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Bottom-Up Splay Trees–Analysis

Bottom-Up Splay Trees–Analysis. Actual complexity of search, insert, delete, and split is O(n) . Actual complexity of join is O(1) . Amortized complexity of search, insert, delete, join, and split is O(log n). Potential Function. size(x) = #nodes in subtree whose root is x .

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Bottom-Up Splay Trees–Analysis

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  1. Bottom-Up Splay Trees–Analysis • Actual complexity of search, insert, delete, and split is O(n). • Actual complexity of join is O(1). • Amortized complexity of search, insert, delete, join, and split is O(log n).

  2. Potential Function • size(x) = #nodes in subtree whose root is x. • rank(x) = floor(log2 size(x)). • P(i) = Sx is a tree node rank(x). • P(i) is potential after i’th operation. • size(x) and rank(x) are computed after i’thoperation. • P(0) = 0. • When join and split operations are done, number of splay trees > 1 at times. • P(i) is obtained by summing over all nodes in all trees.

  3. 20 10 40 6 30 8 Example 6 2 • size(x) is in red. 3 1 2 1 2 1 1 0 1 0 • rank(x) is in blue. • Potential = 5.

  4. 6 2 20 3 1 2 1 10 40 2 1 1 0 6 30 1 0 8 Root Rank • rank(root) <= floor(log2 n) • n is total number of operations

  5. m S B Join • Actual complexity = 1 • Rank of root is <= floor(log2 n). • Other ranks unchanged. • Potential increases by <= floor(log2 n). • amortized complexity = actual complexity + DP <= 1 + floor(log2 n)

  6. Other Operations • actual complexity = # of splay steps • amortized complexity = actual complexity + DP = # of splay steps + DP

  7. Splay Step Potential Change • If q = null or q is the root, do nothing. • DP = 0.

  8. q p a p q c b c a b 1-Level Move • Do a one-level move. • r(x) = rank of x before splay step. • r’(x) = rank of x after splay step.

  9. q p a p q c b c a b 1-Level Move • DP = r’(p) + r’(q) – r(p) – r(q) <= r’(q) – r(q).

  10. q gp a p p d b gp q c c d a b 2-Level Move (case 1) • DP = r’(gp) + r’(p) + r’(q) – r(gp) – r(p) – r(q)

  11. q gp a p p d b gp q c c d a b 2-Level Move (case 1) • r’(gp) <= r’(q) • r (q) <= r(p) • r’(q) = r(gp) • r’(p) <= r’(q)

  12. 2-Level Move (case 1) • DP = r’(gp) + r’(p) + r’(q) – r(gp) – r(p) – r(q) • r’(q) = r(gp) • r’(gp) <= r’(q) • r’(p) <= r’(q) • r (q) <= r(p) • DP <= r’(q) + r’(q) – r(q) – r(q) = 2(r’(q) – r(q))

  13. 2-Level Move (case 1) A more careful analysis reveals that DP <= 3(r’(q) – r(q)) –1

  14. 2-Level Move (case 2) • Similar to Case 1.

  15. Only 1-Level Moves • Suppose that the splay node q is initially at level L > 1. • Let r(q,j) be the rank of q when it is at level j following/preceding a splay step. • When only 1-level moves are made, DP <= S2<=j<=L(r(q,j-1)-r(q,j)) = r(q,1)-r(q,L)

  16. Search With Only 1-Level Moves • amortized complexity = # of splay steps + DP <= L-1+r(q,1)-r(q,L) <= n-1+floor(log2n) = O(n)

  17. Correct Splay • Make as many 2-level moves as possible. A 1-level move is made at the end if necessary. • When L is odd, floor(L/2)2-level moves are made; there is no 1-level move. • Single 2-level move DP <= 3(r’(q) – r(q)) –1 • For complete splay DP <= 3(r(q,1)-r(q,L))-floor(L/2) = 3(r(q,1)-r(q,L))- # of splay steps

  18. Correct Splay • When L is even, L/2-12-level moves are made; there is also one 1-level move. • Single 2-level move DP <= 3(r’(q) – r(q)) –1 • Single 1-level move DP <= r’(q) – r(q) <= 3(r’(q) – r(q)) • For complete splay DP <= 3(r(q,1)-r(q,L))- # of splay steps+1 DP <= S2<=j<=L(r(j-1)-r(j)) = r(1)-r(L)

  19. Correct Splay • So, regardless of the value of L DP <= 3(r(q,1)-r(q,L))- # of splay steps+1 <= 3floor(log2n)-# of splay steps+1

  20. Search With Correct Splay • amortized complexity = # of splay steps + DP <= # of splay steps +3floor(log2n)-# of splay steps+1 = 3floor(log2n)+1 = O(log n)

  21. Insert 7 4 • On every leaf to root path ranks are non-decreasing. • At most floor(log2 n)+1 different ranks on such a path. • Insert may increase the rank of only the last node in each sequence of equal rank nodes (view sequence from bottom to top). • When you insert, potential increases by up to floor(log2 n)+1. 4 4 1 1 0

  22. Insert With Correct Splay • total potential change caused by an insert <=floor(log2n)+1 +3floor(log2n)-# of splay steps+1 • amortized complexity = # of splay steps + DP <= 4floor(log2n)+2 = O(log n)

  23. Delete With Correct Splay • total potential change caused by a delete <=3floor(log2n)-# of splay steps+1 • amortized complexity = # of splay steps + DP <= 3floor(log2n)+1 = O(log n)

  24. Split With Correct Splay • total potential change caused by a split <=3floor(log2n)-# of splay steps+2 • amortized complexity = # of splay steps + DP <= 3floor(log2n)+2 = O(log n)

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