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