Understanding Recursive Equations & Solving Recurrences, Study notes of Algorithms and Programming

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Recursion

and^ RecurrencesCOMP157Sept^10

Recursion

-^ Recursion: When^ a^ function (or

procedure) calls^ itself.^ ‐^ The

Free^ On‐line^ Dictionary

of^ Computing

-^ applies^ to^ both

mathematics^ and

programming

-^ Recursive^ mathematical

equations^ are

called

recurrence^ equations

,^ recurrence^ relations

,^ or

simply^ recurrences

-^ Recurrences

naturally^ occur

in^ the^ analysis

of

recursive^ algorithms.

Solving^ Recurrences • Solving a^ recurrence^ means

to^ find^ an^ explicit (non‐recursive)

formula^ for^ x(n),

or^ to^ prove the^ no^ such^ sequence

exists

-^ In^ algorithm

analysis^ we^ need

this^ solution to^ compare^ functions

or to^ plot^ functions – Recurrences^ can

be^ expensive^ to

Appendix B compute^ directly

Method^ of

Forward^

Substitution

-^ Start^ from^

the^ initial^ condition

-^ Substitute^

into^ recurrence

-^ Repeat^ until

you^ can^ see^

a^ pattern

Forward^ Substitution

Example

-^ x(n)^ =^ 2x(n

‐1)^ +^ 1,^ for^ n^

>^1

x(1)^ =^1 • x(1)^ =^^1 x(2)^ =^ 2x(1)^ +

1 =^ 2*2+1^ =^

x(3)^ =^ 2x(2)^ +

1 =^ 2*3+1^ =^

x(4)^ =^ 2x(3)^ +

1 =^ 2*7+1^ =^

x(5)^ =^ 2x(4)^ +

1 =^ 2*15+1^ =

^31

-^ Pattern:^ x(n)

n^ = 2 – 1,^ for^ n=1^ … 5 Still^ need^ to^ prove

that^ it^ is^ true

for^ n>

Forward^ Substitution

Example

-^ x(n)^ =^ 2x(n

‐1)^ +^ 1,^ for^ n^

>^1

x(1)^ =^1 n^ • x(n)^ =^2 – 1,^

for^ n=1^ … 5

-^ proof^ (by^ substitution

into^ recurrence): x(n)^ =^ 2x(n‐1)

n‐^1 + 1 = 2*(

– 1)^ +^1

n^ = (2– 2)^ +^1 n^ = 2 – 1

-^ proof^ (by^ induction):n+1x(n+1)^ =^2

n^ – 1^ =^ 2*(2^ )^ ‐

n 2 + 1 = 2*(

–1)^ +^1

=^ 2x(n)^ +^1

Backward

Substitution

Example

-^ x(n)^ =^ x(n‐1)

+^ n,^ for^ n^ >^1 x(0) = 0

-^ x(n‐1)^ =

x(n‐2)^ +^ (n‐1),

sub^ into^ x(n): x(n)^ =^ (x(n‐2)

+^ (n‐1))^ +^ n = x(n‐2) +^ (n‐1)^ +^ n x(n‐2) = x(n‐3)^ +^ (n‐2),^

sub^ into^ previous: x(n)^ =^ (x(n‐3)

+^ (n‐2))^ +^ (n‐

1)^ +^ n =^ x(n‐3)^ +^ (n‐

2)^ +^ (n‐1)^ +^ n

-^ Pattern?

Backward

Substitution

Example

-^ x(n)^ =^ x(n‐2)

+^ (n‐1)^ +^ n x(n) = x(n‐3) +^ (n‐2)^ +^ (n‐1)

+^ n

-^ Pattern?^ x(n)^ =^ x(n‐i)

+^ (n‐i+1)^ +^ (n

‐i+2)^ +^ … +^ n when^ n^ =^ i,^ n

‐i =^ 0,^ so^ we^ can

use^ initial condition:x(n)^ =^ x(0)^ +^1

+^2 +^ … +^ n^ =

0 +^1 +^2 +^ … +

n

n = = i ∑= i^0

) (^1) ( + nn 2

Decrease

‐by‐Constant

Recurrences

-^ Decrease‐by

‐constant^ algorithm

use^ the solution^ to^ an

n/b^ size^ problem

to^ solve an^ n^ size^ problem

(typically^ b =2)

-^ Typical^ efficiency

recurrence: T ( n )^ =^ T ( n / b )^

+^ f ( n )

-^ Backward^ substitution

yields: k T ( b )^ =^ T(1)^ +

k i bf )( ∑= i^1

Divide‐and

‐conquer^

Recurrences

-^ Divide‐and

‐conquer^ algorithms

divide^ the problem^ into

several^ smaller

problems,^ which are^ solved^ recursively,

then^ combines

the solutions • Typical^ efficiency

recurrence: T ( n )^ =^ aT ( n / b

)^ +^ f ( n )

-^ Backward^ substitution

yields: T ( n )^ =^

]/)( loglog ) (^1) ([^1 n i a^ i i

abf

b bTn + ∑=

Review:^ Recursive

Algorithms

-^ Give^ a^ recursive

algorithm^ to

compute^ the length^ of^ a^ linked

list base^ case:^ empty^ list

has^ length^ zero recursive^ case:^1 +^ length

of^ list^ with^ head

removed

Review:^ Recursive

Algorithms

-^ Give^ a^ recursive

algorithm^ to

compute^ the length^ of^ a^ linked

list int length(Node*

head) { if^ (head^ ==

NULL)^ return

return^1 +^ length(head

‐>next); }

Review:^ Recursive

Algorithms

-^ Give^ a^ recursive

algorithm^ to

compute^ the sum^ of^ the^ elements

in^ linked^ list double^ sum(Node*

head) { if^ (head^ ==

NULL)^ return

return^ head‐

value^ +^ sum(head

‐>next); }

Review:^ Recursive

Algorithms

-^ Give^ a^ recursive

algorithm^ to

count^ the number^ of^ odd

elements^ in^

a^ linked^ list base^ case:empty^ list^ has^0 odd^ elements recursive^ case:if^ head^ node

element^ is^ odd: 1 + number^ of^ odd^ elements in list with^ head^ removedelse: number of odd^ elements in list with^ head^ removed