| 1 | #include "muscle.h" |
|---|
| 2 | #include "tree.h" |
|---|
| 3 | |
|---|
| 4 | #define TRACE 0 |
|---|
| 5 | |
|---|
| 6 | /*** |
|---|
| 7 | Algorithm to compare two trees, X and Y. |
|---|
| 8 | |
|---|
| 9 | A node x in X and node y in Y are defined to be |
|---|
| 10 | similar iff the set of leaves in the subtree under |
|---|
| 11 | x is identical to the set of leaves under y. |
|---|
| 12 | |
|---|
| 13 | A node is defined to be dissimilar iff it is not |
|---|
| 14 | similar to any node in the other tree. |
|---|
| 15 | |
|---|
| 16 | Nodes x and y are defined to be married iff every |
|---|
| 17 | node in the subtree under x is similar to a node |
|---|
| 18 | in the subtree under y. Married nodes are considered |
|---|
| 19 | to be equal. The subtrees under two married nodes can |
|---|
| 20 | at most differ by exchanges of left and right branches, |
|---|
| 21 | which we do not consider to be significant here. |
|---|
| 22 | |
|---|
| 23 | A node is defined to be a bachelor iff it is not |
|---|
| 24 | married. If a node is a bachelor, then it has a |
|---|
| 25 | dissimilar node in its subtree, and it follows |
|---|
| 26 | immediately from the definition of marriage that its |
|---|
| 27 | parent is also a bachelor. Hence all nodes on the path |
|---|
| 28 | from a bachelor node to the root are bachelors. |
|---|
| 29 | |
|---|
| 30 | We assume the trees have the same set of leaves, so |
|---|
| 31 | every leaf is trivially both similar and married to |
|---|
| 32 | the same leaf in the opposite tree. Bachelor nodes |
|---|
| 33 | are therefore always internal (i.e., non-leaf) nodes. |
|---|
| 34 | |
|---|
| 35 | A node is defined to be a diff iff (a) it is married |
|---|
| 36 | and (b) its parent is a bachelor. The subtree under |
|---|
| 37 | a diff is maximally similar to the other tree. (In |
|---|
| 38 | other words, you cannot extend the subtree without |
|---|
| 39 | adding a bachelor). |
|---|
| 40 | |
|---|
| 41 | The set of diffs is the subset of the two trees that |
|---|
| 42 | we consider to be identical. |
|---|
| 43 | |
|---|
| 44 | Example: |
|---|
| 45 | |
|---|
| 46 | -----A |
|---|
| 47 | -----k |
|---|
| 48 | ----j -----B |
|---|
| 49 | --i -----C |
|---|
| 50 | ------D |
|---|
| 51 | |
|---|
| 52 | |
|---|
| 53 | -----A |
|---|
| 54 | -----p |
|---|
| 55 | ----n -----B |
|---|
| 56 | --m -----D |
|---|
| 57 | ------C |
|---|
| 58 | |
|---|
| 59 | |
|---|
| 60 | The following pairs of internal nodes are similar. |
|---|
| 61 | |
|---|
| 62 | Nodes Set of leaves |
|---|
| 63 | ----- ------------- |
|---|
| 64 | k,p A,B |
|---|
| 65 | i,m A,B,C,D |
|---|
| 66 | |
|---|
| 67 | Bachelors in the first tree are i and j, bachelors |
|---|
| 68 | in the second tree are m and n. |
|---|
| 69 | |
|---|
| 70 | Node k and p are married, but i and m are not (because j |
|---|
| 71 | and n are bachelors). The diffs are C, D and k. |
|---|
| 72 | |
|---|
| 73 | The set of bachelor nodes can be viewed as the internal |
|---|
| 74 | nodes of a tree, the leaves of which are diffs. (To see |
|---|
| 75 | that there can't be disjoint subtrees, note that the path |
|---|
| 76 | from a diff to a root is all bachelor nodes, so there is |
|---|
| 77 | always a path between two diffs that goes through the root). |
|---|
| 78 | We call this tree the "diffs tree". |
|---|
| 79 | |
|---|
| 80 | There is a simple O(N) algorithm to build the diffs tree. |
|---|
| 81 | To achieve O(N) we avoid traversing a given subtree multiple |
|---|
| 82 | times and also avoid comparing lists of leaves. |
|---|
| 83 | |
|---|
| 84 | We visit nodes in depth-first order (i.e., a node is visited |
|---|
| 85 | before its parent). |
|---|
| 86 | |
|---|
| 87 | If either child of a node is a bachelor, we flag it as |
|---|
| 88 | a bachelor. |
|---|
| 89 | |
|---|
| 90 | If both children of the node we are visiting are married, |
|---|
| 91 | we check whether the spouses of those children have the |
|---|
| 92 | same parent in the other tree. If the parents are different, |
|---|
| 93 | the current node is a bachelor. If they have the same parent, |
|---|
| 94 | then the node we are visiting is the spouse of that parent. |
|---|
| 95 | We assign this newly identified married couple a unique integer |
|---|
| 96 | id. The id of a node is in one-to-one correspondence with the |
|---|
| 97 | set of leaves in its subtree. Two nodes have the same set of |
|---|
| 98 | leaves iff they have the same id. Bachelor nodes do not get |
|---|
| 99 | an id. |
|---|
| 100 | ***/ |
|---|
| 101 | |
|---|
| 102 | static void BuildDiffs(const Tree &tree, unsigned uTreeNodeIndex, |
|---|
| 103 | const bool bIsDiff[], Tree &Diffs, unsigned uDiffsNodeIndex, |
|---|
| 104 | unsigned IdToDiffsLeafNodeIndex[]) |
|---|
| 105 | { |
|---|
| 106 | #if TRACE |
|---|
| 107 | Log("BuildDiffs(TreeNode=%u IsDiff=%d IsLeaf=%d)\n", |
|---|
| 108 | uTreeNodeIndex, bIsDiff[uTreeNodeIndex], tree.IsLeaf(uTreeNodeIndex)); |
|---|
| 109 | #endif |
|---|
| 110 | if (bIsDiff[uTreeNodeIndex]) |
|---|
| 111 | { |
|---|
| 112 | unsigned uLeafCount = tree.GetLeafCount(); |
|---|
| 113 | unsigned *Leaves = new unsigned[uLeafCount]; |
|---|
| 114 | GetLeaves(tree, uTreeNodeIndex, Leaves, &uLeafCount); |
|---|
| 115 | for (unsigned n = 0; n < uLeafCount; ++n) |
|---|
| 116 | { |
|---|
| 117 | const unsigned uLeafNodeIndex = Leaves[n]; |
|---|
| 118 | const unsigned uId = tree.GetLeafId(uLeafNodeIndex); |
|---|
| 119 | if (uId >= tree.GetLeafCount()) |
|---|
| 120 | Quit("BuildDiffs, id out of range"); |
|---|
| 121 | IdToDiffsLeafNodeIndex[uId] = uDiffsNodeIndex; |
|---|
| 122 | #if TRACE |
|---|
| 123 | Log(" Leaf id=%u DiffsNode=%u\n", uId, uDiffsNodeIndex); |
|---|
| 124 | #endif |
|---|
| 125 | } |
|---|
| 126 | delete[] Leaves; |
|---|
| 127 | return; |
|---|
| 128 | } |
|---|
| 129 | |
|---|
| 130 | if (tree.IsLeaf(uTreeNodeIndex)) |
|---|
| 131 | Quit("BuildDiffs: should never reach leaf"); |
|---|
| 132 | |
|---|
| 133 | const unsigned uTreeLeft = tree.GetLeft(uTreeNodeIndex); |
|---|
| 134 | const unsigned uTreeRight = tree.GetRight(uTreeNodeIndex); |
|---|
| 135 | |
|---|
| 136 | const unsigned uDiffsLeft = Diffs.AppendBranch(uDiffsNodeIndex); |
|---|
| 137 | const unsigned uDiffsRight = uDiffsLeft + 1; |
|---|
| 138 | |
|---|
| 139 | BuildDiffs(tree, uTreeLeft, bIsDiff, Diffs, uDiffsLeft, IdToDiffsLeafNodeIndex); |
|---|
| 140 | BuildDiffs(tree, uTreeRight, bIsDiff, Diffs, uDiffsRight, IdToDiffsLeafNodeIndex); |
|---|
| 141 | } |
|---|
| 142 | |
|---|
| 143 | void DiffTrees(const Tree &Tree1, const Tree &Tree2, Tree &Diffs, |
|---|
| 144 | unsigned IdToDiffsLeafNodeIndex[]) |
|---|
| 145 | { |
|---|
| 146 | #if TRACE |
|---|
| 147 | Log("Tree1:\n"); |
|---|
| 148 | Tree1.LogMe(); |
|---|
| 149 | Log("\n"); |
|---|
| 150 | Log("Tree2:\n"); |
|---|
| 151 | Tree2.LogMe(); |
|---|
| 152 | #endif |
|---|
| 153 | |
|---|
| 154 | if (!Tree1.IsRooted() || !Tree2.IsRooted()) |
|---|
| 155 | Quit("DiffTrees: requires rooted trees"); |
|---|
| 156 | |
|---|
| 157 | const unsigned uNodeCount = Tree1.GetNodeCount(); |
|---|
| 158 | const unsigned uNodeCount2 = Tree2.GetNodeCount(); |
|---|
| 159 | |
|---|
| 160 | const unsigned uLeafCount = Tree1.GetLeafCount(); |
|---|
| 161 | const unsigned uLeafCount2 = Tree2.GetLeafCount(); |
|---|
| 162 | assert(uLeafCount == uLeafCount2); |
|---|
| 163 | |
|---|
| 164 | if (uNodeCount != uNodeCount2) |
|---|
| 165 | Quit("DiffTrees: different node counts"); |
|---|
| 166 | |
|---|
| 167 | // Allocate tables so we can convert tree node index to |
|---|
| 168 | // and from the unique id with a O(1) lookup. |
|---|
| 169 | unsigned *NodeIndexToId1 = new unsigned[uNodeCount]; |
|---|
| 170 | unsigned *IdToNodeIndex2 = new unsigned[uNodeCount]; |
|---|
| 171 | |
|---|
| 172 | bool *bIsBachelor1 = new bool[uNodeCount]; |
|---|
| 173 | bool *bIsDiff1 = new bool[uNodeCount]; |
|---|
| 174 | |
|---|
| 175 | for (unsigned uNodeIndex = 0; uNodeIndex < uNodeCount; ++uNodeIndex) |
|---|
| 176 | { |
|---|
| 177 | NodeIndexToId1[uNodeIndex] = uNodeCount; |
|---|
| 178 | bIsBachelor1[uNodeIndex] = false; |
|---|
| 179 | bIsDiff1[uNodeIndex] = false; |
|---|
| 180 | |
|---|
| 181 | // Use uNodeCount as value meaning "not set". |
|---|
| 182 | IdToNodeIndex2[uNodeIndex] = uNodeCount; |
|---|
| 183 | } |
|---|
| 184 | |
|---|
| 185 | // Initialize node index <-> id lookup tables |
|---|
| 186 | for (unsigned uNodeIndex = 0; uNodeIndex < uNodeCount; ++uNodeIndex) |
|---|
| 187 | { |
|---|
| 188 | if (Tree1.IsLeaf(uNodeIndex)) |
|---|
| 189 | { |
|---|
| 190 | const unsigned uId = Tree1.GetLeafId(uNodeIndex); |
|---|
| 191 | if (uId >= uNodeCount) |
|---|
| 192 | Quit("Diff trees requires existing leaf ids in range 0 .. (N-1)"); |
|---|
| 193 | NodeIndexToId1[uNodeIndex] = uId; |
|---|
| 194 | } |
|---|
| 195 | |
|---|
| 196 | if (Tree2.IsLeaf(uNodeIndex)) |
|---|
| 197 | { |
|---|
| 198 | const unsigned uId = Tree2.GetLeafId(uNodeIndex); |
|---|
| 199 | if (uId >= uNodeCount) |
|---|
| 200 | Quit("Diff trees requires existing leaf ids in range 0 .. (N-1)"); |
|---|
| 201 | IdToNodeIndex2[uId] = uNodeIndex; |
|---|
| 202 | } |
|---|
| 203 | } |
|---|
| 204 | |
|---|
| 205 | // Validity check. This verifies that the ids |
|---|
| 206 | // pre-assigned to the leaves in Tree1 are unique |
|---|
| 207 | // (note that the id<N check above does not rule |
|---|
| 208 | // out two leaves having duplicate ids). |
|---|
| 209 | for (unsigned uId = 0; uId < uLeafCount; ++uId) |
|---|
| 210 | { |
|---|
| 211 | unsigned uNodeIndex2 = IdToNodeIndex2[uId]; |
|---|
| 212 | if (uNodeCount == uNodeIndex2) |
|---|
| 213 | Quit("DiffTrees, check 2"); |
|---|
| 214 | } |
|---|
| 215 | |
|---|
| 216 | // Ids assigned to internal nodes are N, N+1 ... |
|---|
| 217 | // An internal node id uniquely identifies a set |
|---|
| 218 | // of two or more leaves. |
|---|
| 219 | unsigned uInternalNodeId = uLeafCount; |
|---|
| 220 | |
|---|
| 221 | // Depth-first traversal of tree. |
|---|
| 222 | // The order guarantees that a node is visited before |
|---|
| 223 | // its parent is visited. |
|---|
| 224 | for (unsigned uNodeIndex1 = Tree1.FirstDepthFirstNode(); |
|---|
| 225 | NULL_NEIGHBOR != uNodeIndex1; |
|---|
| 226 | uNodeIndex1 = Tree1.NextDepthFirstNode(uNodeIndex1)) |
|---|
| 227 | { |
|---|
| 228 | #if TRACE |
|---|
| 229 | Log("Main loop: Node1=%u IsLeaf=%d IsBachelor=%d\n", |
|---|
| 230 | uNodeIndex1, |
|---|
| 231 | Tree1.IsLeaf(uNodeIndex1), |
|---|
| 232 | bIsBachelor1[uNodeIndex1]); |
|---|
| 233 | #endif |
|---|
| 234 | |
|---|
| 235 | // Leaves are trivial; nothing to do. |
|---|
| 236 | if (Tree1.IsLeaf(uNodeIndex1) || bIsBachelor1[uNodeIndex1]) |
|---|
| 237 | continue; |
|---|
| 238 | |
|---|
| 239 | // If either child is a bachelor, flag |
|---|
| 240 | // this node as a bachelor and continue. |
|---|
| 241 | unsigned uLeft1 = Tree1.GetLeft(uNodeIndex1); |
|---|
| 242 | if (bIsBachelor1[uLeft1]) |
|---|
| 243 | { |
|---|
| 244 | bIsBachelor1[uNodeIndex1] = true; |
|---|
| 245 | continue; |
|---|
| 246 | } |
|---|
| 247 | |
|---|
| 248 | unsigned uRight1 = Tree1.GetRight(uNodeIndex1); |
|---|
| 249 | if (bIsBachelor1[uRight1]) |
|---|
| 250 | { |
|---|
| 251 | bIsBachelor1[uNodeIndex1] = true; |
|---|
| 252 | continue; |
|---|
| 253 | } |
|---|
| 254 | |
|---|
| 255 | // Both children are married. |
|---|
| 256 | // Married nodes are guaranteed to have an id. |
|---|
| 257 | unsigned uIdLeft = NodeIndexToId1[uLeft1]; |
|---|
| 258 | unsigned uIdRight = NodeIndexToId1[uRight1]; |
|---|
| 259 | |
|---|
| 260 | if (uIdLeft == uNodeCount || uIdRight == uNodeCount) |
|---|
| 261 | Quit("DiffTrees, check 5"); |
|---|
| 262 | |
|---|
| 263 | // uLeft2 is the spouse of uLeft1, and similarly for uRight2. |
|---|
| 264 | unsigned uLeft2 = IdToNodeIndex2[uIdLeft]; |
|---|
| 265 | unsigned uRight2 = IdToNodeIndex2[uIdRight]; |
|---|
| 266 | |
|---|
| 267 | if (uLeft2 == uNodeCount || uRight2 == uNodeCount) |
|---|
| 268 | Quit("DiffTrees, check 6"); |
|---|
| 269 | |
|---|
| 270 | // If the spouses of uLeft1 and uRight1 have the same |
|---|
| 271 | // parent, then this parent is the spouse of uNodeIndex1. |
|---|
| 272 | // Otherwise, uNodeIndex1 is a diff. |
|---|
| 273 | unsigned uParentLeft2 = Tree2.GetParent(uLeft2); |
|---|
| 274 | unsigned uParentRight2 = Tree2.GetParent(uRight2); |
|---|
| 275 | |
|---|
| 276 | #if TRACE |
|---|
| 277 | Log("L1=%u R1=%u L2=%u R2=%u PL2=%u PR2=%u\n", |
|---|
| 278 | uLeft1, |
|---|
| 279 | uRight1, |
|---|
| 280 | uLeft2, |
|---|
| 281 | uRight2, |
|---|
| 282 | uParentLeft2, |
|---|
| 283 | uParentRight2); |
|---|
| 284 | #endif |
|---|
| 285 | |
|---|
| 286 | if (uParentLeft2 == uParentRight2) |
|---|
| 287 | { |
|---|
| 288 | NodeIndexToId1[uNodeIndex1] = uInternalNodeId; |
|---|
| 289 | IdToNodeIndex2[uInternalNodeId] = uParentLeft2; |
|---|
| 290 | ++uInternalNodeId; |
|---|
| 291 | } |
|---|
| 292 | else |
|---|
| 293 | bIsBachelor1[uNodeIndex1] = true; |
|---|
| 294 | } |
|---|
| 295 | |
|---|
| 296 | unsigned uDiffCount = 0; |
|---|
| 297 | for (unsigned uNodeIndex = 0; uNodeIndex < uNodeCount; ++uNodeIndex) |
|---|
| 298 | { |
|---|
| 299 | if (bIsBachelor1[uNodeIndex]) |
|---|
| 300 | continue; |
|---|
| 301 | if (Tree1.IsRoot(uNodeIndex)) |
|---|
| 302 | { |
|---|
| 303 | // Special case: if no bachelors, consider the |
|---|
| 304 | // root a diff. |
|---|
| 305 | if (!bIsBachelor1[uNodeIndex]) |
|---|
| 306 | bIsDiff1[uNodeIndex] = true; |
|---|
| 307 | continue; |
|---|
| 308 | } |
|---|
| 309 | const unsigned uParent = Tree1.GetParent(uNodeIndex); |
|---|
| 310 | if (bIsBachelor1[uParent]) |
|---|
| 311 | { |
|---|
| 312 | bIsDiff1[uNodeIndex] = true; |
|---|
| 313 | ++uDiffCount; |
|---|
| 314 | } |
|---|
| 315 | } |
|---|
| 316 | |
|---|
| 317 | #if TRACE |
|---|
| 318 | Log("Tree1:\n"); |
|---|
| 319 | Log("Node Id Bach Diff Name\n"); |
|---|
| 320 | Log("---- ---- ---- ---- ----\n"); |
|---|
| 321 | for (unsigned n = 0; n < uNodeCount; ++n) |
|---|
| 322 | { |
|---|
| 323 | Log("%4u %4u %d %d", |
|---|
| 324 | n, |
|---|
| 325 | NodeIndexToId1[n], |
|---|
| 326 | bIsBachelor1[n], |
|---|
| 327 | bIsDiff1[n]); |
|---|
| 328 | if (Tree1.IsLeaf(n)) |
|---|
| 329 | Log(" %s", Tree1.GetLeafName(n)); |
|---|
| 330 | Log("\n"); |
|---|
| 331 | } |
|---|
| 332 | Log("\n"); |
|---|
| 333 | Log("Tree2:\n"); |
|---|
| 334 | Log("Node Id Name\n"); |
|---|
| 335 | Log("---- ---- ----\n"); |
|---|
| 336 | for (unsigned n = 0; n < uNodeCount; ++n) |
|---|
| 337 | { |
|---|
| 338 | Log("%4u ", n); |
|---|
| 339 | if (Tree2.IsLeaf(n)) |
|---|
| 340 | Log(" %s", Tree2.GetLeafName(n)); |
|---|
| 341 | Log("\n"); |
|---|
| 342 | } |
|---|
| 343 | #endif |
|---|
| 344 | |
|---|
| 345 | Diffs.CreateRooted(); |
|---|
| 346 | const unsigned uDiffsRootIndex = Diffs.GetRootNodeIndex(); |
|---|
| 347 | const unsigned uRootIndex1 = Tree1.GetRootNodeIndex(); |
|---|
| 348 | |
|---|
| 349 | for (unsigned n = 0; n < uLeafCount; ++n) |
|---|
| 350 | IdToDiffsLeafNodeIndex[n] = uNodeCount; |
|---|
| 351 | |
|---|
| 352 | BuildDiffs(Tree1, uRootIndex1, bIsDiff1, Diffs, uDiffsRootIndex, |
|---|
| 353 | IdToDiffsLeafNodeIndex); |
|---|
| 354 | |
|---|
| 355 | #if TRACE |
|---|
| 356 | Log("\n"); |
|---|
| 357 | Log("Diffs:\n"); |
|---|
| 358 | Diffs.LogMe(); |
|---|
| 359 | Log("\n"); |
|---|
| 360 | Log("IdToDiffsLeafNodeIndex:"); |
|---|
| 361 | for (unsigned n = 0; n < uLeafCount; ++n) |
|---|
| 362 | { |
|---|
| 363 | if (n%16 == 0) |
|---|
| 364 | Log("\n"); |
|---|
| 365 | else |
|---|
| 366 | Log(" "); |
|---|
| 367 | Log("%u=%u", n, IdToDiffsLeafNodeIndex[n]); |
|---|
| 368 | } |
|---|
| 369 | Log("\n"); |
|---|
| 370 | #endif |
|---|
| 371 | |
|---|
| 372 | for (unsigned n = 0; n < uLeafCount; ++n) |
|---|
| 373 | if (IdToDiffsLeafNodeIndex[n] == uNodeCount) |
|---|
| 374 | Quit("TreeDiffs check 7"); |
|---|
| 375 | |
|---|
| 376 | delete[] NodeIndexToId1; |
|---|
| 377 | delete[] IdToNodeIndex2; |
|---|
| 378 | |
|---|
| 379 | delete[] bIsBachelor1; |
|---|
| 380 | delete[] bIsDiff1; |
|---|
| 381 | } |
|---|