Activity for WheatWizardâ€
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #289679 |
1. With a normal form of output (e.g. 1 and 0) the set of problems that can be solved is the set of computable decision problems. With halting/non-halting output, the set of problems that can be solved is the set of limit computable decision problems. This includes famously non-computable problems ... (more) |
— | 6 days ago |
| Comment | Post #289679 |
I'm in opposition to this for two reasons. 1. it is not a trivial change in IO, it is a very fundamental change in what can and can't be done. 2. It is not modular, programs written this way can only be combined as part of larger programs in limited ways.
(more) |
— | 6 days ago |
| Edit | Post #289447 |
Post edited: Stipulated that the intermediate values must be integers. |
— | about 1 month ago |
| Edit | Post #289447 |
Post edited: |
— | about 1 month ago |
| Edit | Post #289447 | Initial revision | — | about 1 month ago |
| Question | — |
Give the fool's fibonacci sequence Recently I asked for tips on improving some code-golf of mine. The code was supposed to output every third value of the Fibonacci sequence starting with 2: ``` 2,8,34,144,610,2584,10946,46368,196418,832040 ``` However, I made a mistake in deriving my formula, and my code output a different s... (more) |
— | about 1 month ago |
| Comment | Post #288535 |
You don't get to assume a maximum on these values. (more) |
— | about 2 months ago |
| Comment | Post #288943 |
The result can be exactly $-1, 0, 1$ if no other residue is one of those three residues (and each of those three occurs at least once).
For example look at the example 6. 6 is 0 mod 2, 0 mod 3, 1 mod 5, -1 mod 7. The value 0 occurs twice, this is how you only have 3 values for high $n$. (more) |
— | about 2 months ago |
| Edit | Post #289152 | Initial revision | — | about 2 months ago |
| Question | — |
Reverse engineer the colors for a layout. At my job we have to sometimes lay out materials. Materials come in large long rolls and are cut into smaller pieces when being laid out. When we order the rolls we draw up a layout document which describes how we are going to cut them and layout the pieces. This tells the manufacturer how we are ... (more) |
— | about 2 months ago |
| Edit | Post #288943 |
Post edited: |
— | 3 months ago |
| Comment | Post #288943 |
No, the set needs to be exactly equal to, not just a subset of. (more) |
— | 3 months ago |
| Edit | Post #288943 | Initial revision | — | 3 months ago |
| Question | — |
Find near miss prime multiples. Given a number $n \geq 3$ as input output the smallest number $k$ such that the modular residues of $k$ by the first $n$ primes is exactly $\{-1,0,1\}$. That is there is a prime in the first $n$ primes that divides $k$, one that divides $k+1$ and one that divides $k-1$, and every prime in the firs... (more) |
— | 3 months ago |
| Comment | Post #288885 |
Ah thanks. I blame me missing that on the fact I was expecting to see a numeral. (more) |
— | 3 months ago |
| Comment | Post #288885 |
You should probably mention that dice have 6 faces, currently it has to be intuited from test cases. (more) |
— | 3 months ago |
| Edit | Post #288689 | Initial revision | — | 3 months ago |
| Question | — |
Prove commutativity on this monoid presentation. Given two binary strings $A$ and $B$ such that $A$ is an anagram of $B$, output a third binary string $S$ such that both $A$ and $B$ can be created by iterated removals of the substring $10101$ from $S$. For example for $A=100$ and $B = 010$, one solution is $S = 10101010$ since $$ (10101)01... (more) |
— | 3 months ago |
| Comment | Post #283305 |
Haskell + [hgl](https://gitlab.com/wheatwizard/haskell-golfing-library), 2 bytes: `lq` (more) |
— | 3 months ago |
| Edit | Post #288684 | Initial revision | — | 3 months ago |
| Answer | — |
A: Cumulative Counts Haskell + hgl, 10 bytes ```haskell mpn$ce<gj ```` Attempt This Online! Explanation - `mpn` map across all non-empty prefixes of the input ... - `ce` count the number of elements in each prefix equal to ... - `gj` the last element of that prefix. Reflection I'm happy with this. ... (more) |
— | 3 months ago |
| Comment | Post #288683 |
That big O notation simplifies to $O(n^2)$. (more) |
— | 3 months ago |
| Edit | Post #288653 | Initial revision | — | 3 months ago |
| Answer | — |
A: Build a replacement ball in regex. Haskell + hgl, 30 bytes ```haskell ic"|"<pST<eL<P1 ``` This version is much longer, but in my opinion it has better potential to be shorter than the above version if hgl were improved. Explanation This adds a dummy character `?` to the front of the list then gets all ways to partition t... (more) |
— | 3 months ago |
| Comment | Post #288548 |
Ok, I should have seen that. I'm going to suggest adding a rounded up dummy score for the time being, and then to fix the leaderboard. I'll edit my answer in a bit. (more) |
— | 3 months ago |
| Edit | Post #288622 |
Post edited: Better wording. |
— | 3 months ago |
| Edit | Post #288622 |
Post edited: |
— | 3 months ago |
| Edit | Post #288622 | Post undeleted | — | 3 months ago |
| Edit | Post #288622 | Post deleted | — | 3 months ago |
| Edit | Post #288622 | Initial revision | — | 3 months ago |
| Question | — |
Are these reduced forms of the same thing? Task You are going to take three strings as input $A$, $B$ and $X$. And your goal is to determine if there exists a third string $S$ such that both $A$ and $B$ can be formed by iteratively removing contiguous instances of $X$ in $S$. For example if $X = 10101$ then both $10$ and $01$ can be forme... (more) |
— | 3 months ago |
| Edit | Post #288563 |
Post edited: |
— | 3 months ago |
| Edit | Post #288563 | Initial revision | — | 3 months ago |
| Question | — |
Build a replacement ball in regex. In this challenge you will take a number $n$ and a string $X$ of length $\geq n$, and produce a regular expression which matches all strings that are withing $n$ character substitutions of $X$. Specifically you will take $X$ and $n$ and ouptut all the ways to replace exactly $n$ characters with th... (more) |
— | 3 months ago |
| Edit | Post #288555 |
Post edited: |
— | 3 months ago |
| Edit | Post #288555 | Initial revision | — | 3 months ago |
| Answer | — |
A: Efficient censorship Haskell + hgl, 14 bytes ```haskell xBl<fn<iw ```` Attempt This Online! Explanation `ss` gets all substrings of the input `fn` filters out the substrings that don't ... `iw` checks if the forbidden word is a contiguous substring `xBl` gets the largest result Reflection There... (more) |
— | 3 months ago |
| Comment | Post #288548 |
Actually the leaderboard currently floors the fractional score so there's no use in making a dummy score. I've changed my suggestion to "update the leaderboard to support how people are scoring". (more) |
— | 3 months ago |
| Edit | Post #288548 |
Post edited: |
— | 3 months ago |
| Comment | Post #288548 |
In the last section I'm suggesting that if the leaderboard is not changed, we include a dummy count so the leaderboard scores it predictably. I'll try to reword it. (more) |
— | 3 months ago |
| Comment | Post #288548 |
Maybe I am no expressing myself most clearly. I am saying that the in effect disalllowing and allowing fractional bytes are not tangibly different. I assume the leaderboard doesn't change because it illustrates this point most expressly. (more) |
— | 3 months ago |
| Edit | Post #288548 |
Post edited: |
— | 3 months ago |
| Edit | Post #288548 | Initial revision | — | 3 months ago |
| Answer | — |
A: Codidact Fractional Byte Consensus I'm going to reiterate and build off ideas I laid out on this post on PPCG. First let's point out that at current the leader board does not support fractional scores. In this answer I am going to assume this is unchanged. That may prove to be false, in fact I think it is likely a false assumption,... (more) |
— | 3 months ago |
| Edit | Post #288535 | Initial revision | — | 3 months ago |
| Question | — |
Efficient censorship You are a low-level censor working for the Ministry of Media Accuracy. Part of your job is to make sure that certain words don't appear in publications. Every morning you get a fresh stack of next week's newspapers and its your job to comb through them and fix any of the words that were mistakenly i... (more) |
— | 3 months ago |
| Comment | Post #288526 |
Thanks I fixed it. And it seems the new alignments are golfier since the fixed program comes out 2 bytes shorter. (more) |
— | 3 months ago |
| Edit | Post #288526 |
Post edited: Fixed alignment |
— | 3 months ago |
| Edit | Post #288526 |
Post edited: Improved score. |
— | 3 months ago |
| Edit | Post #288526 |
Post edited: |
— | 3 months ago |
| Edit | Post #288526 | Initial revision | — | 3 months ago |
| Answer | — |
A: Connect the corners without 4 in a row Haskell + hgl, 134 bytes ```haskell k=cy"X.XX" x#1=[4,0,9,9]!x 2#y=8 3#y=[8,9,4,4]!y x#3=[0,3]!x #=0 x?y|(n,j)<-fvD 4$x%4#(y%4)=tk y$dr j$tk x<dr n<cy[dr2 k,k,cy".X",k] ```` Attempt This Online! I first set up a pretty dense background pattern which doesn't break any rules: ``` X... (more) |
— | 3 months ago |
| Edit | Post #288516 |
Post edited: |
— | 3 months ago |
| Edit | Post #288516 | Initial revision | — | 3 months ago |
| Question | — |
Calculate the Z-array Task Given a list of numbers $X$ produce a second list of numbers $Y$ such that $Yi$ is the length of the longest common prefix of $X$ and $X$ with the first $i$ elements removed. For example if the input is ``` [1,2,2,1,1,2,1,2,2,1,2,1] ``` The output should be ``` [12,0,0,1,2,0,4,... (more) |
— | 3 months ago |
| Edit | Post #288398 |
Post edited: More examples. |
— | 3 months ago |
| Edit | Post #288503 | Initial revision | — | 3 months ago |
| Question | — |
Count polyomino bisections An polyomino is a non-empty connected subset of the square tiling consisting of squares joined along their edges. We will not require that polyominos be simply connected, that is they can have holes. A bisection of a polyomino $X$ is a pair of polyominos such that they can be joined together witho... (more) |
— | 3 months ago |
| Edit | Post #288398 |
Post edited: Holes. |
— | 3 months ago |
| Comment | Post #288398 |
The current definition of polyomino doesn't disallow holes. I'll try to work that explicitly into the text. (more) |
— | 3 months ago |
| Edit | Post #288398 |
Post edited: |
— | 3 months ago |
| Comment | Post #288398 |
Ah I think you are misunderstanding, the you are dividing it into k-polyominos. There are no additional requirements on these polyominos. So
```text
X XX
XXXXX
```
can be subdivided into a T-tetramino and a O-tetramino. It doesn't matter that these are not the same 4-polyomino. You just ... (more) |
— | 3 months ago |
| Comment | Post #288398 |
I'm not sure what you mean. There's no canonical way to select an enantiomorph, so yes it is possible for to mirror images to be inputs, if that's what you are asking. No part of this question involves determining if two polyominos are equal so it shouldn't matter whether you consider chiral pairs t... (more) |
— | 3 months ago |
| Comment | Post #288398 |
Input a polyomino is intended to be flexible. I'd be certainly be willing to let you take a list of coordinates, and I believe if taking a list it is acceptable to also take its length. So that would be one way to take the input with the number of squares. (more) |
— | 3 months ago |
| Edit | Post #288398 | Initial revision | — | 3 months ago |
| Question | — |
Determine if a polyomino is "prime" An $n$-polyomino is a connected subset of the square tiling consisting of $n$ squares. We will not require that polyominos be simply connected, that is they can have holes. We will say a $n$-polyomino is prime if it cannot be disected into disjoint $k$-polyominos for any 1<$k$<$n$. For example th... (more) |
— | 3 months ago |