New answers tagged matrix
2
Google Sheets, 90 88
Closing parens discounted.
Input matrix starts at A2:
A1: =COUNTA(2:2), gets number of columns (assume square)
A2: =SUM(ArrayFormula(OFFSET(A2,,,A1,A1)+TRANSPOSE(ArrayFormula(OFFSET(A2,,,A1,A1)))))
That was fun!
How it Works:
Add the matrix to its negative transpose. If the resulting matrix is all 0's, then the sum of all elements is 0,...
1
Java (JDK), 89 bytes
m->{for(int i=0;++i<m.length;)for(int j=0;++j<i;)if(m[i][j]!=-m[j][i])return 0;return 1;}
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I cheated a bit by returning 0 for false and 1 for true instead of the actual boolean/Boolean values.
0
Ruby, 91 bytes
->a{[*1..a.length-1].map{|n|("%0#{a.length}b"%2**n).reverse.split(//).map(&:to_i)}.push(a)}
Try it online!
0
Google Sheets, 52
Closing Parens discounted.
Input cells is Row 1, starting in column B.
A2 - =COUNTA(1:1). Rules say that we can take this as input too, so I have discounted this as well. (Our "k")
A3 - =ArrayFormula(IFERROR(0^MOD(SEQUENCE(A2-1,A2)-1,A2+1)))
The output matrix starts in B1.
How it works
Since this is a spreadsheet, the input ...
0
R, 179 bytes
(or 175 bytes without labelling 'T' and 'L')
function(m,n,t,z=function(l,t){while(t>1){F=c(F,g<-l%/%t);l=l-g;t=t-1};list(l=l,f=F[F>0])})
list(T=z(m,b<-order(sapply(1:t,function(f)z(m,f)$l*z(n,t%/%f)$l))[1])$f,L=z(n,t%/%b)$f)
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Outputs a list of ['T' = horizontal folds from top, 'L' = vertical folds from left].
How does ...
0
Haskell, 49 bytes
import Data.List
f x=x==transpose(map(map(0-))x)
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My first Haskell.
Function tacking a matrix and checking if input is equal to input mapped to (0-value) and transposed
1
05AB1E, 7 bytes
āDδQ`\)
Outputs reversed in both dimensions.
Try it online or verify all test cases.
Explanation:
ā # Push a list in the range [1,length] (without popping the implicit input-list)
D # Duplicate it
δ # Apply double-vectorized:
Q # Check if it's equal
# (this results in an L by L matrix filled with 0s, ...
3
Julia 1.0, 9 bytes
A->A==-A'
A straightforward anonymous function checking the equality.
Try it online!
2
Scala, 32 bytes
l=>l.transpose==l.map(_.map(-1*))
Finally, something that Scala has a builtin for!
The function's pretty straightforward - it compares the transpose of a List[List[Int]](doesn't have to be a List, could be any Iterable) to the negative, found by mapping each list inside l and using - to make it negative.
Try it in Scastie
3
Wolfram Mathematica, 20, 7 bytes
There is a built-in function for this task:
AntisymmetricMatrixQ
But one can simply write a script with less byte counts:
#==-#ᵀ&
The ᵀ character, as it is displayed in notebooks, stands for transpose. But if you copy this into tio, it won't be recognized because these characters are only supported by Mathematica ...
2
R, 425 bytes
p=function(m,t){
d=dim(m);r=d[1];c=d[2]
l=apply(matrix(c(seq(l=r-1),rep(0,r+c-2),seq(l=c-1)),,2),1,function(f){n=array(0,pmax(g<-(f-1)%%d+1,h<-(d-f-1)%%d+1))
`if`(f,n[1:g[1],]<-m[g[1]:1,],n[,1:g[2]]<-m[,g[2]:1])
n[1:h[1],1:h[2]]=n[1:h[1],1:h[2]]+m[(i=g%%d+1)[1]:r,i[2]:c]
if(max(n)<=t)cbind(c(T=f[1],L=f[2]),p(n,t))})
if(!is.null(l)...
1
gorbitsa-ROM, 8 bytes
r1 R A1 B0 T
This is an awful abuse of rule
Input and output can assume whatever forms are most convenient.
If input takes form of "arr[i][j] arr[j][i]", the problem becomes "is sum = 0?".
This code takes pairs of values and outputs their sum if it's not 0
Thus if you provide matrix as previously mentioned pairs, ...
3
Charcoal, 10 bytes
⁼θＥθＥ豧λκ
Try it online! Link is to verbose version of code. Outputs a Charcoal boolean, i.e. - if the matrix is antisymmetric, nothing if not. Explanation:
Ｅθ Map over input matrix rows (should be columns, but it's square)
Ｅθ Map over input matrix rows
§λκ Cell of transpose
± Negated
⁼θ Does ...
5
JavaScript (ES6), 42 bytes
Returns false for antisymmetric or true for non-antisymmetric.
m=>m.some((r,y)=>r.some((v,x)=>m[x][y]+v))
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9
R, 23 bytes
function(m)!any(m+t(m))
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Checks whether there are any non-zero elements in \$M+M^T\$.
7
C (gcc), 67 64 bytes
-3 thanks to AZTECCO
i,j;f(m,s)int**m;{for(i=j=0;i=i?:s--;)j|=m[s][--i]+m[i][s];m=j;}
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Returns 0 if the matrix is antisymmetric, and a nonzero value otherewise.
1
Ruby, 40 bytes
->a{a==a.transpose.map{|r|r.map{|c|-c}}}
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5
Brachylog, 5 bytes
5 bytes seems to be the right length for this (unless you're Jelly). Actually, this would be three bytes if Brachylog implicitly vectorized predicates like negation.
\ṅᵐ²?
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Explanation
\ Transpose
ṅᵐ² Map negation at depth 2
? Assert that the result is the same as the input
1
Pip, 5 bytes
Z_=-_
A function submission; pass a nested list as its argument. Try it online!
Explanation
Z_ The argument, zipped together
= Equals
-_ The argument, negated
4
Pyth, 5 bytes
qC_MM
Try it online!
Explanation
qC_MM
q : Check if input equals
C : Transpose of
_MM : Negated input
6
Octave, 19 bytes
@(a)isequal(a',-a);
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The semicolon doesn't need to be there, but it outputs the function otherwise, so I'll take the one-byte hit to my score for now.
Explanation
It's pretty straightforward - it checks to see if the matrix of the transpose is equal to the negative matrix
3
MATL, 5 bytes
!_GX=
Try it online!
Explanation
!_GX=
// Implicit input on top of stack
! // Replace top stack element with its transpose
_ // Replace top stack element with its negative
G // Push input onto stack
X= // Check for equality
2
Japt, 5 bytes
eUy®n
Try it
e compare input with :
Uy columns of input
®n with each element negated
Previous version ÕeËËn didn't work, corrected using the ® symbol
10
Python 2, 45 bytes
lambda A:A==[[-x for x in R]for R in zip(*A)]
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4
Io, 69 bytes
method(x,x map(i,v,v map(I,V,V+x at(I)at(i)))flatten unique==list(0))
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Explanation
For all a[x][y], it checks whether all a[x][y]+a[y][x]==0.
method(x, // Input x.
x map(i,v, // Map all x's rows (index i):
v map(I,V, // Foreach the ...
13
APL (Dyalog Unicode), 3 bytes
-≡⍉
Try it online!
This is exactly an APLcart entry on "antisymmetric". Basically it checks if the input's negative - matches ≡ the input's transpose ⍉.
4
Jelly, 3 bytes
Z⁼N
Try it online!
Explanation
Z Transposition
⁼ Equals
N Negative
4
05AB1E (legacy), 3 bytes
ø(Q
Try it online!
Explanation
ø The input transposed,
( Negated,
Q Is equal to the orginal input
2
Python 2, 46 bytes
lambda l,k:[l]+zip(*[iter(([1]+[0]*k)*~-k)]*k)
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Takes input as a tuple l and number of terms k, and outputs with both rows and columns reversed.
The idea is to use the zip/iter trick to create an identity-like matrix by splitting a repeating list into chunks. The is similar to my solution to construct the identity matrix ...
1
R, 34 bytes
function(r,k)rbind(diag(k)[-1,],r)
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Takes the length as well; the TIO link has a k=length(r) argument so you can just input the recurrence relation.
2
APL (Dyalog Unicode), 10 bytes
⊢⍪¯1↓⍋∘.=⍋
Try it online!
Tacit function taking the list of coefficients on the right.
Explanation
⊢⍪¯1↓⍋∘.=⍋
⍋ ⍋ ⍝ Grade up to obtain a list of k distinct values
∘.= ⍝ Outer product with operation `equals` (identity matrix)
¯1↓ ⍝ Drop the last row
⊢⍪ ⍝ Prepend the list of coefficients
2
Io, 56 bytes
method(a,a map(i,v,if(i<1,a,a map(I,v,if(I==i-1,1,0)))))
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Explanation
method(a, ) // Input an array.
a map(i,v, ) // Map. i = index, v = value
if(i<1, ) // If the indice is 0,
...
0
C (gcc), 90 89 80 bytes
Saved 9 bytes thanks to ceilingcat!!!
i;j;f(a,k)int*a;{for(i=k;i--;puts(""))for(j=k;j--;)printf("%d ",i?i-1==j:a[j]);}
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Inputs an array of coefficients (in forward order) along with its length.
Prints a matrix that represents the recurrence relation.
5
JavaScript (ES6), 36 bytes
a=>a.map((_,i)=>i?a.map(_=>+!--i):a)
Try it online!
Returns:
$$
\begin{bmatrix} a_1 & a_2 & a_3 & \cdots & a_{k-1} & a_k \\ 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 &...
0
Jelly, 8 bytes
W;J⁼þṖ$$
A monadic Link accepting a list which yields a list of lists in the reversed rows & columns permutation.
Try it online!
How?
W;J⁼þṖ$$ - Link: list A e.g. [5,2,5,4]
W - wrap (A) in a list [[5,2,5,4]]
$ - last two links as a monad - f(A):
J - range of length (A) ...
8
MATL, 8 7 bytes
-1 byte thanks to @LuisMendo
Xy4LY)i
Takes the coefficients in reverse order
Try it online!
Explanation
Xy4LY)i
Xy : Create an identity matrix of size equal to input
4LY) : Remove the first row
i : Insert input onto the stack
1
Charcoal, 12 bytes
ＩＥθ⎇κＥθ⁼⊖κμθ
Try it online! Link is to verbose version of code. Produces the "reversed in both directions" output. Works by replacing the first row of a shifted identity matrix with the input. Explanation:
Ｅθ Map over input list
⎇κ If this is not the first row then
Ｅθ Map over input list
...
6
J, 10 8 bytes
Returns the matrix reversed in both dimensions.
,}:@=@/:
Try it online!
How it works
,}:@=@/: input: 3 _1 19
/: indices that sort: 1 0 2
(just to get k different numbers)
=@ self-classify: 1 0 0
0 1 0
0 0 1
}:@ drop last row: ...
1
Python 3, 60 58 bytes
-2 bytes thanks to @JonathanAllan
lambda a,k:[map(i.__eq__,range(k))for i in range(1,k)]+[a]
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Takes the coefficients in reverse order
3
APL (Dyalog Unicode), 37 35 bytes
{⌽∘⍉@⍺⊢⍵}/⌽(⊂,{(4|⎕)/,⊢∘⊂⌺2 2⍳⍴⍵})⎕
Try it online!
Switched to a more straightforward method after I realized @ also accepts a matrix of coordinates. Then we don't need to fiddle with the coordinate order; we extract the submatrix coordinates with ⊢∘⊂⌺2 2, and just rotate them directly using ⌽∘⍉.
APL (Dyalog Unicode), 37 ...
2
APL (Dyalog Unicode), 70 66 bytes
D←⊢-1∘⌽
+/,{16=×/|D⍵/⍨×⍵×D⍵}¨(⍉1↓⌽)⍣4×(⊢-{⊂(,⍵)[8(-,⊢)3,⍳3]}⌺3 3)⎕
Try it online!
-4 bytes thanks to @Bubbler and @ngn
Full program that requires ⎕IO←0
⍝ Helper function D: cyclic differences of a list
D←⊢-1∘⌽
⊢ ⍝ Each element
- ⍝ Subtract
1∘⌽ ⍝ The one after
⍝ Main code
+/,{16=×/|D⍵/⍨×⍵×D⍵}¨¯2 ¯2↓1⊖1⌽×(⊢-{⊂...
0
Charcoal, 81 bytes
ＦθＦι⊞υκＵＭθκ≔ＬθηＦυＦ⁺⁺⪪ＥυληＥθ⁺λ×θη⟦×θ⊕η×⊕θ⊖η⟧«≔Ｅκ§υλι¿⁼¹№ι⁰§≔υ§κ⌕ι⁰⁻÷×⊕×ηηη²Σι»Ｉ⪪υη
Try it online! Link is to verbose version of code. Uses zero as the "blank" marker. Explanation:
ＦθＦι⊞υκ
Flatten the input array.
ＵＭθκ
Replace the original array with a range from 0 to n-1.
≔Ｌθη
Also the length of the array is used a lot so ...
1
Jelly, 25 bytes
ZṚ,⁸;Jị"$€$§FE
²Œ!ṁ€ÇƇ=ÐṀ
A full program taking n and a list-of-lists-formatted representation of the incomplete square which prints the result in the same format.
Try it online! - too slow for TIO's 60s limit
...so, Try a limited space one which only considers the first 150K permutations - three magic squares two of which match at two ...
3
R, 169 180 142 135 bytes
Edits: +11 bytes to rotate the magic square back to it's original orientation, -38 bytes by wrapping "replace-only-missing-element" into a function, -7 bytes by various golf obfuscations
function(m,n){while(F%%4|sum(!m)){m[n:1,]=apply(m,1,f<-function(v){if(sum(!v)<2)v[!v]=(n^3+n)/2-sum(v);v})
m[d]=f(m[d<-!0:n])
F=...
4
JavaScript (ES7), 143 142 140 bytes
Expects (n)(m), where unknown cells in m are filled with 0's.
n=>g=m=>[0,1,2,3].some(d=>m.some((r,i)=>m.map((R,j)=>t^(t-=(v=d?R:r)[x=[j,i,j,n+~j][d]])||(e--,X=x,V=v),e=1,t=n**3+n>>1)&&!e))?g(m,V[X]=t):m
Try it online!
Commented
n => // outer function taking n
g ...
4
APL (Dyalog Unicode), 60 bytes
{(⍵,m+.×1+⍺*2)⌹(∘.(×⊢×=)⍨⍵)⍪2×m←(⍪↑c(⌽c))⍪(⊢⍪⍴⍴⍉)⍺/c←∘.=⍨⍳⍺}
Try it online!
Not likely the shortest approach, but anyway here is one with Matrix Divide ⌹, a.k.a. Solve Linear Equation. This works because all the cells are uniquely determined by the horizontal/vertical/diagonal sums when joined with the givens. ⌹ has no ...
2
Wolfram Language (Mathematica), 100 bytes
#/.Solve[Tr/@Flatten[{#,Thread@#,{(d=Diagonal)@#,d@Reverse@#}},1]==Table[(l^3+l)/2,2(l=Tr[1^#])+2]]&
Try it online!
6
JavaScript, 559 551 bytes
Fast and methodical.
B=Boolean,f=((e,r)=>(v=r*((r**2+1)/2),e.forEach(e=>e.filter(B).length==r-1?e[e.findIndex(e=>!e)]=v-e.reduce((e,f)=>!(e+=f)||e):0),e[0].reduce((f,l,n)=>!(f[0].push(e[n][n])+f[1].push(e[n][r-1-n]))||f,[[],[]]).forEach((f,l)=>{f.filter(B).length==r-1&&(z=f.findIndex(e=>!e),e[z][l?r-1-z:...
9
MATL, 36 bytes
`nZ@[]etGg)GXz-yt!hs&ytXdwPXdhsh&-ha
The input is an \$ n \times n\$ matrix, with \$0\$ for the unknown numbers.
The code keeps generating random \$ n \times n\$ matrices formed by the numbers \$1, \dots, n^2\$ until one such matrix meets the required conditions. This procedure is guaranteed to finish with probability one.
This is a ...
2
05AB1E, 43 41 30 bytes
-2 bytes by replacing Dgt with ¹ to get first input back
-11 bytes thanks to Kevin Cruijssen!
nLœʒ¹ôD©ø®Å\®Å/)O˜Ë}ʒøε¬_sË~}P
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Takes input as (n, flattened square), where zeros represent blanks, like
3
[4,9,2,3,0,0,0,0,0]
Works by generating all permutations of the numbers from 1 to n2, filtering to only keep those that ...
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