Matrix chain multiplication
Using the most straightfoward algorithm (which we assume here), computing the product of two matrices of dimensions (n1,n2) and (n2,n3) requires n1*n2*n3 FMA operations. The number of operations required to compute the product of matrices A1, A2... An depends on the order of matrix multiplications, hence on where parens are put. Remember that the matrix product is associative, but not commutative, hence only the parens can be moved.
You are encouraged to solve this task according to the task description, using any language you may know.
- 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 number, and grows exponentially with the number of factors.
Here is an example of computation of the total cost, for matrices A(5,6), B(6,3), C(3,1):
- AB costs 5*6*3=90 and produces a matrix of dimensions (5,3), then (AB)C costs 5*3*1=15. The total cost is 105.
- BC costs 6*3*1=18 and produces a matrix of dimensions (6,1), then A(BC) costs 5*6*1=30. The total cost is 48.
In this case, computing (AB)C requires more than twice as many operations as A(BC). The difference can be much more dramatic in real cases.
- Task
Write a function which, given a list of the successive dimensions of matrices A1, A2... An, of arbitrary length, returns the optimal way to compute the matrix product, and the total cost. Any sensible way to describe the optimal solution is accepted. The input list does not duplicate shared dimensions: for the previous example of matrices A,B,C, one will only pass the list [5,6,3,1] (and not [5,6,6,3,3,1]) to mean the matrix dimensions are respectively (5,6), (6,3) and (3,1). Hence, a product of n matrices is represented by a list of n+1 dimensions.
Try this function on the following two lists:
- [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
- [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]
To solve the task, it's possible, but not required, to write a function that enumerates all possible ways to parenthesize the product. This is not optimal because of the many duplicated computations, and this task is a classic application of dynamic programming.
See also Matrix chain multiplication on Wikipedia.
C
<lang c>#include <stdio.h>
- include <limits.h>
- include <stdlib.h>
int **m; int **s;
void optimal_matrix_chain_order(int *dims, int n) {
int len, i, j, k, temp, cost;
n--;
m = (int **)malloc(n * sizeof(int *));
for (i = 0; i < n; ++i) {
m[i] = (int *)calloc(n, sizeof(int));
}
s = (int **)malloc(n * sizeof(int *));
for (i = 0; i < n; ++i) {
s[i] = (int *)calloc(n, sizeof(int));
}
for (len = 1; len < n; ++len) {
for (i = 0; i < n - len; ++i) {
j = i + len;
m[i][j] = INT_MAX;
for (k = i; k < j; ++k) {
temp = dims[i] * dims[k + 1] * dims[j + 1];
cost = m[i][k] + m[k + 1][j] + temp;
if (cost < m[i][j]) {
m[i][j] = cost;
s[i][j] = k;
}
}
}
}
}
void print_optimal_chain_order(int i, int j) {
if (i == j)
printf("%c", i + 65);
else {
printf("(");
print_optimal_chain_order(i, s[i][j]);
print_optimal_chain_order(s[i][j] + 1, j);
printf(")");
}
}
int main() {
int i, j, n;
int a1[4] = {5, 6, 3, 1};
int a2[13] = {1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2};
int a3[12] = {1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10};
int *dims_list[3] = {a1, a2, a3};
int sizes[3] = {4, 13, 12};
for (i = 0; i < 3; ++i) {
printf("Dims : [");
n = sizes[i];
for (j = 0; j < n; ++j) {
printf("%d", dims_list[i][j]);
if (j < n - 1) printf(", "); else printf("]\n");
}
optimal_matrix_chain_order(dims_list[i], n);
printf("Order : ");
print_optimal_chain_order(0, n - 2);
printf("\nCost : %d\n\n", m[0][n - 2]);
for (j = 0; j <= n - 2; ++j) free(m[j]);
free(m);
for (j = 0; j <= n - 2; ++j) free(s[j]);
free(s);
}
return 0;
}</lang>
- Output:
Dims : [5, 6, 3, 1] Order : (A(BC)) Cost : 48 Dims : [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2] Order : ((((((((AB)C)D)E)F)G)(H(IJ)))(KL)) Cost : 38120 Dims : [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10] Order : (A((((((BC)D)(((EF)G)H))I)J)K)) Cost : 1773740
C#
<lang csharp>using System;
class MatrixChainOrderOptimizer {
private int[,] m; private int[,] s;
void OptimalMatrixChainOrder(int[] dims) {
int n = dims.Length - 1;
m = new int[n, n];
s = new int[n, n];
for (int len = 1; len < n; ++len) {
for (int i = 0; i < n - len; ++i) {
int j = i + len;
m[i, j] = Int32.MaxValue;
for (int k = i; k < j; ++k) {
int temp = dims[i] * dims[k + 1] * dims[j + 1];
int cost = m[i, k] + m[k + 1, j] + temp;
if (cost < m[i, j]) {
m[i, j] = cost;
s[i, j] = k;
}
}
}
}
}
void PrintOptimalChainOrder(int i, int j) {
if (i == j)
Console.Write((char)(i + 65));
else {
Console.Write("(");
PrintOptimalChainOrder(i, s[i, j]);
PrintOptimalChainOrder(s[i, j] + 1, j);
Console.Write(")");
}
}
static void Main() {
var mcoo = new MatrixChainOrderOptimizer();
var dimsList = new int[3][];
dimsList[0] = new int[4] {5, 6, 3, 1};
dimsList[1] = new int[13] {1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2};
dimsList[2] = new int[12] {1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10};
for (int i = 0; i < dimsList.Length; ++i) {
Console.Write("Dims : [");
int n = dimsList[i].Length;
for (int j = 0; j < n; ++j) {
Console.Write(dimsList[i][j]);
if (j < n - 1)
Console.Write(", ");
else
Console.WriteLine("]");
}
mcoo.OptimalMatrixChainOrder(dimsList[i]);
Console.Write("Order : ");
mcoo.PrintOptimalChainOrder(0, n - 2);
Console.WriteLine("\nCost : {0}\n", mcoo.m[0, n - 2]);
}
}
}</lang>
- Output:
Dims : [5, 6, 3, 1] Order : (A(BC)) Cost : 48 Dims : [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2] Order : ((((((((AB)C)D)E)F)G)(H(IJ)))(KL)) Cost : 38120 Dims : [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10] Order : (A((((((BC)D)(((EF)G)H))I)J)K)) Cost : 1773740
Fortran
This is a translation of the Python iterative solution.
<lang fortran>module optim_m
implicit none
contains
subroutine optim(a)
implicit none
integer :: a(:), n, i, j, k
integer(8) :: c
integer, allocatable :: u(:, :)
integer(8), allocatable :: v(:, :)
n = ubound(a, 1) - 1
allocate (u(n, n), v(n, n))
v = -1
u(:, 1) = -1
v(:, 1) = 0
do j = 2, n
do i = 1, n - j + 1
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)
if (v(i, j) < 0 .or. c < v(i, j)) then
u(i, j) = k
v(i, j) = c
end if
end do
end do
end do
write (*, "(I0,' ')", advance="no") v(1, n)
call aux(1, n)
print *
deallocate (u, v)
contains
recursive subroutine aux(i, j)
integer :: i, j, k
k = u(i, j)
if (k < 0) then
write (*, "(I0)", advance="no") i
else
write (*, "('(')", advance="no")
call aux(i, k)
write (*, "('*')", advance="no")
call aux(i + k, j - k)
write (*, "(')')", advance="no")
end if
end subroutine
end subroutine
end module
program optim_p
use optim_m 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])
end program</lang>
Output
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))
Go
The first for loop is based on the pseudo and Java code from the
Wikipedia article.
<lang Go>package main
import "fmt"
// PrintMatrixChainOrder prints the optimal order for chain // multiplying matrices. // Matrix A[i] has dimensions dims[i-1]×dims[i]. func PrintMatrixChainOrder(dims []int) { n := len(dims) - 1 m, s := newSquareMatrices(n)
// m[i,j] will be minimum number of scalar multiplactions // needed to compute the matrix A[i]A[i+1]…A[j] = A[i…j]. // Note, m[i,i] = zero (no cost). // s[i,j] will be the index of the subsequence split that // achieved minimal cost. for lenMinusOne := 1; lenMinusOne < n; lenMinusOne++ { for i := 0; i < n-lenMinusOne; i++ { j := i + lenMinusOne m[i][j] = -1 for k := i; k < j; k++ { cost := m[i][k] + m[k+1][j] + dims[i]*dims[k+1]*dims[j+1] if m[i][j] < 0 || cost < m[i][j] { m[i][j] = cost s[i][j] = k } } } }
// Format and print result. const MatrixNames = "ABCDEFGHIJKLMNOPQRSTUVWXYZ" var subprint func(int, int) subprint = func(i, j int) { if i == j { return } k := s[i][j] subprint(i, k) subprint(k+1, j) fmt.Printf("%*s -> %s × %s%*scost=%d\n", n, MatrixNames[i:j+1], MatrixNames[i:k+1], MatrixNames[k+1:j+1], n+i-j, "", m[i][j], ) } subprint(0, n-1) }
func newSquareMatrices(n int) (m, s [][]int) { // Allocates two n×n matrices as slices of slices but // using only one [2n][]int and one [2n²]int backing array. m = make([][]int, 2*n) m, s = m[:n:n], m[n:] tmp := make([]int, 2*n*n) for i := range m { m[i], tmp = tmp[:n:n], tmp[n:] } for i := range s { s[i], tmp = tmp[:n:n], tmp[n:] } return m, s }
func main() { cases := [...][]int{ {1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2}, {1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10}, } for _, tc := range cases { fmt.Println("Dimensions:", tc) PrintMatrixChainOrder(tc) fmt.Println() } }</lang>
- Output:
Dimensions: [1 5 25 30 100 70 2 1 100 250 1 1000 2]
AB -> A × B cost=125
ABC -> AB × C cost=875
ABCD -> ABC × D cost=3875
ABCDE -> ABCD × E cost=10875
ABCDEF -> ABCDE × F cost=11015
ABCDEFG -> ABCDEF × G cost=11017
IJ -> I × J cost=25000
HIJ -> H × IJ cost=25100
ABCDEFGHIJ -> ABCDEFG × HIJ cost=36118
KL -> K × L cost=2000
ABCDEFGHIJKL -> ABCDEFGHIJ × KL cost=38120
Dimensions: [1000 1 500 12 1 700 2500 3 2 5 14 10]
BC -> B × C cost=6000
BCD -> BC × D cost=6012
EF -> E × F cost=1750000
EFG -> EF × G cost=1757500
EFGH -> EFG × H cost=1757506
BCDEFGH -> BCD × EFGH cost=1763520
BCDEFGHI -> BCDEFGH × I cost=1763530
BCDEFGHIJ -> BCDEFGHI × J cost=1763600
BCDEFGHIJK -> BCDEFGHIJ × K cost=1763740
ABCDEFGHIJK -> A × BCDEFGHIJK cost=1773740
Julia
Module: <lang julia>module MatrixChainMultiplications
using OffsetArrays
function optim(a)
n = length(a) - 1
u = fill!(OffsetArray{Int}(0:n, 0:n), 0)
v = fill!(OffsetArray{Int}(0:n, 0:n), typemax(Int))
u[:, 1] .= -1
v[:, 1] .= 0
for j in 2:n, i in 1:n-j+1, k in 1:j-1
c = v[i, k] + v[i+k, j-k] + a[i] * a[i+k] * a[i+j]
if c < v[i, j]
u[i, j] = k
v[i, j] = c
end
end
return v[1, n], aux(u, 1, n)
end
function aux(u, i, j)
k = u[i, j]
if k < 0
return sprint(print, i)
else
return sprint(print, '(', aux(u, i, k), '×', aux(u, i + k, j - k), ")")
end
end
end # module MatrixChainMultiplications</lang>
Main: <lang julia>println(MatrixChainMultiplications.optim([1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2])) println(MatrixChainMultiplications.optim([1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]))</lang>
- Output:
(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))")
Kotlin
This is based on the pseudo-code in the Wikipedia article. <lang scala>// Version 1.2.31
lateinit var m: List<IntArray> lateinit var s: List<IntArray>
fun optimalMatrixChainOrder(dims: IntArray) {
val n = dims.size - 1
m = List(n) { IntArray(n) }
s = List(n) { IntArray(n) }
for (len in 1 until n) {
for (i in 0 until n - len) {
val j = i + len
m[i][j] = Int.MAX_VALUE
for (k in i until j) {
val temp = dims[i] * dims [k + 1] * dims[j + 1]
val cost = m[i][k] + m[k + 1][j] + temp
if (cost < m[i][j]) {
m[i][j] = cost
s[i][j] = k
}
}
}
}
}
fun printOptimalChainOrder(i: Int, j: Int) {
if (i == j)
print("${(i + 65).toChar()}")
else {
print("(")
printOptimalChainOrder(i, s[i][j])
printOptimalChainOrder(s[i][j] + 1, j)
print(")")
}
}
fun main(args: Array<String>) {
val dimsList = listOf(
intArrayOf(5, 6, 3, 1),
intArrayOf(1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2),
intArrayOf(1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10)
)
for (dims in dimsList) {
println("Dims : ${dims.asList()}")
optimalMatrixChainOrder(dims)
print("Order : ")
printOptimalChainOrder(0, s.size - 1)
println("\nCost : ${m[0][s.size - 1]}\n")
}
}</lang>
- Output:
Dims : [5, 6, 3, 1] Order : (A(BC)) Cost : 48 Dims : [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2] Order : ((((((((AB)C)D)E)F)G)(H(IJ)))(KL)) Cost : 38120 Dims : [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10] Order : (A((((((BC)D)(((EF)G)H))I)J)K)) Cost : 1773740
MATLAB
<lang matlab>function [r,s] = optim(a)
n = length(a)-1;
u = zeros(n,n);
v = ones(n,n)*inf;
u(:,1) = -1;
v(:,1) = 0;
for j = 2:n
for i = 1:n-j+1
for k = 1:j-1
c = v(i,k)+v(i+k,j-k)+a(i)*a(i+k)*a(i+j);
if c<v(i,j)
u(i,j) = k;
v(i,j) = c;
end
end
end
end
r = v(1,n);
s = aux(u,1,n);
end
function s = aux(u,i,j)
k = u(i,j);
if k<0
s = sprintf("%d",i);
else
s = sprintf("(%s*%s)",aux(u,i,k),aux(u,i+k,j-k));
end
end</lang>
Output
<lang matlab>[r,s] = optim([1,5,25,30,100,70,2,1,100,250,1,1000,2])
r =
38120
s =
"((((((((1*2)*3)*4)*5)*6)*7)*(8*(9*10)))*(11*12))"
[r,s] = optim([1000,1,500,12,1,700,2500,3,2,5,14,10])
r =
1773740
s =
"(1*((((((2*3)*4)*(((5*6)*7)*8))*9)*10)*11))"</lang>
Perl
<lang perl>sub matrix_mult_chaining {
my(@dimensions) = @_; my(@cp,@path);
# a matrix never needs to be multiplied with itself, so it has cost 0 $cp[$_][$_] = 0 for keys @dimensions;
my $n = $#dimensions;
for my $chain_length (1..$n) {
for my $start (0 .. $n - $chain_length - 1) {
my $end = $start + $chain_length;
$cp[$end][$start] = 10e10;
for my $step ($start .. $end - 1) {
my $new_cost = $cp[$step][$start]
+ $cp[$end][$step + 1]
+ $dimensions[$start] * $dimensions[$step+1] * $dimensions[$end+1];
if ($new_cost < $cp[$end][$start]) {
$cp[$end][$start] = $new_cost; # cost
$cp[$start][$end] = $step; # path
}
}
}
}
find_path(0, $n-1, @cp);
}
sub find_path {
my($start,$end,@cp) = @_; my $result;
if ($start == $end) {
$result .= 'A' . ($start + 1);
} else {
$result .= '(' .
find_path($start, $cp[$start][$end], @cp) .
find_path($cp[$start][$end] + 1, $end, @cp) .
')';
}
return $result;
}
print matrix_mult_chaining(<1 5 25 30 100 70 2 1 100 250 1 1000 2>) . "\n"; print matrix_mult_chaining(<1000 1 500 12 1 700 2500 3 2 5 14 10>) . "\n";</lang>
- Output:
((((((((A1A2)A3)A4)A5)A6)A7)(A8(A9A10)))(A11A12)) (A1((((((A2A3)A4)(((A5A6)A7)A8))A9)A10)A11))
Perl 6
<lang perl6># Reference:
- "Find the optimal way to multiply a chain of matrices. " was the first tasks at
- "masak's Perl 6 Coding Contest 2010". Here is the task definition
- http://strangelyconsistent.org/p6cc2010/p1.zip
- There are five finalist entries on the task, here are the relevant codes and contest host's comments
- 1) http://strangelyconsistent.org/p6cc2010/p1-colomon/
- 2) http://strangelyconsistent.org/p6cc2010/p1-fox/
- 3) http://strangelyconsistent.org/p6cc2010/p1-moritz/
- 4) http://strangelyconsistent.org/p6cc2010/p1-matthias/
- 5) http://strangelyconsistent.org/p6cc2010/p1-util/
- All entries passed perl6.d -c but during run time only the 3) entry (from moritz) produced the correct results.
- !/usr/bin/env perl6
use v6;
sub matrix-mult-chaining(@dimensions) {
# @cp has a dual function: the upper triangle of the diagonal matrix # stores the cost (c) for mulplying matrices $i and $j in @cp[$j][$i], # $j > $i # the lower triangle stores the path (p) that was used for the lowest cost # multiplication to get from $i to $j. # # wikipedia this program # m[i][j] @cp[$j][$i] # s[i][j] @cp[$i][$j] # # it makes the code harder to read, but hey, it saves memory! my @cp; # a matrix never needs to be multiplied with itself, # so it has cost 0 @cp[$_][$_] = 0 for @dimensions.keys; my @path;
my $n = @dimensions.end;
for 1 .. $n -> $chain-length {
for 0 .. $n - $chain-length - 1 -> $start {
my $end = $start + $chain-length;
@cp[$end][$start] = Inf; # until we find a better connection
for $start .. $end - 1 -> $step {
my $new-cost = @cp[$step][$start]
+ @cp[$end][$step + 1]
+ [*] @dimensions[$start, $step+1, $end+1];
if $new-cost < @cp[$end][$start] {
# cost
@cp[$end][$start] = $new-cost;
# path
@cp[$start][$end] = $step;
}
}
}
}
sub find-path(Int $start, Int $end) {
if $start == $end {
take 'A' ~ ($start + 1);
} else {
take '(';
find-path($start, @cp[$start][$end]);
find-path(@cp[$start][$end] + 1, $end);
take ')';
}
}
return gather { find-path(0, $n - 1) }.join;
} multi MAIN {
my $line = get;
my @dimensions = $line.comb: /\d+/;
if @dimensions < 2 {
say "Input must consist of at least two integers.";
} else {
say matrix-mult-chaining(@dimensions);
}
} multi MAIN ('test') {
use Test; plan *; is matrix-mult-chaining(<10 30 5 60>), '((A1A2)A3)', 'wp example'; # http://www.cs.auckland.ac.nz/~jmor159/PLDS210/mat_chain.html is matrix-mult-chaining(<10 100 5 50>), '((A1A2)A3)', 'auckland'; # an example verified by http://modoogle.com/matrixchainorder/ is matrix-mult-chaining(<6 2 5 1 6 3 9 1 25 18>), '((A1((A2A3)((A4A5)(A6A7))))(A8A9))', 'modoogle'; # done_testing;
}</lang>
Output: <lang perl6> ./mcm3.p6 1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2 ((((((((A1A2)A3)A4)A5)A6)A7)(A8(A9A10)))(A11A12)) </lang> <lang perl6> ./mcm3.p6 1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10 (A1((((((A2A3)A4)(((A5A6)A7)A8))A9)A10)A11)) </lang>
Python
We will solve the task in three steps:
1) Enumerate all ways to parenthesize (using a generator to save space), and for each one compute the cost. Then simply look up the minimal cost.
2) Merge the enumeration and the cost function in a recursive cost optimizing function. The computation is roughly the same, but it's much faster as some steps are removed.
3) The recursive solution has many duplicates computations. Memoize the previous function: this yields a dynamic programming approach.
Enumeration of parenthesizations
<lang python>def parens(n):
def aux(n, k):
if n == 1:
yield k
elif n == 2:
yield [k, k + 1]
else:
a = []
for i in range(1, n):
for u in aux(i, k):
for v in aux(n - i, k + i):
yield [u, v]
yield from aux(n, 0)</lang>
Example (in the same order as in the task description)
<lang python>for u in parens(4):
print(u)
[0, [1, [2, 3]]] [0, [[1, 2], 3]] [[0, 1], [2, 3]] [[0, [1, 2]], 3] [[[0, 1], 2], 3]</lang>
And here is the optimization step:
<lang python>def optim1(a):
def cost(k):
if type(k) is int:
return 0, a[k], a[k + 1]
else:
s1, p1, q1 = cost(k[0])
s2, p2, q2 = cost(k[1])
assert q1 == p2
return s1 + s2 + p1 * q1 * q2, p1, q2
cmin = None
n = len(a) - 1
for u in parens(n):
c, p, q = cost(u)
if cmin is None or c < cmin:
cmin = c
umin = u
return cmin, umin</lang>
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.
<lang python>def optim2(a):
def aux(n, k):
if n == 1:
p, q = a[k:k + 2]
return 0, p, q, k
elif n == 2:
p, q, r = a[k:k + 3]
return p * q * r, p, r, [k, k + 1]
else:
m = None
p = a[k]
q = a[k + n]
for i in range(1, n):
s1, p1, q1, u1 = aux(i, k)
s2, p2, q2, u2 = aux(n - i, k + i)
assert q1 == p2
s = s1 + s2 + p1 * q1 * q2
if m is None or s < m:
m = s
u = [u1, u2]
return m, p, q, u
s, p, q, u = aux(len(a) - 1, 0)
return s, u</lang>
Memoized recursive call
The only difference between optim2 and optim3 is the @memoize 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 changing Python's recursion limit).
<lang python>def memoize(f):
h = {}
def g(*u):
if u in h:
return h[u]
else:
r = f(*u)
h[u] = r
return r
return g
def optim3(a):
@memoize
def aux(n, k):
if n == 1:
p, q = a[k:k + 2]
return 0, p, q, k
elif n == 2:
p, q, r = a[k:k + 3]
return p * q * r, p, r, [k, k + 1]
else:
m = None
p = a[k]
q = a[k + n]
for i in range(1, n):
s1, p1, q1, u1 = aux(i, k)
s2, p2, q2, u2 = aux(n - i, k + i)
assert q1 == p2
s = s1 + s2 + p1 * q1 * q2
if m is None or s < m:
m = s
u = [u1, u2]
return m, p, q, u
s, p, q, u = aux(len(a) - 1, 0)
return s, u</lang>
Putting all together
<lang python>import time
u = [[1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2],
[1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]]
for a in u:
print(a)
print()
print("function time cost parens ")
print("-" * 90)
for f in [optim1, optim2, optim3]:
t1 = time.clock()
s, u = f(a)
t2 = time.clock()
print("%s %10.3f %10d %s" % (f.__name__, 1000 * (t2 - t1), s, u))
print()</lang>
Output (timings are in milliseconds)
[1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2] function time cost parens ------------------------------------------------------------------------------------------ optim1 838.636 38120 [[[[[[[[0, 1], 2], 3], 4], 5], 6], [7, [8, 9]]], [10, 11]] optim2 80.628 38120 [[[[[[[[0, 1], 2], 3], 4], 5], 6], [7, [8, 9]]], [10, 11]] optim3 0.373 38120 [[[[[[[[0, 1], 2], 3], 4], 5], 6], [7, [8, 9]]], [10, 11]] [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10] function time cost parens ------------------------------------------------------------------------------------------ optim1 223.186 1773740 [0, [[[[[[1, 2], 3], [[[4, 5], 6], 7]], 8], 9], 10]] optim2 27.660 1773740 [0, [[[[[[1, 2], 3], [[[4, 5], 6], 7]], 8], 9], 10]] optim3 0.307 1773740 [0, [[[[[[1, 2], 3], [[[4, 5], 6], 7]], 8], 9], 10]]
A mean on 1000 loops to get a better precision on the optim3, yields respectively 0.365 ms and 0.287 ms.
Iterative solution
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.
<lang python>def optim4(a):
global u n = len(a) - 1 u = [None] * n u[0] = None, 0 * n for j in range(1, n): v = [None] * (n - j) for i in range(n - j): m = None for k in range(j): s1, c1 = u[k][i] s2, c2 = u[j - k - 1][i + k + 1] c = c1 + c2 + a[i] * a[i + k + 1] * a[i + j + 1] if m is None or c < m: s = k m = c v[i] = [s, m] u[j] = v def aux(i, j): s, c = u[j][i] if s is None: return i else: return [aux(i, s), aux(i + s + 1, j - s - 1)] return u[n - 1][0][1], aux(0, n - 1)
print(optim4([1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]))
print(optim4([1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]))</lang>
Output
(38120, [[[[[[[[0, 1], 2], 3], 4], 5], 6], [7, [8, 9]]], [10, 11]]) (1773740, [0, [[[[[[1, 2], 3], [[[4, 5], 6], 7]], 8], 9], 10]])
Stata
Recursive solution
Here is the equivalent of optim3 in Python's solution. Memoization is done with an associative array. Multiple results are returned in a structure. The same effect as optim2 can be achieved by removing the asarray machinery.
<lang stata>mata struct ans { real scalar p,q,s string scalar u }
struct ans scalar function aux(n,k) { external dim,opt struct ans scalar r,r1,r2 real scalar s,i
if (n==1) { r.p = dim[k] r.q = dim[k+1] r.s = 0 r.u = strofreal(k) return(r) } else if (n==2) { r.p = dim[k] r.q = dim[k+2] r.s = r.p*r.q*dim[k+1] r.u = sprintf("(%f*%f)",k,k+1) return(r) } else if (asarray_contains(opt,(n,k))) { return(asarray(opt,(n,k))) } else { r.p = dim[k] r.q = dim[k+n] r.s = . for (i=1; i<n; i++) { r1 = aux(i,k) r2 = aux(n-i,k+i) s = r1.s+r2.s+r1.p*r1.q*r2.q if (s<r.s) { r.s = s r.u = sprintf("(%s*%s)",r1.u,r2.u) } } asarray(opt,(n,k),r) return(r) } }
function optim(a) { external dim,opt struct ans scalar r real scalar t
timer_clear() dim = a opt = asarray_create("real",2) timer_on(1) r = aux(length(a)-1,1) timer_off(1) t = timer_value(1)[1] printf("%10.0f %10.0f %s\n",t*1000,r.s,r.u) }
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)) end</lang>
Output
0 38120 ((((((((1*2)*3)*4)*5)*6)*7)*(8*(9*10)))*(11*12))
16 1773740 (1*((((((2*3)*4)*(((5*6)*7)*8))*9)*10)*11))
The timing is in milliseconds, but the time resolution is too coarse to get a usable result. A mean on 1000 loops doing the same computation yields respectively 5.772 ms and 4.430 ms for these two cases. For comparison, the computation was made on the same machine as the Python solution.
Iterative solution
<lang stata>mata function aux(u,i,j) { k = u[i,j] if (k<0) { printf("%f",i) } else { printf("(") aux(u,i,k) printf("*") aux(u,i+k,j-k) printf(")") } }
function optim(a) { n = length(a)-1 u = J(n,n,.) v = J(n,n,.) u[.,1] = J(n,1,-1) v[.,1] = J(n,1,0) for (j=2; j<=n; j++) { for (i=1; i<=n-j+1; i++) { for (k=1; k<j; k++) { c = v[i,k]+v[i+k,j-k]+a[i]*a[i+k]*a[i+j] if (c<v[i,j]) { u[i,j] = k v[i,j] = c } } } } printf("%f ",v[1,n]) aux(u,1,n) printf("\n") }
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)) end</lang>
Output
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))
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.
zkl
<lang zkl>fcn optim3(a){ // list --> (int,list)
aux:=fcn(n,k,a){ // (int,int,list) --> (int,int,int,list)
if(n==1){
p,q := a[k,2]; return(0,p,q,k);
}
if(n==2){
p,q,r := a[k,3]; return(p*q*r, p, r, T(k,k+1));
}
m,p,q,u := Void, a[k], a[k + n], Void;
foreach i in ([1..n-1]){
#if 0 // 0.70 sec for both tests
s1,p1,q1,u1 := self.fcn(i,k,a); s2,p2,q2,u2 := self.fcn(n - i, k + i, a);
#else // 0.33 sec for both tests
s1,p1,q1,u1 := memoize(self.fcn, i,k,a); s2,p2,q2,u2 := memoize(self.fcn, n - i, k + i, a);
#endif
_assert_(q1==p2); s:=s1 + s2 + p1*q1*q2; if((Void==m) or (s<m)) m,u = s,T(u1,u2);
}
return(m,p,q,u);
};
h=Dictionary(); // reset memoize s,_,_,u := aux(a.len() - 1, 0,a); return(s,u);
}
var h; // a Dictionary, set/reset in optim3() fcn memoize(f,n,k,a){
key:="%d,%d".fmt(n,k); // Lists make crappy keys if(r:=h.find(key)) return(r); r:=f(n,k,a); h[key]=r; return(r);
}</lang> <lang zkl>fcn pp(u){ // pretty print a list of lists
var letters=["A".."Z"].pump(String);
u.pump(String,
fcn(n){ if(List.isType(n)) String("(",pp(n),")") else letters[n] })
} fcn prnt(s,u){ "%-9,d %s\n\t-->%s\n".fmt(s,u.toString(*,*),pp(u)).println() }</lang> <lang zkl>s,u := optim3(T(1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2)); prnt(s,u);
s,u := optim3(T(1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10)); prnt(s,u);
optim3(T(5,6,3,1)) : prnt(_.xplode());</lang>
- Output:
38,120 L(L(L(L(L(L(L(L(0,1),2),3),4),5),6),L(7,L(8,9))),L(10,11)) -->(((((((AB)C)D)E)F)G)(H(IJ)))(KL) 1,773,740 L(0,L(L(L(L(L(L(1,2),3),L(L(L(4,5),6),7)),8),9),10)) -->A((((((BC)D)(((EF)G)H))I)J)K) 48 L(0,L(1,2)) -->A(BC)