Divide-&-Conquer Algorithms & Recurrence Relations: Selected Exercises

2

10

Find f(n) when n = 2k, where f satisfies the recurrence

relation f(n) = f(n/2) + 1, with f(1) = 1.

3

10 Solution

We are asked for the value of f.

We are given that

f(n) = f(n/2) + 1, with f(1) = 1, where n = 2k.

f(2k) = f(2k-1) + 1, with f(20) = 1.

Answer: f(2k) = k + 1. (Proof would be inductive.)

4

Master Theorem

Let f be an increasing function that such that

f(n) = a.f(n/b) + c.nd

whenever n = bk, where k Z+, a ≥ 1, b >1 is an

integer, and c, d R with c > 0 & d ≥ 0.

If ( a < bd ) then f(n) is O( nd )

else if ( a = bd ) then f(n) is O( ndlogn )

else if ( a > bd ) then f(n) is O( nlogb

a ).

5

10 Solution (2)

• If we were interested only in the asymptotic growth rate, we could use the Master Theorem.

• For f(n) = f(n/2) + 1 = 1f(n/2) + 1n0 a = 1b = 2d = 0So, a = bd

So, f(n) is O(ndlog n) = O(log n).

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20 (a)

Set up a divide-&-conquer recurrence relation for the # of

modular multiplies (MM) required to compute

an mod m, where a, m, n Z+, using the recursive algorithm

from Example 3 in Section 4.4.

Let b = a mod m.

The algorithm then is:

• an mod m = ( an/2 mod m )2 mod m, for even n (1 MM)

• an mod m = ([(a(n-1)/2 mod m)2 mod m] . b) mod n, for odd n (2 MMs)

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20 (a) Solution

The algorithm is:

an mod m = ( an/2 mod m )2 mod m, for even n (1 MM)

an mod m = ([(a(n-1)/2 mod m)2 mod m] . b) mod n, for odd n (2 MMs)

Let f(n) denote the # of multiplies.

For even n, f(n/2) + 1 MM is used.

For odd n, f(n/2) + 2 MMs are used.

Worst case, f(n) = f(n/2) + 2.

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20 (b)

Using f(n) = f(n/2) + 2, construct a big-O estimate for

the # of modular multiplies used to compute

an mod m using the recursive algorithm.

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20 (b) Solution

Use the Master Theorem: f(n) = a.f(n/b) + c.nd

f(n) = f(n/2) + 2 = 1f(n/2) + 2n0

a = 1, b = 2, d = 0.

So, a = bd

So, f(n) is O(ndlog n) = O(log n).

10

30

Exercise 30 assumes that f satisfies the conditions of the Master Theorem.

Use Exercise 29 to show that, for f(n) = a.f(n/b) + c.nd

if a = bd, then f(n) is O( nd log n ).

Exercise 29 gives us:

If a = bd & n = bk

then f(n) = f(1)nd + cnd logbn.

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30 Solution

To show:

f(n) = f(1)nd + cnd logbn is O( nd log n ).

So:

1. f(n) growth clearly is dominated by cnd logbn.

2. c is a constant.

3. logbn differs from log2n by only a constant factor.

( log2b ) . logbn = log2( b logbn ) = log2n.

4. cnd logbn therefore is O( nd log n ).

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13

Theorem 1

Let f be an increasing function such that

f(n) = a.f(n/b) + c

whenever b | n, where a ≥ 1, b >1 are integers, & c > 0 is real.

If ( a > 1 ) then f(n) is O( nlogb

a )

If ( a = 1 ) then f(n) is O( logn ).

When n = bk where k > 0 is integer, and a > 1,

f(n) = C1 nlogb

a + C2,

where C1 = f(1) + c/(a – 1) and C2 = – c/(a – 1).