# What is the Sandbox?

This "Sandbox" is a place where Code Golf users can get feedback on prospective challenges they wish to post to the main page. This is useful because writing a clear and fully specified challenge on the first try can be difficult. There is a much better chance of your challenge being well received if you post it in the Sandbox first.

See the Sandbox FAQ for more information on how to use the Sandbox.

## Get the Sandbox Viewer to view the sandbox more easily

To add an inline tag to a proposal use shortcut link syntax with a prefix: [tag:king-of-the-hill]

Help Indiana Jones and his crew cross the bridge!

This codegolf will solve the Bridge and Torch problem. In this problem, there are multiple people (I'm thinking four) who must all cross a weak bridge to escape an evil dragon as quickly as possible. Because the bridge is weak, only two people can cross the bridge at a time. The whole crew is armed with one torch, which is necessary for 1 or 2 people to cross the bridge. Furthermore, each person takes a certain, integral amount of time to cross the bridge. When two people cross together, they must run at the rate of the slower person. The whole crew needs to quickly figure out how to get all the people across the bridge in the least amount of time to maximize their chances of survival.

Input A list of names of the crew (one word, a-zA-Z) and how long they take to cross the bridge alone.

Output An explanation of who crosses the bridge in which order so that the total time is minimized, and the total time.

Example Input: Indiana 5 Jones 10

Output:

Indiana, Jones 10

Input: A 1 B 2 C 5 D 8

Output:

A, B A C, D B A, B 15

I'm thinking of either just solving this problem with any number of people, or another version in which anyone not at the end (ie. on the first side or on the bridge) after the time limit dies, and the goal is to minimize the number of deaths.

• If it's code-golf it's easier for you as the question poster. If you put a time limit, then different computers will achieve different amounts in that time limit, so you would need to run all the answers on your own computer to give them an official score. – trichoplax Jul 3 '15 at 16:45

A malapropism is the substitution of one word for another that sounds similar, often as a way to make something sound unintentionally humorous. For example, "He's a wolf in cheap clothing" is a malapropism, since the expected word, "sheep's", got replaced by "cheap", which sounds similar but means something different.

A modern version of malapropism is to take scenes from movies and redub the dialog with different words that match the actors mouth movements as a parody. There is a YouTube channel called "Bad Lip Reading" that uses this technique.

I would like to apply the process to some old videos with subtitle files, then watch the videos with the sound turned down to turn it into a long series of malapropisms.

Using mouth movements gives a more flexible range of malapropism so there is some flexibility between different sounds.

### Challenge

Create a malapropism generator. I want to be able to feed text from subtitles into the generator and have text which is different, but still matches the mouth movements of the actors.

### Input

A string of English words, (already processed to remove punctuation and forced to uppercase).

### Output

A string of English words, (same format as input).

### Notes

To simplify the challenge, all input words are in upper case, separated by whitespace, with all punctuation except apostrophes removed.

Lets agree to constraint what words "sound like", to be based on the CMUDICT. You may scrub the data so you don't have to worry about comments or special punctuation entries and remove stress numbers.

Lets also agree on the mouth-movements associated with the sounds, called "visemes". Here is a mapping used by Microsoft's SAPI library, which is itself based on Disney animation rules. Microsoft uses the same set of phonemes from ARPABet as CMUDict data.

#   ARPAbet Phoneme

1   AE AH
2   AA
3   AO
4   EH EY UH
5   ER
6   IH IY Y
7   UW W
8   OW
9   AW
10  OY
11  AY
12  HH
13  R
14  L
15  S  Z
16  CH JH SH ZH
17  DH TH
18  F  V
19  D  N  T
20  G  K  NG
21  B  M  P


You should be able to take each English word, convert to a set of phonemes, then to a set of visemes, then produce a list of words with matching visemes and randomly select one of the other words from the list. If a word doesn't match, or doesn't have any alternative, the original word should be copied to output.

### Examples

• "HELLO WORLD" "HALLOW WHIRLED"
• "I SEE DEAD PEOPLE" "EYE SEA TED PEOPLE"
• "I WILL BE BACK" "EYE WHEEL PEA BAG"
• "IT WAS BEAUTY KILLED THE BEAST" "INN WAS PUNY GILT THE MIST"
• "MAY THE FORCE BE WITH YOU" "MAY THE FOURS BEE WITH YEW"

### Test Cases

Since the words are random, I have selected some pairs of words with unique pronunciation. Your function or program should always return one word when presented with the other:

Input            Output
"AMUSING"        "ABUSING"
"APOGEE"         "APACHE"
"BACKDATING"     "MAGNETIC"
"BALLOONING"     "POLLUTING"
"INVISIBLE"      "INFEASIBLE"
"LAMPS"          "LUMPS"
"SCORN"          "SCORED"
"WEPT"           "WEBBED"


### Rules

• You can write a full program or function.
• Input should be taken from stdin or function parameters. Output should be printed through stdout or returned.

### Scoring

This is [code-golf]. Submission with least number of bytes, (not including data file(s)) wins.

• Interesting challenge, but I have some small suggestions. 1. The example in the first paragraph (moths vs moss) is inconsistent with the definition which follows, and that could confuse. Maybe borrow one of the test cases, or mention The Importance of being Earnest? 2. Why not link to the YouTube channel? 3. For subheadings, ### is better than *. 4. Since the input is nicely cleaned, I can't see why you specify all input words are in the same case rather than all input words are in upper case. The latter would be more useful, since it matches the CMUDICT file. – Peter Taylor Aug 20 '15 at 17:22
• 5. The mapping of ARPAbet to viseme ID is very long and offputting, and it's not really necessary. If you replace it by a list of groups (AE AH newline ... newline B M P) in preformatted text (indent by four spaces) then it will convey all the necessary information while taking a lot less space. 6. Talking of which, it might be worth mentioning ARPAbet to pre-empt questions in the comments. – Peter Taylor Aug 20 '15 at 17:22
• 7. It's a good idea to be explicit that when you say randomly you mean with equal probability or (easier for those whose libraries give them random floats) with a difference between the greater and least probabilities of no more than 1%. 8. The test cases aren't very useful for testing. The most useful test cases will be where most of the words have one or two possible outputs. E.g. MINT CONDITION STAMPS becomes PINNED CONDITION STUMPS. CATERER is also a good test case because it has a repeated phoneme at the end of the word, which could catch some buggy regex-based approaches. – Peter Taylor Aug 20 '15 at 17:23

# Cake

Concerns:

• I'm not sure how well single language questions do. It automatically limits the question to those that already know the language, and those that are willing to learn just for this question. There's no need to limit to Chef with some of the magic this community can produce!

• "selfish/demanding" sounds quite strong and could sound like a bit of a personal attack on Beta Decay. It certainly isn't intended like that, and I expect it will be accepted as intended, but is it better to err on the side of caution?

• Is there any particular reason to limit the challenge to Chef? Writing code that mimics a real world cake recipe could be entertaining in other languages as well. – Dennis Sep 9 '15 at 18:36
• @Dennis, I suppose not! I wouldn't have thought it possible, but I've seen some of the magic this community can produce! – James Webster Sep 10 '15 at 11:39
• I feel like the best tasting submission might just be an actual cake recipe which acts as a no-op, with the actual printing merely tacked on somewhere (eg in Foo) – Sp3000 Sep 11 '15 at 4:12
• @Sp3000, that's possible. I added that rule because with Chef, the ratios for ingredients are often way off. – James Webster Sep 11 '15 at 6:12

## Source code ecological footprint

You've just been hired by a German car manufacturing company. Your first task, as an engineer, is to write a program that computes the ecological footprint of source code.

The ecological footprint of a character is computed as follows (you can assume the source code is ASCII-encoded):

Write the character's ASCII code in binary, and count the number of 1's.

For example, A has a footprint of 2, but O is dirtier with a footprint of 5.

The global footprint of a program is the sum of the footprints of its characters.

Your program must accept a string as parameter, compute its ecological footprint, and output it.

There is a subtlety though. As you wish to enter a new, more restrictive market, you need to tune your program so that it behaves differently in "test mode". Thus:

The program should output 0 when it receives the string test as parameter.

Scoring

The source code with the smaller ecological footprint wins (and yes, the answer test is forbidden!)

• This anonymous company doesn't happen to be named Volt's Wagons, does it? – Geobits Sep 25 '15 at 3:20
• @Zach Sorry, I rolled back your edits - the horizontal lines were really harming readability ? and the fact that we output 0 only when we receive test is important ? – Arnaud Sep 25 '15 at 3:33
• You could have removed the hr tags.. Also, I added a lot of punctuation and grammar fixes.. I didn't removed the "output 0 only when we receive test" detail either? I only removed the redundant clause. – Zach Gates Sep 25 '15 at 3:35
• What's the expected output for an empty input string? I'd normally say 0, but "(and only when)" seems to disqualify that. – Geobits Sep 25 '15 at 3:36
• @Zach I've re-reported your changes. Thanks for your help! – Arnaud Sep 25 '15 at 3:45
• Can we assume that the input will never be empty? – Zach Gates Sep 25 '15 at 6:08
• @Zach I would accept empty input and output 0, as the footprint of an empty program is zero ? – Arnaud Sep 25 '15 at 6:50

## Group Students Into Pairs

Synopsis: A graph theory matching problem. Given a list of which students like and dislike each other, pair the students up to maximise the overal happiness of the classroom.

## Introduction

If you've ever been a teacher before, you've likely encountered the extremely frustrating experience that is getting students to do pair work. More often than not no one is happy with their partner, and the mood of the classroom suffers.

In this challenge, we're going to solve the enduring problem of pair work in classrooms. Here's how our system will work:

1. Each student shall write down what students they are happy to work with, and which ones they are not happy to work with.
2. We will take those lists, and generate student pairs such that the total happiness of the classroom is at its maximum.
3. The students are now (mostly) happy!

So how is the happiness of the classroom calculated? For each student in each pair of students:

• If a student's partner is in their "happy to work with" list, increase the happiness of the classroom by 1.
• If a student's partner is in their "not happy to work with" list, decrease the happiness of the classroom by 1.
• If a student is neutral towards their partner (not in either list), the happiness of the classroom does not change.

Here's a useful table that summarises the possible changes in the happiness with each pair of students:

+-------------------+-------------------+--------------------+
| Student 1 Feeling | Student 2 Feeling | Happiness Modifier |
+-------------------+-------------------+--------------------+
| Happy!            | Happy!            | +2                 |
| Happy!            | Neutral           | +1                 |
| Happy!            | Not happy...      | 0                  |
| Neutral           | Happy!            | +1                 |
| Neutral           | Neutral           | 0                  |
| Neutral           | Not happy...      | -1                 |
| Not happy...      | Happy!            | 0                  |
| Not happy...      | Neutral           | -1                 |
| Not happy...      | Not happy...      | -2                 |
+-------------------+-------------------+--------------------+


## Input

First line of input is a positive integer n indicating the number of students to match up. n is always even. Following that are n lines, each line which has the format:

name likes dislikes


where name is the name of the student (which will only contain characters from [A-Za-z]), likes is a comma separated list of people he or she would be happy to work with, and dislikes is a comma separated list of people he or she would not be happy to work with. For instance:

Clarence Tiger,Anna,Jamal Amelia,James


This would indicate that Clarence is happy to work with Tiger, Anna, or Jamal; he is not happy to work with Amelia or James; and towards any other student he is neutral.

Another line:

Ilya Amelia,Clarence,Anna,Tiger,Jamal _


This would indicate that Ilya is happy to work with Amelia, Clarence, Anna, Tiger, and Jamal. For dislikes, we've used the special keyword _, which indicates that Ilya doesn't have anyone he is not happy to work with. Any remaining students Ilya would be neutral towards.

## Output

Output space separated comma separated pairs of names, where each name corresponds to a pair of students. The pairs should be such that the happiness of the classroom is maximised. Every student should be in exactly one pair.

## Example Inputs and Outputs

Writing reference implementation. This is one of those types of questions where you can't really judge the correct output for an input "by eye".

## Sandbox Questions

1. This problem is quite difficult compared to typical code-golf fare, requiring two different algorithms to solve the weighted maximum matching problem. Is it too difficult for a code-golf? And if it is, what could make a better winning criterion?
• It's certainly not too difficult for a code golf at present, because there are no time constraints so I could brute-force it. With time constraints I'd have to do some research before tackling it, but it's good to have the occasional code golf problem which can't be solved in 30 characters. – Peter Taylor Jun 25 '15 at 17:32
• I would think the challenge would actually do well with a loose time constraint to prevent brute force. While you'd definitely get fewer answers, those answers would definitely be more interesting – Nathan Merrill Jun 27 '15 at 1:55
• "Jamal Amelia" -- Where is the comma? And I think that this would be fine for code-golf. BTW, add which all characters a name could have in the challenge. – Spikatrix Jun 27 '15 at 9:42
• @CoolGuy There's no space because Jamal is the end of the likes and Amelia is the start of the dislikes. I added in which characters a name can contain. – absinthe Jun 27 '15 at 11:29
• Cool Guy's question raises the point that you should be explicit about cases where a person has no likes, no dislikes, or neither. I.e. can trailing spaces be omitted? Can the entire line be omitted if the person has neither likes nor dislikes but is liked or disliked by someone else? – Peter Taylor Jun 27 '15 at 14:29

# The /\/\aze of Mirrors

sniffs...something smells...fishy.

The reason being that there's actually a fish! But it's at the end of a maze of mirrors that you have control over. Your task is to rotate the mirrors until you can see that delicious-smelling ><>. Luckily, there are only nine mirrors, so it doesn't take you long to find a solution.

However, if you encounter this again in the future, you don't want to have to actually do the work yourself. Computers are so much better at it. So you decide to write some code to solve it for you. The shorter, the better, because you don't want that fish to rot!

## The situation

Your program must accept a single positive integer n as input. It then has to find a solution for an n by n mirror maze with the following conditions:

• You, the observer (>), is seated next to one corner.
• The fish (F) is placed next to the opposite corner, on the adjacent wall if n is even and on the opposite wall if n is odd.
• The maze is toroidal because its designers installed cameras and screens.
• However, there is a wall behind the fish, so you can only see it from inside the maze.
• You must use all of the mirrors. No shortcuts!
• Bonus: take an additional non-negative integer x as input and solve the maze for when the fish is x spaces away from the corner (where x=0 is its usual place), flipping to the other wall if x is odd.

## Example solutions

1
+-+
>/F
+-+

2
+--+
>/\|
|//|
+-F+

2 1
+--+
>/\F
|\/|
+--+

3
+---+
>\/\|
|//\|
|\/\F
+---+

4
+----+
>/\/\|
|//\\|
|\\//|
|//\\|
+---F+


## Rules

• No standard loopholes, as usual.
• Your program must output the solution in ASCII like what I've shown above.
• Scoring is in bytes, unless you have the bonus: multiply by 0.8.

I look forward to seeing all the amazeing solutions! Ha! Sorry, that was bad.

• Can I adopt this abandoned proposal? – programmer5000 Jun 9 '17 at 12:20
• @programmer5000: Yes, you may. – El'endia Starman Jun 10 '17 at 3:49

# Approximate phi using Fibonacci numbers

As shown by Johannes Kepler, the quotient of successive Fibonacci numbers approaches phi.

(where F(n) = the nth Fibonacci number)

The fine folks at Wikipedia also say that if you choose two different starting values for the Fibonacci sequence, this expression still holds.

Last, but not least, if you use a later number in the series in the numerator, you get some interesting properties:

## The challenge

• Create a full program or function that approximates phi using Fibonacci numbers.
• The program or function will take a single integer, n (unless a bonus is applied). It will output to STDOUT as follows, stopping when it reaches n:

 1/1=1 2/1=2 3/2=1.5 5/3=1.66666666 ... F(n+1)/F(n)=...

• Fibonacci numbers must be calculated using iteration or recursion.
• No built-ins / standard loopholes, and neither the square root function nor phi should appear in the source code.
• It is OK if floating point limitation/integer overflow prevent accuracy beyond 4 or 5 digits, but you should use the most precise primitive data type (yes, double is one character longer than float).

There are two bonuses that you can earn, for a total of a 25% lower score:

• Input two additional integers as starting values for the Fibonacci sequence. This will still converge on phi, though it may take a bit longer. Reward: -10%.
• Input one additional integer for a in the above formula. This will result in the sequence converging to phi^a. Reward: -15%.

## Scoring

Total score is the size of the program, in bytes, minus any bonuses. Since this is , lowest score wins.

(Insert leaderboard snippet) (Insert example snippet)

Example implementation snippet:

document.getElementById("button").addEventListener("click", process, false);

function process() {
//Store field values to vars
var itrNum = parseInt(document.getElementById("itrNum").value);
var prevNum = parseInt(document.getElementById("start1").value);
var curNum = parseInt(document.getElementById("start2").value);
var aNum = parseInt(document.getElementById("aVal").value);
document.write("(Click \"Run\" to reset)<br><br>");

//Iterate through each fibbonacci number
for (var i = 0; i < itrNum; i++) {

dispNum = getFutureItr(prevNum, curNum, aNum);
document.write(dispNum + " / " + prevNum + " = " + dispNum / prevNum + "<br>");
//prepare numbers for next iteration
var tempNum = curNum + prevNum;
prevNum = curNum;
curNum = tempNum;
}
}

//Helper function for F(n+a)
function getFutureItr(prevNum,
curNum,
aNum) {
var tempNum = curNum;
for (var i = 1; i < aNum; i++) {
tempNum = curNum + prevNum;
prevNum = curNum;
curNum = tempNum;
}
return curNum;
}
<!DOCTYPE html>
<html>

<body>
Iteration number:
<input type="number" min=0 id="itrNum" value=20>
<hr>Starting value 1:
<input type="number" min=1 id="start1" value=1>
<br>
<br>Starting value 2:
<input type="number" min=1 id="start2" value=1>
<hr>Value of a:
<input type="number" min=1 id="aVal" value=1>
<hr>
<button id="button">Calculate</button>
</body>

</html>

• There's a fibonacci tag. It's generally worth making it clear which of the two indexing conventions answers to use: F(0) = 0 or F(0) = 1? – Peter Taylor Nov 2 '15 at 14:37
• Does the output need to be in the exact format 3/2=1.5? – xnor Nov 4 '15 at 21:54

## Compete with awk

The goal of this challenge is to see if the assertion often see in question on SO is true:

This would be faster in whatever

# Test Input

A 10 million lines file generated by this script line:

awk 'BEGIN{for (i=1;i<=10000000;i++) print (i%5?"miss":"hit"),i,"  third\t \tfourth"}' > file


$head -10 file miss 1 third fourth miss 2 third fourth miss 3 third fourth miss 4 third fourth hit 5 third fourth miss 6 third fourth miss 7 third fourth miss 8 third fourth miss 9 third fourth hit 10 third fourth  # Goal: The main goal is to compete with awk, to benchmark it will be on this challenge description: • Split each line on any number of spaces (regex [ ]+) • If the line start by hit X with X being an even number • print fields: 4th 1st and 3rd (in this order) Example output (first 5 lines): fourth hit third fourth hit third fourth hit third fourth hit third fourth hit third  Restrictions: The idea being to compete with awk, your program must behave on the same way: • The file must be read. • The match and output order must be modifiable as an awk program can be (Needing a simple code update is OK, it has not to be a command line parameter). • The number of line in input file should be considered unknown. • As long as your output can be piped to awk, the submission is valid. # Validation: Pipe the output of your program to awk '!seen[$0]++{unq++;r=$0} END{print ((unq==1) && (seen[r]==1000000) && (r=="fourth hit third")) ? "PASS" : "FAIL"}'  If it show PAST, you're in :) # Winner The fastest code would win, to bench yourself here is a mawk version really competitive (which won't count toward answers as I think it's pretty hard to beat): awk '/^hit +[0-9]*[02468] / { print$4, $1,$3 }' file


I've build a test suite on github here which build a result table here.

Answers will be integrated within it to bench all answer on the same machine, if there's a specific way to launch your program, make it appear in the answer so I don't penalize your answer by the way it is launched.

### Side challenge:

If you wish to save me time and craft a pull request to include your code in the test system, you're welcome.

• 1. "Split each line on spaces" would IMO split the line miss 1 third fourth into 15 fields. A possible rewording might be "split each line around the regex [ ]+" (where the otherwise unnecessary [] prevent MarkDown from trimming the space). 2. The question as worded is a bit too special-cased, as is the awk code you provide. If it's permissible to optimise the regex to /hit [0-9]*0/ because you know a priori the form of the input, why would it not be permissible to optimise further and write print "fourth hit third\n"*1E6? – Peter Taylor Nov 19 '15 at 21:02
• @PeterTaylor Thanks for the feeback, I reworded a little to be more specific on restrictions and avoid too much test input optimizations. – Tensibai Nov 20 '15 at 10:20
• I still think your reference answer is cheating with /hit [0-9]*0 /. To match the challenge description it should use /^hit +[0-9]*[02468] / – Peter Taylor Nov 20 '15 at 11:18
• @Peter ok, it does no difference in timing at end anyway (at least nothing I've been able to see on slow machines when benchmarking the codes present on GH) – Tensibai Nov 20 '15 at 11:24

# Comedy of exceptions

Your goal is to demonstrate the shortest program or function that can throw 10 different types of errors/exceptions (I use the two interchangeably here) at runtime.

This question is only open to languages which have an object-oriented type system for errors. Furthermore, in the case of each error, a message including a clear identification of the exception's type must be printed. Evidently, it must also have at least 10 error types. Other systems such as an integer error code identifying the error, or all errors represented by strings are not acceptable.

Input is an integer from 0 to 9 inclusive. The program should reliably throw a different type of error for each input.

Methods which allow exceptions of arbitrary type to be thrown may not be used. E.g, throw ... .

### Determining whether two errors are of different type (TODO: fill in detail of output)

Java: x and y print different stack traces, but are the same type, so they can't be used.

Python: UnboundLocalError ..... and NameError ....... would count as two types of error, since, although UnboundLocalError is a subclass of NameError, type() would give different results for the two.

C++: stoi("aaaa") printing ......... would be valid since the message gives std::invalid_argument. Integer division by zero would not since it crashes the program without a message informing of the error type

### Sandbox

Please comment if you are aware of a language that is borderline on eligibility, blurs boundaries between error types, or has other loopholes.

• Would exceptions from standard modules be allowed (e.g. from urllib.error import *)? How about third-party modules/libraries (e.g. django, boost)? Or even custom exception subclasses? – grc Nov 26 '15 at 11:17
• I don't know if I can find 10 different ones, but CJam can print a number of errors. Some of these are CJam-specific errors (empty stack, type mismatch; these tend to be displayed as "RuntimeException" but with a distinguishing error message) and some of them, I think, are just crashes happening in the interpreter exiting with a Java exception. How would that be treated? (I think it's also different between the Java and the JavaScript interpreter.) – Martin Ender Nov 26 '15 at 15:01
• To make this more specific, here are 11 different errors (with different error messages), but some are backed by the same exception type within the interpreter. Would these still count as different? pastebin.com/VJCiVYMf (Also note that CJam itself has no concept of exceptions... if there's an error it crashes... these are just the exceptions types used by the interpreter.) – Martin Ender Nov 26 '15 at 15:14
• I can think of a language which has exceptions but not an object-oriented type system: SML. (Haskell almost certainly has some exception monad too...). – Peter Taylor Nov 26 '15 at 21:05
• @grc I think I will allow standard libraries, but not 3rd-party ones. Any method of defining and throwing a custom exception type should fall under the rule "Methods which allow exceptions of arbitrary type to be thrown may not be used." – feersum Nov 27 '15 at 0:28
• @MartinBüttner It appears that the are Java exceptions and that CJam does not have its own exception type system, meaning it cannot be used. – feersum Nov 27 '15 at 0:29
• @PeterTaylor HM-based type systems should be OK. Maybe I can find a better description than "object-oriented". – feersum Nov 27 '15 at 0:31
• @feersum Then you should probably clarify what you mean by "error" (because the first sentence says that both errors and exceptions are allowed... just because there's no way to catch the error doesn't mean that language doesn't have errors... and most of these errors are actually thrown by the interpreter on purpose, and not raised by Java due to the interpreter having a bug). – Martin Ender Nov 27 '15 at 14:18
• @sysreq The input is a number when determines which type of exception you should throw. – feersum Dec 10 '15 at 23:24

# Count up by factor keys

TODO: Better title

### Information

Let the factor key of a number n be the result of the following process:

• Compute the prime factorization of n: n = p1e1 * p2e2 * p3e3 * ... where all pi are prime, distinct and in increasing order, and all ei are positive integers.

• Compute each piei; call these factors the fi. So n = f1 * f2 * f3 *....

• Convert each fi to a string.

• Concatenate all the strings together.

• Evaluate as a number.

Examples:

The factor key of 2376 is 82711, since 2376 = 23 * 33 * 11 = 8 * 27 * 11.

The factor key of 931 is 4919, since 931 = 72 * 19 = 49 * 19.

As a special case, the factor key of 1 is 0.

### Challenge

Output all positive integers, starting from 1, in increasing orders of their factor keys. If two numbers have the same factor key, output the smaller number first. The output must be separated by a delimiter, but this delimiter can be any non-numeric character. If you choose spaces, your output should start with the following:

1 2 3 4 5 7 8 9 11 13 16 17 19 6 23 10 25 14 27 29 31 37 41 12 43 28 47 36


This is , so the shortest solution in bytes wins.

• In the case (I'm not sure whether it occurs, but you haven't stated that it doesn't) that two numbers have the same factor key, can they be output in any order or do you want to define a tie-break? – Peter Taylor Nov 30 '15 at 12:35
• @PeterTaylor It does occur (see 36 and 49), but the question states "If two numbers have the same factor key, output the smaller number first." – Mego Nov 30 '15 at 20:48
• By reverse engineering, it seems that the e_i must be non-zero. That's worth stating in the spec, and backing up with an example (i.e. any odd number greater than 1). – Peter Taylor Nov 30 '15 at 22:18
• @PeterTaylor Done. – lirtosiast Nov 30 '15 at 22:25

## Juggling without collisions code-golf

Determine if a siteswap juggling pattern is valid, meaning that no two balls land on the same beat. Fewest bytes wins.

Input: A non-empty list of positive integers.

Output: A consistent Truthy value for valid or Falsey value for invalid.

Siteswap is a notation for juggling patterns. Dividing time into units called beats, a siteswap pattern says how high to throw the ball each beat, measured in beats that it's airborne until you catch it. For example, the pattern

4 2 3


says

• On the first beat t=1, throw the ball 4 beats high, so you catch it at t=5.
• On the second beat t=2, throw the ball 2 beats high, so you catch it at t=4.
• On the third beat t=3, throw the ball 3 beats high, so you catch it at t=6.

If we treat this pattern as a periodic sequence

... 4 2 3 4 2 3 4 2 3 4 2 3 4 2 3 4 2 3 ...


note that on every beat, you catch exactly one ball. But, in

 5 1 3 4 2 ...


the ball thrown for 5 beats at t=1 and the ball thrown for 3 beats at t=3 both come down at t=6, which we don't want (let's ignore that we have two hands). This happens whenever two numbers in the sequence have the same different as the number of beats between them, with the bigger on first. Note that this may happen across repetitions, like in 8 1 2, where the marked balls collide at *.

 v             v *
8 1 2 8 1 2 8 1 2 8 1 2 ...


TODO: Test cases.

Sandbox: Does the explanation make sense? Is it too long for the challenge?

• Related. :) – Martin Ender Dec 21 '15 at 11:26
• @MartinBüttner Those are some nice ASCII diagrams, mind if I steal them? – xnor Dec 21 '15 at 11:32
• Sure, if you don't mind them blowing up your spec. ;) – Martin Ender Dec 21 '15 at 11:32
• I think this challenge should still be happening. :) – Martin Ender Jun 19 '16 at 21:16
• @MartinEnder I've had this in the sandbox way too long, and don't have the energy to polish it up now. Do you want to take it? – xnor Feb 15 '18 at 5:35

# Elements with 100 ≤ Z < 1000

The International Union of Pure and Applied Chemistry (IUPAC) decided that it is necessary to have a systematic naming for the elements, even for those which had not been discovered. The rules for naming are:

1. The name is derived directly from the atomic number of the element using the following Latin numerical roots:

Number     Root
------     ----
0          nil
1          un
2          bi
3          tri
5          pent
6          hex
7          sept
8          oct
9          enn

2. The roots are put together in the order of the digits which make up the atomic number and terminated by 'ium' to spell out the name. The final 'n' of 'enn' is omitted when it occurs before 'nil', and the final 'i' of 'bi' and of 'tri' when it occurs before 'ium'.

3. The symbol of the element is composed of the initial letters of the numerical roots which make up the name.

For this challenge, let us consider the elements having a three digit atomic number.

Examples:

100 --> Unnilnilium
101 --> Unnilunium
111 --> Unununium
150 --> Unpentnilium
200 --> Binilnilium
500 --> Pentnilnilium
999 --> Ennennennium


# Input

• Input can be taken from one of the following

• stdin
• Command-line arguments
• Function arguments (One argument, as a string)
• Input will contain either a valid IUPAC name of an element, or a valid IUPAC symbol of an element.

# Output

• Output the corresponding
• IUPAC name if the input is a symbol of an element.
• symbol if the input is an IUPAC name of an element.

# Rules

• The input will contain a valid IUPAC name/symbol.
• The first character of the symbol/name in input will be in upper-case and should remain capitalized in the output.
• You are permitted to write a full program or a function.
• There should be no unnecessary characters except an optional trailing newline character.

# Test Cases

Unu            --> Unnilunium
Eee            --> Ennennennium
Enn            --> Ennilnilium
Bnb            --> Binilbium
Ubt            --> Unbitrium

Septenntrium   --> Set
Hexunbium      --> Hub
Octennilium    --> Oen
Bibibium       --> Bbb
Ununennium     --> Uue


# Scoring

This is , so the shortest submission (in bytes) wins.

Tags:

• For the title, maybe "Elements after Fermium", since that's element 100. – isaacg Dec 26 '15 at 11:14
• Omit the line "not relevant in this challenge" – isaacg Dec 26 '15 at 11:15
• – lirtosiast Dec 26 '15 at 23:13
• @ThomasKwa Oh. Thanks! Didn't know that a similar challenge existed! – Spikatrix Dec 27 '15 at 4:40
• @isaacg Thanks! I made some changes now. – Spikatrix Dec 27 '15 at 4:41

# Rational Number to Repeating Numeral Conversion

As I'm sure you know, the decimal expansion of every rational number is either terminating—consisting of a finite number of digits, or repeating—consisting of infinitely many digits, but ending with a finite pattern that repeats itself indefinitely: for example, the decimal expansion of the rational number 1/6 is 0.1666..., where the sixes repeat forever. One way to represent this decimal expansion finitely and unambiguously, is to write the repeating part, called the repetend, enclosed in parentheses: going back to the previous example, under this scheme the number 1/6 is written as 0.1(6). We call this representation a repeating decimal. Of course, none of this is specific to base 10. More generally, we call such a representation, in any base, a repeating numeral. Note that, for the sake of this challenge, we use the term "repeating numeral" (or simply, "numeral") to refer to any numeral written using this scheme, whether or not it actually has a repeating part.

## Challenge

Write a program or a function taking a pair of integers, p and q, and returning a repeating numeral representing the rational number p/q. You may assume that p ≥ 0, and q > 0, so that p/q is never negative. The resulting numeral should be in base 10, unless you go for the relevant bonus below.

There is more than one possible numeral representing a given rational number. For example, the rational number 1/1 can be represented, among (infinitely many) other options, as 1, 1.0, 1.(0), 0.(9), and so on... For this challenge, however, we'd like the output to be unique. The following set of rules, which your program must follow, takes care of that:

• When the integer part of the numeral is zero, there should be a single 0 before the radix point. For example, given the input 1/2, the output should be 0.5, and not .5.

• The repetend, if exists, must begin after the radix point; that is, it shouldn't apply to the integer part. For example, given the input 10/9, the output should be 1.(1), and not (1).

• The output should be finite and minimal, under the above rules. This is the most significant rule controlling the output, so it's worth highlighting some of its consequences:

• There should be no leading or trailing zeroes. For example, given the input 3/2, the output should be 1.5, and not 01.5 or 1.50.

• If the fractional part is zero, it should be omitted. For example, given the input 1/1, the output should be 1, and not 1.0.

• If the repetend is zero, it should be omitted. For example, given the input 1/2, the output should be 0.5, and not 0.5(0).

• When the input admits a terminating numeral, there shouldn't be an unnecessary repetend. For example, given the input 1/1, the output should be 1, and not 0.(9).

• The repetend, and the rest of the fractional part, should not be superfluous; there shouldn't be any repetition in the repetend itself, nor in the repetend and the rest of the fractional part. For example, given the input 1/99, the output should be 0.(01), and not 0.01(01), 0.(0101), or 0.0(10).

As usual, you may not use any built-in or library functions aimed specifically at this problem.

## Input and Output

You may take the input through the command line, through STDIN, as function arguments, or using an equivalent method. You may use any convenient format for the input, but make sure to specify it in your post. You may assume that p and q are no greater than 10,000,000, to the extent that it helps you to avoid overflow.

You may write the output to STDOUT, return it as the function's result or through an output parameter, as a string, or use an equivalent method.

## Score

This is code-golf. The shortest answer, in bytes, combined with the any of the bonuses, wins.

## Bonuses

If, in addition to the requirements listed above, your program satisfies the requirements for any of the following bonuses, multiply your score by the specified amount.

×0.8 Bonus Support other bases
Your program should take a third parameter, b, which is an integer between 2 and 36, inclusive, and return the corresponding numeral in base b, instead of base 10. The letters of the alphabet, either in lowercase or uppercase, should be used as digits above 9. For example, given the input 1/2 and b=3, the output should be 0.(1). Note that base b applies only to the output—if your program takes its input in string form, it should interpret it in base 10.

×0.6 Bonus O(1) space complexity
Your program's space complexity should be O(1), where the unit of space is the amount of space required to hold the input. In other words, the amount of memory required by your program should, on a large scale, be proportional to the amount of memory required by the input. You may not use the fact that the input range is bounded when reasoning about your program's space complexity, other than to the extent of establishing the amount of space required by the input. If you go for this bonus, it is strongly advised that you include at least a brief explanation of your program, so that others can verify that it meets the criterion.

If your program does not print the output directly, but rather returns a string, you may ignore the space occupied by the string as long as your program only appends to the string, and doesn't modify or read from it otherwise. (Note that something like s=s+"0" is fine, even though it technically involves reading from the string.)

## Test Cases

Each of the test cases below lists the input, p/q, on the first line, and the corresponding output on the second line. Some of the tests list a third parameter, b, which specifies a different base for the output. These tests are only applicable for the relevant bonus.

Short-Output Tests

Your program should solve each of the following tests in a matter of seconds.

0/1
0

123/123
1

12/3
4

1/2
0.5

1/3
0.(3)

7/6
1.1(6)

3/11
0.(27)

1234/9999
0.(1234)

7/12
0.58(3)

22/7
3.(142857)

123/456
0.269(736842105263157894)

7124771/4545450
1.56(745118)

1/68
0.01(4705882352941176)

678/2345
0.2(891257995735607675906183368869936034115138592750533049040511727078)

7683/238
32.2(815126050420168067226890756302521008403361344537)

123631/99999
1.(23632)

4576/2345
1.9(513859275053304904051172707889125799573560767590618336886993603411)

2239231/4950000
0.45236(98)

673/23430
0.0(2872385830132309005548442168160478019632949210413999146393512590695689)

8984/2318
3.(875754961173425366695427092320966350301984469370146678170836928386540120793787748058671268334771354616048317515099223468507333908541846419327006039689387402933563416738567730802415)

1/2 2
0.1

214/5467 13
0.(067ccb5145334200a07b6253394bc5ca12bc650017b87998acc2c516a7993810702ba1)

214 5467 11
0.0(481119a34a86245a0053a1022114643a64131367585a8095498942556925833303993816590371976aa2a6952238969a61490900a792044229187a18262724060a517a986785002750566607877532070733842a95a27a4476828a12971701a4740884573649355153)

330420/335923 36
0.(zero)

9322181/306936 28
12.ab(cd)

123/6573 29
0.(0flb8c3hblol)

123/6573 21
0.0(85669592cjj9ca173fij6g67hhg7g940b5efk6j2bb7276c1ch25ikf7dd070jf4h96hek41ji102h3ea6kb2ic5hc3089b19kjc8dfhaf9147c48kgg4k5ab463d1fd5k86eigacciai54ihj3k2e7aghdbcfeebfbi811b8ajdh521e4ed334d4bgk9f650e1i99dide8j83if205d77kdk15g3be360gj12jki3h6ae09i28f38hkcb9jb018c753a5bjgd8gc044g0fa9geh7j57f0ce624a882a2fg231h0i6da4379)


Long-Output Tests

The following tests produce significantly longer output. Instead of listing their full output, only the MD5 checksum of the output if given. The checksum is calculated without a trailing newline, and using lowercase letters for digits above 9 (in the relevant test cases). If you want to see the full output for these tests, or to calculate the checksum under different conditions, you can use the test snippet below. Your program should solve each of the following tests in no more than a few minutes.

343/677472
e1810c85500a97c36f2ba5dcc94a2aa6

234/62332

57628/7894211
81222aaa93f192c6d8334fcde6381a74

67332/1267232
3ea190c83a735a8bafc922cbb177e6f4

954/3684332
d4ef940c9986f9c2ac52ee822d94d6a7

783232/3462241
7724d3444b5540b8b4f07d13317e8da3

83/7657124
b295369267f48e642e202e0364898c82

764231/54646224
ecf7f3dd12f2d495b31a178fc03662d7

6743/1231234
9f447fd27447abb881795fc5e2f814e5

764/9343249

345/2234448 22
667d437e35c3825696f64f1b27e6a827

63451/2343324 3
5d86d77856cc0a15ac8a51b46c2f22d6

93457/546464 2

678541/453524 9
5bb54bddaab7bd933258d4d78a83cdba

4572/2341198 15
39af2bc67ab58438068f672ed3a8357d

1/4821466 6
57ba79a48c438b7fd21aa39ee3575c8d

145/5821417 21
009f087ae661cfd4690cc51ba71a827f

5472/2333645 33


## Test Snippet

If you want to further test your submission, you can use the following test snippet to find the repeating numeral representation of arbitrary rational numbers, or to find the rational number corresponding to arbitrary repeating numerals. Note that the result can get very big, very fast, which can be a little much for your browser; in this case, you might want to uncheck the "Result" checkbox, and check the "MD5" checkbox, to only get the MD5 checksum of the output. Note also that you can resize the input and output boxes.

<style>* { font-family: sans-serif; }table { border-collapse: collapse; }table > tbody > tr > td { padding: 0px; }#status { display: none; }#status[loading] { display: initial; font-style: italic; }#main { display: none; width:100%; border-right:.7em solid transparent; font-size: 95%; }#main[loaded] { display: table; }#main > tbody > tr + tr > td { padding-top: .25em; }#main > tbody > tr > td + td { padding-left: .5em; }#main > tbody > tr > td:first-of-type { width: 4.25em; }.label { white-space: pre; }#flags { padding-left: 2.4em; font-size: smaller; }#flags > table > tbody > tr > td + td { padding-left: 1em; }#error { display: none; font-size: smaller; color: #880000; }#error[error] { display: initial; }#job_cancel, #job_throbber { display: none; vertical-align: bottom; }#job_cancel[working], #job_throbber[working] { display: initial; }#job_throbber > img { height: 1em; }#job_throbber[working] { padding-left: .3em; }#job_cancel[working] { padding-left: 1em; }#job_cancel > input { height: 1.8em; font-size: small; }input[type="text"], input[type="number"], textarea { padding: 0.25em; height: 1.4em; font-family: monospace; }textarea { width: 100%; }#base { width: 2.8em; text-align: right; }.output { background-color: #e4e4e4; border: none; }#length { width: 8em; text-align: right; }#md5 { width: 20em; text-align: center; }#result_container[active="false"] { display: none; }#md5_container[active="false"] { display: none; }</style></head><body><div id="status" loading>Loading...</div><table id="main" onkeydown="handle_key_down(event)"><tr><td id="input_label" class="label">Fraction:</td><td><textarea id="input" oninput="update(true)" spellcheck="false">123/456</textarea></td></tr><tr><td class="label">Base:</td><td><table><tr><td><input id="base" type="number" value="10" spellcheck="false" oninput="update(true)"></td><td id="flags"><table><tr><td><label><table><tr><td><input id="flag_result" type="checkbox" checked onchange="handle_flag('result')"></td><td>Result</td></tr></table></label></td><td><label><table><tr><td><input id="flag_md5" type="checkbox" onchange="handle_flag('md5')"></td><td>MD5</td></tr></table></label></td><td><label><table><tr><td><input id="flag_uppercase" type="checkbox" onchange="handle_flag('uppercase')"></td><td>Uppercase</td></tr></table></label></td><td><label><table><tr><td><input id="flag_trailing_newline" type="checkbox" onchange="handle_flag('trailing_newline')"></td><td>Trailing Newline</td></tr></table></label></td><td><label><table><tr><td><input id="flag_progressive_output" type="checkbox" onchange="handle_flag('progressive_output')"></td><td>Progressive Output</td></tr></table></label></td></tr></table></td></tr></table></td></tr><tr><td></td><td><span id="error"></span></td></tr><tr><td class="label">Length:</td><td><table id="length_table"><tr><td><input id="length" class="output" type="text" spellcheck="false" readonly></td><td id="job_throbber"><img src="http://i.stack.imgur.com/sHKZY.gif"></td><td id="job_cancel"><input type="button" value="Cancel" onclick="cancel_job()"><td></tr></table></td></tr><tr id="md5_container"><td class="label">MD5:</td><td><input id="md5" class="output" type="text" spellcheck="false" readonly></td></tr><tr id="result_container"><td id="result_label" class="label">Numeral:</td><td class="full_width"><textarea id="result" class="output" spellcheck="false" readonly></textarea></td></tr></table><script async type="text/javascript" src="https://gist.githack.com/anonymous/49705525fd01ba66c1ad/raw/c35226ad89444e3af07d0e505f7163df8574b860/repnum.js"></script>

• Isn't space complexity technically going to be at least O(log (p+q))? – Martin Ender Dec 22 '15 at 15:20
• @MartinBüttner Technically, yes. I'm going to define the unit of space as the amount of space required to hold the input, thus implicitly dividing by O(log n). – Ell Dec 22 '15 at 15:25
• Ah, that's a very good idea. Much better than the usual attempts of "but ignore the memory used to store the input...". – Martin Ender Dec 22 '15 at 15:26
• I completely ignored your actual question: no, I wouldn't consider it a duplicate. ;) – Martin Ender Dec 22 '15 at 16:28
• I like the simplicity of the idea and would suggest keeping it simple by not requiring other bases or having a complexity limit. – xnor Dec 23 '15 at 6:37
• I agree with @xnor; I also think the bonuses are unnecessary. – lirtosiast Dec 30 '15 at 21:44
• @ThomasKwa Keep in mind that xnor's comment was written before the actual spec, when the bonuses were proposed as hard requirements (see the edit history.) I'm not likely to get rid of the O(1) space thing entirely—I think it makes for a much more interesting challenge (there's also a closely related challenge [edit history] that I don't want to get too close to.) TBH, I think the different bases bonus is mostly interesting in conjunction with the O(1) space bonus; I agree that in itself it doesn't add too much to the challenge. – Ell Dec 30 '15 at 22:17
• I'm not a fan of the bonuses. Also, I'm concerned there will be disagreements about the space complexity of built-ins. In some languages, implementation is not apecified, only result. It's also hard to tell n log n from linear because of constants. – xnor Dec 30 '15 at 23:25
• @xnor I'm going to go with common sense with regard to builtins: If the most obvious implementation of a builtin is O(1), then, unless someone knows otherwise, I'm fine with it. If the builtin does something nontrivial, that might not be O(1), then it's the poster's responsibility to show it's O(1), or not use it. I agree that reasoning about complexity is tricky, and complicates things: that's part of the challenge, and that's why I think it deserves a bonus. – Ell Dec 31 '15 at 13:12
• The space complexity required to hold the input is O(n), not O(1), since n is conventionally the input length, and O(1) is a constant that does not depend on the input. – xnor Dec 31 '15 at 17:14
• @xnor We can use whatever "frame of reference" we want when measuring complexity. If we define the unit of space to be the amount of space required by the input, then a complexity of O(1) simply means that space usage has to be proportional to the space occupied by the input, which might (in "absolute" terms) be fixed, or might depend on the actual value of p and q. Specifying it this way allows us to define the complexity requirement uniformly, without worrying about the details. – Ell Dec 31 '15 at 19:40
• @xnor, ThomasKwa, Anyway, having thought about it some more, I might end up dropping the bonuses, and doing the complexity thing out-of-band as a bounty. I'm going to leave this in the sandbox for a while longer, at least until next year... Happy new year! – Ell Dec 31 '15 at 19:42
• No, that's like saying we define 2 to be 3 and then 2+2=6. O(1) means O(1). 1 is not a variable. Readers will get confused if you redefine existing notation. – xnor Dec 31 '15 at 19:46
• How if my program allocates p+q+n space with n constant and unit space is byte? – Akangka Jan 1 '16 at 2:08
• Make all bonus mandatory. – Akangka Jan 1 '16 at 2:08

This challenge was closed, because it was way too broad, but, in the comments, SuperJedi224 suggested a slightly different challenge that I thought was a really good idea, so I'm stealing it and posting it here.

# Write a text editor in 2000 bytes or less

Write a text editor in 2000 bytes or less. It should support loading and saving files (or something else if your language doesn't support that, maybe?), modifying text, and displaying the contents of a file that's currently open in some format.

## Sandbox questions

• I'm not sure if 2000 bytes is the right number. SuperJedi224 originally suggested 10000, but that seems to me like too many.
• Should some features be required, or should it just be by votes? Will votes take care of possible submissions that aren't actually text editors?
• Should there be some kind of bonus for shorter submissions? Maybe an extra point for every 20 or so bytes you don't use? No, there probably shouldn't be.
• Should languages that don't support file operations be allowed?
• This is still very broad, and I think I would vote to close as off-topic, as this is essentially a UX contest, which isn't much different from an art contest. – FryAmTheEggman Jan 6 '16 at 15:55
• I would recommend either an upper limit on bytes, or plain code golf. Trying to be somewhere in between with a bonus seems awkward. – trichoplax Jan 7 '16 at 0:02
• @trichoplax I agree. Plain code golf won't work unless we can create a much more precise specification, and that doesn't seem as interesting as a popularity contest, so I guess it should just be an upper limit. – KSFT Jan 7 '16 at 0:42
• I think that the more precise you make the requirements, the more the voting can reflect how well the code fulfills these, rather than just being about arbitrary popularity. The more open ended the popularity criteria, the more likely the question will be closed. – trichoplax Jan 7 '16 at 0:56
• That's a good point. Do you have any suggestions for criteria? – KSFT Jan 7 '16 at 0:58
• Whatever limit you choose, there will be some languages which are excluded by this and other languages for which the limit is too high to provide any restriction. So instead of trying to suit all languages, it might be easiest to just think about the readers of the answers - how much code do you want them to have to read through for each answer? – trichoplax Jan 7 '16 at 0:59
• I guess you can add criteria gradually while it's here in the sandbox until it seems ready. What's the bare minimum it should be able to do? Do you just want keyboard input or mouse input too? Are there any shortcuts that should be included? – trichoplax Jan 7 '16 at 1:01
• Not sure if this will be relevant, but just as a rough guide, this list of small text editors includes Emacs in 2000 lines of C. I'm guessing 2000 bytes will be more than enough for a very basic text editor in most languages. – trichoplax Jan 7 '16 at 1:22
• I recommend doing this as a code-golf challenge, an upper limit just discourages the use languages where you are not sure whether you make it under that limit. Then also please make precise requirements, my suggestion: accept filename via stdin, file should contain printable ascii characters only, display blinking(or static) cursor, display should have a certain width (given number of characters) and enforces line breaks if the lines are longer, navigation via arrows, enable backspace (and if you want delete) – flawr Jan 7 '16 at 21:13

### Stitch a Picture

A few weeks ago, I asked Stitch a picture. Since then:

• The question has received several upvotes and no downvotes - hopefully an indication that the community thinks this is as interesting as I do
• Several interesting comments - in particular this one from @bazzargh

In short it looks like this question is a lot harder than I though it would be. It turns out from @bazzargh's comment that I have stumbled into an area of current research, and that perfect solutions are not so easily attainable as I had assumed.

With that in mind, I think its time I took this back to the sandbox to make this into the decent question that I think the subject deserves - and I think I need some community help with that.

I am posting several possibilities I am considering as comments. Vote for these comments as you feel appropriate, and/or add your own suggestions.

I'm going to Yosemite this weekend (woohoo!) so will probably pick this up again next week.

### Update

I have relinquished control of this question by putting it in Secret Santa's Sandbox.

• @Will pointed out that padding the original picture with black boundaries possibly makes the problem harder. Instead we can have the original picture be cropped down to the nearest 100x100 pixel multiple and not worry about black boundaries at all – Digital Trauma Oct 24 '14 at 17:49
• @MartinBüttner suggested turning this into a code-challenge and scoring based on the number of correctly placed tiles. Given that perfect solutions appear to be hard to come by in general, this is probably a better scoring criteria – Digital Trauma Oct 24 '14 at 17:50
• Briefly reviewing the paper cited in @bazzargh's comment I see that my problem is even more complex, in that the tiles are also randomly rotated. Since this is complicating the problem beyond current research, I think it would be good to remove the rotation part completely. – Digital Trauma Oct 24 '14 at 17:50
• Note that I didn't suggest scoring by correctly placed tiles but by correctly aligned edges. That's a huge difference, and in your version you could basically correctly assemble almost the entire picture, but still get 0 score if the chunk you've assembled correctly is off by a tile (since there's no indication where it goes unless you manage to get W tiles next to or H tiles on top of each other. – Martin Ender Oct 24 '14 at 23:44
• @MartinBüttner Yes, that is a better idea. – Digital Trauma Oct 24 '14 at 23:46
• @DigitalTrauma I suggest only allowing translation (no rotation) of the tiles, which simplifies it somewhat, but I suspect still difficult enough to be a challenge. – flawr Jan 8 '16 at 17:23
• – Christopher Jun 5 '17 at 23:20

# Addition of doubles... without doubles fastest-codegolf-cpu

For this challenge, you must write a GOLF assembly program that takes two IEEE 754 double-precision numbers and returns their sum.

### The Format

IEEE 754 doubles use 8 bytes to store a floating-point value. The memory is laid out like this:

For the purposes of this challenge, all numbers will be in big-endian, and you won't have to worry about overflows or underflows. Addition works roughly like this (I won't go into all of the details):

1. Align the two numbers to a common exponent.
2. Add the fractional parts, taking into account the sign.
3. If the addition overflows, then change the exponent to match.

### I/O

Your program will receive 16 bytes of input. The first 8 represent the (big-endian) first number, and the rest represent the second number. Your program will return 8 bytes representing the sum of the two.

### Rules

• This is , so the program that completes the task with the least cycles wins.
• The score of an answer is the mean number of cycles required for adding uniformly distributed numbers in [-10^64, 10^64].
• You must test with at least 2500 trials.
• Your code doesn't have to handle infinities/NaNs.
• Your GOLF binary (after assembling) must fit in 4096 bytes.

## Meta Questions

• Do I have too many/too few trials?
• Is there something that I overlooked?
• Is there a more fitting title?
• How much explanation should I provide? I don't want this question to turn into a Wikipedia article, but questions requiring external resources are generally frowned upon.

# HTML Obfuscation

Write a program or function that takes HTML as a string as input and outputs "obfuscated" HTML. Leave the text between the <s and >s unchanged, but escape all other text.

• You can assume that the < and > characters will only be used to open/close HTML tags.
• You can safely leave any text between these characters alone. However, any text not between those characters should take on the form <ampersand><pound><ascii-encoding><semi-colon>.
• You do NOT have to support self closing tags.

## Test cases:

Input:

<div>Hello World</div>


Output:

<div>&#72;&#101;&#108;&#108;&#111;&#32;&#87;&#111;&#114;&#108;&#100;</div>


Input:

ABC<input type="text">DEF<div>


Output:

&#65;&#66;&#67;<input type="text">&#68;&#69;&#70;<div>


Note: it's ok that the HTML is invalid

Input:

eee<div data-bracket=">">eee


Output:

&#101;&#101;&#101;<div data-bracket=">&#34;&#62;&#101;&#101;&#101;


Note: here your program would mess up the HTML. That's ok for the purposes of the challenge.

This is code golf, so the shortest program wins.

• I think you should be more specific about what exact subset of HTML needs to be handled correctly (and include test cases that cover all of it). – Martin Ender Feb 12 '16 at 15:24
• @MartinBüttner thanks for the feedback. Do you think the edits clarify the challenge? – sudo rm -rf slash Feb 12 '16 at 15:43
• Basically: take a character and return the hexadecimal HTML entity. – Ismael Miguel Feb 13 '16 at 20:08

# Play Flippy Bit

There's a simple game called "Flippy Bit and the Attack of the Hexadecimals from Base 16", originally made as an April Fool's prank in 2014.

The premise is simple: Enemies enter the field from the top. Each one has a hexadecimal number, and are destroyed when the player's binary number matches. For example, an enemy with number 2Ah = 00101010b would be destroyed as soon as the user's number is 00101010. Of course, enemies can only be destroyed after they enter the screen.

The player manipulates their binary number by flipping individual bits with the qwertyui keys, where q is the most significant bit. Each press flips the bit at that location. Therefore, the 2A enemy above will explode if, starting from 0, the player presses etu in any order. When an enemy explodes, the player's number is reset to 0.

## Challenge

Given a list of enemies, determine a sequence of keypresses that can destroy all enemies in the order given. This isn't as simple as it seems: if you're trying to destroy 3 but there's another enemy called 2 on the board, you'll need to press i before u. Your algorithm must terminate within a bounded time.

### Input

A list of strings of hex digits 0123456789ABCDEF, with each string 1 or 2 digits long, and where the 2-digit strings have no leading zero. You may also take the input in a reasonable format other than a list; for example, newline-separated. The input will always represent a solvable configuration.

### Output

A string representing the sequence of keypresses.

### Test cases

Note that there are multiple possible outputs for most inputs.

2A 01 02 04 08 10 20 40 A0 88
quetqwertyui


Verify your solutions with the following Python:

[add verification code]


This is , so the shortest solution in bytes wins.

• This may or may not be an excuse to award a bounty to the highest Flippy Bit score. – lirtosiast Jan 20 '16 at 18:13
• I had no idea that the qwertyui keys would flip the bits. This would actually make kind of a cool esoteric language, however. – Conor O'Brien Jan 21 '16 at 3:49
• The statement that "destroy[ing] all enemies in order ... isn't as simple as it seems" puzzled me for a while, because I couldn't see what was wrong with the simple approach of setting the bits from the least significant to the most significant. Having thought about it, I presume that what's missing is a statement that the order in question is the order in which the enemies are presented in the input. But then an input like 03 01 02 is impossible to solve, so you also need to address insoluble inputs. – Peter Taylor Jan 25 '16 at 12:15
• @PeterTaylor Actually, it's not! quiq. The point of this is some sort of search. – lirtosiast Jan 25 '16 at 14:58
• ff 01 02 04 08 10 20 40 80 then. (Although I see you've addressed the issue, so this comment is only for completeness). – Peter Taylor Jan 25 '16 at 16:06

### Challenge

Write a program, taking an integer n as input from -999,999 to 999,999 inclusive, that returns a string representing a valid expression evaluating to that number in your language. The catch: this expression must NOT evaluate to any other number when evaluated if any single character is removed from it.

### Restrictions

Your program must be <= 1024 bytes in length.

The outputted expressions must be in the same language as the generating program.

### Scoring

Your score is the number of bytes in the largest string generated by your program for the entire range of valid numbers. Tiebreakers go to shortest code.

### Examples of Valid Outputs (in JavaScript)

• Math.PI|0||3 for n=3, because:

• ath.PI|0||3, Mth.PI|0||3, Mah.PI|0||3, and Mat.PI|0||3 cause ReferenceErrors,

• MathPI|0||3 causes a ReferenceError,

• Math.P|0||3, Math.I|0||3, and Math.PI0||3 result in 3,

• Math.PI|||3 causes a SyntaxError,

• Math.PI|0|3 results in 3, and

• Math.PI|0|| causes a SyntaxError.

• 9*(8) is valid for n=72

• (n=9e5)>96?n:9e5 for n=900,000

• 1 for n=1

### Examples of Invalid Outputs (in JavaScript)

• 1e2 for n=100 because 12 is a valid expression that evaluates to a number other than 100.
• "cause an error or continue to evaluate". Those are two very different things, I'd say to just pick that the expression should continue to evaluate – Downgoat Feb 27 '16 at 0:38
• @Downgoat My intent was that the expression could do anything except evaluate to a different number in the range. I thought it would make the challenge more approachable in different languages to allow errors. – jrich Feb 27 '16 at 1:02
• I get 0 for Math.P|0. – feersum Feb 27 '16 at 5:37
• @feersum Whoops, typed this too quickly. Fixed – jrich Feb 27 '16 at 16:09
• For a lot of languages this probably boils down to one or two special cases and the rest of them are max(expr,expr) for numbers greater than 9, min(expr,expr) for numbers less than 0. So it's borderline dupe of this. – Peter Taylor Mar 1 '16 at 15:58
• @PeterTaylor True. Didn't think of that approach. In this case there would be incentive to golf the code to only employ the strategies necessary to achieve the minimal score, so it's not a complete dupe, but I'm not sure that the challenge in its current form is as interesting as I initially thought... – jrich Mar 1 '16 at 23:45

# Challenge

Output the points (in order) for the following fractal of size n:

(The above example is for n = 4)

We begin with the Binary Sierpinski Triangle, which can be generated recursively, by a number of cellular automata (including Rule 90), and by Pascal's Triangle.

As a refresher, Pascal's triangle is:

   |  0 |  1 |  2 |  3 |  4 |  5 |  6 |  7|
---+----+----+----+----+----+----+----+---+
0 |  1 |  0 |  0 |  0 |  0 |  0 |  0 |  0|
1 |  1 |  1 |  0 |  0 |  0 |  0 |  0 |  0|
2 |  1 |  2 |  1 |  0 |  0 |  0 |  0 |  0|
3 |  1 |  3 |  3 |  1 |  0 |  0 |  0 |  0|
4 |  1 |  4 |  6 |  4 |  1 |  0 |  0 |  0|
5 |  1 |  5 | 10 | 10 |  5 |  1 |  0 |  0|
6 |  1 |  6 | 15 | 20 | 15 |  6 |  1 |  0|
7 |  1 |  7 | 21 | 35 | 35 | 21 |  7 |  1|


Here rows are i, columns are j, and the values are i choose j.

Using Pascal's triangle, we replace each value with its remainder mod 2 to get:

   |  0 |  1 |  2 |  3 |  4 |  5 |  6 |  7|
---+----+----+----+----+----+----+----+---+
0 |  1 |  0 |  0 |  0 |  0 |  0 |  0 |  0|     O
1 |  1 |  1 |  0 |  0 |  0 |  0 |  0 |  0|     OO
2 |  1 |  0 |  1 |  0 |  0 |  0 |  0 |  0|     O O
3 |  1 |  1 |  1 |  1 |  0 |  0 |  0 |  0|     OOOO
4 |  1 |  0 |  0 |  0 |  1 |  0 |  0 |  0|     O   O
5 |  1 |  1 |  0 |  0 |  1 |  1 |  0 |  0|     OO  OO
6 |  1 |  0 |  1 |  0 |  1 |  0 |  1 |  0|     O O O O
7 |  1 |  1 |  1 |  1 |  1 |  1 |  1 |  1|     OOOOOOOO


With the result "plotted" to the right, the n = 1 iteration of the Sierpinski Triangle.

Now, to keep the "centers" well defined, we take the center of each triangle to be its NE corner like so:

'              ^
''             |
'*'            N
''''       <--E W-->
'  *'          S
''  ''         |
'*' '*'        v
''''''''


Here is pseudocode to generate the fractal:

fractal(n):
position = nth center
while SE possible:
go SE
mark position
while not all full:
if current level not full:
rotate counterclockwise on current level
mark position
else:
go NW until reach non-full level
rotate counterclockwise on current level
while SE possible:
mark position
go SE
return marked positions


Here's what that looks like for n = 2:

 '
' '
' 12'
' ' ' '
'      9'
' '     ' '
' 10'   ' 11'
' ' ' ' ' ' ' '
'               '
' '             ' '
'  4'           '  8'
' ' ' '         ' ' ' '
'      1'       '      5'
' '     ' '     ' '     ' '
'  2'   '  3'   '  6'   '  7'
' ' ' ' ' ' ' ' ' ' ' ' ' ' ' '


Giving the following points marked (in order):

[[3, 12], [1, 14], [5, 14], [1, 10], [11, 12], [9, 14], [13, 14], [9, 10], [3, 4], [1, 6], [5,6], [1,2]]


# Rules

• Entry may be either a function or full program.
• Input is a non-negative integer.
• Output is the (properly ordered) list of points for the fractal above.
• You may use something other than a list to output the results if it allows the easy reading of the points, in order. For example, newline separated is fine.
• Output must contain exactly (3^n - 3)/2 points.
• This is code golf so shortest wins!

# Questions

• Is everything well defined?
• Should I reformat the ASCII triangles? Or even replace everything with images?

Test cases if anyone is interested or I post this.

This is a challenge that I enjoyed solving so I hope I can smooth out the corners relatively quickly.

• I don't quite understand what the diagram with the ' and * represents. Would it be possible to explain the construction without having to read pseudocode? – Martin Ender Mar 13 '16 at 13:46
• @MartinBüttner I believe so. I considered making a better animation but hoped it would be clear without one. The options I see are: animation, representative list of steps, or general improvement/refactoring of what I have so far. What do you think? – Michael Klein Mar 13 '16 at 20:58
• – Michael Klein Mar 30 '16 at 9:55
• I've been playing with this one recently, have a gif: gifyu.com/image/pumE – Michael Klein Dec 10 '17 at 19:31

Tags:

# Introduction

Magic numbers are a common problem in software development, since they tend to make software hard to maintain. Best practise is to use named constants instead of raw numbers in the source code. In this challenge you have to clean up some code that makes extensive use of magic numbers.

# The Challenge

Given a piece of code, replace every number that is not 0, 1 or 2 with named constants. If you encounter a number that you already created a constant for, use the existing constant instead of adding another one. The replacement is done by replacing the number with an uppercase letter [A-Z] and prepending an assignment in the form of A=5 to the initial input.

# Example

x=input()
if x == 2:
print 42
elif x == 1337:
print 666
print 42


becomes

C=666
B=1337
A=42
x=input()
if x == 2:
print A
elif x == B:
print C
print A


# Notes

• The order of the assinments that get prepended to the input does not matter.
• The order of replacements does not matter as well. You don't have to replace the first magic number with an A for example.
• You may take the input as a list of strings instead of a multiline string.
• You will never have to use more than 26 constants.
• A number that has to be replaced matches the regex [0-9]{2,}|[3-9].

# Rules

Happy Coding!

• Cases to consider - 5.3, 0x30, a12 (simple replacement would corrupt the code), A (already existing in the code), 04 (same as 4 or not)? – ugoren Apr 25 '16 at 7:15
• Also, your regex implies that 1337 does't need to be replaced. – ugoren Apr 25 '16 at 7:16
• @ugoren Yea that regex is definetly wrong. Gonna correct this and also add clarifications for the edge cases. – Denker Apr 25 '16 at 10:13

# Word Squares

Create a function or program that given a list of words and size n as arguments or standard input, outputs a word square with dimensions n by n. The output can be either formatted in a similar shape to a word square, or as a list of the words used in order by rows from the top and then by columns from the right.

## Explanation

Word squares are a grid of letters that form words when read horizontally and vertically. The size of the square refers to the number of letters in each word.

## Example

This is a size 5 word square. The words formed by each row are heart, ember, abuse, resin, and trend. In this example, the same words are formed by each column, but this is not a necessary condition.

H E A R T
E M B E R
A B U S E
R E S I N
T R E N D


Another example is this word square with size 4. The words formed by each row in this case are different from the words formed by each column.

L A C K
I R O N
M E R E
B A K E


## Rules

• This is so the shortest solution wins.
• Builtins that solve this and the standard loopholes are not allowed.
• The words must be from this dictionary. You can take the filename of the dictionary or the contents as input or as an argument. (Recommend a dictionary here that consists of only English alphabet letters, in all lowercase or all uppercase.)
• If no such word square can be found using the given dictionary, no output or a false (or falsey) value can be returned.

# Test Case

n = 4, dictionary = ...
L A C K
I R O N
M E R E
B A K E
-- or --
LACK
IRON
MERE
BAKE
-- or --
["lack", "iron", "mere", "bake", "limb", "area", "cork", "knee"]
-- or --
[["lack", "iron", "mere", "bake"], ["limb", "area", "cork", "knee"]]
-- or --
Similar to the above but with swapped cases.
The case is not important as long as it remains consistent.

• Nice one. Plus 1. – Arjun May 8 '16 at 22:07
• I don't think it's necessary to mention the standard loopholes, but if you do I'd recommend making that a link. – trichoplax May 9 '16 at 13:19
• @trichoplax Yeah but it's more for informing new visitors of standard rules for code-golf. I do agree an actual link to the meta post would be more beneficial since a new visitor may not find it. – miles May 9 '16 at 16:01
• May a word square contain a word multiple times as horizontally or vertically? Will all words in the dictionary have the right length? – xnor May 16 '16 at 10:05
• @xnor I think the dictionary input should only contain words with length n matching the input size. As for repeating words, I don't see any reason to add it as an additional constraint as it's not a rule for standard word squares. – miles May 16 '16 at 20:41

## Draw the Cool S code-golfascii-art

Given a number n≥3, print or output the Cool S made with n vertical bars. The outputs for 3,4,5,6 are:

  / \
/   \
|  |  |
|  |  |
\  \/
/\  \
|  |  |
|  |  |
\   /
\ /

/\
/  \
/    \
/  /\  \
|  |  |  |
|  |  |  |
|  |  |  |
\  \  \/
/\  \  \
|  |  |  |
|  |  |  |
|  |  |  |
\  \/  /
\    /
\  /
\/

/ \
/   \
/     \
/  / \  \
/  /   \  \
|  |  |  |  |
|  |  |  |  |
|  |  |  |  |
|  |  |  |  |
\  \  \  \/
/\  \  \  \
|  |  |  |  |
|  |  |  |  |
|  |  |  |  |
|  |  |  |  |
\  \   /  /
\  \ /  /
\     /
\   /
\ /

/\
/  \
/    \
/  /\  \
/  /  \  \
/  /    \  \
/  /  /\  \  \
|  |  |  |  |  |
|  |  |  |  |  |
|  |  |  |  |  |
|  |  |  |  |  |
|  |  |  |  |  |
\  \  \  \  \/
/\  \  \  \  \
|  |  |  |  |  |
|  |  |  |  |  |
|  |  |  |  |  |
|  |  |  |  |  |
|  |  |  |  |  |
\  \  \/  /  /
\  \    /  /
\  \  /  /
\  \/  /
\    /
\  /
\/


Any invisible whitespace is optional (trailing spaces and trailing/leading newlines).

## How to draw the Cool S

Draw the Cool S just like you've done since childhood, except the number of vertical bars can be more than 3.

• Draw two rows of vertical bars.
• Connect them with slanted lines, tucking away the two remaining bars.
• Pair of the bars with slanted lines at the top and bottom.

Here's how we do this for ASCII art. Let's look at n=5.

A row has n vertical bars with two spaces between them

|  |  |  |  |


Draw two groups of n-1 rows, leaving two empty lines between them

|  |  |  |  |
|  |  |  |  |
|  |  |  |  |
|  |  |  |  |

|  |  |  |  |
|  |  |  |  |
|  |  |  |  |
|  |  |  |  |


Connect each top bar to the bottom bar one position right with a slanted line of two \, then close off the two unused bars with a /.

|  |  |  |  |
|  |  |  |  |
|  |  |  |  |
|  |  |  |  |
\  \  \  \/
/\  \  \  \
|  |  |  |  |
|  |  |  |  |
|  |  |  |  |
|  |  |  |  |


Complete the S on the top and bottom by drawing lines sloping inwards that meet at the center. When n is odd, a gap of one space is left and the center bar isn't used.

     / \
/   \
/     \
/  / \  \
/  /   \  \
|  |  |  |  |
|  |  |  |  |
|  |  |  |  |
|  |  |  |  |
\  \  \  \/
/\  \  \  \
|  |  |  |  |
|  |  |  |  |
|  |  |  |  |
|  |  |  |  |
\  \   /  /
\  \ /  /
\     /
\   /
\ /


Spun off from this Sandboxed challenge by Beta Decay.

• Golfing my childhood – Alex A. May 13 '16 at 22:40
• I've never drawn a "Cool S", whether in my childhood or since. Does that mean that I'm disqualified from answering this question? – Peter Taylor May 14 '16 at 19:19

## Strip Iterated Prisoner's Dilemma

Inspired by https://xkcd.com/696/, of course.

The Prisoner's Dilemma is a classic game (more in the game theory sense than the family fun sense) where two agents - two accomplices to a crime, in the original formulation - must choose whether to sell out the other.

If each player chooses not to betray the other, both win. If one player chooses treachery and the other does not, it wins, but if both betray each other, neither wins. The "iterated" variation is where the game is played multiple times with the same players, and both players know all the decisions each player has made in the past.

Of course, that's all a bit dull, and has been done before besides. We're going to tart it up a bit.

(Unfortunately due to the nature of the test all solutions must be in the same language, and one that supports executing strings. I've chosen Python 3, since it's my favorite and I have to write the runner.)

The Game:

Submit a program - specifically a Python 3 function - that plays Iterated Prisoner's Dilemma against another such program. If both functions betray each other, each one gets one character deleted from the end of its source code. If one function betrays the other, the betrayed function gets two characters deleted from the end of it, and the traitor gets a # appended to it, immunizing it against one future loss. If neither turns traitor, both go unharmed. Functions will be restored to their original text between each contest with a new opponent.

Submissions are scored based on failure rate and length; specifically, score = round( (number_of_trials_failed / total_number_of_trials) * ( length / 2 ) ). The submission with the lowest score wins.

The length of a submission is the number of non-whitespace characters in the submission. Comments are not counted, but are also removed before the contest begins, so a commented out "guard" at the end will not protect your code from deletions. Length also does not count the function specifier (the def submission(y, o, t): which should not be included in your submission text anyway.)

Your function will be called with three parameters, y (your history), o(opponent's history), and t (text of the opponent itself). t is the very same text that the game runner will execute as your function's opponent, which you may analyze or otherwise use to run simulations. It will then return True if it wishes to betray its opponent or False if it does not.

Every possible 2-combination of submissions will be contested against each other - including each submission against itself. Each contest consists of 2,500 trials.

Other Rules:

• All submissions must be in Python 3. (Specifically, they should run under Python 3.4.4)

• The submission text must be the only code that is run to produce the result; you may not import any libraries, ask for user input, read from /dev/urandom/or equivalent, and of course you can't pull results from some webserver (which is already a violation of the standard loophole rules.) You MAY execute the given opponent text, and of course you're allowed to call all the builtins.

• The submission must terminate and return an answer within 1 second (this will be run on a 12GB i7 gaming computer, so this should be plenty of time).

• A submission that emits an uncaught exception or returns an invalid value loses that round.

• Submissions that do not return in one second or less are immediately disqualified.

• Submissions will be closed on [date posted + 8 days] and programmatically judged, results will be appended to the challenge. There will be a "trial run" in the evening of [date posted + 4 days].

API:

The "submission text" is valid, 4-space-indented Python 3 source code that will have the function specifier line def submission(y, o, t):\n appended to the beginning and a single 4-space-indent added to the beginning of each line. So:

if len(y) == 0: return False
else: return (not y[-1])


...is run as...

def submission(y, o, t):
if len(y) == 0: return False
else: return (not y[-1])


The function should return True to betray its opponent or False to trust its opponent. Returning any other value (such as if your return statement has been deleted and you return None) is an error and will result in an automatic loss of that round.

o is a list of True-es or False-es, representing the opponent's previous moves when playing against you; o[n] equals the decision your opponent made in round n, and of course this list will be empty in the first trial. y is similar except it's your previous moves. t is the text of the opponent submission, formatted as specified above.

This is the program that will perform the contest:

[said code here]

Example Submissions: These will also be included in the actual contest.

Chronic Villain Syndrome:

#always choose 'betray'
return True


The Patsy:

#always choose 'trust'
return False


Do Onto Others...:

#always choose what opponent chose last round
if o:
return o[-1]
return True


Professor X:

me="""
#Read opponent's mind and choose optimally.
exec("def e(y, o, t):\n"+t)
return False e(o, y, me)
#Opponent's choice this turn if we betray...
tb = e(o,y+[True], me)
#if we trust...
tt = e(o,y+[False], me)

#Opponents choice NEXT turn if we betray then betray...
tbb = e(o+[tb], y+[True], me)
#... betray then trust...
tbt = e(o+[tb], y+[False], me)
#... trust then betray...
ttb = e(o+[tt], y+[True], me)
#... trust then trust...
ttt = e(o+[tt], y+[False], me)

#Maps (your_choice,opponent_choice) to desirability
#Betrayed = -2, mutual betrayal = -1,
#mutual cooperation = 1, betray opp. = 2
v = {(False, True): -2, (True, True): -1,
(False, False): 1, (True, False): 2}
#Best outcome next turn of trusting now
ftv = max(v[ttt], v[ttb])
#... of betraying now
fbv = max(v[tbt], v[tbb])
#value of betraying now
bv = v[tb] + fbv
#... of trusting now
tb = b[tt] + ftv

#Tuple comparison is done l to r, so
#this returns True if tv >= bv,
#False if bv > tv.
return max( (bv, True), (tv, False) )
"""
exec(me)


(p.s. this one gets disqualified every time - why is left as an exercise to the reader)

• So this is a KOTH? – Rɪᴋᴇʀ May 13 '16 at 22:07
• And why is the professor X bot disqualified? – Rɪᴋᴇʀ May 13 '16 at 22:10
• Finally, should there be a rule against running the opponents code to pick the best output for yourself? – Rɪᴋᴇʀ May 13 '16 at 22:12
• It's a KOTH, yes, with a little bit of golf mixed in. "Professor X" gets disqualified because it will eventually play against itself, which means unbounded recursion, which means never halting and thus getting kicked out by the "must return in one second" rule. So any bot that tries to run its opponent is going to have to at least have some way of stopping its opponent from running it without corrupting the answer. All in all that kind of "mind reading" is cool enough and easy enough to defeat that I think it should be left in. – Schilcote May 14 '16 at 1:26
• Modified quines aren't hard, and neither are narcissist programs. I think you should add a rule against that. I'll see if I can create a bot like that. – Rɪᴋᴇʀ May 14 '16 at 2:03
• @EᴀsᴛᴇʀʟʏIʀᴋ Right, but it should be pretty easy to pull stuff like having side effects (store something in a global for example) or unwinding the stack to see who your caller is. – Schilcote May 14 '16 at 2:41
• Can't you do something like this? Wouldn't that almost always win? – Rɪᴋᴇʀ May 14 '16 at 3:23
• @No, for a few reasons; one, there'd probably be other "mind reader" variations (that's the whole reason Professor X is in there) which means you'd have to check for ALL of them, which would mean a massive character count. For another, I can think of at least two ways (stack unwinding & using globals to check for two calls that are given same sized y) to check if your opponent has done that and alter your decision accordingly. I don't want to make "mind readers" impossible, I just want to put interesting challenges in their way, and I think this ruleset does that. – Schilcote May 14 '16 at 17:33
• @Shilcote okay, forgot that more mind readers could exist. – Rɪᴋᴇʀ May 14 '16 at 18:14
• What does number_of_trials_failed mean? The number where you were disqualified? And why 4-space-indented? That guarantees that when you lose a space from the indentation you will lose the next three rounds due to parse errors. – Peter Taylor May 14 '16 at 19:14
• @PeterTaylor No, when you're disqualified you're totally out, you don't even get a score. number_of_trials_failed is the number of times you got betrayed and lost. Good thought on the indentation; maybe it should delete one non-whitespace character? Or better yet, one statement? – Schilcote May 15 '16 at 0:29
• This is likely going to be a one-up challenge. Given a set of programs, you'd always be able to write another program that beats all previously existing programs. – Nathan Merrill May 16 '16 at 18:37
• I'm sure there'll be a lot of that, but I don't think it'll be exclusively that... – Schilcote May 17 '16 at 1:10

# Do my algebra homework for me!

Inspired by a true story

Ugh, my maths teacher gave me so much homework on quadratics.

Two whole worksheets, filled with equations like Expand (x+1)(x-6) and Factorize x^2-2x+1. I already hate it.

Suddenly, a lightbulb appeared out of nowhere, and landed about five centimetres above my head.

A stroke of thought went through me: Why not write a piece of code down instead of all those stupid answers? It would save me tons of time!

So here's the task: Make an algebra solver for me! To save my hand from breaking, the code needs to be as short as possible.

## Instructions:

• Given an input in the form expand/factorize equation, return the algebraic equation:
• Expanded, if the keyword (the first word) is expand (e.g. expand (x+1)(x+6) would return x^2+7x+6), or
• Factorized, if the keyword is factorize (e.g. factorize x^2+9x+14 would return (x+2)(x+7).
• The equation will be quadratic, in the form ax^2+bx+c or (x+d)(x+e) (a to e are all placeholders, while x is the variable - the variable can be any letter from a to z - so an equation like expand (a+3)(b+6) would still hold).
• With an expansion equation that contains two or more unknowns (e.g. expand (a+3)(b+6)), place the letters in the summands with two or more letters in alphabetical order, and place the remaining summands in alphabetical order (so the previous example would equal ab+6a+3b+18).

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

## Meta

• Any dupes?
• I probably need a better title. Any suggestions?
• Duplicate of expansion. – Leaky Nun May 29 '16 at 12:01
• – Leaky Nun May 29 '16 at 12:02
• Drat, foiled by previous questions! So, there's nothing I can really do, right? – clismique May 30 '16 at 11:31

# Capture the Flag

Do you ever wonder why we're here?

## Objective

Capture the enemy team's flag and return it to your base. The first team to 3 captures wins the round. The player with the most wins across all (number TBD) rounds wins the game.

## Teams

There will be two teams each round. Teams will be randomly, evenly assigned to all submissions at the start of each round.

## Playing

The game will be turn-based. At the start of each turn, each player will be given the current map. All players will move simultaneously. Players shall submit their move as a one or two character ASCII string, composed from the following options:

• First character: wait, move, stab (if not holding flag), drop (if holding flag), or pick-up (if standing on a flag)
• Second character: north, east, south, or west, or nothing if waiting, dropping, or picking-up

Moving a direction will result in the player moving one square in that direction if possible, otherwise standing still (for example, if the player is attempting to move into a wall, or into an occupied square). Stabbing in a direction will result in killing the player standing in the adjacent square in that direction, unless the player is a teammate (no teamkilling). Stabs are processed before moves each turn. Dropping the flag results in it being placed on the ground beneath the player. Drop, pick-up, and wait commands ignore the second character. Invalid commands are interpreted as waiting. Case is ignored, so W and w are the same command. Either team may pick up either flag.

If two or more players attempt to move onto the same square, one of the players will be randomly selected to successfully move, and the rest will not move.

If a player is killed, they will drop the flag they are holding (if they are holding one), and will respawn in 3 turns in an unoccupied square in their home base. If there is no unoccupied square in their home base, they will respawn in the nearest square to the base. Respawns happen after stabs, but before moves.

Inside each base, there will be a 5x5 square room, with doorways in the middle of each wall, and the team's flag in the center of the room. Players who spend 5 consecutive turns inside their team's flag room (this includes the 4 doorways), while no enemies are present in the flag room and they are not holding a flag, will be killed at the conclusion of the 5th turn, to discourage camping. Successfully placing the enemy's flag on top of your flag's stand (in the center of the room), either by dropping it or being killed on top of the flag stand, will result in a point being scored for your team and the enemy's flag immediately returning to their flag stand.

## The Map

(work in progress)

The world map will be a single level (no upstairs or downstairs), represented as such:

# : wall, cannot be moved into
. : an empty space
F : the enemy team's flag
! : a flag stand (with no flag on it)
@ : you
\$ : you, carrying the enemy flag
% : you, carrying your flag
p : one of your teammates
P : an enemy player
c : a teammate, carrying or standing on top of the enemy flag
C : an enemy player, carrying or standing on top of your flag
s : a teammate, carrying or standing on top of your flag
S : an enemy player, carrying or standing on top of the enemy flag


Here is an example map (the actual maps used in the tournament will be posted later):

####################################################
#..................................................#
#..###.###########.######........###......##########
#..#....................#..........................#
#..#.....###.###...................#######.........#
#..#.....#.....#........#..........................#
#..#.....#.....#........#...................#...#..#
#..#........f...........#..........................#
#........#.....#........#....#########.............#
#..#.....#.....#........#..........................#
#..#.....###.###........#......##################..#
#..#...............................................#
#..#.#################.............................#
#.........................###########.........###..#
#..................................................#
####################################...............#
####################################...............#
#..................................................#
#..................................................#
#...............####################################
#...............####################################
#..................................................#
#..###.........###########.........................#
#.............................#################.#..#
#...............................................#..#
#..##################......#........###.###.....#..#
#..........................#........#.....#.....#..#
#.............#########....#........#.....#........#
#..........................#...........F........#..#
#..#...#...................#........#.....#.....#..#
#..........................#........#.....#.....#..#
#.........#######...................###.###.....#..#
#..........................#....................#..#
##########......###........######.###########.###..#
#..................................................#
####################################################


## Controller

The controller and an example map and player are located on the challenge's GitHub project. Once I finish the controller, I'll copy the program here.

## Restrictions

• Bots must be fully deterministic. RNGs may not be used.
• Bots may be written in any language, so long as they support reading ASCII input from STDIN and writing ASCII output to STDOUT. Anything that is written to STDERR will be ignored.
• Bots' processes will be started at the beginning of each turn, and must output their command and terminate within the given 5 seconds.
• Each bot will be able to store up to 1 MiB (1024*1024 bytes) of data on disk per game, for saving any stateful data they desire. The name of the bot's data file will be passed as the first command line argument to the bot process. Should a bot write more than 1 MiB of data during a single game, data from the beginning of the file will be removed to append additional data to the end of the file. At the end of each game, the data files will be wiped.
• Any attempt to tinker with the controller, runtime or other submissions will be disqualified. All submissions should only work with the inputs and storage they are given.
• Bots should not be written to beat or support specific other bots.

## Sandbox notes

Anything missing or unclear (other than the parts specifically marked as TBD)?

• 1. The rules make most sense if turns are sequential rather than simultaneous, but it's nowhere stated which is the case. Simultaneous moves are fairer, but make the rules more complicated. Sequential moves make the assessment of what to do more complicated, because unless you track a lot of state from last time you don't know who's already moved. 2. Respawning in an unoccupied square requires there to be an occupied square. What if there isn't? 3. What stops camping just outside the door, in such a way that no-one can pass? – Peter Taylor Mar 4 '16 at 10:40
• @PeterTaylor I've addressed all 3 of these points in the latest edit. – Mego Mar 4 '16 at 20:41
• No RNG? :( That's not as much fun. – Conor O'Brien Mar 6 '16 at 0:36
• @CᴏɴᴏʀO'Bʀɪᴇɴ Trust me, you'll be glad for that. Nobody wants a Caboose on their team. – Mego Mar 9 '16 at 7:00
• So they can't stab themselves or take their own flag? – Rɪᴋᴇʀ Mar 9 '16 at 16:41
• @RikerW Correct on the stabbing bit. I can't imagine why a player would want to stab themselves, though. Either team can pick up either flag; I need to fix that. – Mego Mar 9 '16 at 21:56
• @Mego Every heard of EmoWolf? >.< – Rɪᴋᴇʀ Mar 10 '16 at 0:02
• @RikerW That is exactly why suicides are not allowed. A rule is not needed, because that's a standard loophole. – Mego Mar 10 '16 at 0:03
• You should make a rule that submissions can't lose on purpose by sacrificing their own flag – Rɪᴋᴇʀ Mar 10 '16 at 0:03
• You never define what a "base" is. Is it half of the board? – MegaTom Jun 13 '16 at 15:25

## Determine the winner of Beggar my Neighbour

The card game Beggar my Neighbour is boring in that the final outcome is entirely determined by the initial arrangement of the deck, so long as certain rules are followed for the order in which cards are picked up from the playing field and moved to decks.

## The Game

1. Both players are dealt 26 cards.

2. Players play their top card alternately, starting with the player who won the previous stack or Player 1 at the beginning of the game.

3. Play is interrupted when either player plays a picture card. In that case their opponent must play a number of cards equal to the value of the picture card above 10, i.e. Jack = 1, Queen = 2, King = 3, Ace = 4. The player then wins all the played cards which are returned to the bottom of their hand, unless the opponent themselves plays a picture card, in which case this rule interrupts their play.

4. If at any point one of the players needs to draw a card from their deck, but their deck is empty, they immediately lose the game.

## Example play

Player 1 starts with 7; Player 2 plays 3; subsequent plays are 9; 9; T; A; 6, J; A; 2, 3, 7, 6: Player 2 adds the cards 7399TA6JA2376 to his deck.
Player 2 starts: J; K; 9, 4, A; 5, 2, J; 2: Player 1 adds the cards JK94A52J2 to his deck.
Player 1 starts: 6; T; T; 5; 9; A; A; 3, K; K; 7, Q; 5, T: Player 1 adds the cards 6TT59AA3KK7Q5T to his deck.

## The Challenge

Given two lists of cards in the players' decks, in any convenient format, output a truthy value if Player 1 wins, and a falsey value if Player 2 wins.

For convenience, a 10 card will be represented with a T, and face cards will be abbreviated (Ace -> A, King -> K, Queen -> Q, Jack -> J), so that all cards are one character long. Alternatively, ranks may be represented with decimal integers 2-14 (Jack -> 11, Queen -> 12, King -> 13, Ace -> 14) or hex digits 2-E (10 -> A, Jack -> B, Queen -> C, King -> D, Ace -> E). Since suits don't matter, suit information will not be given.

You may assume that all games will terminate at some point (though it may take a very long time), and one player will always run out of cards before the other.

There are variations for more than two players but they will not be considered here.

• – Bálint Jun 7 '16 at 10:20

# Introduction

A Pythagorean triple is a tuple of three positive integers a, b and c so that a² + b² = c². One example of that is (3, 4, 5).
One subset of those are primitive Pythagorean triples which require a, b and c to also be coprimes, so their only common divisior is 1. One example is (5, 12, 13)

# The Challenge

Given three numbers representing a triple, output a truthy value if there is a triple representation of them that form a primitive Pythagorean triple and a falsy value if not.

# Test cases

### Truthy

Coming Soon


## Falsy

Coming soon


### TODO

• What about zero as input?
• Test cases
• Example for the different triple configurations?
• More descriptive title
• Unless I'm bad at maths, can't you check for these two conditions totally separately? Since if gcd(a,b,c) = G then we could instead write (Gx)^2 + (Gy)^2 = (Gz)^2 where x,y,z are a,b,c divided by G. Since that doesn't change the equality, wouldn't you just have to check for it being a triple and that they are all coprimes? It's fine if that's what you want (I also may have misunderstood) but it feels... disconnected? – FryAmTheEggman Jun 15 '16 at 14:11
• @FryAmTheEggman, no need to check them all. If a and b have a common factor, it's also a factor of c. So it's a GCD test and a Pythagorean test. @DenkerAffe, "if there is a triple representation of them that form" would be much easier to understand as "if they are". Zero as input is fine: (0, 1, 1) is a (degenerate) primitive Pythagorean triple. I wouldn't worry about built-ins: GCD has been asked before, and a built-in for "Is this a Pythagorean triple" is rather too specific to exist except as a 30-character Mathematica function. – Peter Taylor Jun 15 '16 at 14:25
• @PeterTaylor You're right, but what I was trying to get at was that the challenge felt like two different tests with an and slapped in the middle. While this isn't exactly a problem (particularly because the primitive triples are actually studied) I was trying to suggest the challenge might be better if it felt more like they were connected. However, I have no idea how to accomplish that... – FryAmTheEggman Jun 15 '16 at 14:55
• Also, here is a test script I wrote, if it helps. It only works on sorted inputs, currently. – FryAmTheEggman Jun 15 '16 at 14:55
• @FryAmTheEggman I definetly see you point. Just read about it and wanted to make a challenge about it ^^. It should be fine, since it not just two random things with an AND between them, but I will think about it. – Denker Jun 15 '16 at 17:24
• @PeterTaylor I agree the wording is a bit off. Will clarify later. Thanks for your other thoughts as well! – Denker Jun 15 '16 at 17:26

# Join the dots without crossing the line

Given a collection of distinct points in the unit square, output the points in order. This can be any order such that a closed polygon formed by straight line segments joining each point to the next (and the last back to the first) has no two lines crossing.

# Input

1. There will be between 4 and 255 points.

2. Each one is represented by an ordered pair (x, y)

3. The coordinates will have entries in the range [0, 1), that is 0 <= x < 1

4. Each entry may have up to 8 decimal places, so the range is 0 to 0.99999999.

5. You may choose to accept integers instead, in which case the range will be 0 to 99999999.

6. You may take input in any reasonable format. For example:

• (0.1, 0.2), (0.3, 0.4), (0.5, 0.7)
• 10000000 20000000 30000000 40000000 50000000 70000000

# Output

The output format need not match the input format as long as both are unambiguous.

# Impossible cases

The code does not need to work for impossible cases, such as all points being colinear. Neither does it need to report such cases - it can simply not work. Behaviour is undefined.

# Random algorithms

Your code must be deterministic. That is, it must always give the same output for the same input. For this purpose, the same points in a different order will be counted as distinct inputs and need not have the same output.

You may use pseudo random number generators provided the output is still consistent. If this requires setting a seed, that seed must be zero.

# Time limit

Your code does not have to be particularly efficient, but it must finish for the 255 point test case in under 5 minutes.

The requirement for the code to be deterministic is so that this time limit can be checked with a single run. If your random number generator of choice cannot give consistent behaviour by default, then you will need to seed it with zero. If a random number generator does not allow for seeding and does not give consistent behaviour then you may not use that generator.

# Scoring

This is code golf. Shortest code in bytes wins.

Note that this is not a Traveling Salesman Problem. There is no requirement for the tour to be short, only for it to be non-intersecting.

• It would be good to be explicit about whether non-deterministic solutions are accepted, and if so then what that means for the time limit. – Peter Taylor Jun 30 '16 at 9:47
• I think that it would be reasonable to say that RNGs which seed automatically and can't be reseeded with value 0 may not be used. This would mean that e.g. CJam answers have to be deterministic, but there's a perfectly good deterministic approach anyway. (The obvious approach IMO is to find the convex hull, then remove those points and recurse). – Peter Taylor Jul 1 '16 at 21:04

# The Fast and The Fourier

Implement the Discrete Fourier Transform (DFT) for a sequence of any length using a Fast Fourier Transform algorithm (FFT). This may implemented as either a function or a program and the sequence can be given as either an argument or using standard input. A DFT has time complexity of O(n2) whereas a FFT has time complexity of O(n log n).

The algorithm will compute a result based on standard DFT in the forward direction. The input sequence has length n and consists of the complex values {x0, x1, ..., xn-1}. The output sequence will have the same length and consists of {y0, y1, ..., yn-1} is defined by the relation below.

## Bluestein's algorithm

One algorithm that meets these requirements Bluestein's algorithm. It is a special case of the Chirp-Z transform and is able to compute the FFT for a sequence of any length n by transforming it in order to solve it as a cyclic convolution which can be solved with a time complexity of O(n log n).

Keep in mind that it is not required that you only use this algorithm in your implementation. If you know a better way, feel free to use it.

First, an identity is used to rewrite the initial DFT in a form where a convolution can easily be recognized.

You can obtain two sequences from this new form

which allow you to write the DFT as a convolution of two sequences.

### Sample

Get the input
x = [1, 2, 3, 4, 5]

Get the length of the input
n = 5

Compute the 'a' sequence
a = [1, 1.618 - 1.176j, -2.427 - 1.763j, 3.236 + 2.351j, -4.045 + 2.939j]

Compute the 'b' sequence
b = [1, 0.809 + 0.588j, -0.809 + 0.588j, 0.809 - 0.588j, -0.809 - 0.588j]

Compute the convolution of 'a' and 'b' (using summation)
y[0] = (a[0]*b[0] + a[1]*b[1] + a[2]*b[2] + a[3]*b[3] + a[4]*b[4]) / b[0]
= 15 / 1 = 15

y[1] = (a[1]*b[0] + a[0]*b[1] + a[2]*b[1] + a[3]*b[2] + a[4]*b[3]) / b[1]
= (-4.045 + 1.314j) / (0.809 + 0.588j) = -2.5 + 3.441j

y[2] = (a[2]*b[0] + a[1]*b[1] + a[3]*b[1] + a[0]*b[2] + a[4]*b[2]) / b[2]
= (1.545 - 2.127j) / (-0.809 + 0.588j) = -2.5 + 0.813j

y[3] = (a[3]*b[0] + a[2]*b[1] + a[4]*b[1] + a[1]*b[2] + a[0]*b[3]) / b[3]
= (-2.5 + 0.813j) / (0.809 - 0.588j) = -2.5 - 0.813j

y[4] = (a[4]*b[1] + a[3]*b[1] + a[2]*b[2] + a[1]*b[3] + a[0]*b[4]) / b[4]
= 4.253j / (-0.809 - 0.588j) = -2.5 - 3.441j

The Fourier tranform of x
y = [15, -2.5 + 3.441j, -2.5 + 0.813j, -2.5 - 0.813j, -2.5 - 3.441j]


## Rules

• This is so the shortest solution wins.
• Builtins that compute FFT in forward or backward (also known as inverse) directions are not allowed.
• Builtins that compute the convolution are not allowed. (Most will have not been allowed by the previous rule as they use FFT internally.)
• Your solution must have time complexity of O(n log n) where n is the length of the input sequence.
• Floating-point inaccuracies will not be counted against you.

## Test Cases

FFT([1, 1, 1, 1]) = [4, 0, 0, 0]
FFT([1, 0, 2, 0, 3, 0, 4, 0]) = [10, -2+2j, -2, -2-2j, 10, -2+2j, -2, -2-2j]
FFT([1, 2, 3, 4, 5]) = [15, -2.5+3.44j, -2.5+0.81j, -2.5-0.81j, -2.5-3.44j]
FFT([5-3.28571j, -0.816474-0.837162j, 0.523306-0.303902j, 0.806172-3.69346j, -4.41953+2.59494j, -0.360252+2.59411j, 1.26678+2.93119j] = [2, -3j, 5, -7j, 11, -13j, 17]
`

### Related

• Compute the Discrete Fourier Transform - This contains some implementations for the standard DFT algorithm which has time complexity O(n2). You'll want to understand how to implement this before trying FFT.
• Too Fast, Too Fourier: FFT Code Golf - This previous challenge is the precursor to the current challenge here. Before, you only had to consider sequences where the length n was a power of 2 which allowed for simpler recursive implementations. The difference here is that you now have to implement an FFT algorithm that will work for sequences with any length.
• nice choice for the title – Abr001am May 7 '16 at 20:47
• – Luis Mendo Jun 16 '16 at 7:43
• @LuisMendo That challenge is for sequences where the length is a power of two, but this is for sequences of any length. – miles Jun 16 '16 at 7:48
• Oh, sorry then. I suggest you add a link in your challenge explaining the difference. I think many people may get confused as I did. Apart from this, if length is not a power of 2 there's another potential problem: what algorithms count as a FFT? – Luis Mendo Jun 16 '16 at 7:52
• @LuisMendo I had a short snippet stating the difference but I deleted it in my previous edit for some reason. I want it to be so that any FFT algorithm that has a time complexity better than the naive DFT - O(n^2) - will be accepted. It would probably be best to explain an algorithm, ie Cooley-Tukey FFT, that has time complexity O(n log n), and work through a specific example using it. Applying other algorithms would be left to the solver. – miles Jun 16 '16 at 8:02
• I think the sentence your solution must have time complexity of O(n log n) covers that. Sorry, I missed that again! – Luis Mendo Jun 16 '16 at 8:04
• @LuisMendo Nevertheless, it'd probably be a good idea to explain one approach for solving this. I'll try to add one later. – miles Jun 16 '16 at 8:14
• Is there such a thing as a slow Fourier Transform? – luser droog Jun 24 '16 at 3:18