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2872 Answers 2872

75 76
78 79

4B/5B Encoding

Tags: , , ,

To transmit data in binary there's a technique called NRZI where a transition from +V to -V or vice-versa encodes a 1 and the absence of such a transition a 0.

In case there are a lot of 1 s this is a good method because there will be a lot of transmissions and the risk of getting out of sync is smaller, however it's still tricky for long runs of 0. One solution is 4B/5B encoding which takes 4bits and encodes it in 5bits in a way such that there are at most 3 consecutive bits. The encoding is as follows:

0000 -> 11110
0001 -> 01001
0010 -> 10100
0011 -> 10101
0100 -> 01010
0101 -> 01011
0110 -> 01110
0111 -> 01111
1000 -> 10010
1001 -> 10011
1010 -> 10110
1011 -> 10111
1100 -> 11010
1101 -> 11011
1110 -> 11100
1111 -> 11101


Given an bytestring, you will need to encode it with the aforementioned encoding.


  • Input will be a bytestring
    • to avoid padding-issues, it will always have a length that is a multiple of 4 bytes
  • Output will be of the same a bytestring
  • You may not assume that input (nor output) is printable


Note: You need to handle both printable and unprintable examples!

ASCII examples

SSePCcGG -> ]WW-~U]U=O
aHAHaHAHaHAHaHAHaHaO -> rU%%RrU%%RrU%%RrU%%RrU'%]
S#KPOCFWcSaOCR!@sz!\q9OKqY!L -> ]iU]~WUU9ouWW%]UWJ%^}_j%zzk5uWzW:%Z

Unprintables in hexadecimal

c0 ff ee 00 -> d7 bb de 73 de
66 6f 6f 00 -> 73 9d d7 77 de
50 50 43 47 -> 5f 97 e5 55 4f
48 65 6c 6c 6f 2c 20 57 6f 72 6c 64 21 00 00 00 -> 54 9c b7 69 da 77 69 aa 79 6f 77 5f 47 69 ca a2 7d ef 7b de
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Drat! This was actually run three years ago! See Implement INTERCAL's Binary Operators

Mingle and Select

The programming language INTERCAL has two operators, "mingle" or "interleave" (represented by $), and "select" (represented by ~).

Mingle takes two 16-bit values and produces a 32-bit value by taking single bits alternately from its left and right operands, and concatenating them together - for example, 65535 $ 0 will yield -1431655766 as follows: 65535 is 0xFFFF, or 0b1111111111111111; 0 is 0b0000000000000000. Taking the bits alternately gives 0b10101010101010101010101010101010, or 0xAAAAAAAA, which evaluates to the signed integer -1431655766.

A simplified implementation of Select takes two 32-bit operands, and produces a 32-bit value by comparing the bits in the two operands, and wherever the right operand has a 1 bit, take the corresponding bit from the left operand. The resulting bits are compressed to the right, and zero-filled on the left - for example, an 8-bit version of select, given 79 ~ 42, would return 3, as follows: 79 is 0b01001111, and 42 is 0b00101010. Numbering the bits from the left starting with 1, we need to take the third, fifth, and seventh bits of 79, which are 0, 1, and 1, respectively. We then compress them to the right - 011 - and zero-fill, yielding 0b00000011, or 3.

The challenge

Without using INTERCAL, and without using in any other language any operator builtin that amounts to either mingle or select, implement both operators. You may implement them for 16- or 32-bit integers, and as either operators or functions, but input and output must be UNsigned decimal integers. (I'm not going to require input or output in INTERCAL-style insanity!)

Further constraint: The minimum bit width of the operands must be 16 bits for mingle (no constraint for select), and if they are unequal in width, the shorter is zero-filled on the left to match the width of the longer (zero fill applies to both mingle and select).

test cases to be written and inserted here

This is , so shortest solution in each language 'wins'. (Naturally, the standard loopholes are forbidden.)

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  • \$\begingroup\$ "You may implement them for 16- or 32-bit integers, and as either operators or functions, but input and output must be signed decimal integers." This puts languages such as CJam which only have unbounded integer types at a disadvantage. Is that intended? \$\endgroup\$ – Peter Taylor Aug 13 '18 at 11:03
  • \$\begingroup\$ @PeterTaylor - The intent here is not to exclude or disadvantage any particular language (other than INTERCAL), but to 'force' zero-fill on select, and return negative values for some cases of mingle. Would it be better to say that values returned must be some power of 2 times 16-bits wide? \$\endgroup\$ – Jeff Zeitlin Aug 13 '18 at 11:32
  • \$\begingroup\$ I'm not sure why an integer width is necessary for select: just say that it's zero-filled and make sure that the corner case of a ~ -1 (the only way that there's nothing to fill) is covered by test cases for a positive, zero, and negative. \$\endgroup\$ – Peter Taylor Aug 13 '18 at 11:49
  • \$\begingroup\$ For mingle, it seems to me that if the operands are really 16-bit values then the example should be -1 $ 0, and if you require the output to be negative iff the first operand is negative then it easily generalises to unbounded integer types. However, the result of -1 $ 0 in an unbounded type is unbounded, so there is a genuine problem there. I suggest mentioning that briefly as a rationale for why languages without bounded integer types must nevertheless implement a bounded mingle. \$\endgroup\$ – Peter Taylor Aug 13 '18 at 11:52
  • \$\begingroup\$ @PeterTaylor - but zero-filled to what width? One could argue that if an arbwidth is allowed, then no zeros need to be added, and 79 ~ 10 could validly be argued to be -1. \$\endgroup\$ – Jeff Zeitlin Aug 13 '18 at 11:52
  • \$\begingroup\$ To infinite width. The most significant bit is effectively a sign bit. \$\endgroup\$ – Peter Taylor Aug 13 '18 at 11:57
  • \$\begingroup\$ @PeterTaylor - Good catch on 65535 vs -1. I think that this and the arbwidth problem can be addressed by simply changing the 'signed' constraint to 'unsigned'. \$\endgroup\$ – Jeff Zeitlin Aug 13 '18 at 12:06
  • \$\begingroup\$ But how do I tell the (selected) MSB from presumed zero-filling? Is 79~10 equal to 3, or to -1? \$\endgroup\$ – Jeff Zeitlin Aug 13 '18 at 12:07
  • \$\begingroup\$ UPDATE: Constraints changed. \$\endgroup\$ – Jeff Zeitlin Aug 13 '18 at 12:19
  • \$\begingroup\$ "But how do I tell the (selected) MSB from presumed zero-filling?" The answer would have been that you can only get a negative result if both operands are negative, but now that you've changed to unsigned this is no longer an issue. \$\endgroup\$ – Peter Taylor Aug 13 '18 at 13:44
  • \$\begingroup\$ @PeterTaylor - It's been rendered academic; the challenge has already been run - see codegolf.stackexchange.com/questions/54412/… \$\endgroup\$ – Jeff Zeitlin Aug 13 '18 at 18:30

Lucas and Fibonacci are in pair

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Write a program of at most 150 bytes length that produces different outputs (to stdout) if you insert token-delimiters into the source code.


The program with the highest number of different outputs wins.


For this challenge, we assume that the source code is tokenized prior to compilation/ interpretation. A delimiter is any character which, when inserted at a given position, changes the way the source code is tokenized, but is not part of a token on its own.



a = "0"
b = "0"

This prints 12. If you insert two spaces, you get:

a = "0"
b = "0"
print(a in b)

which prints "True"

This gives a total score of 2 for 2 (known) different outputs

Rule clarifications

  • Only program versions that exit with zero error code count towards the score.
  • All versions have to be listed explicitly in the answer.
  • Put the score into the title of the answer.
  • If your code produces different outputs in different programming languages, they count.
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Write in Ge'ez!

Ge'ez is a now-dead language used for liturgy in the Ethiopian Orthodox Church. It's traditionally written in a very distinctive script with many loops and curves. But for ease of reading, it's often transliterated into Latin letters.

Your task here is to take some writing in Latin characters and convert it to the Ge'ez script. For example, the input gəʿəz (the name of the language) should give the output ግዕዝ.


Every letter in the Ge'ez script represents both a consonant and the following vowel. There are eight vowels

ä u i a e ə o wa

and twenty-six consonants

h l ḥ m ś r s ḳ b t ḫ n ʾ k w ʿ z y d g ṭ p̣ ṣ ḍ f p

which give 208 combinations. (In reality, some of these combinations like "wwa" don't actually happen, and there are some extra letters like "mya" not covered by this system, but ignore all that for the purposes of the challenge.)

The combinations are encoded in a convenient way in Unicode. The consonant defines the starting position, and the vowel defines the offset.

C    start (hex)
h    1200
l    1208
ḥ    1210
m    1218
ś    1220
r    1228
s    1230
q    1240
b    1260
t    1270
x    1280
n    1290
ʾ    12A0
k    12A8
w    12C8
ʿ    12D0
z    12D8
y    12E8
d    12F0
g    1308
ṭ    1320
p̣    1330
ṣ    1338
ḍ    1340
f    1348
p    1350

V    offset
ä    +0
u    +1
i    +2
a    +3
e    +4
ə    +5
o    +6
wa   +7

(Again, reality is a bit more complicated, but ignore that for the challenge.)

Your task is to take a string written in Latin letters, and output the corresponding Ge'ez syllables. If a consonant is followed by a vowel, add the consonant's start point to the vowel's offset, and print the corresponding Unicode character. If a consonant is not followed by a vowel, use the start point with no offset (as if the vowel were ä).

Input is a Unicode string, or a list of Unicode codepoints as integers. Output is a Unicode string, or a list of Unicode codepoints as integers. A "Unicode string" here can be in any official encoding (UTF-7, UTF-8, UTF-16, UTF-32…). NFC and NFD are both acceptable.

This is code golf; shortest code (in bytes) in each language wins. Standard loopholes forbidden.

Test cases:

gəʿəz    ግዕዝ
[TODO more to come here]
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  • 1
    \$\begingroup\$ One option would be replacing the Unicode characters in the input with ASCII digraphs. Would this make the challenge better? \$\endgroup\$ – Draconis Aug 25 '18 at 3:15

Cleanup after Easter

As most people (probably) know, Easter was about a month ago. But around the community park, there're still some eggs hidden! Can you build a (Javascript only, please) bot to help us find them?

The Course:

The bots will compete on a 201 by 201 field, with coordinates ranging from (-100,-100) to (100,100). All bots will start at (0,0) and will begin. The course has 3 types of areas: wall, egg, and grass.

Finding Eggs:

You will build a function to find the Easter eggs. To see, use the function checkZone(x,y). The ouputs are:


To find a player, use findPlayer(name) (Returns coordinates array) or detectPlayers(x,y) (Returns name or null). If your bot lands on a spot with an egg, it will not be able to pick it up until it uses the function fetchEgg(). If there is not an egg, your turn will stop.


Moving is done by returning a string. This string can be north, south, east, west, or none. Assume that directions are like a compass placed on a Cartesian plane:



When all eggs are found, the game is over. If you move North to an egg, you must pick it up on the next turn in order to obtain it. If you try to move into a wall, nothing will happen. To get your current position, use getLocation(), which returns an array [x,y]. Use getBasket() to get how many eggs you have. The course will be designed so that all eggs are accessible, but some may be hard to get. If an egg is found by two bots in the same turn, each bot get 0.5 added to their basket, as they split the reward.

Functions list:

checkZone(x,y): Returns an integer, takes 2 integers as parameters between -100 and 100
findPlayer(name): Takes string as parameter, returns player's position as an array of integers or null if player is nonexistant
detectPlayers(x,y):  Returns an array of player names, takes 2 integers as parameters between -100 and 100
fetchEgg(): Fetches egg in current position if exists
getLocation(): Returns array with x and y coordinates
getBasket(): Returns number of eggs in basket
move functions: Returns true or ends turn (Ends turn if cannot move)


function rabbitInDisguise() {
    var loc = getLocation();
    if (getLocation(loc[0], loc[1]) == 2) {
    } else {
        if (getLocation(loc[0], loc[1] + 1)) {
        } else {

UPDATE 1: Added findPlayer(name) and detectPlayers(x,y) functions, made locating functions "background", and added simultaneous egg finding rules

UPDATE 2: Added functions list

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  • \$\begingroup\$ 1. I don't see the competition here. I find egg, and I grab it as fast as I can. 2. You don't describe what other players look like. \$\endgroup\$ – Nathan Merrill Apr 18 '18 at 15:05
  • \$\begingroup\$ @NathanMerrill Whichever bot is best at searching for the eggs will win. If Bot A moves in circles around the board, but Bot B scans a 5 x 5 area around it and makes a map, who do you think would win? Also, what do you mean by "You don't describe what other players look like." \$\endgroup\$ – Redwolf Programs Apr 18 '18 at 15:08
  • \$\begingroup\$ @user202729 Mistype...I'm about to fix it \$\endgroup\$ – Redwolf Programs Apr 18 '18 at 15:09
  • \$\begingroup\$ And how does program interact with each other? Is there any limit on how many (library) function call can be done in each function? \$\endgroup\$ – user202729 Apr 18 '18 at 15:12
  • \$\begingroup\$ it cannot run more than one function per turn \$\endgroup\$ – Redwolf Programs Apr 18 '18 at 15:16
  • \$\begingroup\$ 2. You said that 0=grass, 1=Wall, and 2=Egg. What does a player look like? 1. Right, let me check all the locations nearest me, and walk to the closest one. (You can slightly improve on this by identifying eggs that other players will go for). it cannot run more than one function per turn: Your example bot breaks that. \$\endgroup\$ – Nathan Merrill Apr 18 '18 at 15:17
  • \$\begingroup\$ Players are invisible, since it could complicate things. I'll edit to address the problems, though. \$\endgroup\$ – Redwolf Programs Apr 18 '18 at 15:18
  • \$\begingroup\$ If players are invisible, then this is not a KotH. This is a code-challenge to retrieve the eggs as fast as possible. \$\endgroup\$ – Nathan Merrill Apr 18 '18 at 15:19
  • \$\begingroup\$ @NathanMerrill Fixed. Players can be found using findPlayer() and detectPlayers() \$\endgroup\$ – Redwolf Programs Apr 18 '18 at 15:27
  • \$\begingroup\$ So they can investigate the whole board in one turn? Sounds more interesting then. \$\endgroup\$ – user202729 Apr 18 '18 at 15:40
  • \$\begingroup\$ @user202729 I think moveRandom() would be fine (It just makes it easier to program), but I'll add more info on the functions. \$\endgroup\$ – Redwolf Programs Apr 18 '18 at 15:45
  • \$\begingroup\$ @user202729 Good point. Feature eliminated. \$\endgroup\$ – Redwolf Programs Apr 18 '18 at 15:51
  • \$\begingroup\$ Any other problems you can think of? \$\endgroup\$ – Redwolf Programs Apr 18 '18 at 15:54
  • 2
    \$\begingroup\$ @PeterTaylor (what does KISS stand for?) \$\endgroup\$ – user202729 Apr 19 '18 at 13:58
  • 1
    \$\begingroup\$ @user202729, softwareengineering.stackexchange.com/a/87/13258 \$\endgroup\$ – Peter Taylor Apr 19 '18 at 14:02

Compute strings with a fixed distance

Consider a binary string of total length N. For example 000111000111 with \$N=12\$. We can store this string more efficiently by just recording the indices where bits are first flipped. In this case \$(3, 6, 9)\$ is sufficient information to reproduce the string. Here are some more examples using this representation:

10101010 is represented inefficiently by (0, 1, 2, 3, 4, 5, 6, 7)
1111100000 is represented by (0, 5)
0000011111 is represented by (5)

Let us call this representation of a string its compressed representation (even though sometimes the compression is worse than doing nothing).

Now recall the Levenshtein distance between two strings. This is the minimum number of single character insertions, deletions and substitutions needed to transform one string into the other.


Given a compressed representation of a string \$S\$ of uncompressed length \$N\$ and two positive integers \$k\$ and \$d\$, the task is to output the compressed representation of \$k\$ distinct strings of uncompressed length \$N\$, each with Levenshtein distance exactly \$d\$ from \$S\$. If there are not that many distinct strings possible, simply output as many as possible.

Your code must run with \$N = 1000000\$ and \$k, d < 20\$ in less than a minute on a normal desktop PC.

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  • \$\begingroup\$ I feel like I'm missing something here. Everything above "Task" is clear, but I don't understand what's going on below. If S = "0000", k = 1, and d = 1, then I simply print out one of "1000", "0100", "0010","0001" (in compressed format)? \$\endgroup\$ – Nathan Merrill Aug 30 '18 at 20:09
  • \$\begingroup\$ @NathanMerrill Yes. It’s less trivial with other starting strings and larger d. \$\endgroup\$ – Anush Aug 30 '18 at 20:26
  • \$\begingroup\$ Right. There seems to be two parts to this challenge then: Iterating through Levenshtien mutations, and converting to/from the compressed representation. Unless I'm missing how they are related, I'd argue that you remove one of the parts. \$\endgroup\$ – Nathan Merrill Aug 30 '18 at 20:32
  • \$\begingroup\$ It appears that iterating Levenshtien mutations would be a duplicate \$\endgroup\$ – Nathan Merrill Aug 30 '18 at 20:33
  • \$\begingroup\$ @NathanMerrill Thanks for the possible dupe link. For your first comment, the idea was to do it without first uncompressing. I will try to improve the question. \$\endgroup\$ – Anush Aug 31 '18 at 9:10

Find the fixed point prime


Let f(n) be the sum of the sums of each possible factor permutation of n. Determine (Y f)(n) (a fixed-point combinator) for all n from 2 to the input i (where i is an integer greater than 2).


A quick example of f(n) first:

f(8) = (2 + 2 + 2) + (2 + 4) + (4 + 2) + (8)
f(8) = 26

An example of (Y f)(n):

(Y f)(8)
  f(8) = 26 (shown above)
  f(26) = (2 + 13) + (26) = 41
  f(41) = (41) = 41
(Y f)(8) = 41

Let g(i) define a function which implements the described problem.





This is a challenge, which means that the solution with the lowest asymptotic time complexity (in Θ(...)) wins!

In the case of a tie, the winner will be determined by rules, so please submit an implementation of your code along with the algorithm. Tie-break will occur on a Google Compute Engine n1-standard-8 instance.

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  • 1
    \$\begingroup\$ In your example of f(8), why do you include 1 in (8+1)? Should it not be just (8)? I could make (2+2+2) into (2+2+2+1). \$\endgroup\$ – Don Thousand Sep 3 '18 at 16:22
  • \$\begingroup\$ @RushabhMehta You're entirely correct! My bad, let me fix that. \$\endgroup\$ – Addison Crump Sep 3 '18 at 16:27
  • \$\begingroup\$ I'm also pretty sure that in your system, the only fixed points are prime numbers. Is that desired functionality? \$\endgroup\$ – Don Thousand Sep 3 '18 at 16:33
  • \$\begingroup\$ @RushabhMehta See the title. :P \$\endgroup\$ – Addison Crump Sep 3 '18 at 16:45
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    \$\begingroup\$ Finally, I would mention that permutations are distinct, just so that its obvious how you are counting all the different factors. \$\endgroup\$ – Don Thousand Sep 3 '18 at 16:47
  • \$\begingroup\$ @RushabhMehta Is that a sufficient change? \$\endgroup\$ – Addison Crump Sep 3 '18 at 16:49
  • 1
    \$\begingroup\$ That should be fine. Nice challenge! \$\endgroup\$ – Don Thousand Sep 3 '18 at 16:52
  • \$\begingroup\$ 1. What's a factor permutation? 2. I doubt anyone can give even a moderately tight analysis of the runtime for an implementation of g without a very tight analysis of the output of f, so I don't think it makes sense to post this without first editing in tight bounds on the output of f. \$\endgroup\$ – Peter Taylor Sep 5 '18 at 10:18

Bijective base banana to decimal

I'm trying to decide between 2 versions of this challenge:

Prefixes version

Let bijective base banana be bijective base 6 with each digit represented by a different prefix of the word "banana" as follows:

1: "b"
2: "ba"
3: "ban"
4: "bana"
5: "banan"
6: "banana"

The input will be a string made up of a concatenation of zero or more of only these strings. This represents a number in bijective base 6 (with most significant digit first).

The output will be the same number in (non-bijective) decimal.

For example: (input : output)

"" : 0
"b" : 1
"bb" : 7
"bbanana" : 12
"bananab" : 37
"bananabanana" : 42

A worked example:

banana ban bana
6      3   4
6*36 + 3*6 + 4*1
216  + 18  + 4

This is code golf, so the winner for each language is the code with the fewest bytes.

Suffixes version

Let bijective base banana be bijective base 6 with each digit represented by a different suffix of the word "banana" as follows:

1: "a"
2: "na"
3: "ana"
4: "nana"
5: "anana"
6: "banana"

The input will be a string made up of a concatenation of zero or more of only these strings. This represents a number in bijective base 6 (with most significant digit first).

The output will be the same number in (non-bijective) decimal.

For example: (input : output)

"" : 0
"a" : 1
"aa" : 7
"abanana" : 12
"bananaa" : 37
"bananabanana" : 42

A worked example:

banana ana nana
6      3   4
6*36 + 3*6 + 4*1
216  + 18  + 4

This is code golf, so the winner for each language is the code with the fewest bytes.

Sandbox questions

  • Is this a duplicate or near enough a duplicate to be worth changing?
  • Are there any flaws that would make the challenge uninteresting?
  • I'm considering changing from prefixes to suffixes following this comment from H.PWiz
  • I currently prefer the suffixes version, but the worked example shows that the interpretation is no longer unique. It would either need to be specified that the match must be greedy (or non-greedy) or a different word would need to be chosen. I lean towards choosing a word which still gives enough ambiguity to force looking ahead, while still keeping the overall conclusion unique. I can't see a way of doing this without making the last character unique though, which reduces to a reversed version of the prefixes version (you can still ignore all other letters).
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  • \$\begingroup\$ This is interesting, but I can't help but feel like this is two problems glued into one. \$\endgroup\$ – Don Thousand Sep 3 '18 at 16:27
  • \$\begingroup\$ It's definitely 2 problems glued onto 1. What I'm wondering is will that make for additional golfing opportunities in at least some languages? I suspect it will, and I'd expect myself to underestimate the possibilities since there are many more insightful people tackling challenges, in many languages. \$\endgroup\$ – trichoplax Sep 3 '18 at 17:58
  • \$\begingroup\$ Hmm, It seems like all that matters is where the bs are. I wonder if it would be a more or less interesting challenge with a different word \$\endgroup\$ – H.PWiz Sep 3 '18 at 18:18
  • \$\begingroup\$ Good point. Suffixes instead of prefixes might give a bit more of a challenge. As long as the output is still guaranteed unique \$\endgroup\$ – trichoplax Sep 3 '18 at 18:34
  • 1
    \$\begingroup\$ @trichoplax I'm pretty sure you can't do suffixes, since its no longer unique. \$\endgroup\$ – Don Thousand Sep 3 '18 at 20:24
  • \$\begingroup\$ @RushabhMehta yes it stops working for the word "banana" then (I added a note about this in my 4th sandbox point). It would still work for words that don't contain any of their own suffixes as non-suffix substrings. Like "aardvark". I'm not sure which way to go yet \$\endgroup\$ – trichoplax Sep 3 '18 at 20:37
  • \$\begingroup\$ @trichoplax Your last bullet was exactly my issue. Suffixes don't help at all. \$\endgroup\$ – Don Thousand Sep 3 '18 at 22:34



Given a non-empty string containing non-numeric printable ASCII characters, compute its facto-run-length encoded version.


Let \$S\$ be a non-empty string of length \$l\in\mathbb{N}^+\$ containing non-numeric printable ASCII characters. For each positive factor \$n|l\$ one can represent \$S\$ as a \$n\times(l/n)\$ matrix of characters. Let \$M^n\$ denote these matrices.
For a given \$n\$ and \$j\in\{1,\dots,l/n\}\$, let \$C^n_j\$ be the string representing the \$j\$-th column of \$M^n\$. Let \$E^n_j := \text{RLE}(C^n_j)\$ denote the array of this string's run-length encoded version.
Let \$R^n := \text{CNCT}(\text{ZIP}(E^n_1,\dots,E^n_{l/n}))\$ denote the string representation of all \$E^n_j\$ zipped together.
The facto-run-length encoding of \$S\$ is defined as the string array \$\text{FRLE}(S):=[R^n:n|l]\$.

For a family of \$k\in\mathbb{N}^+\$ arrays \$A^i\$ with respective lengths \$l^i, i\in\{1,\dots,k\}\$ and elements \$A^i_j,j\in\{1,\dots,l^i\}\$, let \$Z_j:=[A^i_j:i\in\{1,\dots,k\}\land j\leq l_i]\$ denote the array of elements with index \$j\$, if present. Furthermore, define \$\text{ZIP}(A^1,\dots,A^k) := Z_1\Vert\dots\Vert Z_{\max\{l_i\}}\$ as the concatenated array of all \$Z_j\$.
For an array of strings \$A\$ of length \$j\$, define \$\text{CNCT(A)}:=A_1\Vert \dots\Vert A_j\$ as the concatenation of all strings of \$A\$.


Let \$S:=\text{"Hello world!"}\$, therefore \$l=12\$ with positive factors \$\{1,2,3,4,6,12\}\$. $$ M^2=\begin{pmatrix} \text{H}&\text{e}&\text{l}&\text{l}&\text{o}&\text{ }\\ \text{w}&\text{o}&\text{r}&\text{l}&\text{d}&\text{!} \end{pmatrix}, \\C^2_1=\text{"Hw"}, C^2_2=\text{"eo"}, C^2_3=\text{"lr"}, \\C^2_4=\text{"ll"}, C^2_5=\text{"od"}, C^2_6=\text{" !"}, \\E^2_1=[\text{"H"},\text{"w"}], E^2_2=[\text{"e"},\text{"o"}], E^2_3=[\text{"l"},\text{"r"}], \\E^2_4=[\text{"l2"}], E^2_5=[\text{"o"},\text{"d"}], E^2_6=[\text{" "},\text{"!"}], \\R^2=\text{"Hell2o word!"} $$

$$ M^4=\begin{pmatrix} \text{H}&\text{e}&\text{l}\\ \text{l}&\text{o}&\text{ }\\ \text{w}&\text{o}&\text{r}\\ \text{l}&\text{d}&\text{!}\\ \end{pmatrix}, \\C^4_1=\text{"Hlwl"}, C^4_2=\text{"eood"}, C^4_3=\text{"l r!"}, \\E^4_1=[\text{"H"}, \text{"l"}, \text{"w"}, \text{"l"}], E^4_2=[\text{"e"}, \text{"o2"}, \text{"d"}], E^4_1=[\text{"l"}, \text{" "}, \text{"r"}, \text{"!"}],\\ R^4=\text{STR}([\text{"H"},\text{"e"},\text{"l"},\text{"l"},\text{"o2"},\text{" "},\text{"w"},\text{"d"},\text{"r"},\text{"l"},\text{"!"}])=\text{"Hello2 wdrl!"} $$

$$ M^{12}=\begin{pmatrix} \text{H}\\ \text{e}\\ \text{l}\\ \text{l}\\ \text{o}\\ \text{ }\\ \text{w}\\ \text{o}\\ \text{r}\\ \text{l}\\ \text{d}\\ \text{!}\\ \end{pmatrix}, \\C^{12}_1=\text{"Hello world!"}, \\E^{12}_1=[\text{"H"},\text{"e"},\text{"l2"},\text{"o"},\text{" "},\text{"w"},\text{"o"},\text{"r"},\text{"l"},\text{"d"},\text{"!"}], \\R^{12}=\text{"Hel2o world!"} $$

Therefore the following follows. $$ \text{FRLE}(\text{"Hello world!"}) = [\dots, \text{"Hell2o word!"}, \dots, \text{"Hello2 wdrl!"}, \dots, \text{"Hel2o world!"}] $$

Tags , ,

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  • \$\begingroup\$ If you can't think of a way to put it into words, you could put it into code, and then someone here can translate that reference implementation into a description. Plus you can then include the reference implementation in the spec for people who read code better than prose \$\endgroup\$ – trichoplax Sep 4 '18 at 19:13
  • \$\begingroup\$ Okay,I can do that! \$\endgroup\$ – Zacharý Sep 5 '18 at 13:50
  • \$\begingroup\$ @Zacharý I have attempted to write a task definition. If this is not what you intended, you can simply roll back. \$\endgroup\$ – Jonathan Frech Sep 7 '18 at 3:37
  • \$\begingroup\$ By the way, I think giving a factor as input would be a nicer challenge than adding a wrapper to encode for every factor. \$\endgroup\$ – Jonathan Frech Sep 7 '18 at 3:40
  • \$\begingroup\$ @JonathanFrech, there isn't universal agreement on the meaning of zip applied to a non-square 2D array, so you need to define it unambiguously. \$\endgroup\$ – Peter Taylor Sep 7 '18 at 10:13
  • \$\begingroup\$ @PeterTaylor Attempted to do so. I intend to only zip one-dimensional arrays. \$\endgroup\$ – Jonathan Frech Sep 7 '18 at 10:46
  • \$\begingroup\$ Thanks! I think this definition works! \$\endgroup\$ – Zacharý Sep 7 '18 at 14:58

I have an idea for a KOTH challenge (that is marginally similar to the current Reaper KOTH), but none of the expertise required to make it work. The idea is based on the card game Diamant aka Incan Gold, and the rules of the game are:

  • The game is played over 5 expeditions, each with the same structure.
  • At the start of each expedition, a special deck of cards is shuffled, containing gem cards and hazard cards (and in some versions of the game, one artifact card is added to the deck every expedition).
  • Every turn, a card is revealed from the deck.
    • If a hazard card is revealed and another copy of the same hazard has already been revealed in the expedition, the expedition immediately ends and all remaining players lose their accumulated gems. The revealed hazard card is removed from the deck for future expeditions.
    • If a gem card is revealed, the number of gems shown on the card is shared evenly between remaining players, with any remainder placed in a common pool.
  • After the card is revealed, if the expedition has not ended, then players simultaneously decide whether they want to "Stay" or "Go Home".
    • All players who choose "Go Home" share the gems in the common pool between them (leaving any remainder), and then bank all of their gems. They are no longer in the expedition.
    • If a single player chooses "Go Home", they claim all of the gems in the common pool plus any artifacts that were revealed.
  • If there are players remaining in the expedition, then a new turn begins.
  • If there are no players remaining in the expedition, then the expedition ends.
    • If this was the last expedition, then players count up the gems they banked plus bonus gems for artifacts they claimed, and the player with the most gems wins.

My thought is that we could simplify the game for KOTH - one expedition, no artifacts - and run games which pit 4 bots against each other in some configuration, with each bot's overall rating being its average score per game.

The problem is that I have no experience in any meaningful programming languages. The other problem is that I don't know how to run a contest like that. The other other problem is that I don't know whether it's worth running.

Can anyone offer assistance or advice on any of those three problems?

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  • \$\begingroup\$ For problem 1, there are 2 optinos, either ask somebody else to write the bot (if the challenge is interesting enough), or to learn a programming language. \$\endgroup\$ – user202729 Sep 14 '18 at 5:32

Output the answer after yours

Related: Output the answer above yours

This challenge is very similar to the above, but I hope it's different enough to warrant posting. The difference is that you need to post the answer that is posted after yours (the whole answer, not just the <code> block). Also, your output can be the answer text or the HTML code of the answer. When your post is still the newest post, your program should just output nothing.


  1. No URL Shorteners
  2. No programs that need to be run at a specific URL.
  3. No updating your answer with information gained after posting (i.e your answer id, text or answer id of post that comes after yours). Exception: It's OK to hardcode the page number of your answer (on 'oldest' sort).

I think mine is different enough from the linked question, because that has some different rules, and my version closes loopholes used on the other question. Also, no one utilizes the StackAPI on that one.

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Put +, -, * into 1 _ 2 _ 3 ... n to make the expression equal 0, for a given 3 <= n <= 30, and output the valid equation (0=1+2-3). Exactly one operator between each number.

There exists some patterns, such as ((1-2)-(3-4))+((5-6)-(7-8))... and 1+((2-3)-(4-5))+(6-7). Maybe we can find patterns that solve the whole problem?

There are no solutions without * for at least n={5,6,9,10,13,14,17,18,21,22}, so it looks like we need * for n=4x+1 and n=4x+2.

There are no solutions without + for n < 18.

There are no solutions without - (obviously).

Brute-forced examples:

  1. 0=1+2-3
  2. 0=1-2-3+4
  3. 0=1*2-3-4+5
  4. 0=1+2*3+4-5-6
  5. 0=1+2-3-4+5+6-7
  6. 0=1*2*3+4-5-6-7+8
  7. 0=1-2+3*4*5+6-7*8-9
  8. 0=1-2*3+4*5-6*7+8+9+10
  9. 0=1*2+3*4+5*6-7*8-9+10+11
  10. 0=1-2+3*4-5*6-7-8*9+10*11-12
  11. 0=1+2*3+4+5-6-7-8+9+10+11-12-13
  12. 0=1*2+3*4+5*6+7+8*9+10-11*12+13-14
  13. 0=1-2-3-4-5-6-7+8-9+10-11+12-13+14+15
  14. 0=1*2*3*4*5*6+7+8-9*10*11-12+13+14+15*16
  15. 0=1*2*3*4*5*6+7*8*9+10-11*12*13+14*15+16*17
  16. 0=1*2*3*4*5*6-7*8-9-10-11*12-13-14*15-16*17-18
  17. 0=1*2*3*4*5*6-7-8-9*10-11-12-13-14-15*16-17*18-19
  18. 0=1*2*3*4*5*6-7-8-9*10-11-12-13*14-15-16-17-18*19-20
  19. 0=1*2*3*4*5*6-7-8-9-10-11-12*13-14-15-16-17-18-19-20*21
  20. 0=1*2*3*4*5*6-7-8-9-10-11*12-13-14-15-16-17-18-19-20*21-22
  21. 0=1*2*3*4*5*6*7-8*9-10*11-12-13*14-15*16*17-18-19-20-21-22*23
  22. 0=1*2*3*4*5*6*7-8*9*10-11-12-13-14*15*16-17-18-19*20-21*22-23-24
  23. 0=1*2*3*4*5*6*7-8*9*10-11-12*13*14-15*16-17-18*19-20*21-22*23-24*25
  24. 0=1*2*3*4*5*6*7*8-9*10*11-12*13*14-15*16*17-18*19*20-21*22*23-24*25*26
  25. 0=1*2*3*4*5*6*7*8-9*10*11-12-13*14*15-16-17-18*19*20-21-22*23*24-25*26*27
  26. 0=1*2*3*4*5*6*7*8*9-10-11*12*13*14*15-16-17-18-19-20-21*22-23*24-25*26-27*28
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  • \$\begingroup\$ Related Also, do you want this to be code golf (shortest code) or a different type of challenge? \$\endgroup\$ – geokavel Sep 23 '18 at 21:15
  • \$\begingroup\$ Very related. \$\endgroup\$ – user202729 Sep 26 '18 at 15:22
  • \$\begingroup\$ @user202729 yep. Unlike that one, this doesn't allow turning the 1 negative, which does seem to make the problem unsolvable without multiplication. \$\endgroup\$ – Filip Haglund Sep 26 '18 at 15:52

There is a nice task One Ring to rule them all. One String to contain them all. But rule define a string as a linear buffer. A linear buffer is not a Ring :)

My suggestion is:

  • create a new task with title "One Ring to rule them all. One Ring to bring them all"
  • add link to old task in the body of the new task
  • modify an Objective: Output a String which contains every positive integer strictly below 1000 and the String is a Ring.
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  • 1
    \$\begingroup\$ It would be a dupe \$\endgroup\$ – Peter Taylor Sep 28 '18 at 11:32
  • \$\begingroup\$ ok. thanks, Peter \$\endgroup\$ – mazzy Sep 28 '18 at 11:52
  • \$\begingroup\$ I've been thinking about it - no, It is not a duplicate. Main different: 0, 00, 000, ... are different elements in a de Bruijn sequence and same element in a Ring of numbers. This moment make golfed algoritms different. \$\endgroup\$ – mazzy Oct 1 '18 at 8:08
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    \$\begingroup\$ In a de Bruijn sequence containing all 3-digit sequences over 0-9, 000 is an element but 0 and 00 are not. Moreover, if the ring contains 100, 200, etc. then it must contain 00 at least nine times. See my comment of Nov 5 '13 at 11:28 on /q/13088. \$\endgroup\$ – Peter Taylor Oct 1 '18 at 9:51

Make words (need a better title)

Write a program or function that takes a string and an integer \$ n \$ as input and outputs all the \$n\$-letter words formed using only the letters in the string.


The input string will have only small alphabets \$a-z\$.

All the letters in the input string will be unique.

Input integer will be positive.

Winner: This is code-golf so shortest code wins. (Fastest algorithm will also be good in this right?)


output: "aaaaa","aaaab","aaaac",.....,"ddddc","ddddd"

P.S. The output does not have to be sorted.

Any input and output format will do as long as its distinguishable.


input: abcd,5 (ok)
       abcd 5 (ok)
       abcd5 (not allowed)
output: ["aaaaa","aaaab",.... (ok)
        "aaaaa","aaaab",.... (ok)
        "aaaaa""aaaab""aaaac"..... (ok)
        aaaaa aaaab aaaac ..... (ok)
        aaaaaaaaabaaaac...... (not allowed)
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  • \$\begingroup\$ Related, related, related. \$\endgroup\$ – Arnauld Oct 1 '18 at 10:18
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    \$\begingroup\$ For this kind of challenge, neither fastest-algorithm nor fastest-code should be used. For the first one: all answers are likely to have the same time complexity. For the 2nd one: ~99% of CPU time is going to be spent printing the results. \$\endgroup\$ – Arnauld Oct 1 '18 at 10:24
  • \$\begingroup\$ BTW: this is basically count from \$0\$ to \$b^n\$ in custom base \$b\$, which may have already been covered in some other challenge. I failed to find one, though. \$\endgroup\$ – Arnauld Oct 1 '18 at 11:24
  • \$\begingroup\$ It's product in python: docs.python.org/2/library/itertools.html (May help giving a title to the challenge, or searching for duplicates) \$\endgroup\$ – Nathan Merrill Oct 2 '18 at 2:54
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    \$\begingroup\$ @Arnauld It's mostly a duplicate of Cartesian product of a list with itself n times. The only difference is that builtins aren't allowed in that linked challenge, which is probably better since this is 3 bytes in 05AB1E due to an optional parameters requiring an explicit input and even just 1 byte in Jelly (not sure how to pretty-print it as a list in the footer..). \$\endgroup\$ – Kevin Cruijssen Oct 3 '18 at 8:13

Golf A Programming Language

Code golf languages are languages built to complete programming challenges in few lines of code. Instead of creating a code golf language, you will be creating an interpreter for a programming language in as few bytes as possible.


As you are not creating a code golf language, instead your score will be based on the number of bytes in the interpreter.


You may provide a program as input to your interpreter as input using any standard input method

Output provided by a program in your language may be passed on through your interpereter using any standard output method


You must write a program to solve each task in the language you created:

  • Hello World - Output the string "Hello World"

  • Fizz Buzz - List numbers from 1 to 100 replacing multiples of 3 with Fizz, 5 with Buzz, and 15 with FizzBuzz

  • Prime Check - Check if a given number is prime


Your programming language must be implemented yourself. Using builtin functions such as Javascript's eval or CHIQRSX9+'s I on input makes this not very interesting.

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  • \$\begingroup\$ This is too broad, as defining "interpreter" is pretty impossible. The only way this is viable is if you specify the language to be interpreted. \$\endgroup\$ – Nathan Merrill Oct 2 '18 at 2:48
  • \$\begingroup\$ @NathanMerrill Anything can be built for the interpreter. It could have 3 instructions, one for each challenge, but that could mean the interpreter increases length because it has the tasks built in. It might need more tasks, 3 tasks might mean building an actual language takes more code than just hardcoding 3 functions \$\endgroup\$ – pfg Oct 2 '18 at 3:39
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    \$\begingroup\$ Writing a general purpose language will take more characters than providing built-ins for those. In essence, you're looking for a program that returns a program that takes 1 of 3 characters and evals it. You are correct that eval makes this not interesting, but it's also the best way. Unless you make the language significantly harder (and specify the input required to do each of the tasks), I don't see this being interesting. \$\endgroup\$ – Nathan Merrill Oct 2 '18 at 3:44
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    \$\begingroup\$ codegolf.stackexchange.com/questions/111278/… \$\endgroup\$ – user202729 Oct 2 '18 at 5:27

Eeny Meeny Miny Moe

Inspired (somewhat) by this question:

Where should I stand to be captain of my team?


The childhood song Eeny, meeny, miny, moe was often used to select who was a captain when playing some game. Everyone on one team would stand in a line and someone would point at the first person in the line. Although there are several variations, where I grew up, they would sing:

Eeny, meeny, miny, moe,
Catch a tiger by the toe.
If he hollers, let him go,
Eeny, meeny, miny, moe.

adding to the traditional:

My mother told me to pick the very best one and you are not it.

As we sung each word, we pointed at the next person in line, then (if the rhyme wasn't over yet) we would continue the rhyme and start over at the beginning of the line. The person being pointed to when the final "it" was sung would be removed from the line, eliminating them from being captain. Then the rhyme would start over at the beginning of the line to remove the next person. This continued until only one person was left, and that person would be captain.

It didn't take long for me to recognize that if there were two people left (or if we started with just two people), it seemed like the second person would always win. Studying this for a bit I realized that since there are 35 beats this would always happen. I further figured out if there are 3 people, the third would win. So I wondered: could I chart it out for any number? Can I create a formula for it?


Write a program, function or (etc. as standard) where given input of an integer number of starting children greater than one, output what starting child number will be the winner, and where I need to stand so that I can be captain!


Input        Output
  2             2
  3             3
4,5,6or7        4
 8 or 9         5
  10            6
11,12or13       7
  14            8
  15            9
  16           10
17,18,19or20   11 

Winner for each language is that with the least number of bytes. And of course, standard loopholes and all that legal jazz are not allowed as is typically expected in these challenges.

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  • 1
    \$\begingroup\$ aka, the Josephus problem, possibly duplicate \$\endgroup\$ – ngn Oct 2 '18 at 19:49
  • \$\begingroup\$ @ngn Wow! I learn something new every day! Is this a duplicate, then? \$\endgroup\$ – Keeta - reinstate Monica Oct 2 '18 at 19:50
  • \$\begingroup\$ I'm not sure, yours is a special case for k=35 \$\endgroup\$ – ngn Oct 2 '18 at 19:52
  • \$\begingroup\$ by the way, you may find this video enlightening :) \$\endgroup\$ – ngn Oct 2 '18 at 19:53
  • \$\begingroup\$ Different from listed duplicate in several ways. Mine restarts at the beginning after each elimination (which really makes it questionable whether it is Josephus or not). Mine requires only one input. Mine is a firm case for k and mine has a cool backstory :) \$\endgroup\$ – Keeta - reinstate Monica Oct 3 '18 at 12:16

Permutations to the nines

This question is based on an unresolved inquiry which began at Permutations without recursive function call and was followed up at How to improve efficiency of algorithm which generates next lexicographic permutation? (TL;DR), after finding help removing duplicate numbers at Most efficient method to check for range of numbers within number without duplicates, where after performing arithmetic by hand found that for an array having length less than or equal to 9, if we ignore the value held at each index of the initial array and instead convert the indexes to a whole number, the next lexicographic permutation can be determined by adding 9 to the current index as number until a whole number, e.g., [1,2,3] // 123 1-based index or 23 0-based index is reached that satisfies two the conditions

  • Contains only the numbers of the indexes of the original array

  • Does not contain any duplicate numbers

e.g., 123+9=132 // "abc" -> "acb" 132-9=123 // "acb" -> "abc"; the graph for "abcd" is


which does not appear to be linear.

As the linked answers disclose, adding 9 to a whole number is not the most efficient method of determining the next lexicographic permutation, and what found independently by hand has been found by others, consider OEIS A217626

A217626 First differences of A215940, or first differences of permutations of (0,1,2,...,m-1) reading them as decimal numbers, divided by 9 (with 10>=m, and m! > n).

1, 9, 2, 9, 1, 78, 1, 19, 3, 8, 2, 77, 2, 8, 3, 19, 1, 78, 1, 9, 2, 9, 1, 657, 1, 9, 2, 9, 1, 178, 1, 29, 4, 7, 3, 66, 2, 18, 4, 18, 2, 67, 1, 19, 3, 8, 2, 646, 1, 19, 3, 8, 2, 67, 1, 29, 4, 7, 3, 176, 3, 7, 4, 29, 1, 67, 2, 8, 3, 19, 1, 646, 2, 8, 3, 19, 1

Also of interest is that the graph of the specific multiples of 9 to derive all lexicographic permutations of a set can be found by computing only half 1/2 of the total permutations +1, as the declination slope is identical to the inclination slope of multiples of 9, for example, given an input of the string "abc" or array ["a","b","c"], we only need to reach the inverted peak of the graph, that is bca 120 3 18, where the previously calculated values can be reversed and added to the current whole number, 120, to generate the remainder of the lexicographic permutations

abc 012 0
acb 021 1 9
bac 102 2 81
bca 120 3 18
cab 201 4 81
cba 210 5 9

/    \


This challenge is not to determine the most efficient algorithm to calculate permutation 9!, as, again, adding 9 to a whole number to reach that goal can be outperformed by swapping values or other method, as demonstrated at the accepted at the first SO link above.

This challenge is to derive the most efficient method of determining the next lexicographic permutation by calculating the next multiple of 9 that meets the above listed criteria, preferably directly using math, if possible.

The reason am posting this question here and now is because was reminded by the graph of the peaks (particularly the third item) in this question Different combinations possible

Just to give a visual example, they are the following:

   /\     /\      /\/\
/\/  \   /  \/\  /    \

that in spite of performing various calculations using the index of the !n within the resulting array of permutations, e.g.,

var n = N = 1234;
var res = [9,81,18,81,9,702,9,171,27,72,18,693,18,72,27,171,9,702,9,81,18,81,9].map((x)=>n=n+x);
// try to determine the rate of growth
var j = res.map((x)=>x/N);

console.log(res, j);

var tryStuff = 1234*(4321/1234);
console.log(tryStuff, (4321/1234));

am still vexed by and have not been able to independently determine the mathematical formula to derive the next lexicographic permutation directly without adding 9 multiple times to reach the required number; or if that is even mathematically possible for any input set from 2! through 9!. That is, this code

function getNextLexicographicPermutation(arr) {

  for (var l = 1, i = k = arr.length; l < i; k *= l++);

  function checkDigits(min, max, n) {
    var digits = 0;
    while (n) {
      d = (n % 10);
      n = n / 10 >> 0;

      if (d < min || d > max || (digits & (1 << d)))
        return false;
        digits |= 1 << d;
    return true;

  var len = arr.length,
    idx = arr.map(function(_, index) {
      return index
    p = 9,
    min = 0,
    max = len - 1,
    last = Number(idx.slice().reverse().join("")),
    curr = Number(idx.join("")),
    res = [curr],
    diff = [],
    result = [],
    next, times = 0;

  while (res.length < (k / 2) + 1) {
    next = (curr += p);
    if (checkDigits(min, max, next)) res[res.length] = next;
    curr = next;

  for (var i = 0; i < res.length; i++) {
    var item = res[i];
    item = String(item).split("").map(Number);
    item = (item.length < arr.length ? [0].concat(item) : item)
      .map(function(index) {
        return arr[index]

  res.reduce(function(a, b) {
    diff.push(b - a);
    return b

  for (var i = 0, curr = res[res.length - 1], n = diff.length - 2; result.length < k; i++, n--) {
    curr = curr + diff[n];
      .map(function(index) {
        return arr[index]
  return [result, diff, res, times];

var arr = ["a", "b", "c", "d"];


which generates the second half-1 of the permutations from the first half+1 of the permutations checks if the next whole number meets the necessary conditions 210 times for an input array of ["a","b","c","d"] which has a resulting .length of 24. Ideally, we want to generate 24 lexicographic permutations using only 12 or 13 checks; or no checks at all, by determining the irrational number or other mathematical algorithm which will directly calculate (or approximate close enough to determine the next multiple of 9) the next whole number which meets the necessary criteria.

Kindly disregard the length of this post, as am trying to include as much information as consider relevant to the inquiry.


This challenge must use the number 9 (addition, multiplication, division, other mathematical operation) to determine the next lexicographic permutation using the indexes of the current lexicographic permutation converted to a whole number, ideally, in a single operation, else in the least amount of operations necessary to achieve the expected result.

Again, this challenge is not asking how to code the most efficient code which determines the next lexicographic permutation, but what is the most efficient approach is using only the number 9 and math to generate the permutations.

Since we can get the first and last lexicographic permutations by reversing the indexes of the input array, that is not counted as an operation within the program.


An array or sting having .length less than or equal to 9!.


Lexicographically sorted array of permutations of input.


Remaining within the Rules above, determine a mathematical algorithm which directly generates the next lexicographic permutation using only the current indexes of the original input or current permutation. Ideally, directly, without having to add 9 in multiple operations until the listed criteria is met, that is, we want a single algorithm to calculate that we need to add 81 to 132 to get the sum 213 and add 702 to 321 to get the sum 1023 and so forth. Explain the math in the algorithm. Note: The requirement might not be possible. If that is the case, explain why.

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  • 1
    \$\begingroup\$ 1. As I commented elsewhere, it's hard to separate the core challenge from the fluff. 2. I assume there's a typo in the "Input" section, because an array of the permutations of 9! elements would take more memory than is physically possible in this universe. \$\endgroup\$ – Peter Taylor Oct 4 '18 at 21:37
  • \$\begingroup\$ @PeterTaylor Kindly indicate specifically what content in the question that you are referring to. From perspective here, there is no "fluff" in the question. No, there is no typo in the "Input" section. The algorithm MUST be true for 9!->362880 if true for 4!->24. We can check the result of the algorithm by incremental sections and by hand. Simplified core challenge: Find the multiple of 9 required to precisely equal the next p lexicographic permutation (as a whole number) from current p number in least math computations. \$\endgroup\$ – guest271314 Oct 4 '18 at 22:06
  • \$\begingroup\$ @PeterTaylor E.g., if we have input ['a','b','c'] we ignore values and "see" 12, 012 (or, as described at this question for simplification 1-based indexing 123). The next lexicographic permutation is one computation 123+(9*1)=132 from we can "see" ['a','c','b']. In JavaScript we could even use Math.pow(9,1), but that would be a false-positive for a method or pattern to use, because the next multiple is 132+(9*9)=213 (we can "see" ['b','a','c']). How to get from 132 to 213 without adding 9 nines times (Task: least mathematical computations in this step) to 132? \$\endgroup\$ – guest271314 Oct 4 '18 at 22:33
  • \$\begingroup\$ Should we take an array of items (['a','b','c']), an array of indices ([1,2,3]) or its value converted to decimal (123) as input? And what should be the output -- 132, ['1','3','2'] or 9? \$\endgroup\$ – user202729 Oct 5 '18 at 13:08
  • \$\begingroup\$ @user202729 The input is a decimal derived from the first lexicographic permutation or an array of items from which the decimal will be derived; e.g., 123 from ['a','b','c'], or 123456789 (1-based indexing) or 012345678 (0-based indexing) derived from ["a","b","c","d","e","f","g","h","i" ]. The maximum number is the reverse order of the decimal number, 987654321 or 876543210, which is the last lexicographic permutation (as a decimal number) of the input number or array [8,7,6,5,4,3,2,1,0] (0-based) or [9,8,7,6,5,4,3,2,1] (1-based). \$\endgroup\$ – guest271314 Oct 5 '18 at 20:48
  • \$\begingroup\$ @user202729 We can derive all lexicographic permutations of input up to 9! by adding 9 to the first lexicographic permutation as decimal and following the two conditions at the question; that is, by adding 9 to the initial decimal number where no duplicate numbers are found within the resulting number and each number is found within the original decimal number. Example "abc"->curr=123->123+9=132->"acb"; curr becomes 132. We are trying to reduce the number of additions to meet the two conditions by creating a formula to determine the precise number of 9's to add to curr. \$\endgroup\$ – guest271314 Oct 5 '18 at 21:09
  • \$\begingroup\$ @user202729 Is the question clear to you now? \$\endgroup\$ – guest271314 Oct 5 '18 at 21:43
  • \$\begingroup\$ @user202729 The output is n<=9! permutations. We can already achieve that by the code at the question. However the challenge is to reduce the mathematical steps required to reach the result consider "How to get from 132 to 213 without adding 9 nines times (Task: least mathematical computations in this step) to 132". When curr is 132, we do not necessarily know that we will have to add 9 to 132 nine times (81) to reach the sum 213. We find that out now by nine computations, adding. The challenge is to reach the sum 213 using what we have in less than nine computations. \$\endgroup\$ – guest271314 Oct 5 '18 at 23:27
  • \$\begingroup\$ @user202729 There are other mathematical relationships between the indexes of the lexicographically sorted permutations. For example, if we have the input ['a','b','c'] or "abc", when we ignore the values, using 1-based indexing, we can "see" 123 at index 1. 123+90=213 our lexicographically sorted permutation at index 3, 132 at index 2 is 132+90=231; max number is always input reversed. We only need to find 132 (3 operations) to derive all 6 permutations, as the declination slope of [123,132,213] is the reverse order of each number; this relationship differs for 1234. \$\endgroup\$ – guest271314 Oct 5 '18 at 23:46
  • \$\begingroup\$ What is n in that case? /// Consider a specific example. If the input is 132 (1-indexing) what should the output be? \$\endgroup\$ – user202729 Oct 6 '18 at 11:13
  • \$\begingroup\$ @user202729 The output is always n! lexicographic permutations. We already are able to achieve that. Whether the input is 123 or 132 makes no substantial difference here; we "see" that 132 is greater than its constituent parts, 123 and thus know that we are at index 2 (1-indexing), 321 is always the max, and n! is 6. What this challenge is asking is how to use this specific method of determining permutations, that is, adding 9 to the indexes as a whole number until the listed conditions are met, in less mathematical operations that is currently being used - simple addition. \$\endgroup\$ – guest271314 Oct 7 '18 at 16:55
  • \$\begingroup\$ Let me see. So, if 132 is given, the output should be [123,132,213,231,312,321], but with a restriction -- only addition, subtraction, multiplication, and division is allowed. Correct? \$\endgroup\$ – user202729 Oct 7 '18 at 16:59
  • \$\begingroup\$ @user202729 There is no restriction on the mathematical methods which can be used. You can use calculus if that will generate 132 from 123 in the least amount of calculations - using 9 within the calculations. If you can use mathematical relationships between the expected resulting set which uses other numbers, then those methods should be described in detail. The code at the question uses only the number 9 to add to the indexes represented as a whole number, though for some 90 or 711 could be used. The challenge is to use the least number of total math computations. \$\endgroup\$ – guest271314 Oct 7 '18 at 17:05
  • \$\begingroup\$ @user202729 We are trying to generate the next lexicographic permutation using only 1) the current permutation; 2) the maximum possible number (our last permutation); 3) mathematics; as our resources. We begin with a number less than 321 and know that our first permutation must be 123, our last 321, and that between 321 and 123 there are relationships between the numbers which allow us to derive each number that meets our criteria. In the code at the question 9 is suitable to achieve that goal by adding 9 to the number where the sum contains no duplicate and only input numbers. \$\endgroup\$ – guest271314 Oct 7 '18 at 17:12

Qwixx KoTH

In this King of the Hill, our bots will be playing Qwixx. In Qwixx each player will try to get as many points by picking the sum of two dices and marking that number on your own scoresheet. The scoresheet has four rows with different colors (red, yellow, green, blue) with numbers ranging from 2 to 12, where green and blue have the numbers ranked in descending order. You can only mark out a number if you haven't marked a number to the right of it, i.e. after marking red 3 you cannot mark red 2 anymore.


All of the rules of Qwixx are:

  • Red and yellow have numbers 2, 3, ..., 12, and a padlock.
  • Green and blue have numbers 12, 11, ..., 2, and a padlock.
  • There are 6 dices: red, yellow, green, blue and two white dices.
  • On your turn you will throw all the dices, except for deleted ones.
  • Everyone can mark one number on one color equal to the sum of the white dices. Also if it is not your turn.
  • If it is your turn, you can also mark one number in a specific color, with the sum of one colored dice and one white dice. The colored dice that you pick denotes in which color you need to mark the number.
  • You can only mark numbers to the right of already marked numbers.
  • The last number can only be marked if you have already marked at least 5 numbers before. If you do this, you will automatically also mark the padlock, which will give you extra points. If you do this, that colored dice is not thrown anymore and nobody can mark a number in that color. This dice is deleted from the game (thrown as 0).
  • Only on your turn you need to mark at least one number. If you don't do this, you will automatically mark a penalty box, which costs you 5 points.
  • The amount of points is equal to the sum of 1 up to the amount of marks in each color, minus 5 times the amount of penalties. I.e.: 4 in red, 3 in green and 2 penalties is 4+3+2+1 + 3+2+1 = 16 - 2 * 5 = 6 points.
  • Each turn is simultanouesly.
  • The game ends when the second dice is deleted or when a scoresheet contains 4 penalties.


The tournament rules are as follows:

  • The tournament consists of 10000 * amount of bots games.
  • Each game will be played with 5 random bots, in a randomized order.
  • If multiple bots end up with the same amount of points in one game, each bot gets a win.
  • The winner of the tournament is the bot who has the highest win percentage.


The bot needs to be defined in Python and needs to have two functions:

  • __init__(self, index) with the index in scoresheets, so that you know which scoresheet is yours
  • turn(self, scoresheets, dices, my_turn), with the current scoresheets, the dices of this turn and a boolean being True when it is your turn. This function needs to return a tuple of two tuples containing (color, number) with what your bot wants to mark, where the first tuple is for the white dices, and the second tuple only if its your turn.

The simplest bot can be defined as:

class DoNothing:
    def __init__(self, index):
    def turn(self, scoresheets, dices, my_turn):
        return ([],[])

Which does absolutely nothing.

A randomized bot that tries to always mark something when possible is:

class RandomAllowedCombinations:
    def __init__(self, index):
        self.index = index
    def turn(self, scoresheets, dices, my_turn):
        (any_combs, turn_combs) = scoresheets[self.index].allowed_combinations(my_turn, dices)

        any_comb = []
        if any_combs:
            any_comb = any_combs[random.randint(0, len(any_combs)-1)]

        turn_comb = []
        if my_turn and turn_combs:
            turn_comb = turn_combs[random.randint(0, len(turn_combs)-1)]

        return (any_comb, turn_comb)

This also shows some of the utility functions: you can use scoresheets[self.index].allowed_combinations(my_turn, dices) to check what you can mark with the white dices, and if its your turn with one of the colored dices as well.

Both classes (Dices and Scoresheet) have several utility functions. You can check themselves to improve your bot. Note that Dices also contains the indices for each color. And the value of a dice is 0 if it is deleted (i.e. when someone in the game marked the last number and thus the padlock in one of the previous turns).


The controller contains 5 example bots. These will not be used in the tournament, unless there are too few participants.

import random
import numpy

### Bots ###
class DoNothing:
    def __init__(self, index):
    def turn(self, scoresheets, d, my_turn):
        return ([],[])

class AlwaysWhitesInRed:
    def __init__(self, index):
    def turn(self, scoresheets, d, my_turn):
        return ((d.red, d.dices[d.white1] + d.dices[d.white2]),[])

class AlwaysBlueOnlyTurn:
    def __init__(self, index):
    def turn(self, scoresheets, d, my_turn):
        return ([],(d.blue, d.dices[d.blue] + d.dices[d.white2]))

class RandomAllowedCombinations:
    def __init__(self, index):
        self.index = index
    def turn(self, scoresheets, dices, my_turn):
        (any_combs, turn_combs) = scoresheets[self.index].allowed_combinations(my_turn, dices)

        any_comb = []
        if any_combs:
            any_comb = any_combs[random.randint(0, len(any_combs)-1)]

        turn_comb = []
        if my_turn and turn_combs:
            turn_comb = turn_combs[random.randint(0, len(turn_combs)-1)]

        return (any_comb, turn_comb)

class RandomButOnlyOwnTurn:
    def __init__(self, index):
        self.index = index
    def turn(self, scoresheets, dices, my_turn):
        (any_combs, turn_combs) = scoresheets[self.index].allowed_combinations(my_turn, dices)

        turn_comb = []
        if my_turn and turn_combs:
            turn_comb = turn_combs[random.randint(0, len(turn_combs)-1)]

        return ([], turn_comb)

### Classes ###
class Dices:
    def __init__(self):
        self.dices = []
        self.deleted = []
        self.red = 0
        self.yellow = 1
        self.green = 2
        self.blue = 3
        self.white1 = 4
        self.white2 = 5
        self.colors = [self.red, self.yellow, self.green, self.blue]
        self.whites = [self.white1, self.white2]
        self.diff_dices = self.colors + self.whites

    def roll_dice(self):
        self.dices = [0,0,0,0,0,0]
        for i in range(6):
            if i not in self.deleted:
                self.dices[i] = random.randint(1, 6)

    def delete_dice(self, color):
        if not color in self.deleted:
        return len(self.deleted) > 1

    def combinations(self):
        combs = [(color, white) for color in self.colors for white in self.whites]

        any_combs = [(color, self.dices[self.white1] + self.dices[self.white2]) for color in self.colors if self.dices[color] >= 1]
        turn_combs = [(color, self.dices[color] + self.dices[white]) for (color, white) in combs if self.dices[color] >= 1]

        return (any_combs, turn_combs)

class Game:
    def __init__(self, bots):
        self.bots = [bots[i](i) for i in range(len(bots))]
        self.scoresheets = [Scoresheet() for i in self.bots]
        self.dices = Dices()
        self.turn_number = 2

    def round(self):
        (any_combs, turn_combs) = self.dices.combinations()

        finished = False

        for i, bot in enumerate(self.bots):
            my_turn = self.whos_turn() == i
            moves = bot.turn(self.scoresheets, self.dices, my_turn)
            deleted_colors = self.scoresheets[i].mark(moves, my_turn, any_combs, turn_combs, self.dices)
            if deleted_colors:
                finished = finished or any([self.dices.delete_dice(deleted_color) for deleted_color in deleted_colors]) 
            finished = finished or self.scoresheets[i].too_many_penalties()

        self.turn_number += 1

        return finished

    def whos_turn(self):
        return (self.turn_number - 1) % len(self.bots)

    def runGame(self):
        finished = False
        while not finished:
            finished = self.round()
        #for i, scoresheet in enumerate(self.scoresheets):
            #print self.bots[i], scoresheet.values, scoresheet.penalties, scoresheet.points()

class Scoresheet:
    def __init__(self):
        self.red = []
        self.yellow = []
        self.green = []
        self.blue = []
        self.penalties = 0
        self.values = [self.red, self.yellow, self.green, self.blue]

    def points(self):
        points = self.penalties * -5
        for color in self.values:
            points += sum(range(len(color)+1))
        return points

    def allowed_combinations(self, my_turn, dices):
        (any_combs, turn_combs) = dices.combinations()
        any_combs = [(color, number) for (color, number) in any_combs if self.allowed(color, number, any_combs, dices)]
        turn_combs = [(color, number) for (color, number) in turn_combs if self.allowed(color, number, turn_combs, dices)]

        if not my_turn:
            turn_combs = []

        return (list(set(any_combs)), list(set(turn_combs)))

    def allowed(self, color, number, combs, d):
        if color not in d.deleted:
            if (color, number) in combs:
                if color == d.red or color == d.yellow:
                    if not self.values[color] or max(self.values[color]) < number:
                        if number < 12 or self.end_number(color, number, d):
                            return True
                if color == d.green or color == d.blue:
                    if not self.values[color] or min(self.values[color]) > number:
                        if number > 2 or self.end_number(color, number, d):
                            return True
        return False

    def end_number(self, color, number, d):
        if len(self.values[color]) > 5:
            return (number == 2 and (color == d.green or color == d.blue)) or (number == 12 and (color == d.red or color == d.yellow))
        return False

    def mark_one(self, color, number, combs, deleted, d):
        if self.allowed(color, number, combs, d):
            if self.end_number(color, number, d):
        return deleted

    def mark(self, moves, my_turn, any_combs, turn_combs, d):
        beforePoints = self.points()

        deleted = []
        if moves[0]:
            deleted = self.mark_one(moves[0][0], moves[0][1], any_combs, deleted, d)
        if my_turn and moves[1]:
            deleted = self.mark_one(moves[1][0], moves[1][1], turn_combs, deleted, d)

        if my_turn and beforePoints == self.points():
            self.penalties += 1

        return deleted

    def too_many_penalties(self):
        return self.penalties >= 4

class Qwixx:
    def __init__(self, bots, bots_per_game):
        self.bots = bots
        self.games = [0 for i in range(len(self.bots))]
        self.points = [0 for i in range(len(self.bots))]
        self.wins = [0 for i in range(len(self.bots))]
        self.amountOfGames = len(bots) * 1000
        self.bots_per_game = bots_per_game

    def run_tournament(self):
        for g in range(1, self.amountOfGames + 1):
            bot_indices = numpy.random.choice(range(len(self.bots)), self.bots_per_game, replace=False)
            bots_this_game = [self.bots[i] for i in bot_indices]

            game = Game(bots_this_game)

            maxPoints = max([game.scoresheets[i].points() for i in range(len(bot_indices))])
            for bot_index_game, bot_index_total in enumerate(bot_indices):
                points = game.scoresheets[bot_index_game].points()
                self.points[bot_index_total] += points
                self.games[bot_index_total] += 1
                if points == maxPoints:
                    self.wins[bot_index_total] += 1
                    print "GAME", g, "is",  self.wins[bot_index_total], "th win by", self.bots[bot_index_total].__name__

    def tournament(self):

        win_rates = {i: self.wins[i] / float(self.games[i]) for i in range(len(self.bots))}
        format_result = '{:>30}: {:.4f}  {:>6} {:>8} {:>8}' 
        format_header = '{:>30}: {:>6}  {:>6} {:>8} {:>8}' 
        print(format_header.format('Name', 'Win %', 'Wins', 'Games', 'Points'))
        for j, i in enumerate(sorted(win_rates, key=lambda i: win_rates[i], reverse=True)):
            print(format_result.format(self.bots[i].__name__, win_rates[i], self.wins[i], self.games[i], self.points[i]))

### List of all bots ###
all_bots = [DoNothing, BadRandom, RandomAllowedCombinations, Random50, RandomButOnlyOwnTurn]

### Run tournament ###
Qwixx(all_bots, 5).tournament()


  • If someone can check my Python programming capabilities that would be great. I'm no native Python programmer.
  • There are quite some rules. Maybe I should simplify it? These are the official rules, but for example I could change that even on your turn you can only move once instead of twice.
  • This is my first challenge, so any feedback is welcome.
  • Note that I have not yet tested my controller thoroughly, I will do this before posting it as an actual challenge.
| |
  • 1
    \$\begingroup\$ 1. Simplification of the rules is definitely needed. This post is hard to understand and even harder to come up with strategies about. Under "Rules", I'd immediately start explaining the actions bots can perform and their consequenses. 2. Why are there two types of turns (only-white turn, vs color turn)? How does this affect strategies? 3. What's the purpose of marking the last box to prevent other players? How does this affect strategies? 4. I don't understand the scoring, please expound. \$\endgroup\$ – Nathan Merrill Oct 8 '18 at 16:56
  • \$\begingroup\$ 1. It's not entirely clear to me whether on my turn I can choose to make my "colour mark" before my "white mark" if they're on the same colour. 2. Dice is an irregular noun in English: the singular is die and the plural is dice. 3. 10000 * number of bots games is a lot, so it's going to be slow to run the tournament, but at the same time with 13 bots it's too few to run every possible combination once. This might need a more complicated tournament structure with elimination rounds. \$\endgroup\$ – Peter Taylor Oct 11 '18 at 14:20

Roulette Bots II

Last week I posted Roulette Bots, which was a lot of fun. I think the basic format has potential to be even more interesting, so I've been shopping some tweaks in my head and I'd like to hear feedback before I give it another shot. In a nutshell, the first iteration of the game was as follows:

Everyone starts with 100 hp. Each round, 2 players are chosen at random from the pool of contestants who have not yet competed in that round. Both players pick a number between 0 and their current hp, and reveal those numbers at the same time. The player who chose the lower number immediately dies. The other player subtracts their chosen number from their remaining hp and goes on to the next round.

From the bracket of contestants, 2 are chosen at random. They face off, and one or both of them dies. A player dies if:

  1. They choose a number smaller than that of their opponent
  2. Their hp drops to or below zero
  3. They tie three times in a row with their opponent

In the case of ties, both players simply generate new numbers, up to 3 times. After the faceoff, the survivor (if any) is moved to the pool for the next round, and the process repeats until we have exhausted the current round pool. If there is an odd number in the pool, then the odd one out moves on to the next round for free.

Players were given as input their own hp, and their opponent's complete betting history for that round. The idea being that players must balance resource consumption and long-term performance against the danger of dying right now if they don't bid high enough.

There are two things I'd like feedback on:

  1. The minimal set of restrictions on user submissions required to make an interesting game without limiting creativity
  2. Format tweaks that will allow for more creativity within the scope of the rules

The rule set I ended up on for the last game was that you were not allowed to take any action that unambiguously identified your opponent. This was for the purpose of avoiding things like players simply reading the stack to figure out exactly who their opponent was, since then you can simply simulate your opponent and more or less guarantee a win. While clever, it's doesn't really fit the spirit of the game. So the first question is whether or not there is a less restrictive rule that could be applied, or if someone can see a glaring loophole in that rule which would allow for game-breaking behavior in future versions of a similar game.

On the subject of format tweaks, the biggest issue I found with the game was that with a relatively small pool of contestants, very little history was actually generated and so bots that tried to use history to predict opponent guesses fared poorly overall. The first tweak I had in mind was to bootstrap a very large (256 or 512, maybe) initial tournament by random resampling of existing bots so that there would be many rounds and a good history pool to work with by the end of it.

Coupled with the larger pool, I intend to change the scoring system to reward longevity, to encourage more careful resource management, as compared to the previous game which rewarded just living through the first round. Something like making participating in the nth rounds worth n points, so that a player which survives to the nth round gets n(n-1)/2 points total. This puts more stringent constraints on resource management, since you need to survive a lot of rounds to do well. I'd probably bump the starting hp up to 1000, so that there is a bit more room to refine bets. and to make ties even less likely.

Second, there were a couple of attempts to make teams of bots that communicated via betting history in order to feed easy wins to one of them. I like the idea, but but because of the limited options presented by betting history they didn't fare very well. I am considering having bots exchange a "greeting" before each fight - a string or integer which they send to their opponent before battling, which could be used to identify teammates or clones of yourself. Does it add enough possibilities to gameplay to be worth the extra effort? Since with a resampled population you would likely be competing with yourself fairly regularly, having a way to recognize yourself (while at the same time avoiding others being able to recognize you from your greeting) seems like it could be interesting, but I worry that it will just degenerate into people updating their greeting at the last second to avoid other people using them, which would make it pointless to use the greeting to recognize opponents and make a lot of work for me. Can anyone suggest a good way to handle this? I want a way for people to recognize an opponent as probably (but not certainly) friendly without making it a game of "who updates last".

Feedback on the proposed gameplay tweaks, or ideas for others that I have not considered, would be appreciated!

| |
  • \$\begingroup\$ Please do not allow the "feed" strategy. We want all of the bots to be competitive, and the feeder bots aren't competitive, and creates an unfair advantage to individual bots (hmm...I wonder who will win 3v1). I also am not a fan of "send a message" actions: They either don't work or they make abuse possible. (The only exception is if you are making a team-based KotH) Scoring based on longevity is a good idea. If you want a longer history, you can also do a lives-based mechanic where a player has 3 lives. If they choose the lower number 3 times, then they are out of the tournament. \$\endgroup\$ – Nathan Merrill Oct 10 '18 at 14:55
  • \$\begingroup\$ @NathanMerrill the idea for the feeding was that if I have to have multiple copies of each bot in the tournament to get a large pool, then it would be useful to recognize yourself. By default, you will always have "teammates" as long as resampling happens. But I understand the objection. How would lives work, practically? If a bot bids lower, how would I handle that? Ignore the result up to 3 times, and just let them through to the next round without cost to either player? \$\endgroup\$ – KBriggs Oct 10 '18 at 15:32
  • \$\begingroup\$ Oh, I like the idea of feeding copies of your bot. Perhaps giving them a single boolean variable that indicates if it's a clone or not (or something along those lines?). If both both Bot A and B have multiple lives, they would both have their HP removed, but the one that bid lower would have a life removed. Perhaps "life" is the wrong term here: Maybe they get a "strike", and 3 strikes and you are out. \$\endgroup\$ – Nathan Merrill Oct 10 '18 at 15:36
  • 1
    \$\begingroup\$ Actually! If you have multiple clones of a bot, there's an easier solution than a 3-strike system: Simply give the opponent the entire history of all of the clones. \$\endgroup\$ – Nathan Merrill Oct 10 '18 at 15:39
  • \$\begingroup\$ I think if I used the 3 strike system what would happen is that everyone would just bid 0 twice in a row to get into the later rounds with full hp and go from there. Realistically I can't stop the feeding strategy, since you can use betting history to set up a recognizable pattern, so I think it's better to build it into the game somehow. Simply telling people if it's a clone is an option, but then it would just advance one to the next round for free without generating a meaningful bid history data point, so why have the clones in the first place? The entire history of clones is a neat idea... \$\endgroup\$ – KBriggs Oct 10 '18 at 15:49
  • \$\begingroup\$ You tell people if they are playing against their own clone. There's not like an "original bot" or a "clone bot". You simply have a variable that is true if it is Bot 1 vs Bot 1 and false if it is Bot 1 vs Bot 2. You are right that there are no systematic ways to stop feeding, but we have rules against that on SE, and so you can use human judgement to disallow those bots. \$\endgroup\$ – Nathan Merrill Oct 10 '18 at 15:58
  • \$\begingroup\$ I know what you mean. But if I tell them that, then they get to advance deterministically with a bet of 1, which is not a useful data point for anyone else predicting their behavior. I would like to find a solution that forces at least a meaningful bet (or that they have to take the risk of mistaken identity to make the 1-bet). If they know for certain they are facing a clone, then there is no value to having clones at all, since their purpose was to make a longer tournament with more history to use in predictions. But I realize as I say this that the "message passing" idea doesn't work either \$\endgroup\$ – KBriggs Oct 10 '18 at 16:01
  • \$\begingroup\$ Yeah. Perhaps you setup the tournament so that clones never actually fight each other. These battles aren't interesting, and only hurt the bot that randomly got assigned itself. The existence of clones would solely be to have a longer history. \$\endgroup\$ – Nathan Merrill Oct 10 '18 at 16:26
  • 1
    \$\begingroup\$ The only really sound way to avoid the who updates last problem is to host it in such a way that people can't see their competitors' code - i.e. somewhere other than PPCG. On this site you'll always have people special-casing specific opponents. \$\endgroup\$ – Peter Taylor Oct 11 '18 at 11:54
  • \$\begingroup\$ @PeterTaylor probably true. Perhaps I could collect submissions by some other means and just ask players to put a placeholder answer which they can edit with code after the deadline. Would that be allowed? I'd have to keep the controller private, and maybe provide a suite of unit tests that entries should pass before submission. \$\endgroup\$ – KBriggs Oct 11 '18 at 15:16
  • \$\begingroup\$ Something vaguely similar has been done once (a Kolmogorov-complexity question where OP asked people to post a hash of their code, and then after the deadline to post their actual code, so that people couldn't just port someone else's answer to a golfier language), but it's a lot of hassle. Really when you're having to fight the site's design that much, you may as well not bother and just accept that some challenges don't work here. \$\endgroup\$ – Peter Taylor Oct 11 '18 at 15:23
  • \$\begingroup\$ Looking at what happened with your previous challenge, though, I would suggest requiring that answers be self-contained so that the deletion of a function defined in someone else's answer won't break them. \$\endgroup\$ – Peter Taylor Oct 11 '18 at 15:24
  • \$\begingroup\$ @PeterTaylor fair enough, that does seem like a lot of unnecessary work. I'll shop the format a bit more and see if I can come up with something a little simpler. Disallowing interaction between entries is certainly doable, but having people post lat-second updates will always be a problem, I think. \$\endgroup\$ – KBriggs Oct 11 '18 at 15:24

Is it a valid snooker break?

In snooker, the objective is to score as many points as possible by potting two types of balls: red balls, 15 of them, each valued 1 point, and colour balls, 6 of them, valued 2, 3, 4, 5, 6 and 7 points. A break is a sequence of potted balls. The balls must be potted in a predetermined order: the break must start with potting a red ball, then a colour ball and continued with alternating in potting red and colour balls until all of the red balls are potted, then potting colour balls in ascending order of their values (if all of the red balls have already been potted, start with the colour ball with the lowest value). However, if a free ball is declared after an opponent's foul, the break may start with any ball, replacing the ball that should be usually potted first (red if some red balls are still on the table, colour with the lowest value if all reds are potted).

Your task is to determine if the given sequence of balls adheres to the rules mentioned above.


A list of integers representing values of snooker balls. The list will not be empty.


A truthy value if the list is a valid snooker break and falsey if it isn't.

Test cases


1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,2,3,4,5,6,7 //"maximum break"


2,5,6,7 //2 is already off the table
1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,2,3,4,5,6,7     //one red too many
7,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,1,7,2,3,4,5,6,7 //as well as here

To sandbox:

I am thinking about including lists with values out of range of snooker balls' values (less than 1 or greater than 7) in which cases the result should be falsey. As this is a decision problem, I think that these cases should be covered as well.

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  • \$\begingroup\$ Maybe you can let the submissions assume that only a list of [1-7]s are inputted. This would be more flexible for the submissions to do. \$\endgroup\$ – Shieru Asakoto Oct 12 '18 at 7:48

Implement JavaScript's Abstract Equality Comparison Algorithm.


As many know, JavaScript has quite confusing equality rules and it's generally recommended to use === instead to avoid the weird corner cases that come with using ==.

The logic behind the == operator is documented as "The Abstract Equality Comparison Algorithm" in the ECMAScript specification. For this particular question we will be implementing from the ECMAScript 5.1 edition specification. The algorithm is documented in section 11.9.3 of this specification.


Implement the abstract equality comparison algorithm in JavaScript without using == or !=.

Input & output

The input is two variables. These variables can be anything that is not a Reference type. The output is a truthy or falsy value that represents the result of performing the abstract equality comparison algorithm.

Test Cases

Since there are a lot of edge cases, I took these test cases from the test262 repository.

Examples that return truthy values:

true == true
false == false
true == 1
false == "0"
0 == false
"1" == true
+0 == -0
-0 == +0
1.0 == 1
"" == ""
" " == " "
"string" == "string"
1 == "1"
1.100 == "+1.10"
255 == "0xff"
0 == ""
"-1" == -1
"-1.100" == -1.10
"5e-324" == 5e-324
undefined == undefined
void 0 == undefined
undefined == eval("var x")
undefined == null
null == void 0
null == null
{ var x, y; x = {}; y = x; x == y; } // two variables pointing to same object
new Boolean(true) == true
new Number(1) == true
new String("1") == true
true == new Boolean(true)
true == new Number(1)
true == new String("+1")
new Boolean(true) == 1
new Number(-1) == -1
new String("-1") == -1
1 == new Boolean(true)
-1 == new Number(-1)
-1 == new String("-1")
new Boolean(true) == "1"
new Number(-1) == "-1"
new String("x") == "x"
"1" == new Boolean(true)
"-1" == new Number(-1)
"x" == new String("x")
{valueOf: function() {return 1}} == true
{valueOf: function() {return 1}, toString: function() {return 0}} == 1
{valueOf: function() {return 1}, toString: function() {return {}}} == "+1"
{valueOf: function() {return "+1"}, toString: function() {throw "error"}} == true
{toString: function() {return "+1"}} == 1
{valueOf: function() {return {}}, toString: function() {return "+1"}} == "+1"
true == {valueOf: function() {return 1}}
1 == {valueOf: function() {return 1}, toString: function() {return 0}}
"+1" == {valueOf: function() {return 1}, toString: function() {return {}}}
true == {valueOf: function() {return "+1"}, toString: function() {throw "error"}}
1 == {toString: function() {return "+1"}}
"+1" == {valueOf: function() {return {}}, toString: function() {return "+1"}}

Examples that return falsy values:

true == false
false == true
Number.NaN == true
Number.NaN ==  1
Number.NaN == Number.NaN
Number.NaN == Number.MAX_VALUE
Number.NaN == Number.MIN_VALUE
Number.NaN == "string"
Number.NaN == new Object()
true == Number.NaN
-1 == Number.NaN
Number.MAX_VALUE == Number.NaN
Number.MIN_VALUE == Number.NaN
"string" == Number.NaN
new Object() == Number.NaN
1 == 0.999999999999
" " == ""
" string" == "string "
"1.0" == "1"
"0xff" == "255"
1 == "true"
"false" == 0
undefined == true
undefined == 0
undefined == "undefined"
undefined == {}
null == false
null == 0
null == "null"
null == {}
false == undefined
Number.NaN == undefined
"undefined" == undefined
{} == undefined
false == null
0 == null
"null" == null
{} == null
new Boolean(true) == new Boolean(true)
new Number(1) == new Number(1)
new String("x") == new String("x")
new Object() == new Object()
new Boolean(true) == new Number(1)
new Number(1) == new String("1")
new String("1") == new Boolean(true)
{valueOf: function() {return {}}, toString: function() {return "+1"}} == "1"
"1" == {valueOf: function() {return {}}, toString: function() {return "+1"}}

Examples that throw an error:

{valueOf: function() {throw "error"}, toString: function() {return 1}} == 1 // throws "error"
{valueOf: function() {return {}}, toString: function() {return {}}} == 1 // throws TypeError
1 == {valueOf: function() {throw "error"}, toString: function() {return 1}} // throws "error"
1 == {valueOf: function() {return {}}, toString: function() {return {}}} // throws TypeError


Standard rules apply. The shortest code in bytes wins.

Sandbox Questions

  • Is it ok that I am restricting the language to JavaScript? Should I allow languages that compile to JavaScript or have the types listed here?
  • Have I got too many test cases here? Should I cut it down? Perhaps instead I could create a tio.run template which runs all the tests instead of listing them all out here.
  • Is the requirement of not using == and != strict enough? I feel that I want to avoid answers where some inbuilt function indirectly calls == and getting around it that way but I don't know how to enforce that.
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  • \$\begingroup\$ I'd say the number of test cases is still acceptable, though creating a TIO template sounds like a good idea anyway. \$\endgroup\$ – Laikoni Oct 14 '18 at 12:43
  • \$\begingroup\$ We still can use === and !==, can't we? \$\endgroup\$ – Shieru Asakoto Oct 15 '18 at 9:07
  • \$\begingroup\$ @ShieruAsakoto yes \$\endgroup\$ – Cameron Aavik Oct 15 '18 at 9:48
  • \$\begingroup\$ @CameronAavik Build a custom Javascript interpreter where calls to == errors out, then use that. \$\endgroup\$ – user202729 Oct 15 '18 at 10:18
  • \$\begingroup\$ Here's a link to a test framework in TIO. I don't think it's possible to overload operators in javascript though \$\endgroup\$ – Jo King Oct 17 '18 at 9:06
  • \$\begingroup\$ (for example when someone edits the interpreter to ban some features, see i.snag.gy/7ELPOt.jpg - the link "this branch" links to github.com/Mego/Seriously/tree/… --- of course that's a lot of work, but that would make sure that people definitely don't cheat) \$\endgroup\$ – user202729 Oct 19 '18 at 1:40

Cologne Phonetics

From Wikipedia: Cologne phonetics is a phonetic algorithm which assigns to words a sequence of digits, the phonetic code. The aim of this procedure is that identical sounding words have the same code assigned to them. The algorithm can be used to perform a similarity search between words. For example, it is possible in a name [collection?] to find entries like "Meier" under different spellings such as "Maier", "Mayer" or "Mayr".

The Cologne phonetics matches each letter of a word to a digit between 0 and 8. To select the appropriate digit, at most one adjacent letter is used as context.


Processing of a word is done in three steps:

  1. Encode letter by letter from left to right according to the conversion rules below.
  2. Remove all digits occurring more than once next to each other.
  3. Remove all code "0" except at the beginning

Conversion rules

  • A, E, I, J, O, U, Y become 0
  • H becomes nothing
  • B becomes 1
  • P becomes 1, unless it's before H
  • D, T become 2, unless they're before C, S, Z
  • F, V, W become 3
  • P becomes 3, if it's before H
  • G, K, Q become 4
  • C becomes 4 in initial position before A, H, K, L, O, Q, R, U, X
  • C becomes 4, if it's before A, H, K, O, Q, U, X, except after S, Z
  • X becomes 48, unless it's after C, K, Q
  • L becomes 5
  • M, N become 6
  • R becomes 7
  • S, Z become 8
  • C becomes 8, if it's after S, Z
  • C becomes 8 in initial position except before A, H, K, L, O, Q, R, U, X
  • C becomes 8, if it's not before A, H, K, O, Q, U, X
  • D, T become 8, if they're before C, S, Z
  • X becomes 8, if it's after C, K, Q


Implement a program or a function that takes a single string as input and returns the phonetic code according to the rules stated above. You may choose your own output format, but keep in mind that the expected output may have leading zeros. The input string is guaranteed to match /^[a-zA-Z ]*$/, this means it may contain characters from the lowercase and uppercase alphabet, as well as spaces.

Test cases

Input                 Output

[empty string]        [empty]
Kolner Phonetik       45673624
Meier                 67
Mayer                 67
Augsburg              048174
Xanten                48626
Telephon              2356
Chemie                46
Muller Ludenscheid    65752682

This is , the shortest code for each language wins.

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  • \$\begingroup\$ Related: Soundex function \$\endgroup\$ – Laikoni Oct 14 '18 at 12:49
  • \$\begingroup\$ What's the difference between "initial sound" and "initial position"? \$\endgroup\$ – Laikoni Oct 14 '18 at 12:58
  • \$\begingroup\$ Apparently there is none, from what I was able to find \$\endgroup\$ – oktupol Oct 15 '18 at 7:11
  • \$\begingroup\$ Ideally there would be a test case for each conversion rule as the challenge is probably mainly about handling those edge cases in the rules. \$\endgroup\$ – Laikoni Oct 15 '18 at 13:27

Keyboard Row Shift

Given an input string of characters, output the number of times we have to shift to another row while typing that string using a qwerty keyboard.

The input string will contain lower case alphabets,numbers and spaces. The newline key(enter) must also be considered at the end (only at the end). All shifts, from any row to any other will be considered as 1 shift only.


asdfghjkl enter


"sdkflsd" -> 0
"asdwexc" -> 3 # to end the string one enter has to pressed which is in the middle
"poierlkdjfpoeirldskjf" ->3
"123 jkjk" -> 2
"llsdkfj ldkfj" -> 2
"lkasdjmnbcv " -> 3
"jnjn 5" -> 6

This is code-golf, so shortest code wins.

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  • \$\begingroup\$ You need some test cases that contain a space and numbers. I'd also make it clear in your description that a shift can go any distance (going from the top row to the bottom is only 1 shift) \$\endgroup\$ – Nathan Merrill Oct 16 '18 at 12:48
  • \$\begingroup\$ Done, @NathanMerrill \$\endgroup\$ – Vedant Kandoi Oct 17 '18 at 5:36
  • \$\begingroup\$ How flexible is input? Is uppercase OK? How about an array of character strings or even an array that mixes digits and character strings? \$\endgroup\$ – Shaggy Oct 18 '18 at 21:47
  • 1
    \$\begingroup\$ I have mentioned only lower case, input has to be string only not an array. \$\endgroup\$ – Vedant Kandoi Oct 19 '18 at 5:15

Not All Sums Are Equal


When computing the sum of a list of floating point values, the order in which we add up the summands matters. A good way to reduce errors is beginning with the summand with the least modulus and working your way up to the numbers with a greater modulus. As an illustrative example, let us assume our computer uses floating point numbers that comprise 2 (decimal) digits \$m_1,m_2\$ for the mantissa (and a sufficiently large number of digits for the exponent \$e\$.) These represent a number \$0.m_1m_2 \cdot 10^e\$. Let us consider the list

[0.1, 0.009, 0.009]

The exact result would obviously be \$0.118\$, and the best approximation in our computer therefore \$0.12\$. But if we actually tell the computer to sum those numbers, it will truncate all intermediate results, so it will get:

 (0.1 + 0.009) + 0.009 = 0.1 + 0.009 = 0.1

If we use the recipe from above and sort the list by the modulus first, we get

(0.009 + 0.009) + 0.1 = 0.018 + 0.1 = 0.11

which is already closer to the exact result. So we see that any permutation in the order of the summation might produce a different sum. In order to estimate the possible error we're interested in the difference between the greatest and the least sum we can get by just using a permutations of the original list. This is the task of this challenge.


The following examples assume that you use IEEE-754 double numbers in your implementation. You can of course use other implementations, but the results might be different:

[0.0140534017661288 0.50942429920538 1 1.96300019759529 71.1571487560538]
[-0.447570988743782 -0.30397649549556 0 0.765285061877593 3.30425088014391]
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  • \$\begingroup\$ 1. The decimal example only works because it's using a bad rounding mode. With a good rounding mode which gives an answer with half an ULP, both orders would give 0.12. 2. The example is not very clear on what the inputs and outputs are. \$\endgroup\$ – Peter Taylor Oct 22 '18 at 19:15

Are these strings character wise translatable?

Explanation with example:

Suppose you are given two strings of the same length, e.g. "code" and "golf". Try to find a set of translation rules for characters (like "All cs are replaced by gs" , or c -> g for short) such that applying those rules on "code" yields "golf" and applying them to "golf" again yields "code".

If there is such a set of translation rules, return a truthy value. Otherwise, if no such set can exist, return a falsy value.

In our example, the translation is possible with the rules g -> c, d -> l, e -> f, c -> g, l -> d and f -> e, so a truthy value should be returned.

However, if the two strings were "code" and "meta", no such rules exist: To get from "code" to "meta" we need the rule e -> a, but to get from "meta" to "code" we would need e -> o. But e cannot be translated into two different characters! Therefore, a falsy value should be returned.

Consise mathy explanation:

Given two strings \$s\$ and \$t\$ of the same length \$n\$, decide whether the following relation is bijective: $$\{(s_i,t_i)\ |\ 1 \leqslant i \leqslant n\} \cup \{(t_i,s_i)\ |\ 1 \leqslant i \leqslant n\}$$ where \$s_i\$ denotes the \$i^{th}\$ character of string \$s\$.

Test cases


"code", "golf"
"a", "a"
"abdabdcdacabcabd", "bacbacdcbdbadbac"
TODO: Add some more


"code", "meta"
"aa", "ab"
"abc", "bca"
TODO: Add some more
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  • \$\begingroup\$ This seems to be very similar to Check if words are isomorphs. \$\endgroup\$ – Dennis Oct 28 '18 at 2:39
  • \$\begingroup\$ @Dennis It is definitely related. However, as far as I see most answers there work by creating some sort of fingerprint and checking whether it is the same for both inputs. This approach alone won't work here as you also need to check the symmetry of substitutions. \$\endgroup\$ – Laikoni Oct 28 '18 at 11:33
  • 1
    \$\begingroup\$ Not saying this makes it a dupe, because there might be golfier ways, but s and t are character wise translatable if and only if s++t and t++s are isomorphs. \$\endgroup\$ – Dennis Oct 28 '18 at 12:54

Guess the Q in the code

A popular puzzle known to me as Codewords takes a Crossword-like grid and consistently replaces each letter with a code value, usually a number from 1 to 26, and the solver then has to work out which code value represents each letter. There are a number of approaches to this:

  • There is a web site which can search for words given the pattern of repeating letters. Since repeated letters map to repeated code values, you can sometimes use the patten of repeated code values to determine the original word. For instance, the only word with the pattern 12.312..3... is churchwarden.
  • The puzzle normally provides a few of the letters to get you started. This may be enough to use a standard crossword solver to determine the original word. For instance, .h...h...d.. would again point you to the word churchwarden.
  • You could use frequency analysis to guess which letters are likely to be the popular letters such as e or t.
  • You could look for common prefixes or suffixes such as ally or ness. (I picked those two as they have repeating letters which are easier to spot.)
  • At least in all of the Codewords puzzles I have seen, q is always followed by a u and at least one letter. This means that if there is a code that is never the last or penultimate letter of a word, and always appears followed by the same code, then this could be a q.

However, checking all the codes to see how many distinct codes follow them is laborious. We need an automatic solution for this!

Please write a program or function that will guess which code(s) could represent the letter q. The input to the function will be an array of code values in a standard format. You will need to support 27 consistent distinct values, 26 for the codes themselves and a value for the background. These values can be integers but you could also use characters e.g. 26 letters and a space, in which case you can join them into strings, and even join the strings with a 28th character.

Your output will be all of the code values that have the property that, for each occurrence in the grid, either:

  • The previous and following cells are background or would extend past the side of the grid, or
  • There are two non-background following cells, and the first of those cells always contains the same value.

These rules apply in both the horizontal and vertical direction.

This is , so the shortest solution that breaks no standard loopholes wins!

Examples needed, I guess...

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Hack g-code parser

  • \$\begingroup\$ This could be seen as asking for malicious code, which is not on-topic. \$\endgroup\$ – Laikoni Oct 25 '18 at 17:55
  • \$\begingroup\$ @Laikoni this is not malicious\harmful code, because it don't causes harm to user. It is educational programming puzzle for white-hat hackers \$\endgroup\$ – Евгений Новиков Oct 25 '18 at 18:58
  • \$\begingroup\$ @Laikoni rm -rf / is harmful, but alert("pwned") is not \$\endgroup\$ – Евгений Новиков Oct 25 '18 at 18:59
  • 3
    \$\begingroup\$ Right, I think misread the challenge. If it is only about cracking this piece of code rather than actual implementations it's probably fine as a challenge. Any way, you need an objective winning criterion. code-golf in the sense of "the shortest input to open the alert prompt" seems like a good candidate. \$\endgroup\$ – Laikoni Oct 25 '18 at 19:47

Find the pattern v2

I feel tired to do "find the pattern" exercise such as

1 2 3 4 (5)
1 2 4 8 (16)
1 2 3 5 8 (13)

Please write a program that finds the pattern for me.

Here, we define the pattern as a recurrence relation that fits the given input, with the smallest score. If there are multiple answers with the same smallest score, using any one is fine.

Let the \$k\$ first terms be initial terms for the recurrence relation, and the \$i\$'th term be \$f(i)\$ (\$i>k,i\in\mathbb N\$).

  • A non-negative integer \$x\$ adds\$1\$ to the score
  • The current index \$i\$ adds \$1\$ to the score
  • +, -, *, / (round up, down, towards zero, go further to zero, as you decide) and mod (a mod b always equal to a-b*(a/b)) each add \$1\$ to the score
  • For each initial term \$x\$, add \$1\$ to the score
  • \$f(i-n)\$ (with \$n\le k\$) adds \$1\$ to the score. E.g. Using the latest value \$f(i-1)\$ add \$1\$ to the score, and there must be at least 1 initial term.
  • Using parentheses to change the calculation order doesn't add anything to the score.


input   -> [score]    expression(not optimized)
1 2 3 4     -> [1]    f(i) = i
1 2 4 8     -> [3]    f(1) = 1, f(i) = 2*f(i-1)
1 2 3 5 8   -> [5]    f(1) = 1, f(2) = 2, f(i) = f(i-1)+f(i-2)

Lowest score for worse case wins. If tie, shortest program in each language wins. Your program should run in polynomial time.

If someone's score is lower than 4(currently reachable), the first lowest score will be accepted. (differ from winning criticia)

p.s. See history for v1 sandbox

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  • 1
    \$\begingroup\$ I think the challenge could benefit from a precise defining grammar for a pattern. As it stands, I feel the set of all expressions that may fit is too vast. \$\endgroup\$ – Jonathan Frech Oct 16 '18 at 9:28
  • \$\begingroup\$ @JonathanFrech Looks pretty precise to me. About "too vast" - there are only 6 bullets. \$\endgroup\$ – user202729 Oct 17 '18 at 10:43
  • \$\begingroup\$ @user202729 I am unsure if (round as you decide) includes the possibility of not rounding. \$\endgroup\$ – Jonathan Frech Oct 17 '18 at 15:28
  • \$\begingroup\$ Should "3 0 0 0 0 0 ..." be "f(1)=3, f(n)=0", do "for the recurrence relation" work? \$\endgroup\$ – l4m2 Oct 17 '18 at 16:40
  • \$\begingroup\$ @l4m2 Yes. ----- \$\endgroup\$ – user202729 Oct 17 '18 at 16:56
  • \$\begingroup\$ I don't understand the "fun fact"... \$\endgroup\$ – user202729 Oct 21 '18 at 15:42
  • \$\begingroup\$ @user202729 Do you think it no fun or can't figure it out? \$\endgroup\$ – l4m2 Oct 21 '18 at 19:24
  • \$\begingroup\$ The latter. ---- \$\endgroup\$ – user202729 Oct 22 '18 at 0:15

User ranking in language

Challenge is:

  • You need receive the input of "language". By "language" I mean the at left of the comma in the Title of an answer on the present code-golf site.
  • The source for the data are the answers posted here on code-golf site.
  • You must present results in order. Decreasing or Increasing does not matter, it only needs to be one of those two.
  • Output is a table of two columns {User, Upvotes}. Example: User enters Java and will obtain a list as the following:

user_A 9523 user_B 6000 user_C 120

Do not care very much about the output formatting as long it is clear the separation between the two columns and each line.

The will be no winner as it is a per language question.

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  • \$\begingroup\$ "The source for the data are the answers posted here on code-golf site." The answers for this challenge, or ALL challenges on PPCG? And with your example, does this mean it searches for each user all answers given in Java, and sums their total upvote count? Also, do we differentiate java/Java/Java 7/Java 10 etc as different inputs? \$\endgroup\$ – Kevin Cruijssen Nov 1 '18 at 12:45
  • \$\begingroup\$ @KevinCruijssen 1. Yes, the scope is all answers on PPCG. 2. Yes, it is the total upvote account. 3. This is flexible on uppercase/lowercase, but not on the rest. \$\endgroup\$ – sergiol Nov 8 '18 at 19:58
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