Created
November 3, 2016 00:35
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Generates ILP problem to be solved by glpsol
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from collections import defaultdict | |
def name1(i): | |
return chr(ord('A')+i) | |
def name2(x, n): | |
return ''.join(name1(i) for i in xrange(n) if x&(1<<i)) | |
def print_model(L, U, n, G): | |
for x in range(1, 2**n): | |
print 'var c{}, integer, >=0;'.format(name2(x, n)) | |
for i in range(n): | |
for j in range(i+1, n): | |
a, b = U if G[i][j] else L | |
mult = -1 if G[i][j] else 1 | |
line = [] | |
for x in range(2**n): | |
coeff = 0 | |
if x&(1<<i) and x&(1<<j): coeff += mult * b | |
if x&(1<<i) or x&(1<<j): coeff -= mult * a | |
line.append(coeff) | |
expr = ' + '.join('{}*c{}'.format(x, name2(i, n)) for i,x in enumerate(line) if x) | |
print "s.t. s{}{}: {} <= 0;".format(name1(i), name1(j), expr) | |
for i in range(n): | |
expr = ' + '.join('c{}'.format(name2(x, n)) for x in range(2**n) if x&(1<<i)) | |
print "s.t. m{}: {} >= 1;".format(name1(i), expr) | |
print "minimize z: {};".format(' + '.join('c' + name2(x, n) for x in range(1, 2**n))) | |
print 'solve;' | |
print 'end;' | |
N = 6 | |
G = defaultdict(lambda: defaultdict(lambda: False)) | |
for i in range(0, 3): | |
for j in range(3, 6): | |
G[i][j] = G[j][i] = True | |
print_model((1,3), (1,2), N, G) |
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