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Union-find

Union-find. Union-find. Maintain a collection of disjoint sets under the following two operations S 3 = Union(S 1 ,S 2 ) Find(x) : returns the set containing x. Union-find. We assume there are n fixed elements We start with n sets each containing a single element

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Union-find

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  1. Union-find

  2. Union-find • Maintain a collection of disjoint sets under the following two operations • S3 =Union(S1,S2) • Find(x) : returns the set containing x

  3. Union-find • We assume there are n fixed elements • We start with n sets each containing a single element • Each element has a pointer to its representation in the set containing it

  4. S1 = {e1} S2 = {e2} S3={e3} S4={e4} …… A = Union(S3,S4) S1 = {e1} A= {e2,e3} S4={e4} …… Find(e2)  A B = Union(A,S7) S1 = {e1} B= {e2,e3,e7} S4={e4} ……

  5. Why ? • Suppose we want to maintain an equivalence relation: b y v z s t a x y ≡ z

  6. y, z b v s t a x y ≡ s

  7. y, z, s b v t a x y ≡ s

  8. Queries • Equivalent?(y,a)

  9. Can solve this with union-find • Each equivalence class is a set • y ≡ s  union(find(y),find(s)) • Equivalent?(y,a)  return yes if find(y) = find(a)

  10. Representation • Represent each set by a tree, each node represents an item, the root also represents the set B e2 e7 e3

  11. Concretely B e2 e7 e3

  12. Find(e10) C e6 e9 e11 e10 e8

  13. Find(e10) C e6 e9 e11 e10 e8 Find(x) while (x.parent ≠null) do x ← x.parent return (x)

  14. D=Union(B,C) e6 C e9 e2 B e11 e7 e3 e10 e8

  15. D=Union(B,C) e6 C e9 e2 B e11 e7 e3 e10 e8

  16. D=Union(B,C) e6 D e9 e2 B e11 e7 e3 e10 e8

  17. D=Union(B,C) e2 B e6 C e7 e3 e9 e11 e10 e8

  18. D=Union(B,C) e2 B e6 C e7 e3 e9 e11 e10 e8

  19. D=Union(B,C) e2 D e6 C e7 e3 e9 e11 e10 e8

  20. Link by size • For the find’s sake its better to hang the smaller on the larger

  21. D=Union(B,C) e6 5 C e2 3 B e9 e11 e7 e3 e10 e8 If (C.size > B.size) then B.Parent ← C; C.size = C.size + B.size return (C) Else C.parent ← B; B.size = B.size + C.size return (B)

  22. D=Union(B,C) D e6 8 C e2 3 B e9 e11 e7 e3 e10 e8 If (C.size > B.size) then B.Parent ← C; C.size = C.size + B.size return (C) Else C.parent ← B; B.size = B.size + C.size return (B)

  23. Analysis • Union: O(1) time • Find: O(log(n)) time • The depth of a tree is at most log(n)

  24. Proof By induction on the sequence of operations: For every x Depth(x) ≤ log |T(x)| x

  25. Path compression

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