Modified the symmetry breaking routine that arbitrarily resolves
automorphisms so that it arbitrarily assigns all pairs from circuit1 and circuit2 at once rather than assigning one pair at a time and rerunning to convergence. I'm not sure of the validity of this, other than that I have never seen a circuit fail to match after resolving automorphisms, leading me to believe that the way the symmetry breaking is done is irrelevant.
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@ -5874,11 +5874,23 @@ int ResolveAutomorphisms()
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}
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if (C1 == C2 && C1 != 1) {
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unsigned long newhash;
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Magic(newhash);
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E1->hashval = newhash;
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E2->hashval = newhash;
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/* Diagnostic */
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// Fprintf(stdout, "Equivalencing 1 element out of %d.\n", C2);
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/* Do all of them at once. */
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/* NOTE: If this were to fail for some reason, it would */
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/* probably be necessary to do this the original slow way */
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/* which is to rehash one pair at a time and iterate to */
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/* convergence, and repeat. */
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E1 = E2 = EC->elements;
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while (E1 != NULL) {
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while (E1->graph != Circuit1->file) E1 = E1->next;
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while (E2->graph != Circuit2->file) E2 = E2->next;
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Magic(newhash);
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E1->hashval = newhash;
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E2->hashval = newhash;
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E1 = E1->next;
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E2 = E2->next;
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}
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goto converge;
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}
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}
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@ -5899,11 +5911,23 @@ int ResolveAutomorphisms()
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}
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if (C1 == C2 && C1 != 1) {
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unsigned long newhash;
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Magic(newhash);
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N1->hashval = newhash;
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N2->hashval = newhash;
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/* Diagnostic */
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// Fprintf(stdout, "Equivalencing 1 node out of %d.\n", C2);
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/* Do all of them at once */
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/* NOTE: If this were to fail for some reason, it would */
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/* probably be necessary to do this the original slow way */
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/* which is to rehash one pair at a time and iterate to */
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/* convergence, and repeat. */
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N1 = N2 = NC->nodes;
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while (N1 != NULL) {
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while (N1->graph != Circuit1->file) N1 = N1->next;
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while (N2->graph != Circuit2->file) N2 = N2->next;
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Magic(newhash);
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N1->hashval = newhash;
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N2->hashval = newhash;
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N1 = N1->next;
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N2 = N2->next;
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}
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goto converge;
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}
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}
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