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a935450 Nov 4, 2012
Harold Cooper as of June 26, 2008
executable file 370 lines (326 sloc) 10.4 KB
/************************************************************************
*
* lap.cpp
version 1.0 - 4 September 1996
author: Roy Jonker @ MagicLogic Optimization Inc.
e-mail: roy_jonker@magiclogic.com
Code for Linear Assignment Problem, according to
"A Shortest Augmenting Path Algorithm for Dense and Sparse Linear
Assignment Problems," Computing 38, 325-340, 1987
by
R. Jonker and A. Volgenant, University of Amsterdam.
*
CHANGED 2004-08-13 by Harold Cooper (hbc@mit.edu) for the pyLAPJV Python module:
-- commented out system.h and checklap()
*
*************************************************************************/
//#include "system.h"
#include "gnrl.h"
#include "lap.h"
cost lap(int dim,
cost **assigncost,
col *rowsol,
row *colsol,
cost *u,
cost *v)
// input:
// dim - problem size
// assigncost - cost matrix
// output:
// rowsol - column assigned to row in solution
// colsol - row assigned to column in solution
// u - dual variables, row reduction numbers
// v - dual variables, column reduction numbers
{
boolean unassignedfound;
row i, imin, numfree = 0, prvnumfree, f, i0, k, freerow, *pred, *free;
col j, j1, j2, endofpath, last, low, up, *collist, *matches;
cost min, h, umin, usubmin, v2, *d;
free = new row[dim]; // list of unassigned rows.
collist = new col[dim]; // list of columns to be scanned in various ways.
matches = new col[dim]; // counts how many times a row could be assigned.
d = new cost[dim]; // 'cost-distance' in augmenting path calculation.
pred = new row[dim]; // row-predecessor of column in augmenting/alternating path.
// init how many times a row will be assigned in the column reduction.
for (i = 0; i < dim; i++)
matches[i] = 0;
// COLUMN REDUCTION
for (j = dim-1; j >= 0; j--) // reverse order gives better results.
{
// find minimum cost over rows.
min = assigncost[0][j];
imin = 0;
for (i = 1; i < dim; i++)
if (assigncost[i][j] < min)
{
min = assigncost[i][j];
imin = i;
}
v[j] = min;
if (++matches[imin] == 1)
{
// init assignment if minimum row assigned for first time.
rowsol[imin] = j;
colsol[j] = imin;
}
else
colsol[j] = -1; // row already assigned, column not assigned.
}
// REDUCTION TRANSFER
for (i = 0; i < dim; i++)
if (matches[i] == 0) // fill list of unassigned 'free' rows.
free[numfree++] = i;
else
if (matches[i] == 1) // transfer reduction from rows that are assigned once.
{
j1 = rowsol[i];
min = BIG;
for (j = 0; j < dim; j++)
if (j != j1)
if (assigncost[i][j] - v[j] < min)
min = assigncost[i][j] - v[j];
v[j1] = v[j1] - min;
}
// AUGMENTING ROW REDUCTION
int loopcnt = 0; // do-loop to be done twice.
do
{
loopcnt++;
// scan all free rows.
// in some cases, a free row may be replaced with another one to be scanned next.
k = 0;
prvnumfree = numfree;
numfree = 0; // start list of rows still free after augmenting row reduction.
while (k < prvnumfree)
{
i = free[k];
k++;
// find minimum and second minimum reduced cost over columns.
umin = assigncost[i][0] - v[0];
j1 = 0;
usubmin = BIG;
for (j = 1; j < dim; j++)
{
h = assigncost[i][j] - v[j];
if (h < usubmin)
if (h >= umin)
{
usubmin = h;
j2 = j;
}
else
{
usubmin = umin;
umin = h;
j2 = j1;
j1 = j;
}
}
i0 = colsol[j1];
if (umin < usubmin)
// change the reduction of the minimum column to increase the minimum
// reduced cost in the row to the subminimum.
v[j1] = v[j1] - (usubmin - umin);
else // minimum and subminimum equal.
if (i0 >= 0) // minimum column j1 is assigned.
{
// swap columns j1 and j2, as j2 may be unassigned.
j1 = j2;
i0 = colsol[j2];
}
// (re-)assign i to j1, possibly de-assigning an i0.
rowsol[i] = j1;
colsol[j1] = i;
if (i0 >= 0) // minimum column j1 assigned earlier.
if (umin < usubmin)
// put in current k, and go back to that k.
// continue augmenting path i - j1 with i0.
free[--k] = i0;
else
// no further augmenting reduction possible.
// store i0 in list of free rows for next phase.
free[numfree++] = i0;
}
}
while (loopcnt < 2); // repeat once.
// AUGMENT SOLUTION for each free row.
for (f = 0; f < numfree; f++)
{
freerow = free[f]; // start row of augmenting path.
// Dijkstra shortest path algorithm.
// runs until unassigned column added to shortest path tree.
for (j = 0; j < dim; j++)
{
d[j] = assigncost[freerow][j] - v[j];
pred[j] = freerow;
collist[j] = j; // init column list.
}
low = 0; // columns in 0..low-1 are ready, now none.
up = 0; // columns in low..up-1 are to be scanned for current minimum, now none.
// columns in up..dim-1 are to be considered later to find new minimum,
// at this stage the list simply contains all columns
unassignedfound = FALSE;
do
{
if (up == low) // no more columns to be scanned for current minimum.
{
last = low - 1;
// scan columns for up..dim-1 to find all indices for which new minimum occurs.
// store these indices between low..up-1 (increasing up).
min = d[collist[up++]];
for (k = up; k < dim; k++)
{
j = collist[k];
h = d[j];
if (h <= min)
{
if (h < min) // new minimum.
{
up = low; // restart list at index low.
min = h;
}
// new index with same minimum, put on undex up, and extend list.
collist[k] = collist[up];
collist[up++] = j;
}
}
// check if any of the minimum columns happens to be unassigned.
// if so, we have an augmenting path right away.
for (k = low; k < up; k++)
if (colsol[collist[k]] < 0)
{
endofpath = collist[k];
unassignedfound = TRUE;
break;
}
}
if (!unassignedfound)
{
// update 'distances' between freerow and all unscanned columns, via next scanned column.
j1 = collist[low];
low++;
i = colsol[j1];
h = assigncost[i][j1] - v[j1] - min;
for (k = up; k < dim; k++)
{
j = collist[k];
v2 = assigncost[i][j] - v[j] - h;
if (v2 < d[j])
{
pred[j] = i;
if (v2 == min) // new column found at same minimum value
if (colsol[j] < 0)
{
// if unassigned, shortest augmenting path is complete.
endofpath = j;
unassignedfound = TRUE;
break;
}
// else add to list to be scanned right away.
else
{
collist[k] = collist[up];
collist[up++] = j;
}
d[j] = v2;
}
}
}
}
while (!unassignedfound);
// update column prices.
for (k = 0; k <= last; k++)
{
j1 = collist[k];
v[j1] = v[j1] + d[j1] - min;
}
// reset row and column assignments along the alternating path.
do
{
i = pred[endofpath];
colsol[endofpath] = i;
j1 = endofpath;
endofpath = rowsol[i];
rowsol[i] = j1;
}
while (i != freerow);
}
// calculate optimal cost.
cost lapcost = 0;
for (i = 0; i < dim; i++)
{
j = rowsol[i];
u[i] = assigncost[i][j] - v[j];
lapcost = lapcost + assigncost[i][j];
}
// free reserved memory.
delete[] pred;
delete[] free;
delete[] collist;
delete[] matches;
delete[] d;
return lapcost;
}
//UNUSED by pyLAPJV, so commented out:
/*void checklap(int dim, cost **assigncost,
col *rowsol, row *colsol, cost *u, cost *v)
{
row i;
col j;
cost lapcost = 0, redcost = 0;
boolean *matched;
char wait;
matched = new boolean[dim];
for (i = 0; i < dim; i++)
for (j = 0; j < dim; j++)
if ((redcost = assigncost[i][j] - u[i] - v[j]) < 0)
{
printf("\n");
printf("negative reduced cost i %d j %d redcost %d\n", i, j, redcost);
printf("\n\ndim %5d - press key\n", dim);
scanf("%d", &wait);
break;
}
for (i = 0; i < dim; i++)
if ((redcost = assigncost[i][rowsol[i]] - u[i] - v[rowsol[i]]) != 0)
{
printf("\n");
printf("non-null reduced cost i %d soli %d redcost %d\n", i, rowsol[i], redcost);
printf("\n\ndim %5d - press key\n", dim);
scanf("%d", &wait);
break;
}
for (j = 0; j < dim; j++)
matched[j] = FALSE;
for (i = 0; i < dim; i++)
if (matched[rowsol[i]])
{
printf("\n");
printf("column matched more than once - i %d soli %d\n", i, rowsol[i]);
printf("\n\ndim %5d - press key\n", dim);
scanf("%d", &wait);
break;
}
else
matched[rowsol[i]] = TRUE;
for (i = 0; i < dim; i++)
if (colsol[rowsol[i]] != i)
{
printf("\n");
printf("error in row solution i %d soli %d solsoli %d\n", i, rowsol[i], colsol[rowsol[i]]);
printf("\n\ndim %5d - press key\n", dim);
scanf("%d", &wait);
break;
}
for (j = 0; j < dim; j++)
if (rowsol[colsol[j]] != j)
{
printf("\n");
printf("error in col solution j %d solj %d solsolj %d\n", j, colsol[j], rowsol[colsol[j]]);
printf("\n\ndim %5d - press key\n", dim);
scanf("%d", &wait);
break;
}
delete[] matched;
return;
}*/