Am I a Rude Number?
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For a while now, I've been running into a problem when counting on my fingers, specifically, that I can only count to ten. My solution to that problem has been to count in binary on my fingers, putting up my thumb for one, my forefinger for two, both thumb and forefinger for three, etc. However, we run into a bit of a problem when we get to the number four. Specifically, it requires us to put up our middle finger, which results in a rather unfortunate gesture, which is not typically accepted in society. This type of number is a rude number. We come to the next rude number at 36, when we raise the thumb on our second hand and the middle finger of our first hand. The definition of a rude number is any number that, under this system of counting, results in us putting up only the middle finger of any hand. Once we pass 1023 (the maximum number reachable on one person, with two hands of five fingers each), assume we continue with a third hand, with additional hands added as required.
Your Task:
Write a program or function that receives an input and outputs a truthy/falsy value based on whether the input is a rude number.
Input:
An integer between 0 and 109 (inclusive).
Output:
A truthy/falsy value that indicates whether the input is a rude number.
Test Cases:
Input: Output:
0 ---> falsy
3 ---> falsy
4 ---> truthy
25 ---> falsy
36 ---> truthy
127 ---> falsy
131 ---> truthy
Scoring:
This is code-golf, so the lowest score in bytes wins.
code-golf number decision-problem
edited Feb 27 at 17:35
Glorfindel
199119
asked Feb 27 at 3:21
GryphonGryphon
3,51912165
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show 6 more comments
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For a while now, I've been running into a problem when counting on my fingers, specifically, that I can only count to ten. My solution to that problem has been to count in binary on my fingers, putting up my thumb for one, my forefinger for two, both thumb and forefinger for three, etc. However, we run into a bit of a problem when we get to the number four. Specifically, it requires us to put up our middle finger, which results in a rather unfortunate gesture, which is not typically accepted in society. This type of number is a rude number. We come to the next rude number at 36, when we raise the thumb on our second hand and the middle finger of our first hand. The definition of a rude number is any number that, under this system of counting, results in us putting up only the middle finger of any hand. Once we pass 1023 (the maximum number reachable on one person, with two hands of five fingers each), assume we continue with a third hand, with additional hands added as required.
Your Task:
Write a program or function that receives an input and outputs a truthy/falsy value based on whether the input is a rude number.
Input:
An integer between 0 and 109 (inclusive).
Output:
A truthy/falsy value that indicates whether the input is a rude number.
Test Cases:
Input: Output:
0 ---> falsy
3 ---> falsy
4 ---> truthy
25 ---> falsy
36 ---> truthy
127 ---> falsy
131 ---> truthy
Scoring:
This is code-golf, so the lowest score in bytes wins.
code-golf number decision-problem
edited Feb 27 at 17:35
Glorfindel
199119
asked Feb 27 at 3:21
GryphonGryphon
3,51912165
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42
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assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.
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– Veskah
Feb 27 at 3:24
5
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@Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
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– Gryphon
Feb 27 at 3:26
12
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It's worse if you're British - 6 is rude too then!
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– Matthew
Feb 27 at 7:07
1
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Is it OK to take input in a different base than 10?
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– wastl
Feb 27 at 20:08
1
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5 seems fairly rude too. Not sure anyone would say "Oh she had her thumb out, that's perfectly polite"
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– ale10ander
Mar 1 at 1:43
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show 6 more comments
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For a while now, I've been running into a problem when counting on my fingers, specifically, that I can only count to ten. My solution to that problem has been to count in binary on my fingers, putting up my thumb for one, my forefinger for two, both thumb and forefinger for three, etc. However, we run into a bit of a problem when we get to the number four. Specifically, it requires us to put up our middle finger, which results in a rather unfortunate gesture, which is not typically accepted in society. This type of number is a rude number. We come to the next rude number at 36, when we raise the thumb on our second hand and the middle finger of our first hand. The definition of a rude number is any number that, under this system of counting, results in us putting up only the middle finger of any hand. Once we pass 1023 (the maximum number reachable on one person, with two hands of five fingers each), assume we continue with a third hand, with additional hands added as required.
Your Task:
Write a program or function that receives an input and outputs a truthy/falsy value based on whether the input is a rude number.
Input:
An integer between 0 and 109 (inclusive).
Output:
A truthy/falsy value that indicates whether the input is a rude number.
Test Cases:
Input: Output:
0 ---> falsy
3 ---> falsy
4 ---> truthy
25 ---> falsy
36 ---> truthy
127 ---> falsy
131 ---> truthy
Scoring:
This is code-golf, so the lowest score in bytes wins.
code-golf number decision-problem
edited Feb 27 at 17:35
Glorfindel
199119
asked Feb 27 at 3:21
GryphonGryphon
3,51912165
$endgroup$
For a while now, I've been running into a problem when counting on my fingers, specifically, that I can only count to ten. My solution to that problem has been to count in binary on my fingers, putting up my thumb for one, my forefinger for two, both thumb and forefinger for three, etc. However, we run into a bit of a problem when we get to the number four. Specifically, it requires us to put up our middle finger, which results in a rather unfortunate gesture, which is not typically accepted in society. This type of number is a rude number. We come to the next rude number at 36, when we raise the thumb on our second hand and the middle finger of our first hand. The definition of a rude number is any number that, under this system of counting, results in us putting up only the middle finger of any hand. Once we pass 1023 (the maximum number reachable on one person, with two hands of five fingers each), assume we continue with a third hand, with additional hands added as required.
Your Task:
Write a program or function that receives an input and outputs a truthy/falsy value based on whether the input is a rude number.
Input:
An integer between 0 and 109 (inclusive).
Output:
A truthy/falsy value that indicates whether the input is a rude number.
Test Cases:
Input: Output:
0 ---> falsy
3 ---> falsy
4 ---> truthy
25 ---> falsy
36 ---> truthy
127 ---> falsy
131 ---> truthy
Scoring:
This is code-golf, so the lowest score in bytes wins.
code-golf number decision-problem
code-golf number decision-problem
edited Feb 27 at 17:35
Glorfindel
199119
asked Feb 27 at 3:21
GryphonGryphon
3,51912165
edited Feb 27 at 17:35
Glorfindel
199119
asked Feb 27 at 3:21
GryphonGryphon
3,51912165
edited Feb 27 at 17:35
Glorfindel
199119
edited Feb 27 at 17:35
Glorfindel
199119
edited Feb 27 at 17:35
Glorfindel
199119
199119
asked Feb 27 at 3:21
GryphonGryphon
3,51912165
asked Feb 27 at 3:21
GryphonGryphon
3,51912165
asked Feb 27 at 3:21
GryphonGryphon
3,51912165
3,51912165
42
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assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.
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– Veskah
Feb 27 at 3:24
5
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@Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
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– Gryphon
Feb 27 at 3:26
12
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It's worse if you're British - 6 is rude too then!
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– Matthew
Feb 27 at 7:07
1
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Is it OK to take input in a different base than 10?
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– wastl
Feb 27 at 20:08
1
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5 seems fairly rude too. Not sure anyone would say "Oh she had her thumb out, that's perfectly polite"
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– ale10ander
Mar 1 at 1:43
|
show 6 more comments
42
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assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.
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– Veskah
Feb 27 at 3:24
5
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@Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
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– Gryphon
Feb 27 at 3:26
12
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It's worse if you're British - 6 is rude too then!
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– Matthew
Feb 27 at 7:07
1
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Is it OK to take input in a different base than 10?
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– wastl
Feb 27 at 20:08
1
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5 seems fairly rude too. Not sure anyone would say "Oh she had her thumb out, that's perfectly polite"
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– ale10ander
Mar 1 at 1:43
42
42
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assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.$endgroup$
– Veskah
Feb 27 at 3:24
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assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.$endgroup$
– Veskah
Feb 27 at 3:24
5
5
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@Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
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– Gryphon
Feb 27 at 3:26
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@Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
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– Gryphon
Feb 27 at 3:26
12
12
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It's worse if you're British - 6 is rude too then!
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– Matthew
Feb 27 at 7:07
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It's worse if you're British - 6 is rude too then!
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– Matthew
Feb 27 at 7:07
1
1
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Is it OK to take input in a different base than 10?
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– wastl
Feb 27 at 20:08
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Is it OK to take input in a different base than 10?
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– wastl
Feb 27 at 20:08
1
1
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5 seems fairly rude too. Not sure anyone would say "Oh she had her thumb out, that's perfectly polite"
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– ale10ander
Mar 1 at 1:43
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5 seems fairly rude too. Not sure anyone would say "Oh she had her thumb out, that's perfectly polite"
$endgroup$
– ale10ander
Mar 1 at 1:43
|
show 6 more comments
35 Answers
35
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oldest
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APL (dzaima/APL), 5 bytes
4∊32⊤
Try it online!
4∊ is 4 a member of
32⊤ to-base-32?
answered Feb 27 at 5:42
AdámAdám
28.8k276204
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s/bytes/characters though?
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– tripleee
Feb 27 at 10:34
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@tripleee Thanks, added.
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– Adám
Feb 27 at 11:30
add a comment |
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Regex (ECMAScript), 37 bytes
Input is in unary, as the length of a string of xs.
^((?=(x+)(2{31}x*))3)*(x{32})*x{4}$
Try it online!
^
(
(?=(x+)(2{31}x*)) # 2 = floor(tail / 32); 3 = tool to make tail = 2
3 # tail = 2
)* # Loop the above as many times as necessary to make
# the below match
(x{32})*x{4}$ # Assert that tail % 32 == 4
answered Feb 27 at 4:16
DeadcodeDeadcode
1,8641420
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I thought I knew regex, but apparently not.
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– CT Hall
Feb 28 at 4:16
add a comment |
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JavaScript (SpiderMonkey), 23 bytes
f=x=>x&&x%32==4|f(x>>5)
Try it online!
This is a trivial solution, you just want to convert to base 32 and check if there is a 4 in it.
JavaScript (SpiderMonkey), 26 bytes
x=>x.toString(32).match(4)
Try it online!
It's interesting that /4/.test(...) cost one more byte than ....match(4).
answered Feb 27 at 4:17
tshtsh
9,32511652
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Japt, 5 bytes
sH ø4
Try it online!
Explanation
// Implicit input
sH // To a base-H (=32) string
ø // Contains
4 // 4 (JavaScript interprets this as a string)
answered Feb 27 at 4:20
ASCII-onlyASCII-only
4,3081238
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add a comment |
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APL+WIN, 10 bytes
Prompts for input of integer
4∊(6⍴32)⊤⎕
Noting six hands are required to represent 10^9 converts to vector of 6 elements of the base 32 representation and checks if a 4 exists in any element.
answered Feb 27 at 4:45
GrahamGraham
2,49678
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add a comment |
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Ruby, 36 19 bytes
->n{n.to_s(32)[?4]}
Try it online!
Saved 17 bytes with @tsh's method.
answered Feb 27 at 3:37
Doorknob♦Doorknob
54.9k17115352
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This returns true for 2207, which has a binary representation of100010011111
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– Embodiment of Ignorance
Feb 27 at 4:10
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@EmbodimentofIgnorance That is the correct result, is it not? The second hand is00100.
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– Doorknob♦
Feb 27 at 4:16
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I don't speak Ruby. But why not->n{n.to_s(32)=~/4/}?
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– tsh
Feb 27 at 4:19
1
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@tsh because I'm not as clever as you :)
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– Doorknob♦
Feb 27 at 4:23
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Forgive me if I'm not understanding the question, but isn't the first hand of 220710001, the second00111, and the third11? None of them have their middle finger only up
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– Embodiment of Ignorance
Feb 27 at 4:28
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show 3 more comments
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Perl 6, 16 bytes
{.base(32)~~/4/}
Try it online!
Checks if there is a 4 in the base 32 representation of the number. Returns either Nil as false or a Match containing a 4.
You can prove this by the fact that $2^5 = 32$ so each digit is the state of each hand.
answered Feb 27 at 4:26
Jo KingJo King
24.3k357126
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Python 2, 34 32 bytes
f=lambda a:a%32==4or a>0<f(a/32)
Try it online!
2 bytes thanks to tsh
answered Feb 27 at 6:47
Chas BrownChas Brown
4,9941523
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1
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:| you can edit posts, you know that right
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– ASCII-only
Feb 27 at 7:37
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Yes; I know. It was an accident! An accident, I tell ya!
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– Chas Brown
Feb 27 at 8:52
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@tsh Oh nice, forgot that shortcircuits
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– ASCII-only
Feb 27 at 10:24
add a comment |
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Julia 1.0, 25 bytes
f(n)=n%32==4||n>0<f(n>>5)
Try it online!
Julia 1.0, 26 bytes
Alternative that is 1 character shorter, but 1 byte longer, too bad that ∈ takes 3 bytes in unicode.
n->'4'∈string(n,base=32)
Try it online!
answered Feb 27 at 9:49
Kirill L.Kirill L.
5,0951525
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could you usen->n%32...for your first answer for 2 bytes shorter?
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– Giuseppe
Feb 28 at 22:48
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@Giuseppe, unfortunately no, that function is recursive.
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– Kirill L.
Mar 1 at 7:34
add a comment |
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05AB1E, 5 bytes
32B4å
Port of @Adám's APL (dzaima/APL) answer.
Try it online or verify all test cases.
Explanation:
32B # Convert the (implicit) input to Base-32
4å # And check if it contains a 4
# (output the result implicitly)
answered Feb 27 at 12:38
Kevin CruijssenKevin Cruijssen
39.7k560203
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1
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Too bad₆is 36, not 32.
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– Magic Octopus Urn
Feb 28 at 19:51
add a comment |
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Catholicon, 4 bytes
ǔ?QǑ
Takes a number as a base-256 string.
Try it online!
Test suite
answered Feb 27 at 13:11
OkxOkx
12.9k128102
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2
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Hm, if this is allowed, then is it allowed to accept numbers in base 32 instead?
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– recursive
Feb 27 at 19:47
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@recursive You can surround the numbers in<<and>>and it allows for numbers greater than 255 in those, as shown in the test suite.
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– Okx
Feb 28 at 13:11
1
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It was intended as a question about the challenge but it wasn't very clear.
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– recursive
Feb 28 at 15:44
add a comment |
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C# (Visual C# Interactive Compiler), 31 bytes
n=>{for(;n>0;n/=n%32==4?0:32);}
Outputs by throwing an exception. The way you convert one number from decimal to another base is to divide the decimal number by that base repeatedly and take the remainder as a digit. That is what we do, and we check if any of the digits have a value of 4 in base-32;
Try it online!
answered Feb 27 at 5:39
Embodiment of IgnoranceEmbodiment of Ignorance
1,476123
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27? as a bonus it doesn't output in a weird way
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– ASCII-only
Feb 27 at 6:13
1
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also what is tplig
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– ASCII-only
Feb 27 at 6:15
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@ASCII-onlyn>31->n>0
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– tsh
Feb 27 at 6:17
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25 bytes
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– Kevin Cruijssen
Feb 27 at 12:57
1
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Whether or not a program halts is not an allowed output method. Output via exception is allowed.
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– Deadcode
Feb 28 at 22:58
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show 4 more comments
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x86 Machine Code, 17 bytes
6A 20 59 85 C0 74 09 99 F7 F9 83 FA 04 75 F4 91 C3
The above bytes define a function that takes the number as input in the EAX register, and returns the result as a Boolean value in the EAX register (EAX == 0 if the input is not a rude number; EAX != 0 if the input is a rude number).
In human-readable assembly mnemonics:
; Determines whether the specified number is a "rude" number.
; Input: The number to check, in EAX
; Output: The Boolean result, in EAX (non-zero if rude; zero otherwise)
; Clobbers: ECX, EDX
IsRudeNumber:
push 32 ; standard golfing way to enregister a constant value
pop ecx ; / (in this case: ECX <= 32)
CheckNext:
test eax, eax ; if EAX == 0, jump to the end and return EAX (== 0)
jz TheEnd ; / otherwise, fall through and keep executing
cdq ; zero-out EDX because EAX is unsigned (shorter than XOR)
idiv ecx ; EAX <= (EAX / 32)
; EDX <= (EAX % 32)
cmp edx, 4 ; if EDX != 4, jump back to the start of the loop
jne CheckNext ; / otherwise, fall through and keep executing
xchg eax, ecx ; store ECX (== 32, a non-zero value) in EAX
TheEnd:
ret ; return, with result in EAX
Try it online!
answered Mar 1 at 7:04
Cody GrayCody Gray
1,869415
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1
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Interesting idea to useidiv, though. I don't see any incremental improvements to this. But see my answer : 14 bytes for a shift loop that uses MOV/AND/SUB/JZ to check the low 5 bits for rudeness.
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– Peter Cordes
Mar 4 at 0:25
add a comment |
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J, 12 bytes
4 e.32#.inv]
Try it online!
answered Feb 27 at 7:31
Galen IvanovGalen Ivanov
7,00711034
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add a comment |
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R, 50 48 bytes
any(2^(0:4)%*%matrix(scan()%/%2^(0:34)%%2,5)==4)
Try it online!
Uses a neat matrix-based approach now (courtesy of @Giueseppe). It generates a 5x7 matrix of bits, converts this to a series of base 32 integers, and checks for any 4s.
answered Feb 28 at 0:12
Nick KennedyNick Kennedy
44125
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@Giuseppe Oops, completely missed that. Should work now, though disappointingly 19 bytes longer. I don't think there's an inverse function to strtoi other than for hexadecimal and octal in base R
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– Nick Kennedy
Feb 28 at 17:45
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48 bytes with some matrix magic. I believe the bit conversion is longer thanintToBitsbut then we can work withintsinstead ofrawwhich ends up saving a byte -- see for instance this withintToBits
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– Giuseppe
Feb 28 at 18:04
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@Giuseppe it's a completely different (and neat) solution to mine - do you want me to update mine or you post your own?
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– Nick Kennedy
Feb 28 at 18:18
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you're free to take it. :-)
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– Giuseppe
Feb 28 at 18:24
1
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of course, porting one of the many answers that tests for the presence of a digit4in a base-32 number is, oh, 29 bytes.
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– Giuseppe
Feb 28 at 22:46
add a comment |
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Python 3, 43 bytes
Checks every 5-bit chunk to see if it is rude (equal to 4).
lambda n:any(n>>5*i&31==4for i in range(n))
Try it online!
answered Feb 27 at 5:30
Neil A.Neil A.
1,268120
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*5-bit chunk...
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– ASCII-only
Feb 27 at 7:38
add a comment |
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C (gcc), 34 bytes
f(i){return i?i&31^4?f(i/32):1:0;}
Try it online!
answered Feb 27 at 10:03
celtschkceltschk
4,29521619
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32 bytes by rearranging the expression.
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– Arnauld
Feb 27 at 10:11
3
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or 25 bytes by abusing the way gcc is compiling this.
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– Arnauld
Feb 27 at 10:11
add a comment |
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Charcoal, 6 bytes
№⍘Ｎ³²4
Try it online! Link is to verbose version of code. Outputs -s according to how rude the number is. Explanation:
Ｎ Input as a number
⍘ Convert to base as a string
³² Literal 32
№ Count occurrences of
4 Literal string `4`
I use string base conversion to avoid having to separate the numeric literals for 32 and 4.
answered Feb 27 at 12:38
NeilNeil
81.4k745178
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Tidy, 18 bytes
{x:4∈base(32,x)}
Try it online! Checks if 4 is an element of base(32,x) (base conversion).
answered Feb 27 at 15:06
Conor O'BrienConor O'Brien
30.3k264162
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Haskell, 31 bytes
elem 9.(mapM(:[6..36])[0..5]!!)
Try it online!
answered Feb 28 at 4:07
xnorxnor
91.6k18187443
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add a comment |
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ES6, 31 30 bytes
a=b=>b.toString(32).match(/4/)
Feel free to say ideas on how to reduce this further, if any.
answered Feb 28 at 18:01
elipszilonelipszilon
212
New contributor
elipszilon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Welcome to PPCG!
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– Laikoni
Feb 28 at 19:23
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You don't need to count the name of your function, and although I think you can save a byte by usingtest, you can actually save two bytes by matching against4as a number and lettingmatchconvert that into a string and then a RegExp for you.
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– Neil
Mar 1 at 0:53
add a comment |
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Cubix, 26 bytes
u!@-W14;OIS%/;;,p;?wO@u/s
Try it online!
Wraps onto a cube with edge length 3 as follows
u ! @
- W 1
4 ; O
I S % / ; ; , p ; ? w
O @ u / s . . . . . . .
. . . . . . . . . . . .
. . .
. . .
. . .
Watch it run
A fairly basic implementation, without all the redirects it does :
ISinitiates the program by pushing the input and 32 to the stack
%4-!gets the remainder and checks if it is 4 by subtraction
1O@output 1 if it was 4 and halt
;;,clean up the stack and do integer divide
p;?clean up bottom of the stack and check div result for 0
O@if div result zero output and halt
sswap the top of stack and start back at step 2 above
answered Feb 28 at 21:34
MickyTMickyT
10.2k21637
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MATL, 8 bytes
32YA52=a
Try it online!
answered Feb 28 at 13:57
SanchisesSanchises
6,17712352
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@Luis I can definitely drop theG(not sure why I included it in the first place) but that's only one byte (thanks for spotting that!). Changing32YA52to32_YA4is the same number of bytes right?
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– Sanchises
Mar 1 at 7:06
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Ah, yes, I can't count
$endgroup$
– Luis Mendo
Mar 1 at 10:26
2
$begingroup$
@Luis Count? Who needs to count when you can'32_YA4'n'32YA52'n-
$endgroup$
– Sanchises
Mar 1 at 11:56
add a comment |
$begingroup$
Retina 0.8.2, 31 bytes
.+
$*
+`(1+)1{31}
$1;
b1111b
Try it online! Link includes test cases. Outputs zero unless the number is rude. Works by converting the input to unary and then to unary-encoded base 32 and counting the number of 4s in the result.
answered Feb 27 at 13:28
NeilNeil
81.4k745178
$endgroup$
add a comment |
$begingroup$
Java 8, 40 33 bytes
n->n.toString(n,32).contains("4")
Port of @Adám's APL (dzaima/APL) answer.
Try it online.
Explanation:
n-> // Method with Integer parameter and boolean return-type
n.toString(n,32) // Convert the input to a base-32 String
.contains("4") // And check if it contains a "4"
answered Feb 27 at 12:56
Kevin CruijssenKevin Cruijssen
39.7k560203
$endgroup$
add a comment |
$begingroup$
Batch, 77 45 bytes
@cmd/cset/a"m=34636833,n=%1^m*4,(n-m)&~n&m*16
Based on these bit twiddling hacks. Explanation: Only 6 hands need to be checked due to the limited range (30 bits) of the input that's required to be supported. The magic number m is equivalent to 111111 in base 32, so that the first operation toggles the rude bits in the input number. It then remains to find which of the 6 hands is now zero.
answered Mar 1 at 0:50
NeilNeil
81.4k745178
$endgroup$
add a comment |
$begingroup$
><>, 28 bytes
Outputs 4 for rude numbers throws an exception for non-rude numbers.
:1(?^:" ":
,&-v?=4:%&/
;n<
Try it online!
answered Mar 1 at 10:34
EmignaEmigna
46.7k432142
$endgroup$
1
$begingroup$
An exception is acceptable, the C# answer does it
$endgroup$
– ASCII-only
Mar 1 at 11:22
add a comment |
$begingroup$
Wolfram Language (Mathematica), 37 bytes 36 bytes 29 bytes
-2 bytes by Jonathan Frech
#~IntegerDigits~32~MemberQ~4&
Try it online!
31-byte solution:
MemberQ[IntegerDigits[#,32],4]&
Try it online!
edited Mar 2 at 1:49
lirtosiast
18.2k438109
answered Feb 27 at 9:03
Rainer GlügeRainer Glüge
1213
$endgroup$
$begingroup$
Hello and welcome to PPCG. As it stands, your expression is a single Boolean value. Please fix your answer to either be a full program or a function (...#...&is often used in Mathematica).
$endgroup$
– Jonathan Frech
Feb 27 at 9:22
$begingroup$
Hello. Is this what you mean?
$endgroup$
– Rainer Glüge
Feb 27 at 9:32
$begingroup$
pls use tio.run/#mathematica instead of W|A to make sure it's valid mathematica code :P and you don't need the[n]at the end, just the&. Also since posts have edit history, it's fine to leave out previous entries, and the convention for old scores is<s>40</s> <s>36</s>
$endgroup$
– ASCII-only
Feb 27 at 10:22
$begingroup$
Yes. This is what I meant. 29 bytes.
$endgroup$
– Jonathan Frech
Feb 27 at 13:17
$begingroup$
I guess I have to get used to the functional programming style.
$endgroup$
– Rainer Glüge
Feb 27 at 13:34
add a comment |
$begingroup$
x86 machine code, 14 bytes
(same machine code works in 16-bit, 32-bit, and 64-bit. In 16-bit mode, it uses AX and DI instead of EAX and EDI in 32 and 64-bit mode.)
Algorithm: check low 5 bits with x & 31 == 4, then right-shift by 5 bits, and repeat if the shift result is non-zero.
Callable from C with char isrude(unsigned n); according to the x86-64 System V calling convention. 0 is truthy, non-0 is falsy (this is asm, not C1).
line addr code bytes
num
1 ; input: number in EDI
2 ; output: integer result in AL: 0 -> rude, non-zero non-rude
3 ; clobbers: RDI
4 isrude:
5 .check_low_bitgroup:
6 00000000 89F8 mov eax, edi
7 00000002 241F and al, 31 ; isolate low 5 bits
8 00000004 2C04 sub al, 4 ; like cmp but leaves AL 0 or non-zero
9 00000006 7405 jz .rude ; if (al & 31 == 4) return 0;
10
11 00000008 C1EF05 shr edi, 5
12 0000000B 75F3 jnz .check_low_bitgroup
13 ;; fall through to here is only possible if AL is non-zero
14 .rude:
15 0000000D C3 ret
16 0E size: db $ - isrude
This takes advantage of the short-form op al, imm8 encoding for AND and SUB. I could have used XOR al,4 to produce 0 on equality, but SUB is faster because it can macro-fuse with JZ into a single sub-and-branch uop on Sandybridge-family.
Fun fact: using the flag-result of a shift by more than 1 will be slow on P6-family (front-end stalls until the shift retires), but that's fine.
Footnote 1: This is an assembly language function, and x86 asm has both jz and jnz, so as per meta I can choose either way. I'm not intending this to match C truthy/falsy.
It happened to be convenient to return in AL instead of EFLAGS, so we can describe the function to a C compiler without a wrapper, but my choice of truthy/falsy isn't constrained by using a C caller to test it.
answered Mar 4 at 0:21
Peter CordesPeter Cordes
2,0921018
$endgroup$
add a comment |
$begingroup$
Java 8, 28 22 21 bytes
n->n%32==4|n>>5%32==4
Inspired by @kevin-cruijssen's answer. Only works for 2 hands.
Try it online!
Explanation:
n-> // Method with int parameter and boolean return-type
n%32 // Only consider right 5 bytes (fingers)
==4 // Middle finger
| ... n>>5 // Repeat with shifted bits for other hand
answered 2 days ago
Daniel WiddisDaniel Widdis
1114
New contributor
Daniel Widdis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
1 2
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$begingroup$
APL (dzaima/APL), 5 bytes
4∊32⊤
Try it online!
4∊ is 4 a member of
32⊤ to-base-32?
answered Feb 27 at 5:42
AdámAdám
28.8k276204
$endgroup$
$begingroup$
s/bytes/characters though?
$endgroup$
– tripleee
Feb 27 at 10:34
$begingroup$
@tripleee Thanks, added.
$endgroup$
– Adám
Feb 27 at 11:30
add a comment |
$begingroup$
APL (dzaima/APL), 5 bytes
4∊32⊤
Try it online!
4∊ is 4 a member of
32⊤ to-base-32?
answered Feb 27 at 5:42
AdámAdám
28.8k276204
$endgroup$
$begingroup$
s/bytes/characters though?
$endgroup$
– tripleee
Feb 27 at 10:34
$begingroup$
@tripleee Thanks, added.
$endgroup$
– Adám
Feb 27 at 11:30
add a comment |
$begingroup$
APL (dzaima/APL), 5 bytes
4∊32⊤
Try it online!
4∊ is 4 a member of
32⊤ to-base-32?
answered Feb 27 at 5:42
AdámAdám
28.8k276204
$endgroup$
APL (dzaima/APL), 5 bytes
4∊32⊤
Try it online!
4∊ is 4 a member of
32⊤ to-base-32?
answered Feb 27 at 5:42
AdámAdám
28.8k276204
edited Feb 27 at 11:30
answered Feb 27 at 5:42
AdámAdám
28.8k276204
answered Feb 27 at 5:42
AdámAdám
28.8k276204
answered Feb 27 at 5:42
AdámAdám
28.8k276204
28.8k276204
$begingroup$
s/bytes/characters though?
$endgroup$
– tripleee
Feb 27 at 10:34
$begingroup$
@tripleee Thanks, added.
$endgroup$
– Adám
Feb 27 at 11:30
add a comment |
$begingroup$
s/bytes/characters though?
$endgroup$
– tripleee
Feb 27 at 10:34
$begingroup$
@tripleee Thanks, added.
$endgroup$
– Adám
Feb 27 at 11:30
$begingroup$
s/bytes/characters though?
$endgroup$
– tripleee
Feb 27 at 10:34
$begingroup$
s/bytes/characters though?
$endgroup$
– tripleee
Feb 27 at 10:34
$begingroup$
@tripleee Thanks, added.
$endgroup$
– Adám
Feb 27 at 11:30
$begingroup$
@tripleee Thanks, added.
$endgroup$
– Adám
Feb 27 at 11:30
add a comment |
$begingroup$
Regex (ECMAScript), 37 bytes
Input is in unary, as the length of a string of xs.
^((?=(x+)(2{31}x*))3)*(x{32})*x{4}$
Try it online!
^
(
(?=(x+)(2{31}x*)) # 2 = floor(tail / 32); 3 = tool to make tail = 2
3 # tail = 2
)* # Loop the above as many times as necessary to make
# the below match
(x{32})*x{4}$ # Assert that tail % 32 == 4
answered Feb 27 at 4:16
DeadcodeDeadcode
1,8641420
$endgroup$
13
$begingroup$
I thought I knew regex, but apparently not.
$endgroup$
– CT Hall
Feb 28 at 4:16
add a comment |
$begingroup$
Regex (ECMAScript), 37 bytes
Input is in unary, as the length of a string of xs.
^((?=(x+)(2{31}x*))3)*(x{32})*x{4}$
Try it online!
^
(
(?=(x+)(2{31}x*)) # 2 = floor(tail / 32); 3 = tool to make tail = 2
3 # tail = 2
)* # Loop the above as many times as necessary to make
# the below match
(x{32})*x{4}$ # Assert that tail % 32 == 4
answered Feb 27 at 4:16
DeadcodeDeadcode
1,8641420
$endgroup$
13
$begingroup$
I thought I knew regex, but apparently not.
$endgroup$
– CT Hall
Feb 28 at 4:16
add a comment |
$begingroup$
Regex (ECMAScript), 37 bytes
Input is in unary, as the length of a string of xs.
^((?=(x+)(2{31}x*))3)*(x{32})*x{4}$
Try it online!
^
(
(?=(x+)(2{31}x*)) # 2 = floor(tail / 32); 3 = tool to make tail = 2
3 # tail = 2
)* # Loop the above as many times as necessary to make
# the below match
(x{32})*x{4}$ # Assert that tail % 32 == 4
answered Feb 27 at 4:16
DeadcodeDeadcode
1,8641420
$endgroup$
Regex (ECMAScript), 37 bytes
Input is in unary, as the length of a string of xs.
^((?=(x+)(2{31}x*))3)*(x{32})*x{4}$
Try it online!
^
(
(?=(x+)(2{31}x*)) # 2 = floor(tail / 32); 3 = tool to make tail = 2
3 # tail = 2
)* # Loop the above as many times as necessary to make
# the below match
(x{32})*x{4}$ # Assert that tail % 32 == 4
answered Feb 27 at 4:16
DeadcodeDeadcode
1,8641420
edited Feb 28 at 18:26
answered Feb 27 at 4:16
DeadcodeDeadcode
1,8641420
answered Feb 27 at 4:16
DeadcodeDeadcode
1,8641420
answered Feb 27 at 4:16
DeadcodeDeadcode
1,8641420
1,8641420
13
$begingroup$
I thought I knew regex, but apparently not.
$endgroup$
– CT Hall
Feb 28 at 4:16
add a comment |
13
$begingroup$
I thought I knew regex, but apparently not.
$endgroup$
– CT Hall
Feb 28 at 4:16
13
13
$begingroup$
I thought I knew regex, but apparently not.
$endgroup$
– CT Hall
Feb 28 at 4:16
$begingroup$
I thought I knew regex, but apparently not.
$endgroup$
– CT Hall
Feb 28 at 4:16
add a comment |
$begingroup$
JavaScript (SpiderMonkey), 23 bytes
f=x=>x&&x%32==4|f(x>>5)
Try it online!
This is a trivial solution, you just want to convert to base 32 and check if there is a 4 in it.
JavaScript (SpiderMonkey), 26 bytes
x=>x.toString(32).match(4)
Try it online!
It's interesting that /4/.test(...) cost one more byte than ....match(4).
answered Feb 27 at 4:17
tshtsh
9,32511652
$endgroup$
add a comment |
$begingroup$
JavaScript (SpiderMonkey), 23 bytes
f=x=>x&&x%32==4|f(x>>5)
Try it online!
This is a trivial solution, you just want to convert to base 32 and check if there is a 4 in it.
JavaScript (SpiderMonkey), 26 bytes
x=>x.toString(32).match(4)
Try it online!
It's interesting that /4/.test(...) cost one more byte than ....match(4).
answered Feb 27 at 4:17
tshtsh
9,32511652
$endgroup$
add a comment |
$begingroup$
JavaScript (SpiderMonkey), 23 bytes
f=x=>x&&x%32==4|f(x>>5)
Try it online!
This is a trivial solution, you just want to convert to base 32 and check if there is a 4 in it.
JavaScript (SpiderMonkey), 26 bytes
x=>x.toString(32).match(4)
Try it online!
It's interesting that /4/.test(...) cost one more byte than ....match(4).
answered Feb 27 at 4:17
tshtsh
9,32511652
$endgroup$
JavaScript (SpiderMonkey), 23 bytes
f=x=>x&&x%32==4|f(x>>5)
Try it online!
This is a trivial solution, you just want to convert to base 32 and check if there is a 4 in it.
JavaScript (SpiderMonkey), 26 bytes
x=>x.toString(32).match(4)
Try it online!
It's interesting that /4/.test(...) cost one more byte than ....match(4).
answered Feb 27 at 4:17
tshtsh
9,32511652
edited 2 days ago
answered Feb 27 at 4:17
tshtsh
9,32511652
answered Feb 27 at 4:17
tshtsh
9,32511652
answered Feb 27 at 4:17
tshtsh
9,32511652
9,32511652
add a comment |
add a comment |
$begingroup$
Japt, 5 bytes
sH ø4
Try it online!
Explanation
// Implicit input
sH // To a base-H (=32) string
ø // Contains
4 // 4 (JavaScript interprets this as a string)
answered Feb 27 at 4:20
ASCII-onlyASCII-only
4,3081238
$endgroup$
add a comment |
$begingroup$
Japt, 5 bytes
sH ø4
Try it online!
Explanation
// Implicit input
sH // To a base-H (=32) string
ø // Contains
4 // 4 (JavaScript interprets this as a string)
answered Feb 27 at 4:20
ASCII-onlyASCII-only
4,3081238
$endgroup$
add a comment |
$begingroup$
Japt, 5 bytes
sH ø4
Try it online!
Explanation
// Implicit input
sH // To a base-H (=32) string
ø // Contains
4 // 4 (JavaScript interprets this as a string)
answered Feb 27 at 4:20
ASCII-onlyASCII-only
4,3081238
$endgroup$
Japt, 5 bytes
sH ø4
Try it online!
Explanation
// Implicit input
sH // To a base-H (=32) string
ø // Contains
4 // 4 (JavaScript interprets this as a string)
answered Feb 27 at 4:20
ASCII-onlyASCII-only
4,3081238
answered Feb 27 at 4:20
ASCII-onlyASCII-only
4,3081238
answered Feb 27 at 4:20
ASCII-onlyASCII-only
4,3081238
answered Feb 27 at 4:20
ASCII-onlyASCII-only
4,3081238
4,3081238
add a comment |
add a comment |
$begingroup$
APL+WIN, 10 bytes
Prompts for input of integer
4∊(6⍴32)⊤⎕
Noting six hands are required to represent 10^9 converts to vector of 6 elements of the base 32 representation and checks if a 4 exists in any element.
answered Feb 27 at 4:45
GrahamGraham
2,49678
$endgroup$
add a comment |
$begingroup$
APL+WIN, 10 bytes
Prompts for input of integer
4∊(6⍴32)⊤⎕
Noting six hands are required to represent 10^9 converts to vector of 6 elements of the base 32 representation and checks if a 4 exists in any element.
answered Feb 27 at 4:45
GrahamGraham
2,49678
$endgroup$
add a comment |
$begingroup$
APL+WIN, 10 bytes
Prompts for input of integer
4∊(6⍴32)⊤⎕
Noting six hands are required to represent 10^9 converts to vector of 6 elements of the base 32 representation and checks if a 4 exists in any element.
answered Feb 27 at 4:45
GrahamGraham
2,49678
$endgroup$
APL+WIN, 10 bytes
Prompts for input of integer
4∊(6⍴32)⊤⎕
Noting six hands are required to represent 10^9 converts to vector of 6 elements of the base 32 representation and checks if a 4 exists in any element.
answered Feb 27 at 4:45
GrahamGraham
2,49678
answered Feb 27 at 4:45
GrahamGraham
2,49678
answered Feb 27 at 4:45
GrahamGraham
2,49678
answered Feb 27 at 4:45
GrahamGraham
2,49678
2,49678
add a comment |
add a comment |
$begingroup$
Ruby, 36 19 bytes
->n{n.to_s(32)[?4]}
Try it online!
Saved 17 bytes with @tsh's method.
answered Feb 27 at 3:37
Doorknob♦Doorknob
54.9k17115352
$endgroup$
$begingroup$
This returns true for 2207, which has a binary representation of100010011111
$endgroup$
– Embodiment of Ignorance
Feb 27 at 4:10
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is00100.
$endgroup$
– Doorknob♦
Feb 27 at 4:16
$begingroup$
I don't speak Ruby. But why not->n{n.to_s(32)=~/4/}?
$endgroup$
– tsh
Feb 27 at 4:19
1
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
Feb 27 at 4:23
$begingroup$
Forgive me if I'm not understanding the question, but isn't the first hand of 220710001, the second00111, and the third11? None of them have their middle finger only up
$endgroup$
– Embodiment of Ignorance
Feb 27 at 4:28
|
show 3 more comments
$begingroup$
Ruby, 36 19 bytes
->n{n.to_s(32)[?4]}
Try it online!
Saved 17 bytes with @tsh's method.
answered Feb 27 at 3:37
Doorknob♦Doorknob
54.9k17115352
$endgroup$
$begingroup$
This returns true for 2207, which has a binary representation of100010011111
$endgroup$
– Embodiment of Ignorance
Feb 27 at 4:10
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is00100.
$endgroup$
– Doorknob♦
Feb 27 at 4:16
$begingroup$
I don't speak Ruby. But why not->n{n.to_s(32)=~/4/}?
$endgroup$
– tsh
Feb 27 at 4:19
1
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
Feb 27 at 4:23
$begingroup$
Forgive me if I'm not understanding the question, but isn't the first hand of 220710001, the second00111, and the third11? None of them have their middle finger only up
$endgroup$
– Embodiment of Ignorance
Feb 27 at 4:28
|
show 3 more comments
$begingroup$
Ruby, 36 19 bytes
->n{n.to_s(32)[?4]}
Try it online!
Saved 17 bytes with @tsh's method.
answered Feb 27 at 3:37
Doorknob♦Doorknob
54.9k17115352
$endgroup$
Ruby, 36 19 bytes
->n{n.to_s(32)[?4]}
Try it online!
Saved 17 bytes with @tsh's method.
answered Feb 27 at 3:37
Doorknob♦Doorknob
54.9k17115352
edited Feb 27 at 4:23
answered Feb 27 at 3:37
Doorknob♦Doorknob
54.9k17115352
answered Feb 27 at 3:37
Doorknob♦Doorknob
54.9k17115352
answered Feb 27 at 3:37
Doorknob♦Doorknob
54.9k17115352
54.9k17115352
$begingroup$
This returns true for 2207, which has a binary representation of100010011111
$endgroup$
– Embodiment of Ignorance
Feb 27 at 4:10
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is00100.
$endgroup$
– Doorknob♦
Feb 27 at 4:16
$begingroup$
I don't speak Ruby. But why not->n{n.to_s(32)=~/4/}?
$endgroup$
– tsh
Feb 27 at 4:19
1
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
Feb 27 at 4:23
$begingroup$
Forgive me if I'm not understanding the question, but isn't the first hand of 220710001, the second00111, and the third11? None of them have their middle finger only up
$endgroup$
– Embodiment of Ignorance
Feb 27 at 4:28
|
show 3 more comments
$begingroup$
This returns true for 2207, which has a binary representation of100010011111
$endgroup$
– Embodiment of Ignorance
Feb 27 at 4:10
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is00100.
$endgroup$
– Doorknob♦
Feb 27 at 4:16
$begingroup$
I don't speak Ruby. But why not->n{n.to_s(32)=~/4/}?
$endgroup$
– tsh
Feb 27 at 4:19
1
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
Feb 27 at 4:23
$begingroup$
Forgive me if I'm not understanding the question, but isn't the first hand of 220710001, the second00111, and the third11? None of them have their middle finger only up
$endgroup$
– Embodiment of Ignorance
Feb 27 at 4:28
$begingroup$
This returns true for 2207, which has a binary representation of
100010011111$endgroup$
– Embodiment of Ignorance
Feb 27 at 4:10
$begingroup$
This returns true for 2207, which has a binary representation of
100010011111$endgroup$
– Embodiment of Ignorance
Feb 27 at 4:10
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is
00100.$endgroup$
– Doorknob♦
Feb 27 at 4:16
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is
00100.$endgroup$
– Doorknob♦
Feb 27 at 4:16
$begingroup$
I don't speak Ruby. But why not
->n{n.to_s(32)=~/4/}?$endgroup$
– tsh
Feb 27 at 4:19
$begingroup$
I don't speak Ruby. But why not
->n{n.to_s(32)=~/4/}?$endgroup$
– tsh
Feb 27 at 4:19
1
1
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
Feb 27 at 4:23
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
Feb 27 at 4:23
$begingroup$
Forgive me if I'm not understanding the question, but isn't the first hand of 2207
10001, the second 00111, and the third 11? None of them have their middle finger only up$endgroup$
– Embodiment of Ignorance
Feb 27 at 4:28
$begingroup$
Forgive me if I'm not understanding the question, but isn't the first hand of 2207
10001, the second 00111, and the third 11? None of them have their middle finger only up$endgroup$
– Embodiment of Ignorance
Feb 27 at 4:28
|
show 3 more comments
$begingroup$
Perl 6, 16 bytes
{.base(32)~~/4/}
Try it online!
Checks if there is a 4 in the base 32 representation of the number. Returns either Nil as false or a Match containing a 4.
You can prove this by the fact that $2^5 = 32$ so each digit is the state of each hand.
answered Feb 27 at 4:26
Jo KingJo King
24.3k357126
$endgroup$
add a comment |
$begingroup$
Perl 6, 16 bytes
{.base(32)~~/4/}
Try it online!
Checks if there is a 4 in the base 32 representation of the number. Returns either Nil as false or a Match containing a 4.
You can prove this by the fact that $2^5 = 32$ so each digit is the state of each hand.
answered Feb 27 at 4:26
Jo KingJo King
24.3k357126
$endgroup$
add a comment |
$begingroup$
Perl 6, 16 bytes
{.base(32)~~/4/}
Try it online!
Checks if there is a 4 in the base 32 representation of the number. Returns either Nil as false or a Match containing a 4.
You can prove this by the fact that $2^5 = 32$ so each digit is the state of each hand.
answered Feb 27 at 4:26
Jo KingJo King
24.3k357126
$endgroup$
Perl 6, 16 bytes
{.base(32)~~/4/}
Try it online!
Checks if there is a 4 in the base 32 representation of the number. Returns either Nil as false or a Match containing a 4.
You can prove this by the fact that $2^5 = 32$ so each digit is the state of each hand.
answered Feb 27 at 4:26
Jo KingJo King
24.3k357126
answered Feb 27 at 4:26
Jo KingJo King
24.3k357126
answered Feb 27 at 4:26
Jo KingJo King
24.3k357126
answered Feb 27 at 4:26
Jo KingJo King
24.3k357126
24.3k357126
add a comment |
add a comment |
$begingroup$
Python 2, 34 32 bytes
f=lambda a:a%32==4or a>0<f(a/32)
Try it online!
2 bytes thanks to tsh
answered Feb 27 at 6:47
Chas BrownChas Brown
4,9941523
$endgroup$
1
$begingroup$
:| you can edit posts, you know that right
$endgroup$
– ASCII-only
Feb 27 at 7:37
$begingroup$
Yes; I know. It was an accident! An accident, I tell ya!
$endgroup$
– Chas Brown
Feb 27 at 8:52
$begingroup$
@tsh Oh nice, forgot that shortcircuits
$endgroup$
– ASCII-only
Feb 27 at 10:24
add a comment |
$begingroup$
Python 2, 34 32 bytes
f=lambda a:a%32==4or a>0<f(a/32)
Try it online!
2 bytes thanks to tsh
answered Feb 27 at 6:47
Chas BrownChas Brown
4,9941523
$endgroup$
1
$begingroup$
:| you can edit posts, you know that right
$endgroup$
– ASCII-only
Feb 27 at 7:37
$begingroup$
Yes; I know. It was an accident! An accident, I tell ya!
$endgroup$
– Chas Brown
Feb 27 at 8:52
$begingroup$
@tsh Oh nice, forgot that shortcircuits
$endgroup$
– ASCII-only
Feb 27 at 10:24
add a comment |
$begingroup$
Python 2, 34 32 bytes
f=lambda a:a%32==4or a>0<f(a/32)
Try it online!
2 bytes thanks to tsh
answered Feb 27 at 6:47
Chas BrownChas Brown
4,9941523
$endgroup$
Python 2, 34 32 bytes
f=lambda a:a%32==4or a>0<f(a/32)
Try it online!
2 bytes thanks to tsh
answered Feb 27 at 6:47
Chas BrownChas Brown
4,9941523
edited Feb 27 at 19:39
answered Feb 27 at 6:47
Chas BrownChas Brown
4,9941523
answered Feb 27 at 6:47
Chas BrownChas Brown
4,9941523
answered Feb 27 at 6:47
Chas BrownChas Brown
4,9941523
4,9941523
1
$begingroup$
:| you can edit posts, you know that right
$endgroup$
– ASCII-only
Feb 27 at 7:37
$begingroup$
Yes; I know. It was an accident! An accident, I tell ya!
$endgroup$
– Chas Brown
Feb 27 at 8:52
$begingroup$
@tsh Oh nice, forgot that shortcircuits
$endgroup$
– ASCII-only
Feb 27 at 10:24
add a comment |
1
$begingroup$
:| you can edit posts, you know that right
$endgroup$
– ASCII-only
Feb 27 at 7:37
$begingroup$
Yes; I know. It was an accident! An accident, I tell ya!
$endgroup$
– Chas Brown
Feb 27 at 8:52
$begingroup$
@tsh Oh nice, forgot that shortcircuits
$endgroup$
– ASCII-only
Feb 27 at 10:24
1
1
$begingroup$
:| you can edit posts, you know that right
$endgroup$
– ASCII-only
Feb 27 at 7:37
$begingroup$
:| you can edit posts, you know that right
$endgroup$
– ASCII-only
Feb 27 at 7:37
$begingroup$
Yes; I know. It was an accident! An accident, I tell ya!
$endgroup$
– Chas Brown
Feb 27 at 8:52
$begingroup$
Yes; I know. It was an accident! An accident, I tell ya!
$endgroup$
– Chas Brown
Feb 27 at 8:52
$begingroup$
@tsh Oh nice, forgot that shortcircuits
$endgroup$
– ASCII-only
Feb 27 at 10:24
$begingroup$
@tsh Oh nice, forgot that shortcircuits
$endgroup$
– ASCII-only
Feb 27 at 10:24
add a comment |
$begingroup$
Julia 1.0, 25 bytes
f(n)=n%32==4||n>0<f(n>>5)
Try it online!
Julia 1.0, 26 bytes
Alternative that is 1 character shorter, but 1 byte longer, too bad that ∈ takes 3 bytes in unicode.
n->'4'∈string(n,base=32)
Try it online!
answered Feb 27 at 9:49
Kirill L.Kirill L.
5,0951525
$endgroup$
$begingroup$
could you usen->n%32...for your first answer for 2 bytes shorter?
$endgroup$
– Giuseppe
Feb 28 at 22:48
$begingroup$
@Giuseppe, unfortunately no, that function is recursive.
$endgroup$
– Kirill L.
Mar 1 at 7:34
add a comment |
$begingroup$
Julia 1.0, 25 bytes
f(n)=n%32==4||n>0<f(n>>5)
Try it online!
Julia 1.0, 26 bytes
Alternative that is 1 character shorter, but 1 byte longer, too bad that ∈ takes 3 bytes in unicode.
n->'4'∈string(n,base=32)
Try it online!
answered Feb 27 at 9:49
Kirill L.Kirill L.
5,0951525
$endgroup$
$begingroup$
could you usen->n%32...for your first answer for 2 bytes shorter?
$endgroup$
– Giuseppe
Feb 28 at 22:48
$begingroup$
@Giuseppe, unfortunately no, that function is recursive.
$endgroup$
– Kirill L.
Mar 1 at 7:34
add a comment |
$begingroup$
Julia 1.0, 25 bytes
f(n)=n%32==4||n>0<f(n>>5)
Try it online!
Julia 1.0, 26 bytes
Alternative that is 1 character shorter, but 1 byte longer, too bad that ∈ takes 3 bytes in unicode.
n->'4'∈string(n,base=32)
Try it online!
answered Feb 27 at 9:49
Kirill L.Kirill L.
5,0951525
$endgroup$
Julia 1.0, 25 bytes
f(n)=n%32==4||n>0<f(n>>5)
Try it online!
Julia 1.0, 26 bytes
Alternative that is 1 character shorter, but 1 byte longer, too bad that ∈ takes 3 bytes in unicode.
n->'4'∈string(n,base=32)
Try it online!
answered Feb 27 at 9:49
Kirill L.Kirill L.
5,0951525
edited Feb 27 at 10:00
answered Feb 27 at 9:49
Kirill L.Kirill L.
5,0951525
answered Feb 27 at 9:49
Kirill L.Kirill L.
5,0951525
answered Feb 27 at 9:49
Kirill L.Kirill L.
5,0951525
5,0951525
$begingroup$
could you usen->n%32...for your first answer for 2 bytes shorter?
$endgroup$
– Giuseppe
Feb 28 at 22:48
$begingroup$
@Giuseppe, unfortunately no, that function is recursive.
$endgroup$
– Kirill L.
Mar 1 at 7:34
add a comment |
$begingroup$
could you usen->n%32...for your first answer for 2 bytes shorter?
$endgroup$
– Giuseppe
Feb 28 at 22:48
$begingroup$
@Giuseppe, unfortunately no, that function is recursive.
$endgroup$
– Kirill L.
Mar 1 at 7:34
$begingroup$
could you use
n->n%32... for your first answer for 2 bytes shorter?$endgroup$
– Giuseppe
Feb 28 at 22:48
$begingroup$
could you use
n->n%32... for your first answer for 2 bytes shorter?$endgroup$
– Giuseppe
Feb 28 at 22:48
$begingroup$
@Giuseppe, unfortunately no, that function is recursive.
$endgroup$
– Kirill L.
Mar 1 at 7:34
$begingroup$
@Giuseppe, unfortunately no, that function is recursive.
$endgroup$
– Kirill L.
Mar 1 at 7:34
add a comment |
$begingroup$
05AB1E, 5 bytes
32B4å
Port of @Adám's APL (dzaima/APL) answer.
Try it online or verify all test cases.
Explanation:
32B # Convert the (implicit) input to Base-32
4å # And check if it contains a 4
# (output the result implicitly)
answered Feb 27 at 12:38
Kevin CruijssenKevin Cruijssen
39.7k560203
$endgroup$
1
$begingroup$
Too bad₆is 36, not 32.
$endgroup$
– Magic Octopus Urn
Feb 28 at 19:51
add a comment |
$begingroup$
05AB1E, 5 bytes
32B4å
Port of @Adám's APL (dzaima/APL) answer.
Try it online or verify all test cases.
Explanation:
32B # Convert the (implicit) input to Base-32
4å # And check if it contains a 4
# (output the result implicitly)
answered Feb 27 at 12:38
Kevin CruijssenKevin Cruijssen
39.7k560203
$endgroup$
1
$begingroup$
Too bad₆is 36, not 32.
$endgroup$
– Magic Octopus Urn
Feb 28 at 19:51
add a comment |
$begingroup$
05AB1E, 5 bytes
32B4å
Port of @Adám's APL (dzaima/APL) answer.
Try it online or verify all test cases.
Explanation:
32B # Convert the (implicit) input to Base-32
4å # And check if it contains a 4
# (output the result implicitly)
answered Feb 27 at 12:38
Kevin CruijssenKevin Cruijssen
39.7k560203
$endgroup$
05AB1E, 5 bytes
32B4å
Port of @Adám's APL (dzaima/APL) answer.
Try it online or verify all test cases.
Explanation:
32B # Convert the (implicit) input to Base-32
4å # And check if it contains a 4
# (output the result implicitly)
answered Feb 27 at 12:38
Kevin CruijssenKevin Cruijssen
39.7k560203
answered Feb 27 at 12:38
Kevin CruijssenKevin Cruijssen
39.7k560203
answered Feb 27 at 12:38
Kevin CruijssenKevin Cruijssen
39.7k560203
answered Feb 27 at 12:38
Kevin CruijssenKevin Cruijssen
39.7k560203
39.7k560203
1
$begingroup$
Too bad₆is 36, not 32.
$endgroup$
– Magic Octopus Urn
Feb 28 at 19:51
add a comment |
1
$begingroup$
Too bad₆is 36, not 32.
$endgroup$
– Magic Octopus Urn
Feb 28 at 19:51
1
1
$begingroup$
Too bad
₆ is 36, not 32.$endgroup$
– Magic Octopus Urn
Feb 28 at 19:51
$begingroup$
Too bad
₆ is 36, not 32.$endgroup$
– Magic Octopus Urn
Feb 28 at 19:51
add a comment |
$begingroup$
Catholicon, 4 bytes
ǔ?QǑ
Takes a number as a base-256 string.
Try it online!
Test suite
answered Feb 27 at 13:11
OkxOkx
12.9k128102
$endgroup$
2
$begingroup$
Hm, if this is allowed, then is it allowed to accept numbers in base 32 instead?
$endgroup$
– recursive
Feb 27 at 19:47
$begingroup$
@recursive You can surround the numbers in<<and>>and it allows for numbers greater than 255 in those, as shown in the test suite.
$endgroup$
– Okx
Feb 28 at 13:11
1
$begingroup$
It was intended as a question about the challenge but it wasn't very clear.
$endgroup$
– recursive
Feb 28 at 15:44
add a comment |
$begingroup$
Catholicon, 4 bytes
ǔ?QǑ
Takes a number as a base-256 string.
Try it online!
Test suite
answered Feb 27 at 13:11
OkxOkx
12.9k128102
$endgroup$
2
$begingroup$
Hm, if this is allowed, then is it allowed to accept numbers in base 32 instead?
$endgroup$
– recursive
Feb 27 at 19:47
$begingroup$
@recursive You can surround the numbers in<<and>>and it allows for numbers greater than 255 in those, as shown in the test suite.
$endgroup$
– Okx
Feb 28 at 13:11
1
$begingroup$
It was intended as a question about the challenge but it wasn't very clear.
$endgroup$
– recursive
Feb 28 at 15:44
add a comment |
$begingroup$
Catholicon, 4 bytes
ǔ?QǑ
Takes a number as a base-256 string.
Try it online!
Test suite
answered Feb 27 at 13:11
OkxOkx
12.9k128102
$endgroup$
Catholicon, 4 bytes
ǔ?QǑ
Takes a number as a base-256 string.
Try it online!
Test suite
answered Feb 27 at 13:11
OkxOkx
12.9k128102
edited Feb 27 at 13:19
answered Feb 27 at 13:11
OkxOkx
12.9k128102
answered Feb 27 at 13:11
OkxOkx
12.9k128102
answered Feb 27 at 13:11
OkxOkx
12.9k128102
12.9k128102
2
$begingroup$
Hm, if this is allowed, then is it allowed to accept numbers in base 32 instead?
$endgroup$
– recursive
Feb 27 at 19:47
$begingroup$
@recursive You can surround the numbers in<<and>>and it allows for numbers greater than 255 in those, as shown in the test suite.
$endgroup$
– Okx
Feb 28 at 13:11
1
$begingroup$
It was intended as a question about the challenge but it wasn't very clear.
$endgroup$
– recursive
Feb 28 at 15:44
add a comment |
2
$begingroup$
Hm, if this is allowed, then is it allowed to accept numbers in base 32 instead?
$endgroup$
– recursive
Feb 27 at 19:47
$begingroup$
@recursive You can surround the numbers in<<and>>and it allows for numbers greater than 255 in those, as shown in the test suite.
$endgroup$
– Okx
Feb 28 at 13:11
1
$begingroup$
It was intended as a question about the challenge but it wasn't very clear.
$endgroup$
– recursive
Feb 28 at 15:44
2
2
$begingroup$
Hm, if this is allowed, then is it allowed to accept numbers in base 32 instead?
$endgroup$
– recursive
Feb 27 at 19:47
$begingroup$
Hm, if this is allowed, then is it allowed to accept numbers in base 32 instead?
$endgroup$
– recursive
Feb 27 at 19:47
$begingroup$
@recursive You can surround the numbers in
<< and >> and it allows for numbers greater than 255 in those, as shown in the test suite.$endgroup$
– Okx
Feb 28 at 13:11
$begingroup$
@recursive You can surround the numbers in
<< and >> and it allows for numbers greater than 255 in those, as shown in the test suite.$endgroup$
– Okx
Feb 28 at 13:11
1
1
$begingroup$
It was intended as a question about the challenge but it wasn't very clear.
$endgroup$
– recursive
Feb 28 at 15:44
$begingroup$
It was intended as a question about the challenge but it wasn't very clear.
$endgroup$
– recursive
Feb 28 at 15:44
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 31 bytes
n=>{for(;n>0;n/=n%32==4?0:32);}
Outputs by throwing an exception. The way you convert one number from decimal to another base is to divide the decimal number by that base repeatedly and take the remainder as a digit. That is what we do, and we check if any of the digits have a value of 4 in base-32;
Try it online!
answered Feb 27 at 5:39
Embodiment of IgnoranceEmbodiment of Ignorance
1,476123
$endgroup$
$begingroup$
27? as a bonus it doesn't output in a weird way
$endgroup$
– ASCII-only
Feb 27 at 6:13
1
$begingroup$
also what is tplig
$endgroup$
– ASCII-only
Feb 27 at 6:15
$begingroup$
@ASCII-onlyn>31->n>0
$endgroup$
– tsh
Feb 27 at 6:17
$begingroup$
25 bytes
$endgroup$
– Kevin Cruijssen
Feb 27 at 12:57
1
$begingroup$
Whether or not a program halts is not an allowed output method. Output via exception is allowed.
$endgroup$
– Deadcode
Feb 28 at 22:58
|
show 4 more comments
$begingroup$
C# (Visual C# Interactive Compiler), 31 bytes
n=>{for(;n>0;n/=n%32==4?0:32);}
Outputs by throwing an exception. The way you convert one number from decimal to another base is to divide the decimal number by that base repeatedly and take the remainder as a digit. That is what we do, and we check if any of the digits have a value of 4 in base-32;
Try it online!
answered Feb 27 at 5:39
Embodiment of IgnoranceEmbodiment of Ignorance
1,476123
$endgroup$
$begingroup$
27? as a bonus it doesn't output in a weird way
$endgroup$
– ASCII-only
Feb 27 at 6:13
1
$begingroup$
also what is tplig
$endgroup$
– ASCII-only
Feb 27 at 6:15
$begingroup$
@ASCII-onlyn>31->n>0
$endgroup$
– tsh
Feb 27 at 6:17
$begingroup$
25 bytes
$endgroup$
– Kevin Cruijssen
Feb 27 at 12:57
1
$begingroup$
Whether or not a program halts is not an allowed output method. Output via exception is allowed.
$endgroup$
– Deadcode
Feb 28 at 22:58
|
show 4 more comments
$begingroup$
C# (Visual C# Interactive Compiler), 31 bytes
n=>{for(;n>0;n/=n%32==4?0:32);}
Outputs by throwing an exception. The way you convert one number from decimal to another base is to divide the decimal number by that base repeatedly and take the remainder as a digit. That is what we do, and we check if any of the digits have a value of 4 in base-32;
Try it online!
answered Feb 27 at 5:39
Embodiment of IgnoranceEmbodiment of Ignorance
1,476123
$endgroup$
C# (Visual C# Interactive Compiler), 31 bytes
n=>{for(;n>0;n/=n%32==4?0:32);}
Outputs by throwing an exception. The way you convert one number from decimal to another base is to divide the decimal number by that base repeatedly and take the remainder as a digit. That is what we do, and we check if any of the digits have a value of 4 in base-32;
Try it online!
answered Feb 27 at 5:39
Embodiment of IgnoranceEmbodiment of Ignorance
1,476123
edited Feb 28 at 23:15
answered Feb 27 at 5:39
Embodiment of IgnoranceEmbodiment of Ignorance
1,476123
answered Feb 27 at 5:39
Embodiment of IgnoranceEmbodiment of Ignorance
1,476123
answered Feb 27 at 5:39
Embodiment of IgnoranceEmbodiment of Ignorance
1,476123
1,476123
$begingroup$
27? as a bonus it doesn't output in a weird way
$endgroup$
– ASCII-only
Feb 27 at 6:13
1
$begingroup$
also what is tplig
$endgroup$
– ASCII-only
Feb 27 at 6:15
$begingroup$
@ASCII-onlyn>31->n>0
$endgroup$
– tsh
Feb 27 at 6:17
$begingroup$
25 bytes
$endgroup$
– Kevin Cruijssen
Feb 27 at 12:57
1
$begingroup$
Whether or not a program halts is not an allowed output method. Output via exception is allowed.
$endgroup$
– Deadcode
Feb 28 at 22:58
|
show 4 more comments
$begingroup$
27? as a bonus it doesn't output in a weird way
$endgroup$
– ASCII-only
Feb 27 at 6:13
1
$begingroup$
also what is tplig
$endgroup$
– ASCII-only
Feb 27 at 6:15
$begingroup$
@ASCII-onlyn>31->n>0
$endgroup$
– tsh
Feb 27 at 6:17
$begingroup$
25 bytes
$endgroup$
– Kevin Cruijssen
Feb 27 at 12:57
1
$begingroup$
Whether or not a program halts is not an allowed output method. Output via exception is allowed.
$endgroup$
– Deadcode
Feb 28 at 22:58
$begingroup$
27? as a bonus it doesn't output in a weird way
$endgroup$
– ASCII-only
Feb 27 at 6:13
$begingroup$
27? as a bonus it doesn't output in a weird way
$endgroup$
– ASCII-only
Feb 27 at 6:13
1
1
$begingroup$
also what is tplig
$endgroup$
– ASCII-only
Feb 27 at 6:15
$begingroup$
also what is tplig
$endgroup$
– ASCII-only
Feb 27 at 6:15
$begingroup$
@ASCII-only
n>31 -> n>0$endgroup$
– tsh
Feb 27 at 6:17
$begingroup$
@ASCII-only
n>31 -> n>0$endgroup$
– tsh
Feb 27 at 6:17
$begingroup$
25 bytes
$endgroup$
– Kevin Cruijssen
Feb 27 at 12:57
$begingroup$
25 bytes
$endgroup$
– Kevin Cruijssen
Feb 27 at 12:57
1
1
$begingroup$
Whether or not a program halts is not an allowed output method. Output via exception is allowed.
$endgroup$
– Deadcode
Feb 28 at 22:58
$begingroup$
Whether or not a program halts is not an allowed output method. Output via exception is allowed.
$endgroup$
– Deadcode
Feb 28 at 22:58
|
show 4 more comments
$begingroup$
x86 Machine Code, 17 bytes
6A 20 59 85 C0 74 09 99 F7 F9 83 FA 04 75 F4 91 C3
The above bytes define a function that takes the number as input in the EAX register, and returns the result as a Boolean value in the EAX register (EAX == 0 if the input is not a rude number; EAX != 0 if the input is a rude number).
In human-readable assembly mnemonics:
; Determines whether the specified number is a "rude" number.
; Input: The number to check, in EAX
; Output: The Boolean result, in EAX (non-zero if rude; zero otherwise)
; Clobbers: ECX, EDX
IsRudeNumber:
push 32 ; standard golfing way to enregister a constant value
pop ecx ; / (in this case: ECX <= 32)
CheckNext:
test eax, eax ; if EAX == 0, jump to the end and return EAX (== 0)
jz TheEnd ; / otherwise, fall through and keep executing
cdq ; zero-out EDX because EAX is unsigned (shorter than XOR)
idiv ecx ; EAX <= (EAX / 32)
; EDX <= (EAX % 32)
cmp edx, 4 ; if EDX != 4, jump back to the start of the loop
jne CheckNext ; / otherwise, fall through and keep executing
xchg eax, ecx ; store ECX (== 32, a non-zero value) in EAX
TheEnd:
ret ; return, with result in EAX
Try it online!
answered Mar 1 at 7:04
Cody GrayCody Gray
1,869415
$endgroup$
1
$begingroup$
Interesting idea to useidiv, though. I don't see any incremental improvements to this. But see my answer : 14 bytes for a shift loop that uses MOV/AND/SUB/JZ to check the low 5 bits for rudeness.
$endgroup$
– Peter Cordes
Mar 4 at 0:25
add a comment |
$begingroup$
x86 Machine Code, 17 bytes
6A 20 59 85 C0 74 09 99 F7 F9 83 FA 04 75 F4 91 C3
The above bytes define a function that takes the number as input in the EAX register, and returns the result as a Boolean value in the EAX register (EAX == 0 if the input is not a rude number; EAX != 0 if the input is a rude number).
In human-readable assembly mnemonics:
; Determines whether the specified number is a "rude" number.
; Input: The number to check, in EAX
; Output: The Boolean result, in EAX (non-zero if rude; zero otherwise)
; Clobbers: ECX, EDX
IsRudeNumber:
push 32 ; standard golfing way to enregister a constant value
pop ecx ; / (in this case: ECX <= 32)
CheckNext:
test eax, eax ; if EAX == 0, jump to the end and return EAX (== 0)
jz TheEnd ; / otherwise, fall through and keep executing
cdq ; zero-out EDX because EAX is unsigned (shorter than XOR)
idiv ecx ; EAX <= (EAX / 32)
; EDX <= (EAX % 32)
cmp edx, 4 ; if EDX != 4, jump back to the start of the loop
jne CheckNext ; / otherwise, fall through and keep executing
xchg eax, ecx ; store ECX (== 32, a non-zero value) in EAX
TheEnd:
ret ; return, with result in EAX
Try it online!
answered Mar 1 at 7:04
Cody GrayCody Gray
1,869415
$endgroup$
1
$begingroup$
Interesting idea to useidiv, though. I don't see any incremental improvements to this. But see my answer : 14 bytes for a shift loop that uses MOV/AND/SUB/JZ to check the low 5 bits for rudeness.
$endgroup$
– Peter Cordes
Mar 4 at 0:25
add a comment |
$begingroup$
x86 Machine Code, 17 bytes
6A 20 59 85 C0 74 09 99 F7 F9 83 FA 04 75 F4 91 C3
The above bytes define a function that takes the number as input in the EAX register, and returns the result as a Boolean value in the EAX register (EAX == 0 if the input is not a rude number; EAX != 0 if the input is a rude number).
In human-readable assembly mnemonics:
; Determines whether the specified number is a "rude" number.
; Input: The number to check, in EAX
; Output: The Boolean result, in EAX (non-zero if rude; zero otherwise)
; Clobbers: ECX, EDX
IsRudeNumber:
push 32 ; standard golfing way to enregister a constant value
pop ecx ; / (in this case: ECX <= 32)
CheckNext:
test eax, eax ; if EAX == 0, jump to the end and return EAX (== 0)
jz TheEnd ; / otherwise, fall through and keep executing
cdq ; zero-out EDX because EAX is unsigned (shorter than XOR)
idiv ecx ; EAX <= (EAX / 32)
; EDX <= (EAX % 32)
cmp edx, 4 ; if EDX != 4, jump back to the start of the loop
jne CheckNext ; / otherwise, fall through and keep executing
xchg eax, ecx ; store ECX (== 32, a non-zero value) in EAX
TheEnd:
ret ; return, with result in EAX
Try it online!
answered Mar 1 at 7:04
Cody GrayCody Gray
1,869415
$endgroup$
x86 Machine Code, 17 bytes
6A 20 59 85 C0 74 09 99 F7 F9 83 FA 04 75 F4 91 C3
The above bytes define a function that takes the number as input in the EAX register, and returns the result as a Boolean value in the EAX register (EAX == 0 if the input is not a rude number; EAX != 0 if the input is a rude number).
In human-readable assembly mnemonics:
; Determines whether the specified number is a "rude" number.
; Input: The number to check, in EAX
; Output: The Boolean result, in EAX (non-zero if rude; zero otherwise)
; Clobbers: ECX, EDX
IsRudeNumber:
push 32 ; standard golfing way to enregister a constant value
pop ecx ; / (in this case: ECX <= 32)
CheckNext:
test eax, eax ; if EAX == 0, jump to the end and return EAX (== 0)
jz TheEnd ; / otherwise, fall through and keep executing
cdq ; zero-out EDX because EAX is unsigned (shorter than XOR)
idiv ecx ; EAX <= (EAX / 32)
; EDX <= (EAX % 32)
cmp edx, 4 ; if EDX != 4, jump back to the start of the loop
jne CheckNext ; / otherwise, fall through and keep executing
xchg eax, ecx ; store ECX (== 32, a non-zero value) in EAX
TheEnd:
ret ; return, with result in EAX
Try it online!
answered Mar 1 at 7:04
Cody GrayCody Gray
1,869415
answered Mar 1 at 7:04
Cody GrayCody Gray
1,869415
answered Mar 1 at 7:04
Cody GrayCody Gray
1,869415
answered Mar 1 at 7:04
Cody GrayCody Gray
1,869415
1,869415
1
$begingroup$
Interesting idea to useidiv, though. I don't see any incremental improvements to this. But see my answer : 14 bytes for a shift loop that uses MOV/AND/SUB/JZ to check the low 5 bits for rudeness.
$endgroup$
– Peter Cordes
Mar 4 at 0:25
add a comment |
1
$begingroup$
Interesting idea to useidiv, though. I don't see any incremental improvements to this. But see my answer : 14 bytes for a shift loop that uses MOV/AND/SUB/JZ to check the low 5 bits for rudeness.
$endgroup$
– Peter Cordes
Mar 4 at 0:25
1
1
$begingroup$
Interesting idea to use
idiv, though. I don't see any incremental improvements to this. But see my answer : 14 bytes for a shift loop that uses MOV/AND/SUB/JZ to check the low 5 bits for rudeness.$endgroup$
– Peter Cordes
Mar 4 at 0:25
$begingroup$
Interesting idea to use
idiv, though. I don't see any incremental improvements to this. But see my answer : 14 bytes for a shift loop that uses MOV/AND/SUB/JZ to check the low 5 bits for rudeness.$endgroup$
– Peter Cordes
Mar 4 at 0:25
add a comment |
$begingroup$
J, 12 bytes
4 e.32#.inv]
Try it online!
answered Feb 27 at 7:31
Galen IvanovGalen Ivanov
7,00711034
$endgroup$
add a comment |
$begingroup$
J, 12 bytes
4 e.32#.inv]
Try it online!
answered Feb 27 at 7:31
Galen IvanovGalen Ivanov
7,00711034
$endgroup$
add a comment |
$begingroup$
J, 12 bytes
4 e.32#.inv]
Try it online!
answered Feb 27 at 7:31
Galen IvanovGalen Ivanov
7,00711034
$endgroup$
J, 12 bytes
4 e.32#.inv]
Try it online!
answered Feb 27 at 7:31
Galen IvanovGalen Ivanov
7,00711034
answered Feb 27 at 7:31
Galen IvanovGalen Ivanov
7,00711034
answered Feb 27 at 7:31
Galen IvanovGalen Ivanov
7,00711034
answered Feb 27 at 7:31
Galen IvanovGalen Ivanov
7,00711034
7,00711034
add a comment |
add a comment |
$begingroup$
R, 50 48 bytes
any(2^(0:4)%*%matrix(scan()%/%2^(0:34)%%2,5)==4)
Try it online!
Uses a neat matrix-based approach now (courtesy of @Giueseppe). It generates a 5x7 matrix of bits, converts this to a series of base 32 integers, and checks for any 4s.
answered Feb 28 at 0:12
Nick KennedyNick Kennedy
44125
$endgroup$
$begingroup$
@Giuseppe Oops, completely missed that. Should work now, though disappointingly 19 bytes longer. I don't think there's an inverse function to strtoi other than for hexadecimal and octal in base R
$endgroup$
– Nick Kennedy
Feb 28 at 17:45
$begingroup$
48 bytes with some matrix magic. I believe the bit conversion is longer thanintToBitsbut then we can work withintsinstead ofrawwhich ends up saving a byte -- see for instance this withintToBits
$endgroup$
– Giuseppe
Feb 28 at 18:04
$begingroup$
@Giuseppe it's a completely different (and neat) solution to mine - do you want me to update mine or you post your own?
$endgroup$
– Nick Kennedy
Feb 28 at 18:18
$begingroup$
you're free to take it. :-)
$endgroup$
– Giuseppe
Feb 28 at 18:24
1
$begingroup$
of course, porting one of the many answers that tests for the presence of a digit4in a base-32 number is, oh, 29 bytes.
$endgroup$
– Giuseppe
Feb 28 at 22:46
add a comment |
$begingroup$
R, 50 48 bytes
any(2^(0:4)%*%matrix(scan()%/%2^(0:34)%%2,5)==4)
Try it online!
Uses a neat matrix-based approach now (courtesy of @Giueseppe). It generates a 5x7 matrix of bits, converts this to a series of base 32 integers, and checks for any 4s.
answered Feb 28 at 0:12
Nick KennedyNick Kennedy
44125
$endgroup$
$begingroup$
@Giuseppe Oops, completely missed that. Should work now, though disappointingly 19 bytes longer. I don't think there's an inverse function to strtoi other than for hexadecimal and octal in base R
$endgroup$
– Nick Kennedy
Feb 28 at 17:45
$begingroup$
48 bytes with some matrix magic. I believe the bit conversion is longer thanintToBitsbut then we can work withintsinstead ofrawwhich ends up saving a byte -- see for instance this withintToBits
$endgroup$
– Giuseppe
Feb 28 at 18:04
$begingroup$
@Giuseppe it's a completely different (and neat) solution to mine - do you want me to update mine or you post your own?
$endgroup$
– Nick Kennedy
Feb 28 at 18:18
$begingroup$
you're free to take it. :-)
$endgroup$
– Giuseppe
Feb 28 at 18:24
1
$begingroup$
of course, porting one of the many answers that tests for the presence of a digit4in a base-32 number is, oh, 29 bytes.
$endgroup$
– Giuseppe
Feb 28 at 22:46
add a comment |
$begingroup$
R, 50 48 bytes
any(2^(0:4)%*%matrix(scan()%/%2^(0:34)%%2,5)==4)
Try it online!
Uses a neat matrix-based approach now (courtesy of @Giueseppe). It generates a 5x7 matrix of bits, converts this to a series of base 32 integers, and checks for any 4s.
answered Feb 28 at 0:12
Nick KennedyNick Kennedy
44125
$endgroup$
R, 50 48 bytes
any(2^(0:4)%*%matrix(scan()%/%2^(0:34)%%2,5)==4)
Try it online!
Uses a neat matrix-based approach now (courtesy of @Giueseppe). It generates a 5x7 matrix of bits, converts this to a series of base 32 integers, and checks for any 4s.
answered Feb 28 at 0:12
Nick KennedyNick Kennedy
44125
edited Feb 28 at 18:47
answered Feb 28 at 0:12
Nick KennedyNick Kennedy
44125
answered Feb 28 at 0:12
Nick KennedyNick Kennedy
44125
answered Feb 28 at 0:12
Nick KennedyNick Kennedy
44125
44125
$begingroup$
@Giuseppe Oops, completely missed that. Should work now, though disappointingly 19 bytes longer. I don't think there's an inverse function to strtoi other than for hexadecimal and octal in base R
$endgroup$
– Nick Kennedy
Feb 28 at 17:45
$begingroup$
48 bytes with some matrix magic. I believe the bit conversion is longer thanintToBitsbut then we can work withintsinstead ofrawwhich ends up saving a byte -- see for instance this withintToBits
$endgroup$
– Giuseppe
Feb 28 at 18:04
$begingroup$
@Giuseppe it's a completely different (and neat) solution to mine - do you want me to update mine or you post your own?
$endgroup$
– Nick Kennedy
Feb 28 at 18:18
$begingroup$
you're free to take it. :-)
$endgroup$
– Giuseppe
Feb 28 at 18:24
1
$begingroup$
of course, porting one of the many answers that tests for the presence of a digit4in a base-32 number is, oh, 29 bytes.
$endgroup$
– Giuseppe
Feb 28 at 22:46
add a comment |
$begingroup$
@Giuseppe Oops, completely missed that. Should work now, though disappointingly 19 bytes longer. I don't think there's an inverse function to strtoi other than for hexadecimal and octal in base R
$endgroup$
– Nick Kennedy
Feb 28 at 17:45
$begingroup$
48 bytes with some matrix magic. I believe the bit conversion is longer thanintToBitsbut then we can work withintsinstead ofrawwhich ends up saving a byte -- see for instance this withintToBits
$endgroup$
– Giuseppe
Feb 28 at 18:04
$begingroup$
@Giuseppe it's a completely different (and neat) solution to mine - do you want me to update mine or you post your own?
$endgroup$
– Nick Kennedy
Feb 28 at 18:18
$begingroup$
you're free to take it. :-)
$endgroup$
– Giuseppe
Feb 28 at 18:24
1
$begingroup$
of course, porting one of the many answers that tests for the presence of a digit4in a base-32 number is, oh, 29 bytes.
$endgroup$
– Giuseppe
Feb 28 at 22:46
$begingroup$
@Giuseppe Oops, completely missed that. Should work now, though disappointingly 19 bytes longer. I don't think there's an inverse function to strtoi other than for hexadecimal and octal in base R
$endgroup$
– Nick Kennedy
Feb 28 at 17:45
$begingroup$
@Giuseppe Oops, completely missed that. Should work now, though disappointingly 19 bytes longer. I don't think there's an inverse function to strtoi other than for hexadecimal and octal in base R
$endgroup$
– Nick Kennedy
Feb 28 at 17:45
$begingroup$
48 bytes with some matrix magic. I believe the bit conversion is longer than
intToBits but then we can work with ints instead of raw which ends up saving a byte -- see for instance this with intToBits$endgroup$
– Giuseppe
Feb 28 at 18:04
$begingroup$
48 bytes with some matrix magic. I believe the bit conversion is longer than
intToBits but then we can work with ints instead of raw which ends up saving a byte -- see for instance this with intToBits$endgroup$
– Giuseppe
Feb 28 at 18:04
$begingroup$
@Giuseppe it's a completely different (and neat) solution to mine - do you want me to update mine or you post your own?
$endgroup$
– Nick Kennedy
Feb 28 at 18:18
$begingroup$
@Giuseppe it's a completely different (and neat) solution to mine - do you want me to update mine or you post your own?
$endgroup$
– Nick Kennedy
Feb 28 at 18:18
$begingroup$
you're free to take it. :-)
$endgroup$
– Giuseppe
Feb 28 at 18:24
$begingroup$
you're free to take it. :-)
$endgroup$
– Giuseppe
Feb 28 at 18:24
1
1
$begingroup$
of course, porting one of the many answers that tests for the presence of a digit
4 in a base-32 number is, oh, 29 bytes.$endgroup$
– Giuseppe
Feb 28 at 22:46
$begingroup$
of course, porting one of the many answers that tests for the presence of a digit
4 in a base-32 number is, oh, 29 bytes.$endgroup$
– Giuseppe
Feb 28 at 22:46
add a comment |
$begingroup$
Python 3, 43 bytes
Checks every 5-bit chunk to see if it is rude (equal to 4).
lambda n:any(n>>5*i&31==4for i in range(n))
Try it online!
answered Feb 27 at 5:30
Neil A.Neil A.
1,268120
$endgroup$
$begingroup$
*5-bit chunk...
$endgroup$
– ASCII-only
Feb 27 at 7:38
add a comment |
$begingroup$
Python 3, 43 bytes
Checks every 5-bit chunk to see if it is rude (equal to 4).
lambda n:any(n>>5*i&31==4for i in range(n))
Try it online!
answered Feb 27 at 5:30
Neil A.Neil A.
1,268120
$endgroup$
$begingroup$
*5-bit chunk...
$endgroup$
– ASCII-only
Feb 27 at 7:38
add a comment |
$begingroup$
Python 3, 43 bytes
Checks every 5-bit chunk to see if it is rude (equal to 4).
lambda n:any(n>>5*i&31==4for i in range(n))
Try it online!
answered Feb 27 at 5:30
Neil A.Neil A.
1,268120
$endgroup$
Python 3, 43 bytes
Checks every 5-bit chunk to see if it is rude (equal to 4).
lambda n:any(n>>5*i&31==4for i in range(n))
Try it online!
answered Feb 27 at 5:30
Neil A.Neil A.
1,268120
edited Feb 27 at 7:46
answered Feb 27 at 5:30
Neil A.Neil A.
1,268120
answered Feb 27 at 5:30
Neil A.Neil A.
1,268120
answered Feb 27 at 5:30
Neil A.Neil A.
1,268120
1,268120
$begingroup$
*5-bit chunk...
$endgroup$
– ASCII-only
Feb 27 at 7:38
add a comment |
$begingroup$
*5-bit chunk...
$endgroup$
– ASCII-only
Feb 27 at 7:38
$begingroup$
*5-bit chunk...
$endgroup$
– ASCII-only
Feb 27 at 7:38
$begingroup$
*5-bit chunk...
$endgroup$
– ASCII-only
Feb 27 at 7:38
add a comment |
$begingroup$
C (gcc), 34 bytes
f(i){return i?i&31^4?f(i/32):1:0;}
Try it online!
answered Feb 27 at 10:03
celtschkceltschk
4,29521619
$endgroup$
$begingroup$
32 bytes by rearranging the expression.
$endgroup$
– Arnauld
Feb 27 at 10:11
3
$begingroup$
or 25 bytes by abusing the way gcc is compiling this.
$endgroup$
– Arnauld
Feb 27 at 10:11
add a comment |
$begingroup$
C (gcc), 34 bytes
f(i){return i?i&31^4?f(i/32):1:0;}
Try it online!
answered Feb 27 at 10:03
celtschkceltschk
4,29521619
$endgroup$
$begingroup$
32 bytes by rearranging the expression.
$endgroup$
– Arnauld
Feb 27 at 10:11
3
$begingroup$
or 25 bytes by abusing the way gcc is compiling this.
$endgroup$
– Arnauld
Feb 27 at 10:11
add a comment |
$begingroup$
C (gcc), 34 bytes
f(i){return i?i&31^4?f(i/32):1:0;}
Try it online!
answered Feb 27 at 10:03
celtschkceltschk
4,29521619
$endgroup$
C (gcc), 34 bytes
f(i){return i?i&31^4?f(i/32):1:0;}
Try it online!
answered Feb 27 at 10:03
celtschkceltschk
4,29521619
answered Feb 27 at 10:03
celtschkceltschk
4,29521619
answered Feb 27 at 10:03
celtschkceltschk
4,29521619
answered Feb 27 at 10:03
celtschkceltschk
4,29521619
4,29521619
$begingroup$
32 bytes by rearranging the expression.
$endgroup$
– Arnauld
Feb 27 at 10:11
3
$begingroup$
or 25 bytes by abusing the way gcc is compiling this.
$endgroup$
– Arnauld
Feb 27 at 10:11
add a comment |
$begingroup$
32 bytes by rearranging the expression.
$endgroup$
– Arnauld
Feb 27 at 10:11
3
$begingroup$
or 25 bytes by abusing the way gcc is compiling this.
$endgroup$
– Arnauld
Feb 27 at 10:11
$begingroup$
32 bytes by rearranging the expression.
$endgroup$
– Arnauld
Feb 27 at 10:11
$begingroup$
32 bytes by rearranging the expression.
$endgroup$
– Arnauld
Feb 27 at 10:11
3
3
$begingroup$
or 25 bytes by abusing the way gcc is compiling this.
$endgroup$
– Arnauld
Feb 27 at 10:11
$begingroup$
or 25 bytes by abusing the way gcc is compiling this.
$endgroup$
– Arnauld
Feb 27 at 10:11
add a comment |
$begingroup$
Charcoal, 6 bytes
№⍘Ｎ³²4
Try it online! Link is to verbose version of code. Outputs -s according to how rude the number is. Explanation:
Ｎ Input as a number
⍘ Convert to base as a string
³² Literal 32
№ Count occurrences of
4 Literal string `4`
I use string base conversion to avoid having to separate the numeric literals for 32 and 4.
answered Feb 27 at 12:38
NeilNeil
81.4k745178
$endgroup$
add a comment |
$begingroup$
Charcoal, 6 bytes
№⍘Ｎ³²4
Try it online! Link is to verbose version of code. Outputs -s according to how rude the number is. Explanation:
Ｎ Input as a number
⍘ Convert to base as a string
³² Literal 32
№ Count occurrences of
4 Literal string `4`
I use string base conversion to avoid having to separate the numeric literals for 32 and 4.
answered Feb 27 at 12:38
NeilNeil
81.4k745178
$endgroup$
add a comment |
$begingroup$
Charcoal, 6 bytes
№⍘Ｎ³²4
Try it online! Link is to verbose version of code. Outputs -s according to how rude the number is. Explanation:
Ｎ Input as a number
⍘ Convert to base as a string
³² Literal 32
№ Count occurrences of
4 Literal string `4`
I use string base conversion to avoid having to separate the numeric literals for 32 and 4.
answered Feb 27 at 12:38
NeilNeil
81.4k745178
$endgroup$
Charcoal, 6 bytes
№⍘Ｎ³²4
Try it online! Link is to verbose version of code. Outputs -s according to how rude the number is. Explanation:
Ｎ Input as a number
⍘ Convert to base as a string
³² Literal 32
№ Count occurrences of
4 Literal string `4`
I use string base conversion to avoid having to separate the numeric literals for 32 and 4.
answered Feb 27 at 12:38
NeilNeil
81.4k745178
answered Feb 27 at 12:38
NeilNeil
81.4k745178
answered Feb 27 at 12:38
NeilNeil
81.4k745178
answered Feb 27 at 12:38
NeilNeil
81.4k745178
81.4k745178
add a comment |
add a comment |
$begingroup$
Tidy, 18 bytes
{x:4∈base(32,x)}
Try it online! Checks if 4 is an element of base(32,x) (base conversion).
answered Feb 27 at 15:06
Conor O'BrienConor O'Brien
30.3k264162
$endgroup$
add a comment |
$begingroup$
Tidy, 18 bytes
{x:4∈base(32,x)}
Try it online! Checks if 4 is an element of base(32,x) (base conversion).
answered Feb 27 at 15:06
Conor O'BrienConor O'Brien
30.3k264162
$endgroup$
add a comment |
$begingroup$
Tidy, 18 bytes
{x:4∈base(32,x)}
Try it online! Checks if 4 is an element of base(32,x) (base conversion).
answered Feb 27 at 15:06
Conor O'BrienConor O'Brien
30.3k264162
$endgroup$
Tidy, 18 bytes
{x:4∈base(32,x)}
Try it online! Checks if 4 is an element of base(32,x) (base conversion).
answered Feb 27 at 15:06
Conor O'BrienConor O'Brien
30.3k264162
answered Feb 27 at 15:06
Conor O'BrienConor O'Brien
30.3k264162
answered Feb 27 at 15:06
Conor O'BrienConor O'Brien
30.3k264162
answered Feb 27 at 15:06
Conor O'BrienConor O'Brien
30.3k264162
30.3k264162
add a comment |
add a comment |
$begingroup$
Haskell, 31 bytes
elem 9.(mapM(:[6..36])[0..5]!!)
Try it online!
answered Feb 28 at 4:07
xnorxnor
91.6k18187443
$endgroup$
add a comment |
$begingroup$
Haskell, 31 bytes
elem 9.(mapM(:[6..36])[0..5]!!)
Try it online!
answered Feb 28 at 4:07
xnorxnor
91.6k18187443
$endgroup$
add a comment |
$begingroup$
Haskell, 31 bytes
elem 9.(mapM(:[6..36])[0..5]!!)
Try it online!
answered Feb 28 at 4:07
xnorxnor
91.6k18187443
$endgroup$
Haskell, 31 bytes
elem 9.(mapM(:[6..36])[0..5]!!)
Try it online!
answered Feb 28 at 4:07
xnorxnor
91.6k18187443
answered Feb 28 at 4:07
xnorxnor
91.6k18187443
answered Feb 28 at 4:07
xnorxnor
91.6k18187443
answered Feb 28 at 4:07
xnorxnor
91.6k18187443
91.6k18187443
add a comment |
add a comment |
$begingroup$
ES6, 31 30 bytes
a=b=>b.toString(32).match(/4/)
Feel free to say ideas on how to reduce this further, if any.
answered Feb 28 at 18:01
elipszilonelipszilon
212
New contributor
elipszilon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Welcome to PPCG!
$endgroup$
– Laikoni
Feb 28 at 19:23
$begingroup$
You don't need to count the name of your function, and although I think you can save a byte by usingtest, you can actually save two bytes by matching against4as a number and lettingmatchconvert that into a string and then a RegExp for you.
$endgroup$
– Neil
Mar 1 at 0:53
add a comment |
$begingroup$
ES6, 31 30 bytes
a=b=>b.toString(32).match(/4/)
Feel free to say ideas on how to reduce this further, if any.
answered Feb 28 at 18:01
elipszilonelipszilon
212
New contributor
elipszilon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Welcome to PPCG!
$endgroup$
– Laikoni
Feb 28 at 19:23
$begingroup$
You don't need to count the name of your function, and although I think you can save a byte by usingtest, you can actually save two bytes by matching against4as a number and lettingmatchconvert that into a string and then a RegExp for you.
$endgroup$
– Neil
Mar 1 at 0:53
add a comment |
$begingroup$
ES6, 31 30 bytes
a=b=>b.toString(32).match(/4/)
Feel free to say ideas on how to reduce this further, if any.
answered Feb 28 at 18:01
elipszilonelipszilon
212
New contributor
elipszilon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
ES6, 31 30 bytes
a=b=>b.toString(32).match(/4/)
Feel free to say ideas on how to reduce this further, if any.
answered Feb 28 at 18:01
elipszilonelipszilon
212
New contributor
elipszilon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Feb 28 at 18:08
answered Feb 28 at 18:01
elipszilonelipszilon
212
New contributor
elipszilon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Feb 28 at 18:01
elipszilonelipszilon
212
answered Feb 28 at 18:01
elipszilonelipszilon
212
212
New contributor
elipszilon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
elipszilon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
elipszilon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Welcome to PPCG!
$endgroup$
– Laikoni
Feb 28 at 19:23
$begingroup$
You don't need to count the name of your function, and although I think you can save a byte by usingtest, you can actually save two bytes by matching against4as a number and lettingmatchconvert that into a string and then a RegExp for you.
$endgroup$
– Neil
Mar 1 at 0:53
add a comment |
$begingroup$
Welcome to PPCG!
$endgroup$
– Laikoni
Feb 28 at 19:23
$begingroup$
You don't need to count the name of your function, and although I think you can save a byte by usingtest, you can actually save two bytes by matching against4as a number and lettingmatchconvert that into a string and then a RegExp for you.
$endgroup$
– Neil
Mar 1 at 0:53
$begingroup$
Welcome to PPCG!
$endgroup$
– Laikoni
Feb 28 at 19:23
$begingroup$
Welcome to PPCG!
$endgroup$
– Laikoni
Feb 28 at 19:23
$begingroup$
You don't need to count the name of your function, and although I think you can save a byte by using
test, you can actually save two bytes by matching against 4 as a number and letting match convert that into a string and then a RegExp for you.$endgroup$
– Neil
Mar 1 at 0:53
$begingroup$
You don't need to count the name of your function, and although I think you can save a byte by using
test, you can actually save two bytes by matching against 4 as a number and letting match convert that into a string and then a RegExp for you.$endgroup$
– Neil
Mar 1 at 0:53
add a comment |
$begingroup$
Cubix, 26 bytes
u!@-W14;OIS%/;;,p;?wO@u/s
Try it online!
Wraps onto a cube with edge length 3 as follows
u ! @
- W 1
4 ; O
I S % / ; ; , p ; ? w
O @ u / s . . . . . . .
. . . . . . . . . . . .
. . .
. . .
. . .
Watch it run
A fairly basic implementation, without all the redirects it does :
ISinitiates the program by pushing the input and 32 to the stack
%4-!gets the remainder and checks if it is 4 by subtraction
1O@output 1 if it was 4 and halt
;;,clean up the stack and do integer divide
p;?clean up bottom of the stack and check div result for 0
O@if div result zero output and halt
sswap the top of stack and start back at step 2 above
answered Feb 28 at 21:34
MickyTMickyT
10.2k21637
$endgroup$
add a comment |
$begingroup$
Cubix, 26 bytes
u!@-W14;OIS%/;;,p;?wO@u/s
Try it online!
Wraps onto a cube with edge length 3 as follows
u ! @
- W 1
4 ; O
I S % / ; ; , p ; ? w
O @ u / s . . . . . . .
. . . . . . . . . . . .
. . .
. . .
. . .
Watch it run
A fairly basic implementation, without all the redirects it does :
ISinitiates the program by pushing the input and 32 to the stack
%4-!gets the remainder and checks if it is 4 by subtraction
1O@output 1 if it was 4 and halt
;;,clean up the stack and do integer divide
p;?clean up bottom of the stack and check div result for 0
O@if div result zero output and halt
sswap the top of stack and start back at step 2 above
answered Feb 28 at 21:34
MickyTMickyT
10.2k21637
$endgroup$
add a comment |
$begingroup$
Cubix, 26 bytes
u!@-W14;OIS%/;;,p;?wO@u/s
Try it online!
Wraps onto a cube with edge length 3 as follows
u ! @
- W 1
4 ; O
I S % / ; ; , p ; ? w
O @ u / s . . . . . . .
. . . . . . . . . . . .
. . .
. . .
. . .
Watch it run
A fairly basic implementation, without all the redirects it does :
ISinitiates the program by pushing the input and 32 to the stack
%4-!gets the remainder and checks if it is 4 by subtraction
1O@output 1 if it was 4 and halt
;;,clean up the stack and do integer divide
p;?clean up bottom of the stack and check div result for 0
O@if div result zero output and halt
sswap the top of stack and start back at step 2 above
answered Feb 28 at 21:34
MickyTMickyT
10.2k21637
$endgroup$
Cubix, 26 bytes
u!@-W14;OIS%/;;,p;?wO@u/s
Try it online!
Wraps onto a cube with edge length 3 as follows
u ! @
- W 1
4 ; O
I S % / ; ; , p ; ? w
O @ u / s . . . . . . .
. . . . . . . . . . . .
. . .
. . .
. . .
Watch it run
A fairly basic implementation, without all the redirects it does :
ISinitiates the program by pushing the input and 32 to the stack
%4-!gets the remainder and checks if it is 4 by subtraction
1O@output 1 if it was 4 and halt
;;,clean up the stack and do integer divide
p;?clean up bottom of the stack and check div result for 0
O@if div result zero output and halt
sswap the top of stack and start back at step 2 above
answered Feb 28 at 21:34
MickyTMickyT
10.2k21637
answered Feb 28 at 21:34
MickyTMickyT
10.2k21637
answered Feb 28 at 21:34
MickyTMickyT
10.2k21637
answered Feb 28 at 21:34
MickyTMickyT
10.2k21637
10.2k21637
add a comment |
add a comment |
$begingroup$
MATL, 8 bytes
32YA52=a
Try it online!
answered Feb 28 at 13:57
SanchisesSanchises
6,17712352
$endgroup$
$begingroup$
@Luis I can definitely drop theG(not sure why I included it in the first place) but that's only one byte (thanks for spotting that!). Changing32YA52to32_YA4is the same number of bytes right?
$endgroup$
– Sanchises
Mar 1 at 7:06
$begingroup$
Ah, yes, I can't count
$endgroup$
– Luis Mendo
Mar 1 at 10:26
2
$begingroup$
@Luis Count? Who needs to count when you can'32_YA4'n'32YA52'n-
$endgroup$
– Sanchises
Mar 1 at 11:56
add a comment |
$begingroup$
MATL, 8 bytes
32YA52=a
Try it online!
answered Feb 28 at 13:57
SanchisesSanchises
6,17712352
$endgroup$
$begingroup$
@Luis I can definitely drop theG(not sure why I included it in the first place) but that's only one byte (thanks for spotting that!). Changing32YA52to32_YA4is the same number of bytes right?
$endgroup$
– Sanchises
Mar 1 at 7:06
$begingroup$
Ah, yes, I can't count
$endgroup$
– Luis Mendo
Mar 1 at 10:26
2
$begingroup$
@Luis Count? Who needs to count when you can'32_YA4'n'32YA52'n-
$endgroup$
– Sanchises
Mar 1 at 11:56
add a comment |
$begingroup$
MATL, 8 bytes
32YA52=a
Try it online!
answered Feb 28 at 13:57
SanchisesSanchises
6,17712352
$endgroup$
MATL, 8 bytes
32YA52=a
Try it online!
answered Feb 28 at 13:57
SanchisesSanchises
6,17712352
edited Mar 1 at 7:07
answered Feb 28 at 13:57
SanchisesSanchises
6,17712352
answered Feb 28 at 13:57
SanchisesSanchises
6,17712352
answered Feb 28 at 13:57
SanchisesSanchises
6,17712352
6,17712352
$begingroup$
@Luis I can definitely drop theG(not sure why I included it in the first place) but that's only one byte (thanks for spotting that!). Changing32YA52to32_YA4is the same number of bytes right?
$endgroup$
– Sanchises
Mar 1 at 7:06
$begingroup$
Ah, yes, I can't count
$endgroup$
– Luis Mendo
Mar 1 at 10:26
2
$begingroup$
@Luis Count? Who needs to count when you can'32_YA4'n'32YA52'n-
$endgroup$
– Sanchises
Mar 1 at 11:56
add a comment |
$begingroup$
@Luis I can definitely drop theG(not sure why I included it in the first place) but that's only one byte (thanks for spotting that!). Changing32YA52to32_YA4is the same number of bytes right?
$endgroup$
– Sanchises
Mar 1 at 7:06
$begingroup$
Ah, yes, I can't count
$endgroup$
– Luis Mendo
Mar 1 at 10:26
2
$begingroup$
@Luis Count? Who needs to count when you can'32_YA4'n'32YA52'n-
$endgroup$
– Sanchises
Mar 1 at 11:56
$begingroup$
@Luis I can definitely drop the
G (not sure why I included it in the first place) but that's only one byte (thanks for spotting that!). Changing 32YA52 to 32_YA4 is the same number of bytes right?$endgroup$
– Sanchises
Mar 1 at 7:06
$begingroup$
@Luis I can definitely drop the
G (not sure why I included it in the first place) but that's only one byte (thanks for spotting that!). Changing 32YA52 to 32_YA4 is the same number of bytes right?$endgroup$
– Sanchises
Mar 1 at 7:06
$begingroup$
Ah, yes, I can't count
$endgroup$
– Luis Mendo
Mar 1 at 10:26
$begingroup$
Ah, yes, I can't count
$endgroup$
– Luis Mendo
Mar 1 at 10:26
2
2
$begingroup$
@Luis Count? Who needs to count when you can
'32_YA4'n'32YA52'n-$endgroup$
– Sanchises
Mar 1 at 11:56
$begingroup$
@Luis Count? Who needs to count when you can
'32_YA4'n'32YA52'n-$endgroup$
– Sanchises
Mar 1 at 11:56
add a comment |
$begingroup$
Retina 0.8.2, 31 bytes
.+
$*
+`(1+)1{31}
$1;
b1111b
Try it online! Link includes test cases. Outputs zero unless the number is rude. Works by converting the input to unary and then to unary-encoded base 32 and counting the number of 4s in the result.
answered Feb 27 at 13:28
NeilNeil
81.4k745178
$endgroup$
add a comment |
$begingroup$
Retina 0.8.2, 31 bytes
.+
$*
+`(1+)1{31}
$1;
b1111b
Try it online! Link includes test cases. Outputs zero unless the number is rude. Works by converting the input to unary and then to unary-encoded base 32 and counting the number of 4s in the result.
answered Feb 27 at 13:28
NeilNeil
81.4k745178
$endgroup$
add a comment |
$begingroup$
Retina 0.8.2, 31 bytes
.+
$*
+`(1+)1{31}
$1;
b1111b
Try it online! Link includes test cases. Outputs zero unless the number is rude. Works by converting the input to unary and then to unary-encoded base 32 and counting the number of 4s in the result.
answered Feb 27 at 13:28
NeilNeil
81.4k745178
$endgroup$
Retina 0.8.2, 31 bytes
.+
$*
+`(1+)1{31}
$1;
b1111b
Try it online! Link includes test cases. Outputs zero unless the number is rude. Works by converting the input to unary and then to unary-encoded base 32 and counting the number of 4s in the result.
answered Feb 27 at 13:28
NeilNeil
81.4k745178
answered Feb 27 at 13:28
NeilNeil
81.4k745178
answered Feb 27 at 13:28
NeilNeil
81.4k745178
answered Feb 27 at 13:28
NeilNeil
81.4k745178
81.4k745178
add a comment |
add a comment |
$begingroup$
Java 8, 40 33 bytes
n->n.toString(n,32).contains("4")
Port of @Adám's APL (dzaima/APL) answer.
Try it online.
Explanation:
n-> // Method with Integer parameter and boolean return-type
n.toString(n,32) // Convert the input to a base-32 String
.contains("4") // And check if it contains a "4"
answered Feb 27 at 12:56
Kevin CruijssenKevin Cruijssen
39.7k560203
$endgroup$
add a comment |
$begingroup$
Java 8, 40 33 bytes
n->n.toString(n,32).contains("4")
Port of @Adám's APL (dzaima/APL) answer.
Try it online.
Explanation:
n-> // Method with Integer parameter and boolean return-type
n.toString(n,32) // Convert the input to a base-32 String
.contains("4") // And check if it contains a "4"
answered Feb 27 at 12:56
Kevin CruijssenKevin Cruijssen
39.7k560203
$endgroup$
add a comment |
$begingroup$
Java 8, 40 33 bytes
n->n.toString(n,32).contains("4")
Port of @Adám's APL (dzaima/APL) answer.
Try it online.
Explanation:
n-> // Method with Integer parameter and boolean return-type
n.toString(n,32) // Convert the input to a base-32 String
.contains("4") // And check if it contains a "4"
answered Feb 27 at 12:56
Kevin CruijssenKevin Cruijssen
39.7k560203
$endgroup$
Java 8, 40 33 bytes
n->n.toString(n,32).contains("4")
Port of @Adám's APL (dzaima/APL) answer.
Try it online.
Explanation:
n-> // Method with Integer parameter and boolean return-type
n.toString(n,32) // Convert the input to a base-32 String
.contains("4") // And check if it contains a "4"
answered Feb 27 at 12:56
Kevin CruijssenKevin Cruijssen
39.7k560203
edited Feb 27 at 14:58
answered Feb 27 at 12:56
Kevin CruijssenKevin Cruijssen
39.7k560203
answered Feb 27 at 12:56
Kevin CruijssenKevin Cruijssen
39.7k560203
answered Feb 27 at 12:56
Kevin CruijssenKevin Cruijssen
39.7k560203
39.7k560203
add a comment |
add a comment |
$begingroup$
Batch, 77 45 bytes
@cmd/cset/a"m=34636833,n=%1^m*4,(n-m)&~n&m*16
Based on these bit twiddling hacks. Explanation: Only 6 hands need to be checked due to the limited range (30 bits) of the input that's required to be supported. The magic number m is equivalent to 111111 in base 32, so that the first operation toggles the rude bits in the input number. It then remains to find which of the 6 hands is now zero.
answered Mar 1 at 0:50
NeilNeil
81.4k745178
$endgroup$
add a comment |
$begingroup$
Batch, 77 45 bytes
@cmd/cset/a"m=34636833,n=%1^m*4,(n-m)&~n&m*16
Based on these bit twiddling hacks. Explanation: Only 6 hands need to be checked due to the limited range (30 bits) of the input that's required to be supported. The magic number m is equivalent to 111111 in base 32, so that the first operation toggles the rude bits in the input number. It then remains to find which of the 6 hands is now zero.
answered Mar 1 at 0:50
NeilNeil
81.4k745178
$endgroup$
add a comment |
$begingroup$
Batch, 77 45 bytes
@cmd/cset/a"m=34636833,n=%1^m*4,(n-m)&~n&m*16
Based on these bit twiddling hacks. Explanation: Only 6 hands need to be checked due to the limited range (30 bits) of the input that's required to be supported. The magic number m is equivalent to 111111 in base 32, so that the first operation toggles the rude bits in the input number. It then remains to find which of the 6 hands is now zero.
answered Mar 1 at 0:50
NeilNeil
81.4k745178
$endgroup$
Batch, 77 45 bytes
@cmd/cset/a"m=34636833,n=%1^m*4,(n-m)&~n&m*16
Based on these bit twiddling hacks. Explanation: Only 6 hands need to be checked due to the limited range (30 bits) of the input that's required to be supported. The magic number m is equivalent to 111111 in base 32, so that the first operation toggles the rude bits in the input number. It then remains to find which of the 6 hands is now zero.
answered Mar 1 at 0:50
NeilNeil
81.4k745178
edited Mar 1 at 11:13
answered Mar 1 at 0:50
NeilNeil
81.4k745178
answered Mar 1 at 0:50
NeilNeil
81.4k745178
answered Mar 1 at 0:50
NeilNeil
81.4k745178
81.4k745178
add a comment |
add a comment |
$begingroup$
><>, 28 bytes
Outputs 4 for rude numbers throws an exception for non-rude numbers.
:1(?^:" ":
,&-v?=4:%&/
;n<
Try it online!
answered Mar 1 at 10:34
EmignaEmigna
46.7k432142
$endgroup$
1
$begingroup$
An exception is acceptable, the C# answer does it
$endgroup$
– ASCII-only
Mar 1 at 11:22
add a comment |
$begingroup$
><>, 28 bytes
Outputs 4 for rude numbers throws an exception for non-rude numbers.
:1(?^:" ":
,&-v?=4:%&/
;n<
Try it online!
answered Mar 1 at 10:34
EmignaEmigna
46.7k432142
$endgroup$
1
$begingroup$
An exception is acceptable, the C# answer does it
$endgroup$
– ASCII-only
Mar 1 at 11:22
add a comment |
$begingroup$
><>, 28 bytes
Outputs 4 for rude numbers throws an exception for non-rude numbers.
:1(?^:" ":
,&-v?=4:%&/
;n<
Try it online!
answered Mar 1 at 10:34
EmignaEmigna
46.7k432142
$endgroup$
><>, 28 bytes
Outputs 4 for rude numbers throws an exception for non-rude numbers.
:1(?^:" ":
,&-v?=4:%&/
;n<
Try it online!
answered Mar 1 at 10:34
EmignaEmigna
46.7k432142
edited Mar 1 at 11:29
answered Mar 1 at 10:34
EmignaEmigna
46.7k432142
answered Mar 1 at 10:34
EmignaEmigna
46.7k432142
answered Mar 1 at 10:34
EmignaEmigna
46.7k432142
46.7k432142
1
$begingroup$
An exception is acceptable, the C# answer does it
$endgroup$
– ASCII-only
Mar 1 at 11:22
add a comment |
1
$begingroup$
An exception is acceptable, the C# answer does it
$endgroup$
– ASCII-only
Mar 1 at 11:22
1
1
$begingroup$
An exception is acceptable, the C# answer does it
$endgroup$
– ASCII-only
Mar 1 at 11:22
$begingroup$
An exception is acceptable, the C# answer does it
$endgroup$
– ASCII-only
Mar 1 at 11:22
add a comment |
$begingroup$
Wolfram Language (Mathematica), 37 bytes 36 bytes 29 bytes
-2 bytes by Jonathan Frech
#~IntegerDigits~32~MemberQ~4&
Try it online!
31-byte solution:
MemberQ[IntegerDigits[#,32],4]&
Try it online!
edited Mar 2 at 1:49
lirtosiast
18.2k438109
answered Feb 27 at 9:03
Rainer GlügeRainer Glüge
1213
$endgroup$
$begingroup$
Hello and welcome to PPCG. As it stands, your expression is a single Boolean value. Please fix your answer to either be a full program or a function (...#...&is often used in Mathematica).
$endgroup$
– Jonathan Frech
Feb 27 at 9:22
$begingroup$
Hello. Is this what you mean?
$endgroup$
– Rainer Glüge
Feb 27 at 9:32
$begingroup$
pls use tio.run/#mathematica instead of W|A to make sure it's valid mathematica code :P and you don't need the[n]at the end, just the&. Also since posts have edit history, it's fine to leave out previous entries, and the convention for old scores is<s>40</s> <s>36</s>
$endgroup$
– ASCII-only
Feb 27 at 10:22
$begingroup$
Yes. This is what I meant. 29 bytes.
$endgroup$
– Jonathan Frech
Feb 27 at 13:17
$begingroup$
I guess I have to get used to the functional programming style.
$endgroup$
– Rainer Glüge
Feb 27 at 13:34
add a comment |
$begingroup$
Wolfram Language (Mathematica), 37 bytes 36 bytes 29 bytes
-2 bytes by Jonathan Frech
#~IntegerDigits~32~MemberQ~4&
Try it online!
31-byte solution:
MemberQ[IntegerDigits[#,32],4]&
Try it online!
edited Mar 2 at 1:49
lirtosiast
18.2k438109
answered Feb 27 at 9:03
Rainer GlügeRainer Glüge
1213
$endgroup$
$begingroup$
Hello and welcome to PPCG. As it stands, your expression is a single Boolean value. Please fix your answer to either be a full program or a function (...#...&is often used in Mathematica).
$endgroup$
– Jonathan Frech
Feb 27 at 9:22
$begingroup$
Hello. Is this what you mean?
$endgroup$
– Rainer Glüge
Feb 27 at 9:32
$begingroup$
pls use tio.run/#mathematica instead of W|A to make sure it's valid mathematica code :P and you don't need the[n]at the end, just the&. Also since posts have edit history, it's fine to leave out previous entries, and the convention for old scores is<s>40</s> <s>36</s>
$endgroup$
– ASCII-only
Feb 27 at 10:22
$begingroup$
Yes. This is what I meant. 29 bytes.
$endgroup$
– Jonathan Frech
Feb 27 at 13:17
$begingroup$
I guess I have to get used to the functional programming style.
$endgroup$
– Rainer Glüge
Feb 27 at 13:34
add a comment |
$begingroup$
Wolfram Language (Mathematica), 37 bytes 36 bytes 29 bytes
-2 bytes by Jonathan Frech
#~IntegerDigits~32~MemberQ~4&
Try it online!
31-byte solution:
MemberQ[IntegerDigits[#,32],4]&
Try it online!
edited Mar 2 at 1:49
lirtosiast
18.2k438109
answered Feb 27 at 9:03
Rainer GlügeRainer Glüge
1213
$endgroup$
Wolfram Language (Mathematica), 37 bytes 36 bytes 29 bytes
-2 bytes by Jonathan Frech
#~IntegerDigits~32~MemberQ~4&
Try it online!
31-byte solution:
MemberQ[IntegerDigits[#,32],4]&
Try it online!
edited Mar 2 at 1:49
lirtosiast
18.2k438109
answered Feb 27 at 9:03
Rainer GlügeRainer Glüge
1213
edited Mar 2 at 1:49
lirtosiast
18.2k438109
edited Mar 2 at 1:49
lirtosiast
18.2k438109
edited Mar 2 at 1:49
lirtosiast
18.2k438109
18.2k438109
answered Feb 27 at 9:03
Rainer GlügeRainer Glüge
1213
answered Feb 27 at 9:03
Rainer GlügeRainer Glüge
1213
answered Feb 27 at 9:03
Rainer GlügeRainer Glüge
1213
1213
$begingroup$
Hello and welcome to PPCG. As it stands, your expression is a single Boolean value. Please fix your answer to either be a full program or a function (...#...&is often used in Mathematica).
$endgroup$
– Jonathan Frech
Feb 27 at 9:22
$begingroup$
Hello. Is this what you mean?
$endgroup$
– Rainer Glüge
Feb 27 at 9:32
$begingroup$
pls use tio.run/#mathematica instead of W|A to make sure it's valid mathematica code :P and you don't need the[n]at the end, just the&. Also since posts have edit history, it's fine to leave out previous entries, and the convention for old scores is<s>40</s> <s>36</s>
$endgroup$
– ASCII-only
Feb 27 at 10:22
$begingroup$
Yes. This is what I meant. 29 bytes.
$endgroup$
– Jonathan Frech
Feb 27 at 13:17
$begingroup$
I guess I have to get used to the functional programming style.
$endgroup$
– Rainer Glüge
Feb 27 at 13:34
add a comment |
$begingroup$
Hello and welcome to PPCG. As it stands, your expression is a single Boolean value. Please fix your answer to either be a full program or a function (...#...&is often used in Mathematica).
$endgroup$
– Jonathan Frech
Feb 27 at 9:22
$begingroup$
Hello. Is this what you mean?
$endgroup$
– Rainer Glüge
Feb 27 at 9:32
$begingroup$
pls use tio.run/#mathematica instead of W|A to make sure it's valid mathematica code :P and you don't need the[n]at the end, just the&. Also since posts have edit history, it's fine to leave out previous entries, and the convention for old scores is<s>40</s> <s>36</s>
$endgroup$
– ASCII-only
Feb 27 at 10:22
$begingroup$
Yes. This is what I meant. 29 bytes.
$endgroup$
– Jonathan Frech
Feb 27 at 13:17
$begingroup$
I guess I have to get used to the functional programming style.
$endgroup$
– Rainer Glüge
Feb 27 at 13:34
$begingroup$
Hello and welcome to PPCG. As it stands, your expression is a single Boolean value. Please fix your answer to either be a full program or a function (
...#...& is often used in Mathematica).$endgroup$
– Jonathan Frech
Feb 27 at 9:22
$begingroup$
Hello and welcome to PPCG. As it stands, your expression is a single Boolean value. Please fix your answer to either be a full program or a function (
...#...& is often used in Mathematica).$endgroup$
– Jonathan Frech
Feb 27 at 9:22
$begingroup$
Hello. Is this what you mean?
$endgroup$
– Rainer Glüge
Feb 27 at 9:32
$begingroup$
Hello. Is this what you mean?
$endgroup$
– Rainer Glüge
Feb 27 at 9:32
$begingroup$
pls use tio.run/#mathematica instead of W|A to make sure it's valid mathematica code :P and you don't need the
[n] at the end, just the &. Also since posts have edit history, it's fine to leave out previous entries, and the convention for old scores is <s>40</s> <s>36</s>$endgroup$
– ASCII-only
Feb 27 at 10:22
$begingroup$
pls use tio.run/#mathematica instead of W|A to make sure it's valid mathematica code :P and you don't need the
[n] at the end, just the &. Also since posts have edit history, it's fine to leave out previous entries, and the convention for old scores is <s>40</s> <s>36</s>$endgroup$
– ASCII-only
Feb 27 at 10:22
$begingroup$
Yes. This is what I meant. 29 bytes.
$endgroup$
– Jonathan Frech
Feb 27 at 13:17
$begingroup$
Yes. This is what I meant. 29 bytes.
$endgroup$
– Jonathan Frech
Feb 27 at 13:17
$begingroup$
I guess I have to get used to the functional programming style.
$endgroup$
– Rainer Glüge
Feb 27 at 13:34
$begingroup$
I guess I have to get used to the functional programming style.
$endgroup$
– Rainer Glüge
Feb 27 at 13:34
add a comment |
$begingroup$
x86 machine code, 14 bytes
(same machine code works in 16-bit, 32-bit, and 64-bit. In 16-bit mode, it uses AX and DI instead of EAX and EDI in 32 and 64-bit mode.)
Algorithm: check low 5 bits with x & 31 == 4, then right-shift by 5 bits, and repeat if the shift result is non-zero.
Callable from C with char isrude(unsigned n); according to the x86-64 System V calling convention. 0 is truthy, non-0 is falsy (this is asm, not C1).
line addr code bytes
num
1 ; input: number in EDI
2 ; output: integer result in AL: 0 -> rude, non-zero non-rude
3 ; clobbers: RDI
4 isrude:
5 .check_low_bitgroup:
6 00000000 89F8 mov eax, edi
7 00000002 241F and al, 31 ; isolate low 5 bits
8 00000004 2C04 sub al, 4 ; like cmp but leaves AL 0 or non-zero
9 00000006 7405 jz .rude ; if (al & 31 == 4) return 0;
10
11 00000008 C1EF05 shr edi, 5
12 0000000B 75F3 jnz .check_low_bitgroup
13 ;; fall through to here is only possible if AL is non-zero
14 .rude:
15 0000000D C3 ret
16 0E size: db $ - isrude
This takes advantage of the short-form op al, imm8 encoding for AND and SUB. I could have used XOR al,4 to produce 0 on equality, but SUB is faster because it can macro-fuse with JZ into a single sub-and-branch uop on Sandybridge-family.
Fun fact: using the flag-result of a shift by more than 1 will be slow on P6-family (front-end stalls until the shift retires), but that's fine.
Footnote 1: This is an assembly language function, and x86 asm has both jz and jnz, so as per meta I can choose either way. I'm not intending this to match C truthy/falsy.
It happened to be convenient to return in AL instead of EFLAGS, so we can describe the function to a C compiler without a wrapper, but my choice of truthy/falsy isn't constrained by using a C caller to test it.
answered Mar 4 at 0:21
Peter CordesPeter Cordes
2,0921018
$endgroup$
add a comment |
$begingroup$
x86 machine code, 14 bytes
(same machine code works in 16-bit, 32-bit, and 64-bit. In 16-bit mode, it uses AX and DI instead of EAX and EDI in 32 and 64-bit mode.)
Algorithm: check low 5 bits with x & 31 == 4, then right-shift by 5 bits, and repeat if the shift result is non-zero.
Callable from C with char isrude(unsigned n); according to the x86-64 System V calling convention. 0 is truthy, non-0 is falsy (this is asm, not C1).
line addr code bytes
num
1 ; input: number in EDI
2 ; output: integer result in AL: 0 -> rude, non-zero non-rude
3 ; clobbers: RDI
4 isrude:
5 .check_low_bitgroup:
6 00000000 89F8 mov eax, edi
7 00000002 241F and al, 31 ; isolate low 5 bits
8 00000004 2C04 sub al, 4 ; like cmp but leaves AL 0 or non-zero
9 00000006 7405 jz .rude ; if (al & 31 == 4) return 0;
10
11 00000008 C1EF05 shr edi, 5
12 0000000B 75F3 jnz .check_low_bitgroup
13 ;; fall through to here is only possible if AL is non-zero
14 .rude:
15 0000000D C3 ret
16 0E size: db $ - isrude
This takes advantage of the short-form op al, imm8 encoding for AND and SUB. I could have used XOR al,4 to produce 0 on equality, but SUB is faster because it can macro-fuse with JZ into a single sub-and-branch uop on Sandybridge-family.
Fun fact: using the flag-result of a shift by more than 1 will be slow on P6-family (front-end stalls until the shift retires), but that's fine.
Footnote 1: This is an assembly language function, and x86 asm has both jz and jnz, so as per meta I can choose either way. I'm not intending this to match C truthy/falsy.
It happened to be convenient to return in AL instead of EFLAGS, so we can describe the function to a C compiler without a wrapper, but my choice of truthy/falsy isn't constrained by using a C caller to test it.
answered Mar 4 at 0:21
Peter CordesPeter Cordes
2,0921018
$endgroup$
add a comment |
$begingroup$
x86 machine code, 14 bytes
(same machine code works in 16-bit, 32-bit, and 64-bit. In 16-bit mode, it uses AX and DI instead of EAX and EDI in 32 and 64-bit mode.)
Algorithm: check low 5 bits with x & 31 == 4, then right-shift by 5 bits, and repeat if the shift result is non-zero.
Callable from C with char isrude(unsigned n); according to the x86-64 System V calling convention. 0 is truthy, non-0 is falsy (this is asm, not C1).
line addr code bytes
num
1 ; input: number in EDI
2 ; output: integer result in AL: 0 -> rude, non-zero non-rude
3 ; clobbers: RDI
4 isrude:
5 .check_low_bitgroup:
6 00000000 89F8 mov eax, edi
7 00000002 241F and al, 31 ; isolate low 5 bits
8 00000004 2C04 sub al, 4 ; like cmp but leaves AL 0 or non-zero
9 00000006 7405 jz .rude ; if (al & 31 == 4) return 0;
10
11 00000008 C1EF05 shr edi, 5
12 0000000B 75F3 jnz .check_low_bitgroup
13 ;; fall through to here is only possible if AL is non-zero
14 .rude:
15 0000000D C3 ret
16 0E size: db $ - isrude
This takes advantage of the short-form op al, imm8 encoding for AND and SUB. I could have used XOR al,4 to produce 0 on equality, but SUB is faster because it can macro-fuse with JZ into a single sub-and-branch uop on Sandybridge-family.
Fun fact: using the flag-result of a shift by more than 1 will be slow on P6-family (front-end stalls until the shift retires), but that's fine.
Footnote 1: This is an assembly language function, and x86 asm has both jz and jnz, so as per meta I can choose either way. I'm not intending this to match C truthy/falsy.
It happened to be convenient to return in AL instead of EFLAGS, so we can describe the function to a C compiler without a wrapper, but my choice of truthy/falsy isn't constrained by using a C caller to test it.
answered Mar 4 at 0:21
Peter CordesPeter Cordes
2,0921018
$endgroup$
x86 machine code, 14 bytes
(same machine code works in 16-bit, 32-bit, and 64-bit. In 16-bit mode, it uses AX and DI instead of EAX and EDI in 32 and 64-bit mode.)
Algorithm: check low 5 bits with x & 31 == 4, then right-shift by 5 bits, and repeat if the shift result is non-zero.
Callable from C with char isrude(unsigned n); according to the x86-64 System V calling convention. 0 is truthy, non-0 is falsy (this is asm, not C1).
line addr code bytes
num
1 ; input: number in EDI
2 ; output: integer result in AL: 0 -> rude, non-zero non-rude
3 ; clobbers: RDI
4 isrude:
5 .check_low_bitgroup:
6 00000000 89F8 mov eax, edi
7 00000002 241F and al, 31 ; isolate low 5 bits
8 00000004 2C04 sub al, 4 ; like cmp but leaves AL 0 or non-zero
9 00000006 7405 jz .rude ; if (al & 31 == 4) return 0;
10
11 00000008 C1EF05 shr edi, 5
12 0000000B 75F3 jnz .check_low_bitgroup
13 ;; fall through to here is only possible if AL is non-zero
14 .rude:
15 0000000D C3 ret
16 0E size: db $ - isrude
This takes advantage of the short-form op al, imm8 encoding for AND and SUB. I could have used XOR al,4 to produce 0 on equality, but SUB is faster because it can macro-fuse with JZ into a single sub-and-branch uop on Sandybridge-family.
Fun fact: using the flag-result of a shift by more than 1 will be slow on P6-family (front-end stalls until the shift retires), but that's fine.
Footnote 1: This is an assembly language function, and x86 asm has both jz and jnz, so as per meta I can choose either way. I'm not intending this to match C truthy/falsy.
It happened to be convenient to return in AL instead of EFLAGS, so we can describe the function to a C compiler without a wrapper, but my choice of truthy/falsy isn't constrained by using a C caller to test it.
answered Mar 4 at 0:21
Peter CordesPeter Cordes
2,0921018
answered Mar 4 at 0:21
Peter CordesPeter Cordes
2,0921018
answered Mar 4 at 0:21
Peter CordesPeter Cordes
2,0921018
answered Mar 4 at 0:21
Peter CordesPeter Cordes
2,0921018
2,0921018
add a comment |
add a comment |
$begingroup$
Java 8, 28 22 21 bytes
n->n%32==4|n>>5%32==4
Inspired by @kevin-cruijssen's answer. Only works for 2 hands.
Try it online!
Explanation:
n-> // Method with int parameter and boolean return-type
n%32 // Only consider right 5 bytes (fingers)
==4 // Middle finger
| ... n>>5 // Repeat with shifted bits for other hand
answered 2 days ago
Daniel WiddisDaniel Widdis
1114
New contributor
Daniel Widdis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Java 8, 28 22 21 bytes
n->n%32==4|n>>5%32==4
Inspired by @kevin-cruijssen's answer. Only works for 2 hands.
Try it online!
Explanation:
n-> // Method with int parameter and boolean return-type
n%32 // Only consider right 5 bytes (fingers)
==4 // Middle finger
| ... n>>5 // Repeat with shifted bits for other hand
answered 2 days ago
Daniel WiddisDaniel Widdis
1114
New contributor
Daniel Widdis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Java 8, 28 22 21 bytes
n->n%32==4|n>>5%32==4
Inspired by @kevin-cruijssen's answer. Only works for 2 hands.
Try it online!
Explanation:
n-> // Method with int parameter and boolean return-type
n%32 // Only consider right 5 bytes (fingers)
==4 // Middle finger
| ... n>>5 // Repeat with shifted bits for other hand
answered 2 days ago
Daniel WiddisDaniel Widdis
1114
New contributor
Daniel Widdis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Java 8, 28 22 21 bytes
n->n%32==4|n>>5%32==4
Inspired by @kevin-cruijssen's answer. Only works for 2 hands.
Try it online!
Explanation:
n-> // Method with int parameter and boolean return-type
n%32 // Only consider right 5 bytes (fingers)
==4 // Middle finger
| ... n>>5 // Repeat with shifted bits for other hand
answered 2 days ago
Daniel WiddisDaniel Widdis
1114
New contributor
Daniel Widdis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
answered 2 days ago
Daniel WiddisDaniel Widdis
1114
New contributor
Daniel Widdis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Daniel WiddisDaniel Widdis
1114
answered 2 days ago
Daniel WiddisDaniel Widdis
1114
1114
New contributor
Daniel Widdis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Daniel Widdis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Daniel Widdis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 2
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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.
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42
$begingroup$
assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.$endgroup$
– Veskah
Feb 27 at 3:24
5
$begingroup$
@Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
$endgroup$
– Gryphon
Feb 27 at 3:26
12
$begingroup$
It's worse if you're British - 6 is rude too then!
$endgroup$
– Matthew
Feb 27 at 7:07
1
$begingroup$
Is it OK to take input in a different base than 10?
$endgroup$
– wastl
Feb 27 at 20:08
1
$begingroup$
5 seems fairly rude too. Not sure anyone would say "Oh she had her thumb out, that's perfectly polite"
$endgroup$
– ale10ander
Mar 1 at 1:43