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CS 173: Discrete Mathematical Structures

CS 173: Discrete Mathematical Structures. Cinda Heeren heeren@cs.uiuc.edu Siebel Center, rm 2213 Office Hours: W 12:30-2:30. CS 173 Announcements. Homework #9 available, due 11/13, 8a. Giggle. We want to be able to compute T(n), without having to compute T(n-k).

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CS 173: Discrete Mathematical Structures

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  1. CS 173:Discrete Mathematical Structures Cinda Heeren heeren@cs.uiuc.edu Siebel Center, rm 2213 Office Hours: W 12:30-2:30

  2. CS 173 Announcements • Homework #9 available, due 11/13, 8a. Cs173 - Spring 2004

  3. Giggle. We want to be able to compute T(n), without having to compute T(n-k) To achieve this we “solve” the recurrence. CS 173 Recurrences Linear homogeneous recurrence relations with constant coefficients. Today’s first example: T(n) = 7T(n-1) - 10T(n-2) T(0) = 2, T(1) = 1 • Rewrite: T(n) - 7T(n-1) + 10T(n-2) = 0 • Find characteristic eqn: r2 - 7r + 10 = 0 • Find roots of char eqn: (r-2)(r-5) = 0, r=2,5 • General solution is T(n) = A2n + B5n Cs173 - Spring 2004

  4. CS 173 Recurrences Today’s first example: T(n) = 7T(n-1) - 10T(n-2) T(0) = 2, T(1) = 1 • General solution is T(n) = A2n + B5n Now what? … think about what we know! A20 + B50 = 2 A + B = 2 A = 3, B = -1 A21 + B51 = 1 2A + 5B = 1 • Unique solution is T(n) = 3·2n - 5n Cs173 - Spring 2004

  5. CS 173 Recurrences You try a tricky one: T(n) = T(n-2)/4 T(0) = 1, T(1) = 0 • Rewrite: • Find characteristic eqn: • Find roots of char eqn: • General solution is • Set up system of eqn to get unique soln. Cs173 - Spring 2004

  6. CS 173 Recurrences Suppose the characteristic eqn factors into: (r-2)3(r-3)2(r-5) = 0 Roots are 2,2,2,3,3,5 If you have a non-distinct root r of multiplicity k, then nirn, i=0,1,…k are all solutions. In this case the general solution is: T(n) = A1n22n + A2n2n + A32n + A4n3n + A53n + A65n T(n) = (A1n2 + A2n+ A3)2n + (A4n+ A5)3n + A65n To find the unique particular solution, we would need the 6 boundary values. Solve for A1, …A6 by solving 6 equations in 6 unknowns. Cs173 - Spring 2004

  7. CS 173 Recurrences Here’s one for you to try: an = 4an-1 - 5an-2 + 2an-3, n  3 a0 = 0, a1 = 1, a2 = 2 Cs173 - Spring 2004

  8. CS 173 Recurrences Linear NONhomogeneous recurrence relations with constant coefficients. c0an + c1an-1 + c2an-2 + … + ckan-k = f(n), • constant • polynomial in n • cn for some constant c • cn · polynomial(n) Where f(n) is This is approach is different than the one in your text. Easier and more general. Cs173 - Spring 2004

  9. CS 173 Recurrences First, some notation: A sequence a0, a1, a2, …, is denoted an • 2n = 1,2,4,8,… • n2 = 0,1,4,9,… • n = 0,1,2,3,… Examples Note: if an and bn are sequences, then an + bn represents the sequence an +bn (termwise addition). Cs173 - Spring 2004

  10. CS 173 Recurrences Sequence operators: • Constant multiplication c·an defined to be c·an Ex: 3·2n = 3·2n = 3, 6, 12, 24, 48, … • Shift “E” Ean = an+1 shifts sequence to left Ex: E2n = 2n+1 = 2, 4, 8, 16, … Ex: E3n + 1 = 3(n+1) + 1 = 3n + 4 Cs173 - Spring 2004

  11. CS 173 Recurrences Combining operators: • If A,B are seq ops, then A+B is a seq op: (A+B)an defined to be Aan + Ban Ex: (E+2)2n = E2n + 22n = 2n+1 + 2·2n = 2n+1 + 2n+1 = 2n+1 + 2n+1 = 2·2n+1 = 2n+2 • If A,B are seq ops, then AB is a seq op: (AB)an defined to be A(Ban) Ex: E3an = E·E·Ean = E(E(Ean)) = an+3 Cs173 - Spring 2004

  12. Discrete derivative CS 173 Recurrences An important operator: (E-1) • Example: (E-1)n = En - n = n + 1 - n = n + 1 - n = 1 • More generally: (E-1)an = an+1 - an = an+1 - an • (E-1)n2 = En2 - n2 = (n+1)2 - n2 = n2 + 2n + 1 - n2 = 2n+ 1 Cs173 - Spring 2004

  13. (E-1) “annihilates” c CS 173 Recurrences For any constant sequence c: (E-1)c = 0 • TRICK: to solve a NONhomogeneous linear recurrence with constant coefficients, turn it into a homogeneous recurrence by applying operators to annihilate the right side. Cs173 - Spring 2004

  14. (E-1) “annihilates” c CS 173 Recurrences Example: solve an = 5an-1 - 6an-2 + 4 • Rewrite: an - 5an-1 + 6an-2 = 4 • Rewrite again so n is smallest index: an+2 - 5an+1 + 6an = 4 • Rewrite again as a sequence: an+2 - 5an+1 + 6an = 4 • Rewrite again using operators: (E2 - 5E + 6)an = 4 Cs173 - Spring 2004

  15. (E-1) “annihilates” c Homogeneous!!! CS 173 Recurrences Example: solve an = 5an-1 - 6an-2 + 4 (E2 - 5E + 6)an = 4 • Annihilate right side: (E-1)(E2 - 5E + 6)an = (E-1)4 (E3 - 6E2 + 11E - 6)an = 0 • But that’s just: an+3 - 6an+2 + 11an+1 - 6an = 0 an - 6an-1 + 11an-2 - 6an-3 = 0 Cs173 - Spring 2004

  16. Homogeneous!!! CS 173 Recurrences Example: solve an - 6an-1 + 11an-2 - 6an-3 = 0 • Characteristic equation: (r3 - 6r2 + 11r - 6) = 0 (r-1)(r-2)(r-3) = 0 • General solution: an = A1 + A22n + A33n Cs173 - Spring 2004

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