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3422 Answers 3422

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Extract an integer from another

This is a somewhat interesting problem I ran into while nanboxing: given two integers, compute their bitwise-AND and concatenate the resulting "substrings" into a new integer.

More precisely: you are provided two integers as input — an input integer and a bitmask. As output, you should produce the bitwise-AND of the two such that, given a mask with \$n\$ set bits, the corresponding bits from the input are grouped together in the first \$n\$ bits of the resulting integer.

The following pseudocode is one way to implement the function:

-- x and mask are lists of booleans
function (x, mask)
  local result=list();
  for i=1, min(x.length, mask.length) do
    if mask[i] then
  return result;

Example inputs and outputs

// In binary:
// f(x, mask) == result
f(1011, 1111) == 1011 // Select entire number
f(1010, 1010) == 11   // Select bits 1 and 3, and concatenate them
f(11001100, 01100110) == 1010 // Concatenate substrings [1:2] and [5:6]
f(11111111, 10101010) == 1111 // Concatenate bits at odd indices.

// 16-bit variants in hexadecimal:
f(BEEF, 1111) == 9
f(DEAD, 8888) == F 
f(1337, FF00) == 13
f(CODE, 7777) == 82E

// Two 64-bit variants (in hex):
f(400921FB54442D18, CODE601F15DABE57) == 111DE42C8
f(FFFE0000004010CC, 8003000000000000) == 6

Specific rules

  • Standard loopholes, default IO, etc. apply where not overridden.
  • Input and output values must fulfill \$x \in \{b_{set}, b_{unset}\}^{n}\$ for some \$b_{set} \neq b_{unset}, n \in \mathbb{N}_{\geq 16}\$ of your choice. In other words:
    • You must support integers of at least 16 bits in your representation, but you may otherwise arbitrarily constrain their size - i.e. to 32-bit integers. You may also accept integers of any length through lists, arrays, strings, etc.
    • The "set" and "unset" values do not need to be of the same type or length, but they must be constant and distinct.
    • Most "linear" representations for integers are valid: integers, vectors, arrays, strings, etc.
  • This is , so the shortest answer in bytes wins.
  • Have fun!
  • \$\begingroup\$ Is this the PEXT BMI2 instruction? (also, possible duplicate: codegolf.stackexchange.com/q/37167) \$\endgroup\$ – the default. Jul 30 '20 at 13:00
  • \$\begingroup\$ @mypronounismonicareinstate After reading the referenced challenge I think this is indeed a duplicate of that. The required algorithms are identical and other requirements seem to not affect this too much. \$\endgroup\$ – Shieru Asakoto Aug 3 '20 at 4:07
  • \$\begingroup\$ @mypronounismonicareinstate I've given it a read, and it definitely is the same challenge, bar minor cosmetic differences. Of course, the search did not find it when I tried searching for it... \$\endgroup\$ – CompilerPotato Aug 3 '20 at 11:46

Write an ASPIF (clasp's ASP input format) program to find a maximum cap set (https://en.wikipedia.org/wiki/Cap_set) for 4 dimensions.

Share the code you used to generate the ASPIF rather than ASPIF itself. This may be an ASP program.

Winner is smallest word-count (according to wc) in ASPIF format. For ASP, you can get this by running something like:

clingo capset.asp --mode=gringo | grep -v "\(^1 0 1 [0-9]\+ 0 0$\)\|\(^4\)" | wc -w

(note the grep is for excluding unary rules and #show directives neither of which are necessary for solving. The output of this is still a valid clasp program)

I have an example for four dimensions (but I have a better one I won't share right away because I'm curious to see what other people get).

feature(number, (one; two; three)).
feature(shading, (solid; empty; striped)).
feature(color, (red; green; purple)).
feature(shape, (oval; diamond; squiggly)).
dimension(D) :- feature(D, _).
card(c(N,F,C,S)) :-
    feature(number,N); feature(shading,F); feature(color,C); feature(shape,S).
property(c(N,F,C,S),number,N) :- card(c(N,F,C,S)).
property(c(N,F,C,S),shading,F) :- card(c(N,F,C,S)).
property(c(N,F,C,S),color,C) :- card(c(N,F,C,S)).
property(c(N,F,C,S),shape,S) :- card(c(N,F,C,S)).

{in_capset(X) : card(X)}.
:~ in_capset(X).[-1,X]

settable(D, A, B, C) :-
    feature(D, A); feature(D, B); feature(D, C); A != B; A != C; B != C.
settable(D, A, A, A) :- feature(D, A).

:- in_capset(X); in_capset(Y); in_capset(Z);
    settable(D, A, B, C) :
        dimension(D), property(X, D, A), property(Y, D, B), property(Z, D, C);
    X < Y; Y < Z.

#show in_capset/1.

This grounds to an ASPIF program with 9296 "words"

  • \$\begingroup\$ 4D cap set is already known and has a pattern which was found in a challenge of mine, so it might be too trivial. Why not ask to take n as input and solve for n dimensions (without time and memory limit)? \$\endgroup\$ – Bubbler Aug 7 '20 at 3:32
  • 1
    \$\begingroup\$ Also, most people here are not familiar with ASP or ASPIF. It would be helpful if you include relevant links, so we can do some research before tackling the challenge. \$\endgroup\$ – Bubbler Aug 7 '20 at 3:34

Mega Man

My first polyglot challenge, enjoy!

Validness of a program

In this challenge, a "program" doesn't take an input. This challenge doesn't care of any output, though.

An invalid program, either:

  • Doesn't compile, or

  • Compiles, but the program doesn't halt when executed.

A program is valid otherwise.

The 6 Robot Masters

A robot master is a valid program. Their language can be chosen freely, not necessarily all same or all distinct.

There are 6 Robot Masters in total, namely Cut Man, Elec Man, Ice Man, Fire Man, Bomb Man, and Guts Man.

(Yeah, I wanted to include Time Man and Oil Man as well, but that would make this challenge too hard.)


The robot masters have their distictive weapons. (This doesn't mean the robot masters' source code acts as their weapon, though.) A weapon is an operation on a string.

  • Cut Man's weapon, Rolling Cutter, leaves the target source's first half characters only, rounded down. Example: Hello, world!Hello,

  • Elec Man's weapon, Thunder Beam, eliminates all whitespaces. Example: Hello, world!Hello,world!

  • Ice Man's weapon, Ice Slasher, turns all uppercase ASCII letters small. Example: Hello, world!hello, world!

  • Fire Man's weapon, Fire Storm, turns all lowercase ASCII letters capital. Example: Hello, world!HELLO, WORLD!

  • Bomb Man's weapon, Hyper Bomb, eliminates the target source's first word. The behavior on the surrounding whitespaces is implementation-defined. Example: Hello, world!world!

  • Guts Man's weapon, Super Arm, doubles all characters. Example: Hello, world!HHeelllloo,, wwoorrlldd!!


When a weapon is applied to a robot master's source code, if and only if it hits their weakness, the resulting code must be a valid program in the robot master's language.

  • Rolling Cutter is the weakness of Elec Man.

  • Thunder Beam is the weakness of Ice Man.

  • Ice Slasher is the weakness of Fire Man.

  • Fire Storm is the weakness of Bomb Man.

  • Hyper Bomb is the weakness of Guts Man.

  • Super Arm is the weakness of Cut Man.

Every other combination is not a weakness and must result in an invalid program. This includes a weapon applied to its owner.


This is a code golf. The score is the sum of the byte counts of all 6 source codes. The answer with the least score wins.

  • 3
    \$\begingroup\$ I find the organisation of the text quite confusing. If I've understood correctly (and I had to read it several times), the challenge boils down to this: write 6 programs, each of which is only valid when one of the weapons is applied to it. Is that right? Since the programs aren't required to implement the weapons, I don't really understand why each robot master is associated with a program. Lastly, do you mean 'execute' rather than 'compile'? \$\endgroup\$ – Dingus Aug 17 '20 at 8:28
  • \$\begingroup\$ @Dingus Well, it was pretty hard to make a reference to the game. I've clarified the 'compile' and 'execute'. The challenge boils down to, Write 6 programs, each of which is valid and also valid when a weapon is applied, but is invalid when any other weapon is applied. \$\endgroup\$ – Dannyu NDos Aug 17 '20 at 9:42
  • \$\begingroup\$ Ahhh, that makes more sense - I wasn't aware of the game (living under a rock, maybe). Perhaps you should include some more context about it. It seems a bit strange that hitting a weakness results in a valid program (would have expected the opposite), but the description is clear enough. I would suggest a small tweak to 'if and only if it hits their weakness, the resulting code must still be a valid program'. \$\endgroup\$ – Dingus Aug 17 '20 at 10:20
  • \$\begingroup\$ About 'compile', I meant in the bullet points where you define what an invalid program is. Many languages are not compiled languages. \$\endgroup\$ – Dingus Aug 17 '20 at 10:23

[PuyoPuyo] How long is my combo?


PuyoPuyo is a puzzle game where you and your opponent pile up colored slimes (called puyo) in a vertical (13*6 cells) grid. A puyo is one-cell big, but they come as pairs in the screen. Puyo pairs fall from the top to the bottom of the grid, and you can move and rotate them the Tetris way. The list of possible puyo pairs is the cartesian product of {'red','blue','green','yellow'} with itself. The pair sequence for a game is randomly generated for both players at the start of a round, and will be the same for both of them.

If four puyo (or more) of the same color are next to each other (in line, in square, Z-, S-, T-, J- or L-shaped), they disappear, awarding you points and making all the above puyo to fall. If when those puyo fall, they make another group of four (or more) they will disappear too, awarding you with more points than the first group: it is called a two-hit combo.

When a combo stops, whatever its length (1-hit or more), you will send damage to your opponent. The bigger the combo, the more damage is sent. Those damage are called ojama puyo, grey slimes that disappear only when a group disappears next to it. If your third column from the left is filled before your opponent's is, you lose. So to kill your opponent, you must manage to fill its screen before they fill yours.


With a given sequence of puyo pairs associated with their drop locations, output the length of the combo that has been made by this player. Shortest code in any language wins.


List of puyo pairs and their drop locations:

  • 'r1b1r1r2y2r3...'
  • '001000013102...'
  • [('red','1','blue','1'),('red','1','red','2'),('yellow','2','red','3'),...]
  • any sensible way you want, provided you detail how it works


  • This list will never contain any "column number" outside of [1;6] (or [0;5], if 0-indexed) nor a puyo of a color outside of {'red','blue','green','yellow'} (or any set used for your interpretation).
  • This list will always contain at most one combo sequence.
  • This list will never contain unusable data, like two colors in a row, or two column indices in a row.
  • If both puyo of a pair are dropped on the same column, the first one to come is on the top of the pair (in the example #1, red is dropped on the top of the blue on the column 1).
  • Puyo will never remain floating in the grid, but will fall to the lowest available cell of the column they are in, even if the paired puyo has stopped in its column (tl;dr puyo pairs split).


A single integer indicating the length of the combo.

Test cases:

Test case #1



Test case #2



Test case #3



Test case #4



Test case #5



Test case #6



Test case #7 (click me!)



Test case #8 (click me!)



Standard Loopholes apply.

NB: for purists, I know that the 13th row is supposed to be invisible and that puyo that are in that column are not considered 'linked' to nearby puyo, but I figured this challenge was hard enough as-is.

@Sandbox please comment! I'd love to hear your thoughts about such a challenge. I will finish setting it up soon, adding some extra resources about the game (like this one). Questions:

  • Should I reverse the "top / bottom" puyo of a pair rule? The way it is now, it forces to parse input as pairs. If it is reversed, golfers can take puyo one by one and sort it all by columns, making the challenge easier.
  • What tags should it enter with?
  • I will make other test cases soon enough, but should I include a as a file (via pastebin)?
  • 1
    \$\begingroup\$ "The combo"? Will there be only one combo? \$\endgroup\$ – user202729 Aug 15 '20 at 11:07
  • 1
    \$\begingroup\$ Does the two puyo(s) in a pairstick to each other like in tetris? \$\endgroup\$ – user202729 Aug 15 '20 at 11:07
  • \$\begingroup\$ Thanks for your comments @user202729, I'll be editing the challenge soon. This challenge will let you assume there will only be one combo. And puyo pairs are broken upon drop if need be, so that every single puyo cannot be floating in the grid. \$\endgroup\$ – V. Courtois Aug 18 '20 at 11:51

Sort until overflow - POSTED HERE

  • \$\begingroup\$ Now that this has been posted to main, could you delete this proposal to create more space for new answers? \$\endgroup\$ – Dude coinheringaahing Sep 25 '20 at 0:37

Quineoid Triple Uniqueness Optimization

This is a variant of Quineoid Triple with the same requirements but different scoring.

Write three different programs such that when any one program is provided as input to one of the other two, you get the source of the remaining program as output. More explicitly, given programs \$A\$, \$B\$, and \$C\$, where \$f(g)\$ denotes the output obtained from inputting the text of program \$g\$ into program \$f\$, all of the following must hold:

  • \$ A(B) = C \$
  • \$ A(C) = B \$
  • \$ B(A) = C \$
  • \$ B(C) = A \$
  • \$ C(A) = B \$
  • \$ C(B) = A \$


The goal is to have the three programs be as different as possible.

Your score is the sum of:

  • Number of unique bytes found in program \$A\$, but not \$B\$ or \$C\$
  • Number of unique bytes found in program \$B\$, but not \$A\$ or \$C\$
  • Number of unique bytes found in program \$C\$, but not \$A\$ or \$B\$

The theoretical maximum score is 256.

Additional Rules

  • Standard quine rules apply.
  • Each program can be in any language. Any number of them may share languages or each may use a different language.
  • Use any convenient IO format as long as each program uses a consistent convention.
    • Functions are allowed, as this counts as "any convenient IO".
  • The result of feeding a program its own source code is undefined
  • The result of feeding anything other than program text of either of the other two programs is undefined.
  • Byte encoding should be taken into account for languages with dedicated codepages.

SANDBOX: this is kind of -ish. Should I call it a bowling challenge?


Modular Chain Compression

A common trope in some kolmogorov-complexity challenges is to use repeated application of the modulo operator in order to compress large integers or string hashes into some range. For example, we can squash the numbers \$13,4,16\$ into \$0,1,2\$ by taking each number mod \$7\$ and then mod \$3\$. We call this sequence the modular chain \$7,3\$. In general, reducing a number \$k\$ by the modular chain \$n_1,n_2,\ldots,n_i\$ is equivalent to evaluating

$$(((k\ \text{mod}\ n_1)\ \text{mod}\ n_2)\ \ldots\ )\ \text{mod}\ n_i$$

In this challenge, you will be challenged to compress an arbitrary set of integers in this way. We say the length of the modular chain is the number of elements in the chain (not the number of bytes).


You are given a fixed set of \$100\$ random 32-bit integers. Here they are:

... (more on the actual post) ...


Via whatever means necessary (brute-force, mathematics, etc) design a modular chain to compact these numbers into a small a range as possible. The modular chain must have a distinct output for each input number.


In code-golf challenges, a modular chain is usually desirable if it achieves two things:

  • It compacts the input numbers or hashes into a small range.
  • It is short.

In this spirit, your score is the sum of the maximum output number of your modular chain and the length of your modular chain for the given input numbers.

Note an optimal modular chain could compress the input numbers into the range \$0 \ldots 99\$ with a single modulo operation, making the theoretical minimum score \$99 + 1 = 100\$.

The lowest scoring answer wins. You are encouraged (but not required) to post any code, mathematical background etc. that helped you design your modular chain.

  • \$\begingroup\$ Length in modulo operations or bytes? \$\endgroup\$ – user253751 Aug 25 '20 at 9:41
  • \$\begingroup\$ @user253751 The number of operations. Clarified. \$\endgroup\$ – Sisyphus Aug 25 '20 at 10:28

The fastest code to find a subset-sum

Given k sorted integers from low to high, output n permutations of those integers with sum as close as possible to m but not exceeding m. The output needs to be sorted from highest to lowest sum.


k integers, n, m - as described above. All the k integers and m are positive 31-bit integers. Now since this is NP-complete problem, both k and n are small integers, 20 at most.


One row for each premutation, with integers sorted from high to low. Also, the rows need to be sorted from highest to lowest sum

The fastest code wins

For the performance test, we will use the below input:

  • k = [67, 613, 2111, 2179, 2203, 2269, 3433, 3583, 4219, 5011]
  • m = 14,213
  • n = 10
  • 3
    \$\begingroup\$ 1) You might want to provide multiple test cases, so that the submissions don't use an algorithm that is fast only in some of them and very slow in the others. 2) For fair evaluation of speed, you need to run all submissions on your machine, so you need to provide some information about your machine's OS, RAM, CPU (also GPU if you want to allow using it). 3) Is the sum of all k integers guaranteed to be positive 31-bit integers? Otherwise we might face overflows during calculation. 4) I think you mean subsets instead of permutations. \$\endgroup\$ – Bubbler Aug 28 '20 at 4:42


This challenge is based on a somewhat unusual premise; the goal is to create a rectangular program (box) which is as large as possible, where each row and column will be a solution to a different challenge on this site.

For example, a 3×4 box might look like this:


As all solutions will be read from left to right or top to bottom, this would expand into seven programs:

a 1

For an answer to be valid, each one of these must be a valid solution to a different challenge on this site, in the same language. The box may be padded by whitespace, but must be rectangular (x groups of y bytes, separated by newlines). This is code bowling, so the longest answer in bytes wins.

Additional rules/clarifications:

  • Solutions cannot contain newlines (because otherwise they would be two separate rows/columns)
  • Yes, it will be possible to trivially get very high scores with some languages (like unary), but as with most challenges it's a competition within each language
  • Solutions do not have to be original, but ones copied from other answers should link to them
  • \$\begingroup\$ Unary won't be able to compete because the number of challenges on our site is lower than most Unary programs. It will be very hard to compete even in golfing languages, as the answerer will need to manually find the appropriate challenge to use. I don't really think a challenge that makes use of "all challenges on our site" is fun. \$\endgroup\$ – Bubbler Sep 10 '20 at 1:40
  • \$\begingroup\$ @Bubbler Yeah, now that I think about it I agree. I still like the concept, maybe there's something similar that might actually be fun. \$\endgroup\$ – Redwolf Programs Sep 10 '20 at 2:37
  • \$\begingroup\$ Maybe you could select a subset of challenges - something like what was done for this challenge? \$\endgroup\$ – Dingus Sep 17 '20 at 11:43
  • \$\begingroup\$ @Dingus Good idea! I'll see if I can fix a few other things, too. \$\endgroup\$ – Redwolf Programs Sep 17 '20 at 12:39

Targeted sum and difference of a sequence

You are given two things: a target integer(not necessarily positive) n, and a sorted list/array/etc. of non-negative integers a. (The list will have at least two elements). Your goal is to choose one element of a as your total, and then one by one, take elements of a, and either add or subtract them from your total. Print out all possible combinations(duplicates can be removed, but it is optional) of [a_1]±[a_2]±[a_3] or return them as a list/array/etc. .

In other words, find all solutions to n=[a_1]±[a_2]±[a_3]. The first element of a is not guaranteed to be a_1, nor is the second element guaranteed to be n_2.

Test cases:

n = 10, a = [1,2,3,4] :  10 = 1+2+3+4 and 2+1+3+4 ...
# The output should be 1+2+3+4(and 4+2+3+1, and every combination like that)
n = 1, a = [2,3]: 1 = 3-2, so 3-2. 
# (-2+3 wouldn't work, since it is strictly addition or subtraction of positive integers).
# -2 isn't a part of [2,3]. You should think of the steps as [a_1]±[a_2]±[a_3].
# n = -5, a = [0,5,1]. -5 = 0-5, and nothing else.
n=95, a = [50,50,5]. 95 = 50+50-5, 50-5+50. (A second 50+50-5 is optional)

Criteria: Shortest code wins


Is this clear enough(and what should I do to make this more clear)? Also, has this been done before? Finally, should I remove the restriction on the first number being positive?

Thank You!

  • \$\begingroup\$ What do you mean by "The first element of a is not guaranteed to be a_1"? \$\endgroup\$ – Zgarb Sep 11 '20 at 7:01

A centered hexagonal number is a centered figurate number that represents a hexagon with a dot in the center and all other dots surrounding the center dot in a hexagonal lattice.

Illustration of initial terms:

                                 o o o o
                   o o o        o o o o o
         o o      o o o o      o o o o o o
   o    o o o    o o o o o    o o o o o o o
         o o      o o o o      o o o o o o
                   o o o        o o o o o
                                 o o o o

   1      7          19             37

Write a function that takes an integer \$n\$ and returns "Invalid" if \$n\$ is not a centered hexagonal number or its illustration as a multiline rectangular string otherwise.

Sample Output :-

hexLattice(1) ➞ " o "
// o

hexLattice(7) ➞ "  o o  \n o o o \n  o o  "
//  o o
// o o o
//  o o

hexLattice(19) ➞ "   o o o   \n  o o o o  \n o o o o o \n  o o o o  \n   o o o   "
//   o o o
//  o o o o
// o o o o o
//  o o o o
//   o o o

hexLattice(21) ➞ "Invalid"


Shortest Code Wins!

  • 4
    \$\begingroup\$ I like the challenge concept! Usually, we advise against input validation; that is, rather than outputting "Invalid", solutions should assume the input is valid, though if you do that this challenge is almost a duplicate and so I think in this case it could make for an interesting challenge to leave it in. \$\endgroup\$ – hyper-neutrino Mod Sep 23 '20 at 13:11
  • 4
    \$\begingroup\$ Another thing is usually we encourage flexible input/output formatting; in this case, in addition to a multiline string, I would also allow a list of strings or a matrix of characters as output, and rather than strictly outputting "Invalid", I would suggest allowing solutions to state any reasonable parameters for how they'll indicate invalid input. \$\endgroup\$ – hyper-neutrino Mod Sep 23 '20 at 13:13
  • \$\begingroup\$ Related-ish. There are several other hexagon related challenges but this was the only one I could find that required computing the centred hexagonal numbers. \$\endgroup\$ – FryAmTheEggman Sep 24 '20 at 18:58

Word Length-Sum Multiples

  • \$\begingroup\$ Now that this has been posted, I've edited it down to save space and I'd recommend you delete the proposal \$\endgroup\$ – Dude coinheringaahing Sep 25 '20 at 0:15

Double Prime Words

  • 1
    \$\begingroup\$ tags would be decision-problem, code-golf, primes, I think? \$\endgroup\$ – Giuseppe Sep 8 '20 at 19:07
  • 1
    \$\begingroup\$ I think if and only if x is prime should be if and only if n is prime? \$\endgroup\$ – Giuseppe Sep 8 '20 at 19:08
  • \$\begingroup\$ Additional exampleː Is this word a double primeː Hello Worlds aardvark aalii Aani \$\endgroup\$ – Xwtek Sep 10 '20 at 12:36
  • \$\begingroup\$ @Xwtek Is that 4 separate examples, or 1 long example? \$\endgroup\$ – Sumner18 Sep 10 '20 at 15:09
  • \$\begingroup\$ Now that this has been posted, I've edited the post down to save space and I'd recommend you delete this proposal \$\endgroup\$ – Dude coinheringaahing Sep 25 '20 at 0:19

\$d\times n\$ dimensional word matrices [WIP]

Given two positive integers \$n\$ and \$d\$, and a list of words \$a\$, produce a \$d\$-dimensional matrix \$m\$ with each dimension having length \$n\$, filled with letters, that contains the words from \$a\$ placed such that they form a directly adjacent contiguous path through the dimensions.

For example, given \$d = 1\$, \$n = 3\$ and \$a = \$['cat'] output one of:




Given \$d = 2\$, \$n = 3\$ and \$a = \$['cat', 'hat', 'mat'] output something similar to:


Given \$d = 3\$, \$n = 3\$ and \$a = \$['low', 'complexity'] output something similar to:




or, if it's easier to visualise in an array structure:

    ['c', 'o', 'q'],
    ['i', 'g', 'w'],
    ['t', 'y', 'p'],
    ['k', 'm', 'c'],
    ['x', 'e', 'o'],
    ['b', 'u', 'f'],
    ['k', 'p', 'r'],
    ['d', 'l', 'l'],
    ['s', 'c', 'm'],

Which contains low at nested indices \$m[2][1][2]\$, \$m[1][1][2]\$, \$m[0][1][2]\$ and complexity at \$m[0][0][0]\$, \$m[0][0][1]\$, \$m[1][0][1]\$, \$m[2][0][1]\$, \$m[2][1][1]\$, \$m[1][1][1]\$, \$m[1][1][0]\$, \$m[0][1][0]\$, \$m[0][2][0]\$, \$m[0][2][1]\$.

I'd like to add some more complicated examples beyond three dimensions here.

Test Cases



  • Unused spaces should be filled with randomly selected letters.
  • There will always be enough space in the dimensions provided to allow the words to be added without re-using letters.
  • There is no requirement to ensure the words don't also appear elsewhere in the grid, so for example if the filler letters happen to spell one of the provided words, that is acceptable.

Questions for meta

  • This seems fun to me, any thoughts?
  • Is it too easy/hard?
  • Any other tags that are relevant?
  • As a follow up, I'd like to have a nested matrix provided and have programs solve it - but that might be better as a fastest-code challenge - is this a reasonable precursor?
  • \$\begingroup\$ Is d^n large enough to contain all the words without sharing letters? \$\endgroup\$ – Bubbler Jul 23 '20 at 8:18
  • \$\begingroup\$ Yeah, you won't have to be concerned with that, I'll add that to the rules. \$\endgroup\$ – Dom Hastings Jul 23 '20 at 8:21

Find a 3-Language Polyglot

What I had in mind was that cops would create a polyglot with in 3 languages (languages A, B, and C). When run in A, the program would print the name of language B; when run in language B, the program would print the name of language C; and when run in C, it would print the name of language A.

Cops have to provide the names of these 3 languages, as well as their original polyglot's characters scrambled in no particular order. as well as a valid program in A that has the same behavior as the polyglot (prints the name of B). This program must be able to be created by deleting characters from the original polyglot, i.e., all the letters in it are included in the hidden polyglot.

Given the languages and the scrambled programand the sample program, robbers have to find the polyglot (or a polyglot that has the same behavior as the one the cop wrote).


  • Any language chosen must be able to be run on TIO, repl.it, ideone, or someplace else online. If the language is obscure, please provide a link to some such website.
  • Any language used must have documentation on Esolangs, Wikipedia, GitHub, or someplace else. Unless the language is very commonly used and has tons of tutorials everywhere, such as Java, Haskell, or C, please provide a link to documentation. Any feature used in the program must be included in that documentation - it shouldn't be something people have to dig through layers of source code to find.

Questions for meta:

  • Is this too easy/hard? Should I not include the extra A program? Should I only make it for 2 languages?
  • Is there anything unclear about the instructions? How can I improve the phrasing?
  • Should cops also give the length of their programs as an extra hint?
  • 1
    \$\begingroup\$ one thing to consider: you would need some way of restricting languages that are allowed. Otherwise people could just make up their own languages or use really, really obscure languages. \$\endgroup\$ – thesilican Aug 18 '20 at 23:04
  • 2
    \$\begingroup\$ IIRC, the usual way to limit the language list is to specify "the language should be on at least one of Wikipedia, TIO, or esolangs.org", though esolangs is already crazy these days. \$\endgroup\$ – Bubbler Aug 19 '20 at 5:05
  • \$\begingroup\$ @Bubbler Yup, I've edited my question with some restrictions now \$\endgroup\$ – user Aug 20 '20 at 15:01

Terminal Punch Card

moved because apparently it's not clear enough.

So back in the day, computers didn't have fancy keyboard and mouse inputs, and didn't have your fancy screens. Instead they had punch cards.

Punch card

Punch cards punchers punched (try saying that 10 times fast) a hole out of a card to represent a 1-bit, and left it filled to represent a 0-bit. The cards were some number of holes wide, with each hole representing a bit in a byte.

Recently, you discovered an old mainframe at your local university that accepted punch cards that were 8 holes wide. For this challenge, you will be given data as an input, and your job is to punch a punch card to the terminal output, like this:

Hello, World!

: *  *   :
: **  * *:
: ** **  :
: ** **  :
: ** ****:
:  * **  :
:  *     :
: * * ***:
: ** ****:
: ***  * :
: ** **  :
: **  *  :
:  *    *:

The input will be a string or bytes representing the punched data payload. The output data must include rows, which start and end with a :, and have 8 bits between them, represented as a for 0, or a * for 1. There must be one row for each byte of data.

Here's the catch: The punch card puncher only punched one hole at a time, so in your program, must print (or add to the output string) only one character at a time.

Example of unacceptable method call:

# `binary` is some string with the binary bits.
print(":" + binary.replace("0", " ").replace("1", "*") + ":")

Acceptable method call:

for bit in binary:
   print(bit == "1" ? "*" : " ", "")

Also acceptable method call:

output = ""
for bit in binary:
   output += bit == "1" ? "*" : " "

The challenge is code golf, so least number of bytes wins. Standard rules/loopholes are in effect.

  • 4
    \$\begingroup\$ A word about catches before we even discuss observability and clarity issues: Catches are rarely a good idea for two reasons. The first is that challenge writers frequently add catches because they somehow feel their challenge is deficient or too easy and want to salvage it somehow. This coping mechanism usually fails, they are better off writing a new challenge. \$\endgroup\$ – Wheat Wizard Mod Oct 4 '20 at 12:17
  • 3
    \$\begingroup\$ Note also there are already comments here regarding observability, assuming language features, and assuming implementation details \$\endgroup\$ – Luis Mendo Oct 4 '20 at 12:19
  • 4
    \$\begingroup\$ The second, which I think is more relevant to you, is that structuring your challenge with a catch is often confusing. You have already written what is a complete challenge, but then in the last 10% or so the whole task changes. Some people don't read the whole challenge once they think they have it, miss the last bit or become frustrated when things are pulled out from under them. Regardless of how you feel about these people, It is really just better to phrase your challenge in a straight forward and upfront way. Nothing should seem tacked on if you can avoid it. \$\endgroup\$ – Wheat Wizard Mod Oct 4 '20 at 12:21
  • 3
    \$\begingroup\$ One way you could easily make your catch observable is to take a string and an index and output the character at that index. Of course answerers can just produce the entire string and index it, but they could already do that with your existing version (probably, it's a little unclear). \$\endgroup\$ – Wheat Wizard Mod Oct 4 '20 at 12:26

Socially distanced seating

Lord Lloyd Warbler wants to minimise the harm to his theatre's seating capacity for his hit show, Birds, of maintaining social distancing.

The social distancing rules in Westendland are:

  • Groups may sit together without distancing
  • Between groups there must be at least 2 empty seats along the row, and 1 empty row in front and behind.
  • The closest diagonal permitted is a knight's move - only 1 horizontal space empty.


 A _ _ B


 A _ _
 _ _ B

Given a list of group sizes and the theatre size (rows and columns), can you pack them all into the theatre? Groups can sit in any contiguous (connected) arrangement of seats.

Sample tests

(Rows, cols), [groups] -> canFitBool
(1,1), [1] -> true
(2,2), [1,1] -> false
(2,3), [1,1] -> true   // knight's move
(5,2), [4,4] -> true   // 2x2 in rows a,b, gap in c, 2x2 in d,e


  • Should I just provide a list of cases of varying difficulty? Like, some of these could be pretty difficult.
  • Is the knight's move rule too complicated?
  • Is the contiguous rule too permissive, and therefore complicated? It could mean some edge cases are possible if you have a weird shaped group. Could make it rectangular blocks only?
  • \$\begingroup\$ Groups can sit in any contiguous (connected) arrangement of seats You need to decide, and specify in the text, if contiguous means 4-connectivity (up, down, left, right) or 8-connectivity (diagonals count as connected too) \$\endgroup\$ – Luis Mendo Oct 4 '20 at 12:17

Count the strokes of an ASCII character


Given a printable ASCII character (0x21 – 0x7E), count its strokes (as handwritten), then output it.

Note: The strokes are based on how I write the characters. Those with potential controversy are marked * below.


       ! → 2  " → 2  # → 4  $ → 2* % → 3* & → 1* ' → 1
( → 1  ) → 1  * → 3* + → 2  , → 1  - → 1  . → 1  / → 1
0 → 1* 1 → 1* 2 → 1  3 → 1  4 → 2  5 → 2  6 → 1  7 → 2*
8 → 1* 9 → 1  : → 2  ; → 2  < → 1  = → 2  > → 1  ? → 2
@ → 1  A → 3* B → 2  C → 1  D → 2  E → 3* F → 3  G → 2
H → 3  I → 3* J → 1* K → 2* L → 1  M → 4* N → 3* O → 1
P → 2  Q → 1* R → 2  S → 1  T → 2  U → 1  V → 1  W → 1
X → 2  Y → 2  Z → 2* [ → 1  \ → 1  ] → 1  ^ → 1  _ → 1
` → 1  a → 1  b → 1  c → 1  d → 1* e → 1  f → 2  g → 2*
h → 1  i → 2  j → 2  k → 2  l → 1  m → 1  n → 1  o → 1
p → 1  q → 1  r → 1  s → 1  t → 2  u → 1  v → 1  w → 1
x → 2  y → 2  z → 2* { → 1  | → 1  } → 1  ~ → 1


  • Every character outside of U+0021 – U+007E falls in don't care situation.
  • 1
    \$\begingroup\$ There's no requirement for the *s, because people have to follow your specification anyway. If there's a way to solve this beyond simple compression, then this challenge will be good. \$\endgroup\$ – Razetime Oct 15 '20 at 6:29
  • \$\begingroup\$ Can we take input as an ASCII codepoint? \$\endgroup\$ – pxeger Oct 15 '20 at 16:11
  • \$\begingroup\$ Also can I just say how the hell do you write Q with only one stroke? \$\endgroup\$ – pxeger Oct 15 '20 at 16:17
  • \$\begingroup\$ Does "don't care" mean "assume we won't be given this" or "it doesn't matter what you output"? \$\endgroup\$ – pxeger Oct 15 '20 at 16:17
  • \$\begingroup\$ @pxeger ASCII codepoint is acceptable. "Don't care" means both. For the Q, the tail bisects the bowl, so it can be written in one stroke. \$\endgroup\$ – Dannyu NDos Oct 15 '20 at 21:06

Position my geohashes

The Challenge

This is the reverse challenge of Geohash my positions. Given a Geohash string of length 8, convert it to a latitude and a longitude. The conversion is done by the following algorithm, using u09tunqu as an example input.

  • For each character of the Geohash string, find its 0-indexed position in the map 0123456789bcdefghjkmnpqrstuvwxyz.
    • u09tunqu becomes 26 0 9 25 26 20 22 26
  • Convert each integer into a binary string of length 5.
    • 26 0 9 25 26 20 22 26 becomes 11010 00000 01001 11001 11010 10100 10110 11010
  • Join the binary strings together.
    • 11010 00000 01001 11001 11010 10100 10110 11010 becomes 1101000000010011100111010101001011011010
  • Separate the odd positions in the joined binary string from the even positions. These represent the longitude and latitude, respectively.
    • 1101000000010011100111010101001011011010 becomes 10000001101000011011 (odd positions: longitude) and 11000101011111001100 (even positions: latitude).
  • The latitude should be somewhere in the range (-90, 90). Narrow down the range, based on the first character in the latitude binary string. If the first character is 0, the latitude should converge to the lower half of this range, i.e. (-90, 0). If the first character is 1, the latitude should converge to the upper half of this range, i.e. (0, 90).
    • The 1st character in 11000101011111001100 is 1, so the new range becomes (0, 90).
  • The remaining characters in the binary string are to be processed in the same way, where 0 represents the lower half of the new range and 1 represents the upper half of the new range.
    • The 2nd character in 11000101011111001100 is 1, so the new range becomes (45, 90).
    • The 3rd character in 11000101011111001100 is 0, so the new range becomes (45.0, 67.5).
    • The 4th character in 11000101011111001100 is 0, so the new range becomes (45.0, 56.25).
    • The 5th character in 11000101011111001100 is 0, so the new range becomes (45.0, 50.625).
    • The 6th character in 11000101011111001100 is 1, so the new range becomes (47.8125, 50.625).
    • ...
    • The 20th character in 11000101011111001100 is 0, so the final range becomes (48.85826, 48.85843).
  • The final latitude is the midpoint of the final range.
    • (48.85826, 48.85843) becomes 48.85835
  • Repeat the same process for the longitude, starting from the range (-180, 180)
    • 10000001101000011011 becomes 2.29460
  • Output the final latitude and longitude.
    • u09tunqu becomes 48.85835, 2.29460


A string of length 8, consisting only of the characters 0123456789bcdefghjkmnpqrstuvwxyz.


Two signed floats in the ranges (-90.0, 90.0) and (-180.0, 180.0) representing the corresponding latitude and longitude.

Test cases

u09tunqu → (48.85835, 2.2946)
dr5r7p62 → (40.68933, -74.04459)
stq4s8cf → (29.97525, 31.13783)
75cm2txp → (-22.9519, -43.21043)
usdkfsq8 → (71.17089, 25.78302)
zzzzzzzz → (89.99991, 179.99983)
00000000 → (-89.99991, -179.99983)
ezs42s00 → (42.60507, -5.60286)
7zzzzzzz → (-0.00009, -0.00017)

General remarks


AOG Day 6: Filtering the Playlist

I promise the next one will be better

You've sent the invitations for the party (and made the postperson do a whole lot more work than they should have, smh), made the decorations as interesting as possible (who doesn't love a painstakingly written quine) and made sure that the event won't kill anyone (at least, not due to COVID). The next thing that needs to be planned is the music.

Now, of course, you could go ahead and create a YouTube playlist by hand, but that's way too tedious and, well, predictable. Instead, you've decided to write a program that randomly chooses songs from the music genre (I know...very efficient isn't it).

But of course, there's just one problem with that plan: there's a very small chance that a song selected at random might just ruin the party vibes for everyone (even though people such as myself would consider it a Christmas miracle, others would probably see it as a lame stunt and potentially leave the party).

Thankfully, the magic of code allows us to check the html of the YouTube video before hand to tell if it is indeed a rickroll.

The Challenge

Given a YouTube link as input (not shortened, but a full standard link), retrieve the title and description of the video and output whether or not it is a rickroll. In order for a video to be considered a rickroll, it must have either the unbroken phrase Never Gonna Give You Up or Rickroll in the title or description.

Test Cases

Under construction

  • \$\begingroup\$ How about checking the top 5 comments for the unbroken phrase Never Gonna Give You Up or Rickroll as well? Would that make it inconsistent? \$\endgroup\$ – Razetime Nov 12 '20 at 13:25

English Stroke Count Alphabet

In a Chinese glossary/index for any given book, to find terms that are contained within the book and because Chinese doesn't have an alphabet like in English, they are sorted by stroke count instead. (一畫 = 1 stroke,二畫 = 2 strokes,三畫 = 3 strokes,四畫 = 4 strokes,and so on)

An English glossary, having an alphabet, is naturally sorted alphabetically. For this challenge, we flip that idea to the Chinese manner. And we'll follow some Chinese writing rules to help determine stroke order for the alphabet below.

Take 口 (kou) for example, a simple square. You'd think it is 4 strokes, but it is actually 3. The 1st being the left vertical line, the 2nd being the top horizontal and right vertical in one fluid stroke, and the 3rd being the lower horizontal line. This pattern, among others, holds relatively true across Chinese characters. For sake of simplicity though, and for some diversity in the English Stroke Count Alphabet, this will be the primary pattern used.

First, I need to define stroke count for each letter. For sake of simplicity, and somewhat subjectively, I'll use the characters as they appear below. If there are any arguments why a letter should have a different stroke count, please make your case, but in order to promote diversity in stroke counts, I made some personal judgment calls. These stroke counts could easily change with different fonts.


3 3 1 2 4 3 2 3 3 1 3 2 4 3 1 2 2 3 1 2 1 2 4 2 3 3

a b c d e f g h i j k l m n o p q r s t u v w x y z

2 2 1 2 2 2 2 2 3 2 3 2 3 2 1 2 2 2 1 2 2 2 4 2 2 3

Letters with equal stroke counts should retain the original alphabetic order as before. So the English Stroke Order Alphabet is as follows. (If I made an error, please say as much, there are a lot of examples that I might have to adjust)


c o s a b d e f g h j l n p q r t u v x y i k m z w

The Challenge Given a non-empty string input containing a sentence/series of words, or a list of words, organize all words according to this new English Stroke Count alphabet. Output can be either a string, or a list of properly words is a single string of properly organized words, including duplicates should they exist.

Note 1: If upper and lowercase for the same letter have the same stroke count, uppercase letters take precedence.

  • "Cousin" precedes "cousin"
  • "father" precedes "Father" (because lowercase f is 2 strokes, while the uppercase is 3)
  • "Stop" precedes "soap" (while the o would precede t in stroke count, uppercase S precedes lowercase s)
  • KO precedes kO (K precedes k)
  • kO precedes ko (O precedes o)

Note 2: I've intentionally avoided weird words in input. Inputs such as "WeIrD", "COVID-19". Input will never include any numbers, punctuation, or special characters.

Input / Output

"It was the best of times it was the worst of tImes" / "of of best the the tImes times It it worst was was"

["When", "life", "gives", "you", "lemons", "make", "Lemonade"] / ["gives", "Lemonade", "lemons", "life", "you", "When", "make"]

[The, journey, of, a, thousand, miles, begins, with, one, step,] / [of, one, step, The, a, begins, journey, thousand, miles, with]

"English Stroke Count Alphabet" / "Count Stroke Alphabet English"

"A man a plan a canal panama" / "canal a a panama plan A man"

"Carry on my wayward son" / "Carry on son my wayward"

"Close our store and begin destroying every flower green house just lose no people quietly rather than using vexing xrays yesterday it killed my zoo wombat" / Same as input (If you can write a better sentence than above, I'd be much appreciated.)

["May", "the", "Force", "be", "with", "you"] / ["be", "the", "you", "Force", "May", "with"]

[Im, going, to, make, him, an, offer, he, cant, refuse] / [cant, offer, an, going, he, him, refuse, to, Im, make]

"jello Jello JellO JEllo JELlo JELlO JELLO" / "JellO Jello JELLO JELlO JELlo JEllo jello"

"We suffer more often In imagination than IN reality" / "often suffer reality than In IN imagination more We"

"Code Golf and Coding Challenges" / "Code Coding Challenges Golf and"

["Do", "or", "DO", "not", "there", "is", "no", "try"] / ["or", "DO", "Do", "no", "not", "there", "try", "is"]

"Failure the best teacher is" / "best teacher the Failure is"

"Can you tell that I am a Star Wars fan" / "Can Star a am fan tell that you I Wars"

[enough examples no more words] / [enough examples no more words]

  • \$\begingroup\$ If I is 3, then surely i is 4, no? Similarly for J vs j. \$\endgroup\$ – Adám Nov 2 '20 at 21:32
  • \$\begingroup\$ You should be more lenient about input. E.g. allow a list of words. \$\endgroup\$ – Adám Nov 2 '20 at 21:34
  • \$\begingroup\$ "Uppercase letters take precedence when determining stroke counts, should they be equal." means there's just a single case-sensitive alphabet. It'd be more interesting if uppercase matching lowercase on stroke counts would only be used as a tie breaker. \$\endgroup\$ – Adám Nov 2 '20 at 21:37
  • \$\begingroup\$ @Adám As per point 1, I can see the argument. I can make that change, but it'll likely take a second to get all of the examples in line. Point 2, that input format seems adequate. Point 3, I think I know what you mean. And I'm pretty certain that that's what I was intending, but I had poor wording. \$\endgroup\$ – Sumner18 Nov 2 '20 at 21:49
  • \$\begingroup\$ No rush. I recommend sandboxing for at least one week. \$\endgroup\$ – Adám Nov 2 '20 at 21:52
  • 1
    \$\begingroup\$ You should add KD vs kO and kO vs ko to the examples for Note 1. \$\endgroup\$ – Adám Nov 2 '20 at 21:53
  • \$\begingroup\$ I'm not sure what you mean by "that input format seems adequate", but please familiarise yourself with this. \$\endgroup\$ – Adám Nov 2 '20 at 21:55
  • \$\begingroup\$ @Adám I'd like to add one thing about how stroke count works in Mandarin. 口 (kou) in Mandarin appears to be a square, and you'd think it has 4 strokes, but it actually has 3. The 1st stroke is the left vertical line, the 2nd stroke is the top horizontal and right vertical line, and the 3rd and final stroke is the bottom horizontal line. These patterns hold fairly true across Chinese characters. In the case of the letter i, I actually see 3 strokes instead of your suggested 4. The 1st being the slight horizontal tick and vertical line, the 2nd being the bottom line, and the 3rd for the dot. \$\endgroup\$ – Sumner18 Nov 2 '20 at 21:57
  • 1
    \$\begingroup\$ I see, but then J should be 1, no? 一丿 \$\endgroup\$ – Adám Nov 2 '20 at 21:59
  • \$\begingroup\$ @Adám Correct, I wasn't necessarily thinking of that when I made the challenge, but I'll add the explanation and make the edit. \$\endgroup\$ – Sumner18 Nov 2 '20 at 22:02

Make a die of given number of faces


Given an integer \$n\$ greater than 3, identify an \$n\$-sided die with the "greatest" symmetry, then decompose \$n\$ to numbers of faces grouped up to the symmetry, and then output the decomposition.


A die can have one of the following symmetries, from the greatest and with descending order:

  • \$I_h\$, icosahedral symmetry

  • \$O_h\$, octahedral symmetry

  • \$T_d\$, tetrahedral symmetry

  • \$D_{ph}\$, \$p\$-fold prismatic symmetry, in ascending order on \$p\$, where \$p\$ is an odd prime number

Note that every other symmetry is redundant.


A die of each symmetry can have the following faces:

  • For a die of symmetry \$I_h\$:

  • For a die of symmetry \$O_h\$:

  • For a die of symmetry \$T_d\$:

  • For a die of symmetry \$D_{ph}\$:

    • Optionally \$2\$ faces, those from base faces of a \$p\$-gonal prism

    • Optionally \$p\$ faces, those from side faces of a \$p\$-gonal prism

    • Optionally \$2p\$ faces, those from a \$p\$-gonal bipyramid

    • Zero of more sets \$4p\$ faces, those from a \$2p\$-gonal bipyramid

Note that faces from the Catalan solids that are not mentioned here are redundant.


  • The input and output format doesn't matter. Possible choices of output format include:

    • A list, sorted or unsorted

    • A multiset

  • Invalid inputs fall in don't care situation. Especially, integers that are 3 or less.


  • For \$n=4\$, the die has \$T_d\$ symmetry, so \$n\$ decomposes to \$(4)\$, with the die being a tetrahedron.

  • For \$n=5\$, the die has \$D_{3h}\$ symmetry, so \$n\$ decomposes to \$(2,3)\$, with the die being a triangular prism.

  • For \$n=6\$, the die has \$O_h\$ symmetry, so \$n\$ decomposes to \$(6)\$, with the die being a cube.

  • For \$n=7\$, the die has \$D_{5h}\$ symmetry, so \$n\$ decomposes to \$(2,5)\$, with the die being a pentagonal prism.

  • For \$n=8\$, the die has \$O_h\$ symmetry, so \$n\$ decomposes to \$(8)\$, with the die being an octahedron.

  • For \$n=9\$, the die has \$D_{3h}\$ symmetry, so \$n\$ decomposes to \$(3,6)\$, with the die being a truncated triangular bipyramid. Note that the die won't have \$D_{7h}\$ symmetry because \$D_{3h}\$ is greater.

  • For \$n=10\$, the die has \$T_d\$ symmetry, so \$n\$ decomposes to \$(4,6)\$, with the die being a chamfered tetrahedron. Note that this is different than the usual d10, which is a pentagonal trapezohedron.

  • For \$n=11\$, the die has \$D_{3h}\$ symmetry, so \$n\$ decomposes to \$(2,3,6)\$.

  • For \$n=12\$, the die has \$I_h\$ symmetry, so \$n\$ decomposes to \$(12)\$, with the die being a dodecahedron. Note that due to the greater symmetry, dodecahedron supersedes rhombic dodecahedron and triakis tetrahedron.

  • For \$n=100\$, the die has \$T_d\$ symmetry, so \$n\$ decomposes to \$(4,24,24,24,24)\$. Note that this is different than usual Zocchihedron, which has prismatic symmtery.

Note that, if \$p\$ and \$p+2\$ are twin primes, \$p+2\$ will always decompose to \$(2,p)\$.

Ungolfed solution


This implementation mimics the ReadP parser.

import Control.Monad

type DResult = [([Int], Int)]
type DParser = DResult -> DResult

returnD :: Int -> DResult
returnD n = [([],n)]

pfail :: DParser
pfail _ = []

get :: Int -> DParser
get m results = do
    (ns, n) <- results
    guard (m <= n)
    return (m:ns, n - m)

many :: DParser -> DParser
many p results = let
    results2 = p results
    in case results2 of
        [] -> results
        _  -> results ++ many p results2

optional :: DParser -> DParser
optional p results = p results ++ results

run :: DParser
run = filter ((0==) . snd)

(<++) :: DParser -> DParser -> DParser
(<++) p q results = case p results of
    [] -> q results
    results2 -> results2

decomposeDph :: Int -> DParser
decomposeDph prismFold = run . optional (get 2) . optional (get prismFold) . optional (get (2*prismFold)) . many (get (4*prismFold))

decomposeTd :: DParser
decomposeTd = run . optional (get 4) . optional (get 6) . optional (get 12) . many (get 24)

decomposeOh :: DParser
decomposeOh = run . optional (get 6) . optional (get 8) . optional (get 12) . optional (get 24) . many (get 48)

decomposeIh :: DParser
decomposeIh = run . optional (get 12) . optional (get 20) . optional (get 30) . optional (get 60) . many (get 120)

decomposeDie :: Int -> [Int]
decomposeDie n = fst . head $ foldr (<++) pfail (decomposeIh : decomposeOh : decomposeTd : map decomposeDph [3,5..]) (returnD n)

Topologies on Rational Numbers (WIP)


Construct a subset \$P\$ of \$\mathbb{Q}\$ such that:

  • \$P\$ is neither open nor closed in \$\mathbb{Q}\$ as a subspace of \$\mathbb{R}\$, and

  • \$P\$ is open but not closed in \$\mathbb{Q}\$ as a subspace of \$\mathbb{R}_l\$.

Or, in other words, construct a subset \$P\$ of \$\mathbb{Q}\$ such that:

  • There exists \$p \in P\$ such that, there doesn't exist an open interval \$p \in (a,b) \subset \mathbb{R}\$ such that, \$\mathbb{Q} \cap (a,b) \subset P\$.

  • For every \$p \in P\$, there exists a half-open interval \$p \in [a,b) \subset \mathbb{R}\$ such that, \$\mathbb{Q} \cap [a,b) \subset P\$.

  • There exists \$p \in \mathbb{Q} \setminus P\$ such that, there doesn't exist a half-open interval \$p \in [a,b) \subset \mathbb{R}\$ such that, \$\mathbb{Q} \cap [a,b) \subset \mathbb{Q} \setminus P\$.

Notes and Rules

  • Note that \$P\$ is necessarily infinite, and thus cannot be represented as an associative container. One way of representing \$P\$ is to have a function \$f : \mathbb{Q} → \mathbb{Z}_2\$ that halts for every input, where \$\mathbb{Z}_2\$ is the set of the boolean values. Then \$p \in P\$ shall satisfy iff \$f(p)\$ is true.

  • The representation of \$\mathbb{Q}\$ must be exact. Thus you cannot have floating-point values as an input. Though native rational-number arithmetic will be preferred, you may use two arbitrary-length integers as an input. In this case, the fraction is assumed to be irreducible and to have a positive denominator. Otherwise, the fraction falls in don't care situation.

  • Invalid inputs fall in don't care situation.


An example of such \$P\$ is:

$$ \mathbb{Q} \cap ((0,1) \cup [2,3)) $$

Work in progress due to a trivial example.

  • 1
    \$\begingroup\$ Is there a way you can frame this question without topology? At the moment I think participation would be very low due to the high background knowledge demanded. \$\endgroup\$ – Sisyphus Nov 10 '20 at 9:58

Find the longest streak of Fibonacci numbers on the Ulam spiral

Fibonacci numbers

Fibonacci numbers are a sequence where each element is the sum of the previous two elements. In the original Fibonacci sequence, the first two number are 1. So the sequence goes: 1, 1, 2, 3, 5, 8, 13, 21, .... For the challenge, we will accept any two numbers as the two starting numbers of the series.

Ulam spiral

The Ulam spiral is an arrangement of natural numbers. The spiral goes counter-clockwise and starts with the numbers 1, 2, where the 2 is right of the 1. For this exercise, only the shape of the spiral is relevant.

Ulam spiral


Find the length of the longest streak of generalised Fibonacci numbers (with any two starting numbers) following the Ulam spiral in a given array of integer numbers.


The following 5x5 array has two generalised Fibonacci sequences: one of length 4 (in blue + yellow: 138, 81, 219, 300) and one of length 8 (in green + blue: 24, 57, 81, 138, 219, 357, 576, 933). The answer is thus 8.



  • Your program should at least support arrays up to 65535 * 65535 in size and array elements with values up to 4,294,967,295.
  • Invalid input (non-square arrays, float or negative elements, non-arrays, etc.) may lead to unpredicted output, errors or (un)defined behaviour.
  • Default I/O rules apply and default loopholes are forbidden.

on question 1; see comments below

  • This is , so the shortest answer in bytes wins.

on question 2; see comments below

  • This is , so the fastest answer wins.
  • Fastest code is measured in average user time over 5 different, undisclosed input matrices of sizes 100, 1000 and 10,000, each run 3 times on my late 2013 MacBook Pro with 2,3 GHz quadcore Intel i7 CPU and 16 GB of RAM.

Review questions

  • I plan on publishing this question twice: one time as a codegolf and one time as a fastest-code. See the last section of my question. I think this challenge has interesting but very different optimization strategies for speed and size. Is this be something that would be frowned upon by the CGSE community?
  • Is the challenge clear enough as stated?
  • Should I add more/larger test cases? Or a test case generating Python script?

Isomorphic Modular Arithmetic

For this question, we define \$U(n)\$ as the group consisting of a number below \$n\$ that is coprime to n (1 included but 0 doesn't) and multiplication as group operator.

Your task is to print every integer (so it's an infinite loop, but there should be output during looping) and group it into lines so that:

  1. The line consists of a sequence of sorted number so that the \$U\$ group based on each number is isomorphic
  2. The output itself has to be sorted based on the first number on each line
  3. Every number have to be eventually outputted given enough time.

A pair of groups \$(X,\times_A)\$ and \$(X,\times_B)\$ is called isomorphic if there is a pair of functions \$f : X\to X\$ and \$g:X\to X\$ so that:

\$ f(g(x)) = x = g(f(x)) \text{ (i.e. they are inverses)} \$

\$ f(x) \times_B f(y) = f(x \times_A y) \$

\$ g(x) \times_A g(y) = g(x \times_B y) \$

The shortest code wins.


U(2) consists of only one element 1. U(3) and U(4) is isomorphic because both contains only 2 elements (former 1, 2 and latter 1, 3. 2 is excluded because 2 divides 4) and the table of multiplication is identical aside of replacement of 2 and 3. Less trivially U(8) and U(12) is isomorphic, but not U(5) the elements of U(8) is 1,3,5,7 the elements of U(12) is 1,5,7,11 the elements of U(5) is 1,2,3,4 Look at the table of multiplication:

    U(8)          U(12)          U(5)     
  1 3 5 1       1  5  7 11     1 2 3 4
1 1 3 5 7    1  1  5  7 11   1 1 2 3 4
3 3 1 7 5    5  5  1 11  7   2 2 4 1 3
5 5 7 1 3    7  7 11  1  5   3 3 1 4 2
7 7 5 3 1   11 11  7  5  1   4 4 3 2 1

By replacing 1 <-> 1, 3 <-> 5, 5 <-> 7, 7 <-> 11, the table of multiplication for U(8) and U(12) is identical. U(8) and U(5) is not, as n * n = 1 for any n in U(8), but 22 = 4 and 11 = 1 in U(5). If there is such a pair f : U(8) -> U(5) and g : U(5) -> U(8), then g(22)=g(4) <=> g(2)g(2)=g(4) <=> 1 = g(4) <=> g(1) * g(1) = g(4) <=> g(11) = g(4) <=> f(g(11)) = f(g(4)) <=> 1*1=4 <=> 1=4, which is a contradiction.

Even less obviously U(7) and U(9) are isomorphic, the multiplication table is:

   U(7)           U(9)
1 3 2 5 4 6   1 2 4 5 7 8
3 2 6 1 5 4   2 4 8 1 5 7
2 6 4 3 1 5   4 8 7 2 1 5
5 1 3 4 6 2   5 1 2 7 8 4
4 5 1 6 2 3   7 5 1 8 4 2
6 4 5 2 3 1   8 7 5 4 2 1

Aside of swapping the row and column (has been done) and relabeling, the multiplication table is identical, So, they are symmetric

Output for 2-15

3 4 6
5 10
7 9
  • \$\begingroup\$ Your opening sentence doesn't quite define a group. Do you mean a group generated by the coprime and multiplication? It seems like the most logical choice. \$\endgroup\$ – Wheat Wizard Mod Nov 24 '20 at 14:55
  • \$\begingroup\$ I am starting to doubt that since the result is always isomorphic to the integers under addition, and completely ignores n. Do you maybe mean the group generated by the coprime and multiplication on the cyclic group of order n? This makes more sense, but not a whole ton of sense. I think if you explained the output requirements more directly it would be easier. As it stands I do not understand them at all. \$\endgroup\$ – Wheat Wizard Mod Nov 24 '20 at 15:19
  • \$\begingroup\$ @WheatWizard I mean group generated by coprime and multiplication on the cyclic group of order n \$\endgroup\$ – Xwtek Nov 25 '20 at 10:43
  • \$\begingroup\$ You still haven't actually changed the problem statement. \$\endgroup\$ – Wheat Wizard Mod Nov 28 '20 at 21:04

Use the wrong paradigm

This is just a general idea, I don't know how you would go about scoring it or what the goal would be, but I think it would be fun to see stuff like oop in haskell.


Golfing on a Budget

The Task

Every answer should take a positive number as input, and print/return every number in reverse down to zero (inclusive), in any reasonable format. For example:

10    -> "10 9 8 7 6 5 4 3 2 1 0"
8     -> [8, 7, 6, 5, 4, 3, 2, 1, 0]
1000  -> [["1", "0", "0", "0"], ["9", "9", "9"], ["9", "9", "8"], ...]

The requirements

Every program has $100 to spend.

The cost of each byte is determined by \$2^{t-1}\$, where \$t\$ is the number of times the byte has appeared in previous programs in addition to the current one.

For example, assuming it's the first program, x->x++ would cost $10. Each - and > are $1, and each x and + are $2.

  • 2
    \$\begingroup\$ In some sense, doesn't this become a contest of speed to solution? The earlier I post, the lower my score. It also severely advantages certain languages that don't use ASCII. \$\endgroup\$ – Xcali Dec 3 '20 at 4:17
  • \$\begingroup\$ Also, what's the criteria for winning? What is the reason for the $100? \$\endgroup\$ – Xcali Dec 3 '20 at 4:25
  • \$\begingroup\$ @Xcali This is currently a very rough idea, so I didn't have any winning criterion decided on yet. What I was trying to do was make it so that it would be harder to come up with valid answers as time went on, but I think the execution ended up really bad here. \$\endgroup\$ – Redwolf Programs Dec 3 '20 at 4:27

Perfect radicals

  • Posted

Perfect radicals


Posted to the main forum

  • 1
    \$\begingroup\$ We do have occasional fastest code challenges, and many of them get great answers. For objective scoring, (unfortunately) it is necessary to score it on your machine because the specifics of OS/CPU/RAM/storage device/whatever can change the relative run time of two programs. One suggestion is to use smaller test cases (reduce board size and number of ships) so that programs can be tested without waiting for hours. \$\endgroup\$ – Bubbler Dec 1 '20 at 10:01
  • \$\begingroup\$ I can, for instance. set up a Linux virtual machine for the tests. And yes, it is possible to parameterise so as to have a scale of complexity: highest complexity gives authoritative score but lower ones used for weeding. \$\endgroup\$ – user46773 Dec 1 '20 at 15:07
  • 1
    \$\begingroup\$ I'd suggest posting a valid board as the example and then a separate invalid one to illustrate that point. \$\endgroup\$ – Xcali Dec 3 '20 at 4:21
  • \$\begingroup\$ @Xcali - good point. I will do that. \$\endgroup\$ – user46773 Dec 3 '20 at 13:01

Play the percussion for Ravel's Boléro


The Boléro by Maurice Ravel is a piece of music characterised by a consistent snare drum pattern, repeated over and over, at unchanging tempo. This task is very challenging for a human performer, who must maintain the beat without rushing the tempo, or increasing the volume too quickly.

That's why you're going to write a program to do it!

I want your program to do something at the tempo and rhythm specified. It can output a key to a console, produce a beep, make a window pop up, have a graphical display flash colour; anything at all is acceptable, as long as it is A) observable to a human, and B) does not need human interaction for the program to continue. So if you make a pop-up, you should not require someone to click it away before the next pop-up can appear.

Using a tempo of 60 bpm (a bit slow but Ravel liked it played slowly), the whole rhythm takes exactly 6 seconds to play through. Here are the timings for when your program must send out some kind of signal, the unit being seconds. I'll tolerate divergences from the ideal of up to 0.02 seconds in either direction.


Note that here the final beat, at exactly 6 seconds, is the first beat of the next iteration of the rhythm; the next percussion hit occurs at 6.50, then 6.67, and 6.83, etcetera.

So, make your program produce that rhythm indefinitely. Choose whatever output you want. It's , so the shortest code wins!

For the sandboxers; I have considered using either this question as is, or having it be a popularity contest with added the stipulations that it ends after 170 repetitions and that the output somehow gradually gets intenser throughout that time, like the music does. The con is that popularity contests are a bit messier; but I would love to see creative ways to display the rhythm.

  • 1
    \$\begingroup\$ You should add an explanation for people who cannot read music; not everyone knows what a triplet is, or what bpm means. \$\endgroup\$ – Robin Ryder Dec 6 '20 at 13:50
  • \$\begingroup\$ @RobinRyder No knowledge of sheet music is needed; the picture is just illustrative. The list of timings is accurate. \$\endgroup\$ – KeizerHarm Dec 6 '20 at 15:29
  • \$\begingroup\$ The list of timings is accurate at 60 bpm, but not at other tempos. \$\endgroup\$ – Robin Ryder Dec 6 '20 at 16:16
  • 1
    \$\begingroup\$ @RobinRyder Very well, I will make 60 bpm the only accepted tempo and just have the timings be the final specification. \$\endgroup\$ – KeizerHarm Dec 6 '20 at 20:09
96 97
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