9
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In the examples below, A and B will be 2-by-2 matrices, and the matrices are one-indexed.

A Kronecker product has the following properties:

A⊗B =  A(1,1)*B   A(1,2)*B
        A(2,1)*B   A(2,2)*B

     =  A(1,1)*B(1,1)   A(1,1)*B(1,2)   A(1,2)*B(1,1)   A(1,2)*B(1,2)
        A(1,1)*B(2,1)   A(1,1)*B(2,2)   A(1,2)*B(2,1)   A(1,2)*B(2,2)
        A(2,1)*B(1,1)   A(2,1)*B(1,2)   A(2,2)*B(1,1)   A(2,2)*B(1,2)
        A(2,2)*B(2,1)   A(2,2)*B(1,2)   A(2,2)*B(2,1)   A(2,2)*B(2,2)

A Kronecker sum has the following properties:

A⊕B = A⊗Ib + Ia⊗B

Ia and Ib are the identity matrices with the dimensions of A and B respectively. A and B are square matrices. Note that A and B can be of different sizes.

A⊕B =  A(1,1)+B(1,1)  B(1,2)         A(1,2)         0
        B(2,1)         A(1,1)+B(2,2)  0              A(1,2)
        A(2,1)         0              A(2,2)+B(1,1)  B(1,2)
        0              A(2,1)         B(2,1)         A(2,2)+B(2,2)

Given two square matrices, A and B, calculate the Kronecker sum of the two matrices.

  • The size of the matrices will be at least 2-by-2. The maximum size will be whatever your computer / language can handle by default, but minimum 5-by-5 input (5 MB output).
  • All input values will be non-negative integers
  • Builtin functions that calculate the Kronecker sum or Kronecker products are not allowed
  • In general: Standard rules regarding I/O format, program & functions, loopholes etc.

Test cases:

A =
     1     2
     3     4
B =
     5    10
     7     9

A⊕B =
     6    10     2     0
     7    10     0     2
     3     0     9    10
     0     3     7    13

----

A =
    28    83    96
     5    70     4
    10    32    44
B =
    39    19    65
    77    49    71
    80    45    76

A⊕B =
    67    19    65    83     0     0    96     0     0
    77    77    71     0    83     0     0    96     0
    80    45   104     0     0    83     0     0    96
     5     0     0   109    19    65     4     0     0
     0     5     0    77   119    71     0     4     0
     0     0     5    80    45   146     0     0     4
    10     0     0    32     0     0    83    19    65
     0    10     0     0    32     0    77    93    71
     0     0    10     0     0    32    80    45   120

----

A =
    76    57    54
    76     8    78
    39     6    94
B =
    59    92
    55    29

A⊕B =
   135    92    57     0    54     0
    55   105     0    57     0    54
    76     0    67    92    78     0
     0    76    55    37     0    78
    39     0     6     0   153    92
     0    39     0     6    55   123
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2
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Jelly, 26 21 20 19 bytes

æ*9Bs2¤×€€/€S;"/€;/

Input is a list of two 2D lists, output is a single 2D list. Try it online! or verify all test cases.

How it works

æ*9Bs2¤×€€/€S;"/€;/  Main link.
                     Argument: [A, B] (matrices of dimensions n×n and m×m)

      ¤              Evaluate the four links to the left as a niladic chain.
  9B                 Convert 9 to base 2, yielding [1, 0, 0, 1].
    s2               Split into sublists of length 2, yielding [[1, 0], [0, 1]].
æ*                   Vectorized matrix power.
                     This yields [[A¹, B⁰], [A⁰, B¹]], where B⁰ and A⁰ are the
                     identity matrices of dimensions m×m and n×n.
          /€         Reduce each pair by the following:
        €€             For each entry of the first matrix:
       ×                 Multiply the second matrix by that entry.
            S        Sum the two results, element by element.
                     This yields the Kronecker sum, in form of a n×n matrix of
                     m×m matrices.
               /€    Reduce each row of the outer matrix...
             ;"        by zipwith-concatenation.
                     This concatenates the columns of the matrices in each row,
                     yielding a list of length n of n×nm matrices.
                 ;/  Concatenate the lists, yielding a single nm×nm matrix.
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\$\endgroup\$
  • \$\begingroup\$ So many euros... your program is rich! \$\endgroup\$ – Luis Mendo Apr 27 '16 at 12:11
5
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CJam, 40 39 38 bytes

9Yb2/q~f{.{_,,_ff=?}:ffff*::.+:~}:..+p

Input format is a list containing A and B as 2D lists, e.g.

[[[1 2] [3 4]] [[5 10] [7 9]]]

Output format is a single CJam-style 2D list.

Test suite. (With more readable output format.)

Explanation

This code is an exercise in compound (or infix) operators. These are generally useful for array manipulation, but this challenge exacerbated the need for them. Here is a quick overview:

  • f expects a list and something else on the stack and maps the following binary operator over the list, passing in the other element as the second argument. E.g. [1 2 3] 2 f* and 2 [1 2 3] f* both give [2 4 6]. If both elements are lists, the first one is mapped over and the second one is used to curry the binary operator.
  • : has two uses: if the operator following it is unary, this is a simple map. E.g. [1 0 -1 4 -3] :z is [1 0 1 4 3], where z gets the modulus of a number. If the operator following it is binary, this will fold the operator instead. E.g. [1 2 3 4] :+ is 10.
  • . vectorises a binary operator. It expects two lists as arguments and applies the operator to corresponding pairs. E.g. [1 2 3] [5 7 11] .* gives [5 14 33].

Note that : itself is always a unary operator, whereas f and . themselves are always binary operators. These can be arbitrarily nested (provided they have the right arities). And that's what we'll do...

9Yb      e# Push the binary representation of 9, i.e. [1 0 0 1].
2/       e# Split into pairs, i.e. [[1 0] [0 1]]. We'll use these to indicate
         e# which of the two inputs we turn into an identity matrix.
q~       e# Read and evaluate input, [A B].
f{       e# This block is mapped over the [[1 0] [0 1]] list, also pushing
         e# [A B] onto the stack for each iteration.
  .{     e#   The stack has either [1 0] [A B] or [0 1] [A B]. We apply this
         e#   block to corresponding pairs, e.g. 1 A and 0 B.
    _,   e#     Duplicate the matrix and get its length/height N.
    ,_   e#     Turn into a range [0 1 ... N-1] and duplicate it.
    ff=  e#     Double f on two lists is an interesting idiom to compute an
         e#     outer product: the first f means that we map over the first list
         e#     with the second list as an additional parameter. That means for
         e#     the remaining operator the two arguments are a single integer
         e#     and a list. The second f then maps over the second list, passing
         e#     in the the number from the outer map as the first parameter.
         e#     That means the operator following ff is applied to every possible
         e#     pair of values in the two lists, neatly laid out in a 2D list.
         e#     The operator we're applying is an equality check, which is 1
         e#     only along the diagonal and 0 everywhere else. That is, we've
         e#     created an NxN identity matrix.
    ?    e#     Depending on whether the integer we've got along with the matrix
         e#     is 0 or 1, either pick the original matrix or the identity.
  }
         e#   At this point, the stack contains either [A Ib] or [Ia B]. 
         e#   Note that A, B, Ia and Ib are all 2D matrices.
         e#   We now want to compute the Kronecker product of this pair.
  :ffff* e#   The ffff* is the important step for the Kronecker product (but
         e#   not the whole story). It's an operator which takes two matrices
         e#   and replaces each cell of the first matrix with the second matrix
         e#   multiplied by that cell (so yeah, we'll end up with a 4D list of
         e#   matrices nested inside a matrix).
         e#   The leading : is a fold operation, but it's a bit of a degenerate
         e#   fold operation that is only used to apply the following binary operator
         e#   to the two elements of a list.
         e#   Now the ffff* works essentially the same as the ff= above, but
         e#   we have to deal with two more dimensions now. The first ff maps
         e#   over the cells of the first matrix, passing in the second matrix
         e#   as an additional argument. The second ff then maps over the second
         e#   matrix, passing in the cell from the outer map. We multiply them
         e#   with *.
         e#   Just to recap, we've essentially got the Kronecker product on the
         e#   stack now, but it's still a 4D list not a 2D list.
         e#   The four dimensions are:
         e#     1. Columns of the outer matrix.
         e#     2. Rows of the outer matrix.
         e#     3. Columns of the submatrices.
         e#     4. Rows of the submatrices.
         e#   We need to unravel that into a plain 2D matrix.
  ::.+   e#   This joins the rows of submatrices across columns of the outer matrix.
         e#   It might be easiest to read this from the right:
         e#     +    Takes two rows and concatenates them.
         e#     .+   Takes two matrices and concatenates corresponding rows.
         e#     :.+  Takes a list of matrices and folds .+ over them, thereby
         e#          concatenating the corresponding rows of all matrices.
         e#     ::.+ Maps this fold operation over the rows of the outer matrix.
         e#   We're almost done now, we just need to flatten the outer-most level
         e#   in order to get rid of the distinction of rows of the outer matrix.
  :~     e#   We do this by mapping ~ over those rows, which simply unwraps them.
}
         e# Phew: we've now got a list containing the two Kronecker products
         e# on the stack. The rest is easy, just perform pairwise addition.
:..+     e# Again, the : is a degenerate fold which is used to apply a binary
         e# operation to the two list elements. The ..+ then simply vectorises
         e# addition twice, such that we add corresponding cells of the 2D matrices.
p        e# All done, just pretty-print the matrix.
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\$\endgroup\$
  • \$\begingroup\$ fffffffffff what on earth... I hope that survives golfing so that you explain it eventually :P \$\endgroup\$ – FryAmTheEggman Apr 26 '16 at 14:25
  • \$\begingroup\$ @FryAmTheEggman :ffff* might be the longest (compound) operator I've ever used in CJam... For one more byte one could go even crazier though: 9Yb2/Q~f.{\{,,_ff=}&}::ffff*:::.+::~:..+p (and yeah, will add an explanation when I'm done golfing). \$\endgroup\$ – Martin Ender Apr 26 '16 at 14:27
4
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J - 38 33 31 bytes

i=:=@i.@#
[:,./^:2(*/i)+(*/~i)~

Usage

   f =: [:,./^:2(*/i)+(*/~i)~
   (2 2 $ 1 2 3 4) f (2 2 $ 5 10 7 9)
6 10 2  0
7 10 0  2
3  0 9 10
0  3 7 13
   (3 3 $ 28 83 96 5 70 4 10 32 44) f (3 3 $ 39 19 65 77 49 71 80 45 76)
67 19  65  83   0   0 96  0   0
77 77  71   0  83   0  0 96   0
80 45 104   0   0  83  0  0  96
 5  0   0 109  19  65  4  0   0
 0  5   0  77 119  71  0  4   0
 0  0   5  80  45 146  0  0   4
10  0   0  32   0   0 83 19  65
 0 10   0   0  32   0 77 93  71
 0  0  10   0   0  32 80 45 120
   (3 3 $ 76 57 54 76 8 78 39 6 94) f (2 2 $ 59 92 55 29)
135  92 57  0  54   0
 55 105  0 57   0  54
 76   0 67 92  78   0
  0  76 55 37   0  78
 39   0  6  0 153  92
  0  39  0  6  55 123
|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Using matrix division will fail if one of the matrices is singular. For example, (2 2 $ 1 2 3 4) f (2 2 $ 1 1 1 1) will raise a domain error. \$\endgroup\$ – Dennis Apr 27 '16 at 0:20
  • \$\begingroup\$ @Dennis good catch, I was only testing against random values ? 4 4 $ 100 . I'm not sure if there's a way to make use of dyad compose x f&g y = (g x) f (g y) or something else here. \$\endgroup\$ – miles Apr 27 '16 at 1:18
2
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Julia, 60 59 58 56 bytes

A%B=hvcat(sum(A^0),sum(i->map(a->a*B^i,A'^-~-i),0:1)...)

Try it online!

How it works

  • For matrices A and B, map(a->a*B,A') computes the Kronecker product A⊗B.

    The result is a vector of matrix blocks with the dimensions of B.

    We have to transpose A (with ') since matrices are stored in column-major order.

  • Since bitwise NOT with two's complement satisfies the identity ~n = -(n + 1) for all integers n, we have that -~-n = -(~(-n)) = --((-n) + 1) = 1 - n, so -~-0 = 1 and -~-1 = 0.

    This way the anonymous function i->map(a->a*B^i,A'^-~-i) applies the above map to B⁰ (the identity matrix with B's dimensions) and A¹ = A when i = 0, and to and A⁰ when i = 1.

  • sum(i->map(a->a*B^i,A'^-~-i),0:1) sums over {0,1} with the above anonymous function, computing the Kronecker sum A⊕B as A¹⊗B⁰ + A⁰⊗B¹.

    The result is a vector of matrix blocks with the dimensions of B.

  • sum(A^0) computes the sum of all entries of the identity matrix of A's dimensions. For an n×n matrix A, this yields n.

  • Finally, hvcat(sum(A^0),sum(i->map(a->a*B^i,A'^-~-i),0:1)...) concatenates the matrix blocks that form A⊕B.

    With first argument n, hvcat concatenates n matrix blocks horizontally, and the resulting (larger) blocks vertically.

|improve this answer|||||
\$\endgroup\$
0
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Ruby, 102

->a,b{r=0..-1+a.size*q=b.size
r.map{|i|r.map{|j|(i/q==j/q ?b[i%q][j%q]:0)+(i%q==j%q ?a[i/q][j/q]:0)}}}

In test program

f=->a,b{r=0..-1+a.size*q=b.size
r.map{|i|r.map{|j|(i/q==j/q ?b[i%q][j%q]:0)+(i%q==j%q ?a[i/q][j/q]:0)}}}

aa =[[1,2],[3,4]]
bb =[[5,10],[7,9]]
f[aa,bb].each{|e|p e}
puts

aa =[[28,83,96],[5,70,4],[10,32,44]]
bb =[[39,19,65],[77,49,71],[80,45,76]]
f[aa,bb].each{|e|p e}
puts

aa =[[76,57,54],[76,8,78],[39,6,94]]
bb =[[59,92],[55,29]]
f[aa,bb].each{|e|p e}
puts

Requires two 2D arrays as input and returns a 2D array.

There are probably better ways of doing this: using a function to avoid repetition; using a single loop and printing the output. Will look into them later.

|improve this answer|||||
\$\endgroup\$
0
\$\begingroup\$

JavaScript (ES6), 109

Built upon the answer to the other challenge

(a,b)=>a.map((a,k)=>b.map((b,i)=>a.map((y,l)=>b.map((x,j)=>r.push(y*(i==j)+x*(k==l))),t.push(r=[]))),t=[])&&t

Test

f=(a,b)=>a.map((a,k)=>b.map((b,i)=>a.map((y,l)=>b.map((x,j)=>r.push(y*(i==j)+x*(k==l))),t.push(r=[]))),t=[])&&t

console.log=x=>O.textContent+=x+'\n'

function show(label, mat)
{
  console.log(label)
  console.log(mat.join`\n`)
}

;[ 
  {a:[[1,2],[3,4]], b:[[5,10],[7,9]]},
  {a:[[28,83,96],[5,70,4],[10,32,44]], b:[[39,19,65],[77,49,71],[80,45,76]]},
  {a:[[76,57,54],[76,8,78],[39,6,94]], b:[[59,92],[55,29]]}
].forEach(t=>{
  show('A',t.a)  
  show('B',t.b)
  show('A⊕B',f(t.a,t.b))
  show('B⊕A',f(t.b,t.a))  
  console.log('-----------------')
})
<pre id=O></pre>

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\$\endgroup\$

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