New answers tagged matrix
2
Jelly, 21 bytes
æịþḶFµḢ;ⱮṖŒ!;€ṪƲIỊẠ€S
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Takes forever for n>3 because it tries to generate all permutations of n**2-2 items.
æịþḶFµḢ;ⱮṖŒ!;€ṪƲIỊẠ€S Monadic link. Input: n
æịþḶF Generate complex numbers (1..n)+(0..n-1)i
µḢ;ⱮṖŒ!;€ṪƲ Generate all permutations with 1st and last fixed:
Ḣ;Ɱ ...
1
Python 3, 103 bytes
r=range
f=lambda A:[*zip(*[[max(i*(c[j-i]==i)for i in r(len(c)))for j in r(len(c))]for c in zip(*A)])]
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This is more or less a one-to-one translation of my numpy answer.
The same explanation and caveat apply.
4
Python 3+numpy, 73 bytes
import numpy
def f(A):t=numpy.c_[:len(A)];return(t*(A[t-t.T]==t)).max(1)
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Note: this probably fails for double wrap around EDIT or, more generally, for matrix entries equal to or larger than the matrix height thanks @alephalpha for pointing this out end EDIT. I'm just not sure it is a requirement (this is my first ...
1
Wolfram Language (Mathematica), 63 bytes
MapThread[Max,RotateRight[#/.x_/;x!=i:>0,i]~Table~{i,Max@#},2]&
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3
Python 3+numpy, 77 bytes
import numpy
b=0*a
for i in range(1,a.max()+1):b[numpy.roll(a==i,i,axis=0)]=i
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the roll function does it all, and i stole the zeros_like trick from @ovs
1
Pari/GP, 50 bytes
m->polcoef(vecprod([x^2^i|i<-[0..#m-1]]*m),2^#m-1)
Based on @miles's Mathematica answer.
According to Wolfram MatheWorld, the permanent of a matrix \$A = (a_{ij})\$ is the coefficient of \$x_1 \cdots x_n\$ in the polynomial \$\prod_{i=1}^{n}\sum_{j=1}^na_{ij}x_j\$. This is exactly what miles's answer does.
But in PARI/GP there isn't ...
1
jq, 28 bytes
[transpose[]|sort]|transpose
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1
Charcoal, 23 bytes
ＩＥθＥι⌈Ｅθ∧¬﹪⁻⁺§νμξκＬθ§νμ
Try it online! Link is to verbose version of code. Explanation:
θ Input array
Ｅ Map over rows
ι Current row
Ｅ Map over cells
θ Input array
Ｅ Map over rows
§νμ Inner ...
1
Charcoal, 35 20 bytes
ＷＳ⊞υι↑Ｅθ⭆²Φ⭆υ§νκ⁻Ｉνλ
Try it online! Link is to verbose version of code. Takes input as a list of digit strings. Explanation:
ＷＳ⊞υι
Read the input.
↑Ｅθ⭆²Φ⭆υ§νκ⁻Ｉνλ
Map over each column, taking the transpose and filtering out the 0s and 1s each time, so that the 1s are effectively sorted to the front, but then print the whole lot ...
1
Retina 0.8.2, 39 bytes
+`(?<=(.)*)1(.*¶(?<-1>.)*(?(1)^))0
0$+1
Try it online! Takes input as a list of digit strings although commas can be redundantly included as per the example link. Explanation: .NET's balancing groups are used to ensure that the regular expression matches a 1 directly above a 0; the digits are thus exchanged. $+ is used ...
1
Nim, 138 bytes
proc x(i:int)=
for x in 1..i:
for j in 1..x:
stdout.write($x&" ")
for j in x+1..i:
stdout.write($j&" ")
echo()
2
TI-Basic, 38 bytes
Input N
identity(N→[A]
For(I,1,N
For(J,1,N
max(I,J→[A](I,J
End
End
Output is stored as a matrix in [A].
1
Swift 5.5/Xcode 13.0, 215 bytes
var c=[Int:Int]();let d=m.count;let e=m[0].count;for i in 0..<e{for j in 0..<d{c[i,default:0]+=m[j][i]}};var r=Array(repeating:Array(repeating:1,count:e),count:d);for k in c{for i in 0..<d-k.1{r[i][k.0]=0}};return r
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2
05AB1E, 19 18 14 bytes
˜εQy*yFÁ]øεø€à
-5 bytes thanks to @ovs.
Try it online or verify all test cases.
Explanation:
Z # Get the flattened maximum of the (implicit) input-matrix
L # Pop and push a list in the range [1,max]
˜ # (`ZL` is now `˜`: flatten the (implicit) input-matrix)
ε # Map over each:
Q # Check ...
3
Core Maude, 248 244 236 bytes
load linear
mod D is inc MATRIX{Nat0}*(sort Matrix{Nat0}to M). var A B C D : Nat . op d :
Nat M -> M . eq(A,B)|-> C ;(A,B)|-> D =(A,B)|-> max(C,D). eq d(A,(B,C)|-> D ;
X:M)=(B,(C + D)rem A)|-> D ; d(A,X:M). eq d(A,X:M)= X:M[owise]. endm
Example Session
\||||||||||||||||||/
--- Welcome ...
2
Ruby, 155 bytes
->m{r,c,o=m.size,m[0].size,{}
o.default=0
e="r.times{|y|c.times{|x|"
eval e+"v=m[y][x]
v>o[[(y+v)%r,x]]&&o[[(y+v)%r,x]]=v}}"
eval e+"m[y][x]=o[[y,x]]}}"
m}
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We build an hash object with 2d-index as key.
Then we iterate input and we set the key corresponding to landing position to the ...
5
Python 3 + numpy, 91 bytes
def f(a):r,c=a.nonzero();v=a[r,c];r+=v;i=v.argsort();a*=0;a[r[i]%len(a),c[i]]=v[i];return a
-23 due to @ovs
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Much Different from the first solution. Takes a numpy array
(slightly outdated explanation) Assume the array
[[0 0 2 0 1]
[0 2 1 1 0]
[3 0 2 1 0]
[0 0 0 0 0]]
r,c=nonzero(a) First save the x-indices (rows)...
4
R, 82 bytes
Or R>=4.1, 75 bytes by replacing the word function with \.
function(m,k=nrow(m),t=0*m){t[((row(m)+m-1)%%k+1+col(m)*k-k)[i]]=m[i<-order(m)];t}
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Inspired by @wasif's answer.
Old solution (shorter for R>=4.1):
R, 94 92 89 88 bytes
Or R>=4.1, 74 bytes by replacing two function appearances with \s.
function(m,k=nrow(m))...
1
Python 2, 157 bytes
e=enumerate;l=input();j=[[0]*len(l[0])for _ in l]
for x,y,z in sorted([((n+x)%len(l),m,x)for n,y in e(l)for m,x in e(y)],key=lambda e:e[2]):j[x][y]=z
print j
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-7 thanks to @ykcul
1
RAD, 4 bytes
⍉<⍉⍵
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Transpose the input (⍉⍵), sort each (<), then retranspose ⍉.
1
MY, 11 bytes
⎕⍉86ǵ'ƒ⇹(⍉←
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⎕⍉86ǵ'ƒ⇹(⍉←
⎕ - Input
⍉ - Transpose
86ǵ'ƒ⇹( - Sort each
⍉ - Transpose again
← - Output
7
J, 24 20 bytes
[:>./-@,|."0 2,*,=/]
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-4 thanks to xash!
Solved independently, but seems very similar to ovs's APL solution. This one works for any number, because I didn't notice that the question only required handling 1-9.
The following explanation is slightly out of date, but the high-level concept, which is the interesting part, ...
7
APL(Dyalog Unicode), 17 16 14 bytes SBCS
-1 byte by looking at Jonah's J answer and another -2 bytes by implementing xash's suggestion on the same answer.
⌈/⊂(-⍤⊢⊖⊣×=)¨,
Try it on APLgolf!
I initially implemented a lot more "by hand", which ended up more than 3 times longer:
{{(⊃⍺)@(⊂1↓⍺)⊢⍵}/((↓⊂∘⍒⌷⊢){k,m|(k←≢⍵)0+⊃⍺}⌸⍸⍵),⊂0⍴⍨m←⍴⍵}
Try it on ...
5
JavaScript (ES6), 82 bytes
m=>m.map((r,y)=>r.map((_,x)=>m.map(q=(r,Y)=>q=(Y+(v=r[x]))%m.length-y|v<q?q:v)|q))
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How?
For each position \$(x,y)\$ in the input matrix \$M\$ of height \$h\$, we look for all values \$M_{x,Y}\$ on the same column such that:
$$Y+M_{x,Y}\equiv y\pmod h$$
and keep the highest one.
6
Jelly, 16 12 bytes
I thought 16 seemed a bit much!
Zµ=Ɱ×ṙ"N»/)Z
A monadic Link that accepts a list of lists of integers and yields a list of lists of integers.
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How?
Zµ=Ɱ×ṙ"N»/)Z - Link: matrix, M
Z - transpose -> columns of M (top to bottom)
µ ) - for each (c in columns of M):
Ɱ - map (across v ...
1
TI-Basic, 54 bytes
Input [A]
dim([A]
For(I,1,Ans(2
Matr►list([A],I,A
SortA(ʟA
For(J,1,dim(ʟA
ʟA(J→[A](J,I
End
End
Output is stored in [A], which was the inputted matrix.
1
Pari/GP, 15 bytes
a->vecsort(a~)~
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4
Python 3 + NumPy, 18 bytes
lambda a:a.sort(0)
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Input a np.array, output by modify it in-place.
1
JavaScript, 142 Bytes
for(j=0,l=a.length,w=a[0].length,c=new Array(l),e=1;j<w;j++){for(i=0;i<l;i++){if(a[i][j]){p=l-e++;!c[p]&&(c[p]=new Array(w));c[p][j]=1}}e=1;}
Where a is the input array and c is the output array.
Might be cheating since I am pulling a vacuum on all the 0s rather then applying gravity to all the 1s.
It works similar to the ...
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