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CS322

Week 6 - Wednesday. CS322. Last time. What did we talk about last time ? Examples of induction Strong induction Recurrence relations. Questions?. Logical warmup. On the planet Og , there are green people and red people Likewise, there are northerners and southerners

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CS322

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  1. Week 6 - Wednesday CS322

  2. Last time • What did we talk about last time? • Examples of induction • Strong induction • Recurrence relations

  3. Questions?

  4. Logical warmup • On the planet Og, there are green people and red people • Likewise, there are northerners and southerners • Green northerners tell the truth • Red northerners lie • Green southerners lie • Red southerners tell the truth • Consider the following statements by two natives named Ork and Bork: • Ork: Bork is from the north • Bork:Ork is from the south • Ork: Bork is red • Bork:Ork is green • What are the colors and origins of Ork and Bork?

  5. The proof is on fire • Perhaps you believe that you have the correct recurrence relation for perfect squares • Can you prove it? • Hint: Use mathematical induction • Recursion and induction and two sides of the same coin • The right cross and the left jab, if you will, of the computer scientist's arsenal

  6. Tower of Hanoi • Imagine that you have a tower of disks such that each is smaller than the one it rests on • Rules: • There are 3 pegs, and all the disks are on peg 1 • No two disks are the same size • A larger disk may not be placed on a smaller disk • Disks can only be moved one at a time • Move all the disks to peg 3

  7. Power of Hanoi • What's the smallest number of moves needed to move the disks? • Consider the following algorithm for moving k disks from the starting pole to the ending pole: • (Recursively) transfer the top k – 1 disks from the starting pole to the temporary pole • Move the bottom disk from the starting pole to the ending pole • (Recursively) move the top k – 1 disks from the temporary pole to the ending pole

  8. Tower of annoy • How do we represent the running time of this algorithm recursively? • We have to (recursively) move k – 1 disks, then a single disk, then (recursively) another k – 1 disks • mk = mk-1 + 1 + mk-1 or • mk = 2mk-1 + 1 • Clearly, it takes 1 move to move a single disk, so m1 = 1 • Find m2, m3, m4, and m5

  9. Fibonacci • We all know and love Fibonacci by now, but where does it come from? • It is actually supposed to model rabbit populations • For the first month of their lives, they cannot reproduce • After that, they reproduce every single month

  10. Fibonacci rules • From this information about rabbit physiology (which is a simplification, of course) we can think about pairs of rabbits • At time k, rabbits born at time k – 1 will not reproduce • Any rabbits born at k – 2 or earlier will, however • So, we assume that all the rabbits from time k – 2 have doubled between time k – 1 and k • Thus, our recurrence relation is: • Fk = Fk – 1 + Fk – 2, k 2 • Assuming one starting pair of rabbits, our initial conditions are: • F0 = 1 • F1 = 1

  11. Compound interest • It's boring, but useful • Interest is compounded based on some period of time • We can define the value recursively • Let i is the annual percentage rate (APR) of interest • Let m be the number of times per year the interest is compounded • Thus, the total value of the investment at the kth period is • Pk = Pk-1 + Pk-1(i/m), k 1 • P0 = initial principle

  12. Solving Recurrence Relations

  13. Recursion • … is confusing • We don't naturally think recursively (but perhaps you can raise your children to think that way?) • As it turns out, the total number of moves needed to solve the Tower of Hanoi for n disks is 2n – 1 • Likewise, with an interest rate of i, a principle of P0, and m periods per year, the investment will yield Po(i/m + 1)k after k periods

  14. Finding explicit formulas by iteration • Consequently, we want to be able to turn recurrence relations into explicit formulas whenever possible • Often, the simplest way is to find these formulas by iteration • The technique of iteration relies on writing out many expansions of the recursive sequence and looking for patterns • That's it

  15. Iteration example • Find a pattern for the following recurrence relation: • ak = ak-1 + 2 • a0 = 1 • Start at the first term • Write the next below • Do not combine like terms! • Leave everything in expanded form until patterns emerge

  16. Arithmetic sequence • In principle, we should use mathematical induction to prove that the explicit formula we guess actually holds • The previous example (odd integers) shows a simple example of an arithmetic sequence • These are recurrences of the form: • ak = ak-1 + d, for integers k ≥ 1 • Note that these recurrences are always equivalent to • an = a0 + dn, for all integers n ≥ 0 • Let's prove it

  17. Geometric sequence • Find a pattern for the following recurrence relation: • ak = rak-1, k ≥ 1 • a0 = a • Again, start at the first term • Write the next below • Do not combine like terms! • Leave everything in expanded form until patterns emerge

  18. Geometric sequence • It appears that any geometric sequence with the following form • ak = rak-1, k ≥ 1 • is equivalent to • an = a0rn, for all integers n ≥ 0 • This result applies directly to compound interest calculation • Let's prove it

  19. Employing outside formulas • Sure, intelligent pattern matching gets you a long way • However, it is sometimes necessary to substitute in some known formula to simplify a series of terms • Recall • Geometric series: 1 + r + r2 + … + rn = (rn+1 – 1)/(r – 1) • Arithmetic series: 1 + 2 + 3 + … + n = n(n + 1)/2

  20. Tower of Hanoi solution • Find a pattern for the following recurrence relation: • mk = 2mk-1 + 1, k ≥ 2 • m1 = 1 • Again, start at the first term • Write the next below • Do not combine like terms! • Leave everything in expanded form until patterns emerge • Use the arithmetic series or geometric series equations as needed

  21. How many edges are in a complete graph? • In a complete graph, every node is connected to every other node • If we want to make a complete graph with k nodes, we can take a complete graph with k – 1 nodes, add a new node, and add k – 1 edges (so that all the old nodes are connected to the new node • Recursively, this means that the number of edges in a complete graph is • sk = sk-1 + (k – 1), k ≥ 2 • s1 = 0 (no edges in a graph with a single node) • Use iteration to solve this recurrence relation

  22. Mistakes were made! • You might make a mistake when you are solving by iteration • Consider the following recursive definition • ck = 2ck-1 + k, k ≥ 1 • c0 = 1 • Careless analysis might lead you to the explicit formula • cn = 2n + n, n ≥ 0 • How would this be caught in a proof by induction verification?

  23. Solving Second-Order Linear Homogeneous Relations with Constant Coefficients Student Lecture

  24. Solving Second-Order Linear Homogeneous Relations with Constant Coefficients

  25. Second-Order Linear Homogeneous Relations with Constant Coefficients • Second-order linear homogeneous relations with constant coefficients are recurrence relations of the following form: • ak = Aak-1 + Bak-2 where A, B R and B  0 • These relations are: • Second order because they depend on ak-1 and ak-2 • Linear because ak-1 and ak-2 are to the first power and not multiplied by each other • Homogeneous because there is no constant term • Constant coefficients because A and B are fixed values

  26. Why do we care? • I'm sure you're thinking that this is an awfully narrow class of recurrence relations to have special rules for • It's true: There are many (infinitely many) ways to formulate a recurrence relation • Some have explicit formulas • Some do not have closed explicit formulas • We care about this one partly for two reasons • We can solve it • It lets us get an explicit formula for Fibonacci!

  27. Pick 'em out • Which of the following are second-order linear homogeneous recurrence relations with constant coefficients? • ak = 3ak-1 + 2ak-2 • bk = bk-1 + bk-2+ bk-3 • ck = (1/2)ck-1 – (3/7)ck-2 • dk = d2k-1 + dk-1dk-2 • ek = 2ek-2 • fk = 2fk-1 + 1 • gk = gk-1 + gk-2 • hk = (-1)hk-1 + (k-1)hk-2

  28. Characteristic equation • We will use a tool to solve a SOLHRRwCC called its characteristic equation • The characteristic equation of ak = Aak-1 + Bak-2 is: • t2 – At – B = 0 where t 0 • Note that the sequence 1, t, t2, t3, …, tn satisfies ak = Aak-1 + Bak-2 if and only if t satisfies t2 – At – B = 0

  29. Demonstrating the characteristic equation • We can see that 1, t, t2, t3, …, tn satisfies ak = Aak-1 + Bak-2 if and only if t satisfies t2 – At – B = 0 by substituting in t terms for ak as follows: • tk = Atk-1 + Btk-2 • Since t 0, we can divide both sides through by tk-2 • t2 = At + B • t2 – At – B = 0

  30. Using the characteristic equation • Consider ak = ak-1 + 2ak-2 • What is its characteristic equation? • t2 – t – 2 = 0 • What are its roots? • t = 2 and t = -1 • What are the sequences defined by each value of t? • 1, 2, 22, 23, …, 2n, … • 1, -1, 1, -1, …, (-1)n, … • Do these sequences satisfy the recurrence relation?

  31. Finding other sequences • An infinite number of sequences satisfy the recurrence relation, depending on the initial conditions • If r0, r1, r2, … and s0, s1, s2, … are sequences that satisfy the same SOLHRRwCC, then, for any C, D R, the following sequence also satisfies the SOLHRRwCC • an = Crn + Dsn, for integers n ≥ 0

  32. Finding a sequence with specific initial conditions • Solve the recurrence relation ak = ak-1 + 2ak-2 where a0 = 1 and a1 = 8 • The result from the previous slide says that, for any sequence rk and sk that satisfy ak • an = Crn + Dsn, for integers n ≥ 0 • We have these sequences that satisfy ak • 1, 2, 22, 23, …, 2n, … • 1, -1, 1, -1, …, (-1)n, … • Thus, we have • a0 = 1 = C20 + D(-1)0 = C + D • a1 = 8 = C21 + D(-1)1 = 2C – D • Solving for C and D, we get: • C = 3 • D = -2 • Thus, our final result is an = 3∙2n – 2(-1)n

  33. Distinct Roots Theorem • We can generalize this result • Let sequence a0, a1, a2, … satisfy: • ak = Aak-1 + Bak-2 for integers k ≥ 2 • If the characteristic equation t2 – At – B = 0 has two distinct roots r and s, then the sequence satisfies the explicit formula: • an = Crn + Dsn • where C and D are determined by a0 and a1

  34. Now we can solve Fibonacci! • Fibonacci is defined as follows: • Fk = Fk-1 + Fk-2, k ≥ 2 • F0 = 1 • F1 = 1 • What is its characteristic equation? • t2 – t – 1 = 0 • What are its roots? • Thus,

  35. We just need C and D • So, we plug in F0 = 1 and F1 = 1 • Solving for C and D, we get • Substituting in, this yields

  36. Single-Root Case • Our previous technique only works when the characteristic equation has distinct roots r and s • Consider sequence ak = Aak-1 + Bak-2 • If t2 – At – B = 0 only has a single root r, then the following sequences both satisfy ak • 1, r1, r2, r3, … rn, … • 0, r, 2r2, 3r3, … nrn, …

  37. Single root equation • Our old rule said that if r0, r1, r2, … and s0, s1, s2, … are sequences that satisfy the same SOLHRRwCC, then, for any C, D R, the following sequence also satisfies the SOLHRRwCC • an = Crn + Dsn, for integers n ≥ 0 • For the single root case, this means that the explicit formula is: • an = Crn + Dnrn, for integers n ≥ 0 • where C and D are determined by a0 and a1

  38. SOLHRRwCC recap • To solve sequence ak = Aak-1 + Bak-2 • Find its characteristic equation t2 – At – B = 0 • If the equation has two distinct roots r and s • Substitute a0 and a1 into an = Crn + Dsn to find C and D • If the equation has a single root r • Substitute a0 and a1 into an = Crn + Dnrn to find C and D

  39. Quiz

  40. Upcoming

  41. Next time… • General recursive definitions • Introduction to set theory

  42. Reminders • Homework 4 is due on Friday • Read Chapter 6

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