The 465 Arrangement

Here's the challenge. Write some code to output all the integers in a range. Sounds easy, but here's the tricky part. It will start with the lowest number, then the highest. Then the lowest number which isn't yet in the array. Then the highest which isn't yet in it.

Example:

Lets take 1 to 5 as our start

The numbers are [1, 2, 3, 4, 5].

We take the first, so [1]. Remaining numbers are [2, 3, 4, 5].
We take the last, new array is [1, 5]. Remaining numbers are [2, 3, 4].
We take the first, new array is [1, 5, 2]. Remaining numbers are [3, 4].
We take the last, new array is [1, 5, 2, 4]. Remaining numbers are [3].
We take the first, new array is [1, 5, 2, 4, 3]. No numbers remaining, we're done.
Output [1, 5, 2, 4, 3]

Rules:

This is code golf, write it in the fewest bytes, any language.

No standard loopholes.

Links to an online interpreter, please? (E.g. https://tio.run/)

Two inputs, both integers. Low end of range, and high end of range.

I don't mind what the data type of the output is, but it must show the numbers in the correct order.

Examples

Low: 4
High: 6
Result:
4
6
5

Low: 1
High: 5
Result:
1
5
2
4
3

Low: -1
High: 1
Result:
-1
1
0

Low: -1
high: 2
Result:
-1
2
0
1

Low: -50
High: 50
Result:
-50
50
-49
49
-48
48
-47
47
-46
46
-45
45
-44
44
-43
43
-42
42
-41
41
-40
40
-39
39
-38
38
-37
37
-36
36
-35
35
-34
34
-33
33
-32
32
-31
31
-30
30
-29
29
-28
28
-27
27
-26
26
-25
25
-24
24
-23
23
-22
22
-21
21
-20
20
-19
19
-18
18
-17
17
-16
16
-15
15
-14
14
-13
13
-12
12
-11
11
-10
10
-9
9
-8
8
-7
7
-6
6
-5
5
-4
4
-3
3
-2
2
-1
1
0

Happy golfing!

asked Feb 13 at 15:08

AJFaraday

3,47643159

2

$begingroup$
Almost duplicate (the difference being that this one requires reversing the second half before merging).
$endgroup$
– Peter Taylor
Feb 13 at 15:13

$begingroup$
is the input always going to be in the order of low end, high end?
$endgroup$
– Sumner18
Feb 13 at 16:14

1

$begingroup$
@Sumner18 yes. The community here is dead-set against input validation, and I haven’t asked for a reverse-order input, so we can assume it’ll always be low - high.
$endgroup$
– AJFaraday
Feb 13 at 16:21

1

$begingroup$
@Sumner18 How these challenges usually work is that we don't care how invalid inputs are handled. Your code is only judged to be successful by how it deals with valid inputs (i.e. both are integers, the first is lower than the second)
$endgroup$
– AJFaraday
Feb 13 at 16:36

1

$begingroup$
@AJFaraday: you should add a note to the main post indicating that X will be always strictly lower than Y (i.e. X != Y), I missed this comment ;)
$endgroup$
– digEmAll
Feb 13 at 18:23

|
show 2 more comments

Example:

Lets take 1 to 5 as our start

The numbers are [1, 2, 3, 4, 5].

Rules:

This is code golf, write it in the fewest bytes, any language.

No standard loopholes.

Links to an online interpreter, please? (E.g. https://tio.run/)

Two inputs, both integers. Low end of range, and high end of range.

I don't mind what the data type of the output is, but it must show the numbers in the correct order.

Examples

Low: 4
High: 6
Result:
4
6
5

Low: 1
High: 5
Result:
1
5
2
4
3

Low: -1
High: 1
Result:
-1
1
0

Low: -1
high: 2
Result:
-1
2
0
1

Happy golfing!

asked Feb 13 at 15:08

AJFaraday

3,47643159

2

$begingroup$
Almost duplicate (the difference being that this one requires reversing the second half before merging).
$endgroup$
– Peter Taylor
Feb 13 at 15:13

$begingroup$
is the input always going to be in the order of low end, high end?
$endgroup$
– Sumner18
Feb 13 at 16:14

1

$begingroup$
@Sumner18 yes. The community here is dead-set against input validation, and I haven’t asked for a reverse-order input, so we can assume it’ll always be low - high.
$endgroup$
– AJFaraday
Feb 13 at 16:21

1

$begingroup$
@Sumner18 How these challenges usually work is that we don't care how invalid inputs are handled. Your code is only judged to be successful by how it deals with valid inputs (i.e. both are integers, the first is lower than the second)
$endgroup$
– AJFaraday
Feb 13 at 16:36

1

$begingroup$
@AJFaraday: you should add a note to the main post indicating that X will be always strictly lower than Y (i.e. X != Y), I missed this comment ;)
$endgroup$
– digEmAll
Feb 13 at 18:23

|
show 2 more comments

Example:

Lets take 1 to 5 as our start

The numbers are [1, 2, 3, 4, 5].

Rules:

This is code golf, write it in the fewest bytes, any language.

No standard loopholes.

Links to an online interpreter, please? (E.g. https://tio.run/)

Two inputs, both integers. Low end of range, and high end of range.

I don't mind what the data type of the output is, but it must show the numbers in the correct order.

Examples

Low: 4
High: 6
Result:
4
6
5

Low: 1
High: 5
Result:
1
5
2
4
3

Low: -1
High: 1
Result:
-1
1
0

Low: -1
high: 2
Result:
-1
2
0
1

Happy golfing!

asked Feb 13 at 15:08

AJFaraday

3,47643159

Example:

Lets take 1 to 5 as our start

The numbers are [1, 2, 3, 4, 5].

Rules:

This is code golf, write it in the fewest bytes, any language.

No standard loopholes.

Links to an online interpreter, please? (E.g. https://tio.run/)

Two inputs, both integers. Low end of range, and high end of range.

I don't mind what the data type of the output is, but it must show the numbers in the correct order.

Examples

Low: 4
High: 6
Result:
4
6
5

Low: 1
High: 5
Result:
1
5
2
4
3

Low: -1
High: 1
Result:
-1
1
0

Low: -1
high: 2
Result:
-1
2
0
1

Happy golfing!

code-golf

asked Feb 13 at 15:08

AJFaraday

3,47643159

asked Feb 13 at 15:08

AJFaraday

3,47643159

asked Feb 13 at 15:08

AJFaraday

3,47643159

asked Feb 13 at 15:08

AJFaraday

3,47643159

asked Feb 13 at 15:08

AJFaraday

3,47643159

2

$begingroup$
Almost duplicate (the difference being that this one requires reversing the second half before merging).
$endgroup$
– Peter Taylor
Feb 13 at 15:13

$begingroup$
is the input always going to be in the order of low end, high end?
$endgroup$
– Sumner18
Feb 13 at 16:14

1

$begingroup$
@Sumner18 yes. The community here is dead-set against input validation, and I haven’t asked for a reverse-order input, so we can assume it’ll always be low - high.
$endgroup$
– AJFaraday
Feb 13 at 16:21

1

$begingroup$
@Sumner18 How these challenges usually work is that we don't care how invalid inputs are handled. Your code is only judged to be successful by how it deals with valid inputs (i.e. both are integers, the first is lower than the second)
$endgroup$
– AJFaraday
Feb 13 at 16:36

1

$begingroup$
@AJFaraday: you should add a note to the main post indicating that X will be always strictly lower than Y (i.e. X != Y), I missed this comment ;)
$endgroup$
– digEmAll
Feb 13 at 18:23

|
show 2 more comments

2

$begingroup$
Almost duplicate (the difference being that this one requires reversing the second half before merging).
$endgroup$
– Peter Taylor
Feb 13 at 15:13

$begingroup$
is the input always going to be in the order of low end, high end?
$endgroup$
– Sumner18
Feb 13 at 16:14

1

$begingroup$
@Sumner18 yes. The community here is dead-set against input validation, and I haven’t asked for a reverse-order input, so we can assume it’ll always be low - high.
$endgroup$
– AJFaraday
Feb 13 at 16:21

1

$begingroup$
@Sumner18 How these challenges usually work is that we don't care how invalid inputs are handled. Your code is only judged to be successful by how it deals with valid inputs (i.e. both are integers, the first is lower than the second)
$endgroup$
– AJFaraday
Feb 13 at 16:36

1

$begingroup$
@AJFaraday: you should add a note to the main post indicating that X will be always strictly lower than Y (i.e. X != Y), I missed this comment ;)
$endgroup$
– digEmAll
Feb 13 at 18:23

Almost duplicate (the difference being that this one requires reversing the second half before merging).

– Peter Taylor
Feb 13 at 15:13

is the input always going to be in the order of low end, high end?

– Sumner18
Feb 13 at 16:14

@Sumner18 yes. The community here is dead-set against input validation, and I haven’t asked for a reverse-order input, so we can assume it’ll always be low - high.

– AJFaraday
Feb 13 at 16:21

@Sumner18 How these challenges usually work is that we don't care how invalid inputs are handled. Your code is only judged to be successful by how it deals with valid inputs (i.e. both are integers, the first is lower than the second)

– AJFaraday
Feb 13 at 16:36

@AJFaraday: you should add a note to the main post indicating that X will be always strictly lower than Y (i.e. X != Y), I missed this comment ;)

– digEmAll
Feb 13 at 18:23

|
show 2 more comments

50 Answers
50

active

oldest

votes

R, 38 37 36 bytes

function(a,b)rbind(a:b,b:a)[a:b-a+1]

Try it online!

-1 byte thanks to @user2390246

-1 byte thanks to @Kirill L.

Exploiting the fact that R stores matrices column-wise

edited Feb 13 at 20:57

answered Feb 13 at 17:59

digEmAll

3,369415

$begingroup$
Using rbind is much better than my approach, but you can save 1 byte by using [seq(a:b)] instead of unique.
$endgroup$
– user2390246
Feb 13 at 18:03

$begingroup$
You're right, I missed the comment where has been specified that a < b (never equal), so we can use seq(a:b)
$endgroup$
– digEmAll
Feb 13 at 18:06

$begingroup$
@digEmAll My solution was essentially a literal interpretation of the puzzle, I never would have even thought of doing something such as this. Impressive, have an upvote!
$endgroup$
– Sumner18
Feb 13 at 18:07

1

$begingroup$
-1 more
$endgroup$
– Kirill L.
Feb 13 at 20:32

add a comment |

Haskell, 30 bytes

a%b=a:take(b-a)(b:(a+1)%(b-1))

Try it online!

answered Feb 14 at 0:56

xnor

90.9k18186441

$begingroup$
Dammit! I have just found the exact same solution. Oh well
$endgroup$
– proud haskeller
Feb 14 at 13:49

add a comment |

Haskell, 39 bytes

a#b|a>b=a:b#(a-1)|a<b=a:b#(a+1)|1>0=[a]

Try it online!

answered Feb 13 at 17:15

ovs

19k21159

add a comment |

R, 65 64 61 60 bytes

-1 byte thanks to Robert S.

-4 more thanks to digEmAll

x=scan();z=x:x[2];while(sum(z|1)){cat(z[1],"");z=rev(z[-1])}

Try it online!

edited Feb 13 at 18:15

answered Feb 13 at 16:58

Sumner18

5507

$begingroup$
You can replace length(z) with sum(z|1) to save 1 byte :)
$endgroup$
– Robert S.
Feb 13 at 17:09

$begingroup$
I don't understand how that works but I guess it does. sum(z|1) seems like it would always evaluate to at least 1, which would cause the while loop to loop endlessly. but apparently not
$endgroup$
– Sumner18
Feb 13 at 17:27

3

$begingroup$
z is a vector. each element of that vector is |ed with 1. Which is always equal to 1. When you take the sum, you have a vector filled with TRUEs so the result is equal to the length of the vector. If the vector is empty, you the is nothing to | with so the output vector is logical(0). When you take that sum, it's 0
$endgroup$
– OganM
Feb 13 at 23:59

add a comment |

Python 2, 44 bytes

f=lambda a,b:[a]*(a==b)or[a]+f(b,a-cmp(a,b))

Try it online!

answered Feb 13 at 16:58

ovs

19k21159

add a comment |

PowerShell, 59 48 bytes

param($a,$b)(($z=0..($b-$a))|%{$a+$_;$b-$_})[$z]

Try it online!

~~(Seems long...)~~

Takes input $a and $b, constructs the range 0 .. ($b-$a), stores that into $z, then loops through that range. The looping through that range is just used as a counter to ensure we get enough iterations. Each iteration, we put $a and $b on the pipeline with addition/subtraction. That gives us something like 1,5,2,4,3,3,4,2,5,1 so we need to slice into that from 0 up to the $b-$a (i.e., the count) of the original array so we're only left with the appropriate elements. That's left on the pipeline and output is implicit.

-11 bytes thanks to mazzy.

edited Feb 13 at 17:40

answered Feb 13 at 15:51

AdmBorkBork

27.2k466234

$begingroup$
48 bytes
$endgroup$
– mazzy
Feb 13 at 17:29

$begingroup$
@mazzy Ah, I like that $b-$a trick -- that's clever!
$endgroup$
– AdmBorkBork
Feb 13 at 17:40

add a comment |

05AB1E, 6 bytes

ŸDvć,R

Try it online!

Explanation

Ÿ        # push range [min ... max]

 D       # duplicate

  v      # for each element in the copy

   ć,    # extract and print the head of the original list

     R   # and then reverse it

edited Feb 14 at 10:36

answered Feb 13 at 17:02

Emigna

46.3k432141

$begingroup$
Ÿ2ä`R.ι non iterative using interleave, but this is still much better.
$endgroup$
– Magic Octopus Urn
Feb 14 at 2:36

1

$begingroup$
@MagicOctopusUrn: I tried a non-iterative solution first, but it was even worse since I didn't know about .ι ;)
$endgroup$
– Emigna
Feb 14 at 7:36

$begingroup$
Similar as what I had in mind, so obvious +1 from me. I do like your alternative 7-byter as well through, @MagicOctopusUrn. :)
$endgroup$
– Kevin Cruijssen
Feb 14 at 8:28

1

$begingroup$
@KristianWilliams: Seems to be working for me.
$endgroup$
– Emigna
Feb 14 at 9:08

1

$begingroup$
@KevinCruijssen: I switched to a pair instead as that felt more intuitive anyways :)
$endgroup$
– Emigna
Feb 14 at 10:50

|
show 3 more comments

Wolfram Language (Mathematica), 56 54 bytes

This is my first time golfing!

f[a_,b_]:=(c=a~Range~b;Drop[c~Riffle~Reverse@c,a-b-1])

Try it online!

Saved 2 bytes using infix notation.

Explanation:

f[a_,b_]:=                                   function of two variables

c=a~Range~b;                                 list of integers from a to b 

                           Reverse@c         same list in reverse

                  c~Riffle~Reverse@c         interleave the two lists

             Drop[c~Riffle~Reverse@c,a-b-1]  drop last |a-b-1| elements (note a-b-1 < 0)

Alternatively, we could use Take[...,b-a+1] for the same result.

Tests:

f[4, 6]

f[1, 5]

f[-1, 1]

f[-1, 2]

Ouput:

{4, 6, 5}

{1, 5, 2, 4, 3}

{-1, 1, 0}

{-1, 2, 0, 1}

edited Feb 15 at 1:08

answered Feb 14 at 21:50

Kai

1713

New contributor

$begingroup$
The "Try it online" link returns a 403. "Sorry, you do not have permission to access this item."
$endgroup$
– Rohit Namjoshi
Feb 15 at 0:02

$begingroup$
@RohitNamjoshi I updated the link
$endgroup$
– Kai
Feb 15 at 0:24

$begingroup$
note that on TIO you can place header and footer code in the text boxes above and below the actual code box. This makes the code look cleaner, as well as allow you to take advantage the PPCG answer formatter (esc-s-g). Try it online!
$endgroup$
– Jo King
Feb 15 at 0:46

$begingroup$
@JoKing much appreciated, I had never used it before!
$endgroup$
– Kai
Feb 15 at 0:50

add a comment |

Japt, 14 bytes

òV

íUs w)c vUl

Try it online!

answered Feb 13 at 15:42

Luis felipe De jesus Munoz

4,79721465

$begingroup$
13 bytes, using the same approach.
$endgroup$
– Shaggy
Feb 13 at 23:01

add a comment |

Stax, 7 bytes

É╓ÅìΔà▲

Run and debug it

answered Feb 13 at 16:57

recursive

5,3091322

add a comment |

cQuents, 19 bytes

#|B-A+1&A+k-1,B-k+1

Try it online!

Note that it does not work on TIO right now because TIO's interpreter is not up to date.

Explanation

#|B-A+1&A+k-1,B-k+1

                      A is the first input, B is the second input

#|B-A+1               n = B - A + 1

       &              Print the first n terms of the sequence

                      k starts at 1 and increments whenever we return to the first term

        A+k-1,         Terms alternate between A + k - 1 and

              B-k+1     B - k + 1

                       increment k

answered Feb 13 at 18:21

Stephen

7,34823295

add a comment |

Haskell, 39 bytes

f(a:b)=a:f(reverse b)

f x=x

a#b=f[a..b]

Try it online!

answered Feb 13 at 19:09

nimi

31.8k32285

add a comment |

C# (Visual C# Interactive Compiler), 46 bytes

a=>b=>{for(;a<=b;Write(a+(b>a++?b--+"":"")));}

Saved 4 bytes thanks to dana!

Try it online!

C# (Visual C# Interactive Compiler), 65 bytes

void z(int a,int b){if(a<=b){Write(a+(b>a?b+"":""));z(a+1,b-1);}}

Try it online!

edited Feb 14 at 16:25

answered Feb 13 at 16:50

Embodiment of Ignorance

1,100119

$begingroup$
I didn't notice the extra empty string there, thanks!
$endgroup$
– Embodiment of Ignorance
Feb 14 at 16:23

add a comment |

APL (dzaima/APL), 21 bytes

⌈⊢+.5×-+∘(⌽×¯1*)∘⍳1+-

Try it online!

answered Feb 13 at 16:03

dzaima

15.1k21856

add a comment |

Japt, 7 bytes

Takes input as an array.

rõ

ÊÆÔv

Try it or run all test cases

         :Implicit input of array U=[low,high]

r        :Reduce by

 õ       :  Inclusive, reversed range (giving the range [high,low])

n       :Reassign to U

Ê        :Length

 Æ       :Map the range [0,Ê)

  Ô      :  Reverse U

   v     :  Remove the first element

edited Feb 13 at 17:31

answered Feb 13 at 17:20

Shaggy

20.1k21667

add a comment |

R, 51 bytes

function(x,y,z=x:y)matrix(c(z,rev(z)),2,,T)[seq(z)]

Try it online!

Explanation:
For a sequence x:y of length N, create a two-by-N matrix consisting of the sequence x:y in the top row and y:x in the bottom row matrix(c(z,rev(z)),2,,T). If we select the first N elements of the matrix [seq(z)], they will be chosen by column, giving the required output.

Outgolfed by digEmAll

edited Feb 13 at 18:04

answered Feb 13 at 17:59

user2390246

1,264111

1

$begingroup$
I just posted a very similar approach 30 seconds before you :D
$endgroup$
– digEmAll
Feb 13 at 18:01

$begingroup$
@digEmAll Yes, but yours is a lot better!
$endgroup$
– user2390246
Feb 13 at 18:03

add a comment |

MATL, 8 bytes

&:t"1&)P

Try it online!

Explanation

&:      % Take two inputs (implicit). Two-input range

t       % Duplicate

"       % For each

  1&)   %   Push first element, then an array with the rest

  P     %   Reverse array

        % End (implicit). Display (implicit)

edited Feb 13 at 18:11

answered Feb 13 at 17:57

Luis Mendo

74.5k888291

add a comment |

JavaScript, 40 bytes

l=>g=h=>h>l?[l++,h--,...g(h)]:h<l?:[l]

Try It Online!

edited Feb 13 at 20:13

answered Feb 13 at 18:13

Shaggy

20.1k21667

add a comment |

Forth (gforth), 52 bytes

: f 2dup - 1+ 0 do dup . i 2 mod 2* 1- - swap loop ;

Try it online!

Explanation

Loop from 0 to (End - Start). Place End and Start on top of the stack.

Each Iteration:

Output the current number

Add (or subtract) 1 from the current number

Switch the current number with the other number

Code Explanation

: f            start new word definition

  2dup -       get the size of the range (total number of integers)

  1+ 0         add 1 to the size because forth loops are [Inclusive, Exclusive) 

  do           start counted loop from 0 to size+1

    dup .      output the current top of the stack

    i 2 mod    get the index of the loop modulus 2

    2* 1-      convert from 0,1 to -1,1

    -          subtract result from top of stack (adds 1 to lower bound and subtracts 1 from upper)

    swap       swap the top two stack numbers 

  loop         end the counted loop

;              end the word definition

edited Feb 13 at 20:23

answered Feb 13 at 18:02

reffu

61126

add a comment |

Julia 0.7, 29 bytes

f(a,b)=[a:b b:-1:a]'[1:1+b-a]

Try it online!

answered Feb 13 at 20:30

Kirill L.

4,4151523

add a comment |

Haskell, 30 bytes

l%h=l:take(h-l)(h:(l+1)%(h-1))

Usage: 3%7 gives `[3,7,4,6,5]

For the inputs l, h the function calls recursively with the inputs l+1, h-1, and adds l,h to the beggining.
Instead of any halting condition, the code uses take(h-l) to shorten the sequence to the right length (which would otherwise be an infinite sequence of increasing and decreasing numbers).

answered Feb 14 at 13:48

proud haskeller

5,71611536

add a comment |

JVM bytecode (OpenJDK asmtools JASM), 449 bytes

enum b{const #1=Method java/io/PrintStream.print:(I)V;static Method a:(II)V stack 2 locals 4{getstatic java/lang/System.out:"Ljava/io/PrintStream;";astore 3;ldc 0;istore 2;l:iload 2;ldc 1;if_icmpeq t;aload 3;iload 0;invokevirtual #1;iinc 0,1;iinc 2,1;goto c;t:aload 3;iload 1;invokevirtual #1;iinc 1,-1;iinc 2,-1;c:aload 3;ldc 32;i2c;invokevirtual java/io/PrintStream.print:(C)V;iload 0;iload 1;if_icmpne l;aload 3;iload 0;invokevirtual #1;return;}}

Ungolfed (and slightly cleaner)

 enum b {    

    public static Method "a":(II)V stack 5 locals 4 {

        getstatic "java/lang/System"."out":"Ljava/io/PrintStream;";

        astore 3;

        ldc 0;

        istore 2;

    loop:

        iload 2;

        ldc 1;

        if_icmpeq true;

    false:

        aload 3;

        iload 0;

        invokevirtual "java/io/PrintStream"."print":"(I)V";

        iinc 0,1;

        iinc 2,1;

        goto cond;

    true:

        aload 3;

        iload 1;

        invokevirtual "java/io/PrintStream"."print":"(I)V";

        iinc 1,-1;

        iinc 2,-1;

        goto cond;

    cond:

        iload 0;

        iload 1;

        if_icmpne loop;

        aload 3;

        iload 0;

        invokevirtual "java/io/PrintStream"."print":"(I)V";

        return;

    }

}

Standalone function, needs to be called from Java as b.a(num1,num2).

Explanation

This code uses the method parameters as variables, as well as a boolean in local #3 deciding which number to output. Each loop iteration either the left or right is output, and that number is incremented for the left or decremented for the right. Loop continues until both numbers are equal, then that number is output.

...I have a distinct feeling I'm massively outgunned on the byte count

answered Feb 14 at 22:30

Archduke Liamus

1412

add a comment |

Ink, 46 bytes

=h(I,A)

{I<=A:{I} {I<A:{A} ->h(I+1,A-1)}}->->

(I don't think there's an online interpreter for Ink, sorry)

Defines a stitch called h, which takes two arguments I and A, which are the bounds of the range.

Outputs by printing values, separated by spaces, to stdout.

Explanation

=h(min, max) // Define the stitch.

{min <= max:{min}/* print min unless it's greater than max */{min < max: {max} /*Also print max if it's greater than min*/->h(min+1, max-1)/*Then divert, with the arguments changed*/}}

->-> // If we didn't divert earlier, divert to wherever the stitch was called from

answered Feb 13 at 19:01

Sara J

613

add a comment |

Perl 5 `-ln`, 37 bytes

@.=$_..<>;say shift@.,$/,pop@.while@.

Try it online!

answered Feb 13 at 20:02

Xcali

5,335520

$begingroup$
33bytes
$endgroup$
– Nahuel Fouilleul
Feb 14 at 14:47

add a comment |

Java (JDK), 52 bytes

(l,h,o)->{for(int i=0;l<=h;i^=1)o.add(i<1?l++:h--);}

Try it online!

edited Feb 13 at 20:44

answered Feb 13 at 17:25

Olivier Grégoire

9,10511943

add a comment |

Clean, 48 bytes

import StdEnv

$a b|a<>b=[a: $b(a+sign(b-a))]=[a]

Try it online!

answered Feb 14 at 1:29

Οurous

7,12111035

add a comment |

Ruby, 37 36 33 bytes

f=->a,b{a>b ?:[a,b]|f[a+1,b-1]}

Try it online!

Recursive version with 3 bytes saved by G B.

Ruby, 38 bytes

->a,b{d=*c=a..b;c.map{d.reverse!.pop}}

Try it online!

Non-recursive version.

edited Feb 14 at 15:11

answered Feb 14 at 10:56

Kirill L.

4,4151523

add a comment |

Cubix, 16 bytes

;w(.II>sO-?@;)^/

Try it here

Cubified

    ; w

    ( .

I I > s O - ? @

; ) ^ / . . . .

    . .

    . .

Explanation

Basically, this moves the two bounds closer together one step at a time until they meet. Each time through the loop, we swap the bounds, Output, take the difference, and increment with ) or decrement with ( based on the sign.

answered Feb 14 at 16:41

Mnemonic

4,7351730

add a comment |

Pyth, 10 8 bytes

{.iF_B}F

Try it here

Explanation

{.iF_B}F

      }FQ  Generate the range between the (implicit) inputs.

 .iF_B     Interleave it with its reverse.

{          Deduplicate.

edited Feb 14 at 17:44

answered Feb 13 at 22:08

Mnemonic

4,7351730

add a comment |

JavaScript, 35 bytes

f=(a,b,s=1)=>a-b?a+[,f(b,a+s,-s)]:a

Try it online!

Thanks to Arnauld, 1 byte saved.

edited Feb 15 at 1:49

answered Feb 14 at 3:24

tsh

9,06511650

add a comment |

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50 Answers
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50 Answers
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votes

R, 38 37 36 bytes

function(a,b)rbind(a:b,b:a)[a:b-a+1]

Try it online!

-1 byte thanks to @user2390246

-1 byte thanks to @Kirill L.

Exploiting the fact that R stores matrices column-wise

edited Feb 13 at 20:57

answered Feb 13 at 17:59

digEmAll

3,369415

$begingroup$
Using rbind is much better than my approach, but you can save 1 byte by using [seq(a:b)] instead of unique.
$endgroup$
– user2390246
Feb 13 at 18:03

$begingroup$
You're right, I missed the comment where has been specified that a < b (never equal), so we can use seq(a:b)
$endgroup$
– digEmAll
Feb 13 at 18:06

$begingroup$
@digEmAll My solution was essentially a literal interpretation of the puzzle, I never would have even thought of doing something such as this. Impressive, have an upvote!
$endgroup$
– Sumner18
Feb 13 at 18:07

1

$begingroup$
-1 more
$endgroup$
– Kirill L.
Feb 13 at 20:32

add a comment |

R, 38 37 36 bytes

function(a,b)rbind(a:b,b:a)[a:b-a+1]

Try it online!

-1 byte thanks to @user2390246

-1 byte thanks to @Kirill L.

Exploiting the fact that R stores matrices column-wise

edited Feb 13 at 20:57

answered Feb 13 at 17:59

digEmAll

3,369415

$begingroup$
Using rbind is much better than my approach, but you can save 1 byte by using [seq(a:b)] instead of unique.
$endgroup$
– user2390246
Feb 13 at 18:03

$begingroup$
You're right, I missed the comment where has been specified that a < b (never equal), so we can use seq(a:b)
$endgroup$
– digEmAll
Feb 13 at 18:06

$begingroup$
@digEmAll My solution was essentially a literal interpretation of the puzzle, I never would have even thought of doing something such as this. Impressive, have an upvote!
$endgroup$
– Sumner18
Feb 13 at 18:07

1

$begingroup$
-1 more
$endgroup$
– Kirill L.
Feb 13 at 20:32

add a comment |

R, 38 37 36 bytes

function(a,b)rbind(a:b,b:a)[a:b-a+1]

Try it online!

-1 byte thanks to @user2390246

-1 byte thanks to @Kirill L.

Exploiting the fact that R stores matrices column-wise

edited Feb 13 at 20:57

answered Feb 13 at 17:59

digEmAll

3,369415

R, 38 37 36 bytes

function(a,b)rbind(a:b,b:a)[a:b-a+1]

Try it online!

-1 byte thanks to @user2390246

-1 byte thanks to @Kirill L.

Exploiting the fact that R stores matrices column-wise

edited Feb 13 at 20:57

answered Feb 13 at 17:59

digEmAll

3,369415

edited Feb 13 at 20:57

answered Feb 13 at 17:59

digEmAll

3,369415

answered Feb 13 at 17:59

digEmAll

3,369415

answered Feb 13 at 17:59

digEmAll

3,369415

$begingroup$
Using rbind is much better than my approach, but you can save 1 byte by using [seq(a:b)] instead of unique.
$endgroup$
– user2390246
Feb 13 at 18:03

$begingroup$
You're right, I missed the comment where has been specified that a < b (never equal), so we can use seq(a:b)
$endgroup$
– digEmAll
Feb 13 at 18:06

$begingroup$
@digEmAll My solution was essentially a literal interpretation of the puzzle, I never would have even thought of doing something such as this. Impressive, have an upvote!
$endgroup$
– Sumner18
Feb 13 at 18:07

1

$begingroup$
-1 more
$endgroup$
– Kirill L.
Feb 13 at 20:32

add a comment |

$begingroup$
Using rbind is much better than my approach, but you can save 1 byte by using [seq(a:b)] instead of unique.
$endgroup$
– user2390246
Feb 13 at 18:03

$begingroup$
You're right, I missed the comment where has been specified that a < b (never equal), so we can use seq(a:b)
$endgroup$
– digEmAll
Feb 13 at 18:06

$begingroup$
@digEmAll My solution was essentially a literal interpretation of the puzzle, I never would have even thought of doing something such as this. Impressive, have an upvote!
$endgroup$
– Sumner18
Feb 13 at 18:07

1

$begingroup$
-1 more
$endgroup$
– Kirill L.
Feb 13 at 20:32

Using rbind is much better than my approach, but you can save 1 byte by using [seq(a:b)] instead of unique.

– user2390246
Feb 13 at 18:03

You're right, I missed the comment where has been specified that a < b (never equal), so we can use seq(a:b)

– digEmAll
Feb 13 at 18:06

@digEmAll My solution was essentially a literal interpretation of the puzzle, I never would have even thought of doing something such as this. Impressive, have an upvote!

– Sumner18
Feb 13 at 18:07

-1 more

– Kirill L.
Feb 13 at 20:32

add a comment |

Haskell, 30 bytes

a%b=a:take(b-a)(b:(a+1)%(b-1))

Try it online!

answered Feb 14 at 0:56

xnor

90.9k18186441

$begingroup$
Dammit! I have just found the exact same solution. Oh well
$endgroup$
– proud haskeller
Feb 14 at 13:49

add a comment |

Haskell, 30 bytes

a%b=a:take(b-a)(b:(a+1)%(b-1))

Try it online!

answered Feb 14 at 0:56

xnor

90.9k18186441

$begingroup$
Dammit! I have just found the exact same solution. Oh well
$endgroup$
– proud haskeller
Feb 14 at 13:49

add a comment |

Haskell, 30 bytes

a%b=a:take(b-a)(b:(a+1)%(b-1))

Try it online!

answered Feb 14 at 0:56

xnor

90.9k18186441

Haskell, 30 bytes

a%b=a:take(b-a)(b:(a+1)%(b-1))

Try it online!

answered Feb 14 at 0:56

xnor

90.9k18186441

answered Feb 14 at 0:56

xnor

90.9k18186441

answered Feb 14 at 0:56

xnor

90.9k18186441

answered Feb 14 at 0:56

xnor

90.9k18186441

$begingroup$
Dammit! I have just found the exact same solution. Oh well
$endgroup$
– proud haskeller
Feb 14 at 13:49

add a comment |

$begingroup$
Dammit! I have just found the exact same solution. Oh well
$endgroup$
– proud haskeller
Feb 14 at 13:49

Dammit! I have just found the exact same solution. Oh well

– proud haskeller
Feb 14 at 13:49

add a comment |

Haskell, 39 bytes

a#b|a>b=a:b#(a-1)|a<b=a:b#(a+1)|1>0=[a]

Try it online!

answered Feb 13 at 17:15

ovs

19k21159

add a comment |

Haskell, 39 bytes

a#b|a>b=a:b#(a-1)|a<b=a:b#(a+1)|1>0=[a]

Try it online!

answered Feb 13 at 17:15

ovs

19k21159

add a comment |

Haskell, 39 bytes

a#b|a>b=a:b#(a-1)|a<b=a:b#(a+1)|1>0=[a]

Try it online!

answered Feb 13 at 17:15

ovs

19k21159

Haskell, 39 bytes

a#b|a>b=a:b#(a-1)|a<b=a:b#(a+1)|1>0=[a]

Try it online!

answered Feb 13 at 17:15

ovs

19k21159

answered Feb 13 at 17:15

ovs

19k21159

answered Feb 13 at 17:15

ovs

19k21159

answered Feb 13 at 17:15

ovs

19k21159

add a comment |

R, 65 64 61 60 bytes

-1 byte thanks to Robert S.

-4 more thanks to digEmAll

x=scan();z=x:x[2];while(sum(z|1)){cat(z[1],"");z=rev(z[-1])}

Try it online!

edited Feb 13 at 18:15

answered Feb 13 at 16:58

Sumner18

5507

$begingroup$
You can replace length(z) with sum(z|1) to save 1 byte :)
$endgroup$
– Robert S.
Feb 13 at 17:09

$begingroup$
I don't understand how that works but I guess it does. sum(z|1) seems like it would always evaluate to at least 1, which would cause the while loop to loop endlessly. but apparently not
$endgroup$
– Sumner18
Feb 13 at 17:27

3

$begingroup$
z is a vector. each element of that vector is |ed with 1. Which is always equal to 1. When you take the sum, you have a vector filled with TRUEs so the result is equal to the length of the vector. If the vector is empty, you the is nothing to | with so the output vector is logical(0). When you take that sum, it's 0
$endgroup$
– OganM
Feb 13 at 23:59

add a comment |

R, 65 64 61 60 bytes

-1 byte thanks to Robert S.

-4 more thanks to digEmAll

x=scan();z=x:x[2];while(sum(z|1)){cat(z[1],"");z=rev(z[-1])}

Try it online!

edited Feb 13 at 18:15

answered Feb 13 at 16:58

Sumner18

5507

$begingroup$
You can replace length(z) with sum(z|1) to save 1 byte :)
$endgroup$
– Robert S.
Feb 13 at 17:09

$begingroup$
I don't understand how that works but I guess it does. sum(z|1) seems like it would always evaluate to at least 1, which would cause the while loop to loop endlessly. but apparently not
$endgroup$
– Sumner18
Feb 13 at 17:27

3

$begingroup$
z is a vector. each element of that vector is |ed with 1. Which is always equal to 1. When you take the sum, you have a vector filled with TRUEs so the result is equal to the length of the vector. If the vector is empty, you the is nothing to | with so the output vector is logical(0). When you take that sum, it's 0
$endgroup$
– OganM
Feb 13 at 23:59

add a comment |

R, 65 64 61 60 bytes

-1 byte thanks to Robert S.

-4 more thanks to digEmAll

x=scan();z=x:x[2];while(sum(z|1)){cat(z[1],"");z=rev(z[-1])}

Try it online!

edited Feb 13 at 18:15

answered Feb 13 at 16:58

Sumner18

5507

R, 65 64 61 60 bytes

-1 byte thanks to Robert S.

-4 more thanks to digEmAll

x=scan();z=x:x[2];while(sum(z|1)){cat(z[1],"");z=rev(z[-1])}

Try it online!

edited Feb 13 at 18:15

answered Feb 13 at 16:58

Sumner18

5507

edited Feb 13 at 18:15

answered Feb 13 at 16:58

Sumner18

5507

answered Feb 13 at 16:58

Sumner18

5507

answered Feb 13 at 16:58

Sumner18

5507

$begingroup$
You can replace length(z) with sum(z|1) to save 1 byte :)
$endgroup$
– Robert S.
Feb 13 at 17:09

$begingroup$
I don't understand how that works but I guess it does. sum(z|1) seems like it would always evaluate to at least 1, which would cause the while loop to loop endlessly. but apparently not
$endgroup$
– Sumner18
Feb 13 at 17:27

3

$begingroup$
z is a vector. each element of that vector is |ed with 1. Which is always equal to 1. When you take the sum, you have a vector filled with TRUEs so the result is equal to the length of the vector. If the vector is empty, you the is nothing to | with so the output vector is logical(0). When you take that sum, it's 0
$endgroup$
– OganM
Feb 13 at 23:59

add a comment |

$begingroup$
You can replace length(z) with sum(z|1) to save 1 byte :)
$endgroup$
– Robert S.
Feb 13 at 17:09

$begingroup$
I don't understand how that works but I guess it does. sum(z|1) seems like it would always evaluate to at least 1, which would cause the while loop to loop endlessly. but apparently not
$endgroup$
– Sumner18
Feb 13 at 17:27

3

$begingroup$
z is a vector. each element of that vector is |ed with 1. Which is always equal to 1. When you take the sum, you have a vector filled with TRUEs so the result is equal to the length of the vector. If the vector is empty, you the is nothing to | with so the output vector is logical(0). When you take that sum, it's 0
$endgroup$
– OganM
Feb 13 at 23:59

You can replace length(z) with sum(z|1) to save 1 byte :)

– Robert S.
Feb 13 at 17:09

I don't understand how that works but I guess it does. sum(z|1) seems like it would always evaluate to at least 1, which would cause the while loop to loop endlessly. but apparently not

– Sumner18
Feb 13 at 17:27

z is a vector. each element of that vector is |ed with 1. Which is always equal to 1. When you take the sum, you have a vector filled with TRUEs so the result is equal to the length of the vector. If the vector is empty, you the is nothing to | with so the output vector is logical(0). When you take that sum, it's 0

– OganM
Feb 13 at 23:59

add a comment |

Python 2, 44 bytes

f=lambda a,b:[a]*(a==b)or[a]+f(b,a-cmp(a,b))

Try it online!

answered Feb 13 at 16:58

ovs

19k21159

add a comment |

Python 2, 44 bytes

f=lambda a,b:[a]*(a==b)or[a]+f(b,a-cmp(a,b))

Try it online!

answered Feb 13 at 16:58

ovs

19k21159

add a comment |

Python 2, 44 bytes

f=lambda a,b:[a]*(a==b)or[a]+f(b,a-cmp(a,b))

Try it online!

answered Feb 13 at 16:58

ovs

19k21159

Python 2, 44 bytes

f=lambda a,b:[a]*(a==b)or[a]+f(b,a-cmp(a,b))

Try it online!

answered Feb 13 at 16:58

ovs

19k21159

answered Feb 13 at 16:58

ovs

19k21159

answered Feb 13 at 16:58

ovs

19k21159

answered Feb 13 at 16:58

ovs

19k21159

add a comment |

PowerShell, 59 48 bytes

param($a,$b)(($z=0..($b-$a))|%{$a+$_;$b-$_})[$z]

Try it online!

~~(Seems long...)~~

-11 bytes thanks to mazzy.

edited Feb 13 at 17:40

answered Feb 13 at 15:51

AdmBorkBork

27.2k466234

$begingroup$
48 bytes
$endgroup$
– mazzy
Feb 13 at 17:29

$begingroup$
@mazzy Ah, I like that $b-$a trick -- that's clever!
$endgroup$
– AdmBorkBork
Feb 13 at 17:40

add a comment |

PowerShell, 59 48 bytes

param($a,$b)(($z=0..($b-$a))|%{$a+$_;$b-$_})[$z]

Try it online!

~~(Seems long...)~~

-11 bytes thanks to mazzy.

edited Feb 13 at 17:40

answered Feb 13 at 15:51

AdmBorkBork

27.2k466234

$begingroup$
48 bytes
$endgroup$
– mazzy
Feb 13 at 17:29

$begingroup$
@mazzy Ah, I like that $b-$a trick -- that's clever!
$endgroup$
– AdmBorkBork
Feb 13 at 17:40

add a comment |

PowerShell, 59 48 bytes

param($a,$b)(($z=0..($b-$a))|%{$a+$_;$b-$_})[$z]

Try it online!

~~(Seems long...)~~

-11 bytes thanks to mazzy.

edited Feb 13 at 17:40

answered Feb 13 at 15:51

AdmBorkBork

27.2k466234

PowerShell, 59 48 bytes

param($a,$b)(($z=0..($b-$a))|%{$a+$_;$b-$_})[$z]

Try it online!

~~(Seems long...)~~

-11 bytes thanks to mazzy.

edited Feb 13 at 17:40

answered Feb 13 at 15:51

AdmBorkBork

27.2k466234

edited Feb 13 at 17:40

answered Feb 13 at 15:51

AdmBorkBork

27.2k466234

answered Feb 13 at 15:51

AdmBorkBork

27.2k466234

answered Feb 13 at 15:51

AdmBorkBork

27.2k466234

$begingroup$
48 bytes
$endgroup$
– mazzy
Feb 13 at 17:29

$begingroup$
@mazzy Ah, I like that $b-$a trick -- that's clever!
$endgroup$
– AdmBorkBork
Feb 13 at 17:40

add a comment |

$begingroup$
48 bytes
$endgroup$
– mazzy
Feb 13 at 17:29

$begingroup$
@mazzy Ah, I like that $b-$a trick -- that's clever!
$endgroup$
– AdmBorkBork
Feb 13 at 17:40

48 bytes

– mazzy
Feb 13 at 17:29

@mazzy Ah, I like that $b-$a trick -- that's clever!

– AdmBorkBork
Feb 13 at 17:40

add a comment |

05AB1E, 6 bytes

ŸDvć,R

Try it online!

Explanation

Ÿ        # push range [min ... max]

 D       # duplicate

  v      # for each element in the copy

   ć,    # extract and print the head of the original list

     R   # and then reverse it

edited Feb 14 at 10:36

answered Feb 13 at 17:02

Emigna

46.3k432141

$begingroup$
Ÿ2ä`R.ι non iterative using interleave, but this is still much better.
$endgroup$
– Magic Octopus Urn
Feb 14 at 2:36

1

$begingroup$
@MagicOctopusUrn: I tried a non-iterative solution first, but it was even worse since I didn't know about .ι ;)
$endgroup$
– Emigna
Feb 14 at 7:36

$begingroup$
Similar as what I had in mind, so obvious +1 from me. I do like your alternative 7-byter as well through, @MagicOctopusUrn. :)
$endgroup$
– Kevin Cruijssen
Feb 14 at 8:28

1

$begingroup$
@KristianWilliams: Seems to be working for me.
$endgroup$
– Emigna
Feb 14 at 9:08

1

$begingroup$
@KevinCruijssen: I switched to a pair instead as that felt more intuitive anyways :)
$endgroup$
– Emigna
Feb 14 at 10:50

|
show 3 more comments

05AB1E, 6 bytes

ŸDvć,R

Try it online!

Explanation

Ÿ        # push range [min ... max]

 D       # duplicate

  v      # for each element in the copy

   ć,    # extract and print the head of the original list

     R   # and then reverse it

edited Feb 14 at 10:36

answered Feb 13 at 17:02

Emigna

46.3k432141

$begingroup$
Ÿ2ä`R.ι non iterative using interleave, but this is still much better.
$endgroup$
– Magic Octopus Urn
Feb 14 at 2:36

1

$begingroup$
@MagicOctopusUrn: I tried a non-iterative solution first, but it was even worse since I didn't know about .ι ;)
$endgroup$
– Emigna
Feb 14 at 7:36

$begingroup$
Similar as what I had in mind, so obvious +1 from me. I do like your alternative 7-byter as well through, @MagicOctopusUrn. :)
$endgroup$
– Kevin Cruijssen
Feb 14 at 8:28

1

$begingroup$
@KristianWilliams: Seems to be working for me.
$endgroup$
– Emigna
Feb 14 at 9:08

1

$begingroup$
@KevinCruijssen: I switched to a pair instead as that felt more intuitive anyways :)
$endgroup$
– Emigna
Feb 14 at 10:50

|
show 3 more comments

05AB1E, 6 bytes

ŸDvć,R

Try it online!

Explanation

Ÿ        # push range [min ... max]

 D       # duplicate

  v      # for each element in the copy

   ć,    # extract and print the head of the original list

     R   # and then reverse it

edited Feb 14 at 10:36

answered Feb 13 at 17:02

Emigna

46.3k432141

05AB1E, 6 bytes

ŸDvć,R

Try it online!

Explanation

Ÿ        # push range [min ... max]

 D       # duplicate

  v      # for each element in the copy

   ć,    # extract and print the head of the original list

     R   # and then reverse it

edited Feb 14 at 10:36

answered Feb 13 at 17:02

Emigna

46.3k432141

edited Feb 14 at 10:36

answered Feb 13 at 17:02

Emigna

46.3k432141

answered Feb 13 at 17:02

Emigna

46.3k432141

answered Feb 13 at 17:02

Emigna

46.3k432141

$begingroup$
Ÿ2ä`R.ι non iterative using interleave, but this is still much better.
$endgroup$
– Magic Octopus Urn
Feb 14 at 2:36

1

$begingroup$
@MagicOctopusUrn: I tried a non-iterative solution first, but it was even worse since I didn't know about .ι ;)
$endgroup$
– Emigna
Feb 14 at 7:36

$begingroup$
Similar as what I had in mind, so obvious +1 from me. I do like your alternative 7-byter as well through, @MagicOctopusUrn. :)
$endgroup$
– Kevin Cruijssen
Feb 14 at 8:28

1

$begingroup$
@KristianWilliams: Seems to be working for me.
$endgroup$
– Emigna
Feb 14 at 9:08

1

$begingroup$
@KevinCruijssen: I switched to a pair instead as that felt more intuitive anyways :)
$endgroup$
– Emigna
Feb 14 at 10:50

|
show 3 more comments

$begingroup$
Ÿ2ä`R.ι non iterative using interleave, but this is still much better.
$endgroup$
– Magic Octopus Urn
Feb 14 at 2:36

1

$begingroup$
@MagicOctopusUrn: I tried a non-iterative solution first, but it was even worse since I didn't know about .ι ;)
$endgroup$
– Emigna
Feb 14 at 7:36

$begingroup$
Similar as what I had in mind, so obvious +1 from me. I do like your alternative 7-byter as well through, @MagicOctopusUrn. :)
$endgroup$
– Kevin Cruijssen
Feb 14 at 8:28

1

$begingroup$
@KristianWilliams: Seems to be working for me.
$endgroup$
– Emigna
Feb 14 at 9:08

1

$begingroup$
@KevinCruijssen: I switched to a pair instead as that felt more intuitive anyways :)
$endgroup$
– Emigna
Feb 14 at 10:50

Ÿ2ä`R.ι non iterative using interleave, but this is still much better.

– Magic Octopus Urn
Feb 14 at 2:36

@MagicOctopusUrn: I tried a non-iterative solution first, but it was even worse since I didn't know about .ι ;)

– Emigna
Feb 14 at 7:36

Similar as what I had in mind, so obvious +1 from me. I do like your alternative 7-byter as well through, @MagicOctopusUrn. :)

– Kevin Cruijssen
Feb 14 at 8:28

@KristianWilliams: Seems to be working for me.

– Emigna
Feb 14 at 9:08

@KevinCruijssen: I switched to a pair instead as that felt more intuitive anyways :)

– Emigna
Feb 14 at 10:50

|
show 3 more comments

Wolfram Language (Mathematica), 56 54 bytes

This is my first time golfing!

f[a_,b_]:=(c=a~Range~b;Drop[c~Riffle~Reverse@c,a-b-1])

Try it online!

Saved 2 bytes using infix notation.

Explanation:

f[a_,b_]:=                                   function of two variables

c=a~Range~b;                                 list of integers from a to b 

                           Reverse@c         same list in reverse

                  c~Riffle~Reverse@c         interleave the two lists

             Drop[c~Riffle~Reverse@c,a-b-1]  drop last |a-b-1| elements (note a-b-1 < 0)

Alternatively, we could use Take[...,b-a+1] for the same result.

Tests:

f[4, 6]

f[1, 5]

f[-1, 1]

f[-1, 2]

Ouput:

{4, 6, 5}

{1, 5, 2, 4, 3}

{-1, 1, 0}

{-1, 2, 0, 1}

edited Feb 15 at 1:08

answered Feb 14 at 21:50

Kai

1713

New contributor

$begingroup$
The "Try it online" link returns a 403. "Sorry, you do not have permission to access this item."
$endgroup$
– Rohit Namjoshi
Feb 15 at 0:02

$begingroup$
@RohitNamjoshi I updated the link
$endgroup$
– Kai
Feb 15 at 0:24

$begingroup$
note that on TIO you can place header and footer code in the text boxes above and below the actual code box. This makes the code look cleaner, as well as allow you to take advantage the PPCG answer formatter (esc-s-g). Try it online!
$endgroup$
– Jo King
Feb 15 at 0:46

$begingroup$
@JoKing much appreciated, I had never used it before!
$endgroup$
– Kai
Feb 15 at 0:50

add a comment |

Wolfram Language (Mathematica), 56 54 bytes

This is my first time golfing!

f[a_,b_]:=(c=a~Range~b;Drop[c~Riffle~Reverse@c,a-b-1])

Try it online!

Saved 2 bytes using infix notation.

Explanation:

f[a_,b_]:=                                   function of two variables

c=a~Range~b;                                 list of integers from a to b 

                           Reverse@c         same list in reverse

                  c~Riffle~Reverse@c         interleave the two lists

             Drop[c~Riffle~Reverse@c,a-b-1]  drop last |a-b-1| elements (note a-b-1 < 0)

Alternatively, we could use Take[...,b-a+1] for the same result.

Tests:

f[4, 6]

f[1, 5]

f[-1, 1]

f[-1, 2]

Ouput:

{4, 6, 5}

{1, 5, 2, 4, 3}

{-1, 1, 0}

{-1, 2, 0, 1}

edited Feb 15 at 1:08

answered Feb 14 at 21:50

Kai

1713

New contributor

$begingroup$
The "Try it online" link returns a 403. "Sorry, you do not have permission to access this item."
$endgroup$
– Rohit Namjoshi
Feb 15 at 0:02

$begingroup$
@RohitNamjoshi I updated the link
$endgroup$
– Kai
Feb 15 at 0:24

$begingroup$
note that on TIO you can place header and footer code in the text boxes above and below the actual code box. This makes the code look cleaner, as well as allow you to take advantage the PPCG answer formatter (esc-s-g). Try it online!
$endgroup$
– Jo King
Feb 15 at 0:46

$begingroup$
@JoKing much appreciated, I had never used it before!
$endgroup$
– Kai
Feb 15 at 0:50

add a comment |

Wolfram Language (Mathematica), 56 54 bytes

This is my first time golfing!

f[a_,b_]:=(c=a~Range~b;Drop[c~Riffle~Reverse@c,a-b-1])

Try it online!

Saved 2 bytes using infix notation.

Explanation:

f[a_,b_]:=                                   function of two variables

c=a~Range~b;                                 list of integers from a to b 

                           Reverse@c         same list in reverse

                  c~Riffle~Reverse@c         interleave the two lists

             Drop[c~Riffle~Reverse@c,a-b-1]  drop last |a-b-1| elements (note a-b-1 < 0)

Alternatively, we could use Take[...,b-a+1] for the same result.

Tests:

f[4, 6]

f[1, 5]

f[-1, 1]

f[-1, 2]

Ouput:

{4, 6, 5}

{1, 5, 2, 4, 3}

{-1, 1, 0}

{-1, 2, 0, 1}

edited Feb 15 at 1:08

answered Feb 14 at 21:50

Kai

1713

New contributor

Wolfram Language (Mathematica), 56 54 bytes

This is my first time golfing!

f[a_,b_]:=(c=a~Range~b;Drop[c~Riffle~Reverse@c,a-b-1])

Try it online!

Saved 2 bytes using infix notation.

Explanation:

f[a_,b_]:=                                   function of two variables

c=a~Range~b;                                 list of integers from a to b 

                           Reverse@c         same list in reverse

                  c~Riffle~Reverse@c         interleave the two lists

             Drop[c~Riffle~Reverse@c,a-b-1]  drop last |a-b-1| elements (note a-b-1 < 0)

Alternatively, we could use Take[...,b-a+1] for the same result.

Tests:

f[4, 6]

f[1, 5]

f[-1, 1]

f[-1, 2]

Ouput:

{4, 6, 5}

{1, 5, 2, 4, 3}

{-1, 1, 0}

{-1, 2, 0, 1}

edited Feb 15 at 1:08

answered Feb 14 at 21:50

Kai

1713

New contributor

edited Feb 15 at 1:08

answered Feb 14 at 21:50

Kai

1713

New contributor

answered Feb 14 at 21:50

Kai

1713

answered Feb 14 at 21:50

Kai

1713

New contributor

Kai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$begingroup$
The "Try it online" link returns a 403. "Sorry, you do not have permission to access this item."
$endgroup$
– Rohit Namjoshi
Feb 15 at 0:02

$begingroup$
@RohitNamjoshi I updated the link
$endgroup$
– Kai
Feb 15 at 0:24

$begingroup$
note that on TIO you can place header and footer code in the text boxes above and below the actual code box. This makes the code look cleaner, as well as allow you to take advantage the PPCG answer formatter (esc-s-g). Try it online!
$endgroup$
– Jo King
Feb 15 at 0:46

$begingroup$
@JoKing much appreciated, I had never used it before!
$endgroup$
– Kai
Feb 15 at 0:50

add a comment |

$begingroup$
The "Try it online" link returns a 403. "Sorry, you do not have permission to access this item."
$endgroup$
– Rohit Namjoshi
Feb 15 at 0:02

$begingroup$
@RohitNamjoshi I updated the link
$endgroup$
– Kai
Feb 15 at 0:24

$begingroup$
note that on TIO you can place header and footer code in the text boxes above and below the actual code box. This makes the code look cleaner, as well as allow you to take advantage the PPCG answer formatter (esc-s-g). Try it online!
$endgroup$
– Jo King
Feb 15 at 0:46

$begingroup$
@JoKing much appreciated, I had never used it before!
$endgroup$
– Kai
Feb 15 at 0:50

The "Try it online" link returns a 403. "Sorry, you do not have permission to access this item."

– Rohit Namjoshi
Feb 15 at 0:02

@RohitNamjoshi I updated the link

– Kai
Feb 15 at 0:24

note that on TIO you can place header and footer code in the text boxes above and below the actual code box. This makes the code look cleaner, as well as allow you to take advantage the PPCG answer formatter (esc-s-g). Try it online!

– Jo King
Feb 15 at 0:46

@JoKing much appreciated, I had never used it before!

– Kai
Feb 15 at 0:50

add a comment |

Japt, 14 bytes

òV

íUs w)c vUl

Try it online!

answered Feb 13 at 15:42

Luis felipe De jesus Munoz

4,79721465

$begingroup$
13 bytes, using the same approach.
$endgroup$
– Shaggy
Feb 13 at 23:01

add a comment |

Japt, 14 bytes

òV

íUs w)c vUl

Try it online!

answered Feb 13 at 15:42

Luis felipe De jesus Munoz

4,79721465

$begingroup$
13 bytes, using the same approach.
$endgroup$
– Shaggy
Feb 13 at 23:01

add a comment |

Japt, 14 bytes

òV

íUs w)c vUl

Try it online!

answered Feb 13 at 15:42

Luis felipe De jesus Munoz

4,79721465

Japt, 14 bytes

òV

íUs w)c vUl

Try it online!

answered Feb 13 at 15:42

Luis felipe De jesus Munoz

4,79721465

answered Feb 13 at 15:42

Luis felipe De jesus Munoz

4,79721465

answered Feb 13 at 15:42

Luis felipe De jesus Munoz

4,79721465

answered Feb 13 at 15:42

Luis felipe De jesus Munoz

4,79721465

$begingroup$
13 bytes, using the same approach.
$endgroup$
– Shaggy
Feb 13 at 23:01

add a comment |

$begingroup$
13 bytes, using the same approach.
$endgroup$
– Shaggy
Feb 13 at 23:01

13 bytes, using the same approach.

– Shaggy
Feb 13 at 23:01

add a comment |

Stax, 7 bytes

É╓ÅìΔà▲

Run and debug it

answered Feb 13 at 16:57

recursive

5,3091322

add a comment |

Stax, 7 bytes

É╓ÅìΔà▲

Run and debug it

answered Feb 13 at 16:57

recursive

5,3091322

add a comment |

Stax, 7 bytes

É╓ÅìΔà▲

Run and debug it

answered Feb 13 at 16:57

recursive

5,3091322

Stax, 7 bytes

É╓ÅìΔà▲

Run and debug it

answered Feb 13 at 16:57

recursive

5,3091322

answered Feb 13 at 16:57

recursive

5,3091322

answered Feb 13 at 16:57

recursive

5,3091322

answered Feb 13 at 16:57

recursive

5,3091322

add a comment |

cQuents, 19 bytes

#|B-A+1&A+k-1,B-k+1

Try it online!

Note that it does not work on TIO right now because TIO's interpreter is not up to date.

Explanation

#|B-A+1&A+k-1,B-k+1

                      A is the first input, B is the second input

#|B-A+1               n = B - A + 1

       &              Print the first n terms of the sequence

                      k starts at 1 and increments whenever we return to the first term

        A+k-1,         Terms alternate between A + k - 1 and

              B-k+1     B - k + 1

                       increment k

answered Feb 13 at 18:21

Stephen

7,34823295

add a comment |

cQuents, 19 bytes

#|B-A+1&A+k-1,B-k+1

Try it online!

Note that it does not work on TIO right now because TIO's interpreter is not up to date.

Explanation

#|B-A+1&A+k-1,B-k+1

                      A is the first input, B is the second input

#|B-A+1               n = B - A + 1

       &              Print the first n terms of the sequence

                      k starts at 1 and increments whenever we return to the first term

        A+k-1,         Terms alternate between A + k - 1 and

              B-k+1     B - k + 1

                       increment k

answered Feb 13 at 18:21

Stephen

7,34823295

add a comment |

cQuents, 19 bytes

#|B-A+1&A+k-1,B-k+1

Try it online!

Note that it does not work on TIO right now because TIO's interpreter is not up to date.

Explanation

#|B-A+1&A+k-1,B-k+1

                      A is the first input, B is the second input

#|B-A+1               n = B - A + 1

       &              Print the first n terms of the sequence

                      k starts at 1 and increments whenever we return to the first term

        A+k-1,         Terms alternate between A + k - 1 and

              B-k+1     B - k + 1

                       increment k

answered Feb 13 at 18:21

Stephen

7,34823295

cQuents, 19 bytes

#|B-A+1&A+k-1,B-k+1

Try it online!

Note that it does not work on TIO right now because TIO's interpreter is not up to date.

Explanation

#|B-A+1&A+k-1,B-k+1

                      A is the first input, B is the second input

#|B-A+1               n = B - A + 1

       &              Print the first n terms of the sequence

                      k starts at 1 and increments whenever we return to the first term

        A+k-1,         Terms alternate between A + k - 1 and

              B-k+1     B - k + 1

                       increment k

answered Feb 13 at 18:21

Stephen

7,34823295

answered Feb 13 at 18:21

Stephen

7,34823295

answered Feb 13 at 18:21

Stephen

7,34823295

answered Feb 13 at 18:21

Stephen

7,34823295

add a comment |

Haskell, 39 bytes

f(a:b)=a:f(reverse b)

f x=x

a#b=f[a..b]

Try it online!

answered Feb 13 at 19:09

nimi

31.8k32285

add a comment |

Haskell, 39 bytes

f(a:b)=a:f(reverse b)

f x=x

a#b=f[a..b]

Try it online!

answered Feb 13 at 19:09

nimi

31.8k32285

add a comment |

Haskell, 39 bytes

f(a:b)=a:f(reverse b)

f x=x

a#b=f[a..b]

Try it online!

answered Feb 13 at 19:09

nimi

31.8k32285

Haskell, 39 bytes

f(a:b)=a:f(reverse b)

f x=x

a#b=f[a..b]

Try it online!

answered Feb 13 at 19:09

nimi

31.8k32285

answered Feb 13 at 19:09

nimi

31.8k32285

answered Feb 13 at 19:09

nimi

31.8k32285

answered Feb 13 at 19:09

nimi

31.8k32285

add a comment |

C# (Visual C# Interactive Compiler), 46 bytes

a=>b=>{for(;a<=b;Write(a+(b>a++?b--+"":"")));}

Saved 4 bytes thanks to dana!

Try it online!

C# (Visual C# Interactive Compiler), 65 bytes

void z(int a,int b){if(a<=b){Write(a+(b>a?b+"":""));z(a+1,b-1);}}

Try it online!

edited Feb 14 at 16:25

answered Feb 13 at 16:50

Embodiment of Ignorance

1,100119

$begingroup$
I didn't notice the extra empty string there, thanks!
$endgroup$
– Embodiment of Ignorance
Feb 14 at 16:23

add a comment |

C# (Visual C# Interactive Compiler), 46 bytes

a=>b=>{for(;a<=b;Write(a+(b>a++?b--+"":"")));}

Saved 4 bytes thanks to dana!

Try it online!

C# (Visual C# Interactive Compiler), 65 bytes

void z(int a,int b){if(a<=b){Write(a+(b>a?b+"":""));z(a+1,b-1);}}

Try it online!

edited Feb 14 at 16:25

answered Feb 13 at 16:50

Embodiment of Ignorance

1,100119

$begingroup$
I didn't notice the extra empty string there, thanks!
$endgroup$
– Embodiment of Ignorance
Feb 14 at 16:23

add a comment |

C# (Visual C# Interactive Compiler), 46 bytes

a=>b=>{for(;a<=b;Write(a+(b>a++?b--+"":"")));}

Saved 4 bytes thanks to dana!

Try it online!

C# (Visual C# Interactive Compiler), 65 bytes

void z(int a,int b){if(a<=b){Write(a+(b>a?b+"":""));z(a+1,b-1);}}

Try it online!

edited Feb 14 at 16:25

answered Feb 13 at 16:50

Embodiment of Ignorance

1,100119

C# (Visual C# Interactive Compiler), 46 bytes

a=>b=>{for(;a<=b;Write(a+(b>a++?b--+"":"")));}

Saved 4 bytes thanks to dana!

Try it online!

C# (Visual C# Interactive Compiler), 65 bytes

void z(int a,int b){if(a<=b){Write(a+(b>a?b+"":""));z(a+1,b-1);}}

Try it online!

edited Feb 14 at 16:25

answered Feb 13 at 16:50

Embodiment of Ignorance

1,100119

edited Feb 14 at 16:25

answered Feb 13 at 16:50

Embodiment of Ignorance

1,100119

answered Feb 13 at 16:50

Embodiment of Ignorance

1,100119

answered Feb 13 at 16:50

Embodiment of Ignorance

1,100119

$begingroup$
I didn't notice the extra empty string there, thanks!
$endgroup$
– Embodiment of Ignorance
Feb 14 at 16:23

add a comment |

$begingroup$
I didn't notice the extra empty string there, thanks!
$endgroup$
– Embodiment of Ignorance
Feb 14 at 16:23

I didn't notice the extra empty string there, thanks!

– Embodiment of Ignorance
Feb 14 at 16:23

add a comment |

APL (dzaima/APL), 21 bytes

⌈⊢+.5×-+∘(⌽×¯1*)∘⍳1+-

Try it online!

answered Feb 13 at 16:03

dzaima

15.1k21856

add a comment |

APL (dzaima/APL), 21 bytes

⌈⊢+.5×-+∘(⌽×¯1*)∘⍳1+-

Try it online!

answered Feb 13 at 16:03

dzaima

15.1k21856

add a comment |

APL (dzaima/APL), 21 bytes

⌈⊢+.5×-+∘(⌽×¯1*)∘⍳1+-

Try it online!

answered Feb 13 at 16:03

dzaima

15.1k21856

APL (dzaima/APL), 21 bytes

⌈⊢+.5×-+∘(⌽×¯1*)∘⍳1+-

Try it online!

answered Feb 13 at 16:03

dzaima

15.1k21856

answered Feb 13 at 16:03

dzaima

15.1k21856

answered Feb 13 at 16:03

dzaima

15.1k21856

answered Feb 13 at 16:03

dzaima

15.1k21856

add a comment |

Japt, 7 bytes

Takes input as an array.

rõ

ÊÆÔv

Try it or run all test cases

         :Implicit input of array U=[low,high]

r        :Reduce by

 õ       :  Inclusive, reversed range (giving the range [high,low])

n       :Reassign to U

Ê        :Length

 Æ       :Map the range [0,Ê)

  Ô      :  Reverse U

   v     :  Remove the first element

edited Feb 13 at 17:31

answered Feb 13 at 17:20

Shaggy

20.1k21667

add a comment |

Japt, 7 bytes

Takes input as an array.

rõ

ÊÆÔv

Try it or run all test cases

         :Implicit input of array U=[low,high]

r        :Reduce by

 õ       :  Inclusive, reversed range (giving the range [high,low])

n       :Reassign to U

Ê        :Length

 Æ       :Map the range [0,Ê)

  Ô      :  Reverse U

   v     :  Remove the first element

edited Feb 13 at 17:31

answered Feb 13 at 17:20

Shaggy

20.1k21667

add a comment |

Japt, 7 bytes

Takes input as an array.

rõ

ÊÆÔv

Try it or run all test cases

         :Implicit input of array U=[low,high]

r        :Reduce by

 õ       :  Inclusive, reversed range (giving the range [high,low])

n       :Reassign to U

Ê        :Length

 Æ       :Map the range [0,Ê)

  Ô      :  Reverse U

   v     :  Remove the first element

edited Feb 13 at 17:31

answered Feb 13 at 17:20

Shaggy

20.1k21667

Japt, 7 bytes

Takes input as an array.

rõ

ÊÆÔv

Try it or run all test cases

         :Implicit input of array U=[low,high]

r        :Reduce by

 õ       :  Inclusive, reversed range (giving the range [high,low])

n       :Reassign to U

Ê        :Length

 Æ       :Map the range [0,Ê)

  Ô      :  Reverse U

   v     :  Remove the first element

edited Feb 13 at 17:31

answered Feb 13 at 17:20

Shaggy

20.1k21667

edited Feb 13 at 17:31

answered Feb 13 at 17:20

Shaggy

20.1k21667

answered Feb 13 at 17:20

Shaggy

20.1k21667

answered Feb 13 at 17:20

Shaggy

20.1k21667

add a comment |

R, 51 bytes

function(x,y,z=x:y)matrix(c(z,rev(z)),2,,T)[seq(z)]

Try it online!

Outgolfed by digEmAll

edited Feb 13 at 18:04

answered Feb 13 at 17:59

user2390246

1,264111

1

$begingroup$
I just posted a very similar approach 30 seconds before you :D
$endgroup$
– digEmAll
Feb 13 at 18:01

$begingroup$
@digEmAll Yes, but yours is a lot better!
$endgroup$
– user2390246
Feb 13 at 18:03

add a comment |

R, 51 bytes

function(x,y,z=x:y)matrix(c(z,rev(z)),2,,T)[seq(z)]

Try it online!

Outgolfed by digEmAll

edited Feb 13 at 18:04

answered Feb 13 at 17:59

user2390246

1,264111

1

$begingroup$
I just posted a very similar approach 30 seconds before you :D
$endgroup$
– digEmAll
Feb 13 at 18:01

$begingroup$
@digEmAll Yes, but yours is a lot better!
$endgroup$
– user2390246
Feb 13 at 18:03

add a comment |

R, 51 bytes

function(x,y,z=x:y)matrix(c(z,rev(z)),2,,T)[seq(z)]

Try it online!

Outgolfed by digEmAll

edited Feb 13 at 18:04

answered Feb 13 at 17:59

user2390246

1,264111

R, 51 bytes

function(x,y,z=x:y)matrix(c(z,rev(z)),2,,T)[seq(z)]

Try it online!

Outgolfed by digEmAll

edited Feb 13 at 18:04

answered Feb 13 at 17:59

user2390246

1,264111

edited Feb 13 at 18:04

answered Feb 13 at 17:59

user2390246

1,264111

answered Feb 13 at 17:59

user2390246

1,264111

answered Feb 13 at 17:59

user2390246

1,264111

1

$begingroup$
I just posted a very similar approach 30 seconds before you :D
$endgroup$
– digEmAll
Feb 13 at 18:01

$begingroup$
@digEmAll Yes, but yours is a lot better!
$endgroup$
– user2390246
Feb 13 at 18:03

add a comment |

1

$begingroup$
I just posted a very similar approach 30 seconds before you :D
$endgroup$
– digEmAll
Feb 13 at 18:01

$begingroup$
@digEmAll Yes, but yours is a lot better!
$endgroup$
– user2390246
Feb 13 at 18:03

I just posted a very similar approach 30 seconds before you :D

– digEmAll
Feb 13 at 18:01

@digEmAll Yes, but yours is a lot better!

– user2390246
Feb 13 at 18:03

add a comment |

MATL, 8 bytes

&:t"1&)P

Try it online!

Explanation

&:      % Take two inputs (implicit). Two-input range

t       % Duplicate

"       % For each

  1&)   %   Push first element, then an array with the rest

  P     %   Reverse array

        % End (implicit). Display (implicit)

edited Feb 13 at 18:11

answered Feb 13 at 17:57

Luis Mendo

74.5k888291

add a comment |

MATL, 8 bytes

&:t"1&)P

Try it online!

Explanation

&:      % Take two inputs (implicit). Two-input range

t       % Duplicate

"       % For each

  1&)   %   Push first element, then an array with the rest

  P     %   Reverse array

        % End (implicit). Display (implicit)

edited Feb 13 at 18:11

answered Feb 13 at 17:57

Luis Mendo

74.5k888291

add a comment |

MATL, 8 bytes

&:t"1&)P

Try it online!

Explanation

&:      % Take two inputs (implicit). Two-input range

t       % Duplicate

"       % For each

  1&)   %   Push first element, then an array with the rest

  P     %   Reverse array

        % End (implicit). Display (implicit)

edited Feb 13 at 18:11

answered Feb 13 at 17:57

Luis Mendo

74.5k888291

MATL, 8 bytes

&:t"1&)P

Try it online!

Explanation

&:      % Take two inputs (implicit). Two-input range

t       % Duplicate

"       % For each

  1&)   %   Push first element, then an array with the rest

  P     %   Reverse array

        % End (implicit). Display (implicit)

edited Feb 13 at 18:11

answered Feb 13 at 17:57

Luis Mendo

74.5k888291

edited Feb 13 at 18:11

answered Feb 13 at 17:57

Luis Mendo

74.5k888291

answered Feb 13 at 17:57

Luis Mendo

74.5k888291

answered Feb 13 at 17:57

Luis Mendo

74.5k888291

add a comment |

JavaScript, 40 bytes

l=>g=h=>h>l?[l++,h--,...g(h)]:h<l?:[l]

Try It Online!

edited Feb 13 at 20:13

answered Feb 13 at 18:13

Shaggy

20.1k21667

add a comment |

JavaScript, 40 bytes

l=>g=h=>h>l?[l++,h--,...g(h)]:h<l?:[l]

Try It Online!

edited Feb 13 at 20:13

answered Feb 13 at 18:13

Shaggy

20.1k21667

add a comment |

JavaScript, 40 bytes

l=>g=h=>h>l?[l++,h--,...g(h)]:h<l?:[l]

Try It Online!

edited Feb 13 at 20:13

answered Feb 13 at 18:13

Shaggy

20.1k21667

JavaScript, 40 bytes

l=>g=h=>h>l?[l++,h--,...g(h)]:h<l?:[l]

Try It Online!

edited Feb 13 at 20:13

answered Feb 13 at 18:13

Shaggy

20.1k21667

edited Feb 13 at 20:13

answered Feb 13 at 18:13

Shaggy

20.1k21667

answered Feb 13 at 18:13

Shaggy

20.1k21667

answered Feb 13 at 18:13

Shaggy

20.1k21667

add a comment |

Forth (gforth), 52 bytes

: f 2dup - 1+ 0 do dup . i 2 mod 2* 1- - swap loop ;

Try it online!

Explanation

Loop from 0 to (End - Start). Place End and Start on top of the stack.

Each Iteration:

Output the current number

Add (or subtract) 1 from the current number

Switch the current number with the other number

Code Explanation

: f            start new word definition

  2dup -       get the size of the range (total number of integers)

  1+ 0         add 1 to the size because forth loops are [Inclusive, Exclusive) 

  do           start counted loop from 0 to size+1

    dup .      output the current top of the stack

    i 2 mod    get the index of the loop modulus 2

    2* 1-      convert from 0,1 to -1,1

    -          subtract result from top of stack (adds 1 to lower bound and subtracts 1 from upper)

    swap       swap the top two stack numbers 

  loop         end the counted loop

;              end the word definition

edited Feb 13 at 20:23

answered Feb 13 at 18:02

reffu

61126

add a comment |

Forth (gforth), 52 bytes

: f 2dup - 1+ 0 do dup . i 2 mod 2* 1- - swap loop ;

Try it online!

Explanation

Loop from 0 to (End - Start). Place End and Start on top of the stack.

Each Iteration:

Output the current number

Add (or subtract) 1 from the current number

Switch the current number with the other number

Code Explanation

: f            start new word definition

  2dup -       get the size of the range (total number of integers)

  1+ 0         add 1 to the size because forth loops are [Inclusive, Exclusive) 

  do           start counted loop from 0 to size+1

    dup .      output the current top of the stack

    i 2 mod    get the index of the loop modulus 2

    2* 1-      convert from 0,1 to -1,1

    -          subtract result from top of stack (adds 1 to lower bound and subtracts 1 from upper)

    swap       swap the top two stack numbers 

  loop         end the counted loop

;              end the word definition

edited Feb 13 at 20:23

answered Feb 13 at 18:02

reffu

61126

add a comment |

Forth (gforth), 52 bytes

: f 2dup - 1+ 0 do dup . i 2 mod 2* 1- - swap loop ;

Try it online!

Explanation

Loop from 0 to (End - Start). Place End and Start on top of the stack.

Each Iteration:

Output the current number

Add (or subtract) 1 from the current number

Switch the current number with the other number

Code Explanation

: f            start new word definition

  2dup -       get the size of the range (total number of integers)

  1+ 0         add 1 to the size because forth loops are [Inclusive, Exclusive) 

  do           start counted loop from 0 to size+1

    dup .      output the current top of the stack

    i 2 mod    get the index of the loop modulus 2

    2* 1-      convert from 0,1 to -1,1

    -          subtract result from top of stack (adds 1 to lower bound and subtracts 1 from upper)

    swap       swap the top two stack numbers 

  loop         end the counted loop

;              end the word definition

edited Feb 13 at 20:23

answered Feb 13 at 18:02

reffu

61126

Forth (gforth), 52 bytes

: f 2dup - 1+ 0 do dup . i 2 mod 2* 1- - swap loop ;

Try it online!

Explanation

Loop from 0 to (End - Start). Place End and Start on top of the stack.

Each Iteration:

Output the current number

Add (or subtract) 1 from the current number

Switch the current number with the other number

Code Explanation

: f            start new word definition

  2dup -       get the size of the range (total number of integers)

  1+ 0         add 1 to the size because forth loops are [Inclusive, Exclusive) 

  do           start counted loop from 0 to size+1

    dup .      output the current top of the stack

    i 2 mod    get the index of the loop modulus 2

    2* 1-      convert from 0,1 to -1,1

    -          subtract result from top of stack (adds 1 to lower bound and subtracts 1 from upper)

    swap       swap the top two stack numbers 

  loop         end the counted loop

;              end the word definition

edited Feb 13 at 20:23

answered Feb 13 at 18:02

reffu

61126

edited Feb 13 at 20:23

answered Feb 13 at 18:02

reffu

61126

answered Feb 13 at 18:02

reffu

61126

answered Feb 13 at 18:02

reffu

61126

add a comment |

Julia 0.7, 29 bytes

f(a,b)=[a:b b:-1:a]'[1:1+b-a]

Try it online!

answered Feb 13 at 20:30

Kirill L.

4,4151523

add a comment |

Julia 0.7, 29 bytes

f(a,b)=[a:b b:-1:a]'[1:1+b-a]

Try it online!

answered Feb 13 at 20:30

Kirill L.

4,4151523

add a comment |

Julia 0.7, 29 bytes

f(a,b)=[a:b b:-1:a]'[1:1+b-a]

Try it online!

answered Feb 13 at 20:30

Kirill L.

4,4151523

Julia 0.7, 29 bytes

f(a,b)=[a:b b:-1:a]'[1:1+b-a]

Try it online!

answered Feb 13 at 20:30

Kirill L.

4,4151523

answered Feb 13 at 20:30

Kirill L.

4,4151523

answered Feb 13 at 20:30

Kirill L.

4,4151523

answered Feb 13 at 20:30

Kirill L.

4,4151523

add a comment |

Haskell, 30 bytes

l%h=l:take(h-l)(h:(l+1)%(h-1))

Usage: 3%7 gives `[3,7,4,6,5]

answered Feb 14 at 13:48

proud haskeller

5,71611536

add a comment |

Haskell, 30 bytes

l%h=l:take(h-l)(h:(l+1)%(h-1))

Usage: 3%7 gives `[3,7,4,6,5]

answered Feb 14 at 13:48

proud haskeller

5,71611536

add a comment |

Haskell, 30 bytes

l%h=l:take(h-l)(h:(l+1)%(h-1))

Usage: 3%7 gives `[3,7,4,6,5]

answered Feb 14 at 13:48

proud haskeller

5,71611536

Haskell, 30 bytes

l%h=l:take(h-l)(h:(l+1)%(h-1))

Usage: 3%7 gives `[3,7,4,6,5]

answered Feb 14 at 13:48

proud haskeller

5,71611536

answered Feb 14 at 13:48

proud haskeller

5,71611536

answered Feb 14 at 13:48

proud haskeller

5,71611536

answered Feb 14 at 13:48

proud haskeller

5,71611536

add a comment |

JVM bytecode (OpenJDK asmtools JASM), 449 bytes

enum b{const #1=Method java/io/PrintStream.print:(I)V;static Method a:(II)V stack 2 locals 4{getstatic java/lang/System.out:"Ljava/io/PrintStream;";astore 3;ldc 0;istore 2;l:iload 2;ldc 1;if_icmpeq t;aload 3;iload 0;invokevirtual #1;iinc 0,1;iinc 2,1;goto c;t:aload 3;iload 1;invokevirtual #1;iinc 1,-1;iinc 2,-1;c:aload 3;ldc 32;i2c;invokevirtual java/io/PrintStream.print:(C)V;iload 0;iload 1;if_icmpne l;aload 3;iload 0;invokevirtual #1;return;}}

Ungolfed (and slightly cleaner)

 enum b {    

    public static Method "a":(II)V stack 5 locals 4 {

        getstatic "java/lang/System"."out":"Ljava/io/PrintStream;";

        astore 3;

        ldc 0;

        istore 2;

    loop:

        iload 2;

        ldc 1;

        if_icmpeq true;

    false:

        aload 3;

        iload 0;

        invokevirtual "java/io/PrintStream"."print":"(I)V";

        iinc 0,1;

        iinc 2,1;

        goto cond;

    true:

        aload 3;

        iload 1;

        invokevirtual "java/io/PrintStream"."print":"(I)V";

        iinc 1,-1;

        iinc 2,-1;

        goto cond;

    cond:

        iload 0;

        iload 1;

        if_icmpne loop;

        aload 3;

        iload 0;

        invokevirtual "java/io/PrintStream"."print":"(I)V";

        return;

    }

}

Standalone function, needs to be called from Java as b.a(num1,num2).

Explanation

...I have a distinct feeling I'm massively outgunned on the byte count

answered Feb 14 at 22:30

Archduke Liamus

1412

add a comment |

JVM bytecode (OpenJDK asmtools JASM), 449 bytes

enum b{const #1=Method java/io/PrintStream.print:(I)V;static Method a:(II)V stack 2 locals 4{getstatic java/lang/System.out:"Ljava/io/PrintStream;";astore 3;ldc 0;istore 2;l:iload 2;ldc 1;if_icmpeq t;aload 3;iload 0;invokevirtual #1;iinc 0,1;iinc 2,1;goto c;t:aload 3;iload 1;invokevirtual #1;iinc 1,-1;iinc 2,-1;c:aload 3;ldc 32;i2c;invokevirtual java/io/PrintStream.print:(C)V;iload 0;iload 1;if_icmpne l;aload 3;iload 0;invokevirtual #1;return;}}

Ungolfed (and slightly cleaner)

 enum b {    

    public static Method "a":(II)V stack 5 locals 4 {

        getstatic "java/lang/System"."out":"Ljava/io/PrintStream;";

        astore 3;

        ldc 0;

        istore 2;

    loop:

        iload 2;

        ldc 1;

        if_icmpeq true;

    false:

        aload 3;

        iload 0;

        invokevirtual "java/io/PrintStream"."print":"(I)V";

        iinc 0,1;

        iinc 2,1;

        goto cond;

    true:

        aload 3;

        iload 1;

        invokevirtual "java/io/PrintStream"."print":"(I)V";

        iinc 1,-1;

        iinc 2,-1;

        goto cond;

    cond:

        iload 0;

        iload 1;

        if_icmpne loop;

        aload 3;

        iload 0;

        invokevirtual "java/io/PrintStream"."print":"(I)V";

        return;

    }

}

Standalone function, needs to be called from Java as b.a(num1,num2).

Explanation

...I have a distinct feeling I'm massively outgunned on the byte count

answered Feb 14 at 22:30

Archduke Liamus

1412

add a comment |

JVM bytecode (OpenJDK asmtools JASM), 449 bytes

enum b{const #1=Method java/io/PrintStream.print:(I)V;static Method a:(II)V stack 2 locals 4{getstatic java/lang/System.out:"Ljava/io/PrintStream;";astore 3;ldc 0;istore 2;l:iload 2;ldc 1;if_icmpeq t;aload 3;iload 0;invokevirtual #1;iinc 0,1;iinc 2,1;goto c;t:aload 3;iload 1;invokevirtual #1;iinc 1,-1;iinc 2,-1;c:aload 3;ldc 32;i2c;invokevirtual java/io/PrintStream.print:(C)V;iload 0;iload 1;if_icmpne l;aload 3;iload 0;invokevirtual #1;return;}}

Ungolfed (and slightly cleaner)

 enum b {    

    public static Method "a":(II)V stack 5 locals 4 {

        getstatic "java/lang/System"."out":"Ljava/io/PrintStream;";

        astore 3;

        ldc 0;

        istore 2;

    loop:

        iload 2;

        ldc 1;

        if_icmpeq true;

    false:

        aload 3;

        iload 0;

        invokevirtual "java/io/PrintStream"."print":"(I)V";

        iinc 0,1;

        iinc 2,1;

        goto cond;

    true:

        aload 3;

        iload 1;

        invokevirtual "java/io/PrintStream"."print":"(I)V";

        iinc 1,-1;

        iinc 2,-1;

        goto cond;

    cond:

        iload 0;

        iload 1;

        if_icmpne loop;

        aload 3;

        iload 0;

        invokevirtual "java/io/PrintStream"."print":"(I)V";

        return;

    }

}

Standalone function, needs to be called from Java as b.a(num1,num2).

Explanation

...I have a distinct feeling I'm massively outgunned on the byte count

answered Feb 14 at 22:30

Archduke Liamus

1412

JVM bytecode (OpenJDK asmtools JASM), 449 bytes

enum b{const #1=Method java/io/PrintStream.print:(I)V;static Method a:(II)V stack 2 locals 4{getstatic java/lang/System.out:"Ljava/io/PrintStream;";astore 3;ldc 0;istore 2;l:iload 2;ldc 1;if_icmpeq t;aload 3;iload 0;invokevirtual #1;iinc 0,1;iinc 2,1;goto c;t:aload 3;iload 1;invokevirtual #1;iinc 1,-1;iinc 2,-1;c:aload 3;ldc 32;i2c;invokevirtual java/io/PrintStream.print:(C)V;iload 0;iload 1;if_icmpne l;aload 3;iload 0;invokevirtual #1;return;}}

Ungolfed (and slightly cleaner)

 enum b {    

    public static Method "a":(II)V stack 5 locals 4 {

        getstatic "java/lang/System"."out":"Ljava/io/PrintStream;";

        astore 3;

        ldc 0;

        istore 2;

    loop:

        iload 2;

        ldc 1;

        if_icmpeq true;

    false:

        aload 3;

        iload 0;

        invokevirtual "java/io/PrintStream"."print":"(I)V";

        iinc 0,1;

        iinc 2,1;

        goto cond;

    true:

        aload 3;

        iload 1;

        invokevirtual "java/io/PrintStream"."print":"(I)V";

        iinc 1,-1;

        iinc 2,-1;

        goto cond;

    cond:

        iload 0;

        iload 1;

        if_icmpne loop;

        aload 3;

        iload 0;

        invokevirtual "java/io/PrintStream"."print":"(I)V";

        return;

    }

}

Standalone function, needs to be called from Java as b.a(num1,num2).

Explanation

...I have a distinct feeling I'm massively outgunned on the byte count

answered Feb 14 at 22:30

Archduke Liamus

1412

answered Feb 14 at 22:30

Archduke Liamus

1412

answered Feb 14 at 22:30

Archduke Liamus

1412

answered Feb 14 at 22:30

Archduke Liamus

1412

add a comment |

Ink, 46 bytes

=h(I,A)

{I<=A:{I} {I<A:{A} ->h(I+1,A-1)}}->->

(I don't think there's an online interpreter for Ink, sorry)

Defines a stitch called h, which takes two arguments I and A, which are the bounds of the range.

Outputs by printing values, separated by spaces, to stdout.

Explanation

=h(min, max) // Define the stitch.

{min <= max:{min}/* print min unless it's greater than max */{min < max: {max} /*Also print max if it's greater than min*/->h(min+1, max-1)/*Then divert, with the arguments changed*/}}

->-> // If we didn't divert earlier, divert to wherever the stitch was called from

answered Feb 13 at 19:01

Sara J

613

add a comment |

Ink, 46 bytes

=h(I,A)

{I<=A:{I} {I<A:{A} ->h(I+1,A-1)}}->->

(I don't think there's an online interpreter for Ink, sorry)

Defines a stitch called h, which takes two arguments I and A, which are the bounds of the range.

Outputs by printing values, separated by spaces, to stdout.

Explanation

=h(min, max) // Define the stitch.

{min <= max:{min}/* print min unless it's greater than max */{min < max: {max} /*Also print max if it's greater than min*/->h(min+1, max-1)/*Then divert, with the arguments changed*/}}

->-> // If we didn't divert earlier, divert to wherever the stitch was called from

answered Feb 13 at 19:01

Sara J

613

add a comment |

Ink, 46 bytes

=h(I,A)

{I<=A:{I} {I<A:{A} ->h(I+1,A-1)}}->->

(I don't think there's an online interpreter for Ink, sorry)

Defines a stitch called h, which takes two arguments I and A, which are the bounds of the range.

Outputs by printing values, separated by spaces, to stdout.

Explanation

=h(min, max) // Define the stitch.

{min <= max:{min}/* print min unless it's greater than max */{min < max: {max} /*Also print max if it's greater than min*/->h(min+1, max-1)/*Then divert, with the arguments changed*/}}

->-> // If we didn't divert earlier, divert to wherever the stitch was called from

answered Feb 13 at 19:01

Sara J

613

Ink, 46 bytes

=h(I,A)

{I<=A:{I} {I<A:{A} ->h(I+1,A-1)}}->->

(I don't think there's an online interpreter for Ink, sorry)

Defines a stitch called h, which takes two arguments I and A, which are the bounds of the range.

Outputs by printing values, separated by spaces, to stdout.

Explanation

=h(min, max) // Define the stitch.

{min <= max:{min}/* print min unless it's greater than max */{min < max: {max} /*Also print max if it's greater than min*/->h(min+1, max-1)/*Then divert, with the arguments changed*/}}

->-> // If we didn't divert earlier, divert to wherever the stitch was called from

answered Feb 13 at 19:01

Sara J

613

answered Feb 13 at 19:01

Sara J

613

answered Feb 13 at 19:01

Sara J

613

answered Feb 13 at 19:01

Sara J

613

add a comment |

Perl 5 `-ln`, 37 bytes

@.=$_..<>;say shift@.,$/,pop@.while@.

Try it online!

answered Feb 13 at 20:02

Xcali

5,335520

$begingroup$
33bytes
$endgroup$
– Nahuel Fouilleul
Feb 14 at 14:47

add a comment |

Perl 5 `-ln`, 37 bytes

@.=$_..<>;say shift@.,$/,pop@.while@.

Try it online!

answered Feb 13 at 20:02

Xcali

5,335520

$begingroup$
33bytes
$endgroup$
– Nahuel Fouilleul
Feb 14 at 14:47

add a comment |

Perl 5 `-ln`, 37 bytes

@.=$_..<>;say shift@.,$/,pop@.while@.

Try it online!

answered Feb 13 at 20:02

Xcali

5,335520

Perl 5 `-ln`, 37 bytes

@.=$_..<>;say shift@.,$/,pop@.while@.

Try it online!

answered Feb 13 at 20:02

Xcali

5,335520

answered Feb 13 at 20:02

Xcali

5,335520

answered Feb 13 at 20:02

Xcali

5,335520

answered Feb 13 at 20:02

Xcali

5,335520

$begingroup$
33bytes
$endgroup$
– Nahuel Fouilleul
Feb 14 at 14:47

add a comment |

$begingroup$
33bytes
$endgroup$
– Nahuel Fouilleul
Feb 14 at 14:47

33bytes

– Nahuel Fouilleul
Feb 14 at 14:47

add a comment |

Java (JDK), 52 bytes

(l,h,o)->{for(int i=0;l<=h;i^=1)o.add(i<1?l++:h--);}

Try it online!

edited Feb 13 at 20:44

answered Feb 13 at 17:25

Olivier Grégoire

9,10511943

add a comment |

Java (JDK), 52 bytes

(l,h,o)->{for(int i=0;l<=h;i^=1)o.add(i<1?l++:h--);}

Try it online!

edited Feb 13 at 20:44

answered Feb 13 at 17:25

Olivier Grégoire

9,10511943

add a comment |

Java (JDK), 52 bytes

(l,h,o)->{for(int i=0;l<=h;i^=1)o.add(i<1?l++:h--);}

Try it online!

edited Feb 13 at 20:44

answered Feb 13 at 17:25

Olivier Grégoire

9,10511943

Java (JDK), 52 bytes

(l,h,o)->{for(int i=0;l<=h;i^=1)o.add(i<1?l++:h--);}

Try it online!

edited Feb 13 at 20:44

answered Feb 13 at 17:25

Olivier Grégoire

9,10511943

edited Feb 13 at 20:44

answered Feb 13 at 17:25

Olivier Grégoire

9,10511943

answered Feb 13 at 17:25

Olivier Grégoire

9,10511943

answered Feb 13 at 17:25

Olivier Grégoire

9,10511943

add a comment |

Clean, 48 bytes

import StdEnv

$a b|a<>b=[a: $b(a+sign(b-a))]=[a]

Try it online!

answered Feb 14 at 1:29

Οurous

7,12111035

add a comment |

Clean, 48 bytes

import StdEnv

$a b|a<>b=[a: $b(a+sign(b-a))]=[a]

Try it online!

answered Feb 14 at 1:29

Οurous

7,12111035

add a comment |

Clean, 48 bytes

import StdEnv

$a b|a<>b=[a: $b(a+sign(b-a))]=[a]

Try it online!

answered Feb 14 at 1:29

Οurous

7,12111035

Clean, 48 bytes

import StdEnv

$a b|a<>b=[a: $b(a+sign(b-a))]=[a]

Try it online!

answered Feb 14 at 1:29

Οurous

7,12111035

answered Feb 14 at 1:29

Οurous

7,12111035

answered Feb 14 at 1:29

Οurous

7,12111035

answered Feb 14 at 1:29

Οurous

7,12111035

add a comment |

Ruby, 37 36 33 bytes

f=->a,b{a>b ?:[a,b]|f[a+1,b-1]}

Try it online!

Recursive version with 3 bytes saved by G B.

Ruby, 38 bytes

->a,b{d=*c=a..b;c.map{d.reverse!.pop}}

Try it online!

Non-recursive version.

edited Feb 14 at 15:11

answered Feb 14 at 10:56

Kirill L.

4,4151523

add a comment |

Ruby, 37 36 33 bytes

f=->a,b{a>b ?:[a,b]|f[a+1,b-1]}

Try it online!

Recursive version with 3 bytes saved by G B.

Ruby, 38 bytes

->a,b{d=*c=a..b;c.map{d.reverse!.pop}}

Try it online!

Non-recursive version.

edited Feb 14 at 15:11

answered Feb 14 at 10:56

Kirill L.

4,4151523

add a comment |

Ruby, 37 36 33 bytes

f=->a,b{a>b ?:[a,b]|f[a+1,b-1]}

Try it online!

Recursive version with 3 bytes saved by G B.

Ruby, 38 bytes

->a,b{d=*c=a..b;c.map{d.reverse!.pop}}

Try it online!

Non-recursive version.

edited Feb 14 at 15:11

answered Feb 14 at 10:56

Kirill L.

4,4151523

Ruby, 37 36 33 bytes

f=->a,b{a>b ?:[a,b]|f[a+1,b-1]}

Try it online!

Recursive version with 3 bytes saved by G B.

Ruby, 38 bytes

->a,b{d=*c=a..b;c.map{d.reverse!.pop}}

Try it online!

Non-recursive version.

edited Feb 14 at 15:11

answered Feb 14 at 10:56

Kirill L.

4,4151523

edited Feb 14 at 15:11

answered Feb 14 at 10:56

Kirill L.

4,4151523

answered Feb 14 at 10:56

Kirill L.

4,4151523

answered Feb 14 at 10:56

Kirill L.

4,4151523

add a comment |

Cubix, 16 bytes

;w(.II>sO-?@;)^/

Try it here

Cubified

    ; w

    ( .

I I > s O - ? @

; ) ^ / . . . .

    . .

    . .

Explanation

answered Feb 14 at 16:41

Mnemonic

4,7351730

add a comment |

Cubix, 16 bytes

;w(.II>sO-?@;)^/

Try it here

Cubified

    ; w

    ( .

I I > s O - ? @

; ) ^ / . . . .

    . .

    . .

Explanation

answered Feb 14 at 16:41

Mnemonic

4,7351730

add a comment |

Cubix, 16 bytes

;w(.II>sO-?@;)^/

Try it here

Cubified

    ; w

    ( .

I I > s O - ? @

; ) ^ / . . . .

    . .

    . .

Explanation

answered Feb 14 at 16:41

Mnemonic

4,7351730

Cubix, 16 bytes

;w(.II>sO-?@;)^/

Try it here

Cubified

    ; w

    ( .

I I > s O - ? @

; ) ^ / . . . .

    . .

    . .

Explanation

answered Feb 14 at 16:41

Mnemonic

4,7351730

answered Feb 14 at 16:41

Mnemonic

4,7351730

answered Feb 14 at 16:41

Mnemonic

4,7351730

answered Feb 14 at 16:41

Mnemonic

4,7351730

add a comment |

Pyth, 10 8 bytes

{.iF_B}F

Try it here

Explanation

{.iF_B}F

      }FQ  Generate the range between the (implicit) inputs.

 .iF_B     Interleave it with its reverse.

{          Deduplicate.

edited Feb 14 at 17:44

answered Feb 13 at 22:08

Mnemonic

4,7351730

add a comment |

Pyth, 10 8 bytes

{.iF_B}F

Try it here

Explanation

{.iF_B}F

      }FQ  Generate the range between the (implicit) inputs.

 .iF_B     Interleave it with its reverse.

{          Deduplicate.

edited Feb 14 at 17:44

answered Feb 13 at 22:08

Mnemonic

4,7351730

add a comment |

Pyth, 10 8 bytes

{.iF_B}F

Try it here

Explanation

{.iF_B}F

      }FQ  Generate the range between the (implicit) inputs.

 .iF_B     Interleave it with its reverse.

{          Deduplicate.

edited Feb 14 at 17:44

answered Feb 13 at 22:08

Mnemonic

4,7351730

Pyth, 10 8 bytes

{.iF_B}F

Try it here

Explanation

{.iF_B}F

      }FQ  Generate the range between the (implicit) inputs.

 .iF_B     Interleave it with its reverse.

{          Deduplicate.

edited Feb 14 at 17:44

answered Feb 13 at 22:08

Mnemonic

4,7351730

edited Feb 14 at 17:44

answered Feb 13 at 22:08

Mnemonic

4,7351730

answered Feb 13 at 22:08

Mnemonic

4,7351730

answered Feb 13 at 22:08

Mnemonic

4,7351730

add a comment |

JavaScript, 35 bytes

f=(a,b,s=1)=>a-b?a+[,f(b,a+s,-s)]:a

Try it online!

Thanks to Arnauld, 1 byte saved.

edited Feb 15 at 1:49

answered Feb 14 at 3:24

tsh

9,06511650

add a comment |

JavaScript, 35 bytes

f=(a,b,s=1)=>a-b?a+[,f(b,a+s,-s)]:a

Try it online!

Thanks to Arnauld, 1 byte saved.

edited Feb 15 at 1:49

answered Feb 14 at 3:24

tsh

9,06511650

add a comment |

JavaScript, 35 bytes

f=(a,b,s=1)=>a-b?a+[,f(b,a+s,-s)]:a

Try it online!

Thanks to Arnauld, 1 byte saved.

edited Feb 15 at 1:49

answered Feb 14 at 3:24

tsh

9,06511650

JavaScript, 35 bytes

f=(a,b,s=1)=>a-b?a+[,f(b,a+s,-s)]:a

Try it online!

Thanks to Arnauld, 1 byte saved.

edited Feb 15 at 1:49

answered Feb 14 at 3:24

tsh

9,06511650

edited Feb 15 at 1:49

answered Feb 14 at 3:24

tsh

9,06511650

answered Feb 14 at 3:24

tsh

9,06511650

answered Feb 14 at 3:24

tsh

9,06511650

add a comment |

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

draft saved

draft discarded

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

draft saved

draft discarded

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This page is only for reference, If you need detailed information, please check here

The 465 Arrangement

Example:

Rules:

Examples

Example:

Rules:

Examples

Example:

Rules:

Examples

Example:

Rules:

Examples

50 Answers 50

R, 38 37 36 bytes

Haskell, 30 bytes

Haskell, 39 bytes

R, 65 64 61 60 bytes

Python 2, 44 bytes

PowerShell, 59 48 bytes

05AB1E, 6 bytes

Wolfram Language (Mathematica), 56 54 bytes

Japt, 14 bytes

Stax, 7 bytes

cQuents, 19 bytes

Explanation

Haskell, 39 bytes

C# (Visual C# Interactive Compiler), 46 bytes

C# (Visual C# Interactive Compiler), 65 bytes

APL (dzaima/APL), 21 bytes

Japt, 7 bytes

R, 51 bytes

MATL, 8 bytes

Explanation

JavaScript, 40 bytes

Forth (gforth), 52 bytes

Explanation

Code Explanation

Julia 0.7, 29 bytes

Haskell, 30 bytes

JVM bytecode (OpenJDK asmtools JASM), 449 bytes

Explanation

Ink, 46 bytes

Explanation

Perl 5 -ln, 37 bytes

Java (JDK), 52 bytes

Clean, 48 bytes

Ruby, 37 36 33 bytes

Ruby, 38 bytes

Cubix, 16 bytes

Cubified

Explanation

Pyth, 10 8 bytes

Explanation

JavaScript, 35 bytes

Your Answer

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50 Answers 50

50 Answers 50

R, 38 37 36 bytes

R, 38 37 36 bytes

R, 38 37 36 bytes

R, 38 37 36 bytes

Haskell, 30 bytes

Haskell, 30 bytes

Haskell, 30 bytes

Haskell, 30 bytes

Haskell, 39 bytes

Haskell, 39 bytes

Haskell, 39 bytes

Haskell, 39 bytes

R, 65 64 61 60 bytes

R, 65 64 61 60 bytes

R, 65 64 61 60 bytes

R, 65 64 61 60 bytes

Python 2, 44 bytes

Python 2, 44 bytes

Python 2, 44 bytes

50 Answers
50

Perl 5 `-ln`, 37 bytes

50 Answers
50

50 Answers
50