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cd97968b03
This week I expanded that comment into a blog post: https://www.chiark.greenend.org.uk/~sgtatham/quasiblog/findloop/ which improves on the comment in three ways: 1. diagrams 2. adds a further reason why the footpath-dsf algorithm was unsatisfactory, pointed out by a Mastodon comment after I published the original version of the blog post 3. adds the punchline that the loop tracing approach _could_ have been made to work after all! So I've deleted the comment and replaced it with a link to the article.
363 lines
12 KiB
C
363 lines
12 KiB
C
/*
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* Routine for finding loops in graphs, reusable across multiple
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* puzzles.
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*
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* The strategy is Tarjan's bridge-finding algorithm, which is
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* designed to list all edges whose removal would disconnect a
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* previously connected component of the graph. We're interested in
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* exactly the reverse - edges that are part of a loop in the graph
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* are precisely those which _wouldn't_ disconnect anything if removed
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* (individually) - but of course flipping the sense of the output is
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* easy.
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*
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* For some fun background reading about all the _wrong_ ways the
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* Puzzles code base has tried to solve this problem in the past:
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* https://www.chiark.greenend.org.uk/~sgtatham/quasiblog/findloop/
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*/
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#include "puzzles.h"
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struct findloopstate {
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int parent, child, sibling, component_root;
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bool visited;
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int index, minindex, maxindex;
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int minreachable, maxreachable;
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int bridge;
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};
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struct findloopstate *findloop_new_state(int nvertices)
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{
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/*
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* Allocate a findloopstate structure for each vertex, and one
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* extra one at the end which will be the overall root of a
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* 'super-tree', which links the whole graph together to make it
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* as easy as possible to iterate over all the connected
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* components.
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*/
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return snewn(nvertices + 1, struct findloopstate);
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}
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void findloop_free_state(struct findloopstate *state)
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{
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sfree(state);
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}
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bool findloop_is_loop_edge(struct findloopstate *pv, int u, int v)
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{
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/*
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* Since the algorithm is intended for finding bridges, and a
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* bridge must be part of any spanning tree, it follows that there
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* is at most one bridge per vertex.
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*
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* Furthermore, by finding a _rooted_ spanning tree (so that each
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* bridge is a parent->child link), you can find an injection from
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* bridges to vertices (namely, map each bridge to the vertex at
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* its child end).
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*
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* So if the u-v edge is a bridge, then either v was u's parent
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* when the algorithm ran and we set pv[u].bridge = v, or vice
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* versa.
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*/
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return !(pv[u].bridge == v || pv[v].bridge == u);
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}
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static bool findloop_is_bridge_oneway(
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struct findloopstate *pv, int u, int v, int *u_vertices, int *v_vertices)
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{
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int r, total, below;
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if (pv[u].bridge != v)
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return false;
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r = pv[u].component_root;
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total = pv[r].maxindex - pv[r].minindex + 1;
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below = pv[u].maxindex - pv[u].minindex + 1;
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if (u_vertices)
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*u_vertices = below;
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if (v_vertices)
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*v_vertices = total - below;
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return true;
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}
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bool findloop_is_bridge(
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struct findloopstate *pv, int u, int v, int *u_vertices, int *v_vertices)
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{
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return (findloop_is_bridge_oneway(pv, u, v, u_vertices, v_vertices) ||
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findloop_is_bridge_oneway(pv, v, u, v_vertices, u_vertices));
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}
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bool findloop_run(struct findloopstate *pv, int nvertices,
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neighbour_fn_t neighbour, void *ctx)
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{
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int u, v, w, root, index;
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int nbridges, nedges;
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root = nvertices;
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/*
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* First pass: organise the graph into a rooted spanning forest.
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* That is, a tree structure with a clear up/down orientation -
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* every node has exactly one parent (which may be 'root') and
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* zero or more children, and every parent-child link corresponds
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* to a graph edge.
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*
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* (A side effect of this is to find all the connected components,
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* which of course we could do less confusingly with a dsf - but
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* then we'd have to do that *and* build the tree, so it's less
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* effort to do it all at once.)
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*/
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for (v = 0; v <= nvertices; v++) {
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pv[v].parent = root;
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pv[v].child = -2;
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pv[v].sibling = -1;
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pv[v].visited = false;
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}
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pv[root].child = -1;
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nedges = 0;
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debug(("------------- new find_loops, nvertices=%d\n", nvertices));
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for (v = 0; v < nvertices; v++) {
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if (pv[v].parent == root) {
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/*
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* Found a new connected component. Enumerate and treeify
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* it.
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*/
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pv[v].sibling = pv[root].child;
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pv[root].child = v;
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pv[v].component_root = v;
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debug(("%d is new child of root\n", v));
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u = v;
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while (1) {
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if (!pv[u].visited) {
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pv[u].visited = true;
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/*
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* Enumerate the neighbours of u, and any that are
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* as yet not in the tree structure (indicated by
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* child==-2, and distinct from the 'visited'
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* flag) become children of u.
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*/
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debug((" component pass: processing %d\n", u));
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for (w = neighbour(u, ctx); w >= 0;
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w = neighbour(-1, ctx)) {
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debug((" edge %d-%d\n", u, w));
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if (pv[w].child == -2) {
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debug((" -> new child\n"));
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pv[w].child = -1;
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pv[w].sibling = pv[u].child;
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pv[w].parent = u;
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pv[w].component_root = pv[u].component_root;
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pv[u].child = w;
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}
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/* While we're here, count the edges in the whole
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* graph, so that we can easily check at the end
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* whether all of them are bridges, i.e. whether
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* no loop exists at all. */
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if (w > u) /* count each edge only in one direction */
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nedges++;
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}
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/*
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* Now descend in depth-first search.
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*/
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if (pv[u].child >= 0) {
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u = pv[u].child;
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debug((" descending to %d\n", u));
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continue;
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}
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}
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if (u == v) {
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debug((" back at %d, done this component\n", u));
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break;
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} else if (pv[u].sibling >= 0) {
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u = pv[u].sibling;
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debug((" sideways to %d\n", u));
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} else {
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u = pv[u].parent;
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debug((" ascending to %d\n", u));
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}
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}
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}
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}
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/*
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* Second pass: index all the vertices in such a way that every
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* subtree has a contiguous range of indices. (Easily enough done,
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* by iterating through the tree structure we just built and
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* numbering its elements as if they were those of a sorted list.)
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*
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* For each vertex, we compute the min and max index of the
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* subtree starting there.
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*
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* (We index the vertices in preorder, per Tarjan's original
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* description, so that each vertex's min subtree index is its own
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* index; but that doesn't actually matter; either way round would
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* do. The important thing is that we have a simple arithmetic
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* criterion that tells us whether a vertex is in a given subtree
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* or not.)
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*/
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debug(("--- begin indexing pass\n"));
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index = 0;
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for (v = 0; v < nvertices; v++)
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pv[v].visited = false;
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pv[root].visited = true;
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u = pv[root].child;
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while (1) {
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if (!pv[u].visited) {
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pv[u].visited = true;
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/*
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* Index this node.
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*/
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pv[u].minindex = pv[u].index = index;
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debug((" vertex %d <- index %d\n", u, index));
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index++;
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/*
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* Now descend in depth-first search.
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*/
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if (pv[u].child >= 0) {
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u = pv[u].child;
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debug((" descending to %d\n", u));
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continue;
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}
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}
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if (u == root) {
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debug((" back at %d, done indexing\n", u));
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break;
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}
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/*
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* As we re-ascend to here from its children (or find that we
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* had no children to descend to in the first place), fill in
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* its maxindex field.
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*/
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pv[u].maxindex = index-1;
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debug((" vertex %d <- maxindex %d\n", u, pv[u].maxindex));
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if (pv[u].sibling >= 0) {
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u = pv[u].sibling;
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debug((" sideways to %d\n", u));
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} else {
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u = pv[u].parent;
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debug((" ascending to %d\n", u));
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}
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}
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/*
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* We're ready to generate output now, so initialise the output
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* fields.
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*/
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for (v = 0; v < nvertices; v++)
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pv[v].bridge = -1;
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/*
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* Final pass: determine the min and max index of the vertices
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* reachable from every subtree, not counting the link back to
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* each vertex's parent. Then our criterion is: given a vertex u,
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* defining a subtree consisting of u and all its descendants, we
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* compare the range of vertex indices _in_ that subtree (which is
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* just the minindex and maxindex of u) with the range of vertex
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* indices in the _neighbourhood_ of the subtree (computed in this
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* final pass, and not counting u's own edge to its parent), and
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* if the latter includes anything outside the former, then there
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* must be some path from u to outside its subtree which does not
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* go through the parent edge - i.e. the edge from u to its parent
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* is part of a loop.
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*/
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debug(("--- begin min-max pass\n"));
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nbridges = 0;
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for (v = 0; v < nvertices; v++)
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pv[v].visited = false;
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u = pv[root].child;
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pv[root].visited = true;
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while (1) {
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if (!pv[u].visited) {
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pv[u].visited = true;
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/*
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* Look for vertices reachable directly from u, including
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* u itself.
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*/
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debug((" processing vertex %d\n", u));
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pv[u].minreachable = pv[u].maxreachable = pv[u].minindex;
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for (w = neighbour(u, ctx); w >= 0; w = neighbour(-1, ctx)) {
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debug((" edge %d-%d\n", u, w));
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if (w != pv[u].parent) {
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int i = pv[w].index;
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if (pv[u].minreachable > i)
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pv[u].minreachable = i;
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if (pv[u].maxreachable < i)
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pv[u].maxreachable = i;
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}
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}
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debug((" initial min=%d max=%d\n",
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pv[u].minreachable, pv[u].maxreachable));
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/*
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* Now descend in depth-first search.
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*/
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if (pv[u].child >= 0) {
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u = pv[u].child;
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debug((" descending to %d\n", u));
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continue;
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}
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}
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if (u == root) {
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debug((" back at %d, done min-maxing\n", u));
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break;
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}
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/*
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* As we re-ascend to this vertex, go back through its
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* immediate children and do a post-update of its min/max.
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*/
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for (v = pv[u].child; v >= 0; v = pv[v].sibling) {
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if (pv[u].minreachable > pv[v].minreachable)
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pv[u].minreachable = pv[v].minreachable;
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if (pv[u].maxreachable < pv[v].maxreachable)
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pv[u].maxreachable = pv[v].maxreachable;
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}
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debug((" postorder update of %d: min=%d max=%d (indices %d-%d)\n", u,
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pv[u].minreachable, pv[u].maxreachable,
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pv[u].minindex, pv[u].maxindex));
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/*
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* And now we know whether each to our own parent is a bridge.
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*/
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if ((v = pv[u].parent) != root) {
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if (pv[u].minreachable >= pv[u].minindex &&
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pv[u].maxreachable <= pv[u].maxindex) {
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/* Yes, it's a bridge. */
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pv[u].bridge = v;
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nbridges++;
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debug((" %d-%d is a bridge\n", v, u));
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} else {
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debug((" %d-%d is not a bridge\n", v, u));
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}
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}
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if (pv[u].sibling >= 0) {
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u = pv[u].sibling;
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debug((" sideways to %d\n", u));
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} else {
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u = pv[u].parent;
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debug((" ascending to %d\n", u));
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}
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}
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debug(("finished, nedges=%d nbridges=%d\n", nedges, nbridges));
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/*
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* Done.
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*/
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return nbridges < nedges;
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}
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