linear-assignment: a function to solve least-cost assignment problems

The problem solved by the code introduced in this commit goes like this:
given two sets of items, and a cost matrix which says how much it
"costs" to assign any given item of the first set to any given item of
the second, assign all items (except when the sets have different size)
in the cheapest way.

We use the Jonker-Volgenant algorithm to solve the assignment problem to
answer questions such as: given two different versions of a topic branch
(or iterations of a patch series), what is the best pairing of
commits/patches between the different versions?

Signed-off-by: Johannes Schindelin <[email protected]>
Signed-off-by: Junio C Hamano <[email protected]>

Author: dscho

Makefile

@@ -870,6 +870,7 @@ LIB_OBJS += gpg-interface.o
 LIB_OBJS += graph.o
 LIB_OBJS += grep.o
 LIB_OBJS += hashmap.o
+LIB_OBJS += linear-assignment.o
 LIB_OBJS += help.o
 LIB_OBJS += hex.o
 LIB_OBJS += ident.o

linear-assignment.c

@@ -0,0 +1,201 @@
+/*
+ * Based on: Jonker, R., & Volgenant, A. (1987). <i>A shortest augmenting path
+ * algorithm for dense and sparse linear assignment problems</i>. Computing,
+ * 38(4), 325-340.
+ */
+#include "cache.h"
+#include "linear-assignment.h"
+
+#define COST(column, row) cost[(column) + column_count * (row)]
+
+/*
+ * The parameter `cost` is the cost matrix: the cost to assign column j to row
+ * i is `cost[j + column_count * i].
+ */
+void compute_assignment(int column_count, int row_count, int *cost,
+			int *column2row, int *row2column)
+{
+	int *v, *d;
+	int *free_row, free_count = 0, saved_free_count, *pred, *col;
+	int i, j, phase;
+
+	memset(column2row, -1, sizeof(int) * column_count);
+	memset(row2column, -1, sizeof(int) * row_count);
+	ALLOC_ARRAY(v, column_count);
+
+	/* column reduction */
+	for (j = column_count - 1; j >= 0; j--) {
+		int i1 = 0;
+
+		for (i = 1; i < row_count; i++)
+			if (COST(j, i1) > COST(j, i))
+				i1 = i;
+		v[j] = COST(j, i1);
+		if (row2column[i1] == -1) {
+			/* row i1 unassigned */
+			row2column[i1] = j;
+			column2row[j] = i1;
+		} else {
+			if (row2column[i1] >= 0)
+				row2column[i1] = -2 - row2column[i1];
+			column2row[j] = -1;
+		}
+	}
+
+	/* reduction transfer */
+	ALLOC_ARRAY(free_row, row_count);
+	for (i = 0; i < row_count; i++) {
+		int j1 = row2column[i];
+		if (j1 == -1)
+			free_row[free_count++] = i;
+		else if (j1 < -1)
+			row2column[i] = -2 - j1;
+		else {
+			int min = COST(!j1, i) - v[!j1];
+			for (j = 1; j < column_count; j++)
+				if (j != j1 && min > COST(j, i) - v[j])
+					min = COST(j, i) - v[j];
+			v[j1] -= min;
+		}
+	}
+
+	if (free_count ==
+	    (column_count < row_count ? row_count - column_count : 0)) {
+		free(v);
+		free(free_row);
+		return;
+	}
+
+	/* augmenting row reduction */
+	for (phase = 0; phase < 2; phase++) {
+		int k = 0;
+
+		saved_free_count = free_count;
+		free_count = 0;
+		while (k < saved_free_count) {
+			int u1, u2;
+			int j1 = 0, j2, i0;
+
+			i = free_row[k++];
+			u1 = COST(j1, i) - v[j1];
+			j2 = -1;
+			u2 = INT_MAX;
+			for (j = 1; j < column_count; j++) {
+				int c = COST(j, i) - v[j];
+				if (u2 > c) {
+					if (u1 < c) {
+						u2 = c;
+						j2 = j;
+					} else {
+						u2 = u1;
+						u1 = c;
+						j2 = j1;
+						j1 = j;
+					}
+				}
+			}
+			if (j2 < 0) {
+				j2 = j1;
+				u2 = u1;
+			}
+
+			i0 = column2row[j1];
+			if (u1 < u2)
+				v[j1] -= u2 - u1;
+			else if (i0 >= 0) {
+				j1 = j2;
+				i0 = column2row[j1];
+			}
+
+			if (i0 >= 0) {
+				if (u1 < u2)
+					free_row[--k] = i0;
+				else
+					free_row[free_count++] = i0;
+			}
+			row2column[i] = j1;
+			column2row[j1] = i;
+		}
+	}
+
+	/* augmentation */
+	saved_free_count = free_count;
+	ALLOC_ARRAY(d, column_count);
+	ALLOC_ARRAY(pred, column_count);
+	ALLOC_ARRAY(col, column_count);
+	for (free_count = 0; free_count < saved_free_count; free_count++) {
+		int i1 = free_row[free_count], low = 0, up = 0, last, k;
+		int min, c, u1;
+
+		for (j = 0; j < column_count; j++) {
+			d[j] = COST(j, i1) - v[j];
+			pred[j] = i1;
+			col[j] = j;
+		}
+
+		j = -1;
+		do {
+			last = low;
+			min = d[col[up++]];
+			for (k = up; k < column_count; k++) {
+				j = col[k];
+				c = d[j];
+				if (c <= min) {
+					if (c < min) {
+						up = low;
+						min = c;
+					}
+					col[k] = col[up];
+					col[up++] = j;
+				}
+			}
+			for (k = low; k < up; k++)
+				if (column2row[col[k]] == -1)
+					goto update;
+
+			/* scan a row */
+			do {
+				int j1 = col[low++];
+
+				i = column2row[j1];
+				u1 = COST(j1, i) - v[j1] - min;
+				for (k = up; k < column_count; k++) {
+					j = col[k];
+					c = COST(j, i) - v[j] - u1;
+					if (c < d[j]) {
+						d[j] = c;
+						pred[j] = i;
+						if (c == min) {
+							if (column2row[j] == -1)
+								goto update;
+							col[k] = col[up];
+							col[up++] = j;
+						}
+					}
+				}
+			} while (low != up);
+		} while (low == up);
+
+update:
+		/* updating of the column pieces */
+		for (k = 0; k < last; k++) {
+			int j1 = col[k];
+			v[j1] += d[j1] - min;
+		}
+
+		/* augmentation */
+		do {
+			if (j < 0)
+				BUG("negative j: %d", j);
+			i = pred[j];
+			column2row[j] = i;
+			SWAP(j, row2column[i]);
+		} while (i1 != i);
+	}
+
+	free(col);
+	free(pred);
+	free(d);
+	free(v);
+	free(free_row);
+}

linear-assignment.h

@@ -0,0 +1,22 @@
+#ifndef LINEAR_ASSIGNMENT_H
+#define LINEAR_ASSIGNMENT_H
+
+/*
+ * Compute an assignment of columns -> rows (and vice versa) such that every
+ * column is assigned to at most one row (and vice versa) minimizing the
+ * overall cost.
+ *
+ * The parameter `cost` is the cost matrix: the cost to assign column j to row
+ * i is `cost[j + column_count * i].
+ *
+ * The arrays column2row and row2column will be populated with the respective
+ * assignments (-1 for unassigned, which can happen only if column_count !=
+ * row_count).
+ */
+void compute_assignment(int column_count, int row_count, int *cost,
+			int *column2row, int *row2column);
+
+/* The maximal cost in the cost matrix (to prevent integer overflows). */
+#define COST_MAX (1<<16)
+
+#endif