461
\$\begingroup\$

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.

To post to the Sandbox, scroll to the bottom of this page or click on the "Add Proposal" link below, and click "Answer This Question". Click "OK" when it asks if you really want to add another answer. Write your challenge just as you would when actually posting it. You may also add some notes about specific things you would like to clarify before posting it. Other users will help you improve your challenge by rating and discussing it. When you think your challenge is ready for the public, go ahead and post it, replace the post here with a link to the challenge and delete it.

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

The Sandbox works best if you sort posts by "active".

Add Proposal

Search the Sandbox

Browse your pending proposals

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]

\$\endgroup\$
  • \$\begingroup\$ How are tags added to questions? \$\endgroup\$ – guest271314 Jan 9 at 7:51
  • \$\begingroup\$ @guest271314 You can use this markup to create a tag in a draft: [tag:code-golf] \$\endgroup\$ – DJMcMayhem Aug 29 at 15:19
  • \$\begingroup\$ Why no featured anymore? Can't we have it auto-added or something? \$\endgroup\$ – JL2210 Sep 26 at 15:57
  • 1
    \$\begingroup\$ @JL2210 We now have a permanent info box that links to the Sandbox, so the featured tag isn't necessary \$\endgroup\$ – caird coinheringaahing Sep 29 at 13:43

2615 Answers 2615

0
\$\begingroup\$

Find the Least Conforming Word

Given a list of words, find which word follows Zipf's Law the least.

For each distinct word w in this list, we know the following two properties of w in this list:

  • Its frequency in the list (that is, the number of occurrences of it divided by the number of elements in the list), and
  • Its frequency rank; that is, the number of words that occur more often than it.

In the case that two words have the same frequency, we assume the one that comes first when sorted by lexicographical order is first.

If we plot a large list of words with regular natural language distribution on a logarithmic graph with these two properties as the x and y values, we see that the points form nearly a straight line with slope of approximately -1. This is known as Zipf's Law.

The Challenge

You will be given a list of words l. Imagine each distinct word being plotted this way on a logarithmic graph. Then, the line k is the Linear Regression Line for the data set. Then, for every point p except the top-most ranking and most frequent point, find the vertical distance to the line k, and let this distance be d. Then, we scale the vertical distance so that points in the middle of the line will be the most heavily weighted. We do this by the following. We find the x coordinate of the mid-point of the line (let this be x). Then, half of the horizontal span of the line is w, the width. If we let px be the x coordinate of the point p, then we multiply d by 1 - (x - px) / w. This value is the score.

Input

The input will be an unordered list of words in any reasonable format you wish.

Output

The output will be a single word indicating which word conforms the least to the "standard".

Test Case

Input: a a a a a a a a a a b b b b b b b c c c c c d d
Output: b
See an explanation here

\$\endgroup\$
  • 1
    \$\begingroup\$ 1. "Normal distribution" is a technical term: anyone who's studied statistics in English will immediately think of the distribution also known as Gaussian. IMO you should just call it Zipf's law. 2. The description given for the score is a) ambiguous and b) very odd. It's ambiguous because the question in general implies that the absolute difference should be what matters, but nothing explicitly says this. It's very odd because comparing gradients gives a massive bias towards the more frequent values. Why not vertical distance from the line? (And why not use the line of best fit?) \$\endgroup\$ – Peter Taylor Apr 4 '17 at 8:34
  • \$\begingroup\$ @PeterTaylor Thanks for the suggestions. Yes, I forgot that Normal Distribution is actually something. And yes, I should explicitly state absolute difference. The reason that I compare gradients is because vertical distance makes a larger difference near the ends of the line. So I will change the specs to compare vertical distance, but with a specific weighting. By the way, how is line of best fit calculated exactly? I can only find definitions where you draw a line that looks like it fits the system best. \$\endgroup\$ – user42649 Apr 4 '17 at 13:20
  • 1
    \$\begingroup\$ en.wikipedia.org/wiki/Linear_regression \$\endgroup\$ – Peter Taylor Apr 4 '17 at 13:31
  • \$\begingroup\$ @PeterTaylor Okay, thanks. I've adjusted the specs to include all of your suggestions. \$\endgroup\$ – user42649 Apr 4 '17 at 13:39
0
\$\begingroup\$

Lock the Language!

This is a challenge.

Cop's Thread:

Challenge:

  • Write a program that given two arbitrary integers adds those and prints them to STDOUT
  • In this Thread you post the fully working solution including language
  • The robbers will try to find any other language where an anagram of your source code will also add two integers
    • The language has to be a different one then yours
      • Different versions of the same language don't count as different languages
  • After 5 days you're code is safe and can't be cracked any more
  • If you're code is cracked you mark that in the header of the answer and add a link to the crack in the robbers thread

Winning criteria:

  • The shortest safe program (in bytes) wins.

Robber's Thread

  • The cops wrote an addition program you want to steal
  • Since the cops encrypted it so you can't use the same language as the cop you stole the program from used, so you have to use an different language
    • Different versions of the same language are count as the same language
  • You may use the cops character set in any order (anagram)

Winning criteria:

  • The robber with the most cracked answers wins.
\$\endgroup\$
  • \$\begingroup\$ Does it need to be a "proper" anagram or can I crack an answer by finding a language in which the solution is a polyglot? \$\endgroup\$ – Martin Ender Apr 4 '17 at 14:24
  • \$\begingroup\$ "After 5 days you're code is safe and can't be cracked any more." We usually require the cops to reveal their solution (after the period has passed) before they are considered safe. \$\endgroup\$ – Martin Ender Apr 4 '17 at 14:25
  • \$\begingroup\$ How does it work with languages using different encodings? Is it the actual characters that should be used for the anagram or the byte-values? \$\endgroup\$ – Emigna Apr 4 '17 at 14:26
  • \$\begingroup\$ Related. (Same idea, different task) \$\endgroup\$ – ETHproductions Apr 4 '17 at 14:27
0
\$\begingroup\$

Comment polyglot

From an original idea by Digital Trauma

The challenge

This challenge requires you to write a polyglot which contains a comment in as many languages as possible.

The comment must say This is a polyglot comment.

The program must do nothing at all.

The winner is whoever writes the polyglot with most languages!

The rules

Your submission should consist of a full program whose source contains the text "This is a polyglot comment".

For each language of the polyglot, your code executed in that language should:

  • Take no input
  • Generate no errors.
  • Produce either no output or the same output produced by the empty program in this language.
  • Still comply to the previous rules if any part of This is a polyglot comment is removed.

Addictionally, there can be no superflous code. This means that it should not be possible to remove a subset of characters (outside of the comment) and still have a submission that complies to the previous rules for all the languages of the polyglot.

Ok? Ok. Now please wait a minute before running to see how many languages you can find that use # for comments, because we have one more rule:

We define a character of your source as required for a language if removing that character makes the program change its behavior for that language (this could mean for example changing its output or producing an error). The sets of required characters for each language must all be distinct.

Example submission

#This is a polyglot comment

A polyglot in Python and Brainfuck for a score of 2.

The set of required characters for Brainfuck is empty. This is perfectly valid, as long as no other set is empty.

The set of required characters for Python is composed of just the first #. We can't add Bash to this submission because it would share the same set.

That's all, good luck and have fun!


Sandbox question:

  • Do you see any potential loopholes in the rules?
  • Should I add some other factors to the score (like code length for example) or is the number of languages enough?
  • Anything else I've missed?
\$\endgroup\$
  • \$\begingroup\$ I'm unsure about the definition of required. If I write ///This is a polyglot comment in a language which uses // for comments, is / a required character? If I remove one / nothing changes, but if I remove two then I get an error. Similarly, if my language has both # and // as comments, what is the required set for #///This is a polyglot comment ? \$\endgroup\$ – Peter Taylor Apr 5 '17 at 10:53
  • \$\begingroup\$ a character has to be required for at least one language I think it says \$\endgroup\$ – Destructible Lemon Apr 7 '17 at 3:12
  • \$\begingroup\$ @PeterTaylor you're right, this definition doesn't work. I'm having a hard time to come up with a correct one, I'll update this proposal if something suitable comes to my mind. \$\endgroup\$ – Leo Apr 8 '17 at 12:52
0
\$\begingroup\$

Operation Permutations

We're likely all familiar with the order of operations; the precedence rules that we use to govern basic arithmatic. Most schoolchildren are taught this order using an acronym such as "BODMAS", "BIDMAS", or "PEMDAS". There's no common consensus on the 'right' acronym to use, with the one that you learnt depending on the country you went to school in, or even the teacher who taught you. Your challenge is to determine the validity of an order of operations acronym (O3A), even if it's so unpronounceable that no-one would ever actually use it.

Challenge Specification

Input

Input for this program is guaranteed to be exactly six characters. Your program can require upper-case, require lower-case, or be case agnostic. You may accept this input in any manner that you please.

Each character is guaranteed to be one of the following: (B, P, I, O, E, D, M, A, S). Furthermore, each character is guaranteed to appear at most only once in the input.

The following characters correspond to the following operations:

| Characters | Operation      |
| ---------- | -------------- |
| (B, P)     | Brackets       |
| (I, O, E)  | Exponentiation |
| D          | Division       |
| M          | Multiplication |
| A          | Addition       |
| S          | Subtraction    |

Output

Your program must yield a truthy value for a valid O3A, and a falsy value for an invalid one. Your program is not expected to handle input not matching the above specification.

O3A Validity

The first character in a valid O3A must correspond to 'Brackets'. Likewise, the second character in a valid O3A must correspond to 'Exponentiation'. The next two characters must correspond to 'Division' and 'Multiplication', but either order is valid as these operations are on the same precedence level. The final two characters must correspond to 'Addition' and 'Subtraction', but again, either order is valid. Any other O3As are invalid.

Testing

Reference Implementation

The following Python 3 reference implementation is given for guidance. Note that while this reference implementation is case-agnostic, your program does not need to be.

def validate(acronym):
    acronym = acronym.upper()

    if acronym[0] not in {"B", "P"}:
        return False

    if acronym[1] not in {"I", "O", "E"}:
        return False

    if set(acronym[2:4]) != {"D", "M"}:
        return False

    if set(acronym[4:6]) != {"A", "S"}:
        return False

    return True

Test Input

BIDMAS => True
PEMDAS => True
PIDMSA => True
BOMDAS => True
BMIDAS => False
IBMDAS => False
BODAMS => False

Rules

  • This is , so the shortest code wins counted in bytes.
  • Standard rules and loopholes apply
  • Check meta for consensus

Sandbox Notes

My worry with this challenge is that it would be so trivially regex-able in something like Retina that it wouldn't be fun. I'd like feedback from others about this before going live.

\$\endgroup\$
  • 1
    \$\begingroup\$ So: test whether an input which is known to match [BPIOEDMAS]{6} also matches [BP][IOE](DM|MD)(AS|SA)? It's possible that there's a hash function which would be shorter than an actual check, but... If you're worried about boring answers then I suggest trying to find a solution which beats the boring regex in a language you know before posting. \$\endgroup\$ – Peter Taylor Apr 5 '17 at 11:53
  • \$\begingroup\$ @fəˈnɛtɪk I just checked, and no, it doesn't fail. \$\endgroup\$ – FourOhFour Apr 5 '17 at 12:09
  • \$\begingroup\$ @PeterTaylor I do believe that a shorter solution than a naive regex exists, but I also don't think I'm good enough at golfing to find it. Hence I posted on sandbox in the hope that more experienced golfers would be able to better judge the golf-ability of the challenge. \$\endgroup\$ – FourOhFour Apr 5 '17 at 12:10
  • 1
    \$\begingroup\$ @PeterTaylor You can actually shorten it slightly to ...[AS]+ since the input characters are also guaranteed to be unique (making it even harder to beat the regex solution). \$\endgroup\$ – Martin Ender Apr 5 '17 at 12:16
  • \$\begingroup\$ @MartinEnder, in fact that becomes [BP][IOE][MD]+[AS]+ applying the same reasoning. \$\endgroup\$ – Peter Taylor Apr 5 '17 at 14:58
  • \$\begingroup\$ @PeterTaylor Oh right, good catch. \$\endgroup\$ – Martin Ender Apr 5 '17 at 15:00
  • \$\begingroup\$ I've found a trick which lets me golf 3 chars off the regex-inspired solution in CJam (which doesn't have regexes). Perl will still beat it, but there's some non-trivial interest to the question. \$\endgroup\$ – Peter Taylor Apr 5 '17 at 15:19
  • \$\begingroup\$ @PeterTaylor I was hoping that it would still be an interesting challenge for CJam golfers etc... I might put it up live tomorrow on the basis that it's not completely trivial (so long as it's seen as an per-language challenge). \$\endgroup\$ – FourOhFour Apr 6 '17 at 0:08
0
\$\begingroup\$

Haiku detector in Haiku:

I found a haiku-w detector problem, and a popularity-contest challenge for writing factorial in haiku, but not the combination. Let's fix that:

Input: a string consisting of alphanumeric characters, standard punctuation, and |'s, where the | is used to indicate where the syllable splits are, and newlines (to indicate where the line splits are).

Output: a Truthy value if the string consists of zero or more triplets of lines where the first and third line have 4 |'s and the second line has 6 |'s.

Examples:

The|quick|brown|fox|jumps
O|ver|the|la|zy|dog|while
the|cat|eats|goat|cheese

--> True

This|is|not|a|hai|ku

--> False

A|B|C|d|E
1|2|3|4|5|6|7
A|B|C|D|E
Oops

--> False

Scoring: 1 Point for each triplet (so a whitespace answer will automatically win with a score of 0; I'm open to suggestions on how to mitigate this); this is code golf so lowest score wins.

Determining if code is valid haiku: Must consist of lines of code consisting of 5/7/5 syllables (with an optional empty line in between triplets)

Numbers and Variable names are read as written in standard English. If you want to get fancy and use a word that can be read as different number of syllables (e.g. coop as in chicken coop vs coop as in the short form of cooperative that isn't always rendered using an umlaut), have at it.

Special characters:

  ( is either 2,3 or 4 syllables (paren, left paren, open paren)
  ) is either 2, or 3 syllables (paren, right paren, close paren)
  ! is 1 or 5 syllables (bang, not, or exclamation point)
  . is 1 (dot)
  # is 1 (hash or pound)
  $ is 2 or 3 (dollar or dollar sign)
  % is 2 or 3 (percent or percent sign)
  - is 1 or 2 (dash, hyphen, minus)
  + is 1 or 2 (plus, plus sign)
  ^ is 2 (caret or xor)
  & is 3 (ampersand)
  * is 1 (star)
  [ is 2,3,4,5 (bracket, open bracket, left bracket, square bracket, open square bracket, etc.)
  ], {, and } are likewise 2,3,4,5
  _ is 3 (underscore)
  = is 2 (equals)
  : is 2 (colon)
  ; is 4 (semicolon)
  > is 2 or 3 (greater or greater than)
  < is 2 or 3
  --> is either combination of two -'s and one > or 2 (arrow)
  , is 2 (comma)
  / is 1 or 3 (slash or forward slash)
  \ is 2 (backslash)
  | is 1 or 4 (pipe or vertical bar)
  ? is 2 or 3 (question or question mark)
  @ is 1 (at)
  ~ is 2 (squiggle or tilde)
  ` is 2 (backtick)
  ' is 1 or 3 (quote or single quote)
  " is 3 (double quote)

Note that for everything but quote, if you use the same symbol 2 (or more times), you can instead say double symbol (or triple or whatever), e.g. && can be either 5 or 6 syllables)

I think that's all the punctuation... other unprintable characters are banned unless you can give a compelling argument on how many syllables it should be (and if I missed any pronounciation alternatives or other punctuation, let me know)

\$\endgroup\$
0
\$\begingroup\$

Golf the Stack Exchange code block

The standard Stack Exchange code block button / hot key prepends 4 spaces to each line, unless they all begin with 4 spaces, in which case it removes them. Additionally, it avoids changing the first or last lines if they are empty. But why stop at the first or last lines? Markdown is quite happy if intervening empty lines remain empty. As this saves bytes, I want you to write as short a program (or function) as you can that follows this enhanced behaviour:

  • All empty lines should remain unchanged
  • If all other lines begin with 4 spaces, then remove 4 spaces from them
  • Otherwise, prepend 4 spaces to each of them

This is , so the shortest program that breaks no standard loopholes wins.

Note that should your answer itself contain a blank line, you should ensure that there are no spaces on that line in your formatted code block.

\$\endgroup\$
  • \$\begingroup\$ Related (and currently active). I'd say this challenge should be explicitly described as a followup to that one, because it's basically "much the same task but harder". \$\endgroup\$ – user62131 Apr 9 '17 at 22:14
0
\$\begingroup\$

Swipe over the keyboard

(I'm not a native speaker so any good alternative for swipe is highly appreciated)

You have a keyboard in front of you and you swipe over the keyboard from left to right with your whole hand. Your task is to simulate the output. The keyboard is represented with 4 arrays (here is an example of a german keyboard layout, you are free to choose any other layout):

[1,2,3,4,5,6,7,8,9,0,ß,´]
[q,w,e,r,t,z,u,i,o,p,ü,+]
[a,s,d,f,g,h,j,k,l,ö,ä,#]
[<,y,x,c,v,b,n,m,,,.,-]

Now choose randomly one character of the first column [1,q,a,<]. Pop the chosen key and do the same with the newly created first column. For example the first key was q, the new first column will be [1,w,a,<]. The popped key will be printed out or skipped. Repeat it until your arrays are empty.

The rules are the following:

  • You can represent the keyboard with 4 arrays matching the 4 rows on your keyboard. For example [1,2,3,4,5,6,7,8,9,0,ß,´] would be my first row (german keyboard layout). Choose the keyboard layout you want, but it has to include all letters and numbers
  • There is no need to represent non-printable keys like tabulator, shift, backspace, ... so in general you won't hit the first and last column and the last row (you won't hit space too)
  • Since you are swiping from left to right, the probability to hit a previously located key is 0. So if the key 5 is pressed, the keys 1-4 won't be pressed anymore.
  • Each key needs to have a probability P!=0 to be pressed
  • Each key needs to have a probability P!=0 to be skipped
  • The rows are independent
  • There is no input

Open questions

  • Should I allow only one keyboard type?
  • Are there still rules that need to be added?
\$\endgroup\$
0
\$\begingroup\$

CRC Collision

Your task: Accept a string (one could also include symbols in the string, does not have to be only alphabets and numbers) of any length as input, calculate 16bit CRC for the given string. Output a string of same length which gives you the exact same CRC sum.

Example:

Input Text: Hello World!
Caclculated CRC : 0xF444

Output Text (with same CRC): DW=9Mzk?7P4y

You may use in-built CRC libraries to calculate.

Shortest code that fulfills the requirement wins!

\$\endgroup\$
  • \$\begingroup\$ What's the winning criterion? Can we use any 16-bit CRC or does it have to be a specific one? Which characters/bytes are allowed in input/output?What should happen if there isn't another string of the same length with the same checksum? \$\endgroup\$ – Dennis Apr 12 '17 at 15:53
  • \$\begingroup\$ @Dennis Any 16-bit CRC routine would do! It has to be a string i.e ASCII characters. The input string should be minimum 3 bytes long, there should be a collision right? \$\endgroup\$ – Abel Tom Apr 12 '17 at 16:06
  • \$\begingroup\$ I think so, yes, but I'm not sure if the output can only consist of ASCII characters. \$\endgroup\$ – Dennis Apr 12 '17 at 17:20
  • \$\begingroup\$ So i guess non ASCII characters are acceptable too! \$\endgroup\$ – Abel Tom Apr 13 '17 at 11:04
  • \$\begingroup\$ @Dennis Is there a suggestion to make it a little harder may be? Or can i put it like this as a challenge? \$\endgroup\$ – Abel Tom Apr 22 '17 at 16:55
0
\$\begingroup\$

Display a fractal maze

Based on this Puzzling.SE question: Alice and the Fractal Hedge Maze.

Given a description of a maze as input, output a [graphical / ASCII-art ?] representation of the maze.

Maze layout

For the purpose of this challenge, all inner mazes are assumed to be smaller copies of the original maze. So a top-down view of the above maze would look like this:

top-down view showing recursion

Input:

Input will be given as a list of connections between different ports of the same level of the maze and the inner sub-mazes.

One list of connections for the above maze would be the following:

1,12,a5,a12

This is because, in the maze, ports 1 and 12 are connected to each other, and they are also connected to ports 5 and 12 of sub-maze A.

A full description of the above example maze would be the following:

a3      (note that a connection by itself means this is the starting point)
1,12,a5,a12
2,8,10,a9,c4
3,b1,b4
4,b5,b6
5,d4,d5
6,d8
7,b7,d10
9,c7,c11
11,a10
b8,d1,d3
b11,b12
c2,c12
d11,d12

Flexibility:

You may take each port like above or as a tuple with numbers instead of letters, such as using 0 for the outer maze and numbering the inner mazes sequentially. So the following would all be acceptable for a line:

2,8,10,a9,c4
[2,8,10,(1,9),(3,4)]
[2,8,10,('A',9),('C',4)]
[(0,2),(0,8),(0,10),(1,9),(3,4)]
[(0,2),(0,8),(0,10),(-1,9),(-3,4)]

The starting point will be by itself. You can take it in the list as a port by itself, or as a separate parameter. The starting position will always be an entrance to a sub-maze. Each port will be listed at most once in the input. Note that not all ports have to be listed, as there are some dead ends.

Output:

Output a representation of the maze given. Note that mazes may require bridging of paths (crossing without connection).

Note: or ?

\$\endgroup\$
  • \$\begingroup\$ I think this challenge might be more interesting if it where about solving fractal mazes instead of displaying them. \$\endgroup\$ – Laikoni Apr 17 '17 at 8:51
  • \$\begingroup\$ That was going to be a second, separate challenge. \$\endgroup\$ – mbomb007 Apr 17 '17 at 13:29
0
\$\begingroup\$

Which answer is correct?

I've posted an earlier version of this question already; but now hoping to refine here before correcting the original post. Thanks all in advance for any advise.


I saw this elegant logic puzzle on Mathematics:

Which answer in this list is the correct answer to this question?

 1. All of the below.
 2. None of the below.
 3. All of the above.
 4. One of the above.
 5. None of the above.
 6. None of the above.

One user submitted a nice answer in code (quoted below), which inspired this puzzle.

Task:

Write code which determines the answer to the above logic puzzle according to standard Code Golf rules (i.e. a working solution with the minimum number of characters wins).

The input should be a sequence (i.e. ordered list / array) with each item in that sequence representing one of the given statements. Each items in the list should hold 2 values: - the first representing All, One, or None. - the second representing above, or below.

The result should be a list/array of the indices of any correct answers.

The following test scenarios are considered invalid; i.e. your code would not be expected to handle such data:

  • A statement is the first statement in the sequence and refers to the above, it evaluates to false.
  • A statement is the last statement in teh sequence and refers to the below, it evaluates to false.

We should be optimistic about results. i.e. given the following, the program would return 1, 2:

1. all of the below
2. one of the above

When contradictions occur, neither statement is correct. i.e. the following would return a blank string.

1. none of the below
2. none of the above

Code for Referenced Answer

// gcc ImpredictivePropositionalLogic1.c -o ImpredictivePropositionalLogic1.exe -std=c99 -Wall -O3

/*
Which answer in this list is the correct answer to this question?

(a) All of the below.
(b) None of the below.
(c) All of the above.
(d) One of the above.
(e) None of the above.
(f) None of the above.
*/

#include <stdio.h>
#define iff(x, y) ((x)==(y))

int main() {
  printf("a b c d e f\n");
  for (int a = 0; a <= 1; a++)
  for (int b = 0; b <= 1; b++)
  for (int c = 0; c <= 1; c++)
  for (int d = 0; d <= 1; d++)
  for (int e = 0; e <= 1; e++)
  for (int f = 0; f <= 1; f++) {
    int Ra = iff(a, b && c && d && e && f);
    int Rb = iff(b, !c && !d && !e && !f);
    int Rc = iff(c, a && b);
    int Rd = iff(d, (a && !b && !c) || (!a && b && !c) || (!a && !b && c));
    int Re = iff(e, !a && !b && !c && !d);
    int Rf = iff(f, !a && !b && !c && !d && !e);

    int R = Ra && Rb && Rc && Rd && Re && Rf;
    if (R) printf("%d %d %d %d %d %d\n", a, b, c, d, e, f);
  }
  return 0;
}

Test Cases

To represent the conditions in these test cases I've used the following convention:

  • [...] to represent the entire sequence
  • (match,direction) to represent each item where match and direction have values:

Match:

  • 0 - None of the
  • 1 - One of the
  • 2 - All of the

Direction:

  • 0 - below
  • 1 - above

1. The Original

[
    (2,0)  //all of the below
    ,(0,0) //none of the below
    ,(2,1) //all of the above
    ,(1,1) //one of the above
    ,(0,1) //none of the above
    ,(0,1) //none of the above
]

//expected answer is 5

2. Both

[
    (2,0)  //all of the below
    ,(1,1) //one of the above
]

//expected answer is 1, 2

3. Neither

[
    (0,0)  //none of the below
    ,(0,1) //none of the above
]

//expected answer is (nothing)

4. Simple

[
    (2,0)  //all of the below
    ,(0,1) //none of the above
]

//expected answer is 2
\$\endgroup\$
  • 2
    \$\begingroup\$ You should probably allow the output to be a list of possible answers, rather than requiring a string like you implicitly do at the moment. \$\endgroup\$ – user62131 Apr 13 '17 at 23:50
  • \$\begingroup\$ I believe you get only answer like this: sandbox.onlinephpfunctions.com/code/… First column the possible answer if only one is true as binary in decimal value second column is which answer are true also in binary as decimal value and in the third column only the answer where second column is equal to first column. I like this kind of problems \$\endgroup\$ – Jörg Hülsermann Apr 14 '17 at 1:11
  • \$\begingroup\$ Thanks @ais523; amended \$\endgroup\$ – JohnLBevan Apr 15 '17 at 8:36
0
\$\begingroup\$

Help me discover myself!

Hola, People call me file. At any point of time, all I know is how people compile me. Some use gcc some use gpp and some Java and some even csc. I couldn't really figure out who I am with just this info!

All I can do for you to help me is show a snippet inside me for you to find out the programming language. If you couldn't find with the snippet(the snippet will not have any incomplete line of code), only then can I share how programmers compile me.

And yes, you are brilliant enough to find out the language in which the program in me is written!

I may have...

Example1:

Input1:

int main(){return 0;}

Output1:

c|cpp

Input2:

gpp

Output2: c++

or

Input2:

gcc

Output2:

c

Example 2:

Input1:

#include<iostream>
int main(){return 0;}

Output1:

c++

Example3:

Input1:

main(){}

Output1:

c
  • the above example can't possibly be java because the mandatory return type for the method in java is missing.

Example4:

Input1:

#include<stdio.h>
main(){}

Output1:

c

Example5:

Input1:

class hola{public static void main(){}}

Output1:

java

Example6:

Input1:

class Hola{public static void Main(){}}

Output1:

c#

Oh yeah, I also know few things that might be helpful for you. The snippet that I'm allowed to share is only headers and the main function - first function that is called when program is executed.

And also, there are some conventions.

And then,

  • c - return type for main is not mandatory, standard header is stdio.h
  • c++ return type is mandatory, standard header is iosstream
  • c# - Main is a title - first letter is caps!

Input format Any code snippet it is more than enough if your answer identifies the above given six example snippet.

Output format programming language in which the snippet possibly fits in.

This is code golf and hence shortest code in each language wins!

\$\endgroup\$
  • \$\begingroup\$ How many languages - and which ones - do we need to support, and to which accuracy? Or, is it sufficient to hard-code the output for some given list of possible inputs? \$\endgroup\$ – John Dvorak Apr 17 '17 at 16:12
  • \$\begingroup\$ its enough if the answer supports the given input! And accuracy is 100% because, there is a certain restriction by languages and yeah I'll add that to the question \$\endgroup\$ – Keerthana Prabhakaran Apr 17 '17 at 16:18
  • \$\begingroup\$ I don't think I understand what the challenge is asking. "Assume the input is one of your choosing, and output the corresponding (single, hardcodeable) language"? \$\endgroup\$ – John Dvorak Apr 17 '17 at 16:20
  • \$\begingroup\$ one of six sample inputs and identify the language(c,c++,java or c# - yeah hardcodable). \$\endgroup\$ – Keerthana Prabhakaran Apr 17 '17 at 16:27
  • \$\begingroup\$ so, "hash the input and look up in a six-element table". Not fun. \$\endgroup\$ – John Dvorak Apr 17 '17 at 16:29
  • \$\begingroup\$ not really. There is a trick. If the input contains Class it goes without saying the the language is C# but if the input is int main the language can either be c or c++. It will a set of if and else'. \$\endgroup\$ – Keerthana Prabhakaran Apr 17 '17 at 16:39
  • 1
    \$\begingroup\$ Yeah, I could have a case statement with six regexes, but %w{c java c++ - - - python}[i.size%13] is much shorter. \$\endgroup\$ – John Dvorak Apr 17 '17 at 16:55
  • \$\begingroup\$ sorry I dont get what you mean by "input length % 13" \$\endgroup\$ – Keerthana Prabhakaran Apr 17 '17 at 16:56
  • \$\begingroup\$ I don't know what sources you've been reading about C#, but the capital Class will give a syntax error. \$\endgroup\$ – Peter Taylor Apr 17 '17 at 17:55
  • \$\begingroup\$ @PeterTaylor oh yeah you are right. I dunno why I ever typed that. Its been a while since I programmed using c#. I've edited the post. \$\endgroup\$ – Keerthana Prabhakaran Apr 17 '17 at 19:17
  • \$\begingroup\$ @JanDvorak you have edited your comment, but I still dont get it. \$\endgroup\$ – Keerthana Prabhakaran Apr 17 '17 at 19:21
  • \$\begingroup\$ Answer #1600... \$\endgroup\$ – programmer5000 Apr 17 '17 at 20:07
  • \$\begingroup\$ @programmer5000 don't get it. Can you please be more elaborate! \$\endgroup\$ – Keerthana Prabhakaran Apr 18 '17 at 1:14
  • \$\begingroup\$ @KeerthanaPrabhakaran oh, I was just remarking that we were on our #1600th sandbox submission. Now it's 1602... \$\endgroup\$ – programmer5000 Apr 18 '17 at 1:18
0
\$\begingroup\$

Generate a Voronoi Diagram

Challenge

Given a list of coordinates and colours of points, output a 300px by 300px Voronoi Diagram.

Winning

Shortest code in bytes wins.

To be expanded

\$\endgroup\$
  • 1
    \$\begingroup\$ output a what??? \$\endgroup\$ – Titus Apr 21 '17 at 17:15
  • \$\begingroup\$ @Titus A Voronoi Diagram. I'll add an explanation when I have time \$\endgroup\$ – Beta Decay Apr 21 '17 at 18:34
  • \$\begingroup\$ Five months later... \$\endgroup\$ – Gryphon - Reinstate Monica Sep 27 '17 at 4:38
  • \$\begingroup\$ @Gryphon It's a looong expansion ;) \$\endgroup\$ – Beta Decay Sep 27 '17 at 6:06
0
\$\begingroup\$

Google Autocomplete Fun

Your task is to create a program which, given an input string, will output the first Google autocomplete result for that search. The format of input/output is up to you.

Rules

  1. Your program must take 1 input, a string, and output the top Google autocomplete/auto-fill suggestions result. The format of input/output is up to you. Just be sure to mention what your I/O format is.
  2. Obviously, accessing the Internet is allowed.
  3. URL shorteners (like bit.ly, TinyURL, etc.) are strictly disallowed. You should fetch your autocomplete results from this URL: http://suggestqueries.google.com/complete/search?client=your_browser&q=your_query. You are allowed to assume any browser name for client. In the program I wrote, I assume Chrome. Any browser name or string should work.
  4. Please provide an explanation of how your program works. It is not mandatory, but I strongly advise it.
  5. All standard loopholes are strictly forbidden.

This is , so may the shortest code win and the best programmer prosper...

\$\endgroup\$
  • \$\begingroup\$ I think you technically need to allow internet access in the question. \$\endgroup\$ – Comrade SparklePony Apr 23 '17 at 20:51
  • \$\begingroup\$ @ComradeSparklePony You are so right. :) \$\endgroup\$ – ckjbgames Apr 23 '17 at 21:06
0
\$\begingroup\$

Calculate Call-Changes

Before learning to ring methods, novice bell-ringers usually first learn something called call-changes. This is basically a sequence of permutations (the changes), each one obtained from the previous by exchanging a pair of adjacent bells (the call).

The bells always begin in ascending pitch number order (which is actually descending pitch order). Each exchange (call) can be expressed in one of two different ways, depending on the whim of the conductor. Exchanges always use the bell's pitch number, regardless of the bell's current position.

  1. The exchange can be expressed as a pair of adjacent bell numbers in sequence. The first of the pair is always immediately before the second of the pair. The two bells simply exchange places, so the first of the pair ends up immediately after the second of the pair.

    • The exchange can be expressed as a bell number plus a second bell number which is two places before the first bell. The (unnamed) intervening bell and the first bell exchange places. This also results in the first input bell ending up immediately after the second input bell.
    • The above expression cannot be used to move a bell into first (lead) place, because there is no bell for it to end up after. You therefore need to accept an alternate way of expressing that the first two bells change places. Note that this must still refer to the pitch number of the bell in second place, but this might be achieved e.g. by having a sentinel value for the second bell, or by omitting it completely. (In real call-changes the call is "[number] lead!" rather than "[number] to [number]!".)

Your other input will be the number (not list) of bells in the sequence. You can assume that this will not be more than 9 (but supporting larger numbers may get you more upvotes). Here is a worked example showing the list of calls in both of the above formats and the desired output:

Call 1  Call 2  Output
(5 bells input  12345)
12      20*     21345
34      41      21435
14      42      24135
24      40*     42135
21      14      41235
41      10*     14235
42      21      12435
43      32      12345

*example representation.

Your program or function should output the result after each change. (Showing the initial permutation is permitted but not necessary.) Input and output can be any reasonable method, e.g. nested array of numbers or list of strings (note that if you use strings and want to support more than 9 bells then the next three bells are traditionally numbered 0, E and T). You can assume that all inputs follow the above pattern (or your custom representation for calling the seconds place bell into first place).

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

\$\endgroup\$
  • \$\begingroup\$ I can't work out what the example is showing. Specifically: what input should the function/program take and what output should it produce? \$\endgroup\$ – Peter Taylor Apr 24 '17 at 8:48
  • \$\begingroup\$ @PeterTaylor Except for the first 12345 row, which is only there to try to help demonstrate the mechanics, the left column contains the result. However for each result there are two acceptable inputs (this is just a fact of life, not an attempt to make the question more complex); these are shown in the other two columns. \$\endgroup\$ – Neil Apr 24 '17 at 9:06
  • \$\begingroup\$ So e.g. the input is "42135", "24" and the output is "41235"? \$\endgroup\$ – Peter Taylor Apr 24 '17 at 9:58
  • \$\begingroup\$ No, the input is 5, "12", "34", "14", "24", "21", "41", "42", "43" and the output is "21345"..."42135", "41235"..."12345". \$\endgroup\$ – Neil Apr 24 '17 at 10:50
  • \$\begingroup\$ Could you edit the question to make that clearer? \$\endgroup\$ – Peter Taylor Apr 24 '17 at 13:51
  • \$\begingroup\$ @PeterTaylor Hopefully my rewrite is clearer now. \$\endgroup\$ – Neil Apr 26 '17 at 9:36
0
\$\begingroup\$

Excel NORMSINV function

Implement the excel NORMSINV function in your favourite language, shortest code wins

Build in functions and standard loopholes are not allowed

Loss of precision is also not allowed

Reference C Implementation

/*
 * Original C++ implementation found at http://www.wilmott.com/messageview.cfm?catid=10&threadid=38771
 * C# implementation found at http://weblogs.asp.net/esanchez/archive/2010/07/29/a-quick-and-dirty-implementation-of-excel-norminv-function-in-c.aspx
 *     Compute the quantile function for the normal distribution.
 *
 *     For small to moderate probabilities, algorithm referenced
 *     below is used to obtain an initial approximation which is
 *     polished with a final Newton step.
 *
 *     For very large arguments, an algorithm of Wichura is used.
 *
 *  REFERENCE
 *
 *     Beasley, J. D. and S. G. Springer (1977).
 *     Algorithm AS 111: The percentage points of the normal distribution,
 *     Applied Statistics, 26, 118-121.
 *
 *      Wichura, M.J. (1988).
 *      Algorithm AS 241: The Percentage Points of the Normal Distribution.
 *      Applied Statistics, 37, 477-484.
 */

#include <math.h>

double normsInv(double p, double mu, double sigma)
{
    if (p < 0 || p > 1)
    {
        printf("The probality p must be bigger than 0 and smaller than 1");
        exit(1);
    }
    if (sigma < 0)
    {
        printf("The standard deviation sigma must be positive");
        exit(1);
    }
    if (p == 0)
    {
        return -1e100;
    }
    if (p == 1)
    {
        return 1e100;
    }
    if (sigma == 0)
    {
        return mu;
    }

    double q, r, val;

    q = p - 0.5;

    /*-- use AS 241 --- */
    /* double ppnd16_(double *p, long *ifault)*/
    /*      ALGORITHM AS241  APPL. STATIST. (1988) VOL. 37, NO. 3
            Produces the normal deviate Z corresponding to a given lower
            tail area of P; Z is accurate to about 1 part in 10**16.
    */
    if (fabs(q) <= .425)
    {/* 0.075 <= p <= 0.925 */
        r = .180625 - q * q;
        val =
               q * (((((((r * 2509.0809287301226727 +
                          33430.575583588128105) * r + 67265.770927008700853) * r +
                        45921.953931549871457) * r + 13731.693765509461125) * r +
                      1971.5909503065514427) * r + 133.14166789178437745) * r +
                    3.387132872796366608)
               / (((((((r * 5226.495278852854561 +
                        28729.085735721942674) * r + 39307.89580009271061) * r +
                      21213.794301586595867) * r + 5394.1960214247511077) * r +
                    687.1870074920579083) * r + 42.313330701600911252) * r + 1);
    }
    else
    { /* closer than 0.075 from {0,1} boundary */

        /* r = min(p, 1-p) < 0.075 */
        if (q > 0)
            r = 1 - p;
        else
            r = p;

        r = sqrt(-log(r));
        /* r = sqrt(-log(r))  <==>  min(p, 1-p) = exp( - r^2 ) */

        if (r <= 5)
        { /* <==> min(p,1-p) >= exp(-25) ~= 1.3888e-11 */
            r += -1.6;
            val = (((((((r * 7.7454501427834140764e-4 +
                       .0227238449892691845833) * r + .24178072517745061177) *
                     r + 1.27045825245236838258) * r +
                    3.64784832476320460504) * r + 5.7694972214606914055) *
                  r + 4.6303378461565452959) * r +
                 1.42343711074968357734)
                / (((((((r *
                         1.05075007164441684324e-9 + 5.475938084995344946e-4) *
                        r + .0151986665636164571966) * r +
                       .14810397642748007459) * r + .68976733498510000455) *
                     r + 1.6763848301838038494) * r +
                    2.05319162663775882187) * r + 1);
        }
        else
        { /* very close to  0 or 1 */
            r += -5;
            val = (((((((r * 2.01033439929228813265e-7 +
                       2.71155556874348757815e-5) * r +
                      .0012426609473880784386) * r + .026532189526576123093) *
                    r + .29656057182850489123) * r +
                   1.7848265399172913358) * r + 5.4637849111641143699) *
                 r + 6.6579046435011037772)
                / (((((((r *
                         2.04426310338993978564e-15 + 1.4215117583164458887e-7) *
                        r + 1.8463183175100546818e-5) * r +
                       7.868691311456132591e-4) * r + .0148753612908506148525)
                     * r + .13692988092273580531) * r +
                    .59983220655588793769) * r + 1);
        }

        if (q < 0.0)
        {
            val = -val;
        }
    }

    return mu + sigma * val;
}
\$\endgroup\$
  • 2
    \$\begingroup\$ A reference implementation is not a specification. \$\endgroup\$ – Peter Taylor Apr 26 '17 at 16:20
  • \$\begingroup\$ Yeah there are people who don't speak C. \$\endgroup\$ – Erik the Outgolfer Apr 26 '17 at 16:50
  • \$\begingroup\$ @Peter, do you have any suggestions on how I can create a specification for this? Do you think people will be interested or is this a dead end? \$\endgroup\$ – Johan du Toit Apr 26 '17 at 16:58
  • 1
    \$\begingroup\$ 1. Define the function: what are the inputs? What does the output mean? On the basis of some quick Googling I understand that it's an offset inverse of erf? But there seems to be disagreement about how many input parameters it takes. 2. Be explicit about the numerical analytic requirements. I'm not sure what "Loss of precision is also not allowed" means: that answers must avoid subtractions of values of similar magnitude? 3. With respect to interest, the thematically related codegolf.stackexchange.com/q/9070/194 got a few answers. \$\endgroup\$ – Peter Taylor Apr 27 '17 at 10:43
0
\$\begingroup\$

Find the Calling Pattern

I've written a program to add two numbers a and b:

f=lambda:lambda a:lambda b:lambda:a+b

This has a very specific calling pattern. It must be called as:

f()(a)(b)()

In this case, it was fairly clear, but sometimes the calling pattern can be more complicated:

f(a,None)()("","","")([0,0],b)(())()

In essence, this challenge is about the cops trying to obfuscate this calling pattern, and the robbers trying to find it.

Rules for Cops:

Calling patterns:

  • The function must only need to be called up to 5 times to give the answer.
  • For the purposes of this challenge, a false-unit is None, "", [], 0, or ().
  • A valid argument to the curried function in the calling pattern is either:
    • a false-unit,
    • a list or tuple containing one to five false units, or
    • one of the actual numbers to be added (represented by a and b).
  • Each time it is called, it should only take up to five arguments.
  • The calling pattern can change based on what arguments are passed to it. Be creative!
  • a and b must each appear once in the calling pattern.

Examples of valid calling patterns:

f(a,b)
f("",a)(b,[0,0])
f(a)("",(),[None],[0,()])(b)()

Examples of invalid calling patterns:

f(a,b,0,0,"",[]) # too many arguments
f(a,[(),[],"",(),[],""])(b) # too many elements in [(),[],"",(),[],""]
f(a) # b is not passed
f([a,b]) # a and b may not be contained in a list
f((0,[None,""]),a)(b) # [None,""] is not a false-unit

For your submission, you must post a lambda that, when called according to your calling pattern, evaluates to a + b.

You can assume that a and b are both positive and that a + b < 2147483647 (the highest signed 32-bit integer).

Your submission is safe if it is not cracked by a robber in one week.

Rules for Robbers

To crack a cop's submission, you must post a submission which takes two inputs and returns a string showing how to call the cop's submission to coerce it into adding your two inputs. For example, if this was a cop's submission:

lambda:lambda x,z:lambda y:x+(y*(z==0))

This could be your crack:

lambda a, b: "f()(%s, 0)(%s)" % (a, b)

Sandbox Notes

  • Is there anything I should clarify? Should I add more examples?
  • Should I try to make this language agnostic or restrict it to Python?
  • Is this too weighted towards one side (is it too easy for the cops or the robbers?)
  • Are there any additional restrictions that could make it more fair?
\$\endgroup\$
  • 1
    \$\begingroup\$ Can you give an example of a cops submission that might be hard to crack? \$\endgroup\$ – Leaky Nun Apr 26 '17 at 5:37
  • \$\begingroup\$ Without a language restriction, some may call Currying a non-observable requirement. \$\endgroup\$ – ATaco Apr 26 '17 at 5:41
  • \$\begingroup\$ @ATaco Which is why I was thinking of restricting it to Python. \$\endgroup\$ – Esolanging Fruit Apr 26 '17 at 5:44
  • \$\begingroup\$ often, after a certain time (a week, for example,) A cop's submission is considered "safe", and they reveal their solution (if they want). You may want to mention/implement something like this \$\endgroup\$ – MildlyMilquetoast Apr 26 '17 at 5:46
  • \$\begingroup\$ @MistahFiggins Added. \$\endgroup\$ – Esolanging Fruit Apr 26 '17 at 6:09
  • \$\begingroup\$ I don't think the example works. \$\endgroup\$ – Peter Taylor Apr 27 '17 at 13:28
  • \$\begingroup\$ @PeterTaylor Edited. \$\endgroup\$ – Esolanging Fruit Apr 27 '17 at 15:17
  • \$\begingroup\$ That's still broken except in the special case that y == 0. \$\endgroup\$ – Peter Taylor Apr 27 '17 at 15:37
0
\$\begingroup\$

To Polar and Back

Background

There are 2 main ways to represent a single point on a plane:

  • in Cartesian form, with an x and y value,
  • and in Polar form, with an angle and a magnitude

For example, the point (5, 5) can also be represented with the angle π / 4 and magnitude 5 * sqrt(2)

In this challenge, we will be using radians, and not degrees.

Conversion

This challenge deals with the mapping of Polar form onto a Cartesian grid, by using the angle as the x value, and the magnitude as the y value.

For example, to change the point (sqrt(3), 1) into Polar form and back, we first convert it into polar form, with an angle of π / 6 and magnitude of 2.

Then, these values are mapped into Cartesian form, yielding the Cartesian point (π/6, 2).

The Challenge

Create a program or function that, given a rectangle in Cartesian form, outputs it's area after it has been converted to Polar form and mapped back onto the Cartesian plane by equating angle with x and magnitude with y.

Input

The input will be a rectangle in Cartesian form whose legs are parallel to the x and y axes. Also, all 4 corners will be in the first quadrant, meaning that all x and y values will be strictly positive.

This means that you may take

  • The Cartesian coordinates of all 4 corners
  • The Cartesian coordinates of opposite corners
  • The Cartesian coordinates of one corner, and a width and height value

or some other reasonable form, such as a rectangle object. The important thing is that you don't take Polar coordinates as input.

You can take the points or values in any order you choose.

Please include the input format you used in your answer.

Output

Output will be a single, decimal number rounded to at least the hundredth's place (2 digits after the decimal).

Explained Example

Note: For the sake of clarity, I will use symbols like π or sqrt(3) instead of their decimal equivalents.

Also keep in mind that this is probably one of many ways to go about this problem, and it doesn't matter what method you use to find the area as long as you do. This example is mostly to explain the above by showing it in action.

Anyways, here is out input, in the form of 4 points:

(1, sqrt(3)), (4, sqrt(3)), (1, 1), (4, 1)

First, we might convert each point into polar form, and map that back onto the Cartesian plane:

(x, y)       -> (angle, magnitude)
(1, sqrt(3)) -> (π/3, 2)
(4, sqrt(3)) -> (arctan(sqrt(3)/4), sqrt(19))
(1, 1)       -> (π/4, sqrt(2))
(4, 1)       -> (arctan(1/4), sqrt(17))

This should be relatively easy using trigonometry and the Pythagorean theorem.

These new Cartesian points give us the corners of our new shape, which looks something like this:

courtesy of http://www.meta-calculator.com

Notice that the sides are not straight, meaning you will have to figure out how to get the equations for the sides as well as how to find the area. I'll leave that up to you.

In any case, the area of this shape is about 0.81810689, but you only need to print 0.82 (without the leading 0, if you prefer).

Other Test Cases

The input will be taken in the form of 4 points, in reading order (left to right, then top to bottom)

input -> output
(1, 1.732), (4, 1.732), (1, 1), (4, 1) -> 0.82

Sandbox: I will add more test cases later - any suggestions about certain edge cases I should include would be helpful.

Scoring

This is , so the shortest answer in bytes wins.

Standard loopholes, as always, apply.

Sandbox

  • Any name ideas?
  • Any tags I should add/remove?
  • Suggestions/help with creating test cases?
  • Is the explanation sufficient?
  • Should I change the accuracy requirement to 3 digits after the decimal point, or would that be too hard?

Thanks!

\$\endgroup\$
  • \$\begingroup\$ I'm not quite sure I understand how the edges should be curved. Additionally, I think you should require the program to use the maximum precision that the language can handle, but I'm not sure how well that would work. \$\endgroup\$ – user42649 Apr 27 '17 at 4:30
  • \$\begingroup\$ @HyperNeutrino The edges curve because more points on the straight line lie in a given angle when the segment looked at is farther from the point closest to the origin. In other words, as the line gets farther from the axes, the line gets closer to parallel with a ray extending from the origin to the line. This means that the line on the new graph will be closer to vertical when it gets closer to the axis (at angles 0 or π/2, depending on whether the original line is vertical or horizontal). Hope this helps! Also, not sure about the accuracy yet, so I'm going to leave it as it is for now. \$\endgroup\$ – MildlyMilquetoast Apr 27 '17 at 4:52
  • \$\begingroup\$ Oh, I see. That makes sense. And yeah, I'd do what you're doing and see if other people make suggestions on the accuracy aspect. \$\endgroup\$ – user42649 Apr 27 '17 at 4:53
  • \$\begingroup\$ "Create a program or function that, given a rectangle in Cartesian form, output it's area after it has been converted to polar form and back." Since the conversion is a bijection, converting to polar form and back leaves it unchanged, so it's just a case of finding the area of the rectangle. That doesn't seem to be what the example does, though, so I have to say that the spec is not clear. \$\endgroup\$ – Peter Taylor Apr 27 '17 at 10:24
  • 1
    \$\begingroup\$ @PeterTaylor You're right, this is poorly worded. It should say something like "given a rectangle in Cartesian form, output it's area after it has been converted to Polar form and mapped back onto the Cartesian plane by equating angle with x and magnitude with y" \$\endgroup\$ – MildlyMilquetoast Apr 27 '17 at 15:32
0
\$\begingroup\$

Get the Decimal! (Posted)

\$\endgroup\$
  • \$\begingroup\$ Rounding on the digit, or return the exact digit? (important distinction if people can generate out to x and then take the last digit, for example) \$\endgroup\$ – AdmBorkBork Apr 21 '17 at 15:11
  • 1
    \$\begingroup\$ Please delete this now that it is posted. \$\endgroup\$ – programmer5000 Apr 27 '17 at 17:24
0
\$\begingroup\$

Electron Configurations: Orbitals

Challenge

Given an elements atomic number as input, output its short electronic configuration.

Explanation

http://www.chemguide.co.uk/atoms/properties/elstructs.html

To come

Examples

Sodium: 11

[Ne] 3s1

Hydrogen: 1

1s1

Helium: 2

1s2 

or

[He]

Vanadium: 23

[Ar] 4s2 3d3

Selenium: 34

[Ar] 4s2 3d10 4p4

Winning

Shortest code in bytes wins

\$\endgroup\$
  • \$\begingroup\$ Do we have to worry about unstable isotopes undergoing Beta Decay? ;) Nice challenge btw! \$\endgroup\$ – user42649 Apr 27 '17 at 16:57
  • 1
    \$\begingroup\$ codegolf.stackexchange.com/q/37657/194 \$\endgroup\$ – Peter Taylor Apr 27 '17 at 21:06
0
\$\begingroup\$

Variably Incremented Chained Ranges

Your task will be, given a well-formed input string s, to parse it as a range, with special properties:

  1. A traditional range is defined as a..b = [a, a+1, .., b-1, b].
  2. An incremental range is an extension of the traditional range in the format a.c.b.
    • The format above uses a and b in the traditional sense, but it will use c as the step.
    • That is to say 1.2.5 = [1,3,5] and a.c.b = [a, a+c, .., b iff b-a is divisble by c]
    • This means that 0.2.5 = [0, 2, 4] because b-a % c != 0.
  3. The ranges are chain-able, meaning that a..b..c..d..e..f..g..h..i is valid.
    • Always should be handled left-to-right.
    • The chain should always be continued from the end of the last range.
    • 0.2.5..3 is not [0,2,4,5,4,3] it is [0,2,4,3].
  4. The format of the range may be a.c.b or a..b, absence of c means c = 1.

I & O Gaurantees

  • c <= difference(b, a), you will never see something like 1.100.2.
  • a and b are positive integers or 0, this makes crap easier
  • Notice, however, that neither a > b nor a < b are guaranteed, both are acceptable.
  • a, b and c will always be integer values, floats aren't supported.

This is , lowest byte count wins.

\$\endgroup\$
0
\$\begingroup\$

Give counterexamples to thesis that prime numbers are finite.

It is known since antiquity that there are infinite prime numbers, first proved by Euclid. An unorthodox yet brief formulation of this proof was recently posted on MathExchange:

There are infinitely many primes, if not, multiply all and add 1.

Challenge

Given a prime number N, output a prime number created by multiplying it with all lower primes and adding one, i. e. (2*3*5*...*N) + 1. In other words, if someone claimed that N is the largest finite number, you have to give a counterexample Euclid would likely compute to prove them false.

You don't have to consider input number that is not a prime.

Your representation of a number must have a range of 32-bit signed integers or larger.

You can use floating point representation as long as your code uses int-specific operations (e.g. no double-specific bit-fu).

Shortest code wins, standard loopholes forbidden. You may use a lookup table of primes.

Example Input and Output

2 → 3          // because             2 + 1 = 3
3 → 7          // because           3*2 + 1 = 7
5 → 31         // because         5*3*2 + 1 = 31
13 → 30031     // because 13*11*7*5*3*2 + 1 = 30031
31 → 223092871 // (largest possible output within `int32` range)

Note

This is very similar to, yet not identical with A006862: your solution must output a number belonging to that sequence. The difference is that your solution must work for inputs that are primes, while A006862 recognizes any natural number (including 0).

If your solution is function f(), the following equivalence must be true:

A006862(n) = M <=> f(A000040(n)) = M

A000040 is the sequence of primes, therefore A000040(n) can be read as "n-th prime number".

\$\endgroup\$
  • \$\begingroup\$ In its current state I would have to vote to close this as unclear what you're asking. 1. What should the output be for input 13? 2. Why is the output for 31 not 200560490131? 3. How does it differ from A006862? If these issues are fixed then it likely becomes a duplicate of codegolf.meta.stackexchange.com/a/11638/194 \$\endgroup\$ – Peter Taylor Apr 29 '17 at 18:20
  • \$\begingroup\$ @PeterTaylor, I hope my edit clarifies your doubts. \$\endgroup\$ – Szymon Apr 29 '17 at 22:50
  • 2
    \$\begingroup\$ Do you realise that 30031 = 59 * 509 and is therefore not prime? \$\endgroup\$ – Peter Taylor Apr 29 '17 at 22:59
  • \$\begingroup\$ Oh, that's a game changer: the challenge is obviously still valid, but now it has little meaningful interpretation. I guess I'll leave it abandoned for future reference of other people not understanding Euclid's theorem. \$\endgroup\$ – Szymon Apr 29 '17 at 23:21
0
\$\begingroup\$

Rewrite a number in the Munafo System

Output

The Munafo Alphabetic PT System (created by Robert Munafo) is a system of concisely writing very large numbers such that ordered values become lexicographically ordered when written. A full description (and some examples) of the system can be found in the link, but here's a series of steps to compute the string M(N) from a positive real number N:

  • If N < 10, then M(N) is simply the single digit of N, possibly followed by an factional part; that is, _ followed by any digits after the decimal place of N.
  • If 10 ≤ N < 100, then M(N) is the letter a followed by the two digits of N, possibly followed by an factional part.
  • If 100 ≤ N < 1000, then M(N) is the letter b followed by the three digits of N, possibly followed by an factional part.
  • If 1000 ≤ N < 10 000, then M(N) is the letter c followed by the four digits of N, possibly followed by an factional part.
  • If 10 000 ≤ N < 100 000, then M(N) is the letter d followed by the five digits of N, possibly followed by an factional part.
  • If 100 000 ≤ N < 10300, then M(N) is:

    1. the letter e
    2. exactly three digits, representing the exponent of N
    3. an underscore _
    4. any number of digits, representing the significand of N in standardized form.


    More specifically, a string M(N) = ex1x2x3_y1y2y3... represents the value N = y1.y2y3... × 10x1x2x3.

  • Finally, if 10300N, then to determine M(N), we must first apply the following recursive process:

    1. Set T = 0 and R = N.
    2. If R < 10300, then break out of this process.
    3. Otherwise, set T = T + 1, set R = log10(R), and return to step 2.


    (Effectively, this sets N equal to 1010R with T nested exponents.) Now, with our values of T and R, the value of M(N) is written as follows:

    1. the letter e
    2. the substring M(T)
    3. an underscore _
    4. the substring M(R).

...and that's the entire system. I apologize if anything I said was unclear or vague; in that case, you should definitely check the link for more detail.

META: Is there anything here that can be clarified?

Operation

As input, your program or function is to take three nonnegative numbers X, Y, and Z. X and Z are guaranteed to fit in your language's default integer type, and Y can fit in the default floating point type. These three values represent the number N = 10Y × 10Z, with X initial exponents as before. From here, the challenge is simple: given X, Y, and Z describing a value N, output the Munafo string M(N).

Test cases

META: Still figuring these out. Will probably end up using at least a couple from the link. Suggestions are welcome.

\$\endgroup\$
0
\$\begingroup\$

Same Same But Different

This challenge involves two similar digit-based routines each of which, when given a non-negative integer seed, produce eventually periodic sequences.
The goal is to find the nth seed number (counting up) that does not define the two sequences to be the identical (by virtue of the order of its digits), yet does yield sequences of equal pre-periodic length.


Reverse & Absolute Difference Routine.

A reverse and difference sequence may be defined in any given base for a non-negative seed by iteratively taking the absolute difference between the number and itself with its digits reversed.

An example:

The number 54410 is 10001000002;
The reverse of which is 00000100012, which is 1710;
The difference between the two is 52710;
Repeating the process we have 52710 = 10000011112, reversed is 11110000012 = 96110, and a difference of 43410;
Repeating again, 43410 = 1101100102, reversed is 0100110112 = 15510, and a difference of 27910;
Repeating again yields 18610, then 9310, and then 0, which will continue to yield 0.

  • So, the pre-periodic reversal and difference base-2 trajectory length of 54410 is six.
    - since the iterative procedure may be performed six times before a periodic part, {0}, is reached:

54410, 52710, 43410, 27910, 18610, 9310, {0}

  • Similarly the pre-periodic reversal and difference base-2 trajectory length of 9310 is one
    ...and the pre-periodic reversal and difference base-2 trajectory length of 0 is zero (as it is in any base).

Note: Ignoring the absolute part of this definition, many low bases have their iterations listed at the OEIS: base-2, base-3, base-4, base-5, base-6, base-7, base-8, base-9, base-10, base-11, base-12, base-16, base-20.


Kaprekar Routine.

A similar, but different sequence is the Kaprekar sequence where the absolute difference above is replaced by the difference, and the reversal is preceded by a digit-sort.

An example:

The number 83638110 is CC31D16;
Which, reverse-sorted is DCC3116, which is 90424110;
And forward-sorted is 13CCD16, which is 8110110;
The reverse-sorted less the forward-sorted is then 82314010;
Repeating the process we have 82314010 = C8F6416, reverse-sorted is FC86416 = 103434010, forward sorted is 468CF16 = 28897510, and the subtraction yields 74536510;
Repeating again yields 67983010, then 67167010, then 80682010, then 75352510, then 67167010...

  • Notice that we have hit a periodic part of the sequence, {67167010, 80682010, 75352510}, so the pre-periodic Kaprekar base-16 trajectory of 83638110 is four:

83638110, 82314010, 74536510, 67983010, {67167010, 80682010, 75352510}

Note:: Once again the OEIS lists the iterations for some low bases: base-2, base-3, base-4, base-5, base-6, base-7, base-8, base-9, base-10.
Furthermore some pre-periodic sequences also exist: base-2, base-3, base-4, base-5, base-6, base-7, base-8, base-9, base-10.


Same same...

Any number whose digits are already either forward or reverse sorted in some base will have the same pre-periodic trajectory length for both sequences in that base, since the resulting procedure will be identical.

There are, however, numbers which have the same pre-periodic trajectory length in both sequences even though their digits in the base in question are not already sorted.

The first example in base-10 is 107 which has a pre-periodic trajectory length of three in both sequences:

Reverse & difference: 107, 594, 99, {0}
Kaprekar: 107, 693, 594, {495}

Aside: Any number with less than three digits will always be sorted or reverse-sorted, as will the first three digit number, so the only others to actually check in base-10 before finding 107 would be [102-106].


The challenge

Write a program or function which when provided with a positive* integer, n, and an integer base greater than one**, b, outputs the nth number (in the natural ordering) which has neither sorted nor reverse-sorted digits in the given base, b, but which does have the same pre-periodic trajectory length for both procedures defined above in the given base.

The input and output are as flexible as ever, however the input and output numbers (or string or list representations thereof) are to be in a single unchanging base, consistent between input and output - for most this will probably mean base ten, but any single base is acceptable, as long as it is consistent across any invocations of the code.

* Feel free to have the counting (i.e. n) be 0-based if you would prefer, just say in your submission.

** Note there are no such numbers in base-1, so the code does not need to handle b=1 as an input.


Test cases (one-based n, IO in base-10)

 n     b  result
 1     2      17
 2     2      33
 3     2      51
 4     2      65
 5     2      73
 6     2      85
17     2     297

 1     3      28
 2     3      29
 3     3      32
 4     3      48
 5     3      51
 6     3      55
17     3     101

 1    10     107
 2    10     160
 3    10     161
 4    10     172
 5    10     186
 6    10     187
17    10     329

 1    16     264
 2    16     266
 3    16     355
 4    16     373
 5    16     400
 6    16     401
17    16     522

 1   257   66179
 2   257   98949
 3   257   98951
 4   257   98953
 5   257   98955
 6   257   98957
17   257   98979

 1  1234 1523375
 2  1234 2242181
 3  1234 2243417
 4  1234 2244655
 5  1234 2245897
 6  1234 2247147
17  1234 2260770
\$\endgroup\$
0
\$\begingroup\$

Formula 1 2016 standings

The Formula 1 2017 season is well underway, but I thought that the final standings could be good idea for a challenge.

The task is to print the standings at the end of 2016 season. This is how the table looks like:

1  N Rosberg    385
2  L Hamilton   380
3  D Ricciardo  256
4  S Vettel     212
5  M Verstappen 204
6  K Raikkonen  186
7  S Perez      101
8  V Bottas     85
9  N Hulkenberg 72
10 F Alonso     54
11 F Massa      53
12 C Sainz      46
13 R Grosjean   29
14 D Kvyat      25
15 J Button     21
16 K Magnussen  7
17 F Nasr       2
18 J Palmer     1
19 P Wehrlein   1
20 S Vandoorne  1
21 E Gutierrez  0
22 M Ericsson   0
23 E Ocon       0
24 R Haryanto   0

The table is divided into 3 columns: position at the end of season, driver name and points. In each of the columns, the contents can be aligned to left or to right for each column independently. If all columns are aligned to right, the output will look like this:

 1    N Rosberg 385
 2   L Hamilton 380
 3  D Ricciardo 256
 4     S Vettel 212
 5 M Verstappen 204
 6  K Raikkonen 186
 7      S Perez 101
 8     V Bottas  85
 9 N Hulkenberg  72
10     F Alonso  54
11      F Massa  53
12      C Sainz  46
13   R Grosjean  29
14      D Kvyat  25
15     J Button  21
16  K Magnussen   7
17       F Nasr   2
18     J Palmer   1
19   P Wehrlein   1
20  S Vandoorne   1
21  E Gutierrez   0
22   M Ericsson   0
23       E Ocon   0
24   R Haryanto   0

The columns should be separated exactly by 1 space. Trailing spaces are allowed as well as 1 trailing newline.

This is competition - shortest answer wins.

\$\endgroup\$
  • \$\begingroup\$ Thanks for posting in the Sandbox before posting on the main site ;-) In its current state, the challenge seems like it's just about finding the shortest way to encode all the drivers' names and how many points each had at the end of the season. While this may be an interesting challenge for some, many users on the site may find it fairly boring with not much room for real innovation. (You can tag it kolmogorov-complexity, btw) \$\endgroup\$ – ETHproductions May 2 '17 at 15:30
  • \$\begingroup\$ While reading this, I had an idea for a challenge where you take in a driver's last name and output the three-letter abbreviation, but that would be rather boring because that's currently just the first three letters of each name... \$\endgroup\$ – ETHproductions May 2 '17 at 15:34
  • \$\begingroup\$ @ETHproductions That could be more interesting with MotoGP or Superbike drivers \$\endgroup\$ – AlexRacer May 2 '17 at 16:54
0
\$\begingroup\$

Golf me a golf-lang

In any programing language of your choice write me an interpreter (or compiler targetting a reasonable pre-existing binary) for a golfing language. You will be scored based on a combination of the length of the interpreter and the golfiness of programs in your (user created language). In particular your goal will be to minimize the following formula.

max((Interpreter Length in bytes),512)/16-25*("checkmarks")

A good answer will not only contain an interpreter, but also contain code samples and documentation to encourage use of your language.

The number of "checkmarks" a language has earned is the number of times an answer in your language has beat or tied with the best answer (out of languages which are submissions to this challenge) to an open, positively voted (score of +1 or higher) question (whatever the win criterion) which postdate your language and were not asked by its creator(s) (in the event of a community wiki submission).

Your language will now be in a meta-competition around stack exchange, and you can earn checkmarks if other people choose to win other challenges with your competition.


Note that use of eval (or similar) is banned. You may write your interpreter/compiler in any preexisting programming language, and your language that you implement may be as similar as you wish to currently existing programming languages.

\$\endgroup\$
  • \$\begingroup\$ I'm worried that crafting a good solution to this will take a lot of time (thus meaning that late answers are fairly common) – it is language-design, after all – but the victory condition means that there's a hard limit to how long answers can arrive. Perhaps the languages should be somehow tested on how they fare on questions that postdate the languages themselves? \$\endgroup\$ – user62131 Apr 12 '17 at 21:23
  • \$\begingroup\$ If a challenge is selected which a language can't implement (e.g. one which requires a GUI, but the language only does I/O via stdio), how is it scored? \$\endgroup\$ – Peter Taylor Apr 12 '17 at 21:27
  • \$\begingroup\$ @PeterTaylor I decided to go a completely different way, challenges your language cant complete neither hurt or help you. \$\endgroup\$ – Rohan Jhunjhunwala Apr 12 '17 at 22:26
  • \$\begingroup\$ @ais523 I restructured the challenge, I liked the idea of being tested based on how you fare on languages that post date yourself, this way later answers are at a disadvantage, but they can still compete (and win). \$\endgroup\$ – Rohan Jhunjhunwala Apr 12 '17 at 22:27
  • 2
    \$\begingroup\$ There are plenty of good reasons to not award an accepted answer tick (e.g. some challenges have a reason to preserve the sort order), so not all challenges will have one. Change that to count questions on which your language has the best answer or tied for the best answer (based on the challenge's victory condition), and I like the new scoring. \$\endgroup\$ – user62131 Apr 12 '17 at 22:29
0
\$\begingroup\$

How many times will you have to golf a quine?

For this challenge, you must create a programme which takes an integer x and outputs its source x many times.

Rules

  • This is codegolf, the programme containing the least amount of bytes wins.

  • If you submit a function, the function must take x as a parameter and either return or print the body of the function x many times to STDOUT

  • If you submit a lambda, it is not required for you to assign it to a variable

  • Standard loophole restrictions apply.

  • Empty programmes are forbidden

\$\endgroup\$
  • \$\begingroup\$ There's really no reason not to use code formatting for x. The image doesn't even load for me, I had to check the markdown to see what you're trying to say. \$\endgroup\$ – Rɪᴋᴇʀ May 2 '17 at 17:53
0
\$\begingroup\$

Steal my credit card

recently a lovely man named Boris that I met online sent me a copy of an album that I didn't even know was released yet, the 16kb '.exe' ('EXtremely Excellent-music' apparently) file didn't open, but I did notice that it had managed to make some charges against my credit card, now I'm normally very secure in my banking, I don't write passwords down at all, and I keep my bank details in a '.txt' (Truly eXceedingly proTected) file, so I'm not sure how the hackers managed to get access, to confirm how secure I am:

  • I keep a .txt file named 'info.txt' in my documents folder
  • There's lots of unimportant information in this
  • I store the credit cards in a specific format
  • All of my credit cards have the same number of digits in the number and cvv
  • the 'Type' of card may be different, but it will always be in the format Cardtype - MM/YY
  • I don't format anything else in a similar manner, you can assume that this is the only two lines together.

Example:

Visa - 10/18
4024007172038209/571

(not a real credit card, i'm not an idiot)

which corresponds to:

Type - MM/YY (Expiry date)
CREDIT_CARD_NUM/CVV

The format will always match:

[a-zA-Z]+ - \d{2}/\d{2}\r?\n\d{16}/\d{3}

so it's very difficult to be read by any automatic parsing program.

my insurance doesn't believe me, so I'm here to prove that it's basically impossible, and since we all know programmers are actually just hackers with day jobs, I figured you would all be able to help me.

after adding all of the default Java libraries together and doing some math I believe there's almost no room left for any actual code in the exe file he sent me, so please try and keep the program as short as possible.

The Challenge:

write a program which takes no arguments, and when run will:

  • read in a file named 'info.txt' from the same directory the program is run in.
  • somewhere in the file, two lines will correspond to the above format
  • return only those two lines
  • any amount of leading or trailing white-space is allowed.
  • you can assume one credit card per file
  • you do not need to do any validity checks on the card number etc.

attached is my original info.txt if you want to test against it, but it should work with any credit card number, that way it will be more believable.

[gdrive link to info.txt]

This is so shortest code in each language wins.

Any code, interpreted or compiled, on any system, in any language is valid.

\$\endgroup\$
  • \$\begingroup\$ It's probably best that the victory condition isn't mentioned in just the fluff at the start; I missed it first time I read this. You probably need to confirm what counts as a credit card number or as a CVV. Given the talk about .exe files in the fluff, you should also confirm what languages are allowed (it's likely best if you allow any language, even interpreted languages). \$\endgroup\$ – user62131 May 5 '17 at 12:06
  • \$\begingroup\$ @ais523 added some more clarification there, and a regex to match the two lines. \$\endgroup\$ – colsw May 5 '17 at 12:23
  • \$\begingroup\$ Minor correction: the second Z in the regex should be capitalized (should be [a-zA-Z] not [a-zA-z]) \$\endgroup\$ – Business Cat May 5 '17 at 13:56
  • \$\begingroup\$ @BusinessCat oopsie, thanks. \$\endgroup\$ – colsw May 5 '17 at 15:14
0
\$\begingroup\$

Is the line closed?

Given input consisting of only whitespace characters and -/\|, tell if the line created by -\/| is closed.

For this challenge, \ points one character up and to the left, as well as one character down and to the right. / points up and to the right, and down and to the left. | points one to one character above and below it. - points one character to the left and right of it.

# means it is being pointed to.

#
 \
  #

  #
 /
#

#-#

#
|
#

A line is a bunch of /\-| characters that point to each other.

     /---\
\    |   |
 \---/   \

The characters - and | cannot validly point to each other. Similarly, / and \ cannot point to each other either.

-| = INVALID

|
-  = INVALID

\
 / = INVALID

 /
\  = INVALID

A closed line is one where each of its characters validly point to another of its characters.

/--\
|  |
\--/ = CLOSED

\--\
|  |
\--/ = NOT CLOSED

---- = NOT CLOSED

Your task, as said above, is to write a program that says if a line is closed.

Test cases coming soon.

\$\endgroup\$
  • \$\begingroup\$ Your -/ connections violate your own specification. I think the clearest specification would be to enumerate all possible legal pairs of connected line segments. \$\endgroup\$ – user62131 May 8 '17 at 13:47
  • \$\begingroup\$ @ais523 My mistake, I will fix both this and the other. \$\endgroup\$ – Comrade SparklePony May 8 '17 at 14:14
0
\$\begingroup\$

Determine if a relation is an equivalence relation

(Content to be added.)

Sandbox

\$\endgroup\$
0
\$\begingroup\$

Uncompress a permutation

I have a number of permutations, of which this is an example:

[0] 1 2 3 15 14 26 27 28 29 30 31 32 33 5 4 16 17 18 19 20 21 22 23 35 34 6 7 8 9 10 11 12 13 25 24 36 37 38 39 [40]

(0 is used for 0-indexed permutations and 40 for 1-indexed permutations.) In order to save space, I want to compress the permutation by reducing each ascending run of consecutive numbers to its first element, giving:

0/1 15 14 26 5 4 16 35 34 6 25 24 36

This compression method is invertible, because each missing element from the compressed form must follow its predecessor among the integers.

Your challenge is to write a program or function that takes a compressed permutation as input and returns an uncompressed permutation. Your input will consist of an array or list or other reasonable type representing the compressed permutation, plus an integer number of elements in the uncompressed permutation. Examples:

[0], 2 -> [0, 1] (0-indexed)
[1], 2 -> [1, 2] (1-indexed)
[6, 2, 0], 8 -> [6, 7, 2, 3, 4, 5, 0, 1] (0-indexed)
[7, 4, 1], 9 -> [7, 8, 9, 4, 5, 6, 1, 2, 3] (1-indexed)

In order to make the compression worthwhile I obviously need as short a decompression routine as possible, so this is a challenge.

\$\endgroup\$
  • \$\begingroup\$ I don't think there's actually a specification of the decompression routine, and the one which I would reverse engineer from the examples at the end is not compatible with the first example. \$\endgroup\$ – Peter Taylor May 8 '17 at 14:34
  • \$\begingroup\$ I believe the specification is just "for each number that doesn't appear in the compressed permutation, place it to the right of that number minus 1". It might be worth specifying that in the question. \$\endgroup\$ – user62131 May 8 '17 at 15:11
  • \$\begingroup\$ @PeterTaylor The specification is "Create a permutation of the first n integers which, when compressed by removing consecutive numbers after the first, results in the input array/list". \$\endgroup\$ – Neil May 8 '17 at 18:33
  • \$\begingroup\$ I've made an edit which I think makes that clearer. \$\endgroup\$ – Peter Taylor May 8 '17 at 21:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .