This document describes an algorithm for determining if two trees are isomorphic by hashing the trees. It involves recursively hashing all subtrees, sorting the hashes to ensure uniqueness, and then hashing the overall tree. The time complexity is O(NlogN) where N is the number of vertices. Pseudocode and a C++ implementation are provided. The document also discusses representing trees structurally and handling unrooted trees by trying all possible roots.

1. Tree Isomorphism
郭至軒（KuoE0）
KuoE0.tw@gmail.com
KuoE0.ch

2. Isomorphism
The structures of two trees are equal.

3. so abstract...

4. 7 7
6 1 2 1 2
5 4 3 4 5 3
isomorphism 4 5 6
3 1
2 7 6

5. How to judge two rooted
tree are isomorphic?

7. If the trees are isomorphic, all their sub-trees
are also isomorphic.

8. If the trees are isomorphic, all their sub-trees
are also isomorphic.

9. Hash the tree

10. Hash all sub-tree
recursively

11. single vertex
191 191 191
initial value = 191
191

12. 29247
two-level sub-tree
191 29247 191 191
child = (191)
191
(191 × 701 xor 191) mod 34943 = 29247

13. 33360 29247
three-level sub-tree
191 29247 191 191
child = (191,29247,191)
sort
191 child = (191,191,29247)
(((((191 × 701 xor 191) mod 34943) × 701 xor 191) mod
34943) × 701 xor 29247) mod 34943 = 33360

14. 4687
33360 29247
the total tree
191 29247 191 191
child = (33360,29247)
sort
191 child = (29247,33360)
(((191 × 701 xor 29247) mod 34943) × 701 xor 33360)
mod 34943 = 4687

15. hash value of
the tree is 4687

16. single vertex
initial value = 191 191 191 191
191

17. 29247
two-level sub-tree
child = (191) 191 191 191 29247
191
(191 × 701 xor 191) mod 34943 = 29247

18. 29247 33360
three-level sub-tree
child = (191,191,29247) 191 191 191 29247
sort
child = (191,191,29247) 191
(((((191 × 701 xor 191) mod 34943) × 701 xor 191) mod
34943) × 701 xor 29247) mod 34943 = 33360

19. 4687
29247 33360
the total tree
child = (33360,29247) 191 191 191 29247
sort
child = (29247,33360) 191
(((191 × 701 xor 29247) mod 34943) × 701 xor 33360)
mod 34943 = 4687

20. hash value of
the tree is 4687

21. ≡

22. Algorithm
HASH_TREE(T):
1. hash all sub-trees
2. sort hash value of sub-trees (unique)
3. calculate hash value (any hash function)

23. Time Complexity
O(Nlog2N)
number of vertices
height of tree

24. Source Code
int hash( TREE &now, int root ) {
int value = INIT;
vector< int > sub;
//get all hash value of subtree
for ( int i = 0; i < now[ root ].size(); ++i )
sub.push_back( hash( now, now[ root ]
[ i ] ) );
//sort them to keep unique order
sort( sub.begin(), sub.end() );
//hash this this tree
for ( int i = 0; i < sub.size(); ++i )
value = ( ( value * P1 ) ^ sub[ i ] ) % P2;
return value % P2;
}

25. Representation of Tree
Let the height of left child less than the right one.

26. Representation of Tree
Let the height of left child less than the right one.

27. 4
3 2
1 2 1 1
1

28. 4 4
3 2 2 3
1 2 1 1 1 1 2 1
1 1

29. 4 4 4
3 2 2 3 2 3
1 2 1 1 1 1 2 1 1 1 1 2
1 1 1

30. Algorithm
SORT_CHILD(T):
1. sort all sub-trees
2. compare the height
3. if height is equal, compare child
recursively
4. put the lower at left and the higher at
right

31. How about unrooted tree?

32. find a root

33. eliminate leaves

34. eliminate leaves

35. eliminate leaves

36. eliminate leaves

37. try each root

38. rebuild the tree

39. rebuild the tree

40. rebuild the tree

41. rebuild the tree

42. rebuild the tree

43. rebuild the tree

44. apply the isomorphism detection

45. Practice Now
POJ 1635 - Subway tree systems

46. Thank You for Your
Listening.