1 / 42

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

kyrie
Download Presentation

CS322

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related