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daybs

User since Oct 28, 2022
1 Posts
TypeScript

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strassens algo

0 likes • Oct 13, 2022 • 4 views
TypeScript
// This function takes a matrix and outputs 4 equal size submatrices, starting from top left going to bottom right.
const split = (matrix: number[][]): number[][][] => {
const n: number = matrix.length
const mid: number = n/2
const C11: number[][] = new Array<Array<number>> // first n/2 rows intersection first n/2 columms
const C12: number[][] = new Array<Array<number>> // first n/2 rows intersection last n/2 columns
const C21: number[][] = new Array<Array<number>> // last n/2 rows intersection first n/2 columns
const C22: number[][] = new Array<Array<number>> // last n/2 rows intersection last n/2 columns
for(let i: number = 0; i < mid; i++) {
C11.push([])
C12.push([])
for (let j: number = 0; j < mid; j++) {
C11[i].push(matrix[i][j])
}
for (let j: number = mid; j < n; j++) {
C12[i].push(matrix[i][j])
}
}
for(let i: number = 0; i < mid; i++) {
C21.push([])
C22.push([])
for (let j: number = 0; j < mid; j++) {
C21[i].push(matrix[i+mid][j])
}
for (let j: number = mid; j < n; j++) {
C22[i].push(matrix[i+mid][j])
}
}
return [C11, C12, C21, C22]
}
// This function takes 4 matrices, in order from top left to bottom right, and combines them into one matrix
const combine = (A: number[][], B: number[][], C: number[][], D: number[][]): number[][] => {
const n: number = A.length*2
const combo: number[][] = new Array<Array<number>>
// Populate combo with temporary 0s
for(let i: number = 0; i < A.length; i++) {
combo.push([])
for (let j: number = 0; j < A.length; j++) {
combo[i].push(A[i][j])
}
for (let j: number = A.length; j < n; j++) {
combo[i].push(B[i][j-A.length])
}
}
for(let i: number = A.length; i < n; i++) {
combo.push([])
for (let j: number = 0; j < A.length; j++) {
combo[i].push(C[i-A.length][j])
}
for (let j: number = A.length; j < n; j++) {
combo[i].push(D[i-A.length][j-A.length])
}
}
return combo
}
// Basic matrix addition function
const matrix_add = (A: number[][], B: number[][]): number[][] => {
const matrix_sum: number[][] = [[]]
const n: number = A.length
for(let i: number = 0; i < n; i++) {
matrix_sum[i] = []
for (let j: number = 0; j < n; j++) {
matrix_sum[i][j] = A[i][j] + B[i][j]
}
}
return matrix_sum
}
// Basic matrix subtraction function
const matrix_subtract = (A: number[][], B: number[][]): number[][] => {
const matrix_diff: number[][] = [[]]
const n: number = A.length
for(let i: number = 0; i < n; i++) {
matrix_diff[i] = []
for (let j: number = 0; j < n; j++) {
matrix_diff[i][j] = A[i][j] - B[i][j]
}
}
return matrix_diff
}
const strassen = (A: number[][], B: number[][]): number[][] => {
const n: number = A.length
var C: number[][] = [[]]
// Populate C with temporary 0s
for(let i: number = 0; i < n; i++) {
C[i] = []
for (let j: number = 0; j < n; j++) {
C[i][j] = 0
}
}
// Once the matrices reach a small enough size, compute using the brute force method
if (n <= 2) {
for (let i: number = 0; i < n; i++) {
for (let j: number = 0; j < n; j++) {
for (let z: number = 0; z < n; z++) {
C[i][j] = C[i][j] + (A[i][z]*B[z][j])
}
}
}
return C
}
// Create independent variables from function that returns an array with array destructuring.
const [a, b, c, d] = split(A)
const [e, f, g, h] = split(B)
// Using Strassen's method to calculate 1 less product
const p1: number[][] = strassen(matrix_add(a, d), matrix_add(e,h))
const p2: number[][] = strassen(d, matrix_subtract(g, e))
const p3: number[][] = strassen(matrix_add(a, b), h)
const p4: number[][] = strassen(matrix_subtract(b, d), matrix_add(g, h))
const p5: number[][] = strassen(a, matrix_subtract(f, h))
const p6: number[][] = strassen(matrix_add(c, d), e)
const p7: number[][] = strassen(matrix_subtract(a, c), matrix_add(e, f))
const C11: number[][] = matrix_add(matrix_subtract(matrix_add(p1, p2), p3), p4)
const C12: number[][] = matrix_add(p5, p3)
const C21: number[][] = matrix_add(p6, p2)
const C22: number[][] = matrix_subtract(matrix_subtract(matrix_add(p5, p1), p6), p7)
C = combine(C11, C12, C21, C22)
return C
}
const M1: number[][] = [[1,3],[7,5]]
const M2: number[][] = [[6,8],[4,2]]
const M_Product: number[][] = strassen(M1, M2)
console.log("Matrix 1:")
console.log(M1)
console.log("Matrix 2:")
console.log(M2)
console.log("Product: ")
console.log(M_Product)

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