Sandbox for Proposed Challenges

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.

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Solve Fermat's Last Theorem with matrices

Fermat's Last Theorem states that there is no such triple of positive integers $$\(x, y, z)\$$ such that

$$x^n + y^n = z^n$$

for all integers $$\n > 2\$$. However, this conjecture does not hold for integer matrices:

\begin{align} & \left( \begin{matrix} 1 & 3 \\ 0 & 1 \end{matrix} \right)^3 + \left( \begin{matrix} -1 & 0 \\ 1 & -1 \end{matrix} \right)^3 \\ = & \left( \begin{matrix} 1 & 9 \\ 0 & 1 \end{matrix} \right) + \left( \begin{matrix} -1 & 0 \\ 3 & -1 \end{matrix} \right) \\ = & \left( \begin{matrix} 0 & 9 \\ 3 & 0 \end{matrix} \right) \\ = & \left( \begin{matrix} 0 & 3 \\ 1 & 0 \end{matrix} \right)^3 \\ \end{align}

You are to take two $$\k \times k\$$ integer matrices $$\A\$$ and $$\B\$$, and a positive integer $$\n \ge 2\$$, and output a $$\k \times k\$$ integer matrix $$\C\$$ such that

$$A^n + B^n = C^n$$

You may output any such $$\C\$$. You may also take $$\k\$$ as an input, and you may assume that $$\k \ge 2\$$

This is , so the shortest code in bytes wins

• Is it guaranteed that A, B, n are given so that such a C exists? Why not just make a challenge about taking the nth root of a given matrix (which is arguably the hardest part of the task)? Jul 14 '21 at 6:54

I'm a lizard, cut here!

• I think you mean n≥2
– xnor
Jul 21 '21 at 5:05
• @xnor sure thanks Jul 21 '21 at 5:11

How effective is compression?

Vyxal has a very simple format for compressed integers: a base-255-encoded string of characters wrapped in «.

With base 255, the number to be compressed needs to be 10000 or greater for the compression to have an advantage against ordinary numbers, as with 1 character, you can represent at most 3-digit numbers (up to 254), which ties with 1 + 2 (for the «) = 3 bytes; with 2 you can represent up to 65024 in 4 bytes, beating 5 for n ≥ 10000.

But what if it was another base?

If it were base 99, it would take 3 bytes, since you can only express up to 9800 with 2, which isn't any better than expressing the number. But with 3, you can reach 970299, which has 6 digits

Your challenge is to take a positive integer $$\b\$$ 11 or greater, and find the smallest number $$\n\$$ such that $$\log_b(10^n) + 2 < n\$$. You may output $$\10^n\$$ instead.

This is , shortest wins!

Testcases coming soon.

• Be sure to include the testcases for n=100 and n=1000. The correct answer is floor(1+2/(1-1/log10(n))), but the wrong answer ceil(2/(1-1/log10(n))) gives identical results for all other n. (The correct answer can be also wrong due to floating-point imprecision. Is this allowed?) Jul 29 '21 at 7:45
• @Bubbler Ok, will do. Shouldn't be wrong as it's taking a single logarithm. Jul 29 '21 at 7:56
• Straightforward Python fails because of 1/3, not single logarithm btw Jul 29 '21 at 8:13

KOTH: Prisoners Dilemma Single Elimination king-of-the-hillpython

I was playing around with this and found some interesting strategies. This is sufficiently different from this post because you are playing different bots each game, instead of repeatedly playing the same ones.

How the game works

The basic game is the prisoners dilemma. A bot and it's opponent may choose to defect (1) or cooperate (0). Defecting gives you 1 point, and if your opponent cooperates you gain 2 points. If you defect and your opponent cooperates this adds up to a total of 3 points.

In every round, you play against every opponent (including yourself) in random order. The bot does not know who it is playing against. This process repeats 100 times to form a single round, after each round the player with the lowest score is eliminated, and the scores of the rest of the players is divided are 2. Round continue until either every remaining player always defects, or always cooperates, at which point the best player is decided by their score. (After this points scores converge quite fast, so to avoid noise deciding the winner I measure at the point with the most pronounced differences)

Input

Your bot does not know who it is playing against, not even if it is playing against itself. However, the bot does know what its opponent decided on its last 3 games, and for each of those 3 games what the opponent's opponent decided. Basically, the input will look like this:

[
[0, # Three games ago, your opponent decided not to defect
[1, # Your opponent's opponent from three games ago defected three games before that
0,
0],
[1,  # Your opponent defected in their second to last game
[1, # Your opponent's opponent from two games ago defected the 3 games before that
1,
1]
[0, [0, 0, 0]
]


(With 1 and 0 replaced with True and False)

Any of these players might be the same. Your opponent's opponent could actually be your current opponent if they played against themselves last, or it could be yourself if you played against eachother twice in a row. (Unlikely, but possible at the end of a round)

Your bot does not receive what it's own last decisions are but is free to keep that info in it's own state.

Output

You can output any value, if bool(output) == True your bot defects, otherwise your bot cooperates.

Example Bots

Nice bot never defects:

class NiceBot:
def get_choice(self, last_3_games):
return 0 # Any False-y value will work, for example False or an empty string


Mean bot always defects:

class MeanBot:
def get_choice(self, last_3_games):
return 1


Majority bot chooses whatever it's opponent chose the most often:

class MajorityBot:
def get_choice(self, last_3_games):
choices = [i[0] for i in last_3_games]
return max(set(choices), key=choices.count)


Bandwagon bot will avoid decisions by chosing what the bots opponents chose the most.

class BandwagonBot:
def get_choice(self, last_3_games):
total_true = sum(
i[1].count(True)
for i in last_3_games
)



Bandwagon bot will only play if the number of bots is even.

Other specifics

A bot may implement a reset() method that will be called every time an opponent is eliminated. For bots that keep state this might be a good time to clear it.

Other rules:

• You are not allowed to use any form of randomness.
• You are not allowed to use IO
• Standard loopholes apply, in particular no exploiting the controller
• Standard library imports are allowed for simple functionality that you could implement yourself without breaking any of the other rules eg itertools, functools but not for example sys, time or asyncio. Ask if you are not sure. Definitely no attempts to import parts of the controller or other bots.

Controller

I have a controller written but I would need to simplify it a little to remove some parts that have become redundant. Don't want to put too much effort into making it presentable if it is not a suitable challenge.

Train A Single Perceptron

You must write a program that will train a perceptron to simulate the things described in the test cases section of this question.

What is a perceptron and how to train it

A perceptron takes inputs and returns 1 if the dot product of the inputs and its list of weights is greater than or equal to 0 and it returns 0 otherwise. To take the dot product, multiply the elements of the list and take the sum. In other words, your program must find the line (w⋅x=1) that linearly separates the two possible classes of outputs. Example: dot product of [0,2] and [8,3] is 0+6, which equals 6. Your job is to return the list of weights that will make the perceptron return the correct outputs for a list of inputs 100% of the time.

https://desmos.com/calculator/d2nryjmw2t

https://towardsdatascience.com/perceptron-learning-algorithm-d5db0deab975

Inputs

The program should take as input training inputs and training labels. Each list of inputs corresponds to exactly one output. This data must be used to train the perceptron.

Outputs

The program should output the final list of weights as integers. The dot product of these weights and any valid input to the perceptron should be the correct output of the perceptron 100% of the time.

Training

How you train the perceptron is up to you. To train a perceptron to simulate an AND gate, your program should take training inputs and the correct output for each list of inputs. In the case of the AND gate, your program would take [[0,0],[0,1],[1,0],[1,1]] as the list of training inputs and [0,0,0,1] as the list of what the neuron should output given each input. The program should then output a list of weights, which will output the correct output when used in a perceptron.

Test Cases

1. AND gate:
• Input List: [[0,0],[0,1],[1,0],[1,1]]
• Correct Outputs: [0,0,0,1]
• Your program should output a list of two weights that will solve the problem of behaving like an AND gate when used in a perceptron.
2. Only outputs 1 if the third input is 1:
• Input List: [[0,0,0],[0,0,1],[0,1,0],[0,1,1],[1,0,0],[1,0,1],[1,1,0],[1,1,1]]
• Correct Outputs: [0,1,0,1,0,1,0,1]
3. OR gate:
• Input List: [[0,0],[0,1],[1,0],[1,1]]
• Correct Inputs: [0,1,1,1]
4. Make up your own test case, and try to make it interesting!

Scoring

This is code golf, so the shortest program in bytes wins.

• Could you maybe add a note that this task is to find a line ($w \cdot x=1$) that linearly separates the two classes? That might make the challenge more accessible to people unfamiliar with perceptrons. As a side note, the decision are of linear perceptrons can be visualized quite nicely: desmos.com/calculator/d2nryjmw2t
– ovs
Aug 1 '21 at 10:51
• I guess you changed the threshold to $1$ to avoid the bias? I'm not quite sure this works because now the origin $(0,0)$ is always classified as $w\cdot 0=0$. I think just going with the usual definition should be fine.
– ovs
Aug 1 '21 at 10:55
• Ok, I updated it. Aug 4 '21 at 2:11

Eric Angelini's "1995" puzzle

• your post looks good, tags can be code-golf and sequence. Aug 6 '21 at 14:19

Prove NL=coNL

Description

One of the achievements of complexity theory is showing that NL, the set of problems with solutions that can be verified in logarithmic space, is the same as coNL, the set of problems which can be verified to have no solution in logarithmic space (i.e. the set of problems whose complement is in NL). Your task is to implement a logarithmic-space verifier for the complement of the NL-complete problem "graph connectivity." More precisely, you will be given a directed graph $$\ G \$$ in adjacency matrix format, as well as two vertices $$\ s \$$ and $$\t\$$. Then, you may read some input from STDIN, which you should interpret as a "proof" $$\P\$$ that $$\s\$$ and $$\t\$$ are not connected (i.e. there is no path from $$\s\$$ to $$\t\$$). Finally, you must output whether $$\P\$$ is a valid proof of the fact that $$\s\$$ and $$\t\$$ are disconnected in $$\G\$$. All of this must happen using $$\O(\log n)\$$ extra space, where $$\n\$$ is the size of the adjacency matrix of $$\G\$$.

Rules

• You will write a function that takes $$\(G, s, t)\$$ as inputs. Here, $$\G\$$ is a 2d $$\\ell\times\ell\$$ array, and the vertices are integers less than $$\\ell\$$. If $$\i\$$ and $$\j\$$ are less than $$\\ell\$$, then G[i][j] is truthy (you can decide exactly which truthy value) if there is an edge from $$\i\$$ to $$\j\$$, and falsy otherwise.
• You may not write to the array $$\G\$$, but it does not count towards your memory usage limit of $$\O(\log n)\$$. Be careful about making copies of the array, e.g. G.map(x => ...) in javascript would not be allowed, because it makes a copy of G, which uses too much extra memory.
• You can decide the format of the input from STDIN; there are no restrictions, other than the fact that for every input $$\(G, s, t)\$$, if $$\s\$$ and $$\t\$$ really are disconnected, there must be some proof that works, and conversely, if $$\s\$$ and $$\t\$$ are connected, no proof should be accepted.
• To be clear, when $$\s\$$ and $$\t\$$ really are disconnected, you can think of the proof as "maximally helpful," as though the input wants to prove that $$\s\$$ and $$\t\$$ are disconnected. However, if they are connected, the proof is adversarial. It is trying to trick you into reporting that the vertices are disconnected when they really aren't.
• Pretend that the native integer type of your language is unbounded. On the other hand, for the purposes of counting memory, an integer requires $$\O(\log |i|)\$$ memory, where $$\i\$$ is the value of the integer, because it requires that many bits to store.
• Rather than reading input from STDIN, you may receive the proof $$\P\$$ from user input in another way (e.g. repeated prompt calls in javascript). If your language supports it, you may choose to receive the proof in the form of an "input stream" I, as long as the following restrictions are on the "input stream": you should not be able to have random access to the proof's contents (i.e. access the $$\n\$$th byte without first getting the first $$\n-1\$$ bytes), and once you read a byte or some bytes, they should be "forgotten" unless you explicitly save it (and if you save it, it counts towards your $$\O(\log n)\$$ limit). Thus, for example, an array would not be acceptable, since it has random access, but it would be acceptable to accept a function I which returns the next byte of the "simulated input" every time you call it.
• If your program throws an error, you may count that as equivalent to returning false (i.e. $$\P\$$ is not a valid proof).
• Your program may exceed the space constraints when $$\P\$$ is an invalid proof, since as Anders Kaseorg pointed out, such a program can easily be converted into one which meets the space requirements always.
• Fewest bytes wins!

Algorithm

An algorithm is given in section 7 of here (or is easily found on Google), but I will briefly describe it here for convenience (you do not have to use this algorithm/format):

We will ask the proof to build a sequence $$\r_0, r_1, \dots r_\ell\$$, where $$\\ell\$$ is the number of vertices, such that $$\r_i\$$ gives the number of vertices reacheable from $$\s\$$ in at most $$\i\$$ steps. Initialize $$\r_0 = 1\$$. Then, given $$\r_i\$$, the proof gives a claim of what $$\r_{i+1}\$$ should be. It proves this claim by claiming for each vertex $$\v\$$ whether $$\v\$$ is reachable in $$\i+1\$$ steps from $$\s\$$. In either case, for each vertex, the proof lists all $$\r_i\$$ vertices along with the paths used to get there. We verify that $$\r_i\$$ distinct vertices have been listed, and that at least one of them (or none of them) is connected to $$\v\$$, depending on what was claimed. Then, when we are proving $$\r_\ell\$$, we check that $$\t\$$ is not one of the vertices that can be reached from $$\s\$$.

Test cases

Ideally, you would write a separate program that takes an input $$\(G, s, t)\$$ and outputs a possible proof $$\P\$$ if $$\s\$$ and $$\t\$$ are disconnected; this just makes testing easier, and it does not count towards your total bytes.

G s t --> are_disconnected
[[0,0,0,0],[1,0,1,0],[0,1,0,1],[0,1,1,0]] 0 3 --> true
[[0,1,1],[0,0,1],[1,1,0]] 1 0 --> false
[[0,0],[0,0]] 1 1 --> false
[[0,0,1,1],[0,0,1,0],[0,1,0,1],[0,0,1,0]] 1 0 --> true


Here, true should be interpreted as there is a valid proof that s and t are disconnected, whereas false is interpreted as there is no valid proof.

Which in and out shuffles do I need?

If you watch Matt Parker's latest video as of time of writing, you'll know that for any given start and end position, there is a sequence of six in and out shuffles that will move the card at the start position $$\ p \$$ to the end position $$\ q \$$. One method is to use the 0-indexed formula from the video: Express $$\ 64p-q \$$ in the form $$\ 52t-s \$$ for the smallest positive integers $$\ t \$$ and $$\ s \$$, and then take the bitwise XOR of $$\ s \$$ and $$\ t \$$ as a 6-bit binary integer. For instance, to move the card from position $$\ 16 \$$ to position $$\ 22 \$$, we have $$\ 16 \times 64 - 22 = 1002 = 20 \times 52 - 38 \$$, so the result is the bitwise XOR of $$\ 20 \$$ and $$\ 38 \$$, which is $$\ 50_{10} \$$ or $$\ 110010_2 \$$, so you need two in-shuffles, two out-shuffles, an in-shuffle and an out-shuffle.

Given the start and end position, which you can take as 0- or 1-indexed as you prefer, output the list of shuffles required. You don't have to output a list of six shuffles, but you cannot use more than six shuffles, and if your output is variable length you need to indicate the length in some way. You can use any two distinct symbols to indicate the two types of shuffles, except you can't use 1 or I for out-shuffles or 0 or O for in-shuffles, as that would be too confusing.

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

• Related Aug 10 '21 at 9:21
• I can think of a couple of languages with automatic wins (Lost, Cascade). it's probably impossible in a standard language, and merely very hard in most 2D langs
– Jo King Mod
Aug 25 '21 at 10:07

Fletcher's 16 bit checksum code-golf

Intro

No story this time :( This is something that I needed for a program and I thought that it would make for a fun little challenge. I'd love to see some unusual languages here!

Challenge

Implement Fletcher16. The algorithm is very straight forward, as seen here in pseudocode:

    int sum1, sum2, check1, check2;
char[] message;

sum1 = 0;
sum2 = 0;
for i from 1 to message_length do
sum1 = ( sum1 + message[i] ) modulo 255;
sum2 = ( sum2 + sum1       ) modulo 255;
end for


There are two accumulators sum1 and sum2. We iterate over the message and for every byte inside of it we add the byte to the first sum and then modulo the value by 255. We then add the value of the first accumulator to the second one and modulo that by 255 as well. This is Fletcher's algorithm.

    check1 = 255 - (( sum1 + sum2) modulo 255);
check2 = 255 - (( sum1 + check1 ) modulo 255);


We then calculate the checksum by "simulating" the algorithm for the two resulting values. After subtracting the numbers from 255, we get two values check1 and check2. They are then appended to the input in that order.

To recap:

• Run the first part to get a "raw" checksum
• Run the second part on that raw checksum to get the Fletcher16 checksum.
• Append the Fletch16 checksum to the input.

Evaluating the checksum

• Run the first part on the string with the appended Fletcher16 checksum
• If 0: String is valid, else String isn't valid.
• Note: If we run the second part as well, the result will be 0xFFFF

Input

• Binary data in any format, length is at least 1.
• If needed, the length of the data.

Output

• The Fletcher16 checksum check2 << 8 | check1, since this is what gives the algorithm its name.

Rules

• This is , shortest answer wins
• No standard loopholes
• No builtins. I doubt that there's a language that has one for this, but if so that that's no fun.
• A submissiom may be a program/function/link/lambda/chain/etc.

Test cases

abcde -> 0xC846
abcdef -> 0x2088
abcdefgh -> 0x06D2

abcde\x46\xC8 -> 0xFFFF
abcdef\x88\x20 -> 0xFFFF
abcdefgh\xD2\x06 -> 0xFFFF


You can find a reference implementation in C here

Sandbox

• Are there any mistakes?
• Is something unclear? I'm really not sure about the wording of the input.
• Is the output restriction reasonable?
• The pseudocode is nice but I don't think it is an adequate substitution for an actual explanation of the algorithm.
– Wheat Wizard Mod
Aug 25 '21 at 9:20
• Ok, I'll see what I can do. I'll also add a C reference implementation as soon as I get one working. Aug 25 '21 at 9:24

A phonetic letter

input
a letter of the alphabet, A-Z (upper or lower or a mix, I don't mind)

output
the phonetic (i.e spelled name of the )letter of the one passed in, along with the appropriate indefinite article a or an (again, in any case), and any amount of leading/trailing whitespace you like (i.e. it's fairly flexible)

full set of test cases (lowercase) - see This english.se Answer

input output
a an a
b a bee
c a cee
d a dee
e an e
f an eff (or "an ef")
g a gee
h an aitch
i an i (or "an eye")
j a jay
k a kay
l an el (or "an ell")
m an em
n an en
o an o (or "an oh")
p a pee
q a cue (or "a queue")
r an ar (or "an arrrrr") with an arbitrary number of rs
s an ess (or "an es")
t a tee
u a u (or "a you")
v a vee
w a double-u (or "a double-you", matching u above)
x an ex
y a wy (or "a wye")
z a zed (or "a zee", if you must)

, usual rules and exceptions, shortest bytes wins.

Interpret Tarfish

Tarfish is a more tarpit-style version of ><> that I recently created.

It is two-dimensional, and has a stack (array of numbers that can be popped and pushed) and an instruction pointer, which has x and y coordinates and a direction. The instruction pointer starts at 0,0 (top left corner of program), moving right.

It has the following commands (some are not included because they would overcomplicate this challenge):

Command action
> Set the IP's direction to right
v Set the IP's direction to down
< Set the IP's direction to left
^ Set the IP's direction to up
. Pop stack and output as character
, Pop stack and output as number
x Push x position of the IP to stack
y Push y position of the IP to stack
+ Increment the top item of the stack
- Decrement the top item of the stack
= Pop the top two items from the stack, if they're equal, skip the next instruction.
{ Shift the stack right - put the ToS on the bottom of the stack
} Shift the stack left - put the bottom item on top of the stack
_ Pop the stack

Every tick, the command pointed to by the IP is executed (Unless it was skipped), and the instruction pointer moves one place in its direction.

Despite having so few commands, Tarfish is in theory Turing-Complete.

Your challenge is to interpret this language.

You do not have to implement wrapping - If the instruction poiinter leaves the grid, that is undefined behaviour. If the program tries to pop from an empty stack, that is undefined behaviour.

Scoring

This is , shortest wins!

Testcases

x++++++++++++++++++++++++++++++++:+:+++++++++++:++++++++++++++++++++++++++++:.+++++++++++++++:++++++++++++++:.+++++++:::..+++:.{{{.{.{{.}}.:++++++.:.--------..; => Hello, World!

x+, => 1

v > v > x,
> ^ > ^
=> 8

x:++++++++++v
>+}     :y-=v{,;
^+++++++++{-<
=> 100

• this may be closed as a subset of the existing ><> challenge. Aug 27 '21 at 5:40
• @Razetime About half the commands are different, and this is much simpler. Aug 27 '21 at 8:16
• What's the stack item's range? And what should happen if output as a character is chosen when the top of the stack isn't valid ascii (like x-.)? Sep 8 '21 at 18:23
• Is the input guaranteed to be valid? And what's up with the colons in the example? Sep 8 '21 at 18:26

Counting and so on

• Re: scoring. What if somebody achieves infinity? I suppose it would become impossible if you required different languages for each step. But either require or don't allow at all — merely allowing different languages would create two different challenges in one post. Prescribing a specific shape for each digit would make things less subjective, but then it'll be hard to find something that works both for Jelly and Java :P "And also individual winners for each program too" I suppose you meant language, but anyway, that's not a distinction we usually mention. Aug 18 '21 at 0:22
• the acceptance criteria for the numbers needs to be a lot more specific. Aug 19 '21 at 3:59
• @Razetime What about being able to stretch bits of the number in any orthogonal direction? Or just that it has to be the same shape, but at different scales? Aug 20 '21 at 23:50
• I'd define "X resembles Y" as "Y can be obtained from X by repeatedly removing one out of two consecutive identical rows or columns". Aug 25 '21 at 0:58

Implement the binary operators*

*of INTERCAL

Intro

As we all know, there are 5 different binary operations. Two combine two values, the rest modifies the value given.

Challenge

Your task will be to implement some if not all of the operators. They are defined as followed

• $ (mingle) Given two values, returns a value that consists of the bits of both values alternating. Right argument has the first bit. • ~ (select) Given two values, discard all bits of the right value that aren't 1 in the same place in the second value. Then, pack them to the right. If this explanation is too confusing, refer to this diagram. • &, V and ? (AND, OR and XOR) Given a value, AND/OR/XOR neighbouring the bits. Rules • I/O in any format that isn't a binary representation of the number(s). • Each answer can contain up to five submissions, each handling a different operator. You don't need to use one language for all programs. • A valid submission can be a program/function/link/lambda/chain/etc. • This is . Shortest answer for each operator wins. Test cases  44$     4 = 2224
2 \$     9 =   73

255 ~   341 =   15
4214 ~ 47818 =   96

&  1999 =  967
&    71 =    3

V     4 =    6
V   100 =  118

?    42 =   63
?   428 =  387


Sandbox things

• I'm unsure about what to submit. Is this "up-to-five-programs-one-per-op" fine or should it be changed?
• Anything else unclear?
• Is this question different enough from Implement INTERCAL's Binary Operators ?
• "If this explanation is too confusing, refer to this diagram" I have no idea what's going on in this diagram Aug 19 '21 at 19:05
• "As we all know, there are 5 different binary operations." It'd be good to specify that these are the operations of INTERCAL.
– user
Aug 19 '21 at 19:06
• What does the "(lists of)" provision mean--in what case is a list received as input, and what do you do with it? I'd also note that restricting input formats is generally frowned upon here, but if you're doing it just to block INTERCAL programs from using their native numeric I/O do note that the input is already base 10 ;) Aug 19 '21 at 21:49
• @Dudecoinheringaahing me neither but that's what the manpage shows :) I think that the procedere should be clear if you look at the first test case. Aug 20 '21 at 6:25
• @user The footnote at the end is easy to overlook, I'll change it. Aug 20 '21 at 6:26
• @UnrelatedString I agree, "(lists of)" should be removed to avoid confusion. Bad wording on my part. The intent of the second part was to block I/O from being binary strings. I shoud have worded it like that or drop that entirely Aug 20 '21 at 6:28
• Impractical if your time zone offset from Greenwich is not a whole number multiple of an hour (see e.g. wolframalpha.com/input/?i=Australia+time+zones). Sep 5 '21 at 22:20
• @JonathanFrech Not really. I frequently do LPs with people in Iran which is one of these places. If you live there you know it. But this is a simple code golf challenge and I don't think it serves it to complicate this with that sort of stuff.
– Wheat Wizard Mod
Sep 6 '21 at 8:15
• Yes, I meant your proposed scheme would be impractical if one were to live there; just as a comment. As a string-manipulating golfing challenge I think it has some merit. Sep 7 '21 at 1:24
• @jonathanFrech And I'm saying that while the scheme works best for integer offsets, from actual experience it is perfectly practical for people living in Iran which is UTC+3.5 or UTC+4.5 because if you live there you are aware of the offset and you can easily account for it.
– Wheat Wizard Mod
Sep 7 '21 at 7:12
• No, the Jimmies are on the backs of more Jimmies, and it's Jimmies all the way dow. Sep 10 '21 at 10:19
• What's the center of mass of -- or ----, and does it suffice to have a Jimmy's arm on one side of the central two? Sep 10 '21 at 21:15
• @emanresuA In the first it is at 1 and in the second it's at 2. (relative to the start of the platform) In either case the arm cannot be a place a Jimmy is directly touching so it must be in between two.
– Wheat Wizard Mod
Sep 10 '21 at 21:37

This question has been posted on the main site

• You might want to link to the question from here next time, although it doesn't really matter much.
– user
Sep 26 '21 at 19:08
• Why not allow giving the first n numbers in the sequence?
Sep 19 '21 at 13:29
• @Adám I don't really see any reason why one would want to do that.
– Wheat Wizard Mod
Sep 19 '21 at 13:44

Pythagoras' Golfing Grid

• Does it matter which solution is output? At the very least e and h can be any pair of factors for $f - T$, and you could swap dg with jk, etc. You say "integer" but your example only uses positive integers. Do you mean positive integers? If you do, there might be some T which do not have a solution. Sep 17 '21 at 16:06

Let's go for a rollercodester ride

• also considering using absolute change for scoring instead of curvature Sep 10 '21 at 20:53
• Seems interesting, but I think a program that just switches between something like unary and denser languages like Jelly would be optimal and probably not too difficult. (Although it does seem like it could be a good challenge, making something like that is probably non trivial) Sep 10 '21 at 20:53
• Would it be improved by only allowing each language once? I was also thinking a zip bomb type answer could dominate, but maybe that's not a bad thing. I could divide the score of each step by the sum of the code lengths that went into it, so it would be normalized and always less than 2 (theoretically 0,100,0 becomes +100,-100 becomes -200, from code lengths 0+100+0, leaving 200/100=2. That's the best I can think of). I like that because longer sequences would be the only way to reach higher score values. I guess the minimum sequence would need 3 languages to be scorable. Sep 11 '21 at 2:09
• I'd assumed that each language was only allowed once anyway, so that's probably the best thing to do. A zip bomb type answer probably wouldn't win, because you need more than just a massive increase in size, you need to be able to control what the output is so you can make it do something useful. I think the number of languages should definitely be factored into the score though, since a single massive increase or decrease in size could be the easiest way to win at the moment. Sep 11 '21 at 15:23

Count occurences in Pascal's Triangle

• The challenge itself looks good to post. That said, I don't think there are any interesting strategies beyond generating up to the nth row (starting from the zeroth), flattening and counting Sep 18 '21 at 3:03
• Also, it appears this sequence isn't on OEIS, and that the closest sequence is A006987 Sep 18 '21 at 3:07
• @cairdcoinheringaahing oeis.org/A003016
– tsh
Sep 18 '21 at 8:26

Minimizing flights of stairs climbed

• I think the bonus is the most interesting part and should be the main challenge. For now it's just a simple formula. Also, why the room numbers, not just floor numbers (which are room//100)? Sep 23 '21 at 12:03
• @pajonk I'll make the bonus be the main challenge then. They are room numbers to add a bit of extra challenge for converting into the floor number. Sep 23 '21 at 16:05

Multiplication for geometric algebra

• If you're looking for feedback here in the sandbox, I think it would be nice to remove the challenge from main site for now. Sep 23 '21 at 11:56
• very interesting challenge. Sep 27 '21 at 2:15
• I've edited this down to a stub now that it's been posted to save space Sep 30 '21 at 1:02

Newton Polynomial

A newton polynomial is a interpolation polynomial where the coefficients are found using the Newton's divided differences method. The relevant Wiki is here.

1. Given a set of inputs and outputs to find and display the polynomial that describes the series. The input can be as two lists or a dictionary like structure. e.g. $$[x_1:y_2,\;x_2:y_2,\;...\;, x_{n-1}:y_{n-1},\; x_n:y_n]$$
2. Given any real number, return the interpolated value.

Examples

Presentation of the polynomial must be in a format that is recognisable.

As ever this is code golf: so shortest answer wins.

• A summary of the method would be welcome (in addition to the Wikipedia link). From a super quick skim I think there might be errors in the final example output: shouldn't $(n-1)(n-2)$ be $(n-1)(n-3)$, and similarly $(n-1)(n-2)(n-3)$ be $(n-1)(n-3)(n-7)$? Oct 12 '21 at 4:49
• computers don't use real numbers. but still an interesting challenge. do you care what format the output polynomial is in? Oct 18 '21 at 3:14
• @Dingus thanks, yes working on a summary of the method, wanting to keep it consice though. no every term is included so '(n-1)(n-2)(n-3) etc. Oct 18 '21 at 20:44
• @donbright true, but I haven't had time to look into using complex numbers. I'm currently not interested in using them for this. Id want the polynomial to be readable, what did you have in mind? Oct 18 '21 at 20:47
• no i mean, "given any real number" should be something like "given a 32 bit floating point number in your languages format". as far as the polynomial format i was wondering if you would accept it all expanded out to $a_nx^n + a_{n-1}x^{n-1}...a_0$ form, to the point where it can be represented as an array of numbers, the numbers in the array are coefficients and the position in the array is the exponent. thanks. Oct 19 '21 at 2:09
• @donbright ah I see what you're saying, yes I will amend that then. I'm not sure I quite understand how you want to layout it out. Maybe expecting a particular format is too hard and I should just ask for an array of coefficient, were you thinking of expanding (n-1)(n-2)(n-3)... for example? Oct 19 '21 at 23:06
• yes the final result should be a polynomial, would it be ok to return a string saying 23.4n^4 + 3920n^3 ... Oct 20 '21 at 2:32

Wave square

Given an integer $$\n\$$ with the constraints $$\0 < n \leq 9\$$, output the corresponding square, that for each row shifts the sequence from $$\1\to n\$$ by one, overflowing when needed.

I/O

Input is an integer $$\n\$$, $$\ 0 < n \leq 9\$$. Output can be as a returned String, matrix, array or to STDOUT. You can write a full program or a function.

Examples

5:

12345
23451
34512
45123
51234


2:

12
21


1:

1


9:

123456789
234567891
345678912
456789123
567891234
678912345
789123456
891234567
912345678


meta

Thoughts on extending the challenge to allow for inputs bigger than 9, which would then just be taken mod 10?

• One idea - using letters for numbers greater than 9 (i.e a for 10, b for 11, etc.) would allow for inputs of up to 35 without disrupting the format. Oct 18 '21 at 2:33
• Thanks! I don't know if that'd be enough to make it distinct from another challenge though (someone posted it was a dupe in the nineteenth byte) Oct 18 '21 at 8:22

Largest turing incomplete subset of ASCII

We have challenges about trying to find a short, turing complete number of characters in a language. In this challenge, you'll do the opposite, trying to find the largest subset of characters which is not turing complete.

Rules:

• Like , competition is purely within a language, not between different ones
• Languages with SBCSs or other custom code pages may use those instead of ASCII
• SBCS submissions will not compete against ASCII submissions in the same language
• In this challenge, "ASCII" is not short for "printable ASCII". All of 0x00 to 0x7f is allowed
• So, the highest score possible would be 127 (for non-SBCS submissions), if there was a single character which is always needed for a TC program
• The language you use must be turing complete (and must be TC with full ASCII or its SBCS, whichever you choose)

Possible rule:

You must be able to write a valid program using only (and all of) the chosen subset of ASCII/the SBCS. It doesn't need to do anything useful, it just can't error.

This prevents padding the subset with characters that would never be allowed in a valid program in your language, or not including just one critical character needed in every program. However, strings and comments could be a problem.

• If a language is using SBCS, that would allow for a score higher than 94, wouldn't it? Oct 20 '21 at 15:03
• @AaroneousMiller It would, yes (this is fine since it's a competition within the language) Oct 20 '21 at 15:10
• An empty program is valid in some languages.
– xnor
Oct 20 '21 at 18:09
• @xnor Ah, I meant using all of the characters in the chosen subset. Will fix. Oct 20 '21 at 18:18
• Maybe just allow all 256 bytes 0x00 to 0xff so it's a fair playing field? Oct 20 '21 at 19:00
• I don't think the "possible rule" is needed. The equivalent challenge is to find a small set of chars/bytes such that some of them are absolutely needed for the lang to be TC, and take the set complement. IMO the first part is totally fair across languages (assuming full byte range), and the second part only depends on how you define the full set. So for maximum fairness I'd suggest "every char/byte is on the table, even the ones that cannot appear in any valid program". Oct 20 '21 at 23:30
• @Bubbler To be fair, it's not meant to be fair to all languages. It'll be like code golf where scoring is per language. But I guess I agree the possible rule probably isn't needed. Oct 20 '21 at 23:45

Find all paths

• More test cases would be needed. Is this oeis.org/A121788? Jun 27 '21 at 15:24
• I dont think this is that problem in the link Jun 27 '21 at 15:34
• Anyway, please include more test cases Jun 27 '21 at 16:24
• Is this oeis.org/A001184?
– tsh
Jun 28 '21 at 8:22
• yes it is @tsh updated! Jun 28 '21 at 8:37
• === Looks good! === Jun 28 '21 at 8:43