// Given two fragments that share at least one edge, this will find that edge and construct a new
// edge to make it mutual.
//
// For example, if there is a best edge from aFrg 3' to bFrg 5', this will return that edge in a3,
// and also create the symmetric edge in b5.
//
static
bool
findEdges(ufNode *aFrg, BestEdgeOverlap &a5, BestEdgeOverlap &a3,
ufNode *bFrg, BestEdgeOverlap &b5, BestEdgeOverlap &b3) {
if (OG->isContained(aFrg->ident) ||
OG->isContained(bFrg->ident))
return(false);
// Grab what edges we have.
a5 = *OG->getBestEdgeOverlap(aFrg->ident, false);
a3 = *OG->getBestEdgeOverlap(aFrg->ident, true);
b5 = *OG->getBestEdgeOverlap(bFrg->ident, false);
b3 = *OG->getBestEdgeOverlap(bFrg->ident, true);
// Erase things that aren't correct
if (a5.fragId() != bFrg->ident) a5 = BestEdgeOverlap();
if (a3.fragId() != bFrg->ident) a3 = BestEdgeOverlap();
if (b5.fragId() != aFrg->ident) b5 = BestEdgeOverlap();
if (b3.fragId() != aFrg->ident) b3 = BestEdgeOverlap();
// If we have no edges left, there are no edges!
if ((b5.fragId() != aFrg->ident) && (b3.fragId() != aFrg->ident) &&
(a5.fragId() != bFrg->ident) && (a3.fragId() != bFrg->ident))
return(false);
// If we found TWO edges for any single fragment....that's madness! That means the fragment
// had best dovetail overlaps to the same other fragment off of both ends. We'll complain
// and return failure. Ideally, data like this will be cleaned up by OBT, or filtered from
// our input.
//
if (a5.fragId() == a3.fragId()) {
writeLog("findEdges()-- frag %d has multiple edges to frag %d - a5 %d/%d' a3 %d/%d'\n",
aFrg->ident, a5.fragId(),
a5.fragId(), a5.frag3p() ? 3 : 5,
a5.fragId(), a5.frag3p() ? 3 : 5);
}
if (b5.fragId() == b3.fragId()) {
writeLog("findEdges()-- frag %d has multiple edges to frag %d - b5 %d/%d' b3 %d/%d'\n",
bFrg->ident, b5.fragId(),
b5.fragId(), b5.frag3p() ? 3 : 5,
b5.fragId(), b5.frag3p() ? 3 : 5);
}
if (((a5.fragId() != 0) && (a5.fragId() == a3.fragId())) ||
((b5.fragId() != 0) && (b5.fragId() == b3.fragId()))) {
a5 = BestEdgeOverlap();
a3 = BestEdgeOverlap();
b5 = BestEdgeOverlap();
b3 = BestEdgeOverlap();
return(false);
}
// Now, populate the other edges using whatever we have. Best case is that we have two edges
// (because we're done).
assert(((a5.fragId() == bFrg->ident) +
(a3.fragId() == bFrg->ident) +
(b5.fragId() == aFrg->ident) +
(b3.fragId() == aFrg->ident)) <= 2);
if (((a5.fragId() == bFrg->ident) || (a3.fragId() == bFrg->ident)) &&
((b5.fragId() == aFrg->ident) || (b3.fragId() == aFrg->ident)))
return(true);
// Otherwise, we have exactly one edge, and the other one needs to be created.
assert(((a5.fragId() == bFrg->ident) +
(a3.fragId() == bFrg->ident) +
(b5.fragId() == aFrg->ident) +
(b3.fragId() == aFrg->ident)) == 1);
if (a5.fragId() == bFrg->ident) {
//assert(a5.fragId() == 0);
assert(a3.fragId() == 0);
assert(b5.fragId() == 0);
assert(b3.fragId() == 0);
// Edge off of A's 5' end ('false' below)...
// ...to B's 3' end (so ANTI or NORMAL -- negate the hangs)
// ...to B's 5' end (so INNIE or OUTTIE -- swap the hangs)
if (a5.frag3p())
b3.set(aFrg->ident, false, -a5.ahang(), -a5.bhang());
else
b5.set(aFrg->ident, false, a5.bhang(), a5.ahang());
} else if (a3.fragId() == bFrg->ident) {
assert(a5.fragId() == 0);
//assert(a3.fragId() == 0);
assert(b5.fragId() == 0);
assert(b3.fragId() == 0);
// Edge off of A's 3' end ('true' below)...
// ...to B's 3' end (so INNIE or OUTTIE -- swap the hangs)
// ...to B's 5' end (so ANTI or NORMAL -- negate the hangs)
if (a3.frag3p())
b3.set(aFrg->ident, true, a3.bhang(), a3.ahang());
else
b5.set(aFrg->ident, true, -a3.ahang(), -a3.bhang());
} else if (b5.fragId() == aFrg->ident) {
assert(a5.fragId() == 0);
assert(a3.fragId() == 0);
//assert(b5.fragId() == 0);
assert(b3.fragId() == 0);
if (b5.frag3p())
a3.set(bFrg->ident, false, -b5.ahang(), -b5.bhang());
else
a5.set(bFrg->ident, false, b5.bhang(), b5.ahang());
} else if (b3.fragId() == aFrg->ident) {
assert(a5.fragId() == 0);
assert(a3.fragId() == 0);
assert(b5.fragId() == 0);
//assert(b3.fragId() == 0);
if (b3.frag3p())
a3.set(bFrg->ident, true, b3.bhang(), b3.ahang());
else
a5.set(bFrg->ident, true, -b3.ahang(), -b3.bhang());
} else {
fprintf(stderr, "findEdges()-- Logically impossible!\n");
assert(0);
}
// And now we should have exactly two edges.
assert(((a5.fragId() == bFrg->ident) +
(a3.fragId() == bFrg->ident) +
(b5.fragId() == aFrg->ident) +
(b3.fragId() == aFrg->ident)) == 2);
return(true);
}
