Recurrence: Master Theorem T(n) = aT(n/b) + f (n) where f (n)=cn k 1.a < b k T(n) ~ n k 2.a = b k...

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Recurrence: Master Theorem T(n) = aT(n/b) + f (n) where f (n)=cn k 1. a < b k T(n) ~ n k 2. a = b k T(n) ~ n k log b n 3. a > b k T(n) ~ n log b a

Transcript of Recurrence: Master Theorem T(n) = aT(n/b) + f (n) where f (n)=cn k 1.a < b k T(n) ~ n k 2.a = b k...

Page 1: Recurrence: Master Theorem T(n) = aT(n/b) + f (n) where f (n)=cn k 1.a < b k T(n) ~ n k 2.a = b k T(n) ~ n k log b n 3.a > b k T(n) ~ n log b a.

Recurrence: Master Theorem

T(n) = aT(n/b) + f (n) where f (n)=cnk

1. a < bk T(n) ~ nk

2. a = bk T(n) ~ nk logbn

3. a > bk T(n) ~ nlog b a

Page 2: Recurrence: Master Theorem T(n) = aT(n/b) + f (n) where f (n)=cn k 1.a < b k T(n) ~ n k 2.a = b k T(n) ~ n k log b n 3.a > b k T(n) ~ n log b a.

Derivation of Master Theorem

Recurrence relation:

T(n) = aT(n/b) + f(n)

Assume problem size is bk

T(bk) = aT(bk/b) + f(bk) = aT(bk-1) + f(bk)

= a(aT(bk-1/b)+f(bk-1)) + f(bk)

= a2T(bk-2) + af(bk-1) + f(bk)

= ak[T(1) + ]

k

j

jj abf1

/)(

Page 3: Recurrence: Master Theorem T(n) = aT(n/b) + f (n) where f (n)=cn k 1.a < b k T(n) ~ n k 2.a = b k T(n) ~ n k log b n 3.a > b k T(n) ~ n log b a.

Derivation of Master Theorem

Since k = logbnak=alog

bn

Property of logs: logcx = logcd *logdxalog

bn = a (log

ba * log

an) = a (log

an * log

ba) = (alog

an) * log

ba = nlog

ba

Finally, by letting f(n) = nd and T(1) = constant, we get:

nlogba

nlogba

n

j

jdjb

ablog

1

/

n

j

jdb

ablog

1

)/(

Page 4: Recurrence: Master Theorem T(n) = aT(n/b) + f (n) where f (n)=cn k 1.a < b k T(n) ~ n k 2.a = b k T(n) ~ n k log b n 3.a > b k T(n) ~ n log b a.

Some Useful Things To Remember

• Logarithms:– loga(xy) = logax + logay (previous page)– loga(xy) = y logax (via rule 1, repeated

multiplication)– loga(x/y) = logax-logay (via rule 1 and 2; loga(xy-

1) = logax + logay-1= logax+(-1*logay)– logax = logab*logbx (base conversions)

Let b = as and x = bt. Substituting x = (as)t. Take loga of both sides: logax = loga((as)t)

RHS: t*loga(as) = t*s = logbx*logab

Page 5: Recurrence: Master Theorem T(n) = aT(n/b) + f (n) where f (n)=cn k 1.a < b k T(n) ~ n k 2.a = b k T(n) ~ n k log b n 3.a > b k T(n) ~ n log b a.

Derivation of Master Theorem

nlogb

a

Sum is a geometric series

The sum = logbn if (bd == a)

otherwise sum = (bd/a)

Why this sum?

n

j

jdb

ablog

1

)/(

1)/(

1)/( log

ab

abd

nd b

1

)1(

11

)1(11

1

11)(

11

1

1

0

c

cc

c

cc

c

cc

c

ccc

nnnn

i

nn

i

ii

Page 6: Recurrence: Master Theorem T(n) = aT(n/b) + f (n) where f (n)=cn k 1.a < b k T(n) ~ n k 2.a = b k T(n) ~ n k log b n 3.a > b k T(n) ~ n log b a.

Derivation of Master Theorem

nlogb

a

Sum is a geometric series

The sum = logbn if (bd == a)

If (bd == a) nlogb

a*logbn

nlogb

a * logbn and since bd == a => nd * logbn

n

j

jdb

ablog

1

)/(

Page 7: Recurrence: Master Theorem T(n) = aT(n/b) + f (n) where f (n)=cn k 1.a < b k T(n) ~ n k 2.a = b k T(n) ~ n k log b n 3.a > b k T(n) ~ n log b a.

Derivation of Master Theorem

nlogba

Sum is a geometric series

otherwise sum = (bd/a)

If (bd > a) then bd/a > 1 and sum ~ O((bd/a)logbn)

nlogba* (bd/a)log

bn alog

bn (bd/a)log

bn O(bdlog

bn) O((blog

bn)d)=

O(nd)

If (bd < a), then bd/a < 1 and sum converges ~ O(1) nlog

ba * 1 = O(nlog

ba)

n

j

jdb

ablog

1

)/(

1)/(

1)/( log

ab

abd

nd b

Page 8: Recurrence: Master Theorem T(n) = aT(n/b) + f (n) where f (n)=cn k 1.a < b k T(n) ~ n k 2.a = b k T(n) ~ n k log b n 3.a > b k T(n) ~ n log b a.

Recurrence: Master Theorem

T(n) = aT(n/b) + f (n) where f (n)=cnk

1. a < bk T(n) ~ nk

2. a = bk T(n) ~ nk logbn

3. a > bk T(n) ~ nlog b a


Binomial Coe cients and Generating FunctionsBinomial theorem Theorem 1 (a+b)n = n å k=0 n k akbn k for any integer n >0. Proof. Expanding (a+b)n = (a+b)(a+b) (a+b) yields the sum

Binomial Coe cients and Generating FunctionsBinomial theorem Theorem 1 (a+b)n = n å k=0 n k akbn k for any integer n >0. Proof. Expanding (a+b)n = (a+b)(a+b) (a+b) yields the sum

Lectures on Vedanta B N K Sharma

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O B E K O N Consortium

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