CSC 3323 Notes – Recurrence Relations

Algorithm Analysis

Recurrence Relations

• Applying a “divide and conquer” approach to solving a problem will frequently lead to a recursive algorithm.

• Time complexity for such an algorithm is represented with a recurrence relation.

Algorithm comparison exampleBlock multiplication (BM):

if n ≤ 3 then return SM(x, y)m = n div 2 (rounded up)P1 = BM( X[0..m-1], Y[0..m-1])P2 = BM( X[m..n-1], Y[m..n-1])P3 = BM( Add( X[0..m-1], X[m..n-1]), Add( Y[0..m-1], Y[m..n-1]))D = Sub( Sub( P3, P1), P2)Z[0..2m-1] = P1Z[2m..2n-1] = P2Z[m..2n-1] = Add( D[0..2m+1], Z[m..2n-1])return Z

Recurrence Relations: Block Multiplication

TBM(n) = 3 * TBM(n/2) + c * n

3 * TBM(n/2) = 3 * ( 3TBM(n/4) + c * (n/2))

32 * TBM(n/4) = 32 * ( 3TBM(n/8) + c * (n/4)) …….

3(log2(n)-1) * TBM(2) = 3(log2(n)-1) * ( 3TBM(1)+c*(2)) log2(n)-1

TBM(n)=3log2(n)*TBM(1) + cn∑ (3i/2i)(see formula, p. 8, 6)

= nlog2(3) * TBM(1) +cn(((3/2)log2(n)-1)/(3/2-1))

= nlog2(3) * TBM(1) + c’n((n)log2(3/2)-1)

TBM(n) ε O(nlog2(3)) for 32 bits, 322/321.59 = 4.15

for 64 bits, 642/641.59 = 5.5

Recurrence Relations: Binary Search

TBS(n) = TBS(n/2) + c

TBS(n/2) = TBS(n/4) + c

TBS(n/4) = TBS(n/8) + c …….

TBS(2) = TBS(1)+c

TBS(n) = TBS(1) + c*log2(n)

TBS(n) ε O(log2(n))

Recurrence Relations

• (Section 3.7.1, pp. 137ff)• Generally, for a recurrence relation:

T(n) = a*T(n/b) + c n e

(See Theorem 3.17, p. 139)

T(n) = c’ nlogb(a) + c ne logb(n) if a/be = 1T(n) = c’ nlogb(a) + c (nlogb(a) – ne) otherwise

Recurrence Relations:Fibonacci numbers

• Definition: fib(i) = fib(i-1) + fib(i-2)

T(n) = T(n-1) + T(n-2) + c ≥ 2T(n-2) + c2 T(n-2) ≥ 2(2 T(n-4) + c)

……2n/2-1T(2) ≥ 2n/2-1 (2T(0) + c)

n/2

T(n) ≥ 2n/2T(0)+c ∑ 2i = 2T(0)+c(2n/2 - 1)ε Ω(2n/2) i=0

Recurrence relations

• Sample problems: pp. 141-145

• 3.4, 3.7, 3.8, 3.10, 3.12