29
MATL, 4 bytes
BSXB
Try it at MATL Online
Explanation
% Implicitly grab input as an array
% STACK: [10, 17, 19, 23]
B % Convert each element to binary where each decimal number results in a row
% STACK: [0 1 0 1 0;
% 1 0 0 0 1;
% 1 0 0 1 1;
% 1 0 1 1 1]
S % Sort each column, placing all of the ...
28
Python 3, 124 bytes
def f(n):
while n&4369<n/n:n>>=1
while n&15<1:n>>=4
return'TJLZSIO'["rēȣc63ıGtIJȱᄑ@'̢̑@@@@Ȳq".index(chr(n))%7]
Try it online!
Expects an integer n representing a 4 × 4 binary matrix. Throws if no tetromino is found.
Line 2 slides the shape to the right until a 1 is in the rightmost column. (4369 ...
28
MATL, 21 bytes
3XytPJ*-X*Xj~4&1ZIunq
The input is a matrix with 1 for \ and j (imaginary unit) for /.
Try it online! Or verify all test cases.
With some extra code, you can see the different pieces in random colours. Or increase the resolution for a better looking result.
Explanation
Consider input [1,j; 1,1; j,1; j,j] as an example. This ...
26
MATL, 2 bytes
Xy
A translation of my Octave answer.
Try it online.
A 4 byte version with no built-ins (thanks to Luis Mendo):
:t!=
: take input n and a generate row array [1,2,...n]
t duplicate
! zip
= thread compare over the result
20
TI-BASIC, 2 bytes
identity(Ans
Fun fact: The shortest way to get a list {N,N} is dim(identity(N.
Here's the shortest way without the builtin, in 8 bytes:
randM(Ans,Ans)^0
randM( creates a random matrix with entries all integers between -9 and 9 inclusive (that sounds oddly specific because it is). We then take this matrix to the 0th power.
19
Julia, 9 3 bytes
eye
This is just a built-in function that accepts an integer n and returns an nxn Array{Float64,2} (i.e. a 2D array). Call it like eye(n).
Note that submissions of this form are acceptable per this policy.
17
Wolfram Language (Mathematica), 26 25 bytes
Tr[#//.x_:>#.#.Clip@x]<1&
Try it online!
How it works
Given an adjacency matrix A, we find the fixed point of starting with B=A and then replacing B by A2B, occasionally clipping values larger than 1 to 1. The kth step of this process is equivalent up to the Clip to finding powers A2k+1, in which the (...
16
APL, 5 bytes
∘.=⍨⍳
This is a monadic function train that accepts an integer on the right and returns the identity matrix.
Try it here
16
APL (Dyalog Unicode), 29 28 bytesSBCS
-1 thanks to @Adám
{≢∪∨.∧⍨⍣≡2>+/↑|∘.-⍨⍸⍵}¨⊂,~∘⊂
Try it online!
⊂,~∘⊂ the matrix and its negation
{ }¨ for each of them do
⍸⍵ list of pairs of coords of 1s
+/↑|∘.-⍨ matrix of manhattan distances
2> neighbour matrix
∨.∧⍨⍣≡ transitive closure
≢∪ number of unique rows
16
J, 57 bytes
,&([:(0#@-.~~.@,)](*@[*[:>./((,-)#:i.3)|.!.0])^:_ i.@$)-.
Try it online!
This is one of those where the idea is incredibly simple (and I think fun), but executing it had some mechanical lengthiness which masks the simplicity... eg, shifting the original matrix in all directions with 0 fill is the verbose ((,-)#:i.3) |.!.0.
It's ...
16
APL (Dyalog Unicode), 20 12 bytesSBCS
-6 bytes through dzaima
Full program. Prompts stdin for:
texture as a list of strings
number of rows in shape matrix
number of columns in shape matrix
indices of falsies (to be blanks) in the shape matrix
All the ⎕s are supposed to be rectangles (they're not tofu). They symbolise a computer console which here means ...
14
Octave, 10 4 bytes
@eye
Returns an anonymous function that takes a number n and returns the identity matrix.
14
Python 3, 181 179 171 167 161 159 bytes
Input by UTF-16 little-endian bytes without BOM. First decompose into columns using bit shifts, split by empty column, then hash them into a lookup table.
-2 bytes thanks to ngn.
-5 bytes thanks to Mr. Xcoder.
lambda h,j=''.join:j(' ZAQV;JWP;MBOS;YRKCGXDF;ILHUENT'[int('0'+i,27)%544%135%32]for i in j(chr(64|i&7|...
14
Python 2, 105 bytes
lambda b:sum(b[i+d::d][:(8,7-i%8,i%8)[d%8%5]].find('1')*int(c)>0for i,c in enumerate(b)for d in[1,7,8,9])
Try it online!
Explanation
We take the input as a string of 64 characters '0' or '1'. Using step slices, we cast four "lines of sight" from every queen we encounter. For example, when i = 10 and d = 7, marking the ...
13
Python 2, 275 271 264 249 bytes
Saved four bytes by replacing -1 with H and removing one slicing operation ([:]).
Saved seven bytes thanks to Halvard Hummel; removing yet another slicing operation ([:]), using multiple-target assignment to give a visited entry a value v not in "01" (S=S[1:];M[y][x]=H; -> S=M[y][x]=S[1:];) and switching from a ternary if/...
12
R, 4 bytes
diag
When given a matrix, diag returns the diagonal of the matrix. However, when given an integer n, diag(n) returns the identity matrix.
Try it online
12
Python 2, 42 bytes
lambda n:zip(*[iter(([1]+[0]*n)*n)]*n)[:n]
An anonymous function, produces output like [(1, 0, 0), (0, 1, 0), (0, 0, 1)],
First, creates the list ([1]+[0]*n)*n, which for n=3 looks like
[1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0]
Using the zip/iter trick zip(*[iter(_)]*n to make groups of n gives
[(1, 0, 0), (0, 1, 0), (0, 0, 1), (0, 0, 0)]
...
12
Since this is basically talking about a cellular automaton I give you..
Golly Quicklife rule, 10 bytes
01234/234V
Input the rule, paste the grid into Golly, run pattern. The resulting pattern is the output.
Explanation:
01234 Survive on any number of neighbors
/234 Born on >2 neighbors
V Only directly adjacent neighbors count
Or if you ...
11
16/9 20/11 22/12 28/15 30/16 32/17 34/18 36/19 (Java)
This uses a number of ideas to reduce the search space and cost. View the revision history for more details on earlier versions of the code.
It's clear that wlog we can consider only circulant matrices in which the first row is a Lyndon word: if the word is non-prime then it must have property X, and ...
11
JavaScript (ES6), 148 146 143 bytes
Saved 1 byte thanks to @ngn
s=>[...s].map(c=>g((k=c.charCodeAt()/8)&8|k*8&7)&g(k&7|k/2&8),o=x='',g=n=>x=n?x*27+n:(o+=' DZQGYWXNHJ.CSTIO.AFB.LPVE..KUMR'[x%854%89%35],n))&&o
Test cases
let f =
s=>[...s].map(c=>g((k=c.charCodeAt()/8)&8|k*8&7)&g(k&7|k/2&...
11
Wolfram Language (Mathematica), 16+83=99 bytes
Library import statement (16 bytes):
<<Combinatorica`
Actual function body (83 bytes):
Length@HamiltonianCycle[MakeGraph[#~Position~1~Join~{1>0},##||Norm[#-#2]==1&],All]&
Try it online!
Note that the question just ask for the number of Hamiltonian path in the graph.
However, (for some ...
11
MATL, 5 bytes
JYaZy
Try it online! Or verify all test cases.
Explanation
JYa % Implicit input. Unpad matrix. This removes outer zeros
Zy % Size. Implicit display
10
Jelly, 4 bytes
R=€R
Doesn't use a built-in. Try it online!
How it works
R=€R Main link. Input: n
R Range; yield [1, ..., n].
R Range; yield [1, ..., n].
=€ Compare each.
This compares each element of the left list with the right list, so it
yields [1 = [1, ..., n], ..., n = [1, ..., n]], where comparison is
...
10
J, 4 bytes
=@i.
This is a function that takes an integer and returns the matrix.
10
Jelly, 31 30 bytes
sz0Z
ç€2ZF€ç€8Zœ?@€€-36Ḅ+⁽$ṁỌY
Try it online!
How?
sz0Z - Link 1, split & right-pad with zeros: list, items; number, chunkSize
s - split items into chunks of length chunkSize
z0 - transpose with filler zero
Z - transpose
ç€2ZF€ç€8Zœ?@€€-36Ḅ+⁽$ṁỌY - Main link: list of lists of numbers (1s & 0s), M
ç€2 ...
10
Husk, 15 14 bytes
zȯḋm←CtNCİ□ṁḋN
Try it online!
Continually prints the results as an infinite list.
Explanation
I wonder whether there's a better way to get every nth element from a list than splitting the list into chunks of length n and retrieving the head of each chunk.
tN Get a list of all natural numbers except 1. (A)
...
10
JavaScript (ES7), 86 bytes
Takes input as an array of 64 integers with 254 for a queen and 0 for an empty square.
a=>[s=0,6,7,8].map(d=>a.map(g=(n,p)=>(p%8-(p+=~d)%8)**2<n%4?a[p]?s+=n&1:g(n/2,p):0))|s
Try it online!
This version abuses arithmetic underflow to get a stop condition in the recursive part.
JavaScript (ES7), 89 bytes
Takes ...
10
Jelly, 12 11 bytes
ŒṪŒ!ạƝ€§ÐṂL
Try it online!
Explanation.
ŒṪ Positions of snake blocks.
Œ! All permutations.
For each permutation:
ạƝ€ Calculate the absolute difference for each neighbor pair
§ Vectorized sum.
Now we have a list of Manhattan distance between ...
10
JavaScript (ES6), 154 134 bytes
m=>m.map((r,Y)=>r.map(g=(_,x,y,r=m[y=1/y?y:Y])=>r&&r[x]&&[-1,0,1,2].map(d=>r[r[x]=0,/1/.test(m)?g(_,x+d%2,y+~-d%2):++n,x]=1)),n=0)|n/4
Try it online!
How?
Method
Starting from each possible cell, we flood-fill the matrix, clearing all cells on our way. Whenever the matrix contains no more 1's, ...
10
Clean (with Snappy), 800 785 670 644 bytes
460 402 bytes of code + 360 242-byte string literal
(escaped here and on TIO because it isn't valid UTF-8)
You can verify the length of the literal here.
import StdEnv,Data.List,Data.Maybe,Codec.Compression.Snappy,Text
@a b|b<'~'=b=a
$m x y d=map(@' ')(foldl(\a b=[@u v\\u<-a&v<-b])['~~'..][join['
']k\...
Only top voted, non community-wiki answers of a minimum length are eligible
