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Recurrence Relation

Recurrence Relation. Outline. What is a recurrence relation ? Solving linear recurrence relations Divide-and-conquer algorithms and recurrence relations. What is a recurrence relation ?. Definition.

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Recurrence Relation

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  1. Recurrence Relation

  2. Outline • What is a recurrence relation ? • Solving linear recurrence relations • Divide-and-conquer algorithms and recurrence relations

  3. What is a recurrence relation ?

  4. Definition A recurrence relation for the sequence {an} is an equation that expresses an in terms of one or more of the previous terms of the sequence, namely a0, a1, …, an-1, for all integers n n0, where n0 is a non-negative integer. A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation. Example: Given the recurrence relation an=2an-1-an-2. 3, 5, 7, 9, 11, … satisfies the recurrence relation. 2, 3, 4, 5, 6, … also satisfies the recurrence relation.

  5. Example: Compound interest Let Pn be the amount of money in the account after n years. If you initially deposit Pn=1000 in your account and the bank gives k % per year with interest compounded annually. Pn= (1+k/100) Pn-1 P1= (1+k/100) P0 = (1+k/100) P0 P2= (1+k/100) P1 = (1+k/100)2P0 P3= (1+k/100) P2 = (1+k/100)3P0 Pn= (1+k/100) Pn-1 = (1+k/100)nP0

  6. Example: Fibonacci numbers • A pair of rabbits is placed on an island. After 2 months old, a pair of rabbits produces another pair each month. How many rabbits are there after n months? fn =fn-1 + fn-2

  7. Tower of Hanoi

  8. Example: Tower of Hanoi H(1) = 1 H(n) = 2H(n-1) + 1 H(n) = 2(2H(n-2) +1) +1 = 4H(n-2) + 2 + 1 = 4(2(H(n-3) + 1) + 2 + 1 = 8H(n-3) + 4 + 2 + 1 = 8(2H(n-4) +1) + 4 + 2 + 1 = 16H(n-4) + 8 + 4 + 2 + 1 = 2n-1 + 2n-2 + 2n-3 + … + 1 = 2n - 1 Let x = 2n-1 + 2n-2 + … + 4 + 2 + 1 x +1 = (2n-1 + 2n-2 + … + 4 + 2 + 1) +1 x + 1 = 2n-1 + 2n-2 + … + 4 + 2 + 2 x + 1 = 2n-1 + 2n-2 + … + 4 + 4 x + 1 = 2n-1 + 2n-2 + … + 8 x + 1 = 2n x = 2n - 1

  9. Linear recurrence relation Homogeneous and non-homogeneous

  10. What is a linear recurrence relation ? Linear recurrence relation of degree k an= c1an-1+ c2an-2 +…+ ckan-k + F(n) • ciis a constant, for i= 1, 2, …, k • ck  0 Linear homogeneous recurrence relation • F(n) = 0 Linear non-homogeneous recurrence relation • F(n)  0

  11. Examples Linear homogeneous • an = 1.2 an-1 : degree 1 • fn = fn-1 + fn-2 : degree 2 • an = 3an-3 : degree 3 Non-linear homogeneous • an = a2n-1 + an-2 • an = nan-1 - 2an-2 Linear non-homogeneous • an = an-1 + 2n • hn = 2hn-1 + 1 • an = 3an-1 + n Non-linear non-homogeneous • an = a2n-1 + 2n • an = n2 an-1 + n

  12. Solving linear homogeneous recurrence relation

  13. Theorem • The sequence {an} is a solution of the recurrence relation an = c1an-1 + c2an-2iffan = 1rn1 + 2rn2 , where : • r1and r2are two distinct roots of r2 - c1r - c2 = 0, • a0 = 1 + 2 • a1 = 1r1 + 2r2 • [ 1 = (a1 – a0 r2)/(r1-r2 ) 2 = (a0 r1 –a1)/(r1-r2 ) ]

  14. Informal proof Recurrence relation: an = c1an-1 + c2an-2 Let r1 and r2 be two distinct roots of r2 – c1r – c2= 0. Let  and  be constants. • Show if an = 1rn1 + 2rn2, the sequence {an} is a solution of the recurrence relation. • Show if the sequence {an} is a solution of the recurrence relation, then an = 1rn1 + 2rn2.

  15. Informal proof Show if an = 1rn1 + 2rn2, the sequence {an} is a solution of the recurrence relation. Let an = c1an-1 + c2an-2be a recurrence relation. Let r1 and r2 be two distinct roots of r2 – c1r – c2= 0. Let  and  be constants. Since r1 and r2 are roots of r2 - c1r - c2= 0, r12 - c1r1 - c2 = 0 (r12 = c1r1 + c2) andr22 - c1r2 -c2= 0 (r22 = c1r2 + c2).

  16. Informal proof r12 = c1r1 + c2andr22 = c1r2 + c2. c1an-1 + c2an-2 = c1(1r1n-1+2r2n-1) + c2(1r1n-2+2r2n-2) = c11r1n-1+ c1 2r2n-1 + c21r1n-2+ c2 2r2n-2 = c11r1n-1+ c21r1n-2+c1 2r2n-1 + c2 2r2n-2 = 1r1n-2(c1r1 + c2) + 2r2n-2 (c1r2 + c2) = 1r1n-2 r12 + 2r2n-2 r22= 1 r1n+ 2 r2n =an

  17. Informal proof • Show if the sequence {an} is a solution of the recurrence relation, then there are 1and2 such that an = 1rn1 + 2rn2. • From the initial conditions a0and a1, a0 = 1 + 2and a1 = 1r1 + 2r2 2 = a0 - 1 1r1 = a1 - 2r2 = a1 - (a0-1) r2 = a1 – a0r2 + 1r2 1 (r1 - r2 ) = a1 – a0r2 >> 1 = (a1 – a0r2 )/(r1 - r2 ) 2 =a0-1 =a0-(a1–a0r2 )/(r1-r2 ) = a0r1-a0r2 -a1+a0r2 /(r1-r2 ) 2 = (a0r1-a1) / (r1-r2 )

  18. Example What is the solution of an = an-1 + 2an-2 with a0=2 a1=7 ? We have c1 = 1 andc2 = 2. The characteristic equation is r2 - r - 2 = 0, with roots r1= 2 andr2= -1 . 1 = (a1 – a0r2 )/(r1 - r2 ) = (7+2)/(2+1) = 3 2 = (a0r1 – a1)/(r1- r2 ) = (4-7)/(2+1) = -1 an = 1rn1 + 2rn2 = 32n - (-1)n

  19. Example: Fibonacci number What is the solution of an = an-1 + an-2 with a0=0 a1=1 ? We have c1 = 1 andc2 = 1. The characteristic equation is r2 - r - 1 = 0, with roots r1= (1+5)/2 andr2= (1-5)/2 . 1 = (a1 – a0r2 )/(r1 - r2 ) = (1-0)/5 = 1/5 2 = (a0r1 – a1)/(r1- r2 ) = (0-1)/5 = -1/5 an = 1rn1 + 2rn2 = ((1+5)/2)n /5 - ((1-5)/2)n/5 an = ( (1+5)n - (1-5)n )/(52n )

  20. Theorem • The sequence {an} is a solution of the recurrence relation an = c1an-1 + c2an-2iffan = 1rn0 + 2nrn0 , where : • r0is the only root of r2 - c1r - c2 = 0, • a0 = 1 • a1 = 1r0 + 2r0 = r0(1 + 2) • [ 1 = a0 2 = a1/r0 – a0 ]

  21. Example What is the solution of an= 6an-1 -9an-2 with a0=1 a1=6 ? We have c1 = 6 andc2 = -9. The characteristic equation is r2 - 6r + 9 = 0, with roots r0= 3. 1 = a0= 1 2 = a1/r0 – a0 = 6/3-1 = 1 an = 1rn0 + 2nrn0 = 3n + n 3n = 3n (1+ n).

  22. Theorem • The sequence {an} is a solution of the recurrence relation an = c1an-1 + c2an-2+…+ ckan-kiffan = 1rn1 + 2rn2+…+ krnk, where : • r1, r2 ,…, rkare kdistinct roots of rk -c1rk-1- c2rk-2 -…- ck= 0, • a0 = 1 + 2 +…+ k • a1 = 1r1 + 2r2 +…+ krk • a2 = 1r21 + 2r22 +…+ kr2k • …

  23. Theorem • The sequence {an} is a solution of the recurrence relation an = c1an-1 + c2an-2+…+ ckan-kiffan = 1rn1 + 2rn2+…+ krnk, where : • r1, r2 ,…, rkare kdistinct roots of rk -c1rk-1- c2rk-2 -…- ck= 0, • a0 = 1 + 2 +…+ k • a1 = 1r1 + 2r2 +…+ krk • a2 = 1r21 + 2r22 +…+ kr2k • …

  24. Theorem • The sequence {an} is a solution of the recurrence relation an = c1an-1 + c2an-2+…+ ckan-kiff an = (1,0 + 1,1n + …+ 1,m1nm1-1)rn1 + (2,0 + 2,1n + …+ 2,m2nm2-1)rn2+…+ (t,0 + t,1n + …+ t,mtnmt-1)rnt, where : • r1, r2 ,…, rtare tdistinct roots of rk -c1rk-1- c2rk-2 -…- ck= 0, with multiplicities m1, m2 ,…, mt

  25. Example What is the solution of an=6an-1-11an-2+6an-3 with a0=2, a1=5, a2=15 ? We have c1 = 6, c2 = -11 andc3 = 6. The characteristic equation is r3 - 6r2 + 11r - 6 = 0, with roots r1= 1,r2= 2 andr3= 3 . a0 = 1 + 2 + 3 a1 = 1r1 + 2r2 + 3r3 = 1 + 22 + 33 a2 = 1r12 + 2r22 + 3r32= 1 + 42 + 93 1 = 1, 2 = -1, 3 = 2 an = 1 – 2n + 23n

  26. Solving linear non-homogeneous recurrence relation

  27. Theorem If the sequence {an(p)} is a particular solution of the recurrence relation an = c1an-1 + c2an-2 +…+ ckan-k+ F(n) then every solution is of the form {an(p)+ an(h)} , where {an(h)} is a solution of the associated homogeneous recurrence relation an = c1an-1 + c2an-2 +…+ ckan-k.

  28. Example What are the solutions of an= 3an-1+ 2nwith a1=3 ? We have c1 = 3. The associated characteristic eqn is r-3=0, with root r = 3. Then,an(h) = 3n . Letpn = cn +d. Then, from an= 3an-1+ 2n, cn +d = 3(c(n-1)+d ) + 2n. Thus, c = -1 d = -3/2, and an(p) = -n -3/2 an= an(h) + an(p) = 3n -n -3/2. From a1= 3 = 3 -1 - 3/2,  = 11/6. The solution is an=(11/6)3n -n -3/2.

  29. Example: summation from 1 to n What are the solutions of an= an-1+ nwith a1=1 ? We have c1 = 1. The associated characteristic eqn is r-1=0, with root r = 1. Then,an(h) = 1n =  . Letpn = n(cn+d) = cn2+dn. Then, from an= an-1+ n, cn2 +dn= c(n-1)2+d (n-1) + n. Thus, cn2 +dn= cn2-2cn+c+d n - d + n , c-d + n(1-2c)=0. c-d =0 and 1-2c=0. That is, c=d=1/2. an= an(h) + an(p) =  + n(n+1)/2. From a1= 1 =  +1,  = 0. The solution is an=n(n+1)/2.

  30. Divide-and-conquer Algorithms

  31. Divide-and-Conquer Concept To solve a problem of size n, we solve a sub-problems of size n/b, and combine solutions from sub-problems. n/b n/b n/b n/b X X1 X2 X3 ... Xa n elements

  32. Binary Search • To search for an element a from a sorted list X of size n elements. • If a=m, then stop • If a<m, then search for a from X1. • If a>m, then search for a from X2. X X1 m X2

  33. Fast multiplication of large integers • a = 2nA1 + A0. b = 2nB1 + B0. A1A0B1B0 ab = 22nA1B1 + 2n(A0B1+ A1B0) +A0B0. ab = (22n + 2n) A1B1 + 2n(A1-A0)(B0 – B1) + (2n+1) A0B0. ab = 22n A1B1 + 2nA1B1 + 2nA1 B0 –2nA0B0 +2nA0B1 – 2nA1B1 + 2n A0B0 + A0B0. 10010 00011 10000 10011

  34. Recurrence Relations In Divide-and-conquer algorithms

  35. Divide-and-Conquer Recurrence Relations Let f(n) be the number of operations required to solve a problem of size n. f(n) = a f(n/b) + g(n). n/b n/b n/b n/b X X1 X2 X3 ... Xa n elements

  36. Binary Search • To search for an element a from a sorted list X of size n elements. • If a=m, then stop • If a<m, then search for a from X1. • If a>m, then search for a from X2. Let f(n) be the # of comparison in binary search within n elements. f(n) = f(n/2) + 2. X X1 m X2

  37. Fast multiplication of large integers • a = 2nA1 + A0. b = 2nB1 + B0. A1A0B1B0 ab = (22n + 2n) A1B1 + 2n(A1-A0)(B0 – B1) + (2n+1) A0B0. Let f(n) be the number of operations needed to multiply two n-bitintegers. f(2n) = 3f(n) + Cn. 10010 00011 10000 10011

  38. Theorem Let f be an increasing function that satisfies the recurrence relation f(n) = a f(n/b) + c whenever n is divisible by b, b is an integer greater than 1, a1, c is a positive real number. When n=bk, f(n) = t nlogba + s, for some t and s. When a>1, f(n) is O(nlogba ). When a=1, f(n) is O(log n).

  39. Examples Binary Search: f(n) = f(n/2) + 2. Because a=1, f(n) is O(log n). Find Max. : f(n) = 2f(n/2) + 2. Because a>1, f(n) is O(nlog11 ) = O(n) .

  40. Master Theorem Let f be an increasing function that satisfies the recurrence relation f(n) = a f(n/b) + cnd whenever n=bk for some positive integer k, b is an integer greater than 1, a1, c is a positive real number, d is a non-negative real number. When a<bd, f(n) is O(nd ). When a=bd, f(n) is O(nd log n). When a>bd, f(n) is O(nlogba).

  41. Examples Merge sort: f(n) = 2f(n/2) + n. a = 2, b = 2, c = 1, d = 1 Because a=bd, f(n) is O(nlog n). Integer multiplication: f(n) = 3f(n/2) + Cn a = 3, b = 2, c = C, d = 1 Because a>bd, f(n) is O(nlog23).

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