Matrix chain multiplication
You are encouraged to solve this task according to the task description, using any language you may know.
- Problem
Using the most straightforward 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.
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.
11l
T Optimizer
[Int] dims
[[Int]] m, s
F (dims)
.dims = dims
F findMatrixChainOrder()
V n = .dims.len - 1
.m = [[0] * n] * n
.s = [[0] * n] * n
L(lg) 1 .< n
L(i) 0 .< n - lg
V j = i + lg
.m[i][j] = 7FFF'FFFF
L(k) i .< j
V cost = .m[i][k] + .m[k + 1][j] + .dims[i] * .dims[k + 1] * .dims[j + 1]
I cost < .m[i][j]
.m[i][j] = cost
.s[i][j] = k
F optimalChainOrder(i, j)
I i == j
R String(Char(code' i + ‘A’.code))
E
R ‘(’(.optimalChainOrder(i, .s[i][j]))‘’
‘’(.optimalChainOrder(.s[i][j] + 1, j))‘)’
V Dims1 = [5, 6, 3, 1]
V Dims2 = [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
V Dims3 = [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]
L(dims) [Dims1, Dims2, Dims3]
V opt = Optimizer(dims)
opt.findMatrixChainOrder()
print(‘Dims: ’dims)
print(‘Order: ’opt.optimalChainOrder(0, dims.len - 2))
print(‘Cost: ’opt.m[0][dims.len - 2])
print(‘’)- 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
Ada
This example implements the pseudocode in the reference Wiki page. The pseudocode states that the index values for the array to multiply begin at 0 while the cost and order matrices employ index values beginning at 1. Ada supports this pseudocode directly because Ada allows the programmer to define the index range for any array type. This Ada example is implemented using a simple package and a main procedure. The package specification is:
package mat_chain is
type Vector is array (Natural range <>) of Integer;
procedure Chain_Multiplication (Dims : Vector);
end mat_chain;
The implementation or body of the package is:
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Strings.Unbounded; use Ada.Strings.Unbounded;
package body mat_chain is
type Result_Matrix is
array (Positive range <>, Positive range <>) of Integer;
--------------------------
-- Chain_Multiplication --
--------------------------
procedure Chain_Multiplication (Dims : Vector) is
n : Natural := Dims'Length - 1;
S : Result_Matrix (1 .. n, 1 .. n);
m : Result_Matrix (1 .. n, 1 .. n);
procedure Print (Item : Vector) is
begin
Put ("Array Dimension = (");
for I in Item'Range loop
Put (Item (I)'Image);
if I < Item'Last then
Put (",");
else
Put (")");
end if;
end loop;
New_Line;
end Print;
procedure Chain_Order (Item : Vector) is
J : Natural;
Cost : Natural;
Temp : Natural;
begin
for idx in 1 .. n loop
m (idx, idx) := 0;
end loop;
for Len in 2 .. n loop
for I in 1 .. n - Len + 1 loop
J := I + Len - 1;
m (I, J) := Integer'Last;
for K in I .. J - 1 loop
Temp := Item (I - 1) * Item (K) * Item (J);
Cost := m (I, K) + m (K + 1, J) + Temp;
if Cost < m (I, J) then
m (I, J) := Cost;
S (I, J) := K;
end if;
end loop;
end loop;
end loop;
end Chain_Order;
function Optimal_Parens return String is
function Construct
(S : Result_Matrix; I : Natural; J : Natural)
return Unbounded_String
is
Us : Unbounded_String := Null_Unbounded_String;
Char_Order : Character;
begin
if I = J then
Char_Order := Character'Val (I + 64);
Append (Source => Us, New_Item => Char_Order);
return Us;
else
Append (Source => Us, New_Item => '(');
Append (Source => Us, New_Item => Construct (S, I, S (I, J)));
Append (Source => Us, New_Item => '*');
Append
(Source => Us, New_Item => Construct (S, S (I, J) + 1, J));
Append (Source => Us, New_Item => ')');
return Us;
end if;
end Construct;
begin
return To_String (Construct (S, 1, n));
end Optimal_Parens;
begin
Chain_Order (Dims);
Print (Dims);
Put_Line ("Cost = " & Integer'Image (m (1, n)));
Put_Line ("Optimal Multiply = " & Optimal_Parens);
end Chain_Multiplication;
end mat_chain;
The main procedure is:
with Mat_Chain; use Mat_Chain;
with Ada.Text_IO; use Ada.Text_IO;
procedure chain_main is
V1 : Vector := (5, 6, 3, 1);
V2 : Vector := (1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2);
V3 : Vector := (1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10);
begin
Chain_Multiplication(V1);
New_Line;
Chain_Multiplication(V2);
New_Line;
Chain_Multiplication(V3);
end chain_main;
- Output:
Array Dimension = ( 5, 6, 3, 1) Cost = 48 Optimal Multiply = (A*(B*C)) Array Dimension = ( 1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2) Cost = 38120 Optimal Multiply = ((((((((A*B)*C)*D)*E)*F)*G)*(H*(I*J)))*(K*L)) Array Dimension = ( 1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10) Cost = 1773740 Optimal Multiply = (A*((((((B*C)*D)*(((E*F)*G)*H))*I)*J)*K))
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;
}
- 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#
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]);
}
}
}
- 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++
#include <cstdint>
#include <iostream>
#include <sstream>
#include <string>
#include <vector>
constexpr int32_t MAXIMUM_VALUE = 2'147'483'647;
std::vector<std::vector<int32_t>> cost;
std::vector<std::vector<int32_t>> order;
void print_vector(const std::vector<int32_t>& list) {
std::cout << "[";
for ( uint64_t i = 0; i < list.size() - 1; ++i ) {
std::cout << list[i] << ", ";
}
std::cout << list.back() << "]" << std::endl;
}
int32_t matrix_chain_order(const std::vector<int32_t>& dimensions) {
const uint64_t size = dimensions.size() - 1;
cost = { size, std::vector<int32_t>(size, 0) };
order = { size, std::vector<int32_t>(size, 0) };
for ( uint64_t m = 1; m < size; ++m ) {
for ( uint64_t i = 0; i < size - m; ++i ) {
int32_t j = i + m;
cost[i][j] = MAXIMUM_VALUE;
for ( int32_t k = i; k < j; ++k ) {
int32_t current_cost = cost[i][k] + cost[k + 1][j]
+ dimensions[i] * dimensions[k + 1] * dimensions[j + 1];
if ( current_cost < cost[i][j] ) {
cost[i][j] = current_cost;
order[i][j] = k;
}
}
}
}
return cost[0][size - 1];
}
std::string get_optimal_parenthesizations(const std::vector<std::vector<int32_t>>& order,
const uint64_t& i, const uint64_t& j) {
if ( i == j ) {
std::string result(1, char(i + 65));
return result;
} else {
std::stringstream stream;
stream << "(" << get_optimal_parenthesizations(order, i, order[i][j])
<< " * " << get_optimal_parenthesizations(order, order[i][j] + 1, j) << ")";
return stream.str();
}
}
void matrix_chain_multiplication(const std::vector<int32_t>& dimensions) {
std::cout << "Array Dimension = "; print_vector(dimensions);
std::cout << "Cost = " << matrix_chain_order(dimensions) << std::endl;
std::cout << "Optimal Multiply = "
<< get_optimal_parenthesizations(order, 0, order.size() - 1) << std::endl << std::endl;
}
int main() {
matrix_chain_multiplication({ 5, 6, 3, 1 });
matrix_chain_multiplication({ 1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2 });
matrix_chain_multiplication({ 1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10 });
}
- Output:
Array Dimension = [5, 6, 3, 1] Cost = 48 Optimal Multiply = (A * (B * C)) Array Dimension = [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2] Cost = 38120 Optimal Multiply = ((((((((A * B) * C) * D) * E) * F) * G) * (H * (I * J))) * (K * L)) Array Dimension = [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10] Cost = 1773740 Optimal Multiply = (A * ((((((B * C) * D) * (((E * F) * G) * H)) * I) * J) * K))
EasyLang
This is based on the pseudo-code in the Wikipedia article.
global m[][] s[][] dims[] .
proc mat_chain_order . .
n = len dims[] - 1
m[][] = [ ] ; len m[][] n
s[][] = [ ] ; len s[][] n
for i = 1 to n
len m[i][] n
len s[i][] n
.
for lng = 2 to n
for i = 1 to n - lng + 1
j = i + lng - 1
m[i][j] = 1 / 0
for k = i to j - 1
cost = m[i][k] + m[k + 1][j] + dims[i] * dims[k + 1] * dims[j + 1]
if cost < m[i][j]
m[i][j] = cost
s[i][j] = k
.
.
.
.
.
func$ path a b .
if a = b
return strchar (64 + a)
.
return "(" & path a s[a][b] & path (s[a][b] + 1) b & ")"
.
proc pr_chain_order . .
print "Order : " & path 1 len s[][]
.
dims[][] = [ [ 5 6 3 1 ] [ 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 i to len dims[][]
dims[] = dims[i][]
print "Dims : " & dims[]
mat_chain_order
pr_chain_order
print "Cost : " & m[1][len s[][]]
print ""
.
- 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.
module optim_mod
implicit none
contains
subroutine optim(a)
implicit none
integer :: a(:), n, i, j, k
integer, allocatable :: u(:, :)
integer(8) :: c
integer(8), allocatable :: v(:, :)
n = ubound(a, 1) - 1
allocate (u(n, n), v(n, n))
v = huge(v)
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 (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 matmulchain
use optim_mod
implicit none
call optim([5, 6, 3, 1])
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
Output
48 (1*(2*3)) 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))
FreeBASIC
This is a translation of the Python iterative solution.
Dim Shared As Integer U(), V()
Sub Aux(i As Integer, j As Integer)
Dim As Integer k = U(i, j)
If k < 0 Then
Print Str(i);
Else
Print "(";
Aux(i, k)
Print "*";
Aux(i + k, j - k)
Print ")";
End If
End Sub
Sub Optimize(a() As Integer)
Dim As Integer i, j, k, c
Dim As Integer n = Ubound(a) - 1
Redim U(n, n), V(n, n)
For i = 1 To n
U(i, 1) = -1
V(i, 1) = 0
Next i
For j = 2 To n
For i = 1 To n - j + 1
V(i, j) = &H7FFFFFFF
For k = 1 To j - 1
c = V(i, k) + V(i + k, j - k) + a(i) * a(i + k) * a(i + j)
If c < V(i, j) Then
U(i, j) = k
V(i, j) = c
End If
Next k
Next i
Next j
Print V(1, n); " ";
Aux(1, n)
Print
Erase U, V
End Sub
Dim As Integer A1(1 To 4) = {5, 6, 3, 1}
Optimize(A1())
Dim As Integer A2(1 To 13) = {1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2}
Optimize(A2())
Dim As Integer A3(1 To 12) = {1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10}
Optimize(A3())
Sleep
- Output:
48(1*(2*3)) 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.
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()
}
}
- 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
Haskell
import Data.List (elemIndex)
import Data.Char (chr, ord)
import Data.Maybe (fromJust)
mats :: [[Int]]
mats =
[ [5, 6, 3, 1]
, [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]
]
cost :: [Int] -> Int -> Int -> (Int, Int)
cost a i j
| i < j =
let m =
[ fst (cost a i k) + fst (cost a (k + 1) j) +
(a !! i) * (a !! (j + 1)) * (a !! (k + 1))
| k <- [i .. j - 1] ]
mm = minimum m
in (mm, fromJust (elemIndex mm m) + i)
| otherwise = (0, -1)
optimalOrder :: [Int] -> Int -> Int -> String
optimalOrder a i j
| i < j =
let c = cost a i j
in "(" ++ optimalOrder a i (snd c) ++ optimalOrder a (snd c + 1) j ++ ")"
| otherwise = [chr ((+ i) $ ord 'a')]
printBlock :: [Int] -> IO ()
printBlock v =
let c = cost v 0 (length v - 2)
in putStrLn
("for " ++
show v ++
" we have " ++
show (fst c) ++
" possibilities, z.B " ++ optimalOrder v 0 (length v - 2))
main :: IO ()
main = mapM_ printBlock mats
- Output:
for [5,6,3,1] we have 48 possibilities, z.B (a(bc)) for [1,5,25,30,100,70,2,1,100,250,1,1000,2] we have 38120 possibilities, z.B ((((((((ab)c)d)e)f)g)(h(ij)))(kl)) for [1000,1,500,12,1,700,2500,3,2,5,14,10] we have 1773740 possibilities, z.B (a((((((bc)d)(((ef)g)h))i)j)k)
J
This is no more than a mindless transliteration of the Wikipedia Java code (for moo; for pooc, the author found Go to have the clearest expression for transliteration).
Given J's incredible strengths with arrays and matrices, the author is certain there is a much more succinct and idiomatic approach available, but hasn't spent the time understanding how the Wikipedia algorithm works, so hasn't made an attempt at a more native J solution. Others on RC are welcome and invited to do so.
moo =: verb define
s =. m =. 0 $~ ,~ n=._1+#y
for_lmo. 1+i.<:n do.
for_i. i. n-lmo do.
j =. i + lmo
m =. _ (<i;j)} m
for_k. i+i.j-i do.
cost =. ((<i;k){m) + ((<(k+1);j){m) + */ y {~ i,(k+1),(j+1)
if. cost < ((<i;j){m) do.
m =. cost (<i;j)} m
s =. k (<i;j)} s
end.
end.
end.
end.
m;s
)
poco =: dyad define
'i j' =. y
if. i=j do.
a. {~ 65 + i NB. 65 = a.i.'A'
else.
k =. x {~ <y NB. y = i,j
'(' , (x poco i,k) , (x poco j ,~ 1+k) , ')'
end.
)
optMM =: verb define
'M S' =. moo y
smoutput 'Cost: ' , ": x: M {~ <0;_1
smoutput 'Order: ', S poco 0 , <:#M
)
- Output:
optMM 5 6 3 1
Cost: 48
Order: (A(BC))
optMM 1 5 25 30 100 70 2 1 100 250 1 1000 2
Cost: 38120
Order: ((((((((AB)C)D)E)F)G)(H(IJ)))(KL))
optMM 1000 1 500 12 1 700 2500 3 2 5 14 10
Cost: 1773740
Order: (A((((((BC)D)(((EF)G)H))I)J)K))
Java
Thanks to the Wikipedia page for a working Java implementation.
import java.util.Arrays;
public class MatrixChainMultiplication {
public static void main(String[] args) {
runMatrixChainMultiplication(new int[] {5, 6, 3, 1});
runMatrixChainMultiplication(new int[] {1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2});
runMatrixChainMultiplication(new int[] {1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10});
}
private static void runMatrixChainMultiplication(int[] dims) {
System.out.printf("Array Dimension = %s%n", Arrays.toString(dims));
System.out.printf("Cost = %d%n", matrixChainOrder(dims));
System.out.printf("Optimal Multiply = %s%n%n", getOptimalParenthesizations());
}
private static int[][]cost;
private static int[][]order;
public static int matrixChainOrder(int[] dims) {
int n = dims.length - 1;
cost = new int[n][n];
order = new int[n][n];
for (int lenMinusOne = 1 ; lenMinusOne < n ; lenMinusOne++) {
for (int i = 0; i < n - lenMinusOne; i++) {
int j = i + lenMinusOne;
cost[i][j] = Integer.MAX_VALUE;
for (int k = i; k < j; k++) {
int currentCost = cost[i][k] + cost[k+1][j] + dims[i]*dims[k+1]*dims[j+1];
if (currentCost < cost[i][j]) {
cost[i][j] = currentCost;
order[i][j] = k;
}
}
}
}
return cost[0][n-1];
}
private static String getOptimalParenthesizations() {
return getOptimalParenthesizations(order, 0, order.length - 1);
}
private static String getOptimalParenthesizations(int[][]s, int i, int j) {
if (i == j) {
return String.format("%c", i+65);
}
else {
StringBuilder sb = new StringBuilder();
sb.append("(");
sb.append(getOptimalParenthesizations(s, i, s[i][j]));
sb.append(" * ");
sb.append(getOptimalParenthesizations(s, s[i][j] + 1, j));
sb.append(")");
return sb.toString();
}
}
}
- Output:
Array Dimension = [5, 6, 3, 1] Cost = 48 Optimal Multiply = (A * (B * C)) Array Dimension = [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2] Cost = 38120 Optimal Multiply = ((((((((A * B) * C) * D) * E) * F) * G) * (H * (I * J))) * (K * L)) Array Dimension = [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10] Cost = 1773740 Optimal Multiply = (A * ((((((B * C) * D) * (((E * F) * G) * H)) * I) * J) * K))
jq
Works with gojq, the Go implementation of jq
# Input: array of dimensions
# output: {m, s}
def optimalMatrixChainOrder:
. as $dims
| (($dims|length) - 1) as $n
| reduce range(1; $n) as $len ({m: [], s: []};
reduce range(0; $n-$len) as $i (.;
($i + $len) as $j
| .m[$i][$j] = infinite
| reduce range($i; $j) as $k (.;
($dims[$i] * $dims [$k + 1] * $dims[$j + 1]) as $temp
| (.m[$i][$k] + .m[$k + 1][$j] + $temp) as $cost
| if $cost < .m[$i][$j]
then .m[$i][$j] = $cost
| .s[$i][$j] = $k
else .
end ) )) ;
# input: {s}
def printOptimalChainOrder($i; $j):
if $i == $j
then [$i + 65] | implode #=> "A", "B", ...
else "(" +
printOptimalChainOrder($i; .s[$i][$j]) +
printOptimalChainOrder(.s[$i][$j] + 1; $j) + ")"
end;
def dimsList: [
[5, 6, 3, 1],
[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]
];
dimsList[]
| "Dims : \(.)",
(optimalMatrixChainOrder
| "Order : \(printOptimalChainOrder(0; .s|length - 1))",
"Cost : \(.m[0][.s|length - 1])\n" )- Output:
Dims : [5,6,3,1] Order : (AB) Cost : 90 Dims : [1,5,25,30,100,70,2,1,100,250,1,1000,2] Order : ((((((((AB)C)D)E)F)G)(H(IJ)))K) Cost : 37118 Dims : [1000,1,500,12,1,700,2500,3,2,5,14,10] Order : (A(((((BC)D)(((EF)G)H))I)J)) Cost : 1777600
Julia
Module:
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
Main:
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]))
- 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.
// 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")
}
}
- 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
Lua
-- Matrix A[i] has dimension dims[i-1] x dims[i] for i = 1..n
local function MatrixChainOrder(dims)
local m = {}
local s = {}
local n = #dims - 1;
-- m[i,j] = Minimum number of scalar multiplications (i.e., cost)
-- needed to compute the matrix A[i]A[i+1]...A[j] = A[i..j]
-- The cost is zero when multiplying one matrix
for i = 1,n do
m[i] = {}
m[i][i] = 0
s[i] = {}
end
for len = 2,n do -- Subsequence lengths
for i = 1,(n - len + 1) do
local j = i + len - 1
m[i][j] = math.maxinteger
for k = i,(j - 1) do
local cost = m[i][k] + m[k+1][j] + dims[i]*dims[k+1]*dims[j+1];
if (cost < m[i][j]) then
m[i][j] = cost;
s[i][j] = k; --Index of the subsequence split that achieved minimal cost
end
end
end
end
return m,s
end
local function printOptimalChainOrder(s)
local function find_path(start,finish)
local chainOrder = ""
if (start == finish) then
chainOrder = chainOrder .."A"..start
else
chainOrder = chainOrder .."(" ..
find_path(start,s[start][finish]) ..
find_path(s[start][finish]+1,finish) .. ")"
end
return chainOrder
end
print("Order : "..find_path(1,#s))
end
local dimsList = {{5, 6, 3, 1},{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 k,dim in ipairs(dimsList) do
io.write("Dims : [")
for v=1,(#dim-1) do
io.write(dim[v]..", ")
end
print(dim[#dim].."]")
local m,s = MatrixChainOrder(dim)
printOptimalChainOrder(s)
print("Cost : "..tostring(m[1][#s]).."\n")
end
- Output:
Dims : [5, 6, 3, 1] Order : (A1(A2A3)) Cost : 48 Dims : [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2] Order : ((((((((A1A2)A3)A4)A5)A6)A7)(A8(A9A10)))(A11A12)) Cost : 38120 Dims : [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10] Order : (A1((((((A2A3)A4)(((A5A6)A7)A8))A9)A10)A11)) Cost : 1773740
Mathematica / Wolfram Language
ClearAll[optim, aux]
optim[a_List] := Module[{u, v, n, c, r, s},
n = Length[a] - 1;
u = ConstantArray[0, {n, n}];
v = ConstantArray[\[Infinity], {n, n}];
u[[All, 1]] = -1;
v[[All, 1]] = 0;
Do[
Do[
Do[
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;
]
,
{k, 1, j - 1}
]
,
{i, 1, n - j + 1}
]
,
{j, 2, n}
];
r = v[[1, n]];
s = aux[u, 1, n];
{r, s}
]
aux[u_, i_, j_] := Module[{k},
k = u[[i, j]];
If[k < 0,
i
,
Inactive[Times][aux[u, i, k], aux[u, i + k, j - k]]
]
]
{r, s} = optim[{1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2}];
r
s
{r, s} = optim[{1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10}];
r
s
- 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)
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
- Output:
[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))"
Nim
import sequtils
type Optimizer = object
dims: seq[int]
m: seq[seq[Natural]]
s: seq[seq[Natural]]
proc initOptimizer(dims: openArray[int]): Optimizer =
## Create an optimizer for the given dimensions.
Optimizer(dims: @dims)
proc findMatrixChainOrder(opt: var Optimizer) =
## Find the best order for matrix chain multiplication.
let n = opt.dims.high
opt.m = newSeqWith(n, newSeq[Natural](n))
opt.s = newSeqWith(n, newSeq[Natural](n))
for lg in 1..<n:
for i in 0..<(n - lg):
let j = i + lg
opt.m[i][j] = Natural.high
for k in i..<j:
let cost = opt.m[i][k] + opt.m[k+1][j] + opt.dims[i] * opt.dims[k+1] * opt.dims[j+1]
if cost < opt.m[i][j]:
opt.m[i][j] = cost
opt.s[i][j] = k
proc optimalChainOrder(opt: Optimizer; i, j: Natural): string =
## Return the optimal chain order as a string.
if i == j:
result.add chr(i + ord('A'))
else:
result.add '('
result.add opt.optimalChainOrder(i, opt.s[i][j])
result.add opt.optimalChainOrder(opt.s[i][j] + 1, j)
result.add ')'
when isMainModule:
const
Dims1 = @[5, 6, 3, 1]
Dims2 = @[1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
Dims3 = @[1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]
for dims in [Dims1, Dims2, Dims3]:
var opt = initOptimizer(dims)
opt.findMatrixChainOrder()
echo "Dims: ", dims
echo "Order: ", opt.optimalChainOrder(0, dims.len - 2)
echo "Cost: ", opt.m[0][dims.len - 2]
echo ""
- 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
Perl
use strict;
use feature 'say';
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
}
}
}
}
$cp[$n-1][0] . ' ' . 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;
}
say matrix_mult_chaining(<1 5 25 30 100 70 2 1 100 250 1 1000 2>);
say matrix_mult_chaining(<1000 1 500 12 1 700 2500 3 2 5 14 10>);
- Output:
38120 ((((((((A1A2)A3)A4)A5)A6)A7)(A8(A9A10)))(A11A12)) 1773740 (A1((((((A2A3)A4)(((A5A6)A7)A8))A9)A10)A11))
Phix
As per the wp pseudocode
with javascript_semantics function optimal_chain_order(int i, int j, sequence s) if i==j then return i+'A'-1 end if return "("&optimal_chain_order(i,s[i,j],s) &optimal_chain_order(s[i,j]+1,j,s)&")" end function function optimal_matrix_chain_order(sequence dims) integer n = length(dims)-1 sequence m = repeat(repeat(0,n),n), s = deep_copy(m) for len=2 to n do for i=1 to n-len+1 do integer j = i+len-1 m[i][j] = -1 for k=i to j-1 do atom cost := m[i][k] + m[k+1][j] + dims[i]*dims[k+1]*dims[j+1] if m[i][j]<0 or cost<m[i][j] then m[i][j] = cost; s[i][j] = k; end if end for end for end for return {optimal_chain_order(1,n,s),m[1,n]} end function constant tests = {{5, 6, 3, 1}, {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 i=1 to length(tests) do sequence ti = tests[i] printf(1,"Dims : %s\n",{sprint(ti)}) printf(1,"Order : %s\nCost : %d\n",optimal_matrix_chain_order(ti)) end for
- 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
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
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)
Example (in the same order as in the task description)
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]
And here is the optimization step:
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
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 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
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).
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
Putting all together
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()
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.
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]))
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]])
R
aux <- function(i, j, u) {
k <- u[[i, j]]
if (k < 0) {
i
} else {
paste0("(", Recall(i, k, u), "*", Recall(i + k, j - k, u), ")")
}
}
chain.mul <- function(a) {
n <- length(a) - 1
u <- matrix(0, n, n)
v <- matrix(0, n, n)
u[, 1] <- -1
for (j in seq(2, n)) {
for (i in seq(n - j + 1)) {
v[[i, j]] <- Inf
for (k in seq(j - 1)) {
s <- v[[i, k]] + v[[i + k, j - k]] + a[[i]] * a[[i + k]] * a[[i + j]]
if (s < v[[i, j]]) {
u[[i, j]] <- k
v[[i, j]] <- s
}
}
}
}
list(cost = v[[1, n]], solution = aux(1, n, u))
}
chain.mul(c(5, 6, 3, 1))
# $cost
# [1] 48
# $solution
# [1] "(1*(2*3))"
chain.mul(c(1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2))
# $cost
# [1] 38120
# $solution
# [1] "((((((((1*2)*3)*4)*5)*6)*7)*(8*(9*10)))*(11*12))"
chain.mul(c(1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10))
# $cost
# [1] 1773740
# $solution
# [1] "(1*((((((2*3)*4)*(((5*6)*7)*8))*9)*10)*11))"
Racket
Memoization
#lang racket
(define (memoize f)
(define table (make-hash))
(λ args (hash-ref! table args (thunk (apply f args)))))
(struct $ (cost expl))
(define @ vector-ref)
(define (+: #:combine [combine (thunk* #f)] . xs)
($ (apply + (map $-cost xs)) (apply combine (map $-expl xs))))
(define (min: . xs) (argmin $-cost xs))
(define (compute dims)
(define loop
(memoize
(λ (left right)
(cond
[(= 1 (- right left)) ($ 0 left)]
[else (for/fold ([ans ($ +inf.0 #f)]) ([mid (in-range (add1 left) right)])
(min: ans (+: (loop left mid) (loop mid right)
($ (* (@ dims left) (@ dims mid) (@ dims right)) #f)
#:combine (λ (left-answer right-answer _)
(list left-answer '× right-answer)))))]))))
(loop 0 (sub1 (vector-length dims))))
Main
(define-syntax-rule (echo <x> ...)
(begin (printf "~a: ~a\n" (~a (quote <x>) #:min-width 12) <x>) ...))
(define (solve input)
(match-define-values ((list ($ cost explanation)) _ time _) (time-apply compute (list input)))
(echo input time cost explanation)
(newline))
(solve #(1 5 25 30 100 70 2 1 100 250 1 1000 2))
(solve #(1000 1 500 12 1 700 2500 3 2 5 14 10))
Output (timings are in milliseconds)
input : #(1 5 25 30 100 70 2 1 100 250 1 1000 2) time : 1 cost : 38120 explanation : ((((((((0 × 1) × 2) × 3) × 4) × 5) × 6) × (7 × (8 × 9))) × (10 × 11)) input : #(1000 1 500 12 1 700 2500 3 2 5 14 10) time : 0 cost : 1773740 explanation : (0 × ((((((1 × 2) × 3) × (((4 × 5) × 6) × 7)) × 8) × 9) × 10))
Raku
(formerly Perl 6) This example is based on Moritz Lenz's code, written for Carl Mäsak's Perl 6 Coding Contest, in 2010. Slightly simplified, it fulfills the Rosetta Code task as well.
sub matrix-mult-chaining(@dimensions) {
my @cp;
# @cp has a dual function:
# * the upper triangle of the diagonal matrix stores the cost (c) for
# multiplying matrices $i and $j in @cp[$j][$i], where $j > $i
# * the lower triangle stores the path (p) that was used for the lowest cost
# multiplication to get from $i to $j.
# a matrix never needs to be multiplied with itself, so it has cost 0
@cp[$_][$_] = 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][$start] {
@cp[$end][$start] = $new-cost; # cost
@cp[$start][$end] = $step; # path
}
}
}
}
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, $end);
take ')';
}
}
return @cp[$n-1][0], gather { find-path(0, $n - 1) }.join;
}
say matrix-mult-chaining(<1 5 25 30 100 70 2 1 100 250 1 1000 2>);
say matrix-mult-chaining(<1000 1 500 12 1 700 2500 3 2 5 14 10>);
- Output:
(38120 ((((((((A1A2)A3)A4)A5)A6)A7)(A8(A9A10)))(A11A12))) (1773740 (A1((((((A2A3)A4)(((A5A6)A7)A8))A9)A10)A11)))
Rust
use std::collections::HashMap;
fn main() {
println!("{}\n", mcm_display(vec![5, 6, 3, 1]));
println!(
"{}\n",
mcm_display(vec![1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2])
);
println!(
"{}\n",
mcm_display(vec![1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10])
);
}
fn mcm_display(dims: Vec<i32>) -> String {
let mut costs: HashMap<Vec<i32>, (i32, Vec<usize>)> = HashMap::new();
let mut line = format!("Dims : {:?}\n", dims);
let ans = mcm(dims, &mut costs);
let mut mats = (1..=ans.1.len() + 1)
.map(|x| x.to_string())
.collect::<Vec<String>>();
for i in 0..ans.1.len() {
let mat_taken = mats[ans.1[i]].clone();
mats.remove(ans.1[i]);
mats[ans.1[i]] = "(".to_string() + &mat_taken + "*" + &mats[ans.1[i]] + ")";
}
line += &format!("Order: {}\n", mats[0]);
line += &format!("Cost : {}", ans.0);
line
}
fn mcm(dims: Vec<i32>, costs: &mut HashMap<Vec<i32>, (i32, Vec<usize>)>) -> (i32, Vec<usize>) {
match costs.get(&dims) {
Some(c) => c.clone(),
None => {
let ans = if dims.len() == 3 {
(dims[0] * dims[1] * dims[2], vec![0])
} else {
let mut min_cost = std::i32::MAX;
let mut min_path = Vec::new();
for i in 1..dims.len() - 1 {
let taken = dims[(i - 1)..(i + 2)].to_vec();
let mut rest = dims[..i].to_vec();
rest.extend_from_slice(&dims[(i + 1)..]);
let a1 = mcm(taken, costs);
let a2 = mcm(rest, costs);
if a1.0 + a2.0 < min_cost {
min_cost = a1.0 + a2.0;
min_path = vec![i - 1];
min_path.extend_from_slice(&a2.1);
}
}
(min_cost, min_path)
};
costs.insert(dims, ans.clone());
ans
}
}
}
- Output:
Dims : [5, 6, 3, 1] Order: (1*(2*3)) Cost : 48 Dims : [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2] Order: ((((((((1*2)*3)*4)*5)*6)*7)*(8*(9*10)))*(11*12)) Cost : 38120 Dims : [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10] Order: (1*((((((2*3)*4)*(((5*6)*7)*8))*9)*10)*11)) Cost : 1773740
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.
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
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
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
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.
VBA
Option Explicit
Option Base 1
Dim N As Long, U() As Long, V() As Long
Sub Optimize(A As Variant)
Dim I As Long, J As Long, K As Long, C As Long
N = UBound(A) - 1
ReDim U(N, N), V(N, N)
For I = 1 To N
U(I, 1) = -1
V(I, 1) = 0
Next I
For J = 2 To N
For I = 1 To N - J + 1
V(I, J) = &H7FFFFFFF
For K = 1 To J - 1
C = V(I, K) + V(I + K, J - K) + A(I) * A(I + K) * A(I + J)
If C < V(I, J) Then
U(I, J) = K
V(I, J) = C
End If
Next K
Next I
Next J
Debug.Print V(1, N);
Call Aux(1, N)
Debug.Print
Erase U, V
End Sub
Sub Aux(I As Long, J As Long)
Dim K As Long
K = U(I, J)
If K < 0 Then
Debug.Print CStr(I);
Else
Debug.Print "(";
Call Aux(I, K)
Debug.Print "*";
Call Aux(I + K, J - K)
Debug.Print ")";
End If
End Sub
Sub Test()
Call Optimize(Array(5, 6, 3, 1))
Call Optimize(Array(1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2))
Call Optimize(Array(1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10))
End Sub
Output
48 (1*(2*3)) 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))
Wren
var m = []
var s = []
var optimalMatrixChainOrder = Fn.new { |dims|
var n = dims.count - 1
m = List.filled(n, null)
s = List.filled(n, null)
for (i in 0...n) {
m[i] = List.filled(n, 0)
s[i] = List.filled(n, 0)
}
for (len in 1...n) {
for (i in 0...n-len) {
var j = i + len
m[i][j] = 1/0
for (k in i...j) {
var temp = dims[i] * dims [k + 1] * dims[j + 1]
var cost = m[i][k] + m[k + 1][j] + temp
if (cost < m[i][j]) {
m[i][j] = cost
s[i][j] = k
}
}
}
}
}
var printOptimalChainOrder
printOptimalChainOrder = Fn.new { |i, j|
if (i == j) {
System.write(String.fromByte(i + 65))
} else {
System.write("(")
printOptimalChainOrder.call(i, s[i][j])
printOptimalChainOrder.call(s[i][j] + 1, j)
System.write(")")
}
}
var dimsList = [
[5, 6, 3, 1],
[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 (dims in dimsList) {
System.print("Dims : %(dims)")
optimalMatrixChainOrder.call(dims)
System.write("Order : ")
printOptimalChainOrder.call(0, s.count - 1)
System.print("\nCost : %(m[0][s.count - 1])\n")
}
- 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
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);
}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() }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());- 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)