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

Sandbox FAQ

Posting

To post to the sandbox, scroll to the bottom of this page 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, though you can optionally add a title at the top. 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, and replace the post here with a link to the challenge and delete the sandbox post.

Discussion

The purpose of the sandbox is to give and receive feedback on posts. If you want to, feel free to give feedback to any posts you see here. Important things to comment about can include:

  • Parts of the challenge you found unclear
  • Comments addressing specific points mentioned in the proposal
  • Problems that could make the challenge uninteresting or unfit for the site

You don't need any qualifications to review sandbox posts. The target audience of most of these challenges is code golfers like you, so anything you find unclear will probably be unclear to others.

If you think one of your posts requires more feedback, but it's been ignored, you can ask for feedback in The Nineteenth Byte. It's not only allowed, but highly recommended! Be patient and try not to nag people though, you might have to ask multiple times.

It is recommended to leave your posts in the sandbox for at least several days, and until it receives upvotes and any feedback has been addressed.

Other

Search the sandbox / Browse your pending proposals

The sandbox works best if you sort posts by active.

To add an inline tag to a proposal, use shortcut link syntax with a prefix: [tag:king-of-the-hill]. To search for posts with a certain tag, include the name in quotes: "king-of-the-hill".

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2
  • \$\begingroup\$ What if I posted on the sandbox a long time ago and get no response? \$\endgroup\$
    – None1
    Commented May 15 at 14:05
  • \$\begingroup\$ @None1 If you don't get feedback for a while you can ask in the nineteenth byte \$\endgroup\$
    – mousetail
    Commented May 29 at 13:27

4769 Answers 4769

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posted

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1
  • 1
    \$\begingroup\$ You format inline MathJax here using a backslash before the dollar sign, like this: \$. I have edited your answer to implement this. \$\endgroup\$ Commented Aug 26 at 14:08
0
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Delete every other line from file

Challenge

Delete every other line from file

Example

Input file:

1
2
3
4
5

Output:

1
3
5

Input file:

aa
bbb
cccc
ddddd

Output:

aa
cccc

Rules

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0
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Find the biggest simplex

An n-simplex is a generalization of 'triangleness' in any dimension (specifically, it is the simplest shape requiring n dimensions). Starting with 0 dimensions, the named simplexes are: point, line segment, triangle, tetrahedron, and pentachoron.

If you have a simplex, say this triangle for example:

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

there are many sets of points which are also simplexes:

examples of 1-simplexes in this shape:
abdgk
bm
aem
kmo
hec
ojf
examples of 2-simplexes in this shape:
deh
fijmno
hlm
bgi
cgn
adfkmo

Your challenge for this question is to take in a list of points, and output the size of the most complex simplex that could be made out of them. Complex, for this challenge, is ordered first by dimension (so abc [2 dimensions] is more complex than acfjo [1 dimension]) and then by side length (so abcdef [side length 3] is more complex than abc [side length 2])

It is easiest to explain things with a concrete example to reference, so the remainder of this question will reference this tetrahedron:

   a
  ---
   b
  c d
 -----
   e
  f g
 h i j
-------
   k
  l m
 n o p
q r s t

(imagine the layers stacked like a pyramid of cannonballs, with a on top)

Inputs

Your program is given a list of points, where each point is defined as a list specifying each of the 'layers' which the point is on. For example, [1,1,1] would be point a, [3,2,1] would be point f, or [2,2,2] would be point d. Referencing the picture, the first number represents the distance along the abek direction, the second along the klnq direction, and the third along the qrst direction. Note that, because of how simplexes are structured, this is a strictly positive decreasing list; if you would like to 0-index instead, or take the input in increasing order, you may.

Outputs

You should output both the dimension and side length of the most complex simplex you are able to make out of the given points. You may output these two numbers in any recoverable format, but the examples used in this question will use n as the dimension number and l as the side length. Note that any single point is a valid n=0,l=1 simplex, and any pair of points is a valid n=1,l=2 simplex.

Validity

There will always be at least one point given, and all points will be lists of the same length. All points of a simplex must be evenly spaced in all dimensions.
#TODO: explain how to find simplexes
#note on changes: specifying the points like this means you don't need a structure, since the dimension is the length of the lists and the side length is just the largest number, so I have removed that input.

Examples

#TODO

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3
  • \$\begingroup\$ Maybe I'm just bad at this specific type of thing, but in terms of challenge complexity: merely testing if a set of points is an evenly spaced upright subtriangle of a 2D input triangle in any reasonable I/O format would take me a good while to make an ungolfed solution to, and that's significantly more simplified than what this challenge is asking for. I don't really see a lot of ways in which adding simplifications could outright trivialize the challenge. \$\endgroup\$ Commented Sep 4 at 1:11
  • 1
    \$\begingroup\$ @shapewarriort I was thinking it would trivialize the challenge to much because the way I would approach this would basically be to create that last input format and then look for arithmetic sequences; so taking input like that would solve like half the problem for the golfer. I have been thinking about simplexes for a while now though, so it might just be that I am overestimating the average code golfer's familiarity with geometry. I think I'll rephrase the question so that it's deliberately in that last format, and explicitly mentions arithmetic sequences. \$\endgroup\$
    – guest4308
    Commented Sep 4 at 2:11
  • 1
    \$\begingroup\$ @shapewarriort also, thanks for mentioning 'upright', it looks like I forgot to include any rotated triangles in my first example \$\endgroup\$
    – guest4308
    Commented Sep 4 at 2:12
0
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Raster peekaboo - do you see me or not?

The input is any kind of square raster representation, any side-length (within reasonable* limits), it can be a csv, a string, a bmp, ...;

The input contains the following distinct entities (representations defined by the solution, BYOP-Bring your own protocol, the input protocol can be expected**):

  • Transparent square(>1 ,multiple representations allowed, for instance Space character and 'T')
  • Me-Square(one, one distinct representation, for instance 'M')
  • You-Square(s)- these count as obstacles for visibility towards the others (1-9, individual representations, for instance represented by '1'-'9')
  • Obstacle-Squares (>1, multiple representations allowed, for instance 'O' and '#')

The raster (implicitly from the above minimum 3x3) is a representation of a 2D surface of adjoining squares in a grid, and 'visibility' from one square to the other is defined as any possible idealized line between any point of those two squares that does not touch/intersect with with any obstacle square in between.

The following instance of input would be allowed:

 #  24#;  MO 3;      ;   #  ;      ;    1 

as representing a 2D space like this:

The space represented by the input

The solution now needs to output a visibility map, and a count of non-seers. The visibility-map output needs to adhere to the same protocol that the solution accepted as input. In the output, only the 'you' representations that see 'me' are included, the positions of the non-seers are shown as obstacles(any representation of obstacle that the protocol provides). Additionally a number is output (as either int or chars '1'-'9'), that counts the number of non-seers; The number of non-seers can be anything from 0 to 9.

For the above instance of input(and input protocol), this would be an acceptable output:

'2' and '#TT24#; MO O; ; # ; ; #T' (because two 'you' squares cannot see 'me': '1' and '3' - both have no line of sight towards any part of 'me' that does not touch an obstacle; Squares '2' and '4' both have visibility: Although some possible sight lines between 'me' and those squares intersect an obstacle, there are also unobstructed sight-lines possible; Some transparent squares that were formerly encoded by ' ' are now encoded as 'T' while the non-seers are now represented as either 'O' or '#')

*['reasonable' as in: if your golfing language only can deal with 256 chars in a line, it is reasonable to limit the side-length to ~256; If you have a dirty trick that only works up to rasters of 4x4 , that is not reasonable; also, do not golf the I/O protocol to create artificial reasonability limits]

**[Neither your solution nor your protocol need to enforce the rules in the following way: If you chose to I/O via a .png file, and define Black as 'Me', any non-grey color as 'Transparent', any Grey as 'You' and White as obstacle, then more than 9 'you' entities are possible (16bit grey values), while the rules limit the 'you' entities to 9. That is OK, and your solution does not need to deal with more than 9 'you' entities even though this protocol would allow for more.]

Good to know: If a square A has 180° rotational symmetry around an obstacle towards another square B, A and B have no visibility towards each other.

New contributor
bukwyrm is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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0
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How much does it outlie?

You work for a statistics company and need to check your client’s data.

Given an array of positive integers, return the absolute difference between the median and the mean of the integers.

Example implementation

array = eval(input())

mean = sum(array) / len(array)

print(f"Mean: {mean}")

def median(array):
  if len(array) == 0:
    return None
  index = ((len(array) + 1) / 2) - 1
  if index % 1 == 0.5:
    return sum(sorted(array)[x] for x in map(int, [index - 0.5, index + 0.5])) / 2
  return sorted(array)[int(index)]

mid = median(array)

print(f"Median: {mid}")

print(f"Difference between mean and median of numbers {array} is {abs(mean - mid)}.")

Try it online!

Winning criterion

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0
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ALU: addition, subtraction, multiplication, division

Challenge

Given two integers \$a, b\$ in \$[0, 256)\$ and a number \$c\$ from the set \$\{0, 1, 2, 3\}\$, perform an operation corresponding to \$c\$.

  • If \$c\$ is 0, output \$a + b \pmod{256}\$.
  • If \$c\$ is 1, output \$a - b \pmod{256}\$.
  • If \$c\$ is 2, output \$a \cdot b \pmod{256}\$.
  • If \$c\$ is 3, output \$\mathrm{floor}(a \div b)\$.

Test cases

a, b, c --> Output
2, 2, 0 --> 4
56, 41, 0 --> 97
90, 177, 0 --> 11
2, 2, 1 --> 0
0, 129, 1 --> 127
90, 177, 1 --> 169
2, 2, 2 --> 4
56, 41, 2 --> 248
129, 125, 2 --> 253
2, 2, 3 --> 1
237, 6, 3 --> 39
90, 177, 3 --> 0
144, 0, 3 --> (undefined behavior)

Restrictions

  • You may take \$a, b,\$ and \$c\$ in any order. If they aren't in this order, please specify which one it is.
  • The condition where \$c\$ is 3 and \$b\$ is 0 is undefined behavior.
  • You may use a function or a full program.
  • Standard loopholes and I/O methods apply.

Scoring

This is , so fewest bytes wins.

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1
0
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Count the symmetries

Question: would it be better to exclude the infinity case, on grounds of focusing on one task?

Find the order (size) of the symmetry group of a finite set of integer points in d-dimensional space.

Input

You will be given the coordinates of a finite set of points in d-dimensional space, in any reasonable form. The dimension d may be any positive integer. You can assume that all coordinates are integers, and that all points are distinct. You can optionally take d itself and/or the number of points as parameters.

Output

Output the number of symmetries of the set of points. A symmetry is an isometry of d-dimensional space that maps the set to itself (such as a reflection or a rotation). An isometry can be described by an orthognal matrix A and a vector b, where a point x is mapped to Ax+b. The number of symmetries is always at least 1 (the identity map). The number might be infinity, in which case you must output any one consistent value that is not a positive integer (such as 0 or -1 or the string "infinity"). The symmetries form a group (but you only need to output its size).

Examples

(0,0),(1,0),(1,1)               -> 2
(0,0),(2,1)                     -> 4
(0,0),(1,0),(0,1),(1,1)         -> 8
(1,1),(-1,-1),(1,-1),(-1,1)     -> 8
(0,0),(1,0),(0,2)               -> 1
(0,0)                           -> infinity
(0,0,0),(1,0,0)                 -> infinity
(0,0,5),(1,0,5),(0,1,5),(1,1,5) -> 16
(1,0,0),(0,1,0),(0,0,1)         -> 12
(2,0,0),(0,2,0),(0,0,2),(2,1,0),(0,2,1),(1,0,2)     -> 3
(0,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,0,0),(0,0,0,1)   -> 24
(0),(1),(2)                     -> 2
(0),(1),(3)                     -> 1
(1)                             -> infinity

For instance, (0,0),(1,0),(0,1),(1,1) is a square: the symmetries are 4 reflections (about diagonals and center-lines), 3 rotations, and the identity, making 8. Adding a 3rd coordinate 5 makes it a square in 3 dimensions, so in addition each symmetry can be comoposed with a vertical reflection.

This is . Shortest answer in each language wins.

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0
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Implement a Monad

A monad is a type that wraps another type, and has the following operations, where M is the monad type, and T and U are any type:

  • A type constructor that defines a new type M T that can hold some type T.
  • A type converter that wraps a value and returns an instance of the monad: unit: (T) -> M T
  • A combinator that unwraps the monad, performs a provided operation on the value, then rewraps the value: bind: (M T, (T) → M U) → M U

These operations have the following properties, given some function f and g, and a value x, where <-> is equivalence:

  • unit is a left-identity for bind: bind(unit(x), f) <-> f(x)
  • unit is a right-identity for bind: bind(x, unit) <-> x
  • bind is associative: bind(x, lambda a: bind(f(a),g)) <-> bind(bind(x, f), g)

Your goal is to write the shortest three functions, code snippets, and/or full programs that implements the above 3 operations. You must indicate which function/code snippet/program is which operation, as well as show the three properties above.

Note: If your language does not have the ability to syntactically declare a new type, you may instead replace your type constructor with an explanation on how your monad is represented in your language.

Note: You may use built-ins. However, solutions without built-ins are highly encouraged.

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6
  • \$\begingroup\$ What three operations? I see two, unit and bind \$\endgroup\$
    – Bbrk24
    Commented 2 days ago
  • \$\begingroup\$ @Bbrk24 type constructor, type converter, and combinator. I will make an edit that clarifies that if your language does not have then ability to syntactically declare types, it can be omitted, but it will need an explanation for how it represents the monad. \$\endgroup\$
    – bigyihsuan
    Commented yesterday
  • 2
    \$\begingroup\$ Must these be implemented from scratch? A lot of languages (not just functional ones!) already have monads in one form or another. For example, I think something like #define M std::optional would be a C++ example that covers all three functions, as you can then do M<T> (type constructor), M<T>(value) (wrap value) and M<T>(value).and_then(operation) (combinator). \$\endgroup\$
    – G. Sliepen
    Commented yesterday
  • \$\begingroup\$ @G.Sliepen No, but a no-built-ins restriction will be encouraged. \$\endgroup\$
    – bigyihsuan
    Commented yesterday
  • \$\begingroup\$ So we just need to implement any one specific monad? Is anything stopping us from making a trivial one where unit is the identity, or where the monad type holds only a single value? \$\endgroup\$
    – xnor
    Commented 17 hours ago
  • \$\begingroup\$ @xnor Any one specific monad; both examples you gave are acceptable answers to the challenge. However, you get brownie points if it's a non-trivial or single-valued monad. \$\endgroup\$
    – bigyihsuan
    Commented 16 hours ago
0
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This algebraic idiot!

You want to prove to some mathematicians that it is possible that \$x=x+1\$. Luckily, they’re not so good at maths, so you can just exponentiate both sides.


Plain challenge

Output the real and imaginary parts of one of the possible values of \$x\$, where $$x^n - (x+1)^n = 0$$ which means both sides of the subtraction are equal. \$n\$ is the positive integer that will be input into your program.

If a solution doesn’t exist, or if the value isn’t a positive integer, your program can do whatever it wants. It can throw an error, brick your computer, preform a DDOS attack on the FBI’s IP geolocation service provider, etc.
This is undefined behaviour.

Test cases

To be added. Note that for all even integer values, one possible solution is always \$-\frac{1}{2}\$.


This is , so shortest answer, in bytes obviously, wins!


Meta

  • Any feedback?
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5
  • 1
    \$\begingroup\$ Do you have any examples? Also, there are some values of \$n\$ where this is obviously impossible (like \$n=1\$), what should submissions do then? \$\endgroup\$
    – Bbrk24
    Commented 2 days ago
  • \$\begingroup\$ @Bbrk24 Added. Test cases being added later. \$\endgroup\$ Commented 2 days ago
  • \$\begingroup\$ Will n be a positive integer? \$\endgroup\$
    – xnor
    Commented 17 hours ago
  • \$\begingroup\$ @xnor I’ve decided to say yes, as the solutions would be the same for a negative integer as the corresponding positive integer. \$\endgroup\$ Commented 15 hours ago
  • \$\begingroup\$ default i/o should cover the case for when you should error out. as present, "undefined behavior" allows me to also output an integer that just happens to not be valid (up to you if you want that to be allowed) \$\endgroup\$ Commented 2 hours ago
0
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Vo we l spa ci ng

To da y yo u a re ha vi ng a si lly pro ble m.

To da  y   yo    u     a      re       ha        vi         ng          a           si            lly             pro              ble               m.

Challenge

Given an English text, apply "vo wel spa ci
ng" (see below).

Input

One line of string with these constraints:

  • Consist of alphabet with both cases, space, comma and period.
  • At least 1 letter and at most (at your choice) 15, 16, 31, 32 or 50 letters.
  • Match /^([A-Za-z]+[.,]? )*[A-Za-z]+[.,]?$/ in POSIX ERE; in English, it's a list of space-separated words with just alphabets but no punctuation; some words may contain a punctuation of either a full stop or a comma but not both. No leading spaces, no trailing spaces, separated with exactly one space.

Output

Result of "vo wel spa ci ng". Your output may have trailing spaces but no leading spaces.

"Vo we l spa ci ng" specification

The "Vo we l spa ci ng" is an algorithm who takes a string of input to return a string of output. The procedure is:

  1. For each vowel that is not the last letter of its word, append a space.
  2. For each space (let it N-th space. 1-indexed), replace it with N spaces.
  3. The implementation of this algorithm may append some spaces to the end of the string.
  4. Return the final string.

In the procedure above, I define the following terms. Assuming ASCII character encoding:

  • A vowel is a letter that is one of the following ten letter: AEIOUaeiou.
  • A word is a consequence concatenation of one or more alphabets but nothing else: e.g. "word", "WORD", "wOrD", "w", "o", "r", or "d" but not "word,", "word.", or "w o r d"
  • The last letter of the word is: (a) if it has only a letter, then the letter is the last letter of the word. (b) Otherwise (i.e. it has two or more letters), the last letter of the word without first letter is the last letter of the word. E.g. "word" => "d", "A" => "A".
  • A space is an ASCII character with code point of 0x20.

Test cases

The format is:

Input
=> Output

Here are test cases. Very wide:

Input
=> I npu  t

Output
=> O u  tpu   t

Test cases
=> Te st  ca   se    s

Hello, world.
=> He llo,  wo   rld.

Vowel Spacing
=> Vo we  l   Spa    ci     ng

Today you are having a silly problem.
=> To da  y   yo    u     a      re       ha        vi         ng          a           si            lly             pro              ble               m.

Code Golf and Code Challenge StackExchange
=> Co de  Go   lf    a     nd      Co       de        Cha         lle          nge           Sta            ckE             xcha              nge

AAAAAAAAAAAAAAAAAAAAA
=> A A  A   A    A     A      A       A        A         A          A           A            A             A              A               A                A                 A                  A                   A                    A

BBBBBBBBBBBBBBBBBBBBBB
=> BBBBBBBBBBBBBBBBBBBBBB

QwErTyUiOp AsDfGhJkL zXcVbNm
=> QwE rTyU  i   O    p     A      sDfGhJkL       zXcVbNm

qWeRtYuIoP aSdFgHjKl ZxCvBnM
=> qWe RtYu  I   o    P     a      SdFgHjKl       ZxCvBnM

Given an English text,
=> Gi ve  n   a    n     E      ngli       sh        te         xt,

apply vo we l spa ci ng.
=> a pply  vo   we    l     spa      ci       ng.

rhythm
=> rhythm

FORTRAN LISP COBOL ALGOL
=> FO RTRA  N   LI    SP     CO      BO       L        A         LGO          L

This is code golf. Shortest code wins.
=> Thi s  i   s    co     de      go       lf.        Sho         rte          st           co            de             wi              ns.

I
=> I

am here to, kick out, leading spaces.
=> a m  he   re    to,     ki      ck       o        u         t,          le           a            di             ng              spa               ce                s.

Rules

See also

Example implementation in Perl

sub vo_we__l___spa____ci_____ng {
  my $str = $_[0];
  my $lf;
  if ( ( my $n = chomp $str ) ) {
    $lf = substr $_[0], -$n;
  }

  $str = step1($str);
  $str = step2($str);
  $str = step3($str);
  return $str . $lf;
}

sub step1 {
  my $str = shift;
  my @tokens = tokens($str);
  my $result;
  while ( @tokens ) {
    local $_ = pop @tokens;
    $result .= " " if /^space/;
    $result .= (pop @tokens) if /^punct/;
    if ( /^word/ ) {
      my $word = pop @tokens;
      die unless defined $word;
      $result .= appendSpacesToVowels($word);
    }
  }
  return $result
}

sub tokens {
  local $_ = shift;
  my @result;
  while ( /./ ) {
    push @result, "space" if s/^ //;
    push @result, "word", $& if s/^[A-Za-z]+//;
    push @result, "punct", $& if s/^[.,]//;
  }
  return @result;
}

sub appendSpacesToVowels {
  my $word = shift;
  my $lastLetter = chop $word;
  my $result;
  for ( my $i = 0; $i <= length $word; $i++ ) {
    my $char = substr $word, $i, 1;
    $result .= $char;
    $result .= " " if isVowel($char);
  }
  return $result . $lastLetter;
}

sub isVowel { /[aeiou]/i }

sub step2 {
  my $str = shift;
  my @tokens = tokens($str);
  my $nthSpace = 1;
  my $result;
  while ( @tokens ) {
    my $tokenKind = pop @tokens;
    if ( $tokenKind eq "space" ) {
      $result .= " " x $nthSpace++;
    } else {
      $result .= pop @tokens;
    }
  }
  return $result;
}

sub step3 {
  return $_ . (" "x rand 3);
}

Try it online!

Meta

  • Does title get rendered as monospace?
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2
  • \$\begingroup\$ I don't understand the rules of the challenge at all. How do we go from Input to In pu t? The way I interpret the procedure as currently written would give Input -> I n p u t -> I n p u t. \$\endgroup\$ Commented Sep 4 at 1:27
  • 1
    \$\begingroup\$ @shapewarriort Input to I npu t \$\endgroup\$ Commented Sep 4 at 23:51
-1
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Consecutive Composites

Your task is to write a program or function which, given a positive integer N, finds the first block of N consecutive composite numbers.

This should be the first block of integers which fit the requirements, larger than 0. For example, with an input of 2, the output must be [8, 9], and not [14, 15].

Rules:

  • The numbers in the block should be printed or returned as a list, in any reasonable format.
  • Submissions may be either full programs which perform I/O, or functions - no snippets.
  • You can assume that the block of numbers your program has been request to find is within your language's standard integer range.
  • This is , so the shortest program (in bytes) wins! Standard golfing loopholes apply.

Test Cases

1 -> [1]
2 -> [8, 9]
5 -> [24, 25, 26, 27, 28]
6 -> [90, 91, 92, 93, 94, 95]
10 -> [114, 115, 116, 117, 118, 119, 120, 121, 122, 123]

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3
  • 1
    \$\begingroup\$ This is essentially codegolf.stackexchange.com/q/23844/194 with a tweaked output format. \$\endgroup\$ Commented Feb 6, 2017 at 9:34
  • \$\begingroup\$ One is not a composite number, the smallest is four, so the test case for 1 should be [4]. \$\endgroup\$ Commented Feb 6, 2017 at 20:39
  • \$\begingroup\$ @JonathanAllan i've misused the term composite there, I meant 'non-prime' - regardless, I probably won't post this anyway and Peter pointed out it's basically a dupe. \$\endgroup\$
    – FlipTack
    Commented Feb 6, 2017 at 22:02
-1
\$\begingroup\$

Wifi Puzzle! Crack the router [code-golf] [networking]

SITUATION

Consider that you have three wifi routers in your home , all with different SSIDs and none of them are dualband. You have invited a mischievous friend to your home who had changed the password of each router, without letting you know about it. Now to annoy you more he has set up a programming challenge.

THE CHALLENGE

Your friend has created three .txt files containing a set of passwords with only one correct among them. (i.e. each .txt file contains a correct password while all other are wrong ones. Also one .txt file contains only one correct password) and the .txt files do not specify which one may contain the correct password for a certain router (i.e. you cannot be sure that file1.txt(let us assume it is one of those .txt files) contains the password for router1( say any one of those routers). Now your friend has kept them in a certain directory( say E:\Wifi) and asked you to create a programme or function that would pick up a file and take input from it, try to connect to a random Access point ( out of the three routers) and find which password fits to which router.

Sample Input

Let us consider a file, file1.txt( or any other name you like) be like this

A12e77799U5
Pdc555089rtf
Ds442Y779#1
1&2*fe$996Yt
Uty66%92Gu4

Note that each password contains a capital letter, numbers, special characters, (of a standard keyboard) and each file contains only five unique passwords. Also all the .txt files are in the same directory and there are no subdirectories in the directory concerned. Also each .txt file contains at least one correct password.

Sample Output

Your programme or function must keep a log of its activity in a separate file log.txt which you may put in the same directory concerned or in a different directory. The log file must show which router has been cracked with which password and also the file containing it.

Example: Say that router1 ( SSID of a router) has been cracked by the password A12e77799U5 from file1.txt so the output of the log.txt must be

router1 password A12e77799U5
File: file1.txt

Also you must be sure that all the output goes into the log.txt not seperate files each time a router is cracked. You can create a programme or a function in any programming language.

Keep In Mind

  1. This is code-golf so the shortest answer wins.

  2. Standard loopholes apply as usual.

Discussion I feel to ask this question but the foremost problem I face is how can others test their code. Also strict I/o rules (like the log.txt I mentioned ) are not appreciated here. So please help me out!

\$\endgroup\$
2
  • 2
    \$\begingroup\$ So, you want us to access the network settings programmatically? Even if we disregard the difficulties in testing it, this is not a golfing challenge, but rather a challenge in convincing our OSes to let us fiddle with the settings, and then figuring out how. \$\endgroup\$ Commented Feb 12, 2017 at 19:25
  • \$\begingroup\$ So is there any category I can put it in, I mean any tags. \$\endgroup\$
    – jyoti proy
    Commented Feb 12, 2017 at 20:04
-1
\$\begingroup\$

Portable bitmap checkerboard pattern

Your task is to create a checkerboard pattern and store it in a PBM.

Size of the checkerboard is passed in STDIN as two numbers. Output is written to STDOUT.

Test case:

Input:
5
5

Output:
P1
5 5
0 1 0 1 0
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
0 1 0 1 0

This is so the shortest code wins

\$\endgroup\$
-1
\$\begingroup\$

Bike saddle drawn through a fractal

Based on the Mandelbrot image in every language, and on the observation the 3rd layer (0 indexed) always looks like a bike saddle, I had a little bit different challenge:

  • Language must be capable of graphical output or drawing charts (saving files disallowed)
  • Render a window or control that is resizable by mouse action. As example, it can be a typical GUI Window with the typical frame that allows resizing
  • After resizing the GUI element, the fractal should be updated according to the new pixel space
  • The fractal coordinates range from approximately -2-2i to 2+2i
  • The pixels outside of the 3rd layer (0 indexed) of Mandelbrot set should have one color; the ones inside 3rd and inner layers should have another. The only two colors used should be clearly distinguishable
  • At least 99 iterations
  • ASCII art not allowed

Winning conditions:
Shortest version (size in bytes) for each language will get a mention in this post, ordered by size.
No answer will ever be 'accepted' with the button.

\$\endgroup\$
2
  • \$\begingroup\$ @Mark Jeronimus: credits to you. \$\endgroup\$
    – sergiol
    Commented May 27, 2017 at 8:48
  • \$\begingroup\$ "The only two colors used should be clearly distinguishable" Clarification? \$\endgroup\$
    – Joao-3
    Commented Aug 15, 2023 at 14:30
-1
\$\begingroup\$

Do nothing

Write a program which terminates normally (not in an error), producing no output on the standard output stream (or the language's closest equivalent), nor on the standard error stream, regardless of what content is present on the standard input stream. (Note that this is intentionally overriding the normal I/O defaults; this is a challenge entirely about input/output handling.)

Additionally, your program may not have any other side effects (e.g. writing files, changing persistent state), unless they're an unavoidable consequence of running a program on the operating system you're using (e.g. on Linux, it's OK to change the "next process ID number to be assigned" value inside the kernel, because that happens whenever you run a program).

Finally, to avoid numerous uninteresting 0-byte (or boilerplate-plus-0-byte) solutions, you may not use a language in which the shortest program that does nothing (i.e. complies with the above specification) is also the shortest (or tied for the shortest) program which runs without error (but possibly reacts to input or produces output). In other words, you can't use a language unless doing nothing is more verbose than doing something.

Clarifications

  • Intentionally exiting the program early is permitted. If you do exit the program manually, on a system that uses exit codes, you may do so with any exit code.
  • Crashing the program is not permitted, even if it (for some reason) exits with a "success" code after the crash.
  • "No output" means 0 bytes of output, not even a trailing newline.
  • Likewise, your program must be able to handle any finite sequence of bytes on the standard input stream, even if it isn't, say, made of characters in the current encoding (but rather of arbitrary octets). You do not need to handle infinite input, though (e.g. your program won't be connected to /dev/zero or the like).
  • You don't have to actually read input; it's your choice as to whether you want to read and discard it, or not read it at all.

Victory condition

As a challenge, shorter is better, measured in bytes. (Remember that if you need to run the program in an unusual way, that incurs a byte penalty, under standard PPCG rules.)

Because languages which are particularly suited for this task (such as Perl and Python) are excluded by the rules, there's not much point in talking about the best answer cross-language; rather, the aim is to find the best answer you can in the language which you submit in. (Historically, on this sort of challenge, answers that are more unusual, interesting, or better-explained have tended to get more votes.)

Sandbox questions

Is this too trivial? We were discussing it in chat as a joke, and realised that it's actually possibly more interesting than it sounds. I'm fairly sure the spec's correct (although would definitely appreciate knowing if something's wrong here!), but would appreciate feedback on how much people would hate me if I posted it to main.

\$\endgroup\$
3
  • \$\begingroup\$ you can't use a language unless doing nothing is more verbose than doing something.you can't use a program unless your program is more verbose than any other program which does something. You must provide a shorter program which does something to prove your solutions validity. \$\endgroup\$
    – Adám
    Commented Jun 8, 2017 at 1:03
  • \$\begingroup\$ @Adám: If you did that, people would just add a comment byte or two to create a program of the shortest possible length that was longer than a program that did something. That isn't particularly interesting. \$\endgroup\$
    – user62131
    Commented Jun 8, 2017 at 1:21
  • \$\begingroup\$ So I write a buffer null that accepts input stdin > /dev/null I think it should do nothing. If it produces Moby Dick I will be surprised. \$\endgroup\$
    – Willtech
    Commented Dec 28, 2023 at 8:23
-1
\$\begingroup\$

Plan and Chain a route through OEIS

Your Task is to reach so many OEIS sequences you could make with chaining your last sequence with a operation to a new sequence.

You must avoid last sequence minus last sequence plus first sequence or something similar that your new sequence is based on the first sequence except to make the second sequence.

Your starting OEIS sequence is in every case https://oeis.org/A001477

Given as Input an positive Integer and a Letter that matches [A-Z] or [a-Z]

Example

PHP, 171 bytes

for($a=0;$a<=$argv[1];$a++)$r[]=[$a,$b=$a&1,$c=$a+!$b,$d=(($c-!$b)/2^0)+$b,$A[$b]=$e=$d*$c,$f=$e+$A[!$b],$g=$a?$g*sqrt($f):1,$h=$g%2];echo$r[$argv[1]][ord($argv[2])%32-1];

Try it online!

The example gives back the n value of a OEIS sequence for the following letters. A letter greater h is for this example a invalid input

  • a https://oeis.org/A001477 numbers
    $a Valid first sequence

  • b https://oeis.org/A000035 mod 2
    $b=$a&1 Valid use the variable in the sequence before

  • c https://oeis.org/A109613 odd numbers
    $c=$a+!$b Valid Can use sequences before

  • d https://oeis.org/A110654 a(n) = floor(n/2) + n mod 2
    $d=(($c-!$b)/2^0)+$b Valid an invalid example is $d=(($a/2)^0)+$b cause it not use the sequence before

  • e https://oeis.org/A000217 triangular
    $A[$b]=$e=$d*$c Valid you can create help variables

  • f https://oeis.org/A000290 square
    $f=$e+$A[!$b] Valid use a help variabale and the variable of the sequence before. $f=$A[!$b]+$A[!$b] Invalid causes it makes the same value but use indirectly the variable of the sequence before

  • g https://oeis.org/A000142 factorial $g=$a?$g*sqrt($f):1 Valid cause your condition is not always the case that it have no relationship to the sequence before.

  • h https://oeis.org/A019590 Fermat's Last Theorem $h=$g%2 Valid but now we have the problem to find the next sequence

Could You make a full alphabet? My alphabet ends with the letter h

\$\endgroup\$
9
  • \$\begingroup\$ I'm rather confused as to what is being asked here. It might be helpful to state how one can get from one sequence to another. \$\endgroup\$
    – Wheat Wizard Mod
    Commented Jun 10, 2017 at 20:47
  • \$\begingroup\$ @WheatWizard I could understand you. The problem is at the moment to make rules that avoid that a trivial solution exits. There are too many sequences in OEIS. The way from every sequence to the next should not end in a simple addition or multiplication. But evrything else should be allowed to get more creative solutions \$\endgroup\$ Commented Jun 10, 2017 at 20:56
  • \$\begingroup\$ (1) The first sentence says that the aim is to build the longest chain possible, but the scoring mechanism rewards average code length per element in the chain rather than number of chains. I would think it most likely as it stands that the winner would be a chain of length 1 or at most 2. (2) If you delete everything from the header Example to the end, do you think that the question still makes sense? If not (and I don't think it does), it needs a lot of work. (3) What do the two values in the input mean? Why is the second one a letter rather than a number? \$\endgroup\$ Commented Jun 10, 2017 at 21:10
  • \$\begingroup\$ (4) I'm not sure how feasible it is to write objective rules which forbid "trivial" expressions. (5) It is not clear how to interpret the rule about the 32nd term where either it is not known or the sequence is finite and shorter than 32 terms. \$\endgroup\$ Commented Jun 10, 2017 at 21:12
  • \$\begingroup\$ @PeterTaylor (1) Think you that popularity Contest is a better winning criteria? (2+3) to limit the chaining length to 26. The goal is to show relationsships between two or more sequences. (4+5) Yes it is not easy and I can drop it if I switch to popularity Contest \$\endgroup\$ Commented Jun 10, 2017 at 21:26
  • \$\begingroup\$ @WheatWizard I allow now trivial solutions \$\endgroup\$ Commented Jun 10, 2017 at 21:53
  • \$\begingroup\$ I'm not clear on the purpose of the inputs if we're just supposed to hard code our way from one sequence to the next​. Replacing your PHP example with more generic, more verbose pseudo-code might help. \$\endgroup\$
    – Shaggy
    Commented Jun 11, 2017 at 0:16
  • \$\begingroup\$ @programmer5000 exists a limit of correct tags? \$\endgroup\$ Commented Jun 11, 2017 at 11:39
  • \$\begingroup\$ @Shaggy See it as restriction for ways to code. You must have a chaining to the sequence before. So far I know any working code is a pseudocode \$\endgroup\$ Commented Jun 11, 2017 at 11:48
-1
\$\begingroup\$

Braid Badly Boundlessly


Your program or function must, given a string in any standard input format, output an infinite stream of delimiter-separated strings where each string is determined from the previous by a braiding algorithm. The program starts with printing the input string.

The algorithm is described as follows: Infinitely alternate between

(1) splitting the string into three substrings then swapping the first two substrings and flattening.

and

(2) splitting the string into three substrings then swapping the last two substrings and flattening.

starting with (1).

The three substrings should be of non-increasing length with the maximum length no more than 1 greater than the minimum length of the three substrings. (This means that when the length of the given string is a multiple of three, the three substrings should be the same length. When the length of the given string is one more than a multiple of three, the first substring should be one character longer than each of the last two substrings. When the length of the given string is two more than a multiple of three, the first and second substrings should each be one character longer than the last substring.)

Example

Let the input be "abcdefg". Let the delimiter be a newline.

Then the program would first print "abcdefg".

It applies (1) which splits the string into ["abc","de","fg"] and swaps the first two elements, reaching ["de","abc","fg"]. It flattens to get "deabcfg" which it prints and uses for the next step.

The program applies (2) to "deabcfg" to split into ["dea","bc","fg"] and swaps into ["dea","fg","bc"], flattening to reach "deafgbc".

The program applies (1) to "deafgbc" and the process repeats ad infinitum.

Then the output would be the newline-separated

abcdefg
deabcfg
deafgbc
fgdeabc
fgdbcea
bcfgdea
bcfeagd
eabcfgd
eabgdcf
gdeabcf
gdecfab
cfgdeab
cfgabde
abcfgde
abcdefg
deabcfg
deafgbc
fgdeabc
fgdbcea
bcfgdea
bcfeagd
eabcfgd
eabgdcf
gdeabcf
gdecfab
cfgdeab
cfgabde
abcfgde
abcdefg
[...]

Specifications

  • Note that the string should not be split at the beginning and then only swapped later. The string should be split on each and every iteration
  • The delimiter between lines could be whichever character is convenient. You may assume it does not appear in the input string.
  • The string input shall be at least three characters
  • The input consists solely of printable characters (0x20-0x7F)
  • Of course, standard loopholes are forbidden.

I/O

  • The input and output should be taken in standard I/O methods.
  • The input and output should be taken as string, list of characters, or equivalent.
  • The output should be output continuously, which means you may assume infinite memory.

Test cases

For the test cases, we will assume that the delimiter is a newline. Just the portion before the endless stream is repeats is shown.

input
--
output
-----
abcdefg
--
abcdefg
deabcfg
deafgbc
fgdeabc
fgdbcea
bcfgdea
bcfeagd
eabcfgd
eabgdcf
gdeabcf
gdecfab
cfgdeab
cfgabde
abcfgde
-----
abc
--
abc
bac
bca
cba
cab
acb
-----
abcdefgh
--
abcdefgh
defabcgh
defghabc
ghadefbc
ghabcdef
bcdghaef
bcdefgha
efgbcdha
efghabcd
habefgcd
habcdefg
cdehabfg
cdefghab
fghcdeab
fghabcde
abcfghde
abcdefgh
-----
Braid
--
Braid
aiBrd
aidBr
dBair
dBrai
radBi
raidB
idraB
idBra
Brida
-----
Cycle
--
Cycle
clCye
cleCy
eCcly
eCycl
yceCl
ycleC
leycC
leCyc
Cylec
-----
O Canada!
--
O Canada!
anaO Cda!
anada!O C
da!anaO C
da!O Cana
O Cda!ana
-----
A man, a plan, a canal - panama!
--
A man, a plan, a canal - panama!
an, a canalA man, a pl - panama!
an, a canal - panama!A man, a pl
 - panama!Aan, a canal man, a pl
 - panama!A man, a plan, a canal
 man, a pla - panama!An, a canal
 man, a plan, a canal - panama!A
n, a canal  man, a pla- panama!A
n, a canal - panama!A man, a pla
- panama!A n, a canal man, a pla
- panama!A man, a plan, a canal 
man, a plan- panama!A , a canal 
man, a plan, a canal - panama!A 
, a canal -man, a plan panama!A 
, a canal - panama!A man, a plan
 panama!A m, a canal -an, a plan
 panama!A man, a plan, a canal -
an, a plan, panama!A m a canal -
an, a plan, a canal - panama!A m
 a canal - an, a plan,panama!A m
 a canal - panama!A man, a plan,
panama!A ma a canal - n, a plan,
panama!A man, a plan, a canal - 
n, a plan, panama!A maa canal - 
n, a plan, a canal - panama!A ma
a canal - pn, a plan, anama!A ma
a canal - panama!A man, a plan, 
anama!A mana canal - p, a plan, 
anama!A man, a plan, a canal - p
, a plan, aanama!A man canal - p
, a plan, a canal - panama!A man
 canal - pa, a plan, anama!A man
 canal - panama!A man, a plan, a
nama!A man, canal - pa a plan, a
nama!A man, a plan, a canal - pa
 a plan, a nama!A man,canal - pa
 a plan, a canal - panama!A man,
canal - pan a plan, a ama!A man,
canal - panama!A man, a plan, a 
ama!A man, canal - pana plan, a 
ama!A man, a plan, a canal - pan
a plan, a cama!A man, anal - pan
a plan, a canal - panama!A man, 
anal - panaa plan, a cma!A man, 
anal - panama!A man, a plan, a c
ma!A man, aanal - pana plan, a c
ma!A man, a plan, a canal - pana
 plan, a cama!A man, anal - pana
 plan, a canal - panama!A man, a
nal - panam plan, a caa!A man, a
nal - panama!A man, a plan, a ca
a!A man, a nal - panamplan, a ca
a!A man, a plan, a canal - panam
plan, a cana!A man, a al - panam
plan, a canal - panama!A man, a 
al - panamaplan, a can!A man, a 
al - panama!A man, a plan, a can
!A man, a pal - panamalan, a can
!A man, a plan, a canal - panama
lan, a cana!A man, a pl - panama
lan, a canal - panama!A man, a p
l - panama!lan, a canaA man, a p
l - panama!A man, a plan, a cana
A man, a pll - panama!an, a cana
\$\endgroup\$
-1
\$\begingroup\$

Quick! Tell me all the numbers from 1 to 100,000!

Your task is to write a program or function that, when run, output all the numbers from 1 to 100 thousand as quickly as possible to STDOUT. It's that simple. All answers are tested on an HP Compaq nx9420 with an Intel Core Duo @ 1.83 GHz and 3 gigs of RAM using the time command.


Of course, standard loopholes are strictly forbidden.
This is , so may the fastest code win and the best programmer prosper...

\$\endgroup\$
9
  • 3
    \$\begingroup\$ Have you tried running an example to see if the times are variable enough to be meaningful? As-is, this is going to be strongly dependent upon how fast the code can do I/O, which makes the challenge pretty uninteresting, IMO. \$\endgroup\$ Commented Jul 19, 2017 at 18:16
  • \$\begingroup\$ @AdmBorkBork Might be interesting \$\endgroup\$
    – ckjbgames
    Commented Jul 19, 2017 at 21:12
  • 1
    \$\begingroup\$ As far as I can tell, this takes less than a tenth of a second, which means submissions will probably be differentiated solely by noise on your computer. \$\endgroup\$ Commented Jul 20, 2017 at 2:37
  • \$\begingroup\$ upvoted, though I think the differenciation is really difficult, unless you test it on a raspberry PI (for example) having ONLY the program and its compiler installed. \$\endgroup\$ Commented Jul 20, 2017 at 13:36
  • \$\begingroup\$ @FryAmTheEggman How could I improve on that? \$\endgroup\$
    – ckjbgames
    Commented Jul 20, 2017 at 23:38
  • 1
    \$\begingroup\$ @V.Courtois I do have a Pi, and I think I will use that (it has Raspbian installed). \$\endgroup\$
    – ckjbgames
    Commented Jul 20, 2017 at 23:39
  • \$\begingroup\$ The time is still so small even a basic operating system will have to much noise in process creation, etc, for this to work out. You need to make what we are computing substantially more complicated for this to be reasonable. \$\endgroup\$ Commented Jul 21, 2017 at 0:10
  • \$\begingroup\$ @FryAmTheEggman K \$\endgroup\$
    – ckjbgames
    Commented Jul 21, 2017 at 1:20
  • \$\begingroup\$ @ckjbgames good then :) \$\endgroup\$ Commented Jul 21, 2017 at 5:26
-1
\$\begingroup\$

What's that character? (Part 1)

Recently I ran a command on my laptop that returned a bunch of characters - some printable, some non-printable. I'm having trouble figuring out what those characters are, so I could use some help. Unfortunately, I'm running low on disk space, so you'll have to write me the shortest program you can that I can run.

Challenge

Given a list of ASCII characters, return their names as written on www.asciitable.com, my go-to site for looking up character points.

Input

You may take a string, a list of characters, or a list of ASCII code points (e.g. 'a' -> 97).

You may optionally take the length of the string/list as well. Note that for C, you must take this parameter, since the string could contain NUL bytes, so strlen won't work here.

Output

Output is flexible as usual; you may print or return from a function as you see fit. You should output a list of strings.

The Table

0 NUL
1 SOH
2 STX
3 ETX
4 EOT
5 ENQ
6 ACK
7 BEL
8 BS
9 TAB
10 LF
11 VT
12 FF
13 CR
14 SO
15 SI
16 DLE
17 DC1
18 DC2
19 DC3
20 DC4
21 NAK
22 SYN
23 ETB
24 CAN
25 EM
26 SUB
27 ESC
28 FS
29 GS
30 RS
31 US
32 Space
33 !
34 "
35 #
36 $
37 %
38 &
39 '
40 (
41 )
42 *
43 +
44 ,
45 -
46 .
47 /
48 0
49 1
50 2
51 3
52 4
53 5
54 6
55 7
56 8
57 9
58 :
59 ;
60 
63 ?
64 @
65 A
66 B
67 C
68 D
69 E
70 F
71 G
72 H
73 I
74 J
75 K
76 L
77 M
78 N
79 O
80 P
81 Q
82 R
83 S
84 T
85 U
86 V
87 W
88 X
89 Y
90 Z
91 [
92 \
93 ]
94 ^
95 _
96 `
97 a
98 b
99 c
100 d
101 e
102 f
103 g
104 h
105 i
106 j
107 k
108 l
109 m
110 n
111 o
112 p
113 q
114 r
115 s
116 t
117 u
118 v
119 w
120 x
121 y
122 z
123 {
124 |
125 }
126 ~
127 DEL

Test Cases

[0, 97, 7, 22] -> [NUL, a, BEL, SYN]

More to come...

Meta

  • Would it be more interesting to use the UTF-8 names for the printable characters (0x20 - 0x7E), and the ASCII names for the control characters?
\$\endgroup\$
4
  • \$\begingroup\$ hand copy the table from the website please dont. Try a Google search: theasciicode.com.ar/ascii-codes.txt \$\endgroup\$
    – Stephen
    Commented Jul 23, 2017 at 22:39
  • \$\begingroup\$ @StepHen good call, thanks \$\endgroup\$ Commented Jul 23, 2017 at 23:24
  • \$\begingroup\$ Downvoter: I would much like your feedback rather than just your vote \$\endgroup\$ Commented Jul 24, 2017 at 1:39
  • 1
    \$\begingroup\$ IMO just have take a letter and output the code. Since that part is boilerplate str.chars.map( real program ). Also for ASCII char names NUL is it ok is we output them in lower case? e.g. nul (obviously ascii letters would have fixed case) \$\endgroup\$
    – Downgoat
    Commented Jul 24, 2017 at 1:42
-1
\$\begingroup\$

Lennyface parser and selector

Your mission

Create, in the language of your choice, a program that outputs a randomly selected lennyface (artistic minifigures, see this) from an input - a string composed of numbers and lennyfaces. You will have to : first, parse this input; second, extract a probability mass function f from the parsed input; third, select and output a lennyface respecting f. Read the rules for more details.

Rules

  • Input : A string with lennyfaces and numbers (positive AND negative integers), separated by newlines. You may take input by STDIN or function parameter for example.
  • Output (STDOUT for example) : the randomly-selected lennyface, as a string.
  • The input creates a probability mass function f. If l is a lennyface, then f(l)=(sum of all numbers since the previous lennyface)/x where x is obtained afterwards by summing each of those numerators. @Sandbox : is it clear enough?
  • If (sum of all numbers since the previous lennyface) is equal to zero or negative, you must do as if the numerator is equal to 1 in f's definition.
  • A line with a number contains only this number ; same for a line with a lennyface. So you can assume there will never be a number in a lennyface.
  • If there is nothing on a line (two newlines in a row), you must consider it as a lennyface.
  • You must consider that the last line of the string is directly before its first line. See Test 1 for an example.
  • You can assume there will be at least 1 lennyface in the list; it cannot be composed just by numbers (don't forget that an empty line is a lennyface too).

Example

Given this input list :

( ͡° ͜ʖ ͡°)
2
¯\_ツ_/¯
34
-4
8
└[⸟‿⸟]┘

1

You must have 1/42 chances of outputting ( ͡° ͜ʖ ͡°), 2/42 chances of outputting ¯\_ツ_/¯, 38/42 chances of outputting └[⸟‿⸟]┘ and 1/42 chances of outputting nothing (line 7).

Test cases

Test 1

(⌐■_■)
3

Must output (⌐■_■) with 3/3 chances.

Test 2

ʢ◉ᴥ◉ʡ

Must output ʢ◉ᴥ◉ʡ with 1/1 chance.

Test 3

0
\(ᗝ)/

Must output \(ᗝ)/ with 1/1 chance.

Test 4

( ͡° ͜ʖ ͡°)
2
¯\_ツ_/¯
34
4
☞   ͜ʖ  ☞

0

Must output ( ͡° ͜ʖ ͡°) with 1/42 chance, ¯\_ツ_/¯ with 1/21 chance, ☞  ͜ʖ  ☞ with 19/21 chances and nothing with 1/42 chance.

Test 5

1



( ͡° ͜ʖ ͡°)

Must output ( ͡° ͜ʖ ͡°) with 1/4 chance and nothing with 3/4 chance, since there are 3 empty lines.

Test 6

42

-1
( ͡° ͜ʖ ͡°)

Must output nothing with 43/44 chance and ( ͡° ͜ʖ ͡°) with 1/44 chance.

@Sandbox : should I add test cases?

This is , so shortest code in bytes wins. Standard loopholes apply.

Note : Please do not be discouraged if the parsing is difficult to handle in your language, or if testing is hard because of randomness. Your solution might be very interesting algorithmically, not obviously in terms of golfing. Just please explain in your answer why it works.

Moreover, this is the first code-golf I create, so please let me know if something is not appropriate or if I should give more details on a point. And overall, if you downvote, explain me why so I can improve it.

\$\endgroup\$
16
  • \$\begingroup\$ Yours tests seems a bit contraditory. The number is the chance of the next face (line), so what's the point of the empty line in the example / test 4? By the same logic, the test1 should have a 3/4 of outputting nothing? What is the point of the 0 in the test 4? \$\endgroup\$
    – Rod
    Commented Jul 3, 2017 at 14:03
  • \$\begingroup\$ Why is the chance of outputting ( ͡° ͜ʖ ͡°) 1/42 and not 0 ? (since there are no numbers above it) \$\endgroup\$
    – Dada
    Commented Jul 3, 2017 at 14:04
  • \$\begingroup\$ Sorry ! I forgot to copy paste the fact that the minimal chance is 1! \$\endgroup\$ Commented Jul 3, 2017 at 14:05
  • \$\begingroup\$ Also, a common thing to do on challenges involving randomness, and therefore, hard to test, is to ask people to provide a mandatory explanation, or at least ask them to show why it works. \$\endgroup\$
    – Dada
    Commented Jul 3, 2017 at 14:05
  • \$\begingroup\$ @Dada thanks. I note this. \$\endgroup\$ Commented Jul 3, 2017 at 14:06
  • \$\begingroup\$ @Rod the empty line is a lennyface, as said here : If there is nothing on a line, you must consider it as a lennyface. \$\endgroup\$ Commented Jul 3, 2017 at 14:08
  • \$\begingroup\$ @V.Courtois I meant and empty line without a preceding number \$\endgroup\$
    – Rod
    Commented Jul 3, 2017 at 14:09
  • \$\begingroup\$ As I said, the minimum is one (sorry again for forgetting it). \$\endgroup\$ Commented Jul 3, 2017 at 14:10
  • 1
    \$\begingroup\$ If only positive integers are to be expected, you should write it. Otherwise, give some details and examples about what you consider "numbers". \$\endgroup\$
    – Dada
    Commented Jul 3, 2017 at 14:12
  • \$\begingroup\$ @Dada editing. In fact I said the minimum is 1, but you can have things like 2,-1,-3,17 and then your lennyface ; that means the probability is 15/ total. \$\endgroup\$ Commented Jul 3, 2017 at 14:14
  • \$\begingroup\$ @V.Courtois just a small suggestion, to make the "list as circle" more explicit you could change the value to something else than 0 or 1, this way it would not overlap the "missing number" rule \$\endgroup\$
    – Rod
    Commented Jul 3, 2017 at 14:15
  • \$\begingroup\$ @Rod does it? Sorry if I'm not getting what you are saying, but the list is always a circle, meaning if your list is 2,3,( ͡° ͜ʖ ͡°),4,5,☞  ͜ʖ  ☞,6, you have 6+2+3 chance of getting ( ͡° ͜ʖ ͡°) and 4+5 chance of getting ☞  ͜ʖ  ☞. \$\endgroup\$ Commented Jul 3, 2017 at 14:18
  • \$\begingroup\$ thanks for editing @musicman523 \$\endgroup\$ Commented Jul 4, 2017 at 7:21
  • 5
    \$\begingroup\$ KISS. This is far more complicated than common sense would require. Deliberately overcomplicating things to make it "more difficult" is a guaranteed method to make a bad question. \$\endgroup\$ Commented Jul 4, 2017 at 7:32
  • 1
    \$\begingroup\$ The challenge has two parts as far as I can tell. a) Create a probability mass function from an input by parsing b) sample from the probability mass function. Part a) needs to be rewritten as it is at best ambiguous and at worst just incorrect. \$\endgroup\$
    – user9206
    Commented Jul 5, 2017 at 7:50
-1
\$\begingroup\$

Golf Cubically code

Your task is to optimize Cubically source code using one or more optimizations in this post.

How this challenge works:

  • You will choose one or more optimizations below and write a program (in the language of your choice) that performs those optimizations on a Cubically program.
  • Your program will take a Cubically program as input using any allowed input methods, and output a Cubically program using any allowed output methods.
  • The first answer to successfully perform all optimizations wins!

Optimizations

1. Face turn arguments

Before a face turn is performed, the interpreter calculates turns = turns mod 4. So R5 would be equivalent to R1 which is equivalent to R, R7 is equivalent to R3 which is equivalent to R', etc. Also note that R11111 is equivalent to R5, and R22 is equivalent to nothing at all.

Performing this optimization will mean evaluating all arguments to an R, L, U, D, M, E, or S command and shortening them as much as possible.

Test cases:

Relevant code -> Optimization
R11           -> R2
R1            -> R
L33           -> L2
U22           ->
D222          -> D2
M11111        -> M
E00001        -> E
S9            -> S

2. Repeated face turn

When multiple calls to the same face turn command are present right next to each other, they can clearly be golfed. For example, R2R1 is equivalent to R3. UUU is equivalent to U3. F2F2F2F2 is equivalent F8.

Test cases:

Relevant code -> optimization
R2R2R2        -> R6            (R2 if you also choose optimization 1)
LLL           -> L3
UU            -> UU or U2
D3D2D1        -> D6            (D2 if you also choose optimization 1)

3. "Set notepad to" commands

There are some commands that, instead of adding to/subtracting from/multiplying by/dividing by the notepad, just assign to it. Here are all such commands:

_^=<>⊕«»·|:

When called with multiple arguments, since each argument calls the command separately, only the final argument is relevant. So =123 is equivalent to =3, _00000 is equivalent to _0, and :12345678987654321 is equivalent to 1.

Test cases:

Relevant code -> Optimization
_333          -> _3
=12321        -> =1
+54321        -> +54321
:55           -> :5
/55           -> /55

4. Repeated non-face-turn commands

When multiple face turn commands are present right by each other, their arguments can simply be added together. Commands do not act this way. While R2 calls R with 2, =2 calls = with the face sum of the front face (face index 2).

To perform this optimization, when multiple commands outside of RLUDFBMES appear next to each other, simply remove the duplicated commands without removing the arguments.

Relevant code -> Optimization
_1_1_1_1      -> _1111         (_1 if you also choose optimization 3)
%11%22%33     -> %112233       (%3 if you also choose optimization 3)
+12345+67+8   -> +12345678

5. Nonexistent commands

Go check out the Cubically commands page and you'll see that there are plenty of characters that are not commands. For example, there are no commands that are lowercase letters.

To perform this optimization, remove all nonexistent commands and their arguments from the Cubically source. If the commands also have arguments, you must remove the arguments so that they are not passed to the previous command.

Test cases:

Relevant code -> Optimization
moo cow moo   -> 
moo2cow2moo   -> 
misteR2 FOO   -> R2F
FEAR ME.      -> ERME
u1U2u3U4u5U6  -> U2U4U6   (nothing if you also choose optimization 1, U12 if you also choose optimization 2)

6. Non-implicit commands

There are lots of implicit commands in Cubically (RLUDFBMES()$~&E!), but there are plenty that need to be called with arguments. So %%%% is equivalent to nothing at all while %%2%% is equivalent to %2.

Test cases:

Relevant code -> Optimization
%%%%          -> 
$$$$          -> $$$$
++2++2++2     -> +2+2+2                 (+222 if you also choose optimization 4)
+++>--<-      -> Not Brainf**k, sorry!  (:P)

Sandbox

I'll add more optimizations later.

\$\endgroup\$
2
  • \$\begingroup\$ Clarification on R123: That's the same as R6 and R2, not R3, right? Digits are summed, there are multidigit numbers? That would be better to specify \$\endgroup\$
    – isaacg
    Commented Aug 17, 2017 at 20:13
  • \$\begingroup\$ A few things: first, I can't find the tag "fgitw", is there a typo? Second, does optimization 1 require handling F and B as well, or just the currently listed ones? Third, in optimization 3 most of the listed commands seem invalid because the notepad is used in calculation and then overwritten with the output; for example =11 is not the same as =1 in most circumstances. In fact, I think only _: are valid. Fourth, is the winning answer one which performs all optimizations in a single program, or one which contains a separate program for each optimization? \$\endgroup\$ Commented Aug 18, 2017 at 18:03
-1
\$\begingroup\$

Hungry for Apples?

enter image description here

This challenge is simple, given an integer 0 <= n or 0 < n, output an ASCII-apple with that many bites taken out of it.


No bites (0):

         //
     .-.:|.-.
   .'   ''   '.
   ;          ;
  :            :
  :            :
  :            :
   :          ;
   '.        :
     '-_.._-'

Bite 1:

         //
     .-.:|.-.
   .'   ''   '.
   ;          ;
   '-.         :
     }         :
   .-'         :
   :          ;
   '.        :
     '-_.._-'

Bite 2:

         //
     .-.:|.-.
   .'   ''   '.
   '-.        ;
     }         :
     }         :
     }         :
   .-'        ;
   '.        :
     '-_.._-'

Bite 3:

         //
     .-.:|.-.
   .'   ''   '.
   '-.        ;
     }      .-'
     }      {
     }      '-.
   .-'        ;
   '.        :
     '-_.._-'

Bite 4:

         //
     .-.:|.-.
   .'   ''   '.
   '-.      .-'
     }      {
     }      {
     }      {
   .-'      '.
   '.        :
     '-_.._-'

Bite 5:

         //
     .-.:|.-.
   .'   ''   '.
   '-.      .-'
     }".    {
     } }    {
     } }    {
   .-'"     '.
   '.        :
     '-_.._-'

Bite 6:

         //
     .-.:|.-.
   .'   ''   '.
   '-.      .-'
     }"~".  {
     } } }  {
     } } }  {
   .-'"~"   '.
   '.        :
     '-_.._-'

Bite 7:

         //
     .-.:|.-.
   .'   ''   '.
   '-.      .-'
     }"~"~".{
     } } } }{
     } } } }{
   .-'"~"~" '.
   '.        :
     '-_.._-'

Bite >7:

[empty output]

Rules

  • You may have trailing spaces, make it consistent though.
  • You may have exactly 1 trailing newline.
  • You are NOT doing an animation here, you are taking in n and outputting an apple.
  • You may error on integers less than 0, as the spec provides n > 0.
  • You must have empty output (no error) on n > 7/8.
    • You threw out the core; you didn't error the core into non-existence.

This is

\$\endgroup\$
3
  • 3
    \$\begingroup\$ I feel this would be better if there was some more symmetry in the 5, 6, and 7 bytes so that people could possibly make better compression. \$\endgroup\$ Commented Aug 4, 2017 at 18:26
  • \$\begingroup\$ @AdmBorkBork better? \$\endgroup\$ Commented Aug 22, 2017 at 21:26
  • 1
    \$\begingroup\$ Yes, much better. \$\endgroup\$ Commented Aug 23, 2017 at 12:33
-1
\$\begingroup\$

Proper Kerning

Kerning is the adjustment of spacing between pairs of letters in order to obtain an aesthetic result. When kerning is applied automatically by a program (typically whatever editor you're using), it is said to be automatic. There are two types of automatic kerning. The one used in this challenge is metric kerning. With metric kerning, the amount of space between pairs of letters is dictated by the kerning tables found in the font file.

Given a TrueType font file, output the kerning values for each mapping in the kerning table for ASCII characters 48 - 122 inclusive.

Example

calibri.ttf

l="A" r="C" v="-15"
l="A" r="G" v="-15"
l="A" r="J" v="23"
l="A" r="O" v="-23"
l="A" r="Q" v="-23"
l="A" r="T" v="-160"
l="A" r="U" v="-32"
l="A" r="V" v="-89"
l="A" r="W" v="-80"
l="A" r="Y" v="-150"
l="A" r="t" v="-52"
l="A" r="v" v="-38"
l="A" r="y" v="-41"
l="A" r="?" v="-68"
l="B" r="A" v="-20"
l="B" r="T" v="-48"
l="B" r="V" v="-25"
l="B" r="W" v="-24"
l="B" r="X" v="-44"
l="B" r="Y" v="-57"
l="B" r="Z" v="-20"
l="B" r="f" v="-20"
l="B" r="t" v="-20"
l="B" r="v" v="-20"
l="B" r="x" v="-15"
l="B" r="y" v="-20"
l="C" r="G" v="-18"
l="C" r="J" v="12"
l="C" r="O" v="-18"
l="C" r="Q" v="-18"
l="C" r="T" v="10"
l="D" r="A" v="-30"
l="D" r="J" v="-22"
l="D" r="T" v="-23"
l="D" r="V" v="-24"
l="D" r="W" v="-14"
l="D" r="X" v="-31"
l="D" r="Y" v="-39"
l="D" r="Z" v="-22"
l="E" r="A" v="-22"
l="E" r="C" v="-24"
l="E" r="G" v="-24"
l="E" r="O" v="-32"
l="E" r="Q" v="-32"
l="E" r="S" v="-20"
l="E" r="Z" v="-10"
l="E" r="a" v="-34"
l="E" r="c" v="-28"
l="E" r="d" v="-30"
l="E" r="e" v="-37"
l="E" r="f" v="-64"
l="E" r="o" v="-37"
l="E" r="q" v="-30"
l="E" r="t" v="-24"
l="E" r="v" v="-48"
l="E" r="w" v="-34"
l="E" r="y" v="-48"
l="F" r="A" v="-115"
l="F" r="C" v="-18"
l="F" r="G" v="-18"
l="F" r="J" v="-109"
l="F" r="O" v="-18"
l="F" r="Q" v="-18"
l="F" r="S" v="-29"
l="F" r="X" v="-22"
l="F" r="Z" v="-11"
l="F" r="a" v="-55"
l="F" r="c" v="-28"
l="F" r="d" v="-20"
l="F" r="e" v="-30"
l="F" r="o" v="-28"
l="F" r="q" v="-20"
l="F" r="s" v="-35"
l="G" r="T" v="-10"
l="G" r="V" v="-10"
l="G" r="W" v="-9"
l="G" r="Y" v="-30"
l="G" r="v" v="-29"
l="G" r="w" v="-22"
l="G" r="x" v="-14"
l="G" r="y" v="-30"
l="J" r="A" v="-35"
l="J" r="X" v="-20"
l="K" r="C" v="-78"
l="K" r="G" v="-80"
l="K" r="O" v="-97"
l="K" r="Q" v="-97"
l="K" r="S" v="-18"
l="K" r="U" v="-29"
l="K" r="W" v="-34"
l="K" r="a" v="-34"
l="K" r="c" v="-40"
l="K" r="d" v="-33"
l="K" r="e" v="-37"
l="K" r="f" v="-25"
l="K" r="m" v="-32"
l="K" r="n" v="-32"
l="K" r="o" v="-37"
l="K" r="p" v="-32"
l="K" r="q" v="-33"
l="K" r="r" v="-32"
l="K" r="s" v="-18"
l="K" r="t" v="-38"
l="K" r="u" v="-32"
l="K" r="v" v="-101"
l="K" r="w" v="-95"
l="K" r="y" v="-85"
l="L" r="C" v="-22"
l="L" r="G" v="-47"
l="L" r="J" v="25"
l="L" r="O" v="-45"
l="L" r="Q" v="-45"
l="L" r="T" v="-150"
l="L" r="U" v="-44"
l="L" r="V" v="-147"
l="L" r="W" v="-118"
l="L" r="Y" v="-167"
l="L" r="f" v="-23"
l="L" r="t" v="-38"
l="L" r="v" v="-78"
l="L" r="w" v="-72"
l="L" r="y" v="-79"
l="O" r="A" v="-23"
l="O" r="J" v="-27"
l="O" r="T" v="-55"
l="O" r="V" v="-25"
l="O" r="W" v="-22"
l="O" r="X" v="-64"
l="O" r="Y" v="-55"
l="O" r="Z" v="-38"
l="O" r="x" v="-12"
l="O" r="z" v="-10"
l="P" r="A" v="-151"
l="P" r="J" v="-140"
l="P" r="T" v="-9"
l="P" r="V" v="-10"
l="P" r="X" v="-35"
l="P" r="Y" v="-11"
l="P" r="Z" v="-29"
l="P" r="a" v="-44"
l="P" r="c" v="-43"
l="P" r="d" v="-34"
l="P" r="e" v="-41"
l="P" r="f" v="12"
l="P" r="o" v="-41"
l="P" r="q" v="-34"
l="P" r="s" v="-32"
l="P" r="t" v="12"
l="P" r="y" v="12"
l="Q" r="J" v="41"
l="Q" r="T" v="-47"
l="Q" r="V" v="-25"
l="Q" r="W" v="-12"
l="Q" r="X" v="12"
l="Q" r="Y" v="-46"
l="Q" r="g" v="59"
l="Q" r="j" v="79"
l="Q" r="x" v="31"
l="Q" r=";" v="60"
l="Q" r="]" v="32"
l="R" r="C" v="-18"
l="R" r="G" v="-19"
l="R" r="O" v="-20"
l="R" r="Q" v="-20"
l="R" r="S" v="-27"
l="R" r="T" v="-20"
l="R" r="V" v="-28"
l="R" r="W" v="-18"
l="R" r="Y" v="-30"
l="R" r="e" v="-36"
l="R" r="o" v="-42"
l="R" r="v" v="-26"
l="R" r="w" v="-33"
l="R" r="y" v="-33"
l="S" r="A" v="-15"
l="S" r="J" v="-9"
l="S" r="T" v="-14"
l="S" r="V" v="-14"
l="S" r="W" v="-15"
l="S" r="X" v="-13"
l="S" r="Y" v="-20"
l="S" r="v" v="-23"
l="S" r="w" v="-17"
l="S" r="y" v="-25"
l="T" r="A" v="-160"
l="T" r="C" v="-42"
l="T" r="G" v="-59"
l="T" r="J" v="-65"
l="T" r="O" v="-58"
l="T" r="Q" v="-58"
l="T" r="S" v="-10"
l="T" r="T" v="28"
l="T" r="a" v="-160"
l="T" r="c" v="-177"
l="T" r="d" v="-147"
l="T" r="e" v="-182"
l="T" r="g" v="-151"
l="T" r="m" v="-127"
l="T" r="n" v="-127"
l="T" r="o" v="-182"
l="T" r="p" v="-127"
l="T" r="q" v="-147"
l="T" r="r" v="-127"
l="T" r="s" v="-153"
l="T" r="u" v="-127"
l="T" r="v" v="-92"
l="T" r="w" v="-86"
l="T" r="x" v="-90"
l="T" r="y" v="-93"
l="T" r="z" v="-142"
l="T" r=";" v="-114"
l="T" r=":" v="-134"
l="U" r="A" v="-45"
l="U" r="J" v="-40"
l="V" r="A" v="-96"
l="V" r="C" v="-18"
l="V" r="G" v="-25"
l="V" r="J" v="-80"
l="V" r="O" v="-27"
l="V" r="Q" v="-27"
l="V" r="S" v="-12"
l="V" r="V" v="9"
l="V" r="a" v="-114"
l="V" r="c" v="-103"
l="V" r="d" v="-87"
l="V" r="e" v="-102"
l="V" r="g" v="-100"
l="V" r="m" v="-50"
l="V" r="n" v="-50"
l="V" r="o" v="-86"
l="V" r="p" v="-50"
l="V" r="q" v="-87"
l="V" r="r" v="-50"
l="V" r="s" v="-90"
l="V" r="u" v="-50"
l="V" r="y" v="-35"
l="V" r="z" v="-82"
l="V" r=";" v="-108"
l="V" r=":" v="-73"
l="W" r="A" v="-93"
l="W" r="C" v="-22"
l="W" r="G" v="-22"
l="W" r="J" v="-88"
l="W" r="O" v="-22"
l="W" r="Q" v="-22"
l="W" r="S" v="-10"
l="W" r="X" v="-13"
l="W" r="a" v="-71"
l="W" r="c" v="-78"
l="W" r="d" v="-72"
l="W" r="e" v="-75"
l="W" r="g" v="-54"
l="W" r="m" v="-60"
l="W" r="n" v="-60"
l="W" r="o" v="-86"
l="W" r="p" v="-60"
l="W" r="q" v="-72"
l="W" r="r" v="-60"
l="W" r="s" v="-73"
l="W" r="u" v="-60"
l="W" r="v" v="-34"
l="W" r="y" v="-53"
l="W" r=";" v="-156"
l="X" r="C" v="-57"
l="X" r="G" v="-65"
l="X" r="O" v="-57"
l="X" r="Q" v="-57"
l="X" r="S" v="-20"
l="X" r="d" v="-44"
l="X" r="e" v="-39"
l="X" r="g" v="-9"
l="X" r="o" v="-38"
l="X" r="q" v="-44"
l="X" r="t" v="-31"
l="X" r="u" v="-38"
l="X" r="v" v="-55"
l="X" r="w" v="-49"
l="X" r="y" v="-43"
l="Y" r="A" v="-152"
l="Y" r="C" v="-67"
l="Y" r="G" v="-67"
l="Y" r="J" v="-112"
l="Y" r="O" v="-66"
l="Y" r="Q" v="-66"
l="Y" r="S" v="-17"
l="Y" r="Z" v="-10"
l="Y" r="a" v="-134"
l="Y" r="c" v="-159"
l="Y" r="d" v="-131"
l="Y" r="e" v="-147"
l="Y" r="f" v="-62"
l="Y" r="g" v="-142"
l="Y" r="i" v="-32"
l="Y" r="j" v="-49"
l="Y" r="m" v="-94"
l="Y" r="n" v="-94"
l="Y" r="o" v="-153"
l="Y" r="p" v="-94"
l="Y" r="q" v="-131"
l="Y" r="r" v="-94"
l="Y" r="s" v="-115"
l="Y" r="t" v="-44"
l="Y" r="u" v="-94"
l="Y" r="v" v="-69"
l="Y" r="w" v="-62"
l="Y" r="x" v="-70"
l="Y" r="y" v="-65"
l="Y" r="z" v="-100"
l="Y" r=";" v="-138"
l="Y" r=":" v="-154"
l="Z" r="A" v="-11"
l="Z" r="C" v="-25"
l="Z" r="G" v="-24"
l="Z" r="O" v="-24"
l="Z" r="Q" v="-24"
l="Z" r="W" v="-7"
l="Z" r="Y" v="-7"
l="Z" r="a" v="-10"
l="Z" r="c" v="-12"
l="Z" r="d" v="-18"
l="Z" r="e" v="-31"
l="Z" r="o" v="-29"
l="Z" r="q" v="-18"
l="Z" r="v" v="-45"
l="Z" r="w" v="-38"
l="Z" r="y" v="-37"
l="a" r="f" v="-12"
l="a" r="t" v="-19"
l="a" r="v" v="-34"
l="a" r="w" v="-14"
l="a" r="x" v="-19"
l="a" r="y" v="-38"
l="b" r="f" v="-17"
l="b" r="s" v="-10"
l="b" r="t" v="-9"
l="b" r="v" v="-10"
l="b" r="w" v="-10"
l="b" r="x" v="-41"
l="b" r="y" v="-10"
l="b" r="z" v="-28"
l="c" r="a" v="-17"
l="c" r="o" v="-17"
l="e" r="f" v="-18"
l="e" r="t" v="-11"
l="e" r="v" v="-10"
l="e" r="w" v="-10"
l="e" r="x" v="-31"
l="e" r="y" v="-13"
l="e" r="z" v="-20"
l="f" r="a" v="-40"
l="f" r="c" v="-45"
l="f" r="d" v="-53"
l="f" r="e" v="-51"
l="f" r="f" v="-20"
l="f" r="g" v="-60"
l="f" r="o" v="-43"
l="f" r="q" v="-53"
l="f" r="s" v="-27"
l="f" r="v" v="13"
l="f" r="w" v="6"
l="f" r="y" v="10"
l="f" r="z" v="-20"
l="g" r="a" v="-38"
l="g" r="c" v="-12"
l="g" r="d" v="-19"
l="g" r="e" v="-17"
l="g" r="g" v="19"
l="g" r="o" v="-14"
l="g" r="q" v="-19"
l="g" r="t" v="-31"
l="h" r="f" v="-12"
l="h" r="t" v="-19"
l="h" r="v" v="-34"
l="h" r="w" v="-14"
l="h" r="x" v="-19"
l="h" r="y" v="-38"
l="k" r="a" v="-35"
l="k" r="c" v="-48"
l="k" r="d" v="-56"
l="k" r="e" v="-66"
l="k" r="o" v="-69"
l="k" r="q" v="-56"
l="k" r="s" v="-19"
l="k" r="t" v="-10"
l="k" r="u" v="-26"
l="m" r="f" v="-12"
l="m" r="t" v="-19"
l="m" r="v" v="-34"
l="m" r="w" v="-14"
l="m" r="x" v="-19"
l="m" r="y" v="-38"
l="n" r="f" v="-12"
l="n" r="t" v="-19"
l="n" r="v" v="-34"
l="n" r="w" v="-14"
l="n" r="x" v="-19"
l="n" r="y" v="-38"
l="o" r="v" v="-9"
l="o" r="w" v="-8"
l="o" r="x" v="-40"
l="o" r="y" v="-11"
l="o" r="z" v="-27"
l="p" r="f" v="-17"
l="p" r="s" v="-10"
l="p" r="t" v="-9"
l="p" r="v" v="-10"
l="p" r="w" v="-10"
l="p" r="x" v="-41"
l="p" r="y" v="-10"
l="p" r="z" v="-28"
l="q" r="g" v="10"
l="r" r="a" v="-42"
l="r" r="c" v="-30"
l="r" r="d" v="-28"
l="r" r="e" v="-27"
l="r" r="g" v="-28"
l="r" r="o" v="-33"
l="r" r="q" v="-28"
l="r" r="s" v="-35"
l="r" r="v" v="19"
l="r" r="w" v="11"
l="r" r="y" v="10"
l="s" r="f" v="-19"
l="s" r="t" v="-23"
l="s" r="v" v="-31"
l="s" r="w" v="-10"
l="s" r="x" v="-22"
l="s" r="y" v="-37"
l="s" r="z" v="-18"
l="t" r="a" v="-25"
l="t" r="c" v="-25"
l="t" r="d" v="-23"
l="t" r="e" v="-22"
l="t" r="o" v="-20"
l="t" r="q" v="-23"
l="t" r="t" v="-29"
l="v" r="a" v="-30"
l="v" r="c" v="-25"
l="v" r="d" v="-20"
l="v" r="e" v="-20"
l="v" r="f" v="11"
l="v" r="g" v="-28"
l="v" r="o" v="-19"
l="v" r="q" v="-20"
l="v" r="s" v="-9"
l="v" r="t" v="10"
l="v" r="v" v="12"
l="v" r="w" v="12"
l="v" r="y" v="12"
l="v" r="z" v="-26"
l="w" r="a" v="-23"
l="w" r="c" v="-20"
l="w" r="d" v="-18"
l="w" r="e" v="-18"
l="w" r="f" v="6"
l="w" r="g" v="-18"
l="w" r="o" v="-19"
l="w" r="q" v="-18"
l="w" r="s" v="-18"
l="w" r="t" v="4"
l="w" r="v" v="12"
l="w" r="w" v="8"
l="w" r="y" v="12"
l="w" r="z" v="-17"
l="x" r="a" v="-37"
l="x" r="c" v="-46"
l="x" r="d" v="-44"
l="x" r="e" v="-54"
l="x" r="o" v="-55"
l="x" r="q" v="-44"
l="x" r="s" v="-12"
l="x" r="t" v="6"
l="x" r="u" v="-20"
l="y" r="a" v="-31"
l="y" r="c" v="-26"
l="y" r="d" v="-24"
l="y" r="e" v="-25"
l="y" r="f" v="10"
l="y" r="g" v="-26"
l="y" r="o" v="-24"
l="y" r="q" v="-24"
l="y" r="s" v="-19"
l="y" r="t" v="10"
l="y" r="v" v="12"
l="y" r="w" v="8"
l="y" r="y" v="10"
l="y" r="z" v="-17"
l="z" r="a" v="-34"
l="z" r="c" v="-45"
l="z" r="d" v="-46"
l="z" r="e" v="-46"
l="z" r="f" v="-10"
l="z" r="g" v="-17"
l="z" r="o" v="-45"
l="z" r="q" v="-46"
l="z" r="s" v="-22"
l="z" r="u" v="-10"
l="z" r="v" v="-18"
l="z" r="w" v="-22"
l="z" r="y" v="-18"

Scoring

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

Meta

I know this challenge is going to need a lot of work before it's ready for main. Please hold criticisms for now. Helpful ideas and thoughts are welcome.

\$\endgroup\$
7
  • \$\begingroup\$ I'm not sure that the problem is well defined. There's a reason it's called font hinting: the rendering application is free to take it into account or not, or even to apply more complex logic. E.g. some fonts have multiple sets of font hints for different contexts. There are other complex issues. A font can have Latin and Cyrillic letters and define hints for kerning between pairs of Latin and pairs of Cyrillic but not between Latin and Cyrillic; however, some letters may have identical glyphs, so a judgement on whether the kerning is "correct" might be ambiguous. Then there's antialiasing. \$\endgroup\$ Commented May 24, 2017 at 6:15
  • \$\begingroup\$ @PeterTaylor Good notes. I will likely restrict the character set. I just wanted to start getting ideas down in the sandbox. \$\endgroup\$
    – Poke
    Commented May 24, 2017 at 6:51
  • \$\begingroup\$ Very ambiguous. \$\endgroup\$
    – anna328p
    Commented May 25, 2017 at 17:48
  • \$\begingroup\$ @Mendeleev It's not done yet. I'm aware it's ambiguous. \$\endgroup\$
    – Poke
    Commented May 26, 2017 at 16:10
  • \$\begingroup\$ Looking at developer.apple.com/fonts/TrueType-Reference-Manual/RM06/… I can see a number of issues to address. 16- vs 32-bit entries? Should multiple tables be combined or printed separately? All tables or only tables with certain coverage values? Which of the four defined formats need to be supported? Do you have a test case which covers glyph index differing from codepoint? \$\endgroup\$ Commented Sep 16, 2017 at 17:28
  • \$\begingroup\$ @PeterTaylor I have a proof of concept that I wrote (it's the reason I have taken so long to update this) and I'm planning to address all of your questions. Thanks for doing a bit of research to help me out, though :] \$\endgroup\$
    – Poke
    Commented Sep 16, 2017 at 18:57
  • \$\begingroup\$ Downvoter, why? \$\endgroup\$
    – Poke
    Commented Oct 4, 2017 at 21:03
-1
\$\begingroup\$

Six Flags over HTTP

Let's say you need to transmit six boolean flags in a URL string. Obviously you could do it with six ones or zeroes, but you want better compression. With a little math you can pack them into two characters using 0-7 octal.

How about mapping all six to a single ASCII character? Here we have a problem: you are not allowed to use , / ? : @ & = + $ # or space. Now the range of printable ASCII no longer has 64 valid characters in a row.

In Javascript (or another language that can run from a web page, if any), what is the shortest code for a pair of functions to encode and decode this data, between an array of six booleans and a single character?

\$\endgroup\$
2
  • \$\begingroup\$ -1 language restriction, most languages have HTTP libraries so I think any language should be allowed \$\endgroup\$
    – ASCII-only
    Commented Sep 24, 2017 at 13:11
  • \$\begingroup\$ This challenge could be improved by rephrasing it to: "Write a bijective function between an array of six booleans and a single printable character excluding the characters ,/?:@&=+$# ". Mentioning that the encoder and decoder should be separate programs/functions would be helpful. Also, may the encoder and decoder share code? \$\endgroup\$ Commented Sep 24, 2017 at 22:08
-1
\$\begingroup\$

Count letter frequency

Inspired by question Tweetable hash function challenge, you should take the English dictionary used there and produce a program or function that outputs the the absolute and relative frequency of each character. It is CASE SENSITIVE and the APOSTROPHE is also accountable as a real letter.

Example of a valid output format (but with stupid guessing values):

A      5566    20%
...
Z        60     0.2%
a     27000    30%
...
z       120     0.01%
'       450     3.5%

It is , but no answer will be accepted. Wanna know shortest script for each language.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ -1 (01) Don't rely on another challenge to define yours; include all the information we need in your write-up. (02) Make an effort to come up with some actual test cases - do you honestly expect us to verify our solutions against "stupid guessing values"? \$\endgroup\$
    – Shaggy
    Commented Sep 30, 2017 at 0:55
-1
\$\begingroup\$

Is it a perfect loop?

Your task is to take a GIF or an animated image in any reasonable format as input (including taking the file name of a GIF in the current directory), and output whether it is a "perfect loop" - that is, the frames transition seamlessly from the end to the start, and a human cannot notice where it starts and ends at first glance. Return or print a truthy value if it is a perfect loop, otherwise print or return a falsy value.

Scoring

Winners will be determined from the percentage of test cases they get correct. In the event of a tie, highest votes wins. You can view test cases at https://ghostbin.com/paste/m3yaw. Show your score against the test cases when you post.

Input

If you are not taking input in a GIF, please provide a program that will convert a GIF to your desired format.

Images corresponding to a truthy value have been taken from /r/perfectloops and for falsy test cases, /r/almostperfectloops and /r/gifs.

Restrictions

  • Hard coding is not allowed (violates standard loophole 1 and 2).
  • You must provide consistent results for the same GIF (no randomness)
  • Remember, this is not , so byte count is not needed in your solution. Just post the language name and add the percentage correct when I comment.
\$\endgroup\$
2
  • \$\begingroup\$ I'm not sure it's as simple as comparing the first to the last frame, if it is we'd have duplicate frames. is this challenge allowing HTTP requests? \$\endgroup\$
    – tuskiomi
    Commented Oct 17, 2017 at 21:15
  • \$\begingroup\$ If hashing the inputs is not allowed, then you should clearly define what constitutes a “perfect loop”. It's not good to extrapolate from a handful of test cases where the pass/fail cases are very similar. \$\endgroup\$
    – japh
    Commented Oct 18, 2017 at 14:31
-1
\$\begingroup\$

Highest code size∕output ratio to generate a large executable section inside an elf file.

Your challenge is to create the shortest code in your language of choice or the tools of your choice (like objcopy) that will create an elf file with a the executable section as large as possible.
I mean that if I extract the.text section of the elf binary, the resulting extracted file should be at least 90% of the elf binary.

Requirements

  • The program should takes the desired section size as input.
  • The .text section name needs to corresponds to the executable section.
  • The type of the .text section should bePROGBITSand it should contains instructions.
  • The elf file should have a .shstrtab section.
  • The .text section should be readable and writable.
  • The target architecture should be Pnacl or armelv7 or x86_64.
  • The elf file should be valid and pass Google nacl’s validation whitelist in order to be loaded (but I don´t care if the sandbox segfault).
    If you have no idea about what Google native client is, just create a script that call the patched version of binutils from the nacl_sdk, or make sure the elf file is valid and can be executed on Linux.

Of course, you normally can’t use a compiler because it would takes too much computational years in order to finish.

Winner

The answer with the highest code size∕program output ratio.

\$\endgroup\$
9
  • 1
    \$\begingroup\$ Why not make scoring output size / code size? \$\endgroup\$
    – anna328p
    Commented Apr 4, 2017 at 3:49
  • \$\begingroup\$ Make it a code-challenge \$\endgroup\$
    – anna328p
    Commented Apr 4, 2017 at 3:49
  • \$\begingroup\$ This is essentially the same challenge as this one, and would be closed as a duplicate. Although it's not exactly the same, some answers to the previous question would require very little modification and answers to this question would also require very little modification to be answers to the other one. \$\endgroup\$ Commented Apr 4, 2017 at 8:37
  • \$\begingroup\$ @Alt-F4 : it was a code challenge. \$\endgroup\$ Commented Apr 4, 2017 at 21:52
  • \$\begingroup\$ @PeterTaylor : they were no answer to the previous question. In order to be closed as a duplicate the target needs to be already answered. You known it was closed an unclear, so please suggest change to make this answer clearer. \$\endgroup\$ Commented Apr 4, 2017 at 21:54
  • \$\begingroup\$ Huh? It's open and has 15 answers. \$\endgroup\$ Commented Apr 4, 2017 at 22:09
  • \$\begingroup\$ @PeterTaylor sorry, I thought to an another question that was closed as unclear and didn’t take time to read your link. In that case NO, the aim is to not use the compiler in order to actually build the file. This normally can’t be done with a compiler or an assembler. \$\endgroup\$ Commented Apr 4, 2017 at 22:16
  • \$\begingroup\$ Can't it? Why not? \$\endgroup\$
    – wizzwizz4
    Commented Dec 16, 2017 at 19:55
  • \$\begingroup\$ Wait... shortest code that generate any program? Or what? Don't think this is a good idea... \$\endgroup\$
    – DELETE_ME
    Commented Jan 6, 2018 at 12:10
-1
\$\begingroup\$

Removing a Letter adds a Letter

Your program should output nothing when unaltered, however, when any single character is removed it should have an output length of 1. This extends to any number of characters being removed from the program, as long as there is, at minimum, a single character remaining.


For example, if my program were abcdefg, it should output nothing if unaltered.

However, if I were to remove a and d from this program to get bcefg, it should output any two printable characters that represent 16 bytes of information (2 characters for 2 characters removed).

  • So if bcefg outputs (00,AA,etc...) this is valid.

Taking this further, if we were to remove all but the letter g we'd need an output of 6 characters.

  • So if g outputs ('000000','@$^%@(',etc...) this is valid.

Your program must function for all possible combinations of removals that are possible, that is to say each single letter in your program should be a valid program.


Rules

  • You may "lock" pieces of the code, each locked byte counts for 2-bytes instead of 1-byte.
    • Locked bytes will never be removed.
    • For instance, if my program was abcdefg and bcd is locked, the shortest program we'll get is abcd,bcde,bcdf and bcdg.
    • If bcd was locked in abcdefg it'd be 10 bytes, not 7.
  • The program may output any byte to represent 1 removed character, N-bytes for N removed chars in the code itself.
\$\endgroup\$
4
  • \$\begingroup\$ The rule only leads to totally locked code \$\endgroup\$
    – l4m2
    Commented Mar 13, 2018 at 0:13
  • \$\begingroup\$ @l4m2 hah. I disagree. \$\endgroup\$ Commented Mar 13, 2018 at 0:58
  • \$\begingroup\$ But more constructively, increase the penalty? Limit locked chars? \$\endgroup\$ Commented Mar 13, 2018 at 1:04
  • \$\begingroup\$ Maybe require an unlocked percent? \$\endgroup\$
    – l4m2
    Commented Apr 6, 2018 at 10:52
-1
\$\begingroup\$

Sandbox:

Is this question already available (duplicate)?

Are things too vague?

Does providing the example help or hinder?

Tidy the Pantry (easy)

I hate grocery shopping, particularly the part where I put groceries away--so I'm calling upon the collective hive-mind to handle that.

Challenge

Your challenge is to take a 1D-list of groceries and a 2D pantry as input; and output an newly assorted pantry. The two variables can be of your type choice, and in any order, but please specify what item types your program requires (e.g. string, array, etc.).

Rules & Additional info.

Scoring

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

Rules

  • The pantry should be ordered alphabetically (A - Z, left to right, top to bottom)
    • For simplicity, the pantry is case-insensitive
  • The pantry must retain its horizontal size (but trailing newlines are optional)
  • "Pockets" (empty spaces) should be filled between items (i.e. only the last item is allowed to have a trailing pocket)
  • If the pantry is too small for the incoming groceries, then the pantry must replace older items (Z being the oldest, A the youngest)
    • Z from groceries is younger than A in pantry
  • Standard loopholes are forbidden

Examples ([ and ] are used for readability)

Input (4x4 pantry):

[A][A][ ][ ]
[ ][ ][B][ ]
[C][ ][ ][ ]
[ ][ ][ ][D]

AAD

Output:

[A][A][A][A]
[B][C][D][D]
[ ][ ][ ][ ]
[ ][ ][ ][ ]

Input (2x2 pantry):

[A][B]
[C][D]

XYZ

Output:

[A][X]
[Y][Z]

Example solution

JavaScript ES6 (989 bytes)

// (String, String) -> String
let organise = (pantry, groceries) => {
  let n = pantry.split("\n").sort((a, b) => b.length - a.length); // used at the end of the function for horizontal sizing
  n = n[0].length;

  pantry = pantry
    .replace(/\W/g, "") // get rid of all non-alphanumeric characters
    .split("");         // turn the string into an array

  // we need the properties of the new array
  // so the extra `pantry = pantry` is needed
  pantry = pantry
    .slice(0, pantry.length - groceries.length) // go ahead and remove the last overlapping elements
    .concat(groceries)                          // add the groceries to the pantry
    .join("")                                   // turn into a string
    .split("")                                  // turn into an array
    .sort()                                     // sort the array
    .join("");                                  // turn into a string

    return pantry.replace(RegExp(`(.{${n}})`, 'g'), "$1\n");
};

/** Testing below **/

console.log("Test #2:\n" + organise(
`AJCHDJE
JJ   JA
    ASD
OOQ I U
Q     W
      R`,

'AHJBCJHDHHATTGEH'
))


Test Cases:

Test #1, 4x4 pantry

TVCX <- pantry
ABCD
ATDJ
UAIK

XYXY <- groceries
----
AAAB <- expected output
CCDD
IJKT
XYXY

Test #2, 7x6 pantry

AJCHDJE
JJ   JA
    ASD
OOQ I U
Q     W
      R

AHJBCJHDHHATTGEH
-------
AAAAABC
CDDDEGH
HHHHJJT
T

Test #3, 10x10 pantry

AAAAAAAAAA
AAAAAAAAAA
AAAAAAAAAA
AAAAAAAAAA
AAAAAAAAAA
AAAAAAAAAA
AAAAAAAAAA
AAAAAAAAAA
AAAAAAAAAA
AAAAAAAAAA

ZZZZZZZZZZZZZZZZZZZZ
----------
AAAAAAAAAA
AAAAAAAAAA
AAAAAAAAAA
AAAAAAAAAA
AAAAAAAAAA
AAAAAAAAAA
AAAAAAAAAA
AAAAAAAAAA
ZZZZZZZZZZ
ZZZZZZZZZZ

Test #4, 16x16 pantry pantry

ASDFGHJKLZXCVBNM
QJKAJ  KAKSJD  J
KJASDKFHI YOIER
W   OSDOFJ    DK
E PPPASP     AS
R
TASD 
YAAAAAAAAAAAA
U          JHOLK
IIAUSHODUYOAISUO
OASD  AUSODI 
PIASND JUASJNOIJ
A ASJDH PPOIO 
QHIAIUSOIUOOO
WYYAIUSNNAJSDASD
EAISDUUIOPJPIJPJ
ROQPEWIHRNXCAISD

QWERTYUIOP
----------------
AAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAA
ABCCDDDDDDDDDDDD
DDDEEEEEFFFGHHHH
HHHIIIIIIIIIIIII
IIIIIIJJJJJJJJJJ
JJJJJJKKKKKKKKLL
MNNNNNNOOOOOOOOO
OOOOOOOOOOPPPPPP
PPPPPPQQQQRRRRRS
SSSSSSSSSSSSSSSS
SSSTTUUUUUUUUUUU
UVWWY

Test #5, 2x2 pantry

HE
LO

[no groceries]
--
HE
LO
\$\endgroup\$
6
  • 1
    \$\begingroup\$ why divide the program score? \$\endgroup\$
    – RedClover
    Commented Feb 26, 2018 at 19:03
  • 1
    \$\begingroup\$ I recommend you do count by bytes otherwise someone is just going to encode their entire program in Chinese characters and win. \$\endgroup\$
    – hyper-neutrino Mod
    Commented Feb 26, 2018 at 19:09
  • \$\begingroup\$ @labela--gotoa To get a golfed score (smaller programs get a smaller score), should I change it? \$\endgroup\$ Commented Feb 26, 2018 at 19:13
  • \$\begingroup\$ @EphellonDantzler I don't understand why not just normal scoring...? \$\endgroup\$
    – RedClover
    Commented Feb 26, 2018 at 19:14
  • \$\begingroup\$ LOL, that's why I set in in Sandbox first @labela--gotoa \$\endgroup\$ Commented Feb 26, 2018 at 19:16
  • 1
    \$\begingroup\$ Some notes on your reference implementation: 1 It appears far too soon in the challenge. 2 It's not 1768 bytes. 3 You need to ungolf it and make it readable or it's not much use. 4 As it's JS, create a Snippet for it. 5 Is it necessary? It seems to be thrown in there to try to patch over any holes in the challenge spec. \$\endgroup\$
    – Shaggy
    Commented Feb 26, 2018 at 23:17
-1
\$\begingroup\$

Interpret pseudocode

Wikipedia says pseudocode

is intended for human reading rather than machine reading.

and

A program in pseudocode is not an executable program.

I don't care.

Make a pseudocode interpreter that can run pseudocode that fits the rules described below*. This is based on the IB pseudocode guide, but it is simplified quite a bit to make it fit for the challenge.


Pseudocode specifications

This is a simplified pseudocode to make the challenge less tedious. The pseudocode language has no strings, no arrays, no classes, no methods, and no variables other than integers.

Basic syntax

Comments that start at // and end at a newline (like java one-line comments). // is not necessarily followed by a space, and the comment may be empty. Example:

A = 2 + 3 // I can't write five because my keyboard is broken

Statements are separated by newlines. Lines may be empty (without statements). The exact number of spaces doesn't matter, and spaces are not required. The language is case sensitive.

Variables

All variables are global, and can be accessed anywhere. They do not need to be declared. To keep things simple, all variables can be assumed to be integers. All variable names are UPPERCASE, and consist only of letters. Your program should at least handle integers from -256 to 256. A wider range is not a requirement.

Variables are assigned values using this syntax:

VARIABLE = Expression

Where VARIABLE can be any uppercase name and expression can be any integer expression, as discussed below.

Examples:

A = 5
B = A + 3
NUMBER = A * B

Expressions

An expression can be:

  • An integer, like 42
  • A variable, like NUMBER
  • A binary operation on two other expressions, like NUMBER + 5. There are only four operations: +, -, *, /. Division rounds integers down.

Expressions can be surrounded by parentheses to indicate that they need to be evaluated first. To keep things simple, all expressions are evaluated from left to right no matter what the operations are (unless there are parentheses that specify otherwise), so

A = 2 - RM * 9 + 3 / NUMBER
B = 1 + 2 * (3 - 4) / 6

is equivalent to

A = (((2 - RM) * 9) + 3) / NUMBER
B = ((1 + 2) * (3 - 4)) / 6

Boolean expressions

Boolean expressions can compare two expressions using == (equality), != (not equal to), < (less than), and > (greater than). They are only used for control flow, as discussed below (there are no boolean variables).

Control flow

There are four types of control flow. They can be infinitely nested in all combinations.

If

if (booleanExpression) then
    // statements (discussed below)
endif

If-else

if (booleanExpression) then
    // statements (discussed below)
else
    // other statements
endif

Loop while

loop while (booleanExpression)
    // do stuff
endloop

Where booleanExpressions are boolean expressions. The ifs work the same as in normal programming languages. The while loop is a simple while loop.

The booleanExpressions will always be surrounded by (). The pseudocode is very flexible with spaces, and any number of spaces is valid.

Loop for

loop VARIABLE from Expression1 to Expression2
    // things to do over and over again
end loop

Where Expression1 and Expression2 are expressions that are evaluated before the loop begins and their values are stored until the loop finishes. The content of the loop is executed for every integer from the result of Expression1 to that of Expression1, inclusive. At every iteration, the index variable (VARIABLE in this case) is updated.

Example:

loop I from 3 to 5
    output(I)
endloop

Outputs:

3
4
5

Statements

Output

output(Expression) outputs the evaluated expression. It's like println in programming languages. So:

output(1+1)

prints 2, followed by a newline.

output() with no arguments should print a newline.

Other statements

If the interpreter encounters any other statement that looks like a method call with no arguments, it should pretend it's executing it. For example,

lightsoff()
gohome()

should print (together with a newline):

executing lightsoff
executing gohome

In other words, executing [Method name] should be printed. All statements will be lowercase and will consist entirely of letters.

Keywords cannot be statements. You do not have to deal with the following (it will not appear in the pseudocode): - if() - endif() - loop() - while() - etc.

However, statements that start with keywords are valid. For example, loophole() should print executing loophole, even though loop() itself is not valid.


Challenge rules

  • Your program should take a string as input. It can also take something equivalent, like an array of characters. But you can't take an array of strings; your program must itself separate the lines and tokens. You can also take a file as input.
  • Your program should print the output of the pseudocode in any reasonable form.
  • No standard loopholes.
  • There are no restrictions on what your program should do when given invalid pseudocode.
  • This is code golf. The shortest code in bytes wins.

Example output

1

A = 3
output(A) // prints 3
B = 4 + A * 2
output(B)
helloworld()
output(A + B + 1 * 3)

Should give:

3
14
executing helloworld
54

2

loop NUM from 2 to 20 // cycle through possible prime numbers
    COUNT = 0
    loop DIV from 2 to NUM // cycle through possible divisors
        if(NUM/DIV*DIV == NUM) then // if the number is exactly divisible
            COUNT = COUNT + 1
        endif
    endloop
    if (COUNT == 2) then // if number is prime
        output(NUM)
    endif
endloop

Should give:

2
3
5
7
11
13
17
19

3

Tricky cases that your interpreter should handle:

// empty comment:
//
// empty line:

    // more comment testing // ///
////

if     (3<4) then
    endoftheworld() // a statement
    ifff()
    endifnot()
    // endif in a comment doesn't count
endif
// loops can be empty:
loop I from 0 to 10
endloop
output(I) // variables are global
if(1<2)
    if(3<4) // nesting is ok
        ok()
    endif
endif
// spacing doesn't matter:
output   (2+   8   - 1   )
loop             while(2<1)
    neverhappened()
endloop

Should output:

executing endoftheworld
executing ifff
executing endifnot
10
excecuting ok
9

*Technically, once pseudocode follows rules as strict as those described here, it is arguably not pseudocode anymore. Wikipedia says it's called skeleton code.


Any suggestions?

I double-checked all the specifications, but if anything seems reasonably unclear, please let me know.

\$\endgroup\$
4
  • \$\begingroup\$ Actually that's because the challenge is uninteresting. \$\endgroup\$
    – Xwtek
    Commented May 3, 2018 at 2:11
  • \$\begingroup\$ @Akangka thanks for the feedback. How do you think it could be made more interesting? \$\endgroup\$ Commented May 3, 2018 at 16:47
  • \$\begingroup\$ Unfortunately, there is nothing to improve. You have to find other challenge. Also, it is not pseudo-code. \$\endgroup\$
    – Xwtek
    Commented May 4, 2018 at 3:03
  • \$\begingroup\$ @Ok, thanks. I'll try to think of something. Also, read the *note. :) \$\endgroup\$ Commented May 5, 2018 at 19:39
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