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    <p>This happens when constructing oBDDs. The Idea is: put the tree into a canonical form, and construct a hashtable with an entry for every node. Hash function is a function of the node + the hash functions for the left/right child nodes. Complexity is O(N), but only if one can rely on the hashvalues being unique. The final compare (e.g. for Resolving collisions) will still cost o(N*N) for the recursive subtree &lt;--> subtree compare. <a href="http://en.wikipedia.org/wiki/Binary_decision_diagram" rel="nofollow">More on BDDs</a> or <a href="http://www.cs.cmu.edu/~bryant/pubdir/ieeetc86.pdf" rel="nofollow">the original Bryant paper</a></p> <p>The hashfunction I currently use:</p> <pre><code>#define SHUFFLE(x,n) (((x) &lt;&lt; (n))|((x) &gt;&gt;(32-(n)))) /* a node's hashvalue is based on its value * and (recursively) on it's children's hashvalues. */ #define NODE_HASH2(l,r) ((SHUFFLE((l),5)^SHUFFLE((r),9))) #define NODE_HASH3(v,l,r) ((0x54321u*(v) ^ NODE_HASH2((l),(r)))) </code></pre> <p>Typical usage:</p> <pre><code>void node_sethash(NodeNum num) { if (NODE_IS_NULL(num)) return; if (NODE_IS_TERMINAL(num)) switch (nodes[num].var) { case 0: nodes[num].hash.hash= HASH_FALSE; break; case 1: nodes[num].hash.hash= HASH_TRUE; break; case 2: nodes[num].hash.hash= HASH_FALSE^HASH_TRUE; break; } else if (NODE_IS_NAMED(num)) { NodeNum f,t; f = nodes[num].negative; t = nodes[num].positive; nodes[num].hash.hash = NODE_HASH3 (nodes[num].var, nodes[f].hash.hash, nodes[t].hash.hash); } return ; } </code></pre> <p>Searching the hash table:</p> <pre><code>NodeNum *hash_hnd(NodeNum num, int want_exact) { unsigned slot; NodeNum *ptr, this; if (NODE_IS_NULL(num)) return NULL; slot = nodes[num].hash.hash % COUNTOF(hash_nodes); for (ptr = &amp;hash_nodes[slot]; !NODE_IS_NULL(this= *ptr); ptr = &amp;nodes[this].hash.link) { if (this == num) break; if (want_exact) continue; if (nodes[this].hash.hash != nodes[num].hash.hash) continue; if (nodes[this].var != nodes[num].var) continue; if (node_compare( nodes[this].negative , nodes[num].negative)) continue; if (node_compare( nodes[this].positive , nodes[num].positive)) continue; /* duplicate node := same var+same children */ break; } return ptr; } </code></pre> <p>The recursive compare function:</p> <pre><code>int node_compare(NodeNum one, NodeNum two) { int rc; if (one == two) return 0; if (NODE_IS_NULL(one) &amp;&amp; NODE_IS_NULL(two)) return 0; if (NODE_IS_NULL(one) &amp;&amp; !NODE_IS_NULL(two)) return -1; if (!NODE_IS_NULL(one) &amp;&amp; NODE_IS_NULL(two)) return 1; if (NODE_IS_TERMINAL(one) &amp;&amp; !NODE_IS_TERMINAL(two)) return -1; if (!NODE_IS_TERMINAL(one) &amp;&amp; NODE_IS_TERMINAL(two)) return 1; if (VAR_RANK(nodes[one].var) &lt; VAR_RANK(nodes[two].var) ) return -1; if (VAR_RANK(nodes[one].var) &gt; VAR_RANK(nodes[two].var) ) return 1; rc = node_compare(nodes[one].negative,nodes[two].negative); if (rc) return rc; rc = node_compare(nodes[one].positive,nodes[two].positive); if (rc) return rc; return 0; } </code></pre>
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