30
MATL, 21 bytes
3XytPJ*-X*Xj~4&1ZIunq
The input is a matrix with 1 for \ and j (imaginary unit) for /.
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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 ...
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 1'...
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]
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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
Python 2, 49 bytes
Takes as input a 2D binary matrix \$ a \$, and its size \$ n \$.
lambda a,n:sorted(map(sum,a+zip(*a)))[-2:]==[1,n]
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There may be shorter approaches, but this is what I could find for now. Please let me know if the algorithm is incorrect.
Explanation
Since there is exactly one line, there should be exactly one row/column ...
26
MATL, 2 bytes
Xy
A translation of my Octave answer.
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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
Wolfram Language (Mathematica), 26 25 bytes
Tr[#//.x_:>#.#.Clip@x]<1&
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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 (...
19
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.
18
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
APL (Dyalog Unicode), 29 28 bytesSBCS
-1 thanks to @Adám
{≢∪∨.∧⍨⍣≡2>+/↑|∘.-⍨⍸⍵}¨⊂,~∘⊂
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⊂,~∘⊂ 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
17
J, 57 bytes
,&([:(0#@-.~~.@,)](*@[*[:>./((,-)#:i.3)|.!.0])^:_ i.@$)-.
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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, 5 bytes
∘.=⍨⍳
This is a monadic function train that accepts an integer on the right and returns the identity matrix.
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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 ...
16
J, 41 bytes
1#.*:=1#.2>[:>@~.@,[:+&,&.>/~i.@!<@A.=@i.
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A different take on brute force, which handles up to n=6 on TIO.
Since the idea is more interesting than the J mechanics here, I'll explain that.
I could likely save bytes by translating one of the formulas into J, but that seemed less fun.
the idea
Generate all n! ...
15
MATL, 29 28 24 bytes
z2Y6ttX*,GbZ+5Mz=z]yytvs
Input is a binary matrix with 1 for '+' and 0 for '-'.
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Explanation
Convolution is the key to success
z % Implicit input. Number of nonzeros. This is the number of naive pluses
2Y6 % Push [0 1 0; 1 1 1; 0 1 0] (predefined literal): pattern of double plus
ttX* ...
14
Octave, 10 4 bytes
@eye
Returns an anonymous function that takes a number n and returns the identity matrix.
14
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)]
...
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])
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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/...
13
MATLAB/Octave, 115 109 98* bytes
-6 bytes thanks to Luis Mendo
-7 bytes thanks to Luis Mendo
function p=f(A)
B=A;B(:,2:end)=0;for k=1:nnz(A)
B=conv2(B,mod(magic(3),2),'same').*A;end
p=any(B);
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(end keyword is in footer because it's not required when you define the function in a file)
Takes binary array, outputs:
truthy value as vector full of ...
13
Brachylog, 8 bytes
{z|}≡ᵛ+1
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{z|} Either the rows or the columns
≡ᵛ are all the same list
+1 which sums to 1.
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.
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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 ...
12
JavaScript (ES6), 148 146 143 142 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]))&&o
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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
Husk, 15 14 bytes
zȯḋm←CtNCİ□ṁḋN
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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)
...
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]&
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Note that the question just ask for the number of Hamiltonian path in the graph.
However, (for some ...
11
MATL, 5 bytes
JYaZy
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Explanation
JYa % Implicit input. Unpad matrix. This removes outer zeros
Zy % Size. Implicit display
11
Jelly, 10 bytes
ŒJạ€ŒṪ§Ṃ€Ṁ
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-2 bytes thanks to Sisyphus
Calculate the Manhattan differences from all 0s to all 1s, and the answer is the maximum of the minimums (each row's minimum is the number of stages until it gets infected, so the number of stages needed is the maximum of the stages needed for each person).
Conveniently, if all elements ...
11
Stencil ≢, 2 bytes
×v
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≢ tallies the number of necessary steps (including the initial state) until stability is reached. This command line argument isn't counted towards the byte count, as per Meta consensus.
Each cell's next state is determined by:
× the sign of
v the sum of all values in its von Neumann neighbourhood (including itself)
Only top voted, non community-wiki answers of a minimum length are eligible