Solving recurrence relationsConsider these: # T(n) = 2 T(n/3) + n # T(n) = 3 T(n/3) + n # T(n) = 4 T(n/3) + n 1. recursion diagram is a complete binary tree of depth log3 n. The Ith level from the top has 2i nodes, each with work n/3i. Total work is : n[ 1 + (2/3) + (2/3)2 + ... + (2/3)log3 n] which is O(n). 2. same as above but each node has 3 children, so the Ith level from the top has 3i nodes. Total work is : n[ 1 + (3/3) + (3/3)2 + ... + (3/3)log3 n] which is O(n log n). 3. now each node has 4 children, so total work is : n[ 1 + (4/3) + (4/3)2 + ... + (4/3)log3 n] which is O(n (4/3)log3 n) = O(n1+log3 4/3) = O(n1.26...) . More# T(n) = 8 T(n/3) + n2 # T(n) = 9 T(n/3) + n2 # T(n) = 10 T(n/3) + n2 1. recursion diagram is of depth log3 n. Each node has 8 children. The Ith level from the top has 8i nodes, each with work (n/3i)2 = n2 / 9i . Total work is : n2[ 1 + (8/9) + (8/9)2 + ... + (8/9)log3 n] which is O(n2). 2. Total work is : n2[ 1 + (9/9) + (9/9)2 + ... + (9/9)log3 n] which is O(n2 log n). 3. Total work is : n2[ 1 + (10/9) + (10/9)2 + ... + (10/9)log3 n] which is O(n2 (10/9)log3 n) = O(n2.095...). Repeated Substitution: T(n) = 1.5 T(n/3.1) + n0.5. No recursion diagram!? Try repeated substitution. Use the recurrence relation to replace T(n/3.1) with (1.5 T(n/3.12) + (n/3.1)0.5 : : T(n) = 1.5 T(n/3.1) + n0.5 : = 1.5 [ (1.5 T(n/3.12) + (n/3.1)0.5 ] + n0.5 . Then use the recurrence relation to expand T(n/3.12), and so on, simplifying as you go. After i substitutions, get T(n) = (1.5)i T(n / 3.1i) + (1.5)i-1 [n/3.1i-1]0.5 + ... . Taking i = \log3.1 n, after i substitutions the term left is T(1) (or at least T(O(1))), which we substitute using the base case of the recurrence relation. This gives us : T(n) = n0.5[ 1 + (1.5/3.10.5) + (1.5/3.10.5)2 + ... + (1.5/3.10.5){log1.5 n}] Since 1.5 < 3.10.5, this is O(n0.5). (See GeometricSums .) Proving bounds using induction# T(n) = T(n/3) + T(n/2) + n, T(1) = 1. Can make a recursion diagram for this. Every node has two children, but depth is uneven, and work is not the same for each node on a level. Try guessing, then prove by induction. Guess that T(n) ≤ 10 n. Attempt proof by induction: : Base case: T(1) ≤ 10 . (true) : Inductive step: T(n) = T(n/3) + T(n/2) + n ≤ 10(n/3) + 10(n/2) + n = (10/3+10/2 + 1) n = 9.33.. n ≤ 10 n. (verified) What is the best constant that would work? Guess that T(n) ≤ c n. Attempt proof by induction: : Base case: T(1) ≤ c . (need c ≥ 1 ) : Inductive step: T(n) = T(n/3) + T(n/2) + n ≤ c(n/3) + c(n/2) + n = (c(1/3+1/2) + 1) n ≤ c n ? Need c(1/3+1/2)+1 ≤ c . True iff c ≥ 6 . Conclude T(n) ≤ 6 n . |