ClassS04CS141/Notes 04 09

Recurrence relations, continued

Try these:

1. T(n) = 2 T(n/3) + n
2. T(n) = 3 T(n/3) + n
3. T(n) = 4 T(n/3) + n

1. recurrence tree 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(n^{1+log₃ 4/3}) = O(n^1.26...) .

Oh yeah? Try these!

1. T(n) = 8 T(n/3) + n²
2. T(n) = 9 T(n/3) + n²
3. T(n) = 10 T(n/3) + n²

1. recurrence tree 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(n^2.095...).

Oh YEAH?! TRY THIS:

1. T(n) = 1.5 T(n/3.1) + n^{0.5}.

No recursion tree!?

Try repeated substitution.

After substitutions, get (1.5)ⁱ T(n / 3.1ⁱ) + (1.5)^i-1 [n/3.1^i-1]^0.5 + ... .

Total work is

n^0.5[ 1 + (1.5/3.1^0.5) + (1.5/3.1^0.5)² + ... + (1.5/3.1^0.5)^{log_1.5 n}]

which is O(n² (10/9)^{log₃ n}) = O(n^1.095...).

Since 1.5 < 3.1^0.5, this is O(n^0.5).

ok, what about this one?

1. T(n) = T(n/3) + T(n/2) + n, T(1) = 1.

Can make a recursion tree 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 .