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Optimal Solution: Lowest Common Ancestors & Range Minimum Queries

Learn about finding lowest common ancestors and preprocess arrays for fast range minimum queries in linear time. Discover O(nlog(n)) space algorithms for efficient query times.

orla-herring
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Optimal Solution: Lowest Common Ancestors & Range Minimum Queries

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  1. Lowest common ancestors

  2. 0 Shallowest node 12 8 3 9 1 2 11 4 7 10 5 6 Write an Euler tour of the tree 3 3 2 3 1 4 1 7 1 3 5 6 5 3 5 1 2 4 6 7 LCA(1,5) = 3

  3. 0 minimum 12 8 3 9 1 2 11 4 7 10 5 6 3 5 1 2 4 6 7 3 2 3 1 4 1 7 1 3 5 6 5 3 0 1 0 1 2 1 2 1 0 1 2 1 0

  4. 0 minimum 12 8 3 9 1 2 11 4 7 10 5 6 Range minimum Preprocess an array, such that given i,j you can find the minimum in [i,j] fast 3 5 Reduction takes linear time 1 2 4 6 7 3 2 3 1 4 1 7 1 3 5 6 5 3 0 1 0 1 2 1 2 1 0 1 2 1 0

  5. Trivial algorithms for RMQ

  6. Less trivial algorithms to RMQ • Try to use O(nlog(n)) space to do a query in O(1) time

  7. Optimal solution ... ... ... ... ... Rememeber the min in each block (A’) and its position (B) … ... B[0] B[i] B[2n/logn] … ... A’[0] A’[i] A’[2n/logn]

  8. Example n=16 A[] : 8 0 1 2 3 4 5 6 7 9 10 11 12 13 14 15 10 25 22 7 34 9 2 26 33 12 24 43 5 11 19 27 = 8 10 25 22 7 34 9 … B[] : A’[] : 0 1 2 0 1 2 0 3 5 10 7 9 … …

  9. Preprocess A’ for RMQ using the O(nlog(n)) space algorithm. Since the size of A’ is this would take Space, preprocessing time

  10. How do we answer queries ? i and j might be in the same block, we need some mechanism to answer inside blocks i < j on different blocks, answer the query as follows: • Compute minima from i to end of its block. • Compute minima of all blocks in between i’s and j’s blocks. • Compute minima from the beginning of j’s block to j. Return the index of the minimum of these 3 values.

  11. So what do we do inside blocks?

  12. 0 12 8 3 9 1 2 11 4 7 10 5 6 restriction We need to solve a special case of RMQ 3 5 1 2 4 6 7 3 2 3 1 4 1 7 1 3 5 6 5 3 0 1 0 1 2 1 2 1 0 1 2 1 0

  13. 3 4 5 6 5 4 5 6 5 4 Two subproblems with the same vector of ±1 are equivalent Each subproblem can be described by the first entry and a vector of ±1 0 1 2 3 2 1 2 3 2 1 +1 +1 +1 -1 -1 +1 +1 -1 -1

  14. The bottom line There aren’t too many different subproblems, only For each such subproblem prepare all answers in advance 

  15. Find the solution using i,j: Pick a subproblem:

  16. Summary ... ... ... ... ... Each block know its subproblem (A’’) … ... B[0] B[i] B[2n/logn] … ... A’[0] A’[i] A’[2n/logn] … ... A’’[0] A’’[i] A’’[2n/logn]

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