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4481 Answers 4481

94 95
97 98


The year is 19XX.

You are a spy of some distant country, and your job is to send messages across the globe.

Unfortunately, because, frankly, you suck at being a spy, you need a way to obfuscate your information, so that when the opposition catches you (which they will), they won't know what the heck is written.

How are you going to do this?

Your task is, using two inputs (the first input the encoding "cypher" and the second the message), encode the message.

This is how the encoding works:

  • The message only consists of the lowercase letters and the numerals.
  • Because there are 36 different characters in total, we will convert each individual letter of the message to "base 9" (a = 00, b = 01, c = 02... 8 = 37, 9 = 38). This will be called the FSO, or First-Stage Obfuscation.
    • For example, the message hello1 would then be translated to 07 04 12 12 15 28.
  • Each individual "bit" of the FSO is then stripped of its first part. This will be known as the SSO, or Second-Stage Obfuscation.
    • The example 07 04 12 12 15 28 is then translated to 7 4 2 2 5 8.
  • This is where the encoder comes in handy! The encoder will consist of a string of numbers 0 to 3 (e.g. 1212003).
  • You then add to the start of each digit of the encoder to the corresponding digit of the SSO. This is now the TSO, or Third-Stage Obfuscation.
    • The example 7 4 2 2 5 8 with the encoder 1212003 is then converted to 17 24 12 22 05 08.
    • With a shorter encoder (say 121), this step will "wrap around", so 7 4 2 2 5 8 with the encoder 121 will end up with 17 24 12 12 25 18.
  • We then change the TSO back into readable characters, using the same "base 9" method.
    • The two examples 17 24 12 22 05 08 and 17 24 12 12 25 18 will be converted to QWLUEI and QWLLXR respectively.

So, in summary:

Encoder: 121

Message: hello1

hello1 => 07 04 12 12 15 28 => 7 4 2 2 5 8 => 17 24 12 12 25 18 => QWLLXR

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


  • This is really confusing, and I don't really know how to phrase the "how the encoding works" bit better. Can anyone help me phrase this better? I can offer clarification on certain things if needed.
  • \$\begingroup\$ This is basically a mildly astandard and irreversible Vigenère cipher, so it's virtually a dupe of this question. (I would also say that it violates one of the key criteria for being a good question, which is to have a motivation. The backstory IMO isn't a motivation because it doesn't explain why anyone would want to implement or use what's really a supremely bad hash function). \$\endgroup\$ Jun 23, 2016 at 13:36
  • \$\begingroup\$ @PeterTaylor Yeah, IK. I need a lot of things to fix with this question (I was extremely tired at the time, couldn't think of a nice backstory). Also, a few things about the dupe: ONE, it's 5 years ago, so there's bound to be some new answers out there (if it's even a dupe in the first place), and TWO, how is it a dupe? \$\endgroup\$
    – clismique
    Jun 24, 2016 at 10:55
  • \$\begingroup\$ It's the difference between for (i=0 to n-1) s[i] = handleWrapping(s[i] + k[i % klen]) and for (i=0 to n-1) s[i] = handleWrapping(s[i] % 9 + k[i % klen] * 9). As far as I'm concerned that's a very minor transformation. \$\endgroup\$ Jun 24, 2016 at 11:03
  • \$\begingroup\$ @PeterTaylor Yeah, I kinda understand now. Still some questions, though. ONE: What language is that? TWO: Is there any way I could improve on my explaining? \$\endgroup\$
    – clismique
    Jun 24, 2016 at 11:22
  • \$\begingroup\$ 1. It's pseudocode, because I didn't want to faff around working out what escaping a less-than sign needs in comments. 2. I'm not quite sure what you're asking, but I would ditch this idea completely and try to find something more original. See e.g. meta.codegolf.stackexchange.com/q/1475/194 \$\endgroup\$ Jun 24, 2016 at 11:49
  • \$\begingroup\$ @PeterTaylor Yeah, I can see those, but most of those ideas have been taken already, and I kinda want to move away from numbers for a bit and play around a bit with strings and whatnot. Any suggestions? \$\endgroup\$
    – clismique
    Jun 24, 2016 at 12:36

Regex golf: Match the Thu'um-s


Skyrim is a game made by Bethesda and came out in 2011. One of the objectives of the game is to collect every 27 shouts (or Thu'um-s). In this challenge, you need to match every shout and nothing else.


This is the list of the available shouts in the game:

Raan Mir Tah
Laas Yah Nir
Mid Vur Shaan
Feim Zii Gron
Gol Hah Dov
Od Ah Viing
Hun Kaal Zoor
Lok Vah Koor
Ven Gaar Nos
Zun Haal Viik
Faas Ru Maar
Mul Qah Diiv
Joor Zah Frul
Gaan Lah Haas
Su Grah Dun
Yol Toor Shul
Fo Krah Diin
Liz Slen Nus
Kaan Drem Ov
Krii Lun Aus
Rii Vaaz Zol
Tiid Klo Ul
Strun Bah Qo
Dur Neh Viir
Zul Mey Gut
Fus Ro Dah
Wuld Nah Kest

Additionally, every shout can have 3 levels depending on what the player collected so far, each level adds a new word to each shout, so you need to be able t match the separate words without the full shout.

The words need to be uppercase. A shout should only be matched, if it is a separate word, for example: Golf shouldn't be matched.

The separate words for the same shout in the order as in the list appear next to each other, then they need to be in the same match.

The input strings will only contain ASCII letters and spaces as word separators.

Test cases

The matched text is bold

Hydrogen Sulphur Krah Coal Gaan Lah Haas

one two three four Kaal six seven Joor Zah Frul eleven

red green blue Gol Hah orange purple Qo violet

Ran Miir Taah Raan Mir Tah raan mir tah

Golf Gol Middleage Mid


  • The answer should be a .NET type RegEx and should not contain any other langauge.

  • This is a code-golf, so the shortest answer in bytes wins

  • 2
    \$\begingroup\$ Very closely related. What if a word appears in lower case in the input? What if it appears inside another word like Golf? If the latter should not be matched please clarify what characters can be in the input and which of those are valid word delimiters. All that said, I don't see anyone coming up with a good way to compress the words within the limit framework of regex. \$\endgroup\$ Jun 23, 2016 at 12:18
  • \$\begingroup\$ @MartinEnder I clarified it a but \$\endgroup\$
    – Bálint
    Jun 23, 2016 at 13:16
  • 1
    \$\begingroup\$ Still doesn't say what characters can appear in the input and what characters count as word separators (I'm assuming only letters and spaces, and spaces are separators, but if that's your intention you should say so explicitly, and if not, you should add further test cases). Looks good otherwise. To prevent confusion you might want to say explicitly that people should submit only a regex and which flavours are allowed (and whether flavours like Perl are allowed to make use of their eval features to execute code in the hosting language). \$\endgroup\$ Jun 23, 2016 at 13:21
  • \$\begingroup\$ If only one language is allowed this would limit the participation in your question. I don't understand why it should be only one language. \$\endgroup\$
    – george
    Jun 29, 2016 at 19:38
  • \$\begingroup\$ @george I don't think you know what regex-golf is \$\endgroup\$
    – Bálint
    Jun 29, 2016 at 19:44
  • \$\begingroup\$ @Bálint Whoops I didn't see your title, my mistake. \$\endgroup\$
    – george
    Jun 29, 2016 at 19:47

Find all the Vampire numbers

Shamelessly stolen from https://stackoverflow.com/q/17352108

A "Vampire" number is defined as the product of two numbers of equal length (known as the "fangs") that uses the same digits as the two numbers being multiplied. Examples:

21 * 60 = 1260
15 * 93 = 1395
30 * 51 = 1530

Your task is to find all the Vampire numbers whose fangs have n digits.

This is , so the shortest solution wins.

  • \$\begingroup\$ What about double fangs, triple fangs etc? \$\endgroup\$
    – clismique
    Jun 24, 2016 at 11:30
  • \$\begingroup\$ @DerpfacePython Double fangs? But I suppose I could generalise it to m fangs of n digits if there's enough support (and if there are actually solutions!) \$\endgroup\$
    – Neil
    Jun 24, 2016 at 12:19
  • \$\begingroup\$ Related \$\endgroup\$ Jun 24, 2016 at 13:23
  • \$\begingroup\$ @FryAmTheEggman My bad for not searching first. I might as well delete this. \$\endgroup\$
    – Neil
    Jun 24, 2016 at 14:14
  • \$\begingroup\$ I think all you have to do is add the double/triple fang thing and it probably is different enough? Not sure, but there is probably a way to get it to work. \$\endgroup\$ Jun 24, 2016 at 14:20

Title TBD - Generate Emoticons :)

Your task is to create a program that generates the most emoticons using the least amount of code.


  • The only valid emoticons are listed on Wikipedia, under the heading Western/horizontal emoticons: List of emoticons page (version 722951221)
  • The output can be any format, as long as a delimiter exists between emoticons.


  • Scores will reflect the character (not byte) count.
  • For every emoticon after 10 that is output by your program, subtract 1 from your score. Emoticons over 20, subtract 2, over 30 subtract 3, and so on.


Is the scoring fair/reasonable? Should I edit down the list/create a new list of acceptable emoticons?


Forecast Romantic Dates

Sort of inspired by this.

A Romantic date is a date that, when the year, month and day are converted to Roman Numerals the individual values contain no more than two symbols. For example, in YY-MM-DD format: the Romantic date 20-04-15 would become XX-IV-XV.

For the purpose of this challenge, years will only be tracked by the two least significant digits of the year, as otherwise the last Romantic date was in the 15th century. In addition, they wouldn't add much to the challenge as the omissions of the leap year every 400 years is irrelevant, as February the 29th is not a Romantic date.

Romantic dates

For your convenience, here is a list of all of the two digit numbers that can be represented with two or fewer symbols in Roman numerals:

[1, 2, 4, 5, 6, 9, 10, 11, 15, 20, 40, 50, 51, 55, 60]

These were determined using the "standard method" that negative groups would only be used with the symbols that are powers of ten and only on the values that are five or ten times that symbol's value. So I only combines with V and X for example.

Dates which include only numbers from this list are Romantic dates. For the purpose of this challenge, assume an ideal Western calendar: no dates are ever skipped or repeated, 12 months per year, and more than 20 days per month. Assume there is no year, month or day zero (i.e. year 99 loops to 1 not to 0).


Given a date as input, output that date if it is Romantic, or output the next Romantic date.

Input and Output

You may accept input in any consistent ordering of year, month and day with any consistent separator. You may specify if the input should have the numbers padded to be two digits. If the numbers are padded, you may choose to have no separator. Your output must have the same form as your input.

Test Cases

The following test cases are all in the format YY MM DD, with no padding.

1 1 1 => 1 1 1
20 4 3 => 20 4 4
15 7 1 => 15 9 1
51 9 7 => 51 9 9
70 9 7 => 1 1 1
20 6 24 => 20 9 1
47 12 1 => 50 1 1
60 11 24 => 1 1 1

Here is the script that I used to generate these.


Did I miss any Romantic numbers? I just did that by hand.

Allow unary? I'm unsure about this because it sort of violates the reasoning behind Romantic dates for the values to have >2 symbols...

Should I explain more about parts of dates that are not useful? For example, the length of the months is entirely irrelevant as the later days are all skipped. My concern is that the current one feels clunky already

Should I allow both outputting the input if the date is already Romantic or the strictly next Romantic date (as long as it is consistent)? There doesn't seem to be much different, but I don't know if that'd be too broad? Personally leading towards allowing it.

I'm also somewhat tempted to make use of the silly title a bit more, but I'm not sure if that'd be going overboard.

Too boring / compression based? I've particularly been trying to think of a way for fewer results to wrap back around to 1.

  • \$\begingroup\$ That rule should be spelt out explicitly in the question, because although some people insist on it it's a modern innovation. There are actual Roman inscriptions which do use e.g. IC. \$\endgroup\$ Jun 24, 2016 at 21:05
  • \$\begingroup\$ @PeterTaylor You're right, I originally left out the reasoning because I thought it might clutter up the spec, but I realise now I just left that comment undeleted to prevent people from asking the same question. I don't have time right now but I'll edit it in once I get a chance. Also, I figured it was better with this rule because I thought say VL was rather unintuitive, does that make sense or should I be more laissez-faire about it? \$\endgroup\$ Jun 24, 2016 at 22:04

Stacks and Stacks and Stacks...

Write a program that, with the input as n, finds the first n-gonal and n-gonal pyramid number that is NOT 1.

n is guaranteed to be larger than or equal to 3.


  • n = 3: 10
  • n = 4: 4900
  • n = 2: The output can be nothing, False, or anything that you want, just as long as it can be distinguished from an actual output.

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


  • If your code output both the name of the n-gon and the number: You get a big fat -50% off of your byte count (see below for examples).
    • n = 4: Square 4900
    • n = 3: Triangle 10
    • n = 5: Pentagon ??? (the ??? is a placeholder because I have no idea what the number is)


  • Is the bonus a good idea?
  • \$\begingroup\$ I don't think the bonus is a good idea because 1) the name compression takes away from the original challenge and likely isn't worth it and 2) you haven't defined the naming scheme (e.g. 12-gon vs dodecagon) \$\endgroup\$
    – Sp3000
    Jun 26, 2016 at 7:10
  • \$\begingroup\$ @Sp3000 Ah, OK. I really want to incorporate the use of strings in the challenge, but if the idea sucks, then I'll scrap it. Any further suggestions? \$\endgroup\$
    – clismique
    Jun 26, 2016 at 7:12
  • 2
    \$\begingroup\$ Adding strings just makes it feel like squeezing two challenges into one, unfortunately. I'd recommend posting the shape names as a separate challenge, but I see we have this challenge. As for suggestions, maybe 1) remove the part about n = 2, since you already say n is guaranteed to be at least 3 and 2) maybe make it explicit that the output should be both n-gonal and n-gonal pyramidal (and maybe explain, say, that 10 is the 4th triangular and 3rd triangular pyramid number) \$\endgroup\$
    – Sp3000
    Jun 26, 2016 at 7:16
  • \$\begingroup\$ Hmm my other problem is - what happens if there's no solution for a given n? Note the only reason I'm asking is because the number of solutions for any n could be finite, e.g. A027568. (6 is 946, 8 is 1045, 10 is 175 and 11 is 23725 I believe) \$\endgroup\$
    – Sp3000
    Jun 26, 2016 at 9:28
  • \$\begingroup\$ Yeah, that could be a problem... maybe a time limit? Or maybe check numbers up to a given range. \$\endgroup\$
    – clismique
    Jun 28, 2016 at 6:50
  • \$\begingroup\$ One alternative could be to allow solutions to potentially infinite loop/hit memory or recursion errors in the case of no solution/large solution. Numerical limit to check up to could work too, that'd probably be better than a time limit (since it reduces a dependency) \$\endgroup\$
    – Sp3000
    Jun 28, 2016 at 9:02
  • \$\begingroup\$ @Sp3000 I would probably go with the numerical limit/memory limit thing, whichever comes first. But what about golfing/esoteric languages? There might not necessarily be a memory/numerical bound for those. \$\endgroup\$
    – clismique
    Jun 29, 2016 at 9:04

Pyth Meta-Golf Golf Battery

(now that's a name, isn't it?)

The Challenge

Write a program in a language of your choice that takes an input of code in the same language and outputs a code that does the same thing in Pyth (though not necessarily the shortest code).


Given the test battery set below, you should try to find the best way to minimize your score with the following rules:

  • You must minimize the source code of the submission.
    • You may not used compressed versions of the input source for each test; the Pyth code must be output procedurally.
    • The code must theoretically do the same conversion for any input code, not necessarily just these test cases.
  • You must minimize the input code of each test.
    • If an input is shown to be shortenable by without non-standard libraries of the submitted language, you must change it immediately to the shorter answer. If your answer does not support this change, then you must change your answer to accommodate.
  • You may not submit answers written in Pyth.

Scoring is done with the following equation:

(source code)*((Pyth src out, test 1)/(input src in, test 1)+(same for test 2)+...)

Objectives of each test

Test 1:

Output the string "Hello, world!".

Test 2:

Given an integer input, multiply that input by three and output it.

Test 3:

Quine. (must be a valid quine in the submitted language and in Pyth)

Test 4:

Produce infinite output.

Test 5:

Cat program.

  • 2
    \$\begingroup\$ I'm not a Pyth expert, but I strongly suspect that what this asks is impossible in many languages. E.g. I understand that Python's multi-threading support is extremely limited. Even if true impossibility is not an option, answers in many languages won't fit inside the 30000-character limit. E.g. I don't think the grammar for Java fits inside the limit, let alone a lexer and parser. \$\endgroup\$ Jun 27, 2016 at 9:56
  • 1
    \$\begingroup\$ I agree with Peter, that this seems like unless you use only trivial languages it probably won't be possible to answer. The scoring would also make this confusing: to make sure a solution is correct one would have to not only test 10 programs, you would also have to golf 5 programs. Further, there is a bit of a problem in Pyth's $ operator, which runs literal Python, which means you should probably ban Python as well. I'm not really sure of how to turn this into a good question, too heavily restricting the type of program seems to be the only way, but it also seems to defeat the purpose. \$\endgroup\$ Jun 27, 2016 at 13:09

Don't even think about non-42-related numbers!

Introduction and Credit

We all love our Answer to the Ultimate Question of Life, the Universe, and Everything which, of course is 42. So let's take this unworthy Fibonacci-Sequence thingy and adapt it to be worthy of 42!



The input will be a positive integer.


The output will be either true or false.

What to do?

Your task is to implement the predicate that the number under consideration is an element of the generalized Fibonacci-Sequence given by:

a_1 = 14
a_2 = 42
a_n = 41 * a_{n-1} + 43 * a_{n-2}

Where 42 is the ultimate answer, 41 and 43 are the primes next to it and 14 is the preceding catalan number.

Potential corner cases

The number will always be greater than zero. Your program must pass all (32-bit) test vectors below.

Who wins?

This is code-golf so the shortest answer in bytes wins!
Standard rules apply of course.

Test Vectors

14 -> true
42 -> true
2324 -> true
4080622 -> true
97090 -> true
171480372 -> true
7 -> false
1 -> false
41 -> false
43 -> false
  • \$\begingroup\$ a_3 is already outside the range of 8-bit unsigned integers, a_6 is outside the range of 32-bit signed integers, and a_11 is outside the range of 64-bit signed integers. It would be worth addressing this issue and at least ensuring that languages which are inherently 8-bit can't just special-case the two values which they can handle. \$\endgroup\$ Jun 25, 2016 at 21:58
  • \$\begingroup\$ @PeterTaylor I'm unsure how to formulate this without disqualifying legitimate approaches. Would saying "your program must pass all test vectors" (with 32-bit test vectors) also be considered OK for most people? \$\endgroup\$
    – SEJPM
    Jun 25, 2016 at 22:16
  • \$\begingroup\$ I feel like this is too much two separate challenges: one to test for membership in the sequence, and the other to takeWhile on a condition. I also suspect the sequence has a direct arithmetic membership test like the one where n is a Fibonacci number exactly if either 5*n*n+4 or 5*n*n-4 is a square. \$\endgroup\$
    – xnor
    Jun 25, 2016 at 23:27
  • \$\begingroup\$ What is takewhile? \$\endgroup\$
    – clismique
    Jun 26, 2016 at 7:02
  • \$\begingroup\$ @xnor, I originally only wanted to do the takeWhile, but needed a (mediumly) complex, interesting predicate, so I came up with this one. Of course I'm open to suggestion for more suitable interesting predicates. \$\endgroup\$
    – SEJPM
    Jun 26, 2016 at 10:18
  • \$\begingroup\$ @DerpfacePython, the higher-order functionality described in the second paragraph of "what to do?" is also called takewhile, I've clarified this though. \$\endgroup\$
    – SEJPM
    Jun 26, 2016 at 10:21
  • \$\begingroup\$ @SEJPM I think the other way -- to do a challenge about takeWhile, make the predicate as simple possible. Beware chameleon challenges and needless fluff. One clean condition would be integers being positive. \$\endgroup\$
    – xnor
    Jun 26, 2016 at 10:56

Compute the mincut of a graph

Given a graph, compute a division of the graph such that the edges stranded between the cut.

hayo mouseover readers; leave a comment if you see this!

Red line: a cut

Green line: a mincut


The first line will contain the number of nodes. The rest of the lines will contain pairs of positive integer IDs separated by spaces showing connectedness between the nodes with those IDs. Here's an example; for a graph where 1 is connected to 2 and 2 is connected to 3:

1 2
2 3
  • You may assume that the nodes are numbered consecutively from one to the number of nodes.
  • However, you may not assume that the list of pairs of nodes is in any specific order.


Simply output a comma-separated list of the IDs of the nodes of one of subgraphs created by the cut.

Additional Rules!

  • You cannot implement brute-force search. Other than that, feel free to use Karger's Algorithm* or another algorithm. Remember that Karger's algorithm is likely the easiest to implement.
  • Notice: you must run Karger's algorithm at least this many times to ensure a low chance of failure and a low chance of failure

*Karger's algorithm

For your convenience, I've included a simple description of Karger's algorithm.

  1. find two adjacent nodes and merge them into one node (so that all nodes that where connected to the original two nodes are connected to the new node), concatenating the labels
  2. repeat step one until there are only two nodes
  3. the result is any label of one of the nodes
  4. repeat steps 1-3 at least this many times to ensure a low probability of failure, and choose the result that occurs the most often
  • 1
    \$\begingroup\$ 1. Wouldn't it be better to take the graph as an adjacency matrix or list? 2. Your description of the minimum cut is somewhat confusing. 3. Karger's algorithm is probabilistic, which isn't allowed by our defaults (I don't think). Allowing probabilistic algorithms opens up a whole can of worms (for instance I could write a program that just returns a random cut) -- you should probably make it so that the algorithm must return the minimum cut two-thirds of the time or something similar if you want to allow them. \$\endgroup\$ Jul 5, 2016 at 0:07
  • \$\begingroup\$ @quartata 1. it's an adjacency list 2. yeah I need help with that 3. I made sure you had to repeat it insert some math equation here amount of times \$\endgroup\$ Jul 5, 2016 at 0:13
  • \$\begingroup\$ Sorry I misunderstood the input. \$\endgroup\$ Jul 5, 2016 at 0:14
  • \$\begingroup\$ Generally adjacency lists are done like [node1, connected_node1, connected_node2, ...] and not in pairs like you have it; this is more flexible and you don't have to specify the number of nodes (it is just the length of the input list) \$\endgroup\$ Jul 5, 2016 at 0:35
  • 2
    \$\begingroup\$ "You cannot implement brute-force search" is too vague. What about a basically brute force search that shortcuts some obviously wrong possibilities? I think what you want is a running time bound. \$\endgroup\$
    – xnor
    Jul 5, 2016 at 1:55
  • 1
    \$\begingroup\$ 1. The I/O description seems to assume that all answers will be programs taking input on stdin and writing output to stdout, but our defaults are more flexible. In particular, by default we allow answers to be functions which take arrays and return arrays. Comma-separating is also IMO unnecessarily constrained, especially as the input isn't comma-separated. \$\endgroup\$ Jul 5, 2016 at 7:51
  • 1
    \$\begingroup\$ 2. "Feel free to use Karger's algorithm or another algorithm". There's an implicit licence here to use another non-deterministic algorithm, but although you give an explicit iteration count for Karger's algorithm you don't for e.g. randomised Kruskal's algorithm, which it's based on. 3. Besides which, in general I don't think that questions should tell people which algorithm to use. Specify the task and constraints (e.g. "Randomised algorithms are allowed, but must find the correct answer with probability at least foo. All answers must be polynomial-time"). \$\endgroup\$ Jul 5, 2016 at 7:54
  • \$\begingroup\$ 4. But if you're going to include an algorithm description, be careful to get it right. Karger's algorithm is randomised, but in the description given there's no mention of where the random selection occurs or of what uniformity constraints are required to get the desired behaviour. \$\endgroup\$ Jul 5, 2016 at 7:55
  • 1
    \$\begingroup\$ Food for thought: outputting the value of the min cut instead might lend to more approaches. Also, any rules on min cut/max flow/possibly other optimisation builtins? \$\endgroup\$
    – Sp3000
    Jul 5, 2016 at 10:34
  • \$\begingroup\$ I'm going to add a story to this soon. \$\endgroup\$ Jul 5, 2016 at 13:55

Generate a random Vietnamese syllable


The Vietnamese syllable space is interesting, because it is huge.

TODO: Describe the space and why it is interesting.

Here's how such syllables are made:

The onset matches the regex ^([bcdđghklmnprstvx]|qu|[cgkpt]h|ng|tr)$

The vowel is one of the following massive list:

a à a' ã á a.
â â` â' â~ â´ â.
a. ă` ă' ă~ ă´ ă.
e è e' e~ é e.
ê ê` ê' ê~ ê´ ê.
i ì i' ĩ í i. ia iê
o ò o' õ ó o. oa oă oe
ô ô` ô' ô~ ô´ ô.
o' ò' o" õ' ó' o'.
u ù u' ũ ú u. ua uâ uê ui uô uo' uy
u' ù' u" ũ' ú' u'. u'a u'o'
y y` y' y~ ý y. ya yê

The coda matches the regex ^([iouycptmn]|ch|ng|nh)$

(thanks Peter Taylor!)

The onset, vowel and coda are concatenated to make the result syllable.


The objective is to generate random Vietnamese syllables. Your program has to take no input and as output include only the syllable, with an optional trailing new line.


  • Each syllable must be generated with a non zero probability.

I think it's unclear. Contributions are so much welcome.

  • \$\begingroup\$ 1. I'm not sure what you mean by can be with. 2. You don't mention randomness anywhere excwpt the clarification. 3. Object should probably be Objective. \$\endgroup\$
    – Dennis
    Jul 3, 2016 at 18:44
  • \$\begingroup\$ 1. c can also be with h means that h can follow c as the 2nd stage letter in the syllable. 2. Where should I also mention it? 3. Ah k :) \$\endgroup\$
    – user48538
    Jul 3, 2016 at 18:46
  • \$\begingroup\$ If I'm reading this correcting then it can be vastly simplified by saying that the onset matches the regex ^([bcdđghklmnprstvx]|qu|[cgkpt]h|ng|tr)$, the vowel is one of a massive set of options (I don't see any benefit to splitting that into "stage 2" and "stage 3"), and the coda matches the regex ^([iouycptmn]|ch|ng|nh)$ \$\endgroup\$ Jul 4, 2016 at 16:36

Let's play some Briscola

Briscola is an Italian game, played with a deck of 40 cards, divided in 4 suits - coins (denari - D), swords (spade - S), cups (coppe - C) and clubs (bastoni - B).

The values on the cards range numerically from one through seven, plus 3 special cards - knave (11), knight (12) and king (13).


After the deck is shuffled, each player is dealt three cards. The next card is placed face up on the playing surface, and the remaining deck is placed face down. This card is the Briscola, and represents the trump suit for the game.

First player starts by playing one card face up on the playing surface. Each player subsequently plays a card in turn, until all players have played one card.

The winner of that hand is determined as follows: If any briscola (trump) has been played, the player who played the highest valued trump wins, else the player who played the highest card of the lead suit (suit of the first card played) wins.


Briscola has a special type of ranking:

1   ace
3   three
13  king
12  knight
11  knave


Standard loopholes apply.


As an input, you must accept 5 values (cards), in a reasonable format, for example:

briscola (trump card), 1. card, 2. card, 3. card, 4.card


You must output the winning card

Example input and output:

4S 7D 12B 13B 2S -> 2S
5D 1D 5D 12S 3C -> 1D
3B 2C 4S 5S 7D -> 2C
12D 3S 11B 1B 7S -> 3S
  • \$\begingroup\$ As mentioned in chat, I think this is probably a duplicate of this challenge. Just adding so other people don't have to go looking. \$\endgroup\$ Jul 6, 2016 at 21:15

nth number that multiplies k equals its reverse

Tags: ,

It's quite simple, given n and k, output the nth number such that, if the number is multiplied by k and its digits reversed, it equals the original number. Both input and output are positive numbers.

The challenge originally is from Mego, posted on my broken challenge. Firstly, I used 4 instead of k, but based on my tests, only 1 and 4 values gives output, so I decided to put 4 instead of k, finally I put k back. But the challenge would be ruin with that putting "9"*(n-1) between 2178, so no loopholes will be permitted.

I just posted here for further discussions, suggestions and improvements.

  • \$\begingroup\$ Those numbers are positive right? \$\endgroup\$
    – Fatalize
    Jul 6, 2016 at 7:52
  • \$\begingroup\$ Please add some examples of expected outputs. \$\endgroup\$
    – Fatalize
    Jul 6, 2016 at 7:53
  • \$\begingroup\$ Also you might want to prevent people from hardcoding 2178 in any fashion in their code so that they have to compute the numbers, because it seems they all are of the form 21X...X78 where X...X is a series of nines (except for the first one, which is 0). \$\endgroup\$
    – Fatalize
    Jul 6, 2016 at 7:56
  • \$\begingroup\$ According to the community advises, I'm not allowed to prevent people use methods those work perfectly. \$\endgroup\$
    – Ehsaan
    Jul 6, 2016 at 8:23
  • \$\begingroup\$ Let's wait to see what others think. I personally don't think it's very interesting if people are allowed to hardcode the "format" of those numbers. \$\endgroup\$
    – Fatalize
    Jul 6, 2016 at 8:26
  • \$\begingroup\$ Me neither, I think the challenge isn't interesting at all. \$\endgroup\$
    – Ehsaan
    Jul 6, 2016 at 8:43
  • 1
    \$\begingroup\$ I think there's no good way to prevent hardcoding. Maybe making "4" were an input parameter as well would make solutions actually search for an answer? \$\endgroup\$
    – xnor
    Jul 6, 2016 at 9:01
  • \$\begingroup\$ @xnor You mean make 4 as k input? \$\endgroup\$
    – Ehsaan
    Jul 6, 2016 at 9:33
  • \$\begingroup\$ @Ehsaan Yes, exactly. \$\endgroup\$
    – xnor
    Jul 7, 2016 at 9:09
  • \$\begingroup\$ 9 works too: 1089 * 9 = 9801. \$\endgroup\$
    – Neil
    Jul 10, 2016 at 17:36

Write a program that can determine the median value of a read-only (static, const, immutable) sequence of unsorted numbers (array, list, stream) but minimises storage, without completely sacrificing speed.

The basic bracket is that if we copied all the values into a sorted list and then picked the middle one (or average of the middle pair), it would require storage of the whole sequence, so the storage would be 'n', and the performance would be O(n log n).

The score is the total cost of finding the median of 1 bn values, divided by 1 bn, at a cost of 8 per value stored, 1 per comparison or numerical operation and 1 per read, for the worst case. Thus if our insertion sort costs exactly n*log2(n), the for 1 bn values the total score is 1 for the read, 29.8 for the sort + 8 for the storage, for a total of 37.8.

If instead we skimmed the whole range to get the average (costing 1 for the read and 1 for the summation), we could then only store some portion of the range to sort; but then we would need a second pass to be sure that there were an equal number of values above and below this median (at the cost of another 2).

Lowest score wins, low-level languages (C/C++/D) only so that we can count the actual operations.

  • 3
    \$\begingroup\$ 1. It's not clear to me what counts as a "value stored" or a "read", and I think there are probably gray areas with "comparison or numerical operation" too. (E.g. in C is if (foo) a comparison?) 2. "The score is the total cost ... for the worst case." For any non-trivial algorithm, the full calculation of this score risks being longer than the code. There's a reason that complexity theorists deal with Landau notation rather than exact operation counts. \$\endgroup\$ Jul 7, 2016 at 13:39

Reinventing the Modularization Wheel

In a language of your choice, implement a function or language construct that imports another file of the same language and executes it, making exported values from that file available to the calling file. If one already exists, you may not use it in your implementation.

For example, in Node, you would have to implement require() without using require(), even indirectly. In C, you would implement a function or construct equivalent to #include without using #include in the implementation. In Python, you would implement import. In client-side JavaScript, I suppose the closest equivalent would be <script src="..."></script>. So JavaScript implementations would be restricted to AJAX calls only, since <script> tags would not be allowed in the implementation.

This is not to say that you aren't allowed to use the built-in import at all, but only use them in the implementation. The intention here is to reinvent the wheel.


  • Do not include the built-in modularization in any way in your import implementation.
  • Standard libraries only.
  • Byte-count includes the implementation itself, and any special changes that need to exist on the file being imported, if any.
  • The function or construct accepts a relative file path. As long as this is satisfied, you may extend the functionality of your modularization to have global imports, or even remote imports (like using a URL as input).
  • The imported file must have a construct for denoting values that must be exported. Only these values should be directly accessible from the calling file.
  • Using the built-in export function or construct of your language is acceptable, and if it is a built-in, it does not need to be included in your byte-count.
  • If your language does not have modularization, then implementing a mechanic for exporting should be included in your byte-count.
  • Document the usage of your function or construct.

This is and the shortest answer in bytes wins!

  • \$\begingroup\$ Perhaps just restrict this to languages which support modularization to avoid loopholes \$\endgroup\$
    – Downgoat
    Jul 7, 2016 at 18:45
  • \$\begingroup\$ @Downgoat if people wanted to use a built-in for reading a plaintext file, and then use an eval()-like built-in to execute it in a way that exposes only denoted values (however you define that), I think it would be acceptable. What sort of loopholes do you foresee? \$\endgroup\$ Jul 7, 2016 at 18:49

Print a Pilcrow Scarecrow

Print the following ascii scarecrow using the pilcrow character

    ¶¶¶  ¶
   ¶ ¶ ¶
  ¶¶ ¶ ¶¶
  • Padding must be with (space) and built with
  • Trailing padding is okay
  • Print to stdout
  • This is


It's time to unify!


Wouldn't it be awesome if they whole world would be united and there would be no conflicts and disputes? Now while you can't unify nations, you certainly can unify expressions to resolve their unknown relation and conflicts.
Your mission is simple: Unify the world (of expressions)!
And of course, because you're lazy you want to do this with the least effort (read: code-length) possible.



Your input will be a unification problem. You can format it however you want and need, as long as you don't encode additional information to what is given in the standard / example format. Encoding the number of arguments per function into the input is allowed but not mandatory, you can also just derive this from the input.

Example format:
Your first input will be list of function symbols, which is represented as a list of pairs of strings and non-negative integers.
Your second input will be a list of equalities (you may represent each as a string), which represent the unification problem. They will be represented as a list of strings as well. Anything which is not a parenthesis or an equality sign can be considered a variable. If the number of arguments is 0, parenthesis are omitted.

Example input: [("f",1),("g",2),("h",3),("a",0)], [x=f((g(a,y)),y=h(g(f(a),z),f(z),a)]


The output is either some falsy value or something representing a list of equalities. It is allowed to use the empty list to indicate a falsy value.

What to do?

You need to unify the inputs you got. In the end there must only be variables on the left side of the equality-signs if the you didn't encounter an error. If you did you need to report it (-> false or empty list).

To do the unification, you can - but don't have to - use Martelli and Montanari's algorithm, which goes as follows:

E is always the (complete) set of equalities except the current one
x,y,z are variables, f,g,h are functions, t1,t2, ...,tn,s1,...,sn are arbitrary terms (compositions of functions and variables)
{x=x} E => E, e.g. if you encounter two equivalent variables, discard
{f(t1,...,tn)=f(s1,...,sn)} E => {t1=s1,t2=s2,...,tn=sn} E, e.g. if you encounter the same function on both sides, unify the arguments along with your rest
{f(t1,...,tn}=g(s1,...,sn)} E => Error, if the symbols are different, you can't succeed
{x=f(t1,...,tn)} E => {x=f(t1,...,tn)} E[x -> f(t1,...,tn)], e.g. if you see a variable equals a term, replace the variable with this term in all other expressions
{x=f(t1,...,tn)} E => Error, e.g. if any of the t1,..,tn contain x at some point
{f(t1,...,tn)=x} E => {x=f(t1,...,tn)} E, e.g. if you see a variable "naked" on the right side, swap the sides

Two step-by-step examples are provided below additionally to the test cases.

Corner Cases

You can get an empty list of function symbols, this means you have exclusively variables in the second input.
The input list of equalities will never be empty, your code does not need to handle this case.

Who wins?

This is code-golf so the shortest answer in bytes wins!
Standard rules apply of course.


All these test cases use the functions [("a",0),("b",0),("f",1),("g",1),("h",2)]

[x=b] -> [x=b]
[a=x] -> [x=a]
[a=b] -> []
[y=f(x)] -> [y=f(x)] 
[x=f(x)] -> []
[f(x)=f(y)] -> [x=y] 
[f(x)=g(y)] -> []
[h(x,y)=h(a,b)] -> [x=a,y=b] 
[x=f(z),y=f(a),x=y] -> [x=f(a),y=f(a),z=a]
[h(x,f(y))=z,z=h(f(y),v)] -> [x=f(y),v=f(y),z=h(f(y),f(y))]

Step-By-Step Example

Example 1: Test Case 9
[x=f(z),y=f(a),x=y] => (replace x in third equation with first x)
[x=f(z),y=f(a),f(z)=y] => (replace y in third equation)
[x=f(z),y=f(a),f(z)=f(a)] => (remove f's in third equation)
[x=f(z),y=f(a),z=a] => (replace the z in the first expression)

Example 2:
Let [("f",2),("g",2),("a",0),("b",0)] be your functions
[f(g(a,x),g(y,b)=f(x,g(v,w)),f(x,g(v,w))=f(g(x,a),g(v,b))] => (remove f in second equation)
[f(g(a,x),g(y,b)=f(x,g(v,w)),x=g(x,a),g(v,w)=g(v,b))] => (function symbol missmatch in equation 2)

create a golfed down regexp that matches all substrings

inspired by Determine the "Luck" of a string where I found a way to golf almost 30 bytes at once
(with a falling trick for that challenge, but I still like the idea).

The word "lucky" contains 15 different substrings:

  • lucky
  • luck, ucky
  • luc, uck, cky
  • lu, uc, ck, ky
  • l, u, c, k, y

Create a program or function that, for a given string s, creates the shortest possible regexp using basic PCRE syntax that matches and returns all substrings of s and nothing else.

  • code needs not to be case sensible
  • basic syntax means: alternatives, quantifiers, grouping and custom character classes (e.g. [abc])
  • other features (assertions, backreferences, recursion etc.) may be used, but are not required to qualify
  • the result may include delimiters and modifiers

The result for lucky would be l?ucky?|l?uc?|c?ky?|l|c|y.

  • is the description sufficient?
  • the challenge not too easy, not too hard?
  • any other hints you might have?
  • I will add test cases that expose possible bugs (like silly and digdug)
  • not sure yet if I will go for shortest code yet

Write a Gopher Interpreter

This code golf challenge will task you with writing an interpreter for an esolang I created a while back called Gopher, Details on the language can be found Here

Pass Conditions

This challenge requires you to create an Interpreter (Or you could go a step ahead and create a Compile/Transpiler) however for the code to pass as correct it must meet the following criteria

  • Take in a single input being the Gopher Code
  • Output the result of the Interpreted code

  • Invalid code does not need to be handled, however you may do so if you wish

  • As this is code-golf the smallest byte size wins

Example Input and Output




Hello World
  • \$\begingroup\$ Thanks for using the Sandbox! Anyway, you should add the relevant information on Gopher to the body of this post, as if your github account/repo dies or is changed people still need to be able to answer this question. \$\endgroup\$ Jul 11, 2016 at 17:15

Having had a look, it seems there isn't a challenge for "Given any date, output the day of the week". Is that a challenge worth having?

Something like

"Given an input date, in the form dd/mm/yyyy, output the day of the week"

Shortest code wins

What do we think? perhaps this already exists and I didn't find it.

  • 3
    \$\begingroup\$ Duplicate \$\endgroup\$ Jul 11, 2016 at 14:04
  • 2
    \$\begingroup\$ Glad I checked! \$\endgroup\$
    – Matt
    Jul 11, 2016 at 14:05

Golf your way from (inc|dec)rements to the basic math operations

Write five different functions or programs that do addition, subtraction, multiplication, division and modulo with integers by only using increments, decrements, loops/recursion and comparisons.

  • Assume division & modulo will never receive 0 or negative integers as the divisor/modulus.
  • Modulo's result has the sign of its dividend.
  • Division truncates its quotient, e.g divide(11, 4) returns 2 and divide(-5, 3) returns -1.
  • Programs must print the result to STDOUT. Functions must return the result.
  • Your five functions/programs may invoke each other.
  • All functions/programs must support 32-bit signed integers, i.e everything between -231 and 231-1 (inclusive). Overflow is allowed, i.e it's OK if add(2147483647, 1) returns -2147483648.
  • Explicitly adding/subtracting 1 to/from numbers is allowed, in case you use a programming language that doesn't have built-ins for incrementing and decrementing.
  • Shortest program wins as long as it doesn't exploit standard loopholes!

I seriously have no idea how to make test cases for this.

  • 1
    \$\begingroup\$ Why not one function that returns all of those? You should also specify what you mean by divisions and modulo as they differ slightly from language to language. (E.g. what is -2 mod 5? and what is -1/2?) And only doing increments/decrements, loops/recursion and compraisions is also quite vague. \$\endgroup\$
    – flawr
    Jul 10, 2016 at 16:37
  • 1
    \$\begingroup\$ I don't think test cases would really be necessary since it's just basic arithmetic. You can easily tell if your output is correct or not. Also, I'm assuming that division will truncate the quotient since there isn't really any way to do decimals in this fashion, but that should probably be specified. \$\endgroup\$ Jul 11, 2016 at 14:18
  • 1
    \$\begingroup\$ I already did this because I was bored... \$\endgroup\$
    – univalence
    Nov 19, 2016 at 12:52

ASCII to Unicode equation beautifier

You may well be used to typing equations in ASCII, but with the advent of Unicode we can spruce them up a bit. We can fix

  • Powers (numeric superscripts only)
  • Numeric subscripts
  • Mathematical signs (-, *, / ^ → -, ×, ÷, ↑)


x^3 - 1 = (x - 1)(x^2 + x + 1)  →  x³ − 1 = (x − 1)(x² + x + 1)
g_0 = 3^^3^^3 -= 3^(3^3)        →  g₀ = 3↑↑3↑↑3 = 3↑(3³)
800*600                         →  800×600
1/x                             →  1÷x

You may assume that all digits directly after a ^ or _ are meant to be super/subscripts (and the ^ or _ to be removed) and that all the mathematical signs are to be replaced wherever they appear.

This is , so the shortest solution wins.

  • 2
    \$\begingroup\$ This seems to be two questions crammed into one. The first one is the superscript and subscript transformation, which is mildly interesting; and the second one is the straight substitution of various characters for others, which is completely boring apart from the ambiguity it introduces in the interpretation of ^. I suggest ditching the substitution of minus, times, and divide symbols and giving explicit lists (with copyable characters and Unicode code points in decimal and hex) of the superscript, subscript, and up-arrow characters. \$\endgroup\$ Jul 14, 2016 at 6:39

Autotune a chord

Auto-Tune is a pitch correction program which alters the pitch without changing the length. It can be used to fix off-pitch chords in music, which is good because I have an out of tune piano. The goal of this challenge is given some input waveform which contains a single chord played on my piano, tune each note to the nearest equally tempered note found on a standard piano (see Input for more details).


The input is something which looks like a time-domain audio sample input containing a single chord being played. All data is sampled at 192kHz, with 16-bit PCM (little endian integer), mono channel. The input may come from any source desired (file io, stdio, function parameter, etc.).


The output of your code should be something which looks like a time-domain audio sample containing the tuned chord. It does not need to have the same sample rate or datapoint format as the input, but must be the same length in real time as the original sample (or as close as possible). The output may be to any source desired (file io, stdio, function parameter, etc.).


See this github repo for various inputs and outputs. The provided examples have inputs/outputs in an uncompressed wav file. Feel free to re-encode/gut the data for your inputs.


This is code golf; shortest code wins. Standard loopholes apply. You may use any libraries/builtins so long as they were not designed specifically for performing pitch correction.

Main concern: This challenge seems potentially too difficult, so one alternative I've been considering is changing the piano samples into sine waves at the fundamental frequencies (avoids issues with amplitude decay/harmonics). An even simpler challenge might be to give inputs in the frequency domain (list of fundamental frequencies), though I'm not sure that would make for an interesting challenge as it seems almost too easy at that point.

  • 1
    \$\begingroup\$ It seems very difficult to determine what outputs are considered correct. \$\endgroup\$
    – feersum
    Jul 14, 2016 at 7:44
  • \$\begingroup\$ yeah, that thought had crossed my mind as well. I've considered measures based on the delta of the FFT of user output/expected output, but I'm not sure this is necessarily a good measure of "in tune". \$\endgroup\$ Jul 14, 2016 at 7:47
  • \$\begingroup\$ I suspect that the biggest technical challenge would be phase. The harmonics of each string in isolation should be in phase, because they all derive from a single hammer strike, but the keys of the chord are probably not all struck at exactly the same time, and there will be resonant driving interactions between them which will complicate the signal. I suggest that you explicitly state that people can ignore this issue. \$\endgroup\$ Jul 14, 2016 at 13:39

This will be a challenge. Additional tags are , and .

How fast is your Stack Exchange community?


Your task is to find how fast a Stack Exchange community reacts. "How fast" is here the average of the time elapsed until the first answer or the closing of the question.


  • the Stack Exchange site's name, e.g. stackoverflow, codegolf, codereview etc.
  • optionally the Stack Exchange API URL: https://api.stackexchange.com/2.2/


  • Calculate the average time it takes until the first answer or closing of the question.
  • Take the 1000 latest questions into account, e.q. ten API requests with 100 items each.


  • Output the average time in minutes and seconds, like 01:23 or 1:23.
  • Run your program at least against stackoverflow, codegolf and code review and show the results.
  • Feel free to add results for your other favorite communities as well.


  • You can write a program or a function. If it is an anonymous function, please include an example of how to invoke it.
  • This is so shortest answer in bytes wins.
  • Standard loopholes are disallowed.
  • Leading/trailing whitespaces/newlines are fine.
  • 2
    \$\begingroup\$ How do you count unanswered and unclosed questions? Also, I don't know about the API, but there might be problems with deleted answers. I think you should probably write a reference implementation before posting this. \$\endgroup\$ Jul 14, 2016 at 13:10
  • \$\begingroup\$ @FryAmTheEggman Thanks a lot – all good points. Didn't think that there might be questions that are unclosed and unanswered. Will check the API whether deleted even will be send. Good point with the reference implementation – maybe in JavaScript that it can be run as a stack snippet. What do you think in general about the challenge idea? Boring? Interesting? Too complicated? \$\endgroup\$ Jul 14, 2016 at 13:15
  • 1
    \$\begingroup\$ It's about doing basically one task, so I don't think it is complicated. I think the results are probably more interesting than the challenge (there are only so many ways to average something and to parse html), but it makes sense and isn't trivial, so I wouldn't say it's boring. Seems fine overall. Also note internet, date and, I suppose, math. \$\endgroup\$ Jul 14, 2016 at 13:24
  • \$\begingroup\$ @FryAmTheEggman Thanks again for your feedback and the tag suggestions. I also think that the results are the interesting part. I wanted to try a popularity contest in the first place because of that. But I couldn't come up with the necessary criteria. :) \$\endgroup\$ Jul 14, 2016 at 13:30

Convert a BMP image to grayscale

The images manipulation is a great way to exercise and increase your skills. In my opinion it's also very interesting.

What you must do?

The objective of exercise is much easy: convert an image bmp colorful in an image grey.
You can use every language, the question most appreciated will be that don't use library.
Image stock: http://www.mediafire.com/convkey/c491/p7aya9cxafvfc91zg.jpg Image converted: http://www.mediafire.com/convkey/3903/rcigd79pkwd12qczg.jpg?size_id=3

  • 1
    \$\begingroup\$ What is the winning criterion? It is code-golf, popularity-contest, other criterion? \$\endgroup\$ Jul 15, 2016 at 10:30
  • 1
    \$\begingroup\$ Also, the You can use any language is unnecessary, it's implied here. And you can use ![](<image url>) to show the images. \$\endgroup\$ Jul 15, 2016 at 10:33
  • 4
    \$\begingroup\$ This is very underspecified at present. 1. What weights should be used in the conversion from RGB to greyscale? 2. What bit depths should be supported? 3. Is it required to support all of BMP's features (e.g. ICC colour profiles, CMYK, JPEG, PNG)? If not, what is the minimum feature set which must be supported? \$\endgroup\$ Jul 15, 2016 at 10:53
  • \$\begingroup\$ @Blind To mention someone, you can use @<username>, and please add the tags to your post ([tag:<tag name>]) \$\endgroup\$ Jul 15, 2016 at 12:43
  • \$\begingroup\$ @TùxCräftîñg , It's a code-golf. @ Peter Taylor , It's equal, you can use that weight you want. 2.see 1st. 3.just support BMP. \$\endgroup\$
    – Blind
    Jul 15, 2016 at 16:27
  • 1
    \$\begingroup\$ That doesn't actually answer questions 2 or 3. \$\endgroup\$ Jul 15, 2016 at 19:32
  • \$\begingroup\$ Just to let you know, you can only @mention one person per comment, and it won't work with a space between the @ and the name. \$\endgroup\$ Jul 20, 2016 at 16:20

Do I have an emoji?

Given an input string in your language, return truthy/falsey if the input contains an valid Unicode emoji character.

What is an Emoji?

The word emoji comes from the Japanese:

絵 (e ≅ picture) 文 (mo ≅ writing) 字 (ji ≅ character).

Emojis are pictorial symbols used to represent feelings, actions, or objects.

For this challenge, use the Full Emoji Data list provided by Unicode as a reference to determine which characters are valid Emojis.

Sample test cases:

"" -> 0

"💩" -> 1

"hello💩" -> 1

"hello" -> 0

"!±≡𩸽" -> 0


This seems trivial, but I noticed we didn't have an emoji detection challenge. There might be a concern about the encoding of the input string, but reading the linked meta posts about Strings I feel that this challenge can use whatever String format the language used in the answer supports.

The acceptable output for booleans is also up for discussion. Do we have a meta post on what output formats are acceptable for booleans?

  • 5
    \$\begingroup\$ One question: What exactly is an emoji? I think it should be specified in the challenge. \$\endgroup\$
    – user48538
    Jul 18, 2016 at 16:38
  • 1
    \$\begingroup\$ See meta.ppcg.lol/q/2190, just say truthy/falsey. \$\endgroup\$ Jul 18, 2016 at 16:39
  • \$\begingroup\$ @zyabin101 can I use Unicode's emoji list as a list of valid emoji characters for this challenge? \$\endgroup\$
    – JAL
    Jul 18, 2016 at 16:44
  • \$\begingroup\$ Up to you. [filler text] \$\endgroup\$
    – user48538
    Jul 18, 2016 at 16:51
  • \$\begingroup\$ I've attempted to clarify what an emoji is, at least for this challenge. Hopefully this will make this question more clear and a better fit for the site. \$\endgroup\$
    – JAL
    Jul 19, 2016 at 2:39
  • \$\begingroup\$ A source which gives actual ranges would be more convenient for people writing answers, although unicode.org/Public/emoji/3.0//emoji-data.txt isn't entirely consistent with the other lists. \$\endgroup\$ Jul 19, 2016 at 9:52

Trim trailing spaces in less than O(n²) time

Since s/\s+$// runs in O(n²) time, Stack Overflow needs to replace it with something faster. Please write a code snippet for them. Your score will be the number of bytes in your submission, multiplied by the time taken to process a string of 1000 non-spaces with 1,000,000 leading and trailing spaces, divided by 1000 times the time taken to process a string of 1-non space with 1,000 leading and trailing spaces. (In other words, if your code runs in O(n) time then this should cancel out.)

  • \$\begingroup\$ The fancy scoring seems like it might be confusing/hard to implement. Why not just restrict the complexity to be less than O(n²) like you suggest in the title? \$\endgroup\$ Jul 21, 2016 at 13:18
  • \$\begingroup\$ @FryAmTheEggman That's fairly easily fixed with (?<!\s)\s+$ or using RTL matching in .NET/Retina. (In fact, the latter would probably be a good candidate for winning the challenge even if actual runtime is taken into account.) \$\endgroup\$ Jul 21, 2016 at 15:29
  • \$\begingroup\$ @MartinEnder That is what I had in mind ;) Is there a reason that would be bad? I think that is likely around the best performance you can get, being linear w.r.t. the number of spaces at the end of the string? \$\endgroup\$ Jul 21, 2016 at 15:33
  • \$\begingroup\$ @FryAmTheEggman well it just means there's likely a very simple solution that might win the challenge. \$\endgroup\$ Jul 21, 2016 at 15:35
  • \$\begingroup\$ The scoring mechanism is not going to work well because the times will be so small that there will be more noise in the measurement than signal. Especially for the smaller test case, where the string will fit in L1 cache. \$\endgroup\$ Jul 21, 2016 at 16:15
  • 1
    \$\begingroup\$ The scoring can be exploited by intentionally doing badly on smaller cases. \$\endgroup\$
    – xnor
    Jul 22, 2016 at 2:46

Reverse stdin to stdout, Unicode aware and by grapheme clusters

It's 2016. High time that we were Unicode-aware, don't you think?

Given a UTF-8 string on stdin, reverse it by extended grapheme clusters (as defined by the Unicode consortium; this can be found here, for instance) and place it on stdout. This is seemingly a trivial challenge, but surprisingly difficult in many languages.

You may not use an explicit end-of-file character - if supported, use EOF. If not supported, use an explicit length at the start of the string, and document the format.

Note that the string may be of any length (barring RAM limitations).

You must handle direction override characters "properly", in the sense that the directionality of every character must remain the same after reversing the string.

  • \$\begingroup\$ I need examples for this. \$\endgroup\$
    – TLW
    Jul 21, 2016 at 23:05
  • \$\begingroup\$ Can you give a link to the Unicode definition of grapheme clusters? \$\endgroup\$
    – Steven H.
    Jul 21, 2016 at 23:59
  • \$\begingroup\$ Sure: unicode.org/reports/tr29/#Grapheme_Cluster_Boundaries \$\endgroup\$
    – TLW
    Jul 22, 2016 at 1:18
  • 1
    \$\begingroup\$ I don't understand what you mean about the direction override characters. Could you give some test cases (preferably with hexdumps of input and output)? What other corner cases are there? On a first read-through I can tell that understanding the definition of extended grapheme cluster will take some time, but I'm not clear on whether it covers e.g. emoji type modifiers, emoji families, etc. The definition also makes reference to degenerate cases, and they should be covered by test cases too. \$\endgroup\$ Jul 27, 2016 at 12:05

Given an ordinal A and a nonnegative integer B, compute f_A(B)

A will be expressed in Cantor Normal Form. A number is in Cantor Normal Form if it is 1, or if it is either of the form a+b or w^a, where a and b are in Cantor Normal Form, and w is the ordinal omega.

f_A(B) is the fast-growing hierarchy, specifically, the Wainer Hierarchy. f_0(n)=n+1. f_(a+1)(n)= n iterations of f_a to n. f_a(n)=f_an if a is a limit ordinal. a[n] is the nth member of the fundamental sequence for a.

  • \$\begingroup\$ You'll definitely need to add the definitions of these, as most people won't have any idea how to answer. You can use one of various online TeX -> image services and add those to your post. Sadly we don't have MathJax activated because it borked a bunch of other features. I'd also heavily recommend adding some test cases. \$\endgroup\$ Jul 21, 2016 at 14:38
  • \$\begingroup\$ Test cases are impossible to add. But thanks for formatting advice! \$\endgroup\$ Jul 21, 2016 at 15:14

Difficulty Naming


In the game osu!, players can make their own beatmaps (levels). The harder the beatmap is, the bigger the star rating will be.

Beatmaps have star icons with different colors and letters according to their star rating.

osu! has also four game modes. The shape of the icon will change depending on the game mode.

The Challenge

Given a star rating and a gamemode, output the (generally used) difficulty name.

The Input

  • A positive floating-point number (star rating)
    • You can assume the number will never have more than two decimal places.
  • A positive integer (game mode)
    • 1 is Standard, 2 is Taiko, 3 is Catch the Beat and 4 is Mania (you can use other numbers for the game modes, these are the default.)

The Output

  • A string (difficulty name)

Use this table to determine the output:

diff table http://image.prntscr.com/image/174e34e56dbb40ad81b209fac3fdc5bc.png

(if the star rating is exactly 5.25, it counts as Above 5.25)


  • This is , so the shortest solution wins.
  • Standard loopholes are disallowed.
  • You can take the input either as an array/list, a string separated by commas or spaces, or each number individually.
  • Again, you can use other numbers for the game modes.

Test Cases

2.24, 2 ---> Futsuu
1.1, 4 ---> Basic
3.14, 1 ---> Hard
7.0, 3 ---> Overdose
5.0, 4 ---> Exhaust
31415.0, 2 ---> Ura Oni
0.01, 1 ---> Easy
5.25, 1 ---> Expert


If you see any grammar errors or anything like that, please tell me, I'm not a native speaker.


Could it all add up to nothing?

Given an input number, return a truthy value if it is possible to arrange permutations of + and - signs to make the sum of the range up to that value 0.

Note that you don't have to output which set of combinations is truthy.

Test cases:

1 (1) --> Falsy
2 (1,2) --> Falsy
3 (1,2,3) --> 1+2-3, Truthy
4 (1,2,3,4) --> -1+2+3-4, Truthy
5 (1,...,5) --> Falsy
6 (1,...,6) --> Falsy
7 (1,...,7) --> 1+2-3-4+5+6-7, Truthy
8 (1,...,8) --> 1+2-3-4-5-6+7+8, Truthy

Just one thing - no using eval or similar commands.

This is so the shortest answer in bytes wins!

Sandbox notes

Require a range or a list as input?

  • \$\begingroup\$ For general lists, I think it would be basically the same as this question. For ranges, I suspect there's a direct criterion. \$\endgroup\$
    – xnor
    Jul 25, 2016 at 21:50
  • \$\begingroup\$ @xnor so do I which is why I think it could be more interesting if the pattern could be found. It gets exponentially longer as the lists get longer so is harder to check - better as a fastest algorithm/fastest code? \$\endgroup\$
    – Blue
    Jul 25, 2016 at 21:52
  • \$\begingroup\$ I can show that it's exactly the numbers of form 4*n or 4*n-1. \$\endgroup\$
    – xnor
    Jul 25, 2016 at 21:59
  • \$\begingroup\$ (Ninja'd by xnor) There is probably quite an easy pattern, e.g. if the upper limit is divisible by four, you can always find a zero sum with a "little gauss" argument. \$\endgroup\$
    – flawr
    Jul 25, 2016 at 22:01
  • \$\begingroup\$ @xnor I'd love to see a proof=) \$\endgroup\$
    – flawr
    Jul 25, 2016 at 22:01
  • 4
    \$\begingroup\$ The key is that the pattern +--+ on four consecutive numbers adds to 0. With 4*n, just repeat +--+. With 4*n-1, do +1+2-3, then repeat +--+. With the rest, the sum is odd so it's impossible. \$\endgroup\$
    – xnor
    Jul 25, 2016 at 22:04
  • \$\begingroup\$ So saveable? Sounds trivial now. \$\endgroup\$
    – Blue
    Jul 25, 2016 at 22:31
  • \$\begingroup\$ Well, I'm guessing xnor's solution would be pretty short. \$\endgroup\$
    – mbomb007
    Jul 29, 2016 at 14:35
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