6.1.1 Section 6.1 Recurrence Relations 6.1 RECURRENCE RELATIONS def: A recurrence system is a finite set of initial conditions a 0 = c 0 , a 1 = c 1 , a d = c d . . . , and a formula (called a recurrence relation ) a n = f ( a 0 , . . . , a n − 1 ) that expresses a subscripted variable as a function of lower-indexed values. A sequence = < a n > a 0 , a 1 , a 2 , . . . satisfying the initial conditions and the recur- rence relation is called a solution . The recurrence system with Example 6.1.1: initial condition a 0 = 0 and recurrence relation a n = a n − 1 + 2 n − 1 has the sequence of squares as its solution: = 0 , 1 , 4 , 9 , 16 , 25 , . . . < a n > Coursenotes by Prof. Jonathan L. Gross for use with Rosen: Discrete Math and Its Applic., 5th Ed.
6.1.2 Chapter 6 ADVANCED COUNTING TECHNIQUES N¨ AIVE METHOD OF SOLUTION Step 1. Use the recurrence to calculate a few more values beyond the given initial values. Step 2. Spot a pattern and guess the right answer. Step 3. Prove your answer is correct (by induction). Example 6.1.1, continued: Step 1. Starting from a 0 = 0, we calculate = a 0 + 2 · 1 − 1 = 0 + 1 = 1 a 1 = a 1 + 2 · 2 − 1 = 1 + 3 = 4 a 2 = a 0 + 2 · 3 − 1 = 4 + 5 = 9 a 1 = a 0 + 2 · 4 − 1 = 9 + 7 = 16 a 1 Step 2. Looks like f ( n ) = n 2 . Step 3. BASIS: a 0 = 0 = 0 2 = f (0). IND HYP: Assume that a n − 1 = ( n − 1) 2 . IND STEP: Then a n = a n − 1 + 2 n − 1 from the recursion = ( n − 1) 2 + 2 n − 1 by IND HYP = ( n 2 − 2 n + 1) + 2 n − 1 n 2 = ♦ Coursenotes by Prof. Jonathan L. Gross for use with Rosen: Discrete Math and Its Applic., 5th Ed.
6.1.3 Section 6.1 Recurrence Relations APPLICATIONS Compound Interest Example 6.1.2: Deposit $1 to compound at annual rate r . p 0 = 1 p n = (1 + r ) p n − 1 EARLY TERMS: 1 , 1 + r, (1 + r ) 2 , (1 + r ) 3 , . . . APPARENT PATTERN: p n = (1 + r ) n BASIS: True for n = 0. IND HYP: Assume that p n − 1 = (1 + r ) n − 1 IND STEP: Then p n = (1 + r ) p n − 1 by the recursion = (1 + r )(1 + r ) n − 1 by IND HYP = (1 + r ) n by arithmetic ♦ Coursenotes by Prof. Jonathan L. Gross for use with Rosen: Discrete Math and Its Applic., 5th Ed.
6.1.4 Chapter 6 ADVANCED COUNTING TECHNIQUES Tower of Hanoi Example 6.1.3: RECURRENCE SYSTEM h 0 = 0 h n = 2 h n − 1 + 1 SMALL CASES: 0 , 1 , 3 , 7 , 15 , 31 , . . . APPARENT PATTERN: h n = 2 n − 1 BASIS: h 0 = 0 = 2 0 − 1 IND HYP: Assume that h n − 1 = 2 n − 1 − 1 IND STEP: Then h n = 2 h n − 1 + 1 by the recursion = 2(2 n − 1 − 1) + 1 by IND HYP = 2 n − 1 by arithmetic ♦ Coursenotes by Prof. Jonathan L. Gross for use with Rosen: Discrete Math and Its Applic., 5th Ed.
6.1.5 Section 6.1 Recurrence Relations However, the n¨ aive method has limitations: • It can be non-trivial to spot the pattern. • It can be non-trivial to prove that the apparent pattern is correct. Fibonacci Numbers Example 6.1.4: f 0 = 0 f 1 = 1 f n = f n − 1 + f n − 2 Fibo seq: 0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , . . . . APPARENT PATTERN (ha ha) √ √ 1 � 5) n � 5) n − (1 − 2 n √ f n = (1 + 5 It is possible, but not uncomplicated, to simplify this with the binomial expansion and to then use induction. Coursenotes by Prof. Jonathan L. Gross for use with Rosen: Discrete Math and Its Applic., 5th Ed.
6.1.6 Chapter 6 ADVANCED COUNTING TECHNIQUES Sometimes there is no fixed limit on the number of previous terms used by a recursion. Catalan Recursion Example 6.1.5: c 0 = 1 c n = c 0 c n − 1 + c 1 c n − 2 + · · · + c n − 1 c 0 for n ≥ 1. SMALL CASES c 1 = c 0 c 0 = 1 · 1 = 1 c 2 = c 0 c 1 + c 1 c 0 = 1 · 1 + 1 · 1 = 2 c 3 = c 0 c 2 + c 1 c 1 + c 2 c 0 = 1 · 2 + 1 · 1 + 2 · 1 = 5 c 4 = 1 · 5 + 1 · 2 + 2 · 1 + 5 · 1 = 14 c 5 = 1 · 14 + 1 · 5 + 2 · 2 + 5 · 1 + 14 · 1 = 42 Catalan seq: 1 , 1 , 2 , 5 , 14 , 42 , . . . . � 2 n � 1 SOLUTION: c n = n + 1 n The Catalan recursion counts binary trees and other objects in computer science. ADMONITION • Most recurrence relations have no solution. • Most sequences have no representation as a recurrence relation. (they are random) Coursenotes by Prof. Jonathan L. Gross for use with Rosen: Discrete Math and Its Applic., 5th Ed.
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