zakarya/puzzles
1
0
Fork 0
mirror of git://git.tartarus.org/simon/puzzles.git synced 2025-04-21 08:01:30 -07:00
Code Issues Packages Projects Releases Wiki Activity

New centralised loop-finder, using Tarjan's algorithm.

Browse Source 
In the course of another recent project I had occasion to read up on
Tarjan's bridge-finding algorithm. This analyses an arbitrary graph
and finds 'bridges', i.e. edges whose removal would increase the
number of connected components. This is precisely the dual problem to
error-highlighting loops in games like Slant that don't permit them,
because an edge is part of some loop if and only if it is not a
bridge.

Having understood Tarjan's algorithm, it seemed like a good idea to
actually implement it for use in these puzzles, because we've got a
long and dishonourable history of messing up the loop detection in
assorted ways and I thought it would be nice to have an actually
reliable approach without any lurking time bombs. (That history is
chronicled in a long comment at the bottom of the new source file, if
anyone is interested.)

So, findloop.c is a new piece of reusable library code. You run it
over a graph, which you provide in the form of a vertex count and a
callback function to iterate over the neighbours of each vertex, and
it fills in a data structure which you can then query to find out
whether any given edge is part of a loop in the graph or not.
This commit is contained in:
Simon Tatham
2016-02-24 18:57:03 +00:00
parent 4a0d9ad39b
commit 1add03f7b8
2 changed files with 535 additions and 0 deletions
Download Patch File Download Diff File Expand all files Collapse all files

500
findloop.c Normal file
View File

@ -0,0 +1,500 @@
/*
* Routine for finding loops in graphs, reusable across multiple
* puzzles.
*
* The strategy is Tarjan's bridge-finding algorithm, which is
* designed to list all edges whose removal would disconnect a
* previously connected component of the graph. We're interested in
* exactly the reverse - edges that are part of a loop in the graph
* are precisely those which _wouldn't_ disconnect anything if removed
* (individually) - but of course flipping the sense of the output is
* easy.
*/
#include "puzzles.h"
struct findloopstate {
int parent, child, sibling, visited;
int index, minindex, maxindex;
int minreachable, maxreachable;
int bridge;
};
struct findloopstate *findloop_new_state(int nvertices)
{
/*
* Allocate a findloopstate structure for each vertex, and one
* extra one at the end which will be the overall root of a
* 'super-tree', which links the whole graph together to make it
* as easy as possible to iterate over all the connected
* components.
*/
return snewn(nvertices + 1, struct findloopstate);
}
void findloop_free_state(struct findloopstate *state)
{
sfree(state);
}
int findloop_is_loop_edge(struct findloopstate *pv, int u, int v)
{
/*
* Since the algorithm is intended for finding bridges, and a
* bridge must be part of any spanning tree, it follows that there
* is at most one bridge per vertex.
*
* Furthermore, by finding a _rooted_ spanning tree (so that each
* bridge is a parent->child link), you can find an injection from
* bridges to vertices (namely, map each bridge to the vertex at
* its child end).
*
* So if the u-v edge is a bridge, then either v was u's parent
* when the algorithm ran and we set pv[u].bridge = v, or vice
* versa.
*/
return !(pv[u].bridge == v || pv[v].bridge == u);
}
int findloop_run(struct findloopstate *pv, int nvertices,
neighbour_fn_t neighbour, void *ctx)
{
int u, v, w, root, index;
int nbridges, nedges;
root = nvertices;
/*
* First pass: organise the graph into a rooted spanning forest.
* That is, a tree structure with a clear up/down orientation -
* every node has exactly one parent (which may be 'root') and
* zero or more children, and every parent-child link corresponds
* to a graph edge.
*
* (A side effect of this is to find all the connected components,
* which of course we could do less confusingly with a dsf - but
* then we'd have to do that *and* build the tree, so it's less
* effort to do it all at once.)
*/
for (v = 0; v <= nvertices; v++) {
pv[v].parent = root;
pv[v].child = -2;
pv[v].sibling = -1;
pv[v].visited = FALSE;
}
pv[root].child = -1;
nedges = 0;
debug(("------------- new find_loops, nvertices=%d\n", nvertices));
for (v = 0; v < nvertices; v++) {
if (pv[v].parent == root) {
/*
* Found a new connected component. Enumerate and treeify
* it.
*/
pv[v].sibling = pv[root].child;
pv[root].child = v;
debug(("%d is new child of root\n", v));
u = v;
while (1) {
if (!pv[u].visited) {
pv[u].visited = TRUE;
/*
* Enumerate the neighbours of u, and any that are
* as yet not in the tree structure (indicated by
* child==-2, and distinct from the 'visited'
* flag) become children of u.
*/
debug((" component pass: processing %d\n", u));
for (w = neighbour(u, ctx); w >= 0;
w = neighbour(-1, ctx)) {
debug((" edge %d-%d\n", u, w));
if (pv[w].child == -2) {
debug((" -> new child\n"));
pv[w].child = -1;
pv[w].sibling = pv[u].child;
pv[w].parent = u;
pv[u].child = w;
}
/* While we're here, count the edges in the whole
* graph, so that we can easily check at the end
* whether all of them are bridges, i.e. whether
* no loop exists at all. */
if (w > u) /* count each edge only in one direction */
nedges++;
}
/*
* Now descend in depth-first search.
*/
if (pv[u].child >= 0) {
u = pv[u].child;
debug((" descending to %d\n", u));
continue;
}
}
if (u == v) {
debug((" back at %d, done this component\n", u));
break;
} else if (pv[u].sibling >= 0) {
u = pv[u].sibling;
debug((" sideways to %d\n", u));
} else {
u = pv[u].parent;
debug((" ascending to %d\n", u));
}
}
}
}
/*
* Second pass: index all the vertices in such a way that every
* subtree has a contiguous range of indices. (Easily enough done,
* by iterating through the tree structure we just built and
* numbering its elements as if they were those of a sorted list.)
*
* For each vertex, we compute the min and max index of the
* subtree starting there.
*
* (We index the vertices in preorder, per Tarjan's original
* description, so that each vertex's min subtree index is its own
* index; but that doesn't actually matter; either way round would
* do. The important thing is that we have a simple arithmetic
* criterion that tells us whether a vertex is in a given subtree
* or not.)
*/
debug(("--- begin indexing pass\n"));
index = 0;
for (v = 0; v < nvertices; v++)
pv[v].visited = FALSE;
pv[root].visited = TRUE;
u = pv[root].child;
while (1) {
if (!pv[u].visited) {
pv[u].visited = TRUE;
/*
* Index this node.
*/
pv[u].minindex = pv[u].index = index;
debug((" vertex %d <- index %d\n", u, index));
index++;
/*
* Now descend in depth-first search.
*/
if (pv[u].child >= 0) {
u = pv[u].child;
debug((" descending to %d\n", u));
continue;
}
}
if (u == root) {
debug((" back at %d, done indexing\n", u));
break;
}
/*
* As we re-ascend to here from its children (or find that we
* had no children to descend to in the first place), fill in
* its maxindex field.
*/
pv[u].maxindex = index-1;
debug((" vertex %d <- maxindex %d\n", u, pv[u].maxindex));
if (pv[u].sibling >= 0) {
u = pv[u].sibling;
debug((" sideways to %d\n", u));
} else {
u = pv[u].parent;
debug((" ascending to %d\n", u));
}
}
/*
* We're ready to generate output now, so initialise the output
* fields.
*/
for (v = 0; v < nvertices; v++)
pv[v].bridge = -1;
/*
* Final pass: determine the min and max index of the vertices
* reachable from every subtree, not counting the link back to
* each vertex's parent. Then our criterion is: given a vertex u,
* defining a subtree consisting of u and all its descendants, we
* compare the range of vertex indices _in_ that subtree (which is
* just the minindex and maxindex of u) with the range of vertex
* indices in the _neighbourhood_ of the subtree (computed in this
* final pass, and not counting u's own edge to its parent), and
* if the latter includes anything outside the former, then there
* must be some path from u to outside its subtree which does not
* go through the parent edge - i.e. the edge from u to its parent
* is part of a loop.
*/
debug(("--- begin min-max pass\n"));
nbridges = 0;
for (v = 0; v < nvertices; v++)
pv[v].visited = FALSE;
u = pv[root].child;
pv[root].visited = TRUE;
while (1) {
if (!pv[u].visited) {
pv[u].visited = TRUE;
/*
* Look for vertices reachable directly from u, including
* u itself.
*/
debug((" processing vertex %d\n", u));
pv[u].minreachable = pv[u].maxreachable = pv[u].minindex;
for (w = neighbour(u, ctx); w >= 0; w = neighbour(-1, ctx)) {
debug((" edge %d-%d\n", u, w));
if (w != pv[u].parent) {
int i = pv[w].index;
if (pv[u].minreachable > i)
pv[u].minreachable = i;
if (pv[u].maxreachable < i)
pv[u].maxreachable = i;
}
}
debug((" initial min=%d max=%d\n",
pv[u].minreachable, pv[u].maxreachable));
/*
* Now descend in depth-first search.
*/
if (pv[u].child >= 0) {
u = pv[u].child;
debug((" descending to %d\n", u));
continue;
}
}
if (u == root) {
debug((" back at %d, done min-maxing\n", u));
break;
}
/*
* As we re-ascend to this vertex, go back through its
* immediate children and do a post-update of its min/max.
*/
for (v = pv[u].child; v >= 0; v = pv[v].sibling) {
if (pv[u].minreachable > pv[v].minreachable)
pv[u].minreachable = pv[v].minreachable;
if (pv[u].maxreachable < pv[v].maxreachable)
pv[u].maxreachable = pv[v].maxreachable;
}
debug((" postorder update of %d: min=%d max=%d (indices %d-%d)\n", u,
pv[u].minreachable, pv[u].maxreachable,
pv[u].minindex, pv[u].maxindex));
/*
* And now we know whether each to our own parent is a bridge.
*/
if ((v = pv[u].parent) != root) {
if (pv[u].minreachable >= pv[u].minindex &&
pv[u].maxreachable <= pv[u].maxindex) {
/* Yes, it's a bridge. */
pv[u].bridge = v;
nbridges++;
debug((" %d-%d is a bridge\n", v, u));
} else {
debug((" %d-%d is not a bridge\n", v, u));
}
}
if (pv[u].sibling >= 0) {
u = pv[u].sibling;
debug((" sideways to %d\n", u));
} else {
u = pv[u].parent;
debug((" ascending to %d\n", u));
}
}
debug(("finished, nedges=%d nbridges=%d\n", nedges, nbridges));
/*
* Done.
*/
return nbridges < nedges;
}
/*
* Appendix: the long and painful history of loop detection in these puzzles
* =========================================================================
*
* For interest, I thought I'd write up the five loop-finding methods
* I've gone through before getting to this algorithm. It's a case
* study in all the ways you can solve this particular problem
* wrongly, and also how much effort you can waste by not managing to
* find the existing solution in the literature :-(
*
* Vertex dsf
* ----------
*
* Initially, in puzzles where you need to not have any loops in the
* solution graph, I detected them by using a dsf to track connected
* components of vertices. Iterate over each edge unifying the two
* vertices it connects; but before that, check if the two vertices
* are _already_ known to be connected. If so, then the new edge is
* providing a second path between them, i.e. a loop exists.
*
* That's adequate for automated solvers, where you just need to know
* _whether_ a loop exists, so as to rule out that move and do
* something else. But during play, you want to do better than that:
* you want to _point out_ the loops with error highlighting.
*
* Graph pruning
* -------------
*
* So my second attempt worked by iteratively pruning the graph. Find
* a vertex with degree 1; remove that edge; repeat until you can't
* find such a vertex any more. This procedure will remove *every*
* edge of the graph if and only if there were no loops; so if there
* are any edges remaining, highlight them.
*
* This successfully highlights loops, but not _only_ loops. If the
* graph contains a 'dumb-bell' shaped subgraph consisting of two
* loops connected by a path, then we'll end up highlighting the
* connecting path as well as the loops. That's not what we wanted.
*
* Vertex dsf with ad-hoc loop tracing
* -----------------------------------
*
* So my third attempt was to go back to the dsf strategy, only this
* time, when you detect that a particular edge connects two
* already-connected vertices (and hence is part of a loop), you try
* to trace round that loop to highlight it - before adding the new
* edge, search for a path between its endpoints among the edges the
* algorithm has already visited, and when you find one (which you
* must), highlight the loop consisting of that path plus the new
* edge.
*
* This solves the dumb-bell problem - we definitely now cannot
* accidentally highlight any edge that is *not* part of a loop. But
* it's far from clear that we'll highlight *every* edge that *is*
* part of a loop - what if there were multiple paths between the two
* vertices? It would be difficult to guarantee that we'd always catch
* every single one.
*
* On the other hand, it is at least guaranteed that we'll highlight
* _something_ if any loop exists, and in other error highlighting
* situations (see in particular the Tents connected component
* analysis) I've been known to consider that sufficient. So this
* version hung around for quite a while, until I had a better idea.
*
* Face dsf
* --------
*
* Round about the time Loopy was being revamped to include non-square
* grids, I had a much cuter idea, making use of the fact that the
* graph is planar, and hence has a concept of faces.
*
* In Loopy, there are really two graphs: the 'grid', consisting of
* all the edges that the player *might* fill in, and the solution
* graph of the edges the player actually *has* filled in. The
* algorithm is: set up a dsf on the *faces* of the grid. Iterate over
* each edge of the grid which is _not_ marked by the player as an
* edge of the solution graph, unifying the faces on either side of
* that edge. This groups the faces into connected components. Now,
* there is more than one connected component iff a loop exists, and
* moreover, an edge of the solution graph is part of a loop iff the
* faces on either side of it are in different connected components!
*
* This is the first algorithm I came up with that I was confident
* would successfully highlight exactly the correct set of edges in
* all cases. It's also conceptually elegant, and very easy to
* implement and to be confident you've got it right (since it just
* consists of two very simple loops over the edge set, one building
* the dsf and one reading it off). I was very pleased with it.
*
* Doing the same thing in Slant is slightly more difficult because
* the set of edges the user can fill in do not form a planar graph
* (the two potential edges in each square cross in the middle). But
* you can still apply the same principle by considering the 'faces'
* to be diamond-shaped regions of space around each horizontal or
* vertical grid line. Equivalently, pretend each edge added by the
* player is really divided into two edges, each from a square-centre
* to one of the square's corners, and now the grid graph is planar
* again.
*
* However, it fell down when - much later - I tried to implement the
* same algorithm in Net.
*
* Net doesn't *absolutely need* loop detection, because of its system
* of highlighting squares connected to the source square: an argument
* involving counting vertex degrees shows that if any loop exists,
* then it must be counterbalanced by some disconnected square, so
* there will be _some_ error highlight in any invalid grid even
* without loop detection. However, in large complicated cases, it's
* still nice to highlight the loop itself, so that once the player is
* clued in to its existence by a disconnected square elsewhere, they
* don't have to spend forever trying to find it.
*
* The new wrinkle in Net, compared to other loop-disallowing puzzles,
* is that it can be played with wrapping walls, or - topologically
* speaking - on a torus. And a torus has a property that algebraic
* topologists would know of as a 'non-trivial H_1 homology group',
* which essentially means that there can exist a loop on a torus
* which *doesn't* separate the surface into two regions disconnected
* from each other.
*
* In other words, using this algorithm in Net will do fine at finding
* _small_ localised loops, but a large-scale loop that goes (say) off
* the top of the grid, back on at the bottom, and meets up in the
* middle again will not be detected.
*
* Footpath dsf
* ------------
*
* To solve this homology problem in Net, I hastily thought up another
* dsf-based algorithm.
*
* This time, let's consider each edge of the graph to be a road, with
* a separate pedestrian footpath down each side. We'll form a dsf on
* those imaginary segments of footpath.
*
* At each vertex of the graph, we go round the edges leaving that
* vertex, in order around the vertex. For each pair of edges adjacent
* in this order, we unify their facing pair of footpaths (e.g. if
* edge E appears anticlockwise of F, then we unify the anticlockwise
* footpath of F with the clockwise one of E) . In particular, if a
* vertex has degree 1, then the two footpaths on either side of its
* single edge are unified.
*
* Then, an edge is part of a loop iff its two footpaths are not
* reachable from one another.
*
* This algorithm is almost as simple to implement as the face dsf,
* and it works on a wider class of graphs embedded in plane-like
* surfaces; in particular, it fixes the torus bug in the face-dsf
* approach. However, it still depends on the graph having _some_ sort
* of embedding in a 2-manifold, because it relies on there being a
* meaningful notion of 'order of edges around a vertex' in the first
* place, so you couldn't use it on a wildly nonplanar graph like the
* diamond lattice. Also, more subtly, it depends on the graph being
* embedded in an _orientable_ surface - and that's a thing that might
* much more plausibly change in future puzzles, because it's not at
* all unlikely that at some point I might feel moved to implement a
* puzzle that can be played on the surface of a Mobius strip or a
* Klein bottle. And then even this algorithm won't work.
*
* Tarjan's bridge-finding algorithm
* ---------------------------------
*
* And so, finally, we come to the algorithm above. This one is pure
* graph theory: it doesn't depend on any concept of 'faces', or 'edge
* ordering around a vertex', or any other trapping of a planar or
* quasi-planar graph embedding. It should work on any graph
* whatsoever, and reliably identify precisely the set of edges that
* form part of some loop. So *hopefully* this long string of failures
* has finally come to an end...
*/

35
puzzles.h
View File

@ -467,6 +467,41 @@ void free_combi(combi_ctx *combi);
/* divides w*h rectangle into pieces of size k. Returns w*h dsf. */
int *divvy_rectangle(int w, int h, int k, random_state *rs);
/*
* findloop.c
*/
struct findloopstate;
struct findloopstate *findloop_new_state(int nvertices);
void findloop_free_state(struct findloopstate *);
/*
* Callback provided by the client code to enumerate the graph
* vertices joined directly to a given vertex.
*
* Semantics: if vertex >= 0, return one of its neighbours; if vertex
* < 0, return a previously unmentioned neighbour of whatever vertex
* was last passed as input. Write to 'ctx' as necessary to store
* state. In either case, return < 0 if no such vertex can be found.
*/
typedef int (*neighbour_fn_t)(int vertex, void *ctx);
/*
* Actual function to find loops. 'ctx' will be passed unchanged to
* the 'neighbour' function to query graph edges. Returns FALSE if no
* loop was found, or TRUE if one was.
*/
int findloop_run(struct findloopstate *state, int nvertices,
neighbour_fn_t neighbour, void *ctx);
/*
* Query whether an edge is part of a loop, in the output of
* find_loops.
*
* Due to the internal storage format, if you pass u,v which are not
* connected at all, the output will be TRUE. (The algorithm actually
* stores an exhaustive list of *non*-loop edges, because there are
* fewer of those, so really it's querying whether the edge is on that
* list.)
*/
int findloop_is_loop_edge(struct findloopstate *state, int u, int v);
/*
* Data structure containing the function calls and data specific
* to a particular game. This is enclosed in a data structure so