Pick and Flip

code-golf array

Challenge

Given an array of integers, perform the Pick and Flip algorithm, which is defined like this:

1. Set n to 1 (if 1-indexed) or 0 (if 0-indexed).
2. Set i to the n-th element of the array. If it doesn't exist, stop and return the array.
3. Reverse the i-th element of the array, if it exists. For example, 15 reverses to 51, and 20 reverses to 2.
4. Increase n by 1 and return to step 2.

Test Cases

1-indexed:

  3   5  19   6 720  50    Input
 [3]  5  19   6 720  50    Pick
  3   5 (91)  6 720  50    Flip
  3  [5] 91   6 720  50    Pick
  3   5  91   6 (27) 50    Flip
  3   5 [91]  6  27  50    Pick
  3   5  91   6  27  50    Flip
  3   5  91  [6] 27  50    Pick
  3   5  91   6  27  (5)   Flip
  3   5  91   6 [27]  5    Pick
  3   5  91   6  27   5    Flip
  3   5  91   6  27  [5]   Pick
  3   5  91   6 (72)  5    Flip
  3   5  91   6  72   5    Output

0-indexed:

  2   4  19   5 720  40    Input
 [2]  4  19   5 720  40    Pick
  2   4 (91)  5 720  40    Flip
  2  [4] 91   5 720  40    Pick
  2   4  91   5 (27) 40    Flip
  2   4 [91]  5  27  40    Pick
  2   4  91   5  27  40    Flip
  2   4  91  [5] 27  40    Pick
  2   4  91   5  27  (4)   Flip
  2   4  91   5 [27]  4    Pick
  2   4  91   5  27   4    Flip
  2   4  91   5  27  [4]   Pick
  2   4  91   5 (72)  4    Flip
  2   4  91   5  72   4    Output

Restrictions

• Every integer in the array will be at least 1 (if 1-indexed) or 0 (if 0-indexed).
• The array will not be empty.

Scoring

• You may use a function or a full program.
• Standard loopholes and I/O methods apply.
• This is code-golf, so shortest code in bytes wins.

Get BB(k-1) from BB(k)

Here BB(k) is the longest running step of k-state, 2-symbol deterministic Turing machine. A movement of tape head is required per step, as data source does.

Test cases:

47176870 => 107 # BB(5)=>BB(4)
107 => 21 # BB(4)=>BB(3)

Explain your code as there's technical difficulty to test answers.

List directories needed to exist to reach a path ("mkdir -p --dry-run")

(The challenge has been posted.)

The first occurrence of k zeros in \$^\infty 3\$

Given k, find the first(lowest) occurrence of k continuous zeros in \$^\infty 3\$ in 10-adic.

Here, 10-adic mean base-10 but extend infinitely to left, and \$^\infty 3\$ is defined as limit of sequence \$a_1=3, a_{i+1}=3^{a_i}\$. It can be proven that every digits eventually get stable.

Your program should run in reasonable time and space for k=3.

The value ends with 05496638003222348723967018485186439059104575627262464195387, so you should return 19(0-index) for k=1.

Test cases

• 1 => 19
• 2 => 49
• 3 => 356
• 4 => 3448

code-golf

Notes

Inspired by Are there G(63) continuous zeros in G(64)?, which I'd say G(64) is much larger than G(63) for this purpose and can be treated \$^\infty 3\$.

This may be not fun as fastest-code due to likely unique and well-explored path

Maybe this is better posted as "Output 3^^inf" don't count zeros

Rational quadratics sequences code-golf math sequence

Let's define a rational quadratics sequence \$a_1, a_2, \cdots, a_n\$ of length \$n\$ one that satisfies all of the following:

• The integers from 1 to \$n\$ appear exactly once each.
• For \$1 \le i \le n - 2\$, at least one of the equations of the form \$a_{i} x^2 \pm a_{i+1} x \pm a_{i+2} = 0\$ has rational roots. In more direct terms, at least one of \$a_{i+1}^2 \pm 4 a_i a_{i+2}\$ is a square number.

Given a positive integer \$n\$, find the number of rational quadratics sequences of length \$n\$.

sequence I/O rules apply. This means you can choose to implement one of the following:

• Take \$n\$ and give the answer for \$n\$.
• Take \$n\$ and give the answers for \$1, \cdots, n\$.
• Without input, give all answers for \$1, 2, 3, \cdots\$.

Test cases

1 => 1
2 => 2
3 => 6
4 => 4
5 => 4
6 => 4
7 => 4
8 => 10
9 => 4
10 => 32
11 => 10
12 => 26
13 => 32
14 => 68
15 => 136

Find the paths and cycles

Given a finite digraph with all in-degrees and out-degrees at most 1, find its partition into directed paths and cycles.

The problem is potentially interesting because paths and cycles are slightly different - combining the two may inspire some ingenious approaches.

Input

A sequence of pairs of vertices representing the directed edges of the graph, which might appear in any order. You can assume:

• The vertices are precisely the integers 0,...,n for some n>0.

• Every such integer appears in some pair (so there are no isolated vertices).

• Every such integer appears at most once on the left of a pair and at most once on the right of the pair (so in-degrees and out-degrees are at most 1).

Output

A sequence of lists of vertices representing the directed cycles and maximal directed paths of the graph.

• A path must be represented by its vertices in order.

• A cycle must be represented by its vertices in order starting from an arbitrary vertex with the initial vertex repeated once at the end.

• The paths and cycles must include all vertices between them, with no repeats except for the repeat of the beginning and end of each cycle.

• Thus, the edges of the input should be precisely the ordered consecutive pairs within each list of the output (in some order).

• The paths and cycles can be output in any order. Each cycle can be started from any of its vertices.

Any reasonable representation of "list of lists of integers" is acceptable for input and output.

Examples

This is code-golf, so shortest program in each language wins!

Return the minimum Conjunctive Normal Formula equivalent to the one given in input.

Translated by Google + some my changes

A conjunctive normal formula (abbreviated here CNF) is a conjunction of clauses, which are disjunctions of variables. It is represented by writings such as

(A∨B∨∼C∨∼D)∧(Q∨R∨∼P∨∼K)∧...∧(∼M)

where (A∨B∨∼C∨∼D), (Q∨R∨∼P∨∼K), ..., (∼M) are each the clauses of the CNF above, where A B C D ... M are "variables" of the respective clauses. Each of these variables can take on any value in the set TF={1,0}. Once a value in the set TF has been assigned to each variable, the value of the CNF can be calculated by performing the functions ∨ ∧ ∼ with the binary values ​​0 1 of each variable.

The operation ∨ is defined by

1∨0=1; 0∨1=1; 1∨1=1; 0∨0=0

The operation ∧ is defined by

1∧0=0; 0∧1=0; 1∧1=1; 0∧0=0

The function ∼ is defined by

∼1=0; ∼0=1

Two CNFs A(x1,x2,...xn) and B(y1,y2,...yn), with variables {x1,x2,...xn} and {y1,y2,...yn}, are said to be equivalent if any assignment {x1,x2,...xn} ∪ {y1,y2,...yn} with values ​​in the set TF={1,0} yields the same final Boolean value (performing all calculations with the functions ∨ ∧ ∼) in CNF A and CNF B.

If A and B are two conjunctive normal formulas, we say that A is less than B if the sum of all occurence of ∧ and or ∨ of A is less than the sum of all occurence of the and ∧ and or ∨ of B

Examples: Let's say two CNFs A and B with

A=(a1∨a2)∧(a3∨a4) and B=(a1∨a2)∧a3

then B<A because numbers of ∧ and ∨ of B is 2 that is less than the number of ∧ and ∨ of A that is 3.

Here we require a program that, given a CNF A, returns a CNF B that is as small as possible and equivalent to A. The winner is the one who finds the minimum number of the sum of all the or[∨] and and[∧] that appear in all the CNF results of below exercises.

If two answers are tied, then the winner is the one with the smallest program in terms of number of bytes in the source code.

Exercises (the formulas are the same as those in Prove me wrong!)

(P∨∼Q∨∼R)∧(Q∨R∨∼P)∧(Q∨∼P∨-R)
(∼P)∧(S)∧(R∨∼P)∧(U∨∼Q)∧(X∨∼R)∧(Q∨∼S)∧(∼P∨∼U)∧(W∨∼Q∨∼U)
(∼P∨∼Q)∧(Q∨P)∧(P∨∼Q)∧(Q∨∼P)
(P)∧(∼P∨∼S)∧(P∨T)∧(Q∨∼R)∧(∼R∨∼S)∧(∼P∨∼Q∨��R)∧(∼P)
(T∨∼S)∧(∼S)∧(Q∨∼R)∧(∼R∨∼S)∧(∼P∨∼Q∨∼R)∧(∼P)

These formulas can be represented in the code with maximum freedom, for example,

(P∨∼Q∨∼R)∧(Q∨R∨∼P)∧(Q∨∼P∨-R)

could be represented as an array of 3 strings:

('P-Q-R')('QR-P')('Q-P-R')

Examples:

In finding a smaller equivalent CNF for (P∨∼Q∨∼R)∧(Q∨∼R∨∼P)∧(Q∨∼P∨-R) I would have found (Q∨∼P)∧(P∨∼Q∨∼R). It has 1 and ∧, and 3 or ∨ in total so 4 is the length of this result formula that is more little of the one of input that has length 6 or and 2 and that would be 8.

Could be useful these equivalences

A∨B∨..∨D is equivalent to each permutations of variables A B..D 
A∧B∧..∧D is equivalent to each permutations of variables A B..D 
A∨B∨..∨∼A∨..∨D is equivalent to 1 or always true, 1 is the name of that proposition 
A∧B∧..∧∼A∧..∧D is equivalent to ∼1 or always false or 0 zero
A∨∼1 is equivalent to  A
A∨ 1 is equivalent to  1
A∧∼1 is equivalent to ∼1 (that is 0 too)
A∧ 1 is equivalent to  A
A∧(A∨∼A) is equivalent to A
A∧(B∨∼A) is equivalent to A∧B
(P∨Q)∧(P∨R) is equivalent to P∨(Q∧R) useful when R is ∼Q
(A∨B∨C∨D)∧(Q∨R∨P∨K)∧(A∨B) is equivalent to (Q∨R∨P∨K)∧(A∨B)

tag logic codegolf

SCR max speed calculator

SCR is a train driving game on Roblox. Distances are measured in miles, and speeds in miles per hour. The driving guide dictates specific speeds for specific scenarios:

• The speed should never be higher than either the maximum speed of the train or the imposed speed limit.
• If the train is less than or equal to 0.35 miles from the station, the train should go no more than 35mph.
• If it is a terminating station, it should be no more than 15mph.
• If the signal is a single yellow aspect, it should be no more than 45mph.
• If the signal is a red aspect and the distance to the signal is less than the distance to the station, the train should be at a complete stop (0mph). If the station is closer, normal station approach rules apply.
• Double yellow and green aspects do not carry speed restrictions by themselves.

Given the train’s maximum speed, the current speed limit, the distance to the next station, the distance to the next signal, and the current signal aspect, output the maximum speed the train is allowed to move at in mph. If the next signal aspect is red and the station is closer than the next signal, you may assume that the train is at most 0.35mi away from the next station. Shortest code wins, standard loopholes apply.

Too complex?

Inference a transformer

Can you Shut The Box?

Shut The Box is a simple dice game, where a player attempts to "shut" all of the numbered tiles (traditionally, 1 through 9) by rolling dice. The way the game works:

• The box has 9 numbered and hinged tiles, numbered 1, 2, 3, etc. up to 9. Initially, all 9 begin "open".
• A player rolls two 6-sided dice and total the two dice.
• They then "shut" any combination of tiles that sum to the dice total.
• Repeat until all tiles are shut, or no more open tiles can sum to the dice total.
• The player's final score is the total of all remaining open tiles, with a lower score being better.

Note that there are multiple ways that one dice roll can shut different tiles. For example, rolling a total of 5 can close either the 5 tile, the 4 and 1 tiles, or the 3 and 2 tiles. Importantly, there is only one of each numbered tile, so once a tile has been used, it cannot be used again. Thus, for a roll of 6, the combination of 3 and 3 isn't valid, but the 3, 2 and 1 tiles are.

Given a non-empty list of positive integers ranging from 2 to 12 inclusive, output the lowest possible score for that series of dice rolls, in that order. You may take input in any acceptable and reasonable format and method, but please note that, as order matters, the manner in which you take input must distinguish different orders of the same integers.

This is a code-golf challenge, so the shortest code in bytes in each language wins.

Test cases

Input -> Output
[10, 10, 10, 10, 5] -> 0

Simulate gravity

Given how simple this is, I'm pretty surprised there wasn't a question on here for this already. The task for this question is to simulate n-body gravity!

Simple? I thought the n-body problem was famously unsolved!

Yeah, yeah; whatever. There's (provably) no analytic equation, but gravity really just follows one simple rule:

$$F = G \frac{m_1m_2}{r^2}$$

That's it. There's no big gatchas until you start dealing with relativity, and relativity puts gatchas on everything. We're actually going to simplify our simulation even more, and say all our masses are one kg, and exist in a magical world where the gravitational constant G is 1\$ \text m^3 \over \text{kg} \cdot \text{s}^2 \$, so the equation you're going to simulate is

$$\ddot r_i = \sum_{j\ne i} \frac{\hat r_{ij}}{|r_{ij}|^2} \quad \forall i$$

"Gah," I hear you cry. "You tricked me, what are all these extra symbols you've added! There's dots and sums and hats and bars! And your A's upside down, the pointy bit goes up!"

Don't Panic!
Mathematicians love code golf too, most of these symbols are just fancy loops. We'll go one step at a time, from the outside in.
The upside down A is the closest you get to an actual for loop in math, and it means that this whole equation individually applies to each of our bodies, represented by a variable \$i\$. It's basically equivalent to for i in bodies
The double dot is a little more tricky. In math terms, it's the second derivative with respect to time, but we can keep it simpler than that. A derivative (in time) is a fancy way to say "the amount something changes (per second)," so this double dot equation is defining the change of the change of each radius. Our pseudocode now looks like this:

for i in bodies {
    v[i] += equation
    r[i] += v[i]
}

(for the math buffs that just started complaining, see this video for why we update velocity first)

The next piece is the summation. This is also a for loop, except you just sum all the terms and return the result. The thing on the bottom is the values we're looping over. This means equation now looks like equation = sum(fraction(i,j) for j in bodies if j != i)

Almost there! The ij subscript means the vector from \$r_i\$ to \$r_j\$, which you calculate by taking the vector difference: rij = r[j] - r[i].
The bars mean finding the magnitude, which you do by taking the square root of the sum of the squares of the components. In code that's rij_mag = sqrt(rij[1]^2 + rij[2]^2 + rij[3]^2)
The hat means normalizing a vector, which you do by dividing the magnitude; but if you look closely at our fraction we're already doing that! Dividing one more time just means incrementing the exponent.

Putting that all together, our pseudocode is:

for i in range(num_bodies) {
    v[i] += sum(
        (r[j]-r[i]) / sqrt(
            ( r[j][1]-r[i][1] )^2 + ( r[j][2]-r[i][2] )^2 + ( r[j][3]-r[i][3] )^2
        )^3
        for j in range(num_bodies)
        if j != i
    )
    r[i] += v[i]
}

Requirements

Your task for this code-golf challenge is to simulate one second of gravity. That is, your code should do the equivalent of running the above for loop once. You will be given \$n\$ bodies, where each body has a velocity and position. You may take these \$6n\$ numbers however you like, but you should return the updated \$6n\$ numbers in the same way. (ie, your code should accept its own output as a new input)

You can assume that two bodies will never perfectly overlap (so there's no division by zero), and you don't need to worry about precision (so just use the default floaty numeric type)

Test cases

{velocities}, {positions} -> {v}, {r}

{}, {} -> {}, {}
{[10, 5, 1]}, {[1, 2, 3]} -> {[10, 5, 1]}, {[11, 7, 4]}
{[0,0,0], [0,0,0], [0,0,0]}, {[1,0,0], [0,1,0], [0,0,1]} -> {[-.7071, .3535, .3535], [.3535, -.7071, .3535], [.3535, .3535, -.7071]}, {[.2928, .3535, .3535], [.3535, .2928, .3535], [.3535, .3535, .2928]}

Brownie Points

We simplified some stuff here, but we've done the hardest part of a fully fledged n-body gravity simulation. If you want extra brownie points, you can add those complications back in.
First, we want the gravitational constant G.
Then, we allow for masses on each bodies with a mass[] array
Finally, we'll take t and dt for total time and step time

Our updated pseudocode is:

T=0
while T < t {
    for i in range(num_bodies) {
        v[i] += G * sum(
            mass[j] * (r[j]-r[i]) / norm(r[j] - r[i])^3
            for j in range(num_bodies)
            if j != i
        ) * dt
        r[i] += v[i] * dt
    }
    T += dt
}

Match the BF output code-golf regex brainfuck

Your task is, when given a BF program as input and 256 1's on the line after the BF program, match the amount of 1's equal to the value of the pointer when the program halts.

For a BF tutorial, see here.

Example programs:

+++
(255 1's)

(The above program should match 3 1's.)

++[>+++>+++<<-]>
(255 1's)

(The above program should match 6 1's.)

Specs:

• The tape in question is 10 long, with wrapping.
• There is wrapping for the individual values: if it goes over 256, it goes back to 0.
• There is no , or . in the BF code.
• You are guaranteed that the code will terminate.
• You may use any flavour you want.
• You may match 1's in any position, as long as you match the correct amount of 1's.

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

Meta stuff:

• How hard would this be to pull off? Is it even possible?
• Anything I could improve with explanations?

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 code-golf, 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]

code-golf number-theory primes

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 code-golf so the shortest code wins

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.

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 code-golf 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.

Plan and Chain a route through OEIS

popularity-contest number math number-theory bitwise sequence integer

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

Braid Badly Boundlessly

code-golfstring

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

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

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 fastest-code, so may the fastest code win and the best programmer prosper...

What's that character? (Part 1) code-golf

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?

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 code-golf, 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.

Golf Cubically code code-challenge fgitw

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.

Hungry for Apples?

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 ascii-artcode-golfhungry-for-apples

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 code-golf, 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.

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?

Is it a perfect loop? test-battery decision-problem

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 code-golf, so byte count is not needed in your solution. Just post the language name and add the percentage correct when I comment.

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.

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.

Non-true, non-false JS boolean

Array prototype isn't redefined, input hasn't getters

function magic(input){
  let result = [];
  if(input.boolean != true){result.push("non-true");}
  if(input.boolean != false){result.push("non-false");}
  result.push(input);
  return result.join("\n");
}

returns

non-true
non-false
{"boolean": true}

What is passed to magic function?

Based on real problem :) I spent 30 minutes on this puzzle

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