
1549681956 0x6A bytes
10460353203 0x6B bytes
29831377653 0x6C bytes
24001245795414021635055080475036 0x6D bytes
246496735155742018538946394995 0x6E bytes
958930391420721167499229875147252 0x6F bytes
40679151345 0x70 bytes
5795411844856032 0x71 bytes
41841412812 0x72 bytes
282429536481 0x73 bytes
805447196631 0x74 bytes
648033636476178584146487172825972 0x75 bytes
108 0x76 bytes
6655411849205034500551552664784 0x77 bytes
25891120568359471522479206628975804 0x78 bytes
1098337086315 0x79 bytes
156484875126743748 0x7A bytes
21747074309037 0x7B bytes
17496908184856821771955153666301244 0x7C bytes
108 0x7D bytes
2808 0x7E bytes
27 0x7F bytes
81 0x80 bytes
179696119928535931514891921949087 0x81 bytes
699060255345705731106938578982346708 0x82 bytes
4225121283523411674 0x83 bytes
472416520991134187843376319996477587 0x84 bytes
108 0x85 bytes
2808 0x86 bytes
27 0x87 bytes
75789 0x88 bytes
108 0x89 bytes
729 0x8A bytes
2079 0x8B bytes
27 0x8C bytes
4851795238070470150902081892625349 0x8D bytes
18874626894334054739887341632523361116 0x8E bytes
114078274655132115171 0x8F bytes
12755246066760623071771160639904894849 0x90 bytes
108 0x91 bytes
2808 0x92 bytes
27 0x93 bytes
75789 0x94 bytes
756 0x95 bytes
2046276 0x96 bytes
2916 0x97 bytes
27 0x98 bytes
19656 0x99 bytes
56133 0x9A bytes
130998471427902694074356211100885179 0x9B bytes
854006569949556303994892977044130020645 0x9C bytes
108 0x9D bytes
2808 0x9E bytes
27 0x9F bytes
75789 0xA0 bytes
756 0xA1 bytes
2046276 0xA2 bytes
55269864 0xA3 bytes
78732 0xA4 bytes
756 0xA5 bytes
530685 0xA6 bytes
1515591 0xA7 bytes
3536958728553372740007617699723899860 0xA8 bytes
23058177388638020207862110380191510557388 0xA9 bytes
108 0xAA bytes
2808 0xAB bytes
27 0xAC bytes
75789 0xAD bytes
756 0xAE bytes
2046276 0xAF bytes
20439 0xB0 bytes
55249425 0xB1 bytes
1492286328 0xB2 bytes
2125764 0xB3 bytes
14348907 0xB4 bytes
40920957 0xB5 bytes
622666287378897486676257185943063330345696 0xB6 bytes
108 0xB7 bytes
2808 0xB8 bytes
27 0xB9 bytes
75789 0xBA bytes
756 0xBB bytes
2046276 0xBC bytes
20439 0xBD bytes
55249425 0xBE bytes
1492286328 0xBF bytes
40291730856 0xC0 bytes
57395628 0xC1 bytes
387420489 0xC2 bytes
1104865839 0xC3 bytes
16811989759230232140258944020462709919333792 0xC4 bytes
108 0xC5 bytes
2808 0xC6 bytes
27 0xC7 bytes
75789 0xC8 bytes
756 0xC9 bytes
2046276 0xCA bytes
20439 0xCB bytes
55249425 0xCC bytes
551880 0xCD bytes
1491734448 0xCE bytes
40291730856 0xCF bytes
1087876733112 0xD0 bytes
1549681956 0xD1 bytes
10460353203 0xD2 bytes
29831377653 0xD3 bytes
453923723499216267786991488552493167822012492 0xD4 bytes
2808 0xD5 bytes
27 0xD6 bytes
75789 0xD7 bytes
756 0xD8 bytes
2046276 0xD9 bytes
20439 0xDA bytes
55249425 0xDB bytes
551880 0xDC bytes
1491734448 0xDD bytes
40291730856 0xDE bytes
1087876733112 0xDF bytes
29372671794024 0xE0 bytes
324270949293 0xE1 bytes
805447196631 0xE2 bytes
12255940534478839230248770190917315531194337311 0xE3 bytes
76545 0xE4 bytes
2066715 0xE5 bytes
55801305 0xE6 bytes
1506635235 0xE7 bytes
832318279426764 0xE8 bytes
21747074309037 0xE9 bytes
330910394430928659216716795154767519342247107424 0xEA bytes
2066715 0xEB bytes
55801305 0xEC bytes
1506635235 0xED bytes
22472634223673946 0xEE bytes
8934580649635073798851353469178723609411678244474 0xEF bytes
55801305 0xF0 bytes
1506635235 0xF1 bytes
40679151345 0xF2 bytes
606761124039196515 0xF3 bytes
241233677540146992568986543667825537454115312600825 0xF4 bytes
1506635235 0xF5 bytes
40679151345 0xF6 bytes
6513309293583968799362636679031305893812560835614495 0xF7 bytes
40679151345 0xF8 bytes
1098337086315 0xF9 bytes
175859350926767157582791190333845259132939142561591365 0xFA bytes
1098337086315 0xFB bytes
4748202475022713254735362139013821996589386504264297360 0xFC bytes
29655101330505 0xFD bytes
128201466825613257877854777753373193907913435615136028720 0xFE bytes
3461439604291557962702078999341076235513663562296408699075 0xFF bytes
Each run of identical bytes combined with the previous runs constitutes a prefix, and the full program is all the strings described above concatenated.
This is an optimal score. (The programs print all numbers from -127 to 128 inclusive.)
To create this answer, first, I copied the Java answer here for generating brainfuck programs to print a string. I ran it with this main function:
public static void main(String args) {
for (int i = -127; i <= 128; ++i) generate(""+i);
}
Then, I ran this Ruby script to generate the human-readable text shown here:
#!/usr/bin/ruby
nums = File.open('f').map do |line|
line.chomp.sub(/^+/, '>+').chars.map{|ch|
'+-><.,'.index(ch).to_s(3).rjust(3,?0)
}.join.to_i 3
end
puts ([0]+nums.sort).each_cons(2).map.with_index{|x,i|"#{x[1]-x[0]} 0x#{i.to_s(16).rjust(2,?0).upcase} bytes"}*"n"
The largest program is 3594571896764310192036774345469579167648804468538578264427 bytes long (and prints -90).
answered Feb 11 at 3:17
Doorknob♦Doorknob
54.8k17114350
$endgroup$
Lenguage, 256³ / 16384 = 1024
The source code of the solution can be described as follows:
183662808731984092 0x00 bytes
4775233027031586288 0x01 bytes
27 0x02 bytes
81 0x03 bytes
128931291729852829668 0x04 bytes
108 0x05 bytes
648 0x06 bytes
81 0x07 bytes
2079 0x08 bytes
27 0x09 bytes
3481144876706026401009 0x0A bytes
108 0x0B bytes
2808 0x0C bytes
27 0x0D bytes
17496 0x0E bytes
81 0x0F bytes
2079 0x10 bytes
27 0x11 bytes
56106 0x12 bytes
756 0x13 bytes
93990911671062712827216 0x14 bytes
108 0x15 bytes
2808 0x16 bytes
27 0x17 bytes
75789 0x18 bytes
756 0x19 bytes
474552 0x1A bytes
27 0x1B bytes
81 0x1C bytes
56025 0x1D bytes
756 0x1E bytes
1514835 0x1F bytes
20439 0x20 bytes
2537754615118693246334805 0x21 bytes
108 0x22 bytes
2808 0x23 bytes
27 0x24 bytes
75789 0x25 bytes
756 0x26 bytes
2046276 0x27 bytes
20439 0x28 bytes
12812877 0x29 bytes
108 0x2A bytes
729 0x2B bytes
2079 0x2C bytes
27 0x2D bytes
1512648 0x2E bytes
40920957 0x2F bytes
1689866867948183303767857 0x30 bytes
66829507740256534347823839 0x31 bytes
2808 0x32 bytes
27 0x33 bytes
75789 0x34 bytes
756 0x35 bytes
2046276 0x36 bytes
20439 0x37 bytes
55249425 0x38 bytes
551880 0x39 bytes
345947652 0x3A bytes
2916 0x3B bytes
27 0x3C bytes
19656 0x3D bytes
56133 0x3E bytes
40842225 0x3F bytes
1104865839 0x40 bytes
45626405434600949201732247 0x41 bytes
1804396708986926427391243545 0x42 bytes
27 0x43 bytes
75789 0x44 bytes
756 0x45 bytes
2046276 0x46 bytes
20439 0x47 bytes
55249425 0x48 bytes
551880 0x49 bytes
1491734448 0x4A bytes
9355487364 0x4B bytes
78732 0x4C bytes
756 0x4D bytes
530685 0x4E bytes
1515591 0x4F bytes
1102740075 0x50 bytes
29831377653 0x51 bytes
1231912946734225628446770696 0x52 bytes
48718711142647013539563576444 0x53 bytes
2066715 0x54 bytes
55801305 0x55 bytes
1506635235 0x56 bytes
292874988897 0x57 bytes
2125764 0x58 bytes
14348907 0x59 bytes
40920957 0x5A bytes
29773982025 0x5B bytes
33261649561824092773510005450 0x5C bytes
1315405200851469365568216563988 0x5D bytes
55801305 0x5E bytes
1506635235 0x5F bytes
7948303851537 0x60 bytes
57395628 0x61 bytes
387420489 0x62 bytes
1104865839 0x63 bytes
803897514675 0x64 bytes
898064538169250504884770147177 0x65 bytes
35515940422989672870341847227676 0x66 bytes
1506635235 0x67 bytes
40679151345 0x68 bytes
214604203991472 0x69 bytes
1549681956 0x6A bytes
10460353203 0x6B bytes
29831377653 0x6C bytes
24001245795414021635055080475036 0x6D bytes
246496735155742018538946394995 0x6E bytes
958930391420721167499229875147252 0x6F bytes
40679151345 0x70 bytes
5795411844856032 0x71 bytes
41841412812 0x72 bytes
282429536481 0x73 bytes
805447196631 0x74 bytes
648033636476178584146487172825972 0x75 bytes
108 0x76 bytes
6655411849205034500551552664784 0x77 bytes
25891120568359471522479206628975804 0x78 bytes
1098337086315 0x79 bytes
156484875126743748 0x7A bytes
21747074309037 0x7B bytes
17496908184856821771955153666301244 0x7C bytes
108 0x7D bytes
2808 0x7E bytes
27 0x7F bytes
81 0x80 bytes
179696119928535931514891921949087 0x81 bytes
699060255345705731106938578982346708 0x82 bytes
4225121283523411674 0x83 bytes
472416520991134187843376319996477587 0x84 bytes
108 0x85 bytes
2808 0x86 bytes
27 0x87 bytes
75789 0x88 bytes
108 0x89 bytes
729 0x8A bytes
2079 0x8B bytes
27 0x8C bytes
4851795238070470150902081892625349 0x8D bytes
18874626894334054739887341632523361116 0x8E bytes
114078274655132115171 0x8F bytes
12755246066760623071771160639904894849 0x90 bytes
108 0x91 bytes
2808 0x92 bytes
27 0x93 bytes
75789 0x94 bytes
756 0x95 bytes
2046276 0x96 bytes
2916 0x97 bytes
27 0x98 bytes
19656 0x99 bytes
56133 0x9A bytes
130998471427902694074356211100885179 0x9B bytes
854006569949556303994892977044130020645 0x9C bytes
108 0x9D bytes
2808 0x9E bytes
27 0x9F bytes
75789 0xA0 bytes
756 0xA1 bytes
2046276 0xA2 bytes
55269864 0xA3 bytes
78732 0xA4 bytes
756 0xA5 bytes
530685 0xA6 bytes
1515591 0xA7 bytes
3536958728553372740007617699723899860 0xA8 bytes
23058177388638020207862110380191510557388 0xA9 bytes
108 0xAA bytes
2808 0xAB bytes
27 0xAC bytes
75789 0xAD bytes
756 0xAE bytes
2046276 0xAF bytes
20439 0xB0 bytes
55249425 0xB1 bytes
1492286328 0xB2 bytes
2125764 0xB3 bytes
14348907 0xB4 bytes
40920957 0xB5 bytes
622666287378897486676257185943063330345696 0xB6 bytes
108 0xB7 bytes
2808 0xB8 bytes
27 0xB9 bytes
75789 0xBA bytes
756 0xBB bytes
2046276 0xBC bytes
20439 0xBD bytes
55249425 0xBE bytes
1492286328 0xBF bytes
40291730856 0xC0 bytes
57395628 0xC1 bytes
387420489 0xC2 bytes
1104865839 0xC3 bytes
16811989759230232140258944020462709919333792 0xC4 bytes
108 0xC5 bytes
2808 0xC6 bytes
27 0xC7 bytes
75789 0xC8 bytes
756 0xC9 bytes
2046276 0xCA bytes
20439 0xCB bytes
55249425 0xCC bytes
551880 0xCD bytes
1491734448 0xCE bytes
40291730856 0xCF bytes
1087876733112 0xD0 bytes
1549681956 0xD1 bytes
10460353203 0xD2 bytes
29831377653 0xD3 bytes
453923723499216267786991488552493167822012492 0xD4 bytes
2808 0xD5 bytes
27 0xD6 bytes
75789 0xD7 bytes
756 0xD8 bytes
2046276 0xD9 bytes
20439 0xDA bytes
55249425 0xDB bytes
551880 0xDC bytes
1491734448 0xDD bytes
40291730856 0xDE bytes
1087876733112 0xDF bytes
29372671794024 0xE0 bytes
324270949293 0xE1 bytes
805447196631 0xE2 bytes
12255940534478839230248770190917315531194337311 0xE3 bytes
76545 0xE4 bytes
2066715 0xE5 bytes
55801305 0xE6 bytes
1506635235 0xE7 bytes
832318279426764 0xE8 bytes
21747074309037 0xE9 bytes
330910394430928659216716795154767519342247107424 0xEA bytes
2066715 0xEB bytes
55801305 0xEC bytes
1506635235 0xED bytes
22472634223673946 0xEE bytes
8934580649635073798851353469178723609411678244474 0xEF bytes
55801305 0xF0 bytes
1506635235 0xF1 bytes
40679151345 0xF2 bytes
606761124039196515 0xF3 bytes
241233677540146992568986543667825537454115312600825 0xF4 bytes
1506635235 0xF5 bytes
40679151345 0xF6 bytes
6513309293583968799362636679031305893812560835614495 0xF7 bytes
40679151345 0xF8 bytes
1098337086315 0xF9 bytes
175859350926767157582791190333845259132939142561591365 0xFA bytes
1098337086315 0xFB bytes
4748202475022713254735362139013821996589386504264297360 0xFC bytes
29655101330505 0xFD bytes
128201466825613257877854777753373193907913435615136028720 0xFE bytes
3461439604291557962702078999341076235513663562296408699075 0xFF bytes
Each run of identical bytes combined with the previous runs constitutes a prefix, and the full program is all the strings described above concatenated.
This is an optimal score. (The programs print all numbers from -127 to 128 inclusive.)
To create this answer, first, I copied the Java answer here for generating brainfuck programs to print a string. I ran it with this main function:
public static void main(String args) {
for (int i = -127; i <= 128; ++i) generate(""+i);
}
Then, I ran this Ruby script to generate the human-readable text shown here:
#!/usr/bin/ruby
nums = File.open('f').map do |line|
line.chomp.sub(/^+/, '>+').chars.map{|ch|
'+-><.,'.index(ch).to_s(3).rjust(3,?0)
}.join.to_i 3
end
puts ([0]+nums.sort).each_cons(2).map.with_index{|x,i|"#{x[1]-x[0]} 0x#{i.to_s(16).rjust(2,?0).upcase} bytes"}*"n"
The largest program is 3594571896764310192036774345469579167648804468538578264427 bytes long (and prints -90).
answered Feb 11 at 3:17
Doorknob♦Doorknob
54.8k17114350
answered Feb 11 at 3:17
Doorknob♦Doorknob
54.8k17114350
answered Feb 11 at 3:17
Doorknob♦Doorknob
54.8k17114350
answered Feb 11 at 3:17
Doorknob♦Doorknob
54.8k17114350
54.8k17114350
1
$begingroup$
Props for the speed and practical problem-solving skill!
$endgroup$
– lirtosiast
Feb 11 at 3:30
add a comment |
1
$begingroup$
Props for the speed and practical problem-solving skill!
$endgroup$
– lirtosiast
Feb 11 at 3:30
1
1
$begingroup$
Props for the speed and practical problem-solving skill!
$endgroup$
– lirtosiast
Feb 11 at 3:30
$begingroup$
Props for the speed and practical problem-solving skill!
$endgroup$
– lirtosiast
Feb 11 at 3:30
add a comment |
6
$begingroup$
Javascript, $(111^{3})/2530 = 540.5656126482213$
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(Credit to this interesting utility for the conversion.) The above effectively evaluates into the following code:
(new Proxy({},{get:(a,p,b,n=p.length+3)=>n%2?(n-1)/2:-n/2})).$0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyzÀāłƃDŽȅɆʇˈ̉͊βόЍюҏӐԑՒ֓הٖؕڗۘܙݚޛߜరᡡⱱ㣣䤤奥榦秧訨驩ꪪ뫫쬬흝ﯯ&''|""**"",~{}
...which creates a proxy that converts the length of a property name into a number. The first prefix ends one character after the ., and uses only 6 unique characters, excluding the .$. The next 63 slices append 63 other single-byte characters which are valid in JS object property names. The next 44 slices append two-byte valid characters made up of unused bytes. The final slices append &'', |""**"", and ,~{}. The first slice returns -2, the second 2, then -3, then 3, -4, 4, -5, etc. up to the last three slices which are 0, 1, and -1.
answered Feb 11 at 7:58
Yair RandYair Rand
3415
$endgroup$
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Javascript, $(111^{3})/2530 = 540.5656126482213$
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(Credit to this interesting utility for the conversion.) The above effectively evaluates into the following code:
(new Proxy({},{get:(a,p,b,n=p.length+3)=>n%2?(n-1)/2:-n/2})).$0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyzÀāłƃDŽȅɆʇˈ̉͊βόЍюҏӐԑՒ֓הٖؕڗۘܙݚޛߜరᡡⱱ㣣䤤奥榦秧訨驩ꪪ뫫쬬흝ﯯ&''|""**"",~{}
...which creates a proxy that converts the length of a property name into a number. The first prefix ends one character after the ., and uses only 6 unique characters, excluding the .$. The next 63 slices append 63 other single-byte characters which are valid in JS object property names. The next 44 slices append two-byte valid characters made up of unused bytes. The final slices append &'', |""**"", and ,~{}. The first slice returns -2, the second 2, then -3, then 3, -4, 4, -5, etc. up to the last three slices which are 0, 1, and -1.
answered Feb 11 at 7:58
Yair RandYair Rand
3415
$endgroup$
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Javascript, $(111^{3})/2530 = 540.5656126482213$
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(Credit to this interesting utility for the conversion.) The above effectively evaluates into the following code:
(new Proxy({},{get:(a,p,b,n=p.length+3)=>n%2?(n-1)/2:-n/2})).$0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyzÀāłƃDŽȅɆʇˈ̉͊βόЍюҏӐԑՒ֓הٖؕڗۘܙݚޛߜరᡡⱱ㣣䤤奥榦秧訨驩ꪪ뫫쬬흝ﯯ&''|""**"",~{}
...which creates a proxy that converts the length of a property name into a number. The first prefix ends one character after the ., and uses only 6 unique characters, excluding the .$. The next 63 slices append 63 other single-byte characters which are valid in JS object property names. The next 44 slices append two-byte valid characters made up of unused bytes. The final slices append &'', |""**"", and ,~{}. The first slice returns -2, the second 2, then -3, then 3, -4, 4, -5, etc. up to the last three slices which are 0, 1, and -1.
answered Feb 11 at 7:58
Yair RandYair Rand
3415
$endgroup$
Javascript, $(111^{3})/2530 = 540.5656126482213$
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(Credit to this interesting utility for the conversion.) The above effectively evaluates into the following code:
(new Proxy({},{get:(a,p,b,n=p.length+3)=>n%2?(n-1)/2:-n/2})).$0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyzÀāłƃDŽȅɆʇˈ̉͊βόЍюҏӐԑՒ֓הٖؕڗۘܙݚޛߜరᡡⱱ㣣䤤奥榦秧訨驩ꪪ뫫쬬흝ﯯ&''|""**"",~{}
...which creates a proxy that converts the length of a property name into a number. The first prefix ends one character after the ., and uses only 6 unique characters, excluding the .$. The next 63 slices append 63 other single-byte characters which are valid in JS object property names. The next 44 slices append two-byte valid characters made up of unused bytes. The final slices append &'', |""**"", and ,~{}. The first slice returns -2, the second 2, then -3, then 3, -4, 4, -5, etc. up to the last three slices which are 0, 1, and -1.
answered Feb 11 at 7:58
Yair RandYair Rand
3415
edited Feb 14 at 4:42
answered Feb 11 at 7:58
Yair RandYair Rand
3415
answered Feb 11 at 7:58
Yair RandYair Rand
3415
answered Feb 11 at 7:58
Yair RandYair Rand
3415
3415
add a comment |
add a comment |
3
$begingroup$
Brain-Flak, $4^3/4=16$
For this one we use brain-flak snippets so we are concerned with the value not what they do.
Complete program:
<>(())[]{}
Prefixes and output
0 : <>
1 : <>(())
-2 : <>(())[]
-1 : <>(())[]{}
Try it online!
Explanation
This is probably the best a brain-flak answer can get since we've used up all the character Brain-flak cares about, and we can't cluster the numbers closer to zero.
The first snippet is <>, <> changes the stack but otherwise evaluates to zero, so we start with zero. From here we add (()) this makes one. We could have just done (), because that is also one, but if we push a one now we will be able to recall it later with {} and . This is pretty much the only way {} and can create value so we do that. Since there is one thing on the stack counts as a 1 so we subtract 3 from the total with [] bringing us to -2. Lastly we use {} to pop the one we pushed earlier bringing us up to -1.
answered Feb 11 at 20:54
TRITICIMAGVSTRITICIMAGVS
35.1k10158369
$endgroup$
add a comment |
3
$begingroup$
Brain-Flak, $4^3/4=16$
For this one we use brain-flak snippets so we are concerned with the value not what they do.
Complete program:
<>(())[]{}
Prefixes and output
0 : <>
1 : <>(())
-2 : <>(())[]
-1 : <>(())[]{}
Try it online!
Explanation
This is probably the best a brain-flak answer can get since we've used up all the character Brain-flak cares about, and we can't cluster the numbers closer to zero.
The first snippet is <>, <> changes the stack but otherwise evaluates to zero, so we start with zero. From here we add (()) this makes one. We could have just done (), because that is also one, but if we push a one now we will be able to recall it later with {} and . This is pretty much the only way {} and can create value so we do that. Since there is one thing on the stack counts as a 1 so we subtract 3 from the total with [] bringing us to -2. Lastly we use {} to pop the one we pushed earlier bringing us up to -1.
answered Feb 11 at 20:54
TRITICIMAGVSTRITICIMAGVS
35.1k10158369
$endgroup$
add a comment |
3
3
3
$begingroup$
Brain-Flak, $4^3/4=16$
For this one we use brain-flak snippets so we are concerned with the value not what they do.
Complete program:
<>(())[]{}
Prefixes and output
0 : <>
1 : <>(())
-2 : <>(())[]
-1 : <>(())[]{}
Try it online!
Explanation
This is probably the best a brain-flak answer can get since we've used up all the character Brain-flak cares about, and we can't cluster the numbers closer to zero.
The first snippet is <>, <> changes the stack but otherwise evaluates to zero, so we start with zero. From here we add (()) this makes one. We could have just done (), because that is also one, but if we push a one now we will be able to recall it later with {} and . This is pretty much the only way {} and can create value so we do that. Since there is one thing on the stack counts as a 1 so we subtract 3 from the total with [] bringing us to -2. Lastly we use {} to pop the one we pushed earlier bringing us up to -1.
answered Feb 11 at 20:54
TRITICIMAGVSTRITICIMAGVS
35.1k10158369
$endgroup$
Brain-Flak, $4^3/4=16$
For this one we use brain-flak snippets so we are concerned with the value not what they do.
Complete program:
<>(())[]{}
Prefixes and output
0 : <>
1 : <>(())
-2 : <>(())[]
-1 : <>(())[]{}
Try it online!
Explanation
This is probably the best a brain-flak answer can get since we've used up all the character Brain-flak cares about, and we can't cluster the numbers closer to zero.
The first snippet is <>, <> changes the stack but otherwise evaluates to zero, so we start with zero. From here we add (()) this makes one. We could have just done (), because that is also one, but if we push a one now we will be able to recall it later with {} and . This is pretty much the only way {} and can create value so we do that. Since there is one thing on the stack counts as a 1 so we subtract 3 from the total with [] bringing us to -2. Lastly we use {} to pop the one we pushed earlier bringing us up to -1.
answered Feb 11 at 20:54
TRITICIMAGVSTRITICIMAGVS
35.1k10158369
answered Feb 11 at 20:54
TRITICIMAGVSTRITICIMAGVS
35.1k10158369
answered Feb 11 at 20:54
TRITICIMAGVSTRITICIMAGVS
35.1k10158369
answered Feb 11 at 20:54
TRITICIMAGVSTRITICIMAGVS
35.1k10158369
35.1k10158369
add a comment |
add a comment |
3
$begingroup$
Python 2, 32 34.7 40 $(11^{3})/32 = 41.59375$
After two hours of trial and error I finally got all integers from -4 to +5 (and 7).
0^cmp('','_')&3+1/~9*2-[-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]][==[]]|6<<7<<7>>8>>8and 5
This is a snippet with the following prefix snippets:
# -1
cmp(cmp,())
# +3 = (-1) & 3
cmp(cmp,())&3
# +7 = 3 + 4
cmp(cmp,())&3+4
# +2 = 3 + (-1) = 3 + floor(4/-10)
cmp(cmp,())&3+4/~9
# +1 = 3 + (-2) = 3 + 2*floor(4/-10)
cmp(cmp,())&3+4/~9*2
# -4 = 1 - 5
cmp(cmp,())&3+4/~9*2-1-1-1-1-1
# -2 = (-4) | 6
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6
# -3 = (-4) | 1 = (-4) | floor(6/2) = (-4) | (6>>2)
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8
# 0 = -3 and 0
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 0
# +5 = -3 and 05
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 05
# +4 = 5 ^ 1
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 05^True
Try it online!
answered Feb 12 at 11:32
ovsovs
19k21159
$endgroup$
1
$begingroup$
You can change the2*2to22&7(-3&6 instead of -3&4) to fix that repetition of*.
$endgroup$
– Kevin Cruijssen
Feb 12 at 11:47
$begingroup$
@KevinCruijssen thanks for the suggestion.
$endgroup$
– ovs
Feb 12 at 11:54
$begingroup$
You can free up'and_by usingcmp(cmp,())instead ofcmp('','_')If that is helpful.
$endgroup$
– TRITICIMAGVS
Feb 13 at 3:46
$begingroup$
You could also make the list section a little shorter and a lot easier to understand by doing--[-1-1-1-1-1], but that doesn't really improve your score.
$endgroup$
– TRITICIMAGVS
Feb 13 at 3:54
add a comment |
3
$begingroup$
Python 2, 32 34.7 40 $(11^{3})/32 = 41.59375$
After two hours of trial and error I finally got all integers from -4 to +5 (and 7).
0^cmp('','_')&3+1/~9*2-[-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]][==[]]|6<<7<<7>>8>>8and 5
This is a snippet with the following prefix snippets:
# -1
cmp(cmp,())
# +3 = (-1) & 3
cmp(cmp,())&3
# +7 = 3 + 4
cmp(cmp,())&3+4
# +2 = 3 + (-1) = 3 + floor(4/-10)
cmp(cmp,())&3+4/~9
# +1 = 3 + (-2) = 3 + 2*floor(4/-10)
cmp(cmp,())&3+4/~9*2
# -4 = 1 - 5
cmp(cmp,())&3+4/~9*2-1-1-1-1-1
# -2 = (-4) | 6
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6
# -3 = (-4) | 1 = (-4) | floor(6/2) = (-4) | (6>>2)
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8
# 0 = -3 and 0
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 0
# +5 = -3 and 05
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 05
# +4 = 5 ^ 1
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 05^True
Try it online!
answered Feb 12 at 11:32
ovsovs
19k21159
$endgroup$
1
$begingroup$
You can change the2*2to22&7(-3&6 instead of -3&4) to fix that repetition of*.
$endgroup$
– Kevin Cruijssen
Feb 12 at 11:47
$begingroup$
@KevinCruijssen thanks for the suggestion.
$endgroup$
– ovs
Feb 12 at 11:54
$begingroup$
You can free up'and_by usingcmp(cmp,())instead ofcmp('','_')If that is helpful.
$endgroup$
– TRITICIMAGVS
Feb 13 at 3:46
$begingroup$
You could also make the list section a little shorter and a lot easier to understand by doing--[-1-1-1-1-1], but that doesn't really improve your score.
$endgroup$
– TRITICIMAGVS
Feb 13 at 3:54
add a comment |
3
3
3
$begingroup$
Python 2, 32 34.7 40 $(11^{3})/32 = 41.59375$
After two hours of trial and error I finally got all integers from -4 to +5 (and 7).
0^cmp('','_')&3+1/~9*2-[-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]][==[]]|6<<7<<7>>8>>8and 5
This is a snippet with the following prefix snippets:
# -1
cmp(cmp,())
# +3 = (-1) & 3
cmp(cmp,())&3
# +7 = 3 + 4
cmp(cmp,())&3+4
# +2 = 3 + (-1) = 3 + floor(4/-10)
cmp(cmp,())&3+4/~9
# +1 = 3 + (-2) = 3 + 2*floor(4/-10)
cmp(cmp,())&3+4/~9*2
# -4 = 1 - 5
cmp(cmp,())&3+4/~9*2-1-1-1-1-1
# -2 = (-4) | 6
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6
# -3 = (-4) | 1 = (-4) | floor(6/2) = (-4) | (6>>2)
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8
# 0 = -3 and 0
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 0
# +5 = -3 and 05
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 05
# +4 = 5 ^ 1
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 05^True
Try it online!
answered Feb 12 at 11:32
ovsovs
19k21159
$endgroup$
Python 2, 32 34.7 40 $(11^{3})/32 = 41.59375$
After two hours of trial and error I finally got all integers from -4 to +5 (and 7).
0^cmp('','_')&3+1/~9*2-[-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]-[-[==][==[]]][==[]]][==[]]|6<<7<<7>>8>>8and 5
This is a snippet with the following prefix snippets:
# -1
cmp(cmp,())
# +3 = (-1) & 3
cmp(cmp,())&3
# +7 = 3 + 4
cmp(cmp,())&3+4
# +2 = 3 + (-1) = 3 + floor(4/-10)
cmp(cmp,())&3+4/~9
# +1 = 3 + (-2) = 3 + 2*floor(4/-10)
cmp(cmp,())&3+4/~9*2
# -4 = 1 - 5
cmp(cmp,())&3+4/~9*2-1-1-1-1-1
# -2 = (-4) | 6
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6
# -3 = (-4) | 1 = (-4) | floor(6/2) = (-4) | (6>>2)
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8
# 0 = -3 and 0
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 0
# +5 = -3 and 05
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 05
# +4 = 5 ^ 1
cmp(cmp,())&3+4/~9*2-1-1-1-1-1|6<<7<<7>>8>>8and 05^True
Try it online!
answered Feb 12 at 11:32
ovsovs
19k21159
edited Feb 13 at 10:02
answered Feb 12 at 11:32
ovsovs
19k21159
answered Feb 12 at 11:32
ovsovs
19k21159
answered Feb 12 at 11:32
ovsovs
19k21159
19k21159
1
$begingroup$
You can change the2*2to22&7(-3&6 instead of -3&4) to fix that repetition of*.
$endgroup$
– Kevin Cruijssen
Feb 12 at 11:47
$begingroup$
@KevinCruijssen thanks for the suggestion.
$endgroup$
– ovs
Feb 12 at 11:54
$begingroup$
You can free up'and_by usingcmp(cmp,())instead ofcmp('','_')If that is helpful.
$endgroup$
– TRITICIMAGVS
Feb 13 at 3:46
$begingroup$
You could also make the list section a little shorter and a lot easier to understand by doing--[-1-1-1-1-1], but that doesn't really improve your score.
$endgroup$
– TRITICIMAGVS
Feb 13 at 3:54
add a comment |
1
$begingroup$
You can change the2*2to22&7(-3&6 instead of -3&4) to fix that repetition of*.
$endgroup$
– Kevin Cruijssen
Feb 12 at 11:47
$begingroup$
@KevinCruijssen thanks for the suggestion.
$endgroup$
– ovs
Feb 12 at 11:54
$begingroup$
You can free up'and_by usingcmp(cmp,())instead ofcmp('','_')If that is helpful.
$endgroup$
– TRITICIMAGVS
Feb 13 at 3:46
$begingroup$
You could also make the list section a little shorter and a lot easier to understand by doing--[-1-1-1-1-1], but that doesn't really improve your score.
$endgroup$
– TRITICIMAGVS
Feb 13 at 3:54
1
1
$begingroup$
You can change the
2*2 to 22&7 (-3&6 instead of -3&4) to fix that repetition of *.$endgroup$
– Kevin Cruijssen
Feb 12 at 11:47
$begingroup$
You can change the
2*2 to 22&7 (-3&6 instead of -3&4) to fix that repetition of *.$endgroup$
– Kevin Cruijssen
Feb 12 at 11:47
$begingroup$
@KevinCruijssen thanks for the suggestion.
$endgroup$
– ovs
Feb 12 at 11:54
$begingroup$
@KevinCruijssen thanks for the suggestion.
$endgroup$
– ovs
Feb 12 at 11:54
$begingroup$
You can free up
' and _ by using cmp(cmp,()) instead of cmp('','_') If that is helpful.$endgroup$
– TRITICIMAGVS
Feb 13 at 3:46
$begingroup$
You can free up
' and _ by using cmp(cmp,()) instead of cmp('','_') If that is helpful.$endgroup$
– TRITICIMAGVS
Feb 13 at 3:46
$begingroup$
You could also make the list section a little shorter and a lot easier to understand by doing
--[-1-1-1-1-1], but that doesn't really improve your score.$endgroup$
– TRITICIMAGVS
Feb 13 at 3:54
$begingroup$
You could also make the list section a little shorter and a lot easier to understand by doing
--[-1-1-1-1-1], but that doesn't really improve your score.$endgroup$
– TRITICIMAGVS
Feb 13 at 3:54
add a comment |
2
$begingroup$
Python 2, $11^3/53 = 25.1132075472$
015-8*2+6/3&9%7ifelse 4^ord('b')
Prefixes:
0 # 0
01 # 1
015 # 13 (octal)
015-8 # 5
015-8*2 # -3
015-8*2+6 # 3
015-8*2+6/3 # -1
015-8*2+6/3&9 # 9 (-1 bitwise-AND 9 = 9)
015-8*2+6/3&9%7 # 2
015-8*2+6/3&9%7ifelse 4 # 4
015-8*2+6/3&9%7ifelse 4^ord('b') # 12 (4 bitwise-XOR 8 -- OR would also work)
Not optimal, but the best I could personally find with this particular structure.
answered Feb 11 at 2:17
Aidan F. PierceAidan F. Pierce
833616
$endgroup$
add a comment |
2
$begingroup$
Python 2, $11^3/53 = 25.1132075472$
015-8*2+6/3&9%7ifelse 4^ord('b')
Prefixes:
0 # 0
01 # 1
015 # 13 (octal)
015-8 # 5
015-8*2 # -3
015-8*2+6 # 3
015-8*2+6/3 # -1
015-8*2+6/3&9 # 9 (-1 bitwise-AND 9 = 9)
015-8*2+6/3&9%7 # 2
015-8*2+6/3&9%7ifelse 4 # 4
015-8*2+6/3&9%7ifelse 4^ord('b') # 12 (4 bitwise-XOR 8 -- OR would also work)
Not optimal, but the best I could personally find with this particular structure.
answered Feb 11 at 2:17
Aidan F. PierceAidan F. Pierce
833616
$endgroup$
add a comment |
2
2
2
$begingroup$
Python 2, $11^3/53 = 25.1132075472$
015-8*2+6/3&9%7ifelse 4^ord('b')
Prefixes:
0 # 0
01 # 1
015 # 13 (octal)
015-8 # 5
015-8*2 # -3
015-8*2+6 # 3
015-8*2+6/3 # -1
015-8*2+6/3&9 # 9 (-1 bitwise-AND 9 = 9)
015-8*2+6/3&9%7 # 2
015-8*2+6/3&9%7ifelse 4 # 4
015-8*2+6/3&9%7ifelse 4^ord('b') # 12 (4 bitwise-XOR 8 -- OR would also work)
Not optimal, but the best I could personally find with this particular structure.
answered Feb 11 at 2:17
Aidan F. PierceAidan F. Pierce
833616
$endgroup$
Python 2, $11^3/53 = 25.1132075472$
015-8*2+6/3&9%7ifelse 4^ord('b')
Prefixes:
0 # 0
01 # 1
015 # 13 (octal)
015-8 # 5
015-8*2 # -3
015-8*2+6 # 3
015-8*2+6/3 # -1
015-8*2+6/3&9 # 9 (-1 bitwise-AND 9 = 9)
015-8*2+6/3&9%7 # 2
015-8*2+6/3&9%7ifelse 4 # 4
015-8*2+6/3&9%7ifelse 4^ord('b') # 12 (4 bitwise-XOR 8 -- OR would also work)
Not optimal, but the best I could personally find with this particular structure.
answered Feb 11 at 2:17
Aidan F. PierceAidan F. Pierce
833616
answered Feb 11 at 2:17
Aidan F. PierceAidan F. Pierce
833616
answered Feb 11 at 2:17
Aidan F. PierceAidan F. Pierce
833616
answered Feb 11 at 2:17
Aidan F. PierceAidan F. Pierce
833616
833616
add a comment |
add a comment |
2
$begingroup$
Python 2 $6^3/9 = 24$
False-False**0&1+[<[]][<]^~3>>2
Here are the prefixes and what they output
0 : False-False
-1 : False-False**0
1 : False-False**0&1
2 : False-False**0&1+[<[]][<]
-2 : False-False**0&1+[<[]][<]^~3
-3 : False-False**0&1+[<[]][<]^~3>>2
answered Feb 11 at 20:36
TRITICIMAGVSTRITICIMAGVS
35.1k10158369
$endgroup$
add a comment |
2
$begingroup$
Python 2 $6^3/9 = 24$
False-False**0&1+[<[]][<]^~3>>2
Here are the prefixes and what they output
0 : False-False
-1 : False-False**0
1 : False-False**0&1
2 : False-False**0&1+[<[]][<]
-2 : False-False**0&1+[<[]][<]^~3
-3 : False-False**0&1+[<[]][<]^~3>>2
answered Feb 11 at 20:36
TRITICIMAGVSTRITICIMAGVS
35.1k10158369
$endgroup$
add a comment |
2
2
2
$begingroup$
Python 2 $6^3/9 = 24$
False-False**0&1+[<[]][<]^~3>>2
Here are the prefixes and what they output
0 : False-False
-1 : False-False**0
1 : False-False**0&1
2 : False-False**0&1+[<[]][<]
-2 : False-False**0&1+[<[]][<]^~3
-3 : False-False**0&1+[<[]][<]^~3>>2
answered Feb 11 at 20:36
TRITICIMAGVSTRITICIMAGVS
35.1k10158369
$endgroup$
Python 2 $6^3/9 = 24$
False-False**0&1+[<[]][<]^~3>>2
Here are the prefixes and what they output
0 : False-False
-1 : False-False**0
1 : False-False**0&1
2 : False-False**0&1+[<[]][<]
-2 : False-False**0&1+[<[]][<]^~3
-3 : False-False**0&1+[<[]][<]^~3>>2
answered Feb 11 at 20:36
TRITICIMAGVSTRITICIMAGVS
35.1k10158369
answered Feb 11 at 20:36
TRITICIMAGVSTRITICIMAGVS
35.1k10158369
answered Feb 11 at 20:36
TRITICIMAGVSTRITICIMAGVS
35.1k10158369
answered Feb 11 at 20:36
TRITICIMAGVSTRITICIMAGVS
35.1k10158369
35.1k10158369
add a comment |
add a comment |
2
$begingroup$
Clean, 40$^3$/400 = 160
import StdEnv,Data.GenDefault,Gast.GenLibTest
derive genShow G
derive gDefault G
n = sum[(bitnot o bitnot) one \ n <-: s]-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one
s = (hd o show o d) undef
d :: G -> G
d g = gDefault{|*|} g
:: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`
Try it online!
Defines n which produces -20, -19, ..., 18, 19 depending on the length of the type G.
The smallest prefix of G looks like :: G = G, preceded by the assorted imports and definitions.
The largest prefix of G looks like :: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`, similarly preceded.
answered Feb 12 at 1:35
ΟurousΟurous
7,05211035
$endgroup$
$begingroup$
I think I'm missing something in this challenge, but… this answer (and a handful of others) is reusing "bytes that have appeared in previous prefixes", isn't it?
$endgroup$
– Blackhole
Feb 12 at 1:39
4
$begingroup$
@Blackhole As I (and I assume others) understood it, you can use bytes multiple times in the same prefix, but not in subsequent ones.
$endgroup$
– Οurous
Feb 12 at 1:40
add a comment |
2
$begingroup$
Clean, 40$^3$/400 = 160
import StdEnv,Data.GenDefault,Gast.GenLibTest
derive genShow G
derive gDefault G
n = sum[(bitnot o bitnot) one \ n <-: s]-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one
s = (hd o show o d) undef
d :: G -> G
d g = gDefault{|*|} g
:: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`
Try it online!
Defines n which produces -20, -19, ..., 18, 19 depending on the length of the type G.
The smallest prefix of G looks like :: G = G, preceded by the assorted imports and definitions.
The largest prefix of G looks like :: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`, similarly preceded.
answered Feb 12 at 1:35
ΟurousΟurous
7,05211035
$endgroup$
$begingroup$
I think I'm missing something in this challenge, but… this answer (and a handful of others) is reusing "bytes that have appeared in previous prefixes", isn't it?
$endgroup$
– Blackhole
Feb 12 at 1:39
4
$begingroup$
@Blackhole As I (and I assume others) understood it, you can use bytes multiple times in the same prefix, but not in subsequent ones.
$endgroup$
– Οurous
Feb 12 at 1:40
add a comment |
2
2
2
$begingroup$
Clean, 40$^3$/400 = 160
import StdEnv,Data.GenDefault,Gast.GenLibTest
derive genShow G
derive gDefault G
n = sum[(bitnot o bitnot) one \ n <-: s]-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one
s = (hd o show o d) undef
d :: G -> G
d g = gDefault{|*|} g
:: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`
Try it online!
Defines n which produces -20, -19, ..., 18, 19 depending on the length of the type G.
The smallest prefix of G looks like :: G = G, preceded by the assorted imports and definitions.
The largest prefix of G looks like :: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`, similarly preceded.
answered Feb 12 at 1:35
ΟurousΟurous
7,05211035
$endgroup$
Clean, 40$^3$/400 = 160
import StdEnv,Data.GenDefault,Gast.GenLibTest
derive genShow G
derive gDefault G
n = sum[(bitnot o bitnot) one \ n <-: s]-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one-one
s = (hd o show o d) undef
d :: G -> G
d g = gDefault{|*|} g
:: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`
Try it online!
Defines n which produces -20, -19, ..., 18, 19 depending on the length of the type G.
The smallest prefix of G looks like :: G = G, preceded by the assorted imports and definitions.
The largest prefix of G looks like :: G = GABCFHIJKMNOPQRUVWXYZcjkqxyz_0123456789`, similarly preceded.
answered Feb 12 at 1:35
ΟurousΟurous
7,05211035
answered Feb 12 at 1:35
ΟurousΟurous
7,05211035
answered Feb 12 at 1:35
ΟurousΟurous
7,05211035
answered Feb 12 at 1:35
ΟurousΟurous
7,05211035
7,05211035
$begingroup$
I think I'm missing something in this challenge, but… this answer (and a handful of others) is reusing "bytes that have appeared in previous prefixes", isn't it?
$endgroup$
– Blackhole
Feb 12 at 1:39
4
$begingroup$
@Blackhole As I (and I assume others) understood it, you can use bytes multiple times in the same prefix, but not in subsequent ones.
$endgroup$
– Οurous
Feb 12 at 1:40
add a comment |
$begingroup$
I think I'm missing something in this challenge, but… this answer (and a handful of others) is reusing "bytes that have appeared in previous prefixes", isn't it?
$endgroup$
– Blackhole
Feb 12 at 1:39
4
$begingroup$
@Blackhole As I (and I assume others) understood it, you can use bytes multiple times in the same prefix, but not in subsequent ones.
$endgroup$
– Οurous
Feb 12 at 1:40
$begingroup$
I think I'm missing something in this challenge, but… this answer (and a handful of others) is reusing "bytes that have appeared in previous prefixes", isn't it?
$endgroup$
– Blackhole
Feb 12 at 1:39
$begingroup$
I think I'm missing something in this challenge, but… this answer (and a handful of others) is reusing "bytes that have appeared in previous prefixes", isn't it?
$endgroup$
– Blackhole
Feb 12 at 1:39
4
4
$begingroup$
@Blackhole As I (and I assume others) understood it, you can use bytes multiple times in the same prefix, but not in subsequent ones.
$endgroup$
– Οurous
Feb 12 at 1:40
$begingroup$
@Blackhole As I (and I assume others) understood it, you can use bytes multiple times in the same prefix, but not in subsequent ones.
$endgroup$
– Οurous
Feb 12 at 1:40
add a comment |
2
$begingroup$
Batch, 176³/7744 = 704
@cmd/cset/a%~z0-110
: !"#$&'()*+,.23456789;<=>?ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`bfghijklnopqruvwxy{|} ¡¢£¤¥¦§¨©ª«¬®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ
The first prefix includes the colon and outputs the number -88. Subsequent prefixes output ascending numbers and the final program outputs 87.
answered Feb 12 at 19:51
NeilNeil
80.8k744178
$endgroup$
add a comment |
2
$begingroup$
Batch, 176³/7744 = 704
@cmd/cset/a%~z0-110
: !"#$&'()*+,.23456789;<=>?ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`bfghijklnopqruvwxy{|} ¡¢£¤¥¦§¨©ª«¬®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ
The first prefix includes the colon and outputs the number -88. Subsequent prefixes output ascending numbers and the final program outputs 87.
answered Feb 12 at 19:51
NeilNeil
80.8k744178
$endgroup$
add a comment |
2
2
2
$begingroup$
Batch, 176³/7744 = 704
@cmd/cset/a%~z0-110
: !"#$&'()*+,.23456789;<=>?ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`bfghijklnopqruvwxy{|} ¡¢£¤¥¦§¨©ª«¬®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ
The first prefix includes the colon and outputs the number -88. Subsequent prefixes output ascending numbers and the final program outputs 87.
answered Feb 12 at 19:51
NeilNeil
80.8k744178
$endgroup$
Batch, 176³/7744 = 704
@cmd/cset/a%~z0-110
: !"#$&'()*+,.23456789;<=>?ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`bfghijklnopqruvwxy{|} ¡¢£¤¥¦§¨©ª«¬®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ
The first prefix includes the colon and outputs the number -88. Subsequent prefixes output ascending numbers and the final program outputs 87.
answered Feb 12 at 19:51
NeilNeil
80.8k744178
answered Feb 12 at 19:51
NeilNeil
80.8k744178
answered Feb 12 at 19:51
NeilNeil
80.8k744178
answered Feb 12 at 19:51
NeilNeil
80.8k744178
80.8k744178
add a comment |
add a comment |
2
$begingroup$
Pyth, $(249^{3})/15500 = 996.0$
-C|l"x01x02x03x04x05x06x07x08tnx0bx0crx0ex0fx10x11x12x13x14x15x16x17x18x19x1ax1bx1cx1dx1ex1f !#$%&'()*+,./0123456789:;<=>?@ABDEFGHIJKLMNOPQRSTUVWXYZ^_`abcdefghijkmnopqrstuvwxyz{}~x80x81x82x83x84x85x86x87x88x89x8ax8bx8cx8dx8ex8fx90x91x92x93x94x95x96x97x98x99x9ax9bx9cx9dx9ex9fxa0¡¢£¤¥¦§¨©ª«¬xad®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ
All of the escaped characters are single bytes - I've presented the program in this form for ease of copying and pasting. This link is to the full program without escaped characters; as can be seen there are a lot of placeholders for the unprintable characters used.
The approach used is similar to lirtosiast's TI-BASIC answer, though I wrote my answer before browsing the other entries. The cord works as follows:
-C|l"...
| "|"
C Character code of the above, i.e. 124
- Subtract from the above...
l"... ... the length of the trailing string - the closing quote is inferred
The string is made of every single-byte character other than 0 (NULL), 127 (backspace), and the boilerplate characters (-C|l"). That way, each character of the string can be removed one at a time to produce strings of lengths 248 down to 0. These programs yield results in the range [-124 - 124], with sum of absolute values of 15500.
The code was generated using this program:
+J"-C\|l""sCM-U256+CMJ,Z127
J"-C\|l"" Assign the boilerplate code to J
CMJ Get the character code of each character in J
+ ,Z127 Append 0 and 127
-U256 Get values in range [0-256) which are not in the above
CM Get each character with code points in the above
s Concatenate into string
+J Prepend boilerplate
And the sum of the numbers generated from each of the truncated programs was found with this program:
sm.a.vd>._+J"-C\|l""sCM-U256+CMJ,Z127 5
+J"-C\|l""sCM-U256+CMJ,Z127 The full program string, as above
._ Get the prefixes of the above
> 5 Discard the first 5 (those prefixes which don't include all the boilerplate code)
m Map each of the above to:
.vd Evaluate as Pyth code
.a Take the absolute value of the above
s Take the sum
answered Feb 13 at 11:26
SokSok
4,007925
$endgroup$
add a comment |
2
$begingroup$
Pyth, $(249^{3})/15500 = 996.0$
-C|l"x01x02x03x04x05x06x07x08tnx0bx0crx0ex0fx10x11x12x13x14x15x16x17x18x19x1ax1bx1cx1dx1ex1f !#$%&'()*+,./0123456789:;<=>?@ABDEFGHIJKLMNOPQRSTUVWXYZ^_`abcdefghijkmnopqrstuvwxyz{}~x80x81x82x83x84x85x86x87x88x89x8ax8bx8cx8dx8ex8fx90x91x92x93x94x95x96x97x98x99x9ax9bx9cx9dx9ex9fxa0¡¢£¤¥¦§¨©ª«¬xad®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ
All of the escaped characters are single bytes - I've presented the program in this form for ease of copying and pasting. This link is to the full program without escaped characters; as can be seen there are a lot of placeholders for the unprintable characters used.
The approach used is similar to lirtosiast's TI-BASIC answer, though I wrote my answer before browsing the other entries. The cord works as follows:
-C|l"...
| "|"
C Character code of the above, i.e. 124
- Subtract from the above...
l"... ... the length of the trailing string - the closing quote is inferred
The string is made of every single-byte character other than 0 (NULL), 127 (backspace), and the boilerplate characters (-C|l"). That way, each character of the string can be removed one at a time to produce strings of lengths 248 down to 0. These programs yield results in the range [-124 - 124], with sum of absolute values of 15500.
The code was generated using this program:
+J"-C\|l""sCM-U256+CMJ,Z127
J"-C\|l"" Assign the boilerplate code to J
CMJ Get the character code of each character in J
+ ,Z127 Append 0 and 127
-U256 Get values in range [0-256) which are not in the above
CM Get each character with code points in the above
s Concatenate into string
+J Prepend boilerplate
And the sum of the numbers generated from each of the truncated programs was found with this program:
sm.a.vd>._+J"-C\|l""sCM-U256+CMJ,Z127 5
+J"-C\|l""sCM-U256+CMJ,Z127 The full program string, as above
._ Get the prefixes of the above
> 5 Discard the first 5 (those prefixes which don't include all the boilerplate code)
m Map each of the above to:
.vd Evaluate as Pyth code
.a Take the absolute value of the above
s Take the sum
answered Feb 13 at 11:26
SokSok
4,007925
$endgroup$
add a comment |
2
2
2
$begingroup$
Pyth, $(249^{3})/15500 = 996.0$
-C|l"x01x02x03x04x05x06x07x08tnx0bx0crx0ex0fx10x11x12x13x14x15x16x17x18x19x1ax1bx1cx1dx1ex1f !#$%&'()*+,./0123456789:;<=>?@ABDEFGHIJKLMNOPQRSTUVWXYZ^_`abcdefghijkmnopqrstuvwxyz{}~x80x81x82x83x84x85x86x87x88x89x8ax8bx8cx8dx8ex8fx90x91x92x93x94x95x96x97x98x99x9ax9bx9cx9dx9ex9fxa0¡¢£¤¥¦§¨©ª«¬xad®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ
All of the escaped characters are single bytes - I've presented the program in this form for ease of copying and pasting. This link is to the full program without escaped characters; as can be seen there are a lot of placeholders for the unprintable characters used.
The approach used is similar to lirtosiast's TI-BASIC answer, though I wrote my answer before browsing the other entries. The cord works as follows:
-C|l"...
| "|"
C Character code of the above, i.e. 124
- Subtract from the above...
l"... ... the length of the trailing string - the closing quote is inferred
The string is made of every single-byte character other than 0 (NULL), 127 (backspace), and the boilerplate characters (-C|l"). That way, each character of the string can be removed one at a time to produce strings of lengths 248 down to 0. These programs yield results in the range [-124 - 124], with sum of absolute values of 15500.
The code was generated using this program:
+J"-C\|l""sCM-U256+CMJ,Z127
J"-C\|l"" Assign the boilerplate code to J
CMJ Get the character code of each character in J
+ ,Z127 Append 0 and 127
-U256 Get values in range [0-256) which are not in the above
CM Get each character with code points in the above
s Concatenate into string
+J Prepend boilerplate
And the sum of the numbers generated from each of the truncated programs was found with this program:
sm.a.vd>._+J"-C\|l""sCM-U256+CMJ,Z127 5
+J"-C\|l""sCM-U256+CMJ,Z127 The full program string, as above
._ Get the prefixes of the above
> 5 Discard the first 5 (those prefixes which don't include all the boilerplate code)
m Map each of the above to:
.vd Evaluate as Pyth code
.a Take the absolute value of the above
s Take the sum
answered Feb 13 at 11:26
SokSok
4,007925
$endgroup$
Pyth, $(249^{3})/15500 = 996.0$
-C|l"x01x02x03x04x05x06x07x08tnx0bx0crx0ex0fx10x11x12x13x14x15x16x17x18x19x1ax1bx1cx1dx1ex1f !#$%&'()*+,./0123456789:;<=>?@ABDEFGHIJKLMNOPQRSTUVWXYZ^_`abcdefghijkmnopqrstuvwxyz{}~x80x81x82x83x84x85x86x87x88x89x8ax8bx8cx8dx8ex8fx90x91x92x93x94x95x96x97x98x99x9ax9bx9cx9dx9ex9fxa0¡¢£¤¥¦§¨©ª«¬xad®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ
All of the escaped characters are single bytes - I've presented the program in this form for ease of copying and pasting. This link is to the full program without escaped characters; as can be seen there are a lot of placeholders for the unprintable characters used.
The approach used is similar to lirtosiast's TI-BASIC answer, though I wrote my answer before browsing the other entries. The cord works as follows:
-C|l"...
| "|"
C Character code of the above, i.e. 124
- Subtract from the above...
l"... ... the length of the trailing string - the closing quote is inferred
The string is made of every single-byte character other than 0 (NULL), 127 (backspace), and the boilerplate characters (-C|l"). That way, each character of the string can be removed one at a time to produce strings of lengths 248 down to 0. These programs yield results in the range [-124 - 124], with sum of absolute values of 15500.
The code was generated using this program:
+J"-C\|l""sCM-U256+CMJ,Z127
J"-C\|l"" Assign the boilerplate code to J
CMJ Get the character code of each character in J
+ ,Z127 Append 0 and 127
-U256 Get values in range [0-256) which are not in the above
CM Get each character with code points in the above
s Concatenate into string
+J Prepend boilerplate
And the sum of the numbers generated from each of the truncated programs was found with this program:
sm.a.vd>._+J"-C\|l""sCM-U256+CMJ,Z127 5
+J"-C\|l""sCM-U256+CMJ,Z127 The full program string, as above
._ Get the prefixes of the above
> 5 Discard the first 5 (those prefixes which don't include all the boilerplate code)
m Map each of the above to:
.vd Evaluate as Pyth code
.a Take the absolute value of the above
s Take the sum
answered Feb 13 at 11:26
SokSok
4,007925
edited Feb 13 at 11:48
answered Feb 13 at 11:26
SokSok
4,007925
answered Feb 13 at 11:26
SokSok
4,007925
answered Feb 13 at 11:26
SokSok
4,007925
4,007925
add a comment |
add a comment |
2
$begingroup$
Jelly, 984
LNḂ¡:2µ}“¢¢¢£¤¥¦©¬®½¿€ÆÇÐÑ×ØŒÞßæçðıȷñ÷øœþ !"#$%&'()*+,-./013456789;<=>?@ABCDEFGHIJKMOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|~¶°¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾ƁƇƊƑƓƘⱮƝƤƬƲȤɓƈɗƒɠɦƙɱɲƥʠɼʂƭʋȥẠḄḌẸḤỊḲḶṂṆỌṚṢṬỤṾẈỴẒȦĊḊĖḞĠḢİĿṀṄȮṖṘṠṪẆẊẎŻạḅḍẹḥịḳḷṃṇọṛṣṭ§Äẉỵẓȧḃċḋėḟġḣŀṁṅȯṗṙṡṫẇẋẏż”‘
Try it online!
You can see the prefixes (separated by a double newline) here.
The numbers printed are, in order:
0 1 -1 -2 2 -3 3 -4 4 -5 5 -6 6 -7 7 -8 8 -9 9 -10 10 -11 11 -12 12 -13 13 -14 14 -15 15 -16 16 -17 17 -18 18 -19 19 -20 20 -21 21 -22 22 -23 23 -24 24 -25 25 -26 26 -27 27 -28 28 -29 29 -30 30 -31 31 -32 32 -33 33 -34 34 -35 35 -36 36 -37 37 -38 38 -39 39 -40 40 -41 41 -42 42 -43 43 -44 44 -45 45 -46 46 -47 47 -48 48 -49 49 -50 50 -51 51 -52 52 -53 53 -54 54 -55 55 -56 56 -57 57 -58 58 -59 59 -60 60 -61 61 -62 62 -63 63 -64 64 -65 65 -66 66 -67 67 -68 68 -69 69 -70 70 -71 71 -72 72 -73 73 -74 74 -75 75 -76 76 -77 77 -78 78 -79 79 -80 80 -81 81 -82 82 -83 83 -84 84 -85 85 -86 86 -87 87 -88 88 -89 89 -90 90 -91 91 -92 92 -93 93 -94 94 -95 95 -96 96 -97 97 -98 98 -99 99 -100 100 -101 101 -102 102 -103 103 -104 104 -105 105 -106 106 -107 107 -108 108 -109 109 -110 110 -111 111 -112 112 -113 113 -114 114 -115 115 -116 116 -117 117 -118 118 -119 119 -120 120 -121 121 -122 122 123
answered Feb 13 at 19:26
Erik the OutgolferErik the Outgolfer
31.9k429103
$endgroup$
add a comment |
2
$begingroup$
Jelly, 984
LNḂ¡:2µ}“¢¢¢£¤¥¦©¬®½¿€ÆÇÐÑ×ØŒÞßæçðıȷñ÷øœþ !"#$%&'()*+,-./013456789;<=>?@ABCDEFGHIJKMOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|~¶°¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾ƁƇƊƑƓƘⱮƝƤƬƲȤɓƈɗƒɠɦƙɱɲƥʠɼʂƭʋȥẠḄḌẸḤỊḲḶṂṆỌṚṢṬỤṾẈỴẒȦĊḊĖḞĠḢİĿṀṄȮṖṘṠṪẆẊẎŻạḅḍẹḥịḳḷṃṇọṛṣṭ§Äẉỵẓȧḃċḋėḟġḣŀṁṅȯṗṙṡṫẇẋẏż”‘
Try it online!
You can see the prefixes (separated by a double newline) here.
The numbers printed are, in order:
0 1 -1 -2 2 -3 3 -4 4 -5 5 -6 6 -7 7 -8 8 -9 9 -10 10 -11 11 -12 12 -13 13 -14 14 -15 15 -16 16 -17 17 -18 18 -19 19 -20 20 -21 21 -22 22 -23 23 -24 24 -25 25 -26 26 -27 27 -28 28 -29 29 -30 30 -31 31 -32 32 -33 33 -34 34 -35 35 -36 36 -37 37 -38 38 -39 39 -40 40 -41 41 -42 42 -43 43 -44 44 -45 45 -46 46 -47 47 -48 48 -49 49 -50 50 -51 51 -52 52 -53 53 -54 54 -55 55 -56 56 -57 57 -58 58 -59 59 -60 60 -61 61 -62 62 -63 63 -64 64 -65 65 -66 66 -67 67 -68 68 -69 69 -70 70 -71 71 -72 72 -73 73 -74 74 -75 75 -76 76 -77 77 -78 78 -79 79 -80 80 -81 81 -82 82 -83 83 -84 84 -85 85 -86 86 -87 87 -88 88 -89 89 -90 90 -91 91 -92 92 -93 93 -94 94 -95 95 -96 96 -97 97 -98 98 -99 99 -100 100 -101 101 -102 102 -103 103 -104 104 -105 105 -106 106 -107 107 -108 108 -109 109 -110 110 -111 111 -112 112 -113 113 -114 114 -115 115 -116 116 -117 117 -118 118 -119 119 -120 120 -121 121 -122 122 123
answered Feb 13 at 19:26
Erik the OutgolferErik the Outgolfer
31.9k429103
$endgroup$
add a comment |
2
2
2
$begingroup$
Jelly, 984
LNḂ¡:2µ}“¢¢¢£¤¥¦©¬®½¿€ÆÇÐÑ×ØŒÞßæçðıȷñ÷øœþ !"#$%&'()*+,-./013456789;<=>?@ABCDEFGHIJKMOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|~¶°¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾ƁƇƊƑƓƘⱮƝƤƬƲȤɓƈɗƒɠɦƙɱɲƥʠɼʂƭʋȥẠḄḌẸḤỊḲḶṂṆỌṚṢṬỤṾẈỴẒȦĊḊĖḞĠḢİĿṀṄȮṖṘṠṪẆẊẎŻạḅḍẹḥịḳḷṃṇọṛṣṭ§Äẉỵẓȧḃċḋėḟġḣŀṁṅȯṗṙṡṫẇẋẏż”‘
Try it online!
You can see the prefixes (separated by a double newline) here.
The numbers printed are, in order:
0 1 -1 -2 2 -3 3 -4 4 -5 5 -6 6 -7 7 -8 8 -9 9 -10 10 -11 11 -12 12 -13 13 -14 14 -15 15 -16 16 -17 17 -18 18 -19 19 -20 20 -21 21 -22 22 -23 23 -24 24 -25 25 -26 26 -27 27 -28 28 -29 29 -30 30 -31 31 -32 32 -33 33 -34 34 -35 35 -36 36 -37 37 -38 38 -39 39 -40 40 -41 41 -42 42 -43 43 -44 44 -45 45 -46 46 -47 47 -48 48 -49 49 -50 50 -51 51 -52 52 -53 53 -54 54 -55 55 -56 56 -57 57 -58 58 -59 59 -60 60 -61 61 -62 62 -63 63 -64 64 -65 65 -66 66 -67 67 -68 68 -69 69 -70 70 -71 71 -72 72 -73 73 -74 74 -75 75 -76 76 -77 77 -78 78 -79 79 -80 80 -81 81 -82 82 -83 83 -84 84 -85 85 -86 86 -87 87 -88 88 -89 89 -90 90 -91 91 -92 92 -93 93 -94 94 -95 95 -96 96 -97 97 -98 98 -99 99 -100 100 -101 101 -102 102 -103 103 -104 104 -105 105 -106 106 -107 107 -108 108 -109 109 -110 110 -111 111 -112 112 -113 113 -114 114 -115 115 -116 116 -117 117 -118 118 -119 119 -120 120 -121 121 -122 122 123
answered Feb 13 at 19:26
Erik the OutgolferErik the Outgolfer
31.9k429103
$endgroup$
Jelly, 984
LNḂ¡:2µ}“¢¢¢£¤¥¦©¬®½¿€ÆÇÐÑ×ØŒÞßæçðıȷñ÷øœþ !"#$%&'()*+,-./013456789;<=>?@ABCDEFGHIJKMOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|~¶°¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾ƁƇƊƑƓƘⱮƝƤƬƲȤɓƈɗƒɠɦƙɱɲƥʠɼʂƭʋȥẠḄḌẸḤỊḲḶṂṆỌṚṢṬỤṾẈỴẒȦĊḊĖḞĠḢİĿṀṄȮṖṘṠṪẆẊẎŻạḅḍẹḥịḳḷṃṇọṛṣṭ§Äẉỵẓȧḃċḋėḟġḣŀṁṅȯṗṙṡṫẇẋẏż”‘
Try it online!
You can see the prefixes (separated by a double newline) here.
The numbers printed are, in order:
0 1 -1 -2 2 -3 3 -4 4 -5 5 -6 6 -7 7 -8 8 -9 9 -10 10 -11 11 -12 12 -13 13 -14 14 -15 15 -16 16 -17 17 -18 18 -19 19 -20 20 -21 21 -22 22 -23 23 -24 24 -25 25 -26 26 -27 27 -28 28 -29 29 -30 30 -31 31 -32 32 -33 33 -34 34 -35 35 -36 36 -37 37 -38 38 -39 39 -40 40 -41 41 -42 42 -43 43 -44 44 -45 45 -46 46 -47 47 -48 48 -49 49 -50 50 -51 51 -52 52 -53 53 -54 54 -55 55 -56 56 -57 57 -58 58 -59 59 -60 60 -61 61 -62 62 -63 63 -64 64 -65 65 -66 66 -67 67 -68 68 -69 69 -70 70 -71 71 -72 72 -73 73 -74 74 -75 75 -76 76 -77 77 -78 78 -79 79 -80 80 -81 81 -82 82 -83 83 -84 84 -85 85 -86 86 -87 87 -88 88 -89 89 -90 90 -91 91 -92 92 -93 93 -94 94 -95 95 -96 96 -97 97 -98 98 -99 99 -100 100 -101 101 -102 102 -103 103 -104 104 -105 105 -106 106 -107 107 -108 108 -109 109 -110 110 -111 111 -112 112 -113 113 -114 114 -115 115 -116 116 -117 117 -118 118 -119 119 -120 120 -121 121 -122 122 123
answered Feb 13 at 19:26
Erik the OutgolferErik the Outgolfer
31.9k429103
edited Feb 13 at 19:57
answered Feb 13 at 19:26
Erik the OutgolferErik the Outgolfer
31.9k429103
answered Feb 13 at 19:26
Erik the OutgolferErik the Outgolfer
31.9k429103
answered Feb 13 at 19:26
Erik the OutgolferErik the Outgolfer
31.9k429103
31.9k429103
add a comment |
add a comment |
1
$begingroup$
Python 3, $(6^{3})/12 = 18$
0x1-2+3|True*~4
This is a snippet.
The individual prefix snippets are:
0 # gives 0
0x1 # gives 1 (hex syntax)
0x1-2 # gives -1 (1 minus 2)
0x1-2+3 # gives 2 (-1 plus 3)
0x1-2+3|True # gives 3 (True == 1; 2 bitwise-OR 1 == 3)
0x1-2+3|True*~4 # gives -5 (~4 == -5; True*~4 == -5; 2|-5 == -5)
Python's bitwise operators treat negative integers as two's complement with unlimited 1s to the right, so -5 is treated as ...11111011, which when ORed with ...000010 gives ...11111011 unchanged.
answered Feb 11 at 1:32
pizzapants184pizzapants184
2,674716
$endgroup$
add a comment |
1
$begingroup$
Python 3, $(6^{3})/12 = 18$
0x1-2+3|True*~4
This is a snippet.
The individual prefix snippets are:
0 # gives 0
0x1 # gives 1 (hex syntax)
0x1-2 # gives -1 (1 minus 2)
0x1-2+3 # gives 2 (-1 plus 3)
0x1-2+3|True # gives 3 (True == 1; 2 bitwise-OR 1 == 3)
0x1-2+3|True*~4 # gives -5 (~4 == -5; True*~4 == -5; 2|-5 == -5)
Python's bitwise operators treat negative integers as two's complement with unlimited 1s to the right, so -5 is treated as ...11111011, which when ORed with ...000010 gives ...11111011 unchanged.
answered Feb 11 at 1:32
pizzapants184pizzapants184
2,674716
$endgroup$
add a comment |
1
1
1
$begingroup$
Python 3, $(6^{3})/12 = 18$
0x1-2+3|True*~4
This is a snippet.
The individual prefix snippets are:
0 # gives 0
0x1 # gives 1 (hex syntax)
0x1-2 # gives -1 (1 minus 2)
0x1-2+3 # gives 2 (-1 plus 3)
0x1-2+3|True # gives 3 (True == 1; 2 bitwise-OR 1 == 3)
0x1-2+3|True*~4 # gives -5 (~4 == -5; True*~4 == -5; 2|-5 == -5)
Python's bitwise operators treat negative integers as two's complement with unlimited 1s to the right, so -5 is treated as ...11111011, which when ORed with ...000010 gives ...11111011 unchanged.
answered Feb 11 at 1:32
pizzapants184pizzapants184
2,674716
$endgroup$
Python 3, $(6^{3})/12 = 18$
0x1-2+3|True*~4
This is a snippet.
The individual prefix snippets are:
0 # gives 0
0x1 # gives 1 (hex syntax)
0x1-2 # gives -1 (1 minus 2)
0x1-2+3 # gives 2 (-1 plus 3)
0x1-2+3|True # gives 3 (True == 1; 2 bitwise-OR 1 == 3)
0x1-2+3|True*~4 # gives -5 (~4 == -5; True*~4 == -5; 2|-5 == -5)
Python's bitwise operators treat negative integers as two's complement with unlimited 1s to the right, so -5 is treated as ...11111011, which when ORed with ...000010 gives ...11111011 unchanged.
answered Feb 11 at 1:32
pizzapants184pizzapants184
2,674716
answered Feb 11 at 1:32
pizzapants184pizzapants184
2,674716
answered Feb 11 at 1:32
pizzapants184pizzapants184
2,674716
answered Feb 11 at 1:32
pizzapants184pizzapants184
2,674716
2,674716
add a comment |
add a comment |
1
$begingroup$
Excel, $(6^{3})/12 = 18$
=0 0
=01 1
=01*2 2
=01*2-5 -3
=01*2-5+8 5
=01*2-5+8/4 -1
Prefix challenge does not gel well with Excel's formula structure
answered Feb 11 at 10:12
WernischWernisch
1,628317
$endgroup$
add a comment |
1
$begingroup$
Excel, $(6^{3})/12 = 18$
=0 0
=01 1
=01*2 2
=01*2-5 -3
=01*2-5+8 5
=01*2-5+8/4 -1
Prefix challenge does not gel well with Excel's formula structure
answered Feb 11 at 10:12
WernischWernisch
1,628317
$endgroup$
add a comment |
1
1
1
$begingroup$
Excel, $(6^{3})/12 = 18$
=0 0
=01 1
=01*2 2
=01*2-5 -3
=01*2-5+8 5
=01*2-5+8/4 -1
Prefix challenge does not gel well with Excel's formula structure
answered Feb 11 at 10:12
WernischWernisch
1,628317
$endgroup$
Excel, $(6^{3})/12 = 18$
=0 0
=01 1
=01*2 2
=01*2-5 -3
=01*2-5+8 5
=01*2-5+8/4 -1
Prefix challenge does not gel well with Excel's formula structure
answered Feb 11 at 10:12
WernischWernisch
1,628317
answered Feb 11 at 10:12
WernischWernisch
1,628317
answered Feb 11 at 10:12
WernischWernisch
1,628317
answered Feb 11 at 10:12
WernischWernisch
1,628317
1,628317
add a comment |
add a comment |
0
$begingroup$
Charcoal, 953/2256 = 380.04211
Ｉ⁻⁴⁸Ｌ !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|}~
The first prefix includes the space and outputs the number 47. Try it online! Subsequent prefixes output descending numbers and the final program outputs -47. Try it online!
answered Feb 11 at 10:53
NeilNeil
80.8k744178
$endgroup$
add a comment |
0
$begingroup$
Charcoal, 953/2256 = 380.04211
Ｉ⁻⁴⁸Ｌ !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|}~
The first prefix includes the space and outputs the number 47. Try it online! Subsequent prefixes output descending numbers and the final program outputs -47. Try it online!
answered Feb 11 at 10:53
NeilNeil
80.8k744178
$endgroup$
add a comment |
0
0
0
$begingroup$
Charcoal, 953/2256 = 380.04211
Ｉ⁻⁴⁸Ｌ !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|}~
The first prefix includes the space and outputs the number 47. Try it online! Subsequent prefixes output descending numbers and the final program outputs -47. Try it online!
answered Feb 11 at 10:53
NeilNeil
80.8k744178
$endgroup$
Charcoal, 953/2256 = 380.04211
Ｉ⁻⁴⁸Ｌ !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijklmnopqrstuvwxyz{|}~
The first prefix includes the space and outputs the number 47. Try it online! Subsequent prefixes output descending numbers and the final program outputs -47. Try it online!
answered Feb 11 at 10:53
NeilNeil
80.8k744178
answered Feb 11 at 10:53
NeilNeil
80.8k744178
answered Feb 11 at 10:53
NeilNeil
80.8k744178
answered Feb 11 at 10:53
NeilNeil
80.8k744178
80.8k744178
add a comment |
add a comment |
0
$begingroup$
Runic Enchantments, 1173/3422 = 468.034
>"ul'B-$; !#%&()*+,.:0123456789@<=?ACDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijkmnopqrstvwxyz{|}~€
Try it online!
And the prefix.
Prints [-58,58] inclusive.
Uses a similar tactic as other entries, using a progressively longer string, getting its length, subtracts B (decimal value of 66), and prints the result. Uses a non-printing character (shown in the code block as € because apparently that's what € evaluates to in HTML) in order to utilize byte value 0x80 (and consuming an otherwise unusable 0x00) and score one more value. Only byte values up to 127 that can't be represented in any fashion are 10 and 13, because they would cause the program to be two lines tall and not be read correctly by the instruction pointer.
answered Feb 14 at 3:38
Draco18sDraco18s
1,356619
$endgroup$
add a comment |
0
$begingroup$
Runic Enchantments, 1173/3422 = 468.034
>"ul'B-$; !#%&()*+,.:0123456789@<=?ACDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijkmnopqrstvwxyz{|}~€
Try it online!
And the prefix.
Prints [-58,58] inclusive.
Uses a similar tactic as other entries, using a progressively longer string, getting its length, subtracts B (decimal value of 66), and prints the result. Uses a non-printing character (shown in the code block as € because apparently that's what € evaluates to in HTML) in order to utilize byte value 0x80 (and consuming an otherwise unusable 0x00) and score one more value. Only byte values up to 127 that can't be represented in any fashion are 10 and 13, because they would cause the program to be two lines tall and not be read correctly by the instruction pointer.
answered Feb 14 at 3:38
Draco18sDraco18s
1,356619
$endgroup$
add a comment |
0
0
0
$begingroup$
Runic Enchantments, 1173/3422 = 468.034
>"ul'B-$; !#%&()*+,.:0123456789@<=?ACDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijkmnopqrstvwxyz{|}~€
Try it online!
And the prefix.
Prints [-58,58] inclusive.
Uses a similar tactic as other entries, using a progressively longer string, getting its length, subtracts B (decimal value of 66), and prints the result. Uses a non-printing character (shown in the code block as € because apparently that's what € evaluates to in HTML) in order to utilize byte value 0x80 (and consuming an otherwise unusable 0x00) and score one more value. Only byte values up to 127 that can't be represented in any fashion are 10 and 13, because they would cause the program to be two lines tall and not be read correctly by the instruction pointer.
answered Feb 14 at 3:38
Draco18sDraco18s
1,356619
$endgroup$
Runic Enchantments, 1173/3422 = 468.034
>"ul'B-$; !#%&()*+,.:0123456789@<=?ACDEFGHIJKLMNOPQRSTUVWXYZ[]^_`abcdefghijkmnopqrstvwxyz{|}~€
Try it online!
And the prefix.
Prints [-58,58] inclusive.
Uses a similar tactic as other entries, using a progressively longer string, getting its length, subtracts B (decimal value of 66), and prints the result. Uses a non-printing character (shown in the code block as € because apparently that's what € evaluates to in HTML) in order to utilize byte value 0x80 (and consuming an otherwise unusable 0x00) and score one more value. Only byte values up to 127 that can't be represented in any fashion are 10 and 13, because they would cause the program to be two lines tall and not be read correctly by the instruction pointer.
answered Feb 14 at 3:38
Draco18sDraco18s
1,356619
answered Feb 14 at 3:38
Draco18sDraco18s
1,356619
answered Feb 14 at 3:38
Draco18sDraco18s
1,356619
answered Feb 14 at 3:38
Draco18sDraco18s
1,356619
1,356619
add a comment |
add a comment |
draft saved
draft discarded
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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2
$begingroup$
The Final submission line appears to have swapped the 3 for a 2 (I haven't edited to correct this in case I've misunderstood)
$endgroup$
– trichoplax
Feb 11 at 1:23
3
$begingroup$
It looks to me like you can get a near-optimal score in Python by writing
print 123-len(open(__file__).read())followed by#then one of each byte not used so far (except some control characters), with prefixes chopping those bytes. Adjust 123 to whatever centers the range of outputs around 0. For a better score, the non-comment part of the program can be rewritten using only the 7 distinct characters that are universal for Python. Is this all valid?$endgroup$
– xnor
Feb 11 at 1:37
1
$begingroup$
@JoKing I do include negatives, when the difference is negative.
$endgroup$
– xnor
Feb 11 at 1:41
1
$begingroup$
@JoKing This method also be done in languages where a string can be written without a closing quote.
$endgroup$
– xnor
Feb 11 at 1:47
2
$begingroup$
I can't believe so few other people noticed it and upvoted the earlier comment. :P I just noticed that @trichoplax spotted the typo, too.
$endgroup$
– Peter Cordes
Feb 12 at 11:07