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CSCI2110 – Discrete Mathematics Tutorial 7 Solving Recurrence

CSCI2110 – Discrete Mathematics Tutorial 7 Solving Recurrence. Wong Chung Hoi (Hollis) chwong@cse.cuhk.edu.hk 19-10-2011. Agenda. How to solve recurrence? Solving Second order linear homogeneous recurrence Fibonacci Series Counting Rabbits (Revisit) Counting Strings (Revisit)

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CSCI2110 – Discrete Mathematics Tutorial 7 Solving Recurrence

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  1. CSCI2110 – Discrete MathematicsTutorial 7Solving Recurrence Wong Chung Hoi (Hollis) chwong@cse.cuhk.edu.hk 19-10-2011

  2. Agenda • How to solve recurrence? • Solving Second order linear homogeneous recurrence • Fibonacci Series • Counting Rabbits (Revisit) • Counting Strings (Revisit) • Gambler’s Ruin

  3. How to solve recurrence? • By iteration • By guessing / observation • For second order linear homogenous recurrence • By Distinct Roots Theorem • By Single Root Theorem

  4. Second Order Linear Homogeneous Recurrence • The format: ak = Aak-1 + Bak-2 • Second Order • ak depends on ak-1, ak-2, but not ak-3, ak-4, or k etc • Linear • ak-1, ak-2 has degree 1 • Homogeneous • All terms has the same degree, i.e. no constant term

  5. Second Order Linear Homogenous Recurrence? • ak = 3ak-2 • ak = 2ak-2 + 3ak-3 • ak = 3ak-1 + 2 • ak = 4ak-12+4ak-22 • ak = 93ak-1 + 10-2ak-2 • ak = ak-1 + k • ak = a1ak-1 + a2ak-2, a1 = 1, a2 = 2

  6. Solving Second Order Linear Homogenous Recurrence • To solve: ak = Aak-1 + Bak-2, a1 = C, a2 = D Solve the characteristic equation t2 – At – B = 0 t has 2 roots? t has single root? Single-Root Theorem an = Ern + Fnrn Distinct-Roots Theorem an = Ern + Fsn Solve E, F by a1, a2

  7. Solving Characteristic Equation The characteristic equation for recurrence an = Aan-1 + Ban-2 is given by t2 – At – B = 0 Exercise: Solve for t in terms of A and B.

  8. Distinct-Roots Theorem If the characteristic equation t2 - At - B = 0 has 2 distinct roots s, r for t. Then, the closed form of an is given by an = Ern + Fsn Exercise: With the assumption when n = k-2, k-1, closed form an = Ern + Fsn is equivalent to recurrence an = Aan-1 + Ban-2, show that when n = k, they are equivalent.

  9. Single-Root Theorem If the characteristic equation t2 - At - B = 0 has only single root r for t. Then, the closed form of an is given by an = Ern + Fnrn Exercise: Show that A = -2r, B = r2. With the assumption when n = k-2, k-1, closed form an = Ern + Fnrn is equivalent to recurrence an = Aan-1 + Ban-2, show that when n = k, they are equivalent.

  10. Solving E, F With boundary values a1 = C, a2 = D solve E,F by C = Er + Fs, D = Er2 + Fs2 for distinct roots C = Er + Fr, D = Er2 + F2r2 for single root

  11. Agenda • How to solve recurrence? • Solving Second order linear homogeneous recurrence • Fibonacci Series • Counting Rabbits (Revisit) • Counting Strings (Revisit) • Gambler’s Ruin

  12. Fibonacci Series • an = an-1 + an-2, a0 = 1, a1 = 1 • Characteristic equation t2 – t - 1 = 0 t = (1+51/2)/2 or (1-51/2)/2 • Distinct roots theorem an = E ((1+51/2)/2)n + F ((1-51/2)/2)n • Solving E, F by a0, a1 1 = E + F 1 = E (1+51/2)/2+ F (1-51/2)/2 E = (51/2 + 1) / (2 x 51/2) F = (51/2 - 1) / (2 x 51/2) an = 5-1/2 x((1+51/2)/2)n+1 – 5-1/2 x ((1-51/2)/2)n+1

  13. Fibonacci Series • an = an-1 + an-2, a0 = 2, a1 = 4 • Characteristic equation t2 – t - 1 = 0 t = (1+51/2)/2 or (1-51/2)/2 • Distinct roots theorem an = E ((1+51/2)/2)n + F ((1-51/2)/2)n • Solving E, F by a0, a1 2 = E + F 4 = E (1+51/2)/2+ F (1-51/2)/2 E = (51/2 + 3) / 51/2 F = (51/2 - 3) / 51/2 an = (1+ 3/51/2) ((1+51/2)/2)n + (1 – 3/51/2) ((1-51/2)/2)n

  14. Agenda • How to solve recurrence? • Solving Second order linear homogeneous recurrence • Fibonacci Series • Counting Rabbits (Revisit) • Counting Strings (Revisit) • Gambler’s Ruin

  15. Counting Rabbits - Revisit • Problem • Start with 1 pair of baby rabbit. • Rabbits take 2 months to become an adult. • Adult rabbits gives birth to 3 pairs of baby rabbits every month. • Rabbits NEVER DIE! • Number of pairs of rabbits in the end of the year? • Denote Rn as the number of rabbit after n month. • R0 = 1, R1 = 1, R2 = 1, R3 = 4, R4 = 7 • Rn = 3Rn-3 + Rn-1

  16. Counting Rabbits - Revisit • Problem • Start with 1 pair of baby rabbit. • Rabbits take 1 months to become an adult. • Adult rabbits gives birth to 3 pairs of baby rabbits every month. • Rabbits NEVER DIE! • Number of pairs of rabbits in the end of the year? • Denote Rn as the number of rabbit after n month. • R0 = 1, R1 = 1, R2 = 4, R3 = 7 • Rn = 3Rn-2 + Rn-1

  17. Counting Rabbits - Revisit • Finding the closed form: • Characteristic equation t2 – t – 3 = 0 t = (1-131/2)/2 or (1+131/2)/2 • Distinct-Roots Theorem Rn = E((1-131/2)/2)n + F((1+131/2)/2)n • Solve E, F by R0, R1 1 = E + F 1 = E(1-131/2)/2 + F(1+131/2)/2 E = (131/2 - 1) / (2 x 131/2) F = (131/2 + 1) / (2 x 131/2)

  18. Agenda • How to solve recurrence? • Solving Second order linear homogeneous recurrence • Fibonacci Series • Counting Rabbits (Revisit) • Counting Strings (Revisit) • Gambler’s Ruin

  19. Counting Strings - Revisit • Problem • A string made of 3 types of alphabets “a”, “b”, “c” • Count the number of string with length n and without the pattern “aa” • Let Ln be the number of string with length n and without pattern “aa” • L1 = 3, L2 = 8, L3 = 22… • Ln = 2Ln-1 + 2Ln-2,

  20. Counting Strings - Revisit Find the closed form for Ln = 2Ln-1 + 2Ln-2 • Characteristic Equation t2 – 2t – 2 = 0 t = 1 + 31/2, or 1 – 31/2 • By distinct roots theorem Ln = E(1+31/2)n + F(1-31/2)n • Solve E,F by L1, L2 3 = E(1+31/2) + F(1-31/2) 8 = E(1+31/2)2 + F(1-31/2)2 E = ? F = ?

  21. Counting Strings - Revisit • Solve E,F by L1, L2 3 = E(1+31/2) + F(1-31/2) 8 = E(1+31/2)2 + F(1-31/2)2 This is too complicated! Trick: What if L0 is part of our sequence? L2 = 2L1 + 2L0 → 8 = 2 x 3 + 2 L0 → L0 = 1 Solve E,F by L0, L1 1 = E + F 3 = E(1+31/2) + F(1-31/2) Exercise: Show that Ln = [(31/2+2)(1+31/2)n + (31/2 - 2)(1-31/2)n] / (2 x 31/2)

  22. Agenda • How to solve recurrence? • Solving Second order linear homogeneous recurrence • Fibonacci Series • Counting Rabbits (Revisit) • Counting Strings (Revisit) • Gambler’s Ruin

  23. Gambler’s Ruin • Problem: • Betting $1 on the outcome of a die until he has $300. • +$1 if he won, -$1 if he lost. • Probability that the gambler is ruined when he begin with $n? • Let Pn be the probability when he begin with $n. • P0 = 1, P300 = 0 • Recurrence equation: Pn-1 = 1/6 x Pn + 5/6 x Pn-2

  24. Gambler’s Ruin • Finding the closed form: • Characteristic equation: t2 - 6t + 5 = 0 t = 1 or 5 • Distinct-Roots Theorem Pn = E + F5n • Solve E, F by P0, P300 1 = E + F 0 = E + F5300 E = 5300 / (5300 - 1) F = 1/(1- 5300) Pn = (5300 – 5n) / (5300-1) • Exercise: Verify the closed form.Exercise: Approximate P20.

  25. Gambler’s Ruin 2 • Problem: • Betting $1 on the outcome of a coin until he has $300. • +$1 if he won, -$1 if he lost. • Probability that the gambler is ruined when he begin with $n? • Let Pn be the probability when he begin with $n. • P0 = 1, P300 = 0 • Recurrence equation: Pn-1 = 1/2 x Pn + 1/2 x Pn-2

  26. Gambler’s Ruin 2 • Finding the closed form: • Characteristic equation: t2 - 2t + 1 = 0 t = 1 • Single-Root Theorem Pn = E1n + Fn1n • Solve E, F by P0, P300 1 = E 0 = E + 300F E = 1 F = -1/300 Pn = 1-n/300 • Exercise: Verify the closed form.Exercise: CalculateP20.

  27. Summary • Ways of solving recurrence • Iteration, guessing / observing • What is Second order linear homogenous recurrence? • How to solve?

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