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The assignment problem is to assign jobs to * workers in a way that minimizes the total cost. Since each worker can perform * only one job and each job can be assigned to only one worker the assignments * represent an independent set of the matrix C. * * One way to generate the optimal set is to create all permutations of * the indices necessary to traverse the matrix so that no row and column * are used more than once. For instance, given this matrix (expressed in * Python) * * matrix = [[5, 9, 1], * [10, 3, 2], * [8, 7, 4]] * * You could use this code to generate the traversal indices:: * * def permute(a, results): * if len(a) == 1: * results.insert(len(results), a) * * else: * for i in range(0, len(a)): * element = a[i] * a_copy = [a[j] for j in range(0, len(a)) if j != i] * subresults = [] * permute(a_copy, subresults) * for subresult in subresults: * result = [element] + subresult * results.insert(len(results), result) * * results = [] * permute(range(len(matrix)), results) # [0, 1, 2] for a 3x3 matrix * * After the call to permute(), the results matrix would look like this:: * * [[0, 1, 2], * [0, 2, 1], * [1, 0, 2], * [1, 2, 0], * [2, 0, 1], * [2, 1, 0]] * * You could then use that index matrix to loop over the original cost matrix * and calculate the smallest cost of the combinations * * n = len(matrix) * minval = sys.maxsize * for row in range(n): * cost = 0 * for col in range(n): * cost += matrix[row][col] * minval = min(cost, minval) * * print minval * * While this approach works fine for small matrices, it does not scale. It * executes in O(n!) time: Calculating the permutations for an n×x-matrix * requires n! operations. For a 12×12 matrix, that’s 479,001,600 * traversals. Even if you could manage to perform each traversal in just one * millisecond, it would still take more than 133 hours to perform the entire * traversal. A 20×20 matrix would take 2,432,902,008,176,640,000 operations. At * an optimistic millisecond per operation, that’s more than 77 million years. * * The Munkres algorithm runs in O(n³) time, rather than O(n!). This * package provides an implementation of that algorithm. * * This version is based on * http://www.public.iastate.edu/~ddoty/HungarianAlgorithm.html. * * This version was originally written for Python by Brian Clapper from the (Ada) * algorithm at the above web site (The ``Algorithm::Munkres`` Perl version, * in CPAN, was clearly adapted from the same web site.) and ported to * JavaScript by Hauke Henningsen (addaleax). * * Usage * ===== * * Construct a Munkres object * * var m = new Munkres(); * * Then use it to compute the lowest cost assignment from a cost matrix. Here’s * a sample program * * var matrix = [[5, 9, 1], * [10, 3, 2], * [8, 7, 4]]; * var m = new Munkres(); * var indices = m.compute(matrix); * console.log(format_matrix(matrix), 'Lowest cost through this matrix:'); * var total = 0; * for (var i = 0; i < indices.length; ++i) { * var row = indices[l][0], col = indices[l][1]; * var value = matrix[row][col]; * total += value; * * console.log('(' + rol + ', ' + col + ') -> ' + value); * } * * console.log('total cost:', total); * * Running that program produces:: * * Lowest cost through this matrix: * [5, 9, 1] * [10, 3, 2] * [8, 7, 4] * (0, 0) -> 5 * (1, 1) -> 3 * (2, 2) -> 4 * total cost: 12 * * The instantiated Munkres object can be used multiple times on different * matrices. * * Non-square Cost Matrices * ======================== * * The Munkres algorithm assumes that the cost matrix is square. However, it's * possible to use a rectangular matrix if you first pad it with 0 values to make * it square. This module automatically pads rectangular cost matrices to make * them square. * * Notes: * * - The module operates on a *copy* of the caller's matrix, so any padding will * not be seen by the caller. * - The cost matrix must be rectangular or square. An irregular matrix will * *not* work. * * Calculating Profit, Rather than Cost * ==================================== * * The cost matrix is just that: A cost matrix. The Munkres algorithm finds * the combination of elements (one from each row and column) that results in * the smallest cost. It’s also possible to use the algorithm to maximize * profit. To do that, however, you have to convert your profit matrix to a * cost matrix. The simplest way to do that is to subtract all elements from a * large value. * * The ``munkres`` module provides a convenience method for creating a cost * matrix from a profit matrix, i.e. make_cost_matrix. * * References * ========== * * 1. http://www.public.iastate.edu/~ddoty/HungarianAlgorithm.html * * 2. Harold W. Kuhn. The Hungarian Method for the assignment problem. * *Naval Research Logistics Quarterly*, 2:83-97, 1955. * * 3. Harold W. Kuhn. Variants of the Hungarian method for assignment * problems. *Naval Research Logistics Quarterly*, 3: 253-258, 1956. * * 4. Munkres, J. Algorithms for the Assignment and Transportation Problems. * *Journal of the Society of Industrial and Applied Mathematics*, * 5(1):32-38, March, 1957. * * 5. https://en.wikipedia.org/wiki/Hungarian_algorithm * * Copyright and License * ===================== * * This software is released under a BSD license, adapted from * <http://opensource.org/licenses/bsd-license.php> * * Copyright (c) 2008 Brian M. Clapper * All rights reserved. * * Redistribution and use in source and binary forms, with or without * modification, are permitted provided that the following conditions are met: * * * Redistributions of source code must retain the above copyright notice, * this list of conditions and the following disclaimer. * * * Redistributions in binary form must reproduce the above copyright notice, * this list of conditions and the following disclaimer in the documentation * and/or other materials provided with the distribution. * * * Neither the name "clapper.org" nor the names of its contributors may be * used to endorse or promote products derived from this software without * specific prior written permission. * * THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS" * AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE * ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE * LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR * CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF * SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS * INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN * CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) * ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE * POSSIBILITY OF SUCH DAMAGE. */ /** * A very large numerical value which can be used like an integer * (i. e., adding integers of similar size does not result in overflow). */ var MAX_SIZE = parseInt(Number.MAX_SAFE_INTEGER/2) || ((1 << 26)*(1 << 26)); /** * A default value to pad the cost matrix with if it is not quadratic. */ var DEFAULT_PAD_VALUE = 0; // --------------------------------------------------------------------------- // Classes // --------------------------------------------------------------------------- /** * Calculate the Munkres solution to the classical assignment problem. * See the module documentation for usage. * @constructor */ function Munkres() { this.C = null this.row_covered = [] this.col_covered = [] this.n = 0 this.Z0_r = 0 this.Z0_c = 0 this.marked = null this.path = null } /** * Pad a possibly non-square matrix to make it square. * * @param {Array} matrix An array of arrays containing the matrix cells * @param {Number} [pad_value] The value used to pad a rectangular matrix * * @return {Array} An array of arrays representing the padded matrix */ Munkres.prototype.pad_matrix = function(matrix, pad_value) { pad_value = pad_value || DEFAULT_PAD_VALUE; var max_columns = 0; var total_rows = matrix.length; for (var i = 0; i < total_rows; ++i) if (matrix[i].length > max_columns) max_columns = matrix[i].length; total_rows = max_columns > total_rows ? max_columns : total_rows; var new_matrix = []; for (var i = 0; i < total_rows; ++i) { var row = matrix[i] || []; var new_row = row.slice(); // If this row is too short, pad it while (total_rows > new_row.length) new_row.push(pad_value); new_matrix.push(new_row); } return new_matrix; }; /** * Compute the indices for the lowest-cost pairings between rows and columns * in the database. Returns a list of (row, column) tuples that can be used * to traverse the matrix. * * **WARNING**: This code handles square and rectangular matrices. * It does *not* handle irregular matrices. * * @param {Array} cost_matrix The cost matrix. If this cost matrix is not square, * it will be padded with DEFAULT_PAD_VALUE. Optionally, * the pad value can be specified via options.padValue. * This method does *not* modify the caller's matrix. * It operates on a copy of the matrix. * @param {Object} [options] Additional options to pass in * @param {Number} [options.padValue] The value to use to pad a rectangular cost_matrix * * @return {Array} An array of ``(row, column)`` arrays that describe the lowest * cost path through the matrix */ Munkres.prototype.compute = function(cost_matrix, options) { options = options || {}; options.padValue = options.padValue || DEFAULT_PAD_VALUE; this.C = this.pad_matrix(cost_matrix, options.padValue); this.n = this.C.length; this.original_length = cost_matrix.length; this.original_width = cost_matrix[0].length; var nfalseArray = []; /* array of n false values */ while (nfalseArray.length < this.n) nfalseArray.push(false); this.row_covered = nfalseArray.slice(); this.col_covered = nfalseArray.slice(); this.Z0_r = 0; this.Z0_c = 0; this.path = this.__make_matrix(this.n * 2, 0); this.marked = this.__make_matrix(this.n, 0); var step = 1; var steps = { 1 : this.__step1, 2 : this.__step2, 3 : this.__step3, 4 : this.__step4, 5 : this.__step5, 6 : this.__step6 }; while (true) { var func = steps[step]; if (!func) // done break; step = func.apply(this); } var results = []; for (var i = 0; i < this.original_length; ++i) for (var j = 0; j < this.original_width; ++j) if (this.marked[i][j] == 1) results.push([i, j]); return results; }; /** * Create an n×n matrix, populating it with the specific value. * * @param {Number} n Matrix dimensions * @param {Number} val Value to populate the matrix with * * @return {Array} An array of arrays representing the newly created matrix */ Munkres.prototype.__make_matrix = function(n, val) { var matrix = []; for (var i = 0; i < n; ++i) { matrix[i] = []; for (var j = 0; j < n; ++j) matrix[i][j] = val; } return matrix; }; /** * For each row of the matrix, find the smallest element and * subtract it from every element in its row. Go to Step 2. */ Munkres.prototype.__step1 = function() { for (var i = 0; i < this.n; ++i) { // Find the minimum value for this row and subtract that minimum // from every element in the row. var minval = Math.min.apply(Math, this.C[i]); for (var j = 0; j < this.n; ++j) this.C[i][j] -= minval; } return 2; }; /** * Find a zero (Z) in the resulting matrix. If there is no starred * zero in its row or column, star Z. Repeat for each element in the * matrix. Go to Step 3. */ Munkres.prototype.__step2 = function() { for (var i = 0; i < this.n; ++i) { for (var j = 0; j < this.n; ++j) { if (this.C[i][j] == 0 && !this.col_covered[j] && !this.row_covered[i]) { this.marked[i][j] = 1; this.col_covered[j] = true; this.row_covered[i] = true; } } } this.__clear_covers(); return 3; }; /** * Cover each column containing a starred zero. If K columns are * covered, the starred zeros describe a complete set of unique * assignments. In this case, Go to DONE, otherwise, Go to Step 4. */ Munkres.prototype.__step3 = function() { var count = 0; for (var i = 0; i < this.n; ++i) { for (var j = 0; j < this.n; ++j) { if (this.marked[i][j] == 1) { this.col_covered[j] = true; ++count; } } } return (count >= this.n) ? 7 : 4; }; /** * Find a noncovered zero and prime it. If there is no starred zero * in the row containing this primed zero, Go to Step 5. Otherwise, * cover this row and uncover the column containing the starred * zero. Continue in this manner until there are no uncovered zeros * left. Save the smallest uncovered value and Go to Step 6. */ Munkres.prototype.__step4 = function() { var done = false; var row = -1, col = -1, star_col = -1; while (!done) { var z = this.__find_a_zero(); row = z[0]; col = z[1]; if (row < 0) return 6; this.marked[row][col] = 2; star_col = this.__find_star_in_row(row); if (star_col >= 0) { col = star_col; this.row_covered[row] = true; this.col_covered[col] = false; } else { this.Z0_r = row; this.Z0_c = col; return 5; } } }; /** * Construct a series of alternating primed and starred zeros as * follows. Let Z0 represent the uncovered primed zero found in Step 4. * Let Z1 denote the starred zero in the column of Z0 (if any). * Let Z2 denote the primed zero in the row of Z1 (there will always * be one). Continue until the series terminates at a primed zero * that has no starred zero in its column. Unstar each starred zero * of the series, star each primed zero of the series, erase all * primes and uncover every line in the matrix. Return to Step 3 */ Munkres.prototype.__step5 = function() { var count = 0; this.path[count][0] = this.Z0_r; this.path[count][1] = this.Z0_c; var done = false; while (!done) { var row = this.__find_star_in_col(this.path[count][1]); if (row >= 0) { count++; this.path[count][0] = row; this.path[count][1] = this.path[count-1][1]; } else { done = true } if (!done) { var col = this.__find_prime_in_row(this.path[count][0]); count++; this.path[count][0] = this.path[count-1][0]; this.path[count][1] = col; } } this.__convert_path(this.path, count); this.__clear_covers(); this.__erase_primes(); return 3; }; /** * Add the value found in Step 4 to every element of each covered * row, and subtract it from every element of each uncovered column. * Return to Step 4 without altering any stars, primes, or covered * lines. */ Munkres.prototype.__step6 = function() { var minval = this.__find_smallest(); for (var i = 0; i < this.n; ++i) { for (var j = 0; j < this.n; ++j) { if (this.row_covered[i]) this.C[i][j] += minval; if (!this.col_covered[j]) this.C[i][j] -= minval; } } return 4; }; /** * Find the smallest uncovered value in the matrix. * * @return {Number} The smallest uncovered value, or MAX_SIZE if no value was found */ Munkres.prototype.__find_smallest = function() { var minval = MAX_SIZE; for (var i = 0; i < this.n; ++i) for (var j = 0; j < this.n; ++j) if (!this.row_covered[i] && !this.col_covered[j]) if (minval > this.C[i][j]) minval = this.C[i][j]; return minval; }; /** * Find the first uncovered element with value 0. * * @return {Array} The indices of the found element or [-1, -1] if not found */ Munkres.prototype.__find_a_zero = function() { for (var i = 0; i < this.n; ++i) for (var j = 0; j < this.n; ++j) if (this.C[i][j] == 0 && !this.row_covered[i] && !this.col_covered[j]) return [i, j]; return [-1, -1]; }; /** * Find the first starred element in the specified row. Returns * the column index, or -1 if no starred element was found. * * @param {Number} row The index of the row to search * @return {Number} */ Munkres.prototype.__find_star_in_row = function(row) { for (var j = 0; j < this.n; ++j) if (this.marked[row][j] == 1) return j; return -1; }; /** * Find the first starred element in the specified column. * * @return {Number} The row index, or -1 if no starred element was found */ Munkres.prototype.__find_star_in_col = function(col) { for (var i = 0; i < this.n; ++i) if (this.marked[i][col] == 1) return i; return -1; }; /** * Find the first prime element in the specified row. * * @return {Number} The column index, or -1 if no prime element was found */ Munkres.prototype.__find_prime_in_row = function(row) { for (var j = 0; j < this.n; ++j) if (this.marked[row][j] == 2) return j; return -1; }; Munkres.prototype.__convert_path = function(path, count) { for (var i = 0; i <= count; ++i) this.marked[path[i][0]][path[i][1]] = (this.marked[path[i][0]][path[i][1]] == 1) ? 0 : 1; }; /** Clear all covered matrix cells */ Munkres.prototype.__clear_covers = function() { for (var i = 0; i < this.n; ++i) { this.row_covered[i] = false; this.col_covered[i] = false; } }; /** Erase all prime markings */ Munkres.prototype.__erase_primes = function() { for (var i = 0; i < this.n; ++i) for (var j = 0; j < this.n; ++j) if (this.marked[i][j] == 2) this.marked[i][j] = 0; }; // --------------------------------------------------------------------------- // Functions // --------------------------------------------------------------------------- /** * Create a cost matrix from a profit matrix by calling * 'inversion_function' to invert each value. The inversion * function must take one numeric argument (of any type) and return * another numeric argument which is presumed to be the cost inverse * of the original profit. * * This is a static method. Call it like this: * * cost_matrix = make_cost_matrix(matrix[, inversion_func]); * * For example: * * cost_matrix = make_cost_matrix(matrix, function(x) { return MAXIMUM - x; }); * * @param {Array} profit_matrix An array of arrays representing the matrix * to convert from a profit to a cost matrix * @param {Function} [inversion_function] The function to use to invert each * entry in the profit matrix * * @return {Array} The converted matrix */ function make_cost_matrix (profit_matrix, inversion_function) { if (!inversion_function) { var maximum = -1.0/0.0; for (var i = 0; i < profit_matrix.length; ++i) for (var j = 0; j < profit_matrix[i].length; ++j) if (profit_matrix[i][j] > maximum) maximum = profit_matrix[i][j]; inversion_function = function(x) { return maximum - x; }; } var cost_matrix = []; for (var i = 0; i < profit_matrix.length; ++i) { var row = profit_matrix[i]; cost_matrix[i] = []; for (var j = 0; j < row.length; ++j) cost_matrix[i][j] = inversion_function(profit_matrix[i][j]); } return cost_matrix; } /** * Convenience function: Converts the contents of a matrix of integers * to a printable string. * * @param {Array} matrix The matrix to print * * @return {String} The formatted matrix */ function format_matrix(matrix) { function log10(v) { if (Math.log10) return Math.log10(v); return Math.log(v) / Math.log(10); } var columnWidths = []; for (var i = 0; i < matrix.length; ++i) { for (var j = 0; j < matrix[i].length; ++j) { var entryWidth = String(matrix[i][j]).length; if (!columnWidths[j] || entryWidth >= columnWidths[j]) columnWidths[j] = entryWidth; } } var formatted = ''; for (var i = 0; i < matrix.length; ++i) { for (var j = 0; j < matrix[i].length; ++j) { var s = String(matrix[i][j]); // pad at front with spaces while (s.length < columnWidths[j]) s = ' ' + s; formatted += s; // separate columns if (j != matrix[i].length - 1) formatted += ' '; } if (i != matrix[i].length - 1) formatted += '\n'; } return formatted; } // --------------------------------------------------------------------------- // Exports // --------------------------------------------------------------------------- Eif (typeof exports != 'undefined' && exports) { exports.version = "1.1.0"; exports.format_matrix = format_matrix; exports.make_cost_matrix = make_cost_matrix; exports.Munkres = Munkres; } // --------------------------------------------------------------------------- // Main // --------------------------------------------------------------------------- Eif (typeof require != 'undefined' && typeof module != 'undefined' && require.main == module) { var assert = require('assert'); var matrices = [ // Square [[[400, 150, 400], [400, 450, 600], [300, 225, 300]], 850], // expected cost // Rectangular variant [[[400, 150, 400, 1], [400, 450, 600, 2], [300, 225, 300, 3]], 452], // expected cost // Square [[[10, 10, 8], [9, 8, 1], [9, 7, 4]], 18], // Rectangular variant [[[10, 10, 8, 11], [9, 8, 1, 1], [9, 7, 4, 10]], 15], // All-zero square [[[0, 0, 0], [0, 0, 0], [0, 0, 0]], 0], ]; var m = new Munkres(); for (var k = 0; k < matrices.length; ++k) { var cost_matrix = matrices[k][0]; var expected_total = matrices[k][1]; console.log('cost matrix'); console.log(format_matrix(cost_matrix)); var indices = m.compute(cost_matrix); var total_cost = 0; for (var l = 0; l < indices.length; ++l) { var r = indices[l][0], c = indices[l][1]; var x = cost_matrix[r][c]; total_cost += x; console.log('(' + r + ', ' + c + ') -> ' + x); } console.log('lowest cost = ' + total_cost + ', expected = ' + expected_total); assert.equal(expected_total, total_cost); } } |