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2023-01-21   ES

Title:Determine n matrix connection multiplication A1A2A3 … ancalculation order, so that the number of “number multiplier” required for the calculation matrix is calculated according to this sequence.

• This problem meets the conditions of dynamic planning
• matrix multiplication to satisfy-combined law
• Two matrix multiplication, to meet the number of columns of the left matrix = the number of rows of the right matrix
• two matrix multiplication, the calculation amount is: the number of rows on the left matrixmultipliedThe number of columns on the left matrix (ie: the number of rows on the right matrix)The number of columns on the right matrix
• Question is not actually a multiplication, but justThe order of the matrix multiplication involved in the decision

Thinking:
Bring the matrix multiplication product to {p (i-1), Pi, P (i+1), …, pk, …, pj}:

• p (i-1) is a matrix AIline number
• PI is a matrix AIcolumn number, also the matrix a (i+1)line number
• p (i+1) is the matrix matrix a (i+1)column number
• is: {p (i-1), Pi, P (i+1)} indicates two matrix AIA (i+1) multiplication

Assuming the last part of the optimal calculation order of the matrix multiplication A [i: j] is interrupted between matrix AK and matrix A (k+1),
K Position is only J-1 possibility
calculation quantity: the optimal calculation amount of: A [i: K] + A [K + 1: J] optimal calculation amount + the calculation amount of the two matrix left by the last multiplication operation (as shown below)

recursive algorithm (memorandum of use: solve the problem of overlapping subtrays, improve algorithm efficiency):

``````#include <iostream>

using namespace std;
#Define MaxSize 20

void matrixchain (int *p, int n, int m [] [] [maxsize], int s [] [] [maxsize]) {//
int i, j, r, k;
int T; // t is the middle amount
for (i = 1; i <= n; i ++) {
m [i] [i] = 0; // Set up a single matrix to 0, because the calculation in the figure above will appear, so it is set to 0
}

For (r = 2; r <= n; r ++) {// r means the number of matrix in the connection matrix
For (i = 1; I <= n-r+1; i ++) {// i indicates the first one of the connection matrix
j = i + r -1; // j means the last one of the connection matrix
m [i] [j] = m [i + 1] [j] + p [i-1] * p [i] * p [j]; // Here I give K to K (refer to the calculated in the figure above)
s [i] [j] = i;
for (k = i+1; k <j; k ++) {
t = m [i] [k] + m [k + 1] [j] + p [i-1] * p [k] * p [j]; // (calculated in the figure above)
if (t <m [i] [j]) {
m [i] [j] = t;
s [i] [j] = k;
}
}
}
}
COUT << Endl << "The best multiplication:" << m  [n] << Endl;
}

void traceback (int i, int j, int s [] [] [maxsize]) {// optimal multiplication path path
if (i <j-1) {{
Traceback (i, s [i] [j], s);
Traceback (s [i] [j]+1, j, s);
COUT << "Matrix a" << i << "," << s [i] [j] << "<< s [i] +1 <<", "<< j <", "j <<<< j <" <endl;
}
}

int main () {)
int m [maxsize] [maxsize], // m [i] [j] recorded from I to J required at least the number
s [maxsize] [maxsize]; // s [i] [j] corresponding to the disconnection position of m [i] [j]

int p [maxsize]; // Store the number of lines of matrix, the number of columns
int n;
do {
COUT << "Please enter the number of matrices connected to the matrix (2 to 20) n =";
cin >> n.
} While (n <2 || n> 20);

Cout << "Please enter the number of rows of n matrices in turn and the number of columns of the last matrix (input n+1 number):" << endl;
for (int i = 0; i <n+1; i ++) {
cin >> p [i];
}

Matrixchain (p, n, m, s);
Traceback (1, n, s);
}``````

Reference materials:
China University MOOK algorithm design and analysis 5.3-> 5.5

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