Matrix chain multiplication

From Rosetta Code
Matrix chain multiplication is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
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.

C[edit]

Translation of: Kotlin
#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#[edit]

Translation of: Kotlin
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

Fortran[edit]

Translation of: Python

This is a translation of the Python iterative solution.

module optim_m
implicit none
contains
subroutine optim(a)
implicit none
integer :: a(:), n, i, j, k
integer(8) :: c
integer, allocatable :: u(:, :)
integer(8), allocatable :: v(:, :)
 
n = ubound(a, 1) - 1
allocate (u(n, n), v(n, n))
v = -1
u(:, 1) = -1
v(:, 1) = 0
do j = 2, n
do i = 1, n - j + 1
do k = 1, j - 1
c = v(i, k) + v(i + k, j - k) + int(a(i), 8) * int(a(i + k), 8) * int(a(i + j), 8)
if (v(i, j) < 0 .or. c < v(i, j)) then
u(i, j) = k
v(i, j) = c
end if
end do
end do
end do
write (*, "(I0,' ')", advance="no") v(1, n)
call aux(1, n)
print *
deallocate (u, v)
contains
recursive subroutine aux(i, j)
integer :: i, j, k
 
k = u(i, j)
if (k < 0) then
write (*, "(I0)", advance="no") i
else
write (*, "('(')", advance="no")
call aux(i, k)
write (*, "('*')", advance="no")
call aux(i + k, j - k)
write (*, "(')')", advance="no")
end if
end subroutine
end subroutine
end module
 
program optim_p
use optim_m
implicit none
 
call optim([1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2])
call optim([1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10])
end program

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))

Julia[edit]

Works with: Julia version 0.6
Translation of: MATLAB

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[edit]

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

MATLAB[edit]

Translation of: Fortran
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))"

Python[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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 descibed 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]])

Stata[edit]

Recursive solution[edit]

Translation of: Python

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[edit]

Translation of: Fortran
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.

zkl[edit]

Translation of: Python
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)