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 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.
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
<lang 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(‘’)</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
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: <lang ada> package mat_chain is
type Vector is array (Natural range <>) of Integer; procedure Chain_Multiplication (Dims : Vector);
end mat_chain; </lang> The implementation or body of the package is: <lang ada> 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; </lang> The main procedure is: <lang ada> 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; </lang>
- 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
<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_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</lang>
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.
<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
Haskell
<lang 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</lang>
- 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.
<lang j>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
)</lang>
- Output:
<lang j> 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))</lang>
Java
Thanks to the Wikipedia page for a working Java implementation. <lang java> 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(); } }
} </lang>
- 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
<lang 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" )</lang>
- 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: <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
Lua
<lang 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</lang>
- 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
<lang Mathematica>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}]; uAll, 1 = -1; vAll, 1 = 0; Do[ Do[ Do[ c = vi, k + vi + k, j - k + ai ai + k ai + j; If[c < vi, j, ui, j = k; vi, j = c; ] , {k, 1, j - 1} ] , {i, 1, n - j + 1} ] , {j, 2, n} ]; r = v1, n; s = aux[u, 1, n]; {r, s} ]
aux[u_, i_, j_] := Module[{k},
k = ui, 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</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)
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>
Nim
<lang 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 ""</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
Perl
<lang 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>);</lang>
- 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
<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]])
R
<lang rsplus>aux <- function(i, j, u) {
k <- ui, 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)) { vi, j <- Inf for (k in seq(j - 1)) { s <- vi, k + vi + k, j - k + ai * ai + k * ai + j if (s < vi, j) { ui, j <- k vi, j <- s } } } }
print(v1, n) aux(1, n, u)
}
chain.mul(c(5, 6, 3, 1))
- [1] 48
- [1] "(1*(2*3))"
chain.mul(c(1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2))
- [1] 38120
- [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))
- [1] 1773740
- [1] "(1*((((((2*3)*4)*(((5*6)*7)*8))*9)*10)*11))"</lang>
Racket
Memoization
<lang racket>#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))))</lang>
Main
<lang racket>(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))</lang>
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. <lang perl6>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>);</lang>
- Output:
(38120 ((((((((A1A2)A3)A4)A5)A6)A7)(A8(A9A10)))(A11A12))) (1773740 (A1((((((A2A3)A4)(((A5A6)A7)A8))A9)A10)A11)))
Rust
<lang 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 } }
}</lang>
- 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.
<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.
VBA
<lang vb>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</lang>
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
<lang ecmascript>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")
}</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
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)