Find the Matrix Power

Question

Problem

Create a program or function that can calculate the result of a matrix raised to the n^th power. Your code will take an arbitrary square matrix A and a non-negative integer n, and return a matrix with the value Aⁿ.

Restrictions

Built-in functions that compute the matrix power and matrix product are not allowed.

The rest of the standard rules for code-golf apply.

Explanation

Given a square matrix A, the value of Aⁿ = A A ⋯ A (repeated matrix product of A with itself, n times). If n is positive, the standard just mentioned is used. When n is zero, the identity matrix with the same order of A is the result.

Goal

This is code-golf and the shortest code wins.

Test Cases

Here, A is the input matrix, n is the input integer, and r is the output matrix where r = Aⁿ.

n = 0
A = 62 72
    10 34
r =  1  0
     0  1

n = 1
A = 23 61 47
    81 11 60
    42  9  0
r = 23 61 47
    81 11 60
    42  9  0

n = 2
A =  12  28 -26  3
     -3 -10 -13  0
     25  41   3 -4
    -20 -14  -4 29
r = -650 -1052  -766 227
    -331  -517   169  43
     332   469 -1158 -53
    -878  -990   574 797

n = 4
A = -42 -19  18 -38
    -33  26 -13  31
    -43  25 -48  28
     34 -26  19 -48
r = -14606833  3168904 -6745178  4491946
      1559282  3713073 -4130758  7251116
      8097114  5970846 -5242241 12543582
     -5844034 -4491274  4274336 -9196467

n = 5
A =  7  0 -3  8 -5  6 -6
     6  7  1  2  6 -3  2
     7  8  0  0 -8  5  2
     3  0  1  2  4 -3  4
     2  4 -1 -7 -4 -1 -8
    -3  8 -9 -2  7 -4 -8
    -4 -5 -1  0  5  5 -1
r =  39557  24398 -75256 131769  50575   14153  -7324
    182127  19109   3586 115176 -23305    9493 -44754
    146840  31906 -23476 190418 -38946   65494  26468
     42010 -21876  41060 -13950 -55148   19290   -406
     44130  34244 -35944  34272  22917  -39987 -54864
      1111  40810 -92324  35831 215711 -117849 -75038
    -70219   8803 -61496   6116  45247   50166   2109

Answer 1

Jelly, 17 16 15 bytes

Z×'³S€€
LṬ€z0Ç¡

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Permalinks with grid output: test case 1 | test case 2 | test case 3 | test case 4 | test case 5

How it works

LṬ€z0Ç¡  Main link. Arguments: A (matrix), n (power)

L        Get the length l of A.
 Ṭ€      Turn each k in [1, ..., l] into a Boolean list [0, 0, ..., 1] of length k.
   z0    Zip; transpose the resulting 2D list, padding with 0 for rectangularity.
         This constructs the identity matrix of dimensions k×k.
     Ç¡  Execute the helper link n times.

Z×'³S€€  Helper link. Argument: B (matrix)

Z        Zip; transpose rows and columns of B.
   ³     Yield A.
 ×'      Spawned multiplication; element-wise mutiply each rows of zipped B (i.e.,
         each column of B) with each row of A.
         This is "backwards", but multiplication of powers of A is commutative.
    S€€  Compute the sum of each product of rows.

Answer 2

MATL, 20 bytes

XJZyXyi:"!J2$1!*s1$e

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Explanation

This avoids the matrix multiplication by doing it manually, using element-wise multiplication with broadcast followed by vectorized sum. Specifically, to multiply matrices M and N, both of size s×s:

1. Transpose M. Call the resulting matrix P.
2. Permute the dimensions of N such that N is "turned" with a rotation axis along the first dimension, giving an s×1×s 3D array, say Q.
3. Multiply each element of P times each element of Q, with implicit broadcast. This means that P is automatically replicated s times along the third dimension, and Q is replicated s times along the second, to make them both s×s×s, before the actual element-wise multiplication takes place.
4. Sum along the first dimension to yield a 1×s×s array.
5. Squeeze the leading singleton dimension out, to produce an s×s result.

Commented code:

XJ      % take matrix A. Copy to clipboard J
ZyXy    % generate identity matrix of the same size
i:      % take exponent n. Generate range [1 2 ... n] (possibly empty)
"       % for each element in that range
  !     %   transpose matrix with product accumulated up to now (this is step 1)
  J     %   paste A
  2$1!  %   permute dimensions: rotation along first-dimension axis (step 2)
  *     %   element-wise multiplication with broadcast (step 3)
  s     %   sum along first dimension (step 4)
  1$e   %   squeeze out singleton dimension, i.e. first dimension (step 5)
        % end for. Display

Answer 3

APL, 32 31 chars

{⍺=0:(⍴⍵)⍴1⍨1+≢⍵⋄+.×⍨⍣(⍺-1)⊣⍵}

Left argument the power to raise to, right argument the matrix. The hardest (most space consuming) bit is building the identity matrix for the case where the desired exponent is 0. The actual computation is based on the fact that the generalised inner product (.) with + and × as operands is effectively the matrix product. This combined with the power ⍣ operator ("repeat") forms the meat of the solution.

Answer 4

K (ngn/k), 14 bytes

{y(+/x*)'/=#x}

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-2 bytes thanks to @coltim on the k tree. The inner train multiplies x on the right side of the current matrix (instead of left side).

Why (+/x*)' is also a matmul:

(+/(e f;g h)*)' (a b;c d)
( (+/(e f;g h)*) a b; (+/(e f;g h)*) c d )
( +/ (ae af;bg bh) ; +/ (ce cf;dg dh) )
((ae+bg) (af+bh); (ce+dg) (cf+dh))

K (ngn/k), 16 bytes

{y(+/'x*\:)/=#x}

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Original solution. Takes [matrix; n] and computes matrix^n.

How it works

{y(+/'x*\:)/=#x}  x:matrix; y:n
            =#x   Identity matrix of size of x
 y(       )/      Repeat y times:
      x*\:        Multiply each row of x to each column of current matrix
   +/'            Sum each to get the matrix product

Why +/'*\: is a matmul:

(a b;c d) *\: (e f;g h)
multiply-eachleft expands to
(a b * (e f;g h) ; c d * (e f;g h))
atomic application gives
((ae af
  bg bh)
 (ce cf
  dg dh))
so applying +/' on this will give the result of matmul.

Answer 5

Sage, 112 bytes

lambda A,n:reduce(lambda A,B:[[sum(map(mul,zip(a,b)))for b in zip(*B)]for a in A],[A]*n,identity_matrix(len(A)))

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Explanation:

The inner lambda (lambda A,B:[[sum(map(mul,zip(a,b)))for b in zip(*B)]for a in A]) is a straightforward implementation of matrix multiplication. The outer lambda (lambda A,n:reduce(...,[A]*n,identity_matrix(len(A)))) uses reduce to compute the matrix power by iterated matrix multiplication (using the aforementioned homemade matrix multiplication function), with the identity matrix as the initial value to support n=0.

Answer 6

Julia, 90 86 68 bytes

f(A,n)=n<1?eye(A):[A[i,:][:]⋅f(A,n-1)[:,j]for i=m=1:size(A,1),j=m]

This is a recursive function that accepts a matrix (Array{Int,2}) and an integer and returns a matrix.

Ungolfed:

function f(A, n)
    if n < 1
        # Identity matrix with the same type and dimensions as the input
        eye(A)
    else
        # Compute the dot product of the ith row of A and the jth column
        # of f called on A with n decremented
        [dot(A[i,:][:], f(A, n-1)[:,j]) for i = (m = 1:size(A,1)), j = m]
    end
end

Try it online! (includes all but the last test case, which is too slow for the site)

Saved 18 bytes thanks to Dennis!

Answer 7

Python 2.7, 158 145 bytes

The worst answer here, but my best golf in Python yet. At least it was fun learning how to do matrix multiplication.

Golfed:

def q(m,p):
 r=range(len(m))
 if p<1:return[[x==y for x in r]for y in r]
 o=q(m,p-1)
 return[[sum([m[c][y]*o[x][c]for c in r])for y in r]for x in r]

Explanation:

#accepts 2 arguments, matrix, and power to raise to
def power(matrix,exponent):
 #get the range object beforehand
 length=range(len(matrix))
 #if we are at 0, return the identity
 if exponent<1:
  #the Boolean statement works because Python supports multiplying ints by bools
  return [[x==y for x in length] for y in length]
 #otherwise, recur
 lower=power(matrix,exponent-1)
 #and return the product
 return [[sum([matrix[c][y]*lower[x][c] for c in length]) for y in length] for x in length]

Answer 8

Jelly, 11 bytes

L=þ`Zḋþ³ƊƓ¡

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A more modern update to Dennis' answer, be sure to give that an upvote.

Additionally, this is a 9 byte answer that takes the dimensions of the matrix as the first 2 command line args and the matrix as the third.

Both take the power via STDIN.

How it works

L=þ`Zḋþ³ƊƓ¡ - Main link. Takes A on the left
L           - Length of A, L
   `        - Using L as both arguments:
  þ         -   Create an LxL matrix, where the element at (i,j) is:
 =          -     Does i = j?
               This creates an identity matrix LxL
         Ɠ  - Read an integer from STDIN, n
        Ɗ ¡ - Do the following n times:
    Z       -   Transpose
       ³    -   Yield A
      þ     -   Pair all rows of the transposed matrix with the rows of A and, over each row:
     ḋ      -     Dot product

Answer 9

Julia, 27 24

a$n=round.(exp(log(a)n))

I am not sure what is allowed, but...

julia> a
5×5 Matrix{Int64}:
 35  18  40  37  77
 31   5  45  23  73
 62  67  29  85  97
 20   9  83  70  65
  2  13  53  59  52

julia> round.(exp(log(a)5)) ≈ a^5
true

24 thanks to @MarcMush.

Answer 10

JavaScript (ES6), 123 bytes

(n,a)=>[...Array(n)].map(_=>r=m(i=>m(j=>m(k=>s+=a[i][k]*r[k][j],s=0)&&s)),m=g=>a.map((_,n)=>g(n)),r=m(i=>m(j=>+!(i^j))))&&r

I had a 132 byte version using reduce but I was just mapping over a so often that it turned out to be 9 bytes shorter to write a helper function to do it for me. Works by creating the identity matrix and multiplying it by a longhand n times. There are a number of expressions that return 0 or 1 for i == j but they all seem to be 7 bytes long.

Answer 11

Python 3, 147 bytes

def f(a,n):
 r=range(len(a));r=[[i==j for j in r]for i in r]
 for i in range(n):r=[[sum(map(int.__mul__,x,y))for y in zip(*a)]for x in r]
 return r

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Answer 12

R, 49 bytes

f=function(m,n)`if`(n,m%*%f(m,n-1),diag(nrow(m)))

Recursive function that takes a matrix and the power n to raise it to. Recursively calls %*%, which computes the dot-product. The initial value for the recursion is the identity matrix of the same size as m. Since m %*% m = m %*% m %*% I, this works just fine.

Answer 13

Python 3, 128 bytes

def f(A,n):r=range(len(A));return n and[[sum(A[i][j]*x[i]for i in r)for j in r]for x in f(A,n-1)]or[[i==j for i in r]for j in r]

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Answer 14

Python 2, 131 bytes

f=lambda M,n:n and[[sum(map(int.__mul__,r,c))for c in zip(*f(M,n-1))]for r in M]or[[0]*i+[1]+[0]*(len(M)+~i)for i in range(len(M))]

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Answer 15

05AB1E, 11 bytes

²āDδQ¹Føδ*O

Try it online! Footer formats output as a grid.

²āDδQ pushes the identity matrix to the stack:
²ā range from 1 to the length of the second input
DδQ equality table with itself

¹F iterate first input times:
øδ*O multiply current matrix with input matrix:
ø transpose current matrix
δ* 3d multiplication table with input and transposed matrix
O sum along last axis