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3287 Answers 3287

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Modular distance

You are given 3 non-negative integers: the domain d, the beginning index b, and the ending index e.

What is a modular distance?

Assume d=5 here. First, generate a range from 0 to 5-1:

0 1 2 3 4

We start from the beginning index. Assuming that is 3:

0 1 2 3 4

We continually go right, circling every number we've passed, until we met the ending index e.

0 1 2 O O

If the pointer is at the right end, it wraps around to the left.

Assuming e=0:

O 1 2 O O

We filter out every item we've circled:

0 3 4

Then, find how many items there are in this list:


Subtract it by 1 and it's our result:



  • You can always assume that b<d and e<d.

Test cases

6 2 5 -> 3
5 3 0 -> 2
  • \$\begingroup\$ Do you mean to say "we filter out every item we haven't circled"? \$\endgroup\$ – lyxal May 5 '20 at 8:30
  • 2
    \$\begingroup\$ Is the modular distance just \$ (e - b) \bmod d\$ ? \$\endgroup\$ – dingledooper May 5 '20 at 20:57

Parse vietnamese infinite decimal notation

I wanted to express infinite decimals in text, but overlines are hard.

You need to take a decimal in vietnamese notation, and output the first 10 or more digits of the normal variant.

The notation

The way it works is that you have 0.ab(cd) and it means 0.abcdcdcd.... Of course, you can have any amount of digits in each spot, even zero. You can also omit the infinite part to represent finite decimals.


It's allowed to not accept 0.2 or 0.2() as input, and it's also allowed to output 0.2000000000 if you do accept them as input.

  • 1
    \$\begingroup\$ Can we output the variant infinitely, instead of outputting the first 10 digits? \$\endgroup\$ – user92069 May 6 '20 at 6:52
  • \$\begingroup\$ An existing keyboard-friendly notation is 0.ab(cd) (Wikipedia reference). \$\endgroup\$ – Bubbler May 6 '20 at 6:57
  • \$\begingroup\$ @Λ̸̸ Sure. I'll edit the question. \$\endgroup\$ – PkmnQ May 6 '20 at 8:22
  • \$\begingroup\$ So is the challenge just to split at the ( and then append the first part to the stuff in the brackets repeated 10 times? \$\endgroup\$ – math junkie May 6 '20 at 16:21
  • \$\begingroup\$ Is this a good time to post the question? \$\endgroup\$ – PkmnQ May 10 '20 at 9:22

Arithmetic Square

Note: Credit goes to CCC 2019 S3 for the problem

You are given a \$ 3 \times 3 \$ grid which contains integers. Some of the \$ 9 \$ elements in the grid already have a value, and some of them remain unknown.

Your task is to fill in values for the unknown elements such that for each row, when read left-to-right, produces an arithmetic sequence, and that for each column, when read top-to-bottom, is also an arithmetic sequence.

Recall that an arithmetic sequence of length \$ 3 \$ is a sequence of integers in the form

$$ a, a + d, a + 2d $$

for integer values of \$ a \$ and \$ d \$. Note that \$ d \$ may be any integer, including zero and negatives.

Input Specification

  • You may input the \$ 3 \times 3 \$ grid in any sensible format
  • The unknown values may be represented by any character, so long that it is not a number (i.e. \$ 0-9 \$)

Output Specification

  • The output must be in the same format as the input, with the exception of unknown values becoming integers
  • All rows and columns must form arithmetic sequences
  • There is guaranteed to be at least one solution, and you may output any of them

Test Cases

(This is the only solution)
 8  9 10       8  9 10
16  X 20  ->  16 18 20
24  X 30      24 27 30

(This is one of many solutions)
14  X  X      14 20 26
 X  X 18  ->  18 18 18
 X 16  X      22 16 10

(This is the only solution)
 X -1 -2       0 -1 -2
 5  X  3  ->   5  4  3
 X  X  X      10  9  8

(This is one of many solutions)
 X  X  X       0  0  0
 X  X  X  ->   0  0  0
 X  X  X       0  0  0

This is , so the shortest code in bytes wins!


Generate a "Poem"

Given a strictly positive integer, N, produce an output satisfying the following:

  • Produce an array of length N.
  • Every string (i.e. "word") in the array is of length N.
  • Every letter in the word is unique.
  • Every first letter of the words are unique between each other.
  • The remaining items of each word are equal to each other.

Example output

For an input of e.g. 3:



  • Trailing whitespace is totally allowed.
  • The "letters" don't have to be from the lowercase alphabet, as long as they aren't whitespace.
  • The maximum N you need to support is 13, since there are 26 letters in the lowercase alphabet.
  • The separator of your array can be anything, as long as you will never involve that character for every possible input from 1 to 13. You can also just output a literal array.
  • 1
    \$\begingroup\$ is there a maximum N we need to support? \$\endgroup\$ – lyxal May 12 '20 at 0:56
  • \$\begingroup\$ Yes, the maximum N here is 13. \$\endgroup\$ – user92069 May 15 '20 at 10:15

Is this a Freeman Dyson Number?


From this Popular Mechanics article

One day, in a gathering of top scientists, one of them wondered out loud whether there exists an integer that you could exactly double by moving its last digit to its front. For instance, 265 would satisfy this if 526 were its exact double – which it isn’t. After apparently just five seconds, Dyson responded, “Of course there is, but the smallest such number has 18 digits.”

Challenge Write a program that, when given a base ten number that is at least 18 digits long, moves the last digit to the front and checks if it is doubled as a result.

Input can be any 18 (or longer) digit integer. Any leading digit must be larger than zero.

The original number with the Dyson transform (last digit moved to the front) and any truthy/falsey value (if that's a digit, it must have a delimiter).

Test Cases/Sample I/O

111111111111111111 -> 111111111111111111,false
100000000000000002 -> 210000000000000000 **F**
123456789123456789 -> [912345678912345678,0]
42105263157894736842 -> 24210526315789473684👎
808080808080808080808080808016 - 680808080808080808080808080801-NO
246802468024680246802468024680246802 -> false224680246802468024680246802468024680
105263157894736842 -> true,210526315789473684
315789473684210526 -> (T:5315789473684210526)
26315789473684210526315789473684210 -> 52631578947368421052631578947368421👍


, so shortest answer in bytes (by language) wins.

  • \$\begingroup\$ I would specify that you are talking about decimal digits. \$\endgroup\$ – Jonathan Frech Mar 22 '20 at 17:46
  • \$\begingroup\$ @JonathanFrech, Do you mean base 10? \$\endgroup\$ – ouflak Mar 22 '20 at 19:32
  • \$\begingroup\$ I think one issue here is how to verify that the specific action of moving the digit from back-to-front, and then subsequently checking for doubling, actually happened. Not sure how to get around that. \$\endgroup\$ – ouflak Mar 22 '20 at 20:10
  • \$\begingroup\$ Yes, you should specify that this is in base 10. \$\endgroup\$ – S.S. Anne Mar 22 '20 at 20:46
  • \$\begingroup\$ @ouflak Yes, I mean base ten. One often hears for example "binary digits", so the term "digits" is in my opinion not clearly defined to mean base ten. \$\endgroup\$ – Jonathan Frech Mar 22 '20 at 21:55
  • \$\begingroup\$ @JonathanFrech, @ S.S. Anne, The reason why I haven't immediately made the change is because I hadn't considered the idea of different number bases, and I'm really liking the idea of a challenge that in fact does include either various number bases, or a specific challenge for binary and this separate challenge for base ten. Mulling it over now. This would mean I'd have to figure out some binary test cases.... \$\endgroup\$ – ouflak Mar 23 '20 at 6:44
  • 1
    \$\begingroup\$ In binary doubling a number is adding 0 to the end of it, so unless you allow leading 0s it's not possible, otherwise it's correct iff the number starts with a 0 \$\endgroup\$ – Command Master Mar 24 '20 at 19:43
  • \$\begingroup\$ @CommandMaster, Yes! For binary, you would have to allow, even implicitly, a leading zero. The most obvious example is '1', which is really '01', which when the '1' is moved to the front becomes '10'. Don't see how you can get around that. It would be a different challenge. The number base thing has got me thinking. \$\endgroup\$ – ouflak Mar 25 '20 at 8:35
  • 2
    \$\begingroup\$ Here are some suggestions: Make it a decision-problem (e.g. returning the Dyson transform and the Truthy/Falsey value is a bit unnecessary). Keep the sample IO consistent (I get that you want to show the variation in possible output formats, but it would be easier to verify cases if the format were consistent). \$\endgroup\$ – dingledooper May 12 '20 at 19:55


Tile the plane with squashed hexagons


Compress a grandmaster chess position


Compress a position from a grandmaster chess game to as few bits as possible on average. A strong submission will probably use that these positions come from real games by top players, and so will make chess sense and strategic sense, rather than just being random legal chess positions. As illustration, a study found that grandmasters do well at memorizing positions from real games using "chunking" but with only perform at novice level memorizing random boards.

The is related to but different from Smallest chess board compression, which scores on the worst-case scenario, and Smallest chess game compression, which compresses full games. (Sandbox: Let me know if this is too similar)


You must write a compressor, which maps a chess position onto a sequence of bits, and a decompressor that returns its to the original position. You can vary the length of the bit sequence by position, and this will likely be important to getting a good score.

The position to compress will just be a the placement of pieces on the chess board. You do not to encode whose move it is, castling rights, or en-passant. It will be given in FEN string format with only the piece placement part, for example:


The chess position for this FEN

Each letter corresponds to a piece (pawn="P", knight="N", bishop="B", rook="R", queen="Q", and king="K"). White's pieces use uppercase letters and Black's are lowercase. Slashes separate the descriptions of each of the rows from top to bottom, that is the 8 files doing from 8 (where black's pieces start) to down to 1. Numbers are used for blocks of that many empty spaces that are horizontally adjacent.


You will be scored on the average length of your compressed bit sequence on 10,000 random game positions. They will chosen at random from games played by grandmasters, restricted to move 5 or later. [Will work out more details when generating this data.]

This Pastebin (TODO) contains 10,000 FEN strings to use as a training set that you can use to get a preliminary score. The final score will be based on a separate secret test set of 10,000 FEN strings.

Your code must correctly decode every game in the position. Be sure that it can handle all positions, such as ones with weird underpromotions, which might appear in the test set but not the training set. (Sandbox: How to handle submissions that break this? A default penalty score for games failed? Ask to resubmit?)

Your compression and decompression must complete within 5 minutes on all the games. (Sandbox: Allow to compress all games at once? Do one game at a time but store state to allow "learning"? Include a memory limit?)

The length of your code is immaterial to this challenge.

  • \$\begingroup\$ How long (in number of positions) would a naive program that hardcodes all of the (recorded) existing grandmasters games be? (if that's not large enough, it would make the challenge trivial) \$\endgroup\$ – user202729 May 17 '20 at 11:49
  • \$\begingroup\$ @user202729 That's a good question, I definitely don't want code to be use that the test set comes from an actual database, so I'd either needed to make that non-viable or ban it. \$\endgroup\$ – xnor May 17 '20 at 12:31
  • \$\begingroup\$ Does the time requirement have to be as strict as 30 games per second? (a certain question of the part also mentions "every game in the position", I assume that's a mistake) (I have no idea how to prevent storing a database of grandmaster games though) \$\endgroup\$ – the default. May 17 '20 at 16:00
  • \$\begingroup\$ @myp I think that this requirement means that it can compress each game in 5 minutes. \$\endgroup\$ – user202729 May 18 '20 at 0:50
  • \$\begingroup\$ @user202729 I had meant total for all the games, but I'll probably loosen it. \$\endgroup\$ – xnor May 18 '20 at 2:19

Permutation primes

A permutation prime is a prime such that at least one of its uniquified permutations (not equal to itself) of its digits is a prime.

Given a number, check if this number is a permutation prime.

Reference program

Here is a reference program I made.


The golfing skills are strong with this one


Consider the base string s = "The golfing skills are strong with this one", an adaptation of the quote "The force is strong with this one" by Darth Vader, an infamous character of the Star Wars saga (sandbox, am I correct?).

You have to output the string s with as many characters as there are bytes in your source code. If your code is longer than s, extend s by concatenating it repeatedly as many times as needed.

Your program must be non-empty.


You may or may not take the string s as input for your program. (Sandbox, maybe it is more interesting to not allow the string as input?)


A string as specified in the Task.

  • 1
    \$\begingroup\$ Many languages, 0 bytes. PHP and /// polyglot, 1 byte: T. If the code is too long, is the string really intended to be The golfing skills are strong with this oneThe golfing skills are strong with this oneThe golfing skills are strong with this one? \$\endgroup\$ – the default. May 20 '20 at 16:42
  • \$\begingroup\$ @mypronounismonicareinstate do you see a problem with the string being like that? What would you suggest? Also, probably should not allow the string to be used as input and require a non-empty program \$\endgroup\$ – RGS May 20 '20 at 16:46
  • 1
    \$\begingroup\$ That is perfectly acceptable, it's just that the transitions aren't very smooth (oneThe). If you can take the string as input, solutions won't have to depend on this specific string, so it would probably be a bad idea. (I mean, a[:5] isn't a very interesting answer) \$\endgroup\$ – the default. May 20 '20 at 16:51
  • \$\begingroup\$ @mypronounismonicareinstate I failed to understand what variation you think is a bad idea. Do you think it is a bad idea to accept it as input or a bad idea to not accept it as input? \$\endgroup\$ – RGS May 20 '20 at 16:52
  • \$\begingroup\$ I think it is a bad idea to accept it as input. \$\endgroup\$ – the default. May 20 '20 at 16:53

Halting problem for simplified Brainfuck

Given a simplified Brainfuck program, you must determine whether it halts. Your program must always halt in finite time on valid inputs.

Simplified Brainfuck is a language that operates on a zero-initialized tape that is infinite in both directions. All cells contain integers from 0 to 255, and operations are performed modulo 256. There are the following instructions:

+ increment the current cell
- decrement the current cell
< move 1 cell to the left along the tape
> move 1 cell to the right along the tape
[ if the current cell is zero, skip past the next ]
] go to the previous [

Loops ([]) can't be nested.

This is tagged , so the shortest answer wins.

  • \$\begingroup\$ Do you mean "Given a simplified Brainfuck program and an input of such program"? \$\endgroup\$ – Domenico Modica May 21 '20 at 5:00
  • \$\begingroup\$ @DomenicoModica No, this language has no IO. Do you think I should mention that explicitly? \$\endgroup\$ – the default. May 21 '20 at 5:02
  • \$\begingroup\$ Oh, I don't know, I was too hasty ahahah... Anyway If the tape was finite surely it would be solvable \$\endgroup\$ – Domenico Modica May 21 '20 at 5:07
  • \$\begingroup\$ I think it is indeed solvable with doubly infinite tape, since the region that the pointer touches within an iteration of a loop is limited (which means we have finite number of states in that region). It's pretty hard to describe the algorithm though. \$\endgroup\$ – Bubbler May 21 '20 at 5:56
  • \$\begingroup\$ That is what I had in mind. Handling two loops in different directions is also non-trivial. \$\endgroup\$ – the default. May 21 '20 at 6:00
  • \$\begingroup\$ I don't think handling multiple loops is that non-trivial. Consider first loop first, the answer is false if it is infinite loop, otherwise run it to the end and run all commands before the second loop. Then consider the second loop just like the first, etc. \$\endgroup\$ – Bubbler May 21 '20 at 6:49
  • 1
    \$\begingroup\$ Actually I think the challenge is not very interesting as[code-golf], because it's necessary to simulate the algorithm anyway and it can be proven (I think) that the number of steps the program takes (if it halts) is no more than \$2^{2^{2^{2^n}}}\$ (where n is the program length), so it will be no longer than an interpreter but takes impractically long to run. \$\endgroup\$ – user202729 May 22 '20 at 8:48
  • \$\begingroup\$ @user202729 Is there proof of an upper bound of time? I feel it's unsolvable \$\endgroup\$ – l4m2 May 27 '20 at 1:38
  • 1
    \$\begingroup\$ @l4m2 This is solvable because loops can't be nested. The body of each loops moves by a constant number of steps X, and if the program doesn't halt than it'll either repeat the same state twice (if X==0) or crosses the bound of the written tape part (because there's only a fixed number of written cells and then repeat states (there's only a finite number of cells touched by the loop body) \$\endgroup\$ – user202729 May 27 '20 at 4:07

Posted: Stepping Through Time


Quickly calculate \$ n! \bmod p \$

The idea is extremely simple: Given two positive integers \$ n \$ and \$ p \$, calculate the result of \$ n! \bmod p \$, where \$ p \$ is a prime.


Your score is the highest \$ p \$ you can achieve within \$ 10 \$ seconds, by running the program \$ 10 \$ separate times. More specifically, each run-through will contain two inputs \$ n \$ and \$ p \$. You are to solve \$ n! \bmod p \$, where \$ n \$ is a random number in the range \$[1, p]\$.

You must use this program to generate the \$ 10 \$ test cases. So for example, if \$ p = 13 \$, the test case would look like this:

n, p
9, 13
3, 13
10, 13
13, 13
7, 13
13, 13
8, 13
9, 13
6, 13
4, 13


  • Make sure that each test case is run separately, meaning you are not allowed to make use of previous test cases
  • Multi-threading is disallowed
  • Official times will be tested on my machine; make sure to include specifcations on how to run it

This is , so the highest score wins!


  • Any loopholes that need to be addressed?
  • Is there an easy, trivial solution to this?
  • 1
    \$\begingroup\$ To force independent calculation you should invoke the program 10 separate times. Although it remains possible to store data in files or similar, it would be pretty obvious. \$\endgroup\$ – user202729 May 22 '20 at 8:58
  • 1
    \$\begingroup\$ As far as I know the best-known time complexity is asymptotically \$ \widetilde O (\sqrt p)\$, although the implementation is rather tedious and uninteresting. \$\endgroup\$ – user202729 May 22 '20 at 9:00
  • 1
    \$\begingroup\$ Your example test case has p=10 which isn't prime \$\endgroup\$ – xnor May 22 '20 at 10:31
  • \$\begingroup\$ @xnor Fixed, thanks. \$\endgroup\$ – dingledooper May 22 '20 at 15:53

Cheat activated


The game Grand Theft Auto: San Andreas went down to history also thanks to its wide selection of cheats. They're almost 90 and anyone who has ever touched this game, no doubt he tried them all!
One cheat is activated (on PC) typing in-game a secret keyword, and then boom, a jet pops out of thin air or perhaps all pedestrians look like Elvis Presley or some other rowdy effect...

They always come with this confirmation message:

enter image description here

Rockstar choosed to store them hashed, so due to collision, in addition to the intended ones there are many other strings that trigger every cheat.

Therefore I propose to solve this downside!


Write a full program that prints CHEAT ACTIVATED if and only if the last part of a string is a cheat code.

Cheat codes



  • A string \$s\$ over the alphabet:
    [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]


  • Print CHEAT ACTIVATED if there exist a cheat code \$c\$ such that \$c\$ is a suffix of \$s\$
  • Nothing otherwise

This is , so the shortest code wins.

  • \$\begingroup\$ Do we have to output the specific Cheat activated string, or just a truthy/falsy value indicating whether or not a valid cheat code exists? (I also don't think \$i\$ is a good name for a string; is that intentional?) \$\endgroup\$ – the default. May 21 '20 at 4:27
  • \$\begingroup\$ @my pronoun is monicareinstate Yes, that specific string, it's a little simulation. And yes, you're right \$s\$ is the canonical name, I choose "i" for input, but never mind \$\endgroup\$ – Domenico Modica May 21 '20 at 4:39
  • 1
    \$\begingroup\$ I think that restricting the output to Cheat activated is unnecessary since it doesn't really add anything to the challenge. Consider allowing just a truthy or falsey value as output. \$\endgroup\$ – math junkie May 21 '20 at 19:01
  • \$\begingroup\$ @math junkie I'm aware that in terms of logic it's useless. But that comes if only the "recognition part" is considered being the intended challenge. From my idea, generating Cheat activated it's also a part of it... I worded it badly in the task part \$\endgroup\$ – Domenico Modica May 21 '20 at 20:08
  • 1
    \$\begingroup\$ 18HOLES is in the cheat codes list, yet the input states that there will be no number. Is this intended? \$\endgroup\$ – dingledooper May 22 '20 at 21:20
  • \$\begingroup\$ @dingledooper No, my mistake, I thought I had removed it in the last edit \$\endgroup\$ – Domenico Modica May 22 '20 at 23:07

Convert an integer to Chinese numerals

Your task is to convert an integer from 1 to \$10^{52}-1\$ (inclusive).

The characters from 1 to 10 with their Unicode code points are:

一 1 U+4E00
二 2 U+4E8C
三 3 U+4E09
四 4 U+56DB
五 5 U+4E94
六 6 U+516D
七 7 U+4E03
八 8 U+516B
九 9 U+4E5D
十 10 U+5341

Number greater that that are composed like this:

十一 11
十二 12
二十 20
二十一 21
二十二 22
百 100
百一 101
百十 110
百九十九 199
二百 200
九百九十九 999
千 1000
九千九百九十九 9999
一万 10,000

This is where it gets interesting, because numbers bigger than 10,000 are groups in groups of four, expressed with 十, 百 and 千. These are the powers we're going to use in this challenge:

十 10 U+5341
百 100 U+767E
千 1000 U+5343
万 10^4 U+4E07
億 10^8 U+5104
兆 10^12 U+5146
京 10^16 U+4EAC
垓 10^20 U+5793
秭 10^24 U+79ED
穣 10^28 U+7A63
溝 10^32 U+6E9D
澗 10^36 U+6F97
正 10^40 U+6B63
載 10^44 U+8F09
極 10^48 U+6975

Let's go through an example with 123456789123456789 as the input (other algorithms are possible)

  • identify groups of four digits, starting from the right: 12,3456,7891,2345,6789
  • convert each group: 十二 三千四百五十六 七千八百九十一 二千三百四十五 六千七百八十九
  • insert the appropriate multipliers: 十二京三千四百五十六兆七千八百九十一億二千三百四十五万六千七百八十九


  • A leading ー MAY be dropped before 千 and 百 and MUST be dropped before 十.

IO format

The input can be an integer in any reasonable format. You can use a string/sequence of characters or a number type, if your language supports it. 128-bit numbers are not large enough, by the way.


input output
1 一
2 二
3 三
4 四
5 五
6 六
7 七
8 八
9 九
10 十
15 十五
20 二十
31 三十一
100 百
123 百二十一
1000 千
8346 八千三百四十六
10000 一万
50010 五万十
100000 十万
123456789123456789 十二京三千四百五十六兆七千八百九十一億二千三百四十五万六千七百八十九
1234567891234567891234567891234567891234567891234567 一千二百三十四極五千六百七十八載九千百二十三正四千五百六十七澗八千九百十二溝三千四百五十六穣七千八百九十一禾予二千三百四十五垓六千七百八十九京一千二百三十四兆五千六百七十八億九千百二十三万四千五百六十七

Standard code-golf rules apply. The shortest code in bytes wins.


  • \$\begingroup\$ I assume this is meant to be code-golf? While the tag is technically enough, I think it is better to have a brief inclusion of that in the body of the challenge. \$\endgroup\$ – FryAmTheEggman May 29 '20 at 18:44
  • \$\begingroup\$ @FryAmTheEggman yes it should be codegolf. \$\endgroup\$ – corvus_192 May 29 '20 at 19:41
  • \$\begingroup\$ It'd be helpful if you include the code points for the Unicode characters. \$\endgroup\$ – Surculose Sputum May 29 '20 at 21:13
  • 1
    \$\begingroup\$ As someone who is actively learning Japanese, they're called Chinese numerals because they were taken from China's numeral system. Japanese numerals are a whole different, extremely complicated thing. \$\endgroup\$ – Ethan Slota May 29 '20 at 21:51
  • \$\begingroup\$ "A leading ー MAY be dropped before 千 and 百". They MUST be dropped in Sino-Korean numerals, tho. \$\endgroup\$ – Dannyu NDos Jun 1 '20 at 4:07
  • \$\begingroup\$ What about 恒河沙, 阿僧祇, 那由他, 不可思議, and 無量大數? \$\endgroup\$ – Dannyu NDos Jun 1 '20 at 4:12
  • \$\begingroup\$ Mathematica has IntererName[#, "Words", Language -> "Chinese"]&, but, unfortunately, it can't handle numbers this large :(. \$\endgroup\$ – the default. Jun 2 '20 at 11:55
  • \$\begingroup\$ Can we take input in base 10000? (Usually done when doing big integer multiplication with int32 so seems somehow reasonable) \$\endgroup\$ – l4m2 Jun 19 '20 at 4:26

The Double-Castle Numbers™


Divide into 2 isosceles triangles

Given the measures of two of the interior angles of a triangle (x and y; the other angle can be easily calculated with 180 - x - y), draw a line segment that cuts this triangle into two isosceles triangles. You need to output the angle measures of both of your triangles.

However, because the base angles are the same, you only need to output the list [apex angle, base angle] of the divided triangles for both of the isosceles triangles. You can output the divided triangles in any order.

An example

Say your input is 100, 60.

Let's take a look at the complete triangle first. The triangle looks approximately like this.


60            20

Now we try to divide one of the angles such that two divided triangles are both isosceles triangles.


(40,20)           20

Now our bottom triangle is an isosceles triangle, since both of the base angles
of the bottom triangle are 20. The angle measures of the bottom triangle
looks approximately like this.

20             20

Now, is the top triangle an isosceles triangle?


It is an isosceles triangle, because two of the angle measures are 40.

Therefore, for [100, 60], you need to output [[100, 40], [140, 20]].

Example cases

[20, 40] -> [[140, 20], [120, 40]]
[45, 45] -> [[90, 45], [90, 45]]
[36, 72] -> [[72, 36], [36, 72]]
[108, 36] -> [[108, 36], [36, 72]]

King+queen vs king checkmate

You are given a chess position, represented either in FEN or as a two-dimensional diagram like this (the example test cases will be using the latter format):


In the examples, K represents the white king, Q represents the white queen, k represents the black king and . represents blank space. You may choose different consistent values instead of these characters. You may also input the diagram as a list of lists or in any other way that is allowed by default for two-dimensional arrays.

It is white's move. The position will always be reachable from the starting position by a sequence of valid moves.

You have to find the minimum number of moves White must do to checkmate Black, assuming perfect play by Black.

Test cases

Incomplete: too many test cases for 1 and no test cases for >1.


Output: 1


Output: 1


Output: 1

  • \$\begingroup\$ I know it would be a lot different, but have you considered the more general question that allows any (valid) disposition of the three pieces? Then the task would be to find the minimum number of moves to checkmate... Isn't it a bit "tautological" to input a position of which I already know it only takes one move to checkmate? -I most probably know also what this move is- \$\endgroup\$ – Domenico Modica Jun 8 '20 at 16:02
  • \$\begingroup\$ Moreover if it happens to be only 1 move from checkmate (or also, if you want, if the moves can be all determined), with this broader task you could totally ask what this (these) move is (are). \$\endgroup\$ – Domenico Modica Jun 8 '20 at 16:11
  • 1
    \$\begingroup\$ I think asking for the optimal depth to mate in White moves is a better question (far less simple than this, but still much less complicated than a proper chess engine); I'll change the proposal later; it's late here. \$\endgroup\$ – the default. Jun 8 '20 at 16:30
  • \$\begingroup\$ This is ambitious for code golf! Is the point to build an endgame tablebase? at least as much of it as is needed to solve the given test case positions? Test case: wKa1Qb2 bKf5 WTM wins in 10... \$\endgroup\$ – Rosie F Jul 19 '20 at 19:28

Underfull \hbox (badness 10000)

Every TeX user has been warned many times that their hboxes are terribly underfull or overfull. So much badness! This challenge is to rate how badly underfull or overfull a line of text is for a simplified line wrapper.


You're given a space-separated string or list of words. Output the minimal badness achievable for the first line.

The text needs to be wrapped on a line that's 10 characters wide, but it can only be split on spaces, no in the middle of words. Any letter that spills beyond the width counts for 1000 overfull badness each, and each leftover empty position at the end of the line counts for 1000 underfull badness.


For input "Overfull hbox", we can keep the word "hbox" in the first line for 3000 overfull badness, or wrap it to the second line for 2000 underfull badness which is smaller, so the output is 2000.


Overfull hbox
hbox    ^^

Note that we don't care about badness of the second line.


The input is a space-separated string or a list of words made of letters a-zA-Z. It won't have any words more than 10 letters long, or be more than 20 characters in total. It won't be empty or have any zero-length words.

Test cases


Sandbox: Is it OK to have a multiplier of 1000 for theme? Should the underfull and overfull badness penalties be different, like 1000 vs 2000?

  • \$\begingroup\$ In my opinion, this might be a bit too simple for the big badness theme to be worth it. I feel like most golfing languages might spend nearly half their code multiplying by 1000 (throwing it together in Pyth, I got 1/4 used for multiplying). If it was say, the badness of each of the lines it might feel better. I don't feel terribly strongly about this though. \$\endgroup\$ – FryAmTheEggman Jun 9 '20 at 20:52
  • \$\begingroup\$ @FryAmTheEggman That for the feedback. I'm now thinking the challenge is too simple overall, multiplier or not. What would you think of something like words being able to be broken at certain places in the middle, either explicitly marked or dervied from some property of the letters? \$\endgroup\$ – xnor Jun 9 '20 at 21:14
  • \$\begingroup\$ I think some level of TeX uses hyphens to indicate possible word breaks, but they don't count for the length of words if unused. Adding that may help, while also being on theme? \$\endgroup\$ – FryAmTheEggman Jun 9 '20 at 21:17

Lucky dice rolls

In pen and paper roleplaying games dice are used for various chance calculations. The usual way to describe a roll is \$n\textbf{d}k\$ where \$n\$ is the number of dice and \$k\$ is the number of faces on a die. For example \$3d6\$ means that you need to roll the classical 6-sided die 3 times (or roll 3 dice at the same time). Both \$n\$ and \$k\$ are positive integers.

Usually the values are then summed and they are used for various game mechanics like chance to hit something or damage calculations.

A lucky roll will mean that you have Fortuna's favor on your side (or against you). Luckiness is an integer number that increases (or decreases) the sum in the following way. The roll is modified to \${(n+|luck|)}\textbf{d}{k}\$ and the sum will be the \$n\$ best (or worst) values. Each die is fair, so they will have the same probability for the outcome of the possible values.

The \$luck\$ can be a negative number, in this case you need to get the \$n\$ worst values for the sum.


The integer values for \$n,k,luck\$ in any way.


The expected value for the sum of the (un)lucky roll. The expected value is \$\sum{x_{i} p_{i}}\$ where \$x_{i}\$ is the possible outcome of the sum and \$p_{i}\$ is the probability for \$x_{i}\$ occuring, and \$i\$ indexes all possible outcomes.


n,k,luck    expected value
1,6,0       3.5
2,6,0       7
2,6,-1      5.541666666666667
2,6,1       8.458333333333334
2,10,-1     8.525
2,10,1      13.475


Shortest code in bytes wins.

Good luck! ;)

  • \$\begingroup\$ Probably worth including a definition of expected value. To what precision should the output be determined? \$\endgroup\$ – Dingus Jun 10 '20 at 12:50
  • \$\begingroup\$ @Dingus Is it now a bit more understandable? \$\endgroup\$ – Gábor Fekete Jun 10 '20 at 13:29
  • 1
    \$\begingroup\$ Yes, that's good, though I'd suggest tweaking the wording a bit: 'The expected value is \$\sum x_ip_i\$ where \$x_i\$ is a possible value for the sum, \$p_i\$ is the probability of that sum occurring, and \$i\$ indexes all possible outcomes.' Perhaps I should rephrase my question about precision - what numeric formats are acceptable for output? Floats are obviously allowed, but do you require a certain number of decimal places? Is it acceptable to output rationals (for languages that support this)? What about 2 integers representing numerator and denominator, respectively? \$\endgroup\$ – Dingus Jun 11 '20 at 12:13
  • \$\begingroup\$ hmm, that's a valid point, but I don't know which one it should be. \$\endgroup\$ – Gábor Fekete Jun 11 '20 at 12:19

Iterate diagonally over nxn matrix

Given a matrix of size n, output the matrix into another matrix of size n such that:

  • the outputted matrix, when traversed diagonally,will result in the original matrix.

For example, taking this 3x3 matrix, we arrive at our solution: enter image description here

Which is checked by following the line beginning at 1: enter image description here


  • The matrix will always be square
  • You must output a grid with the same size as you were given (e.g. Not as a triangle)
  • Mark the end of each row with a delimiter such as \n or .


Example 1


1 2 3
4 5 6
7 8 9


1 3 6
2 5 8
4 7 9

We can check the output by iterating over the array diagonally (follow the arrows for steps 1-5), which will give us the original matrix.

  ↗ ↗ ↗
1 ↗ ↗ ↗
2 ↗ ↗ ↗
3  4 5 

Example 2


a b c d
e f g h
i j k l
m n o p


a c f j
b e i m  
d h l o
g k n p

We can check this by iterating the array in steps 1-7 which outputs the given array.

  ↗ ↗ ↗ ↗
1 ↗ ↗ ↗ ↗
2 ↗ ↗ ↗ ↗
3 ↗ ↗ ↗ ↗
4  5 6 7


Looking at the coordinates, we can see a pattern:

(0,0) -> (0, 1) -> (1, 0) -> (0, 2) -> (1, 1) -> (2, 0) -> (1, 2) -> (2, 1) -> (2,2)
  • \$\begingroup\$ Do the matrices always consist of one character per cell? \$\endgroup\$ – Trebor Jun 19 '20 at 0:45
  • \$\begingroup\$ they don't have to, but that can be a specification. Thoughts? \$\endgroup\$ – Peter S Jun 19 '20 at 1:20
  • \$\begingroup\$ Perhaps the title could be a bit more descriptive, like "put the contents of a matrix into its antidiagonals". Then you could add a definition of the antidiagonals, and then a description of how you traverse the matrix to get the ordering for the antidiagonalization. \$\endgroup\$ – Giuseppe Jun 19 '20 at 17:42

I am surely the fastest!... asymptotically


  • 1
    \$\begingroup\$ Slightly related: codegolf.stackexchange.com/questions/182733/… \$\endgroup\$ – the default. Jun 18 '20 at 3:43
  • \$\begingroup\$ @mypronounismonicareinstate The challenge in your link is slightly more difficult than mine, since prime factorization is not NP-complete. (Godel machines are also a completely different solution) \$\endgroup\$ – Trebor Jun 18 '20 at 3:50


  • \$\begingroup\$ It would ease readability if all of the test cases were in a single code block, with empty lines in between. Then you can add the explanations afterward for cases that really need it. \$\endgroup\$ – fireflame241 Jun 23 '20 at 17:43
  • \$\begingroup\$ Is there a limit to the depth of the tree? \$\endgroup\$ – fireflame241 Jun 23 '20 at 17:44
  • \$\begingroup\$ The problem statement involves a bunch of high-level math terms, which can deter some people. If possible, the challenge would be more approachable if you add an alternate definition, e.g. relating isomorphism to permutations (kinda) of the underlying set. For the explanation of the first case, it would be good to rewrite each line using normal infix notation, e.g. (x+y)+z = x+(y+z) and x+(-x) = 0. \$\endgroup\$ – fireflame241 Jun 23 '20 at 17:49
  • \$\begingroup\$ @fireflame241 I hid the technical details. And could you please explain why I should put a limit to the depth? In what ways does that make the challenge better? ;) \$\endgroup\$ – Trebor Jun 24 '20 at 1:25
  • \$\begingroup\$ I was just wondering. A golfer might be able to optimize for a limit depth of 2, but it's more interesting to have an arbitrary rank \$\endgroup\$ – fireflame241 Jun 24 '20 at 1:29
  • \$\begingroup\$ @fireflame241 After some thoughts, it is clear that every theory can be translated to one that has a limit depth of 2 ;) Also every practically interesting case happens at depth 2. So I think I'll add that. \$\endgroup\$ – Trebor Jun 24 '20 at 1:53
  • \$\begingroup\$ Now that this has been posted to main, could you delete this proposal to create more space for new answers? \$\endgroup\$ – caird coinheringaahing Sep 25 '20 at 1:05

Write an expect program

If you're not already familiar, expect is a Tcl extension that makes it easier to script interactions with programs. It allows you to spawn a process, send lines to it, and wait for expected output before continuing.


The aim of this challenge is to write a very simple implementation of expect in as few bytes as possible (code golf). It should parse a script, with commands separated by newlines. Then it should use this script to interact with a program.

Here are the commands for this implementation:

  • spawn <cmd>: spawn a process.
  • write <line>: write a line into the process' input.
  • expect <line>: expect a substring from the process' output. No timeout is necessary, if the line never appears it is OK for the program to hang.
  • print <line>: print something to stdout.

You can assume that only one spawn will be found in the script, and that it will appear before any write or expect. If your language of choice doesn't have the ability to spawn processes, you can write a helper program in a different language that can pipe input and output through your main program. How you do this is left up to you.

Example script:

spawn /bin/bash
write whoami
expect root
write uname -a
expect Linux
print i am root on Linux


this is Linux


spawn /bin/bash
write uname -a
expect Windows
print this is Windows

(no output.)


In order to keep things fresh, the use of the standard expect utility or any libraries that emulate expect functionality (such as pexpect on Python or jest on Node) are not allowed. The idea is that the bulk of the functionality should be written in the program and not handled in a library.

  • \$\begingroup\$ Will the script always have 4 commands, namely those 4 in that order? \$\endgroup\$ – Adám Jun 30 '20 at 3:52
  • 1
    \$\begingroup\$ While prohibiting the standard expect utility is probably unambiguous, prohibiting expect-like libraries could potentially be problematic because there's no objective way to judge if any given feature is expect-like. I could claim that addition is expect-like, and you'd be hard pressed to counter that. \$\endgroup\$ – Adám Jun 30 '20 at 3:55
  • \$\begingroup\$ The script could have more commands. I will update the examples to reflect that change. As far as the library restrictions, I could remove them - I do see where it could lead to being a problem. Perhaps I could make it more unambiguous by mentioning specific libraries (one that comes to mind is python-pexpect.) \$\endgroup\$ – nununoisy Jun 30 '20 at 16:10
  • \$\begingroup\$ Can we make any assumptions about the order of the commands? \$\endgroup\$ – Adám Jun 30 '20 at 18:06
  • \$\begingroup\$ Yes, I forgot to mention that spawn will come before any command that needs the process. It should be updated now. \$\endgroup\$ – nununoisy Jun 30 '20 at 20:23
  • \$\begingroup\$ What should be done if the expect string is not found? \$\endgroup\$ – user202729 Jul 10 '20 at 3:26
  • \$\begingroup\$ It was mentioned above but for clarity: 'No timeout is necessary, if the line never appears it is OK for the program to hang.' \$\endgroup\$ – nununoisy Jul 10 '20 at 19:58

Count faces in ASCII art

Here's a 2x2 ASCII art face:


Here's a 3x3 ASCII art face:

o o

Here's a 4x4 ASCII art face:

o  o

Your task is to count faces in an ASCII art.

Here's something closer to an actual specification.

The bottom of any face must be a contiguous horizontal row of underscores, such that cells to the right and to the left of it do not contain underscores. If the row is considered as the bottom row of an ASCII square, then that square forms a face if and only if its bottom row is all underscores, its upper left and upper right corners are os, and the rest is whitespace.

You may assume all lines in the input to be padded on the right with whitespace to an equal length. Faces cannot be smaller than 2x2.

[todo: more test cases]

o o

Output: 0

Sandbox stuff

Is it clear what is considered a face and what is not?


The almost impossible chessboard puzzle


Prisoner 1 walks in, sees a chessboard (8x8) where each square has a coin on top, flipped either to heads or tails. The warden places the key under one of the squares, which prisoner 1 sees. Before prisoner 1 leaves, he must turn over one and only one coin. Prisoner 2 then walks in and is supposed to be able to figure out which squares the key is in just by looking at the arrangement of coins.

The prisoners are granted a reward if prisoner 2 correctly tells the location of the key.


Write two program/functions:

  • One for prisoner 1, which outputs the location of the coin to flip given the current board state and the location of the key
  • One for prisoner 2, which outputs the location of the key given the board state after prisoner 1 doing the flip.

If both the solutions are function they may share code with an auxiliary function, though the solutions may not share any information.


This is so shortest bytes wins

Heavily inspired by The almost impossible chessboard puzzle and The impossible chessboard puzzle


  • Should I include the tag
  • Any more tags I should add
  • Is something not clear

Also pretty sure this will require a lot of rewording before it can be asked

  • \$\begingroup\$ Can the two solutions share code, for example by having two functions that both call a third auxiliary function? \$\endgroup\$ – Zgarb Jul 7 '20 at 9:20
  • \$\begingroup\$ Yes, the two solutions can share code but they cannot share information. Updated the question to reflect that \$\endgroup\$ – Mukundan314 Jul 7 '20 at 9:30
  • \$\begingroup\$ Can the auxiliary function be one of the programs? Can both of the programs be the same (and only counted once for bytes, having different behavior based on whether a second argument is passed)? \$\endgroup\$ – fireflame241 Jul 9 '20 at 0:25

Infinite Mirrors Quine

This challenge is to create a program that prints out code that prints the original. Basically, this program should take an input, check if it's A, and if so, run section A. Otherwise, if it's B, run section B. Finally, if it's neither, print nothing. Section A should print the code in section B, and section B should print the code in section A. Section A's code and section b's code should not be identical. Shortest code wins!


Continuous Everywhere, Differentiable Nowhere


Build the Weierstrass function \$f(x) = \sum_{n=0}^\infty a^n \cos(b^n n x)\$, where \$a \in (0,1)\$, \$b\$ is an odd positive integer, and \$ab > 1 + 1.5\pi\$.

What's the fuss?

The Weierstrass function is an example of a function that is continuous everywhere, but differentiable nowhere.


Using floating-point number is permitted. Though it will be preferred to use a datatype that is able to represent arbitrary real numbers.


\$a\$ and \$b\$ are up to your choice, as long as they satisfy the conditions.

For every \$x \in \mathbb{R}\$, evaluation of \$f(x)\$ must halt.


The fact that the function above is defined as an infinite series might seem to contradict the rule, but it actually doesn't. The Weierstrass function is computable, implying that it is possible to halt for every input. In particular, if it were to be implemented over floating-point numbers, it suffices to stop summing when the summand becomes denormal.

  • 1
    \$\begingroup\$ I really doubt that the Weierstrass function is computable since its domain is real numbers and computer programs can only compute a select few functions on arbitrary real input. The Weierstrass function is very likely computable on some restricted domains like the rational numbers. Really this question has a some issues with the fact that continuity and differentiabilty are usually discussed in the context of real numbers, but does not require that it work on actual real numbers. \$\endgroup\$ – Wheat Wizard Jul 16 '20 at 1:36
  • 1
    \$\begingroup\$ For example as it currently stands I could do something like just output the floating point zero regardless of input. This is approximates some cewdnw function, in fact it comes arbitrarily close to approximating an infinite number of cewdnw functions. For example just the Weierstrass function multiplied by a really small positive number. \$\endgroup\$ – Wheat Wizard Jul 16 '20 at 1:38
  • 1
    \$\begingroup\$ However if you do restrict it to real numbers you run into the problem that very few functions are computable on real numbers. (The issue here is that a program on real numbers must be ready to accept an infinite string of input). I suggest restricting your domain to something like rational numbers, but note that this alters continuity and differntiability in subtle ways so it is not a simple patch. \$\endgroup\$ – Wheat Wizard Jul 16 '20 at 1:41
  • \$\begingroup\$ Comment: while this problem is solvable in polynomial time, I guess the code-golf submissions are going to take exponential time. \$\endgroup\$ – user202729 Jul 16 '20 at 6:40
  • \$\begingroup\$ @user202729 Do you actually have a polynomial tome algorithm? \$\endgroup\$ – Wheat Wizard Jul 16 '20 at 12:23
  • \$\begingroup\$ Yes. -- -- -- -- -- -- \$\endgroup\$ – user202729 Jul 16 '20 at 13:29
  • \$\begingroup\$ @user202729 What is it? \$\endgroup\$ – Wheat Wizard Jul 16 '20 at 13:34
  • \$\begingroup\$ Iterate over substrings of the string, then check if it satisfies with f(left, right, prefix) = (can eraser[:prefix] be formed from string[left:right] by repeated erase operations?) At most this is O(n^6). \$\endgroup\$ – user202729 Jul 16 '20 at 13:38
  • \$\begingroup\$ I cannot understand your notation so I do not understand your algorithm, but I will say it seems to me that checking whether an eraser erases a string should naïvely take O(2^n) since in strings like "ototoo" it matters which "oto" you erase first thus you have to branch between the possible choices. \$\endgroup\$ – Wheat Wizard Jul 16 '20 at 14:22
  • \$\begingroup\$ The notation is like Python, string[left:right] is character from left..right (inclusive), eraser[:prefix] is eraser[0:prefix]`, characters are 0-indexed. \$\endgroup\$ – user202729 Jul 16 '20 at 14:32
  • \$\begingroup\$ It's possible to compute each f(left, right, prefix) value from O(n) other values (dynamic programming) and there's only O(n^3) possible parameters. \$\endgroup\$ – user202729 Jul 16 '20 at 14:33
  • \$\begingroup\$ @user202729 Ok, It looked like python but it didn't make any sense as python code, might you actually write this in python? It still doesn't make a whole lot of sense and even then feels like it should be O(2^n) because of "can [...] be formed from [...] by repeated erase operations?" seems to be an O(2^n) check to me. \$\endgroup\$ – Wheat Wizard Jul 16 '20 at 14:37
  • \$\begingroup\$ f=lambda left, right, prefix: string[left:right]==eraser[:prefix] or (left!=right and (string[right-1]==eraser[prefix-1] and f(left, right-1, prefix-1) or f(left, right, len(eraser) or any(f(left, middle, prefix) and f(middle, right, 0) for middle in range(left+1, right)))), something like that, with caching. \$\endgroup\$ – user202729 Jul 16 '20 at 14:42
  • \$\begingroup\$ Looks like this problem (or a similar one) has already appeared somewhere else. See codeforces.com/blog/entry/14090 \$\endgroup\$ – user202729 Jul 16 '20 at 14:46
  • \$\begingroup\$ @user202729 Ok so I've spent a little while unpacking that algoirthm in the blog post and it seems to be O(2^n) unless there is some invariant I am missing. I will say I still do not have the slightest understanding of your algorithm. \$\endgroup\$ – Wheat Wizard Jul 16 '20 at 15:22
  • \$\begingroup\$ About the blog: if you understood it then there is no way it can be 2^n because there are only n^2 different states (possible parameter values) of the dp function . \$\endgroup\$ – user202729 Jul 17 '20 at 2:44
  • \$\begingroup\$ @user202729 The issue is that calculating a cell is not contsant time sometimes we are required to solve the entire problem again on a smaller string to fill in a cell. You can make schemes where the number of these cells is linear witht he size of the program, hence exponential time overall. however at this point I have found a dynamic programming algo that does this in O(n^4), so it doesn't matter much to me any more. \$\endgroup\$ – Wheat Wizard Jul 17 '20 at 3:23
  • \$\begingroup\$ Now that this has been posted to main, could you delete this proposal to create more space for new answers? \$\endgroup\$ – caird coinheringaahing Sep 25 '20 at 1:03

Ant Storage Labyrinth


Using a simplified model, the place where ants store their food can be thought of as an \$n\times n\$ matrix. Each entry of the matrix is an integer that encodes how full that specific spot is, according to the following correspondence:

  • 0 denotes an empty spot (the ants can add two more units of food),
  • 1 denotes a half-filled spot (the ants can add one more unit of food),
  • 2 denotes a filled spot (no more food can be stored in there).

Imagine an ant carrying \$f\$ units of food, that enters the "storage room" at a specific position (row \$i\$, column \$j\$ of the matrix). The ant can move one unit left, right, up or down with each step, and it can drop \$2-q\$ units of food at each spot it walks over (where \$q\$ is the initial capacity of that spot – either 0, 1 or 2 as described above). Your task is to find the length of the shortest path the ant can choose in order to store all \$f\$ units of food.


Let's say that the ant carries \$4\$ units and enters the following storage room (\$6\times 6\$ matrix) at position \$(3,3)\$ (1-indexed):


It drops \$1\$ unit right where it starts (\$3\$ left), then it has four optimal choices:

  • 3 moves to the right, and 1 up,


  • 2 moves to the left, and 2 up,

  • 1 move up, 2 left, and one up,

  • 1 move up, 1 left, 1 up, 1 left.

All of these require \$4\$ steps, so the final answer is \$\boxed{4}\$.

Test cases

In progress. I need help coming up with interesting test cases / maybe a verification program.


Error Once, Hello World Twice

Your task here is to write a Hello World program that, (no, this is not Do X Without Y!) contains two exact copies of the same string. to avoid trivial solutions like print "Hello World!"# your program must error out with only one copy.

This is code-golf, so shortest answer in bytes wins.


  • Wording?
  • Tags?
  • Length?
  • Interesting enough to be posted?
  • \$\begingroup\$ I double the source code, you print hello world? (I'm not saying this is a dupe, I'm asking if that accurately summaries the challenge.) \$\endgroup\$ – lyxal Jul 25 '20 at 11:28
  • \$\begingroup\$ @Lyxal Yep. (15chars) \$\endgroup\$ – null Jul 25 '20 at 12:04
  • \$\begingroup\$ What if, rather than erroring out with one copy, make it so that it's valid if it prints out anything other than "Hello world"? \$\endgroup\$ – Beefster Jul 30 '20 at 16:40
  • \$\begingroup\$ @Beefster Maybe... But that's probably not what I'm intending. I might consider later. \$\endgroup\$ – null Jul 31 '20 at 10:25
  • \$\begingroup\$ @HighlyRadioactive this also reminds me of my 2 cats in a quine challenge from a while ago. My main concern here is that "erroring out" is nebulous and different for every language. You could also make it so that the program must print nothing unless it's duplicated. \$\endgroup\$ – Beefster Jul 31 '20 at 16:13
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