Matrix chain multiplication: Difference between revisions

{{task|Discrete math}}

 

;Problem

 

For instance, with four matrices, one can compute A(B(CD)), A((BC)D), (AB)(CD), (A(BC))D, (AB)C)D. The number of different ways to put the parens is a [[Catalan numbers|Catalan number]], and grows exponentially with the number of factors.

 

__TOC__

=={{header|C}}==

{{trans|Kotlin}}

#include <limits.h>

#include <stdlib.h>

}

return 0;

 

{{output}}

</pre>

 

{{trans|Kotlin}}

 

class MatrixChainOrderOptimizer {

}

}

 

{{output}}

Order : (A((((((BC)D)(((EF)G)H))I)J)K))

Cost : 1773740

</pre>

 

=={{header|Fortran}}==

{{trans|Python}}

This is a translation of the Python iterative solution.

 

implicit none

contains

implicit none

integer :: a(:), n, i, j, k

integer, allocatable :: u(:, :)

integer(8), allocatable :: v(:, :)

n = ubound(a, 1) - 1

allocate (u(n, n), v(n, n))

u(:, 1) = -1

v(:, 1) = 0

do k = 1, j - 1

c = v(i, k) + v(i + k, j - k) + int(a(i), 8) * int(a(i + k), 8) * int(a(i + j), 8)

u(i, j) = k

v(i, j) = c

recursive subroutine aux(i, j)

integer :: i, j, k

k = u(i, j)

if (k < 0) then

end module

 

implicit none

 

call optim([1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2])

call optim([1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10])

 

'''Output'''

 

<pre>

38120 ((((((((1*2)*3)*4)*5)*6)*7)*(8*(9*10)))*(11*12))

1773740 (1*((((((2*3)*4)*(((5*6)*7)*8))*9)*10)*11))

</pre>

 

=={{header|Go}}==

The first <code>for</code> loop is based on the pseudo and Java code from the

[[wp:Matrix_chain_multiplication#A_dynamic_programming_algorithm|Wikipedia article]].

 

import "fmt"

fmt.Println()

}

{{out}}

<pre>

ABCDEFGHIJK -> A × BCDEFGHIJK cost=1773740

 

</pre>

 

 

'''Module''':

 

using OffsetArrays

end

 

 

'''Main''':

 

{{out}}

=={{header|Kotlin}}==

This is based on the pseudo-code in the Wikipedia article.

 

lateinit var m: List<IntArray>

println("\nCost : ${m[0][s.size - 1]}\n")

}

 

{{out}}

</pre>

 

{{trans|Fortran}}

 

n = length(a)-1;

u = zeros(n,n);

s = sprintf("(%s*%s)",aux(u,i,k),aux(u,i+k,j-k));

end

 

 

 

r =

s =

 

 

=={{header|Perl}}==

my(@dimensions) = @_;

my(@cp,@path);

for my $step ($start .. $end - 1) {

my $new_cost = $cp[$step][$start]

if ($new_cost < $cp[$end][$start]) {

$cp[$end][$start] = $new_cost; # cost

}

 

}

 

}

 

{{out}}

 

=={{header|Phix}}==

As per the wp pseudocode

{{out}}

<pre>

</pre>

 

We will solve the task in three steps:

 

=== Enumeration of parenthesizations ===

 

def aux(n, k):

if n == 1:

for v in aux(n - i, k + i):

yield [u, v]

 

'''Example''' (in the same order as in the task description)

 

print(u)

 

[[0, 1], [2, 3]]

[[0, [1, 2]], 3]

 

And here is the optimization step:

 

def cost(k):

if type(k) is int:

cmin = c

umin = u

 

=== Recursive cost optimization ===

The previous function optim1 already used recursion, but only to compute the cost of a given parens configuration, whereas another function (a generator actually) provides these configurations. Here we will do both recursively in the same function, avoiding the computation of configurations altogether.

 

def aux(n, k):

if n == 1:

return m, p, q, u

s, p, q, u = aux(len(a) - 1, 0)

 

=== Memoized recursive call ===

The only difference between optim2 and optim3 is the [[:wp:https://en.wikipedia.org/wiki/Memoization|@memoize]] [https://www.python.org/dev/peps/pep-0318/ decorator]. Yet the algorithm is way faster with this. According to Wikipedia, the complexity falls from O(2^n) to O(n^3). This is confirmed by plotting log(time) vs log(n) for n up to 580 (this needs [https://docs.python.org/3/library/sys.html#sys.setrecursionlimit changing Python's recursion limit]).

 

h = {}

def g(*u):

return m, p, q, u

s, p, q, u = aux(len(a) - 1, 0)

 

=== Putting all together ===

 

 

u = [[1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2],

t2 = time.clock()

print("%s %10.3f %10d %s" % (f.__name__, 1000 * (t2 - t1), s, u))

 

'''Output''' (timings are in milliseconds)

In the previous solution, memoization is done blindly with a dictionary. However, we need to compute the optimal products for all sublists. A sublist is described by its first index and length (resp. i and j+1 in the following function), hence the set of all sublists can be described by the indices of elements in a triangular array u. We first fill the "solution" (there is no product) for sublists of length 1 (u[0]), then for each successive length we optimize using what when know about smaller sublists. Instead of keeping track of the optimal solutions, the single needed one is computed in the end.

 

global u

n = len(a) - 1

 

print(optim4([1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]))

 

'''Output'''

</pre>

 

=== Recursive solution ===

{{trans|Python}}

Here is the equivalent of optim3 in Python's solution. Memoization is done with an [https://www.stata.com/help.cgi?mf_asarray associative array]. Multiple results are returned in a [https://www.stata.com/help.cgi?m2_struct structure]. The same effect as optim2 can be achieved by removing the asarray machinery.

 

struct ans {

real scalar p,q,s

optim((1,5,25,30,100,70,2,1,100,250,1,1000,2))

optim((1000,1,500,12,1,700,2500,3,2,5,14,10))

 

'''Output'''

{{trans|Fortran}}

 

function aux(u,i,j) {

k = u[i,j]

optim((1,5,25,30,100,70,2,1,100,250,1,1000,2))

optim((1000,1,500,12,1,700,2500,3,2,5,14,10))

 

'''Output'''

 

This solution is faster than the recursive one. The 1000 loops run now in 0.234 ms and 0.187 ms per loop on average.

 

=={{header|zkl}}==

{{trans|Python}}

aux:=fcn(n,k,a){ // (int,int,list) --> (int,int,int,list)

if(n==1){

h[key]=r;

return(r);

var letters=["A".."Z"].pump(String);

u.pump(String,

fcn(n){ if(List.isType(n)) String("(",pp(n),")") else letters[n] })

}

prnt(s,u);

 

prnt(s,u);

 

{{out}}

<pre>