{"id":34402,"date":"2024-11-01T09:27:43","date_gmt":"2024-11-01T09:27:43","guid":{"rendered":"http:\/\/atmokpo.com\/w\/?p=34402"},"modified":"2024-11-01T11:41:24","modified_gmt":"2024-11-01T11:41:24","slug":"javascript-coding-test-course-finding-the-minimum-number-of-matrix-multiplications","status":"publish","type":"post","link":"https:\/\/atmokpo.com\/w\/34402\/","title":{"rendered":"JavaScript Coding Test Course, Finding the Minimum Number of Matrix Multiplications"},"content":{"rendered":"
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Author: [Author Name] | Date: [Creation Date]<\/p>\n<\/header>\n

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Problem Description<\/h2>\n

\n This problem requires finding the minimum number of operations needed to multiply a given N<\/strong> matrices.
\n The number of operations for matrix multiplication greatly varies based on the order of multiplication, and thus it is necessary to find the correct order.\n <\/p>\n

\n When the size of matrix A<\/strong> is p \u00d7 q<\/strong> and the size of matrix B<\/strong> is q \u00d7 r<\/strong>,
\n the number of operations for multiplying the two matrices is p \u00d7 q \u00d7 r<\/strong>.
\n When multiplying several matrices at once, an efficient multiplication order must be found.\n <\/p>\n

\n For instance, when multiplying three matrices of sizes 10 \u00d7 20<\/strong>, 20 \u00d7 30<\/strong>, 30 \u00d7 40<\/strong>,
\n the operation count for (10\u00d720) * (20\u00d730) * (30\u00d740) is 10 \u00d7 20 \u00d7 30 + 10 \u00d7 30 \u00d7 40 = 6000 + 12000 = 18000<\/strong>.
\n If we multiply in the order of (10\u00d730) * (30\u00d740) * (20\u00d730), the operation count becomes 10 \u00d7 30 \u00d7 40 + 10 \u00d7 20 \u00d7 40 = 12000 + 8000 = 20000<\/strong>.
\n Here, we can see that the order of the first case is more efficient.\n <\/p>\n<\/section>\n

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Input Format<\/h2>\n

\n The first line contains the number of matrices N<\/strong>. (1 \u2264 N<\/strong> \u2264 100)\n <\/p>\n

\n The second line contains N + 1<\/strong> integers separated by spaces.
\n The i-th<\/strong> integer represents the size of the matrix A[i] \u00d7 A[i+1]<\/strong>.
\n (1 \u2264 A[i]<\/strong> \u2264 500)\n <\/p>\n<\/section>\n

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Output Format<\/h2>\n

\n Print the minimum number of operations required as a single integer.\n <\/p>\n<\/section>\n

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Problem Solving Process<\/h2>\n

Step 1: Understanding Dynamic Programming Technique<\/h3>\n

\n This question is an optimization problem for matrix multiplication, which can be solved using dynamic programming.
\n To find the minimum, a 2D array can be used to store the number of operations for each ordering of matrix multiplication.\n <\/p>\n

Step 2: Initializing the Array<\/h3>\n

\n First, receive an array m<\/strong> representing N<\/strong> matrices. The length of this array is N + 1<\/strong>.
\n It stores the dimension information of each matrix.\n <\/p>\n

\nlet m = [10, 20, 30, 40];\n        <\/code><\/pre>\n
\n            Next, declare a 2D array dp<\/strong> to store the number of operations, initialized to Infinity<\/strong>.
\n            The diagonal elements are set to 0.\n        <\/p>\n
\nlet dp = Array.from(Array(n), () => Array(n).fill(Infinity));\nfor (let i = 0; i < n; i++) {\n    dp[i][i] = 0;\n}\n        <\/code><\/pre>\n
Step 3: Completing the Dynamic Programming Table<\/h3>\n
\n            Using a nested loop, partition the matrices and calculate the minimum number of operations for each multiplication.
\n            The outer loop represents the length of the multiplication chain, while the inner loop represents the indices attempting the multiplications.\n        <\/p>\n
\nfor (let len = 2; len <= n; len++) {\n    for (let i = 0; i <= n - len; i++) {\n        let j = i + len - 1;\n        for (let k = i; k < j; k++) {\n            dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k + 1][j] + m[i] * m[k + 1] * m[j + 1]);\n        }\n    }\n}\n        <\/code><\/pre>\n
Step 4: Outputting the Result<\/h3>\n
\n            Finally, print the value of the one-dimensional array dp[0][n – 1]<\/strong> to check the minimum.\n        <\/p>\n
\nconsole.log(dp[0][n - 1]);\n        <\/code><\/pre>\n<\/section>\n
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Complete Code Example<\/h2>\n
\nfunction matrixChainOrder(m) {\n    const n = m.length - 1;\n    let dp = Array.from(Array(n), () => Array(n).fill(Infinity));\n    \n    for (let i = 0; i < n; i++) {\n        dp[i][i] = 0;\n    }\n\n    for (let len = 2; len <= n; len++) {\n        for (let i = 0; i <= n - len; i++) {\n            let j = i + len - 1;\n            for (let k = i; k < j; k++) {\n                dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k + 1][j] + m[i] * m[k + 1] * m[j + 1]);\n            }\n        }\n    }\n\n    return dp[0][n - 1];\n}\n\nlet m = [10, 20, 30, 40]; \/\/ Matrix sizes\nconsole.log(matrixChainOrder(m)); \/\/ Output\n        <\/code><\/pre>\n<\/section>\n