CS 161 Lecture 3 - 1 Solving recurrences
Last class we introduced recurrence relations such as T(n) = 2T(?n/2?) + n. Typically Here a = 9
Discrete Mathematics II (Spring 2015) - 8.2 Solving Linear
Determine what is the degree of the recurrence relation. • Need to know the general solution equations. • Need to find characteristic equation. • Need to find
Recurrences
n = 2k for some constant k and leaving out base case and constant in ?). Methods for solving recurrences. 1. Substitution method. 2. Iteration method.
Q1: Exercise 2.3-3 (10 points) Use mathematical induction to show
recurrence. T(n) =2 ifn=2. 2T(n/2) + n if n = 2k for k > 1 is T(n) = n logn. . Base Step: If n = 2
Master Theorem: Practice Problems and Solutions
The Master Theorem applies to recurrences of the following form: 2. T(n)=4T(n/2) + n2. 3. T(n) = T(n/2) + 2n. 4. T(n)=2nT(n/2) + nn.
GENERATING FUNCTIONS AND RECURRENCE RELATIONS
A recurrence recurrence relation is a set of equations Linear Recurrence. Fibonacci Sequence an = an?1 + an?2 n ? 2. ... n=2 n(n ? 1)xn +.
Solving Recurrences
For example the following recurrence (written in two different but standard ways) describes the identity function f (n) = n:.
1.1 Time complexity and Big-Oh notation: exercises
2. A quadratic algorithm with processing time T(n) = cn2 spends T(N) To have the well-defined recurrence assume that n = km with the integer.
Sample Induction Proofs
(?1)nn(n + 1). 2 . Base case: When n = 1 the left side of (1) is (?1)12 = ?1
3. Recurrence 3.1. Recursive Definitions. To construct a recursively
Example 3.2.1. The factorial function f(n) = n! is defined recursively as follows: 1. Initial Condition: f(0) = 1. 2. Recursion: f(n +1)=(n + 1)f(n).
Recurrence Relations - Hong Kong University of Science and
an= 2 n¡1; n ‚1: Given a recurrence relation for a sequence with initial conditions Solving the recurrence relation means to ?nd a formula to express the general termanof the sequence 2 Homogeneous Recurrence Relations Any recurrence relation of the form
Solving Recurrence Relations - Texas A&M University
n 2 and the base cases of the induction proof (which is not the same as the base case of the recurrence!) are n= 2 and n= 3 (We are allowed to do this because asymptotic notation only requires us to prove our statement for n n 0 and we can set n 0 = 2 ) We choose n= 2 and n= 3 for our base cases because when we expand the recurrence
Recurrence Relations - University of Ottawa
Many sequences can be a solution for the same recurrence relation a n = 2a n 1 a n 2; for n 2 The following sequences are solutions of this recurrence relation: I a n = 3n for all n 0 I a n = 5 for all n 0 The initial conditions for a sequence specify the terms before n 0 (before the recurrence relation takes e ect)
Solving Recurrence Relations - Princeton University
n 2 to be C0 and C1 Since the sequences fang and f 1rn 1 + 2r n 2g satisfy the degree 2 recurrence and agree on the rst two values they must be identical Example: Find the solution to the recurrence relation an = an 1 +an 2 with initial condi-tions a0 = 2 and a1 = 7 Solution: The characteristic equation is r2 r 2 = 0 i e (r 2)(r +1) = 0
Recurrence Relations - University of Illinois Urbana-Champaign
2 Recall that 1=nk= nk and we have rn 2(r25r + 6) = 0 Then we can factor the polynomial into rn 2(r 2)(r 3) = 0 We assumed that r is non-zero so we can divide by the rst factor rn 2 which leaves (r 2)(r 3) = 0 Thus the characteristic function (also: polynomial) remains on the left-hand side
Searches related to 2^n n^2 recurrence PDF
n = 2n for every nonnegative integer n is a solution of the recurrence relation a n = 2a n-1 - a n-2 for n = 234 Assume a 0=1 and a 1=2 Solution: Check initial conditions a 0= 20 = 1 a 1= 2 1 = 2 Check if {2n} satisfies a n = 2a n-1 - a n-2 2a n-1 - a n-2 = 2(2n-1) - 2n-2 = 2n-2(4-1) = 3(2n-2) 2n So {2n} is not a solution for a n = 2a
What is the solution of the recurrence relation?
.Thenasequence (a n ) is a solution of the recurrence relation a n= c 1a n?1+ c 2a n?2 if and only if a n= ? 1r n 1+ ? 2r n 2 for n ? 0 for some constants ? 1,? 2 Idea of the Proof The proof proceeds exactly as in the case of the Fibonacci numbers. Try to prove it yourself!
What is the solution of recurrence relation an = 10an-1 - 25an-2?
The solution of the recurrence relation an = 10an-1 - 25an-2, is: (a0 = 1, a1 = 15) Q8. Solution to recurrence relation T (n) = T (n - 1) + 4 is given by, where n > 0 and T (0) = 3.
What is a 2 nd-order linear recurrence?
This is a subset of the type of 2 nd -order recurrence formulas used by MCS . The Fibonacci number s are the best-known example of a 2 nd -order linear recurrence.
What is the initial condition for a recurrence relation?
So (a_n = 3a_ {n-1} + 2) is our recurrence relation and the initial condition is (a_0 = 1 ext {.}) We are going to try to solve these recurrence relations.
GENERATING FUNCTIONS AND RECURRENCE
RELATIONS
Generating Functions
Recurrence Relations
Supposea0;a1;a2;:::;an;:::;is an infinite sequence. A recurrence recurrence relation is a set of equations a n=fn(an1;an2;:::;ank):(1) The whole sequence is determined by (6) and the values of a0;a1;:::;ak1.Generating Functions
Linear Recurrence
Fibonacci Sequence
a n=an1+an2n2: a0=a1=1.Generating Functions
b n=jBnj=jfx2 fa;b;cgn:aadoes not occur inxgj: b1=3:a b c
b2=8:ab ac ba bb bc ca cb cc
b n=2bn1+2bn2n2:Generating Functions b n=2bn1+2bn2n2: Let B n=B(b) n[B(c) n[B(a) n whereB() n=fx2Bn:x1=gfor=a;b;c.NowjB(b)
nj=jB(c) nj=jBn1j. The mapf:B(b) n!Bn1, f(bx2x3:::xn) =x2x3:::xnis a bijection. B (a) n=fx2Bn:x1=aandx2=borcg. The map g:B(a) n!B(b) n1[B(c) n1, g(ax2x3:::xn) =x2x3:::xnis a bijection.Hence,jB(a)
nj=2jBn2j.Generating FunctionsTowers of HanoiPeg 1Peg 2 Peg 3
Hnis the minimum number of moves needed to shift
n rings from Peg 1 to Peg 2. One is not allowed to place a larger ring on top of a smaller ring.Generating Functions xxxHn-1moves1 moveHn-1 movesGenerating Functions
We see thatH1=1and Hn=2Hn1+1f orn2.
So, H n2 nHn12 n1=12 n:Summing these equations give
H n2 nH12 =12 n+12 n1++14 =12 12 n: So H n=2n1:Generating Functions Ahasndollars. EverydayAbuys one of a Bun (1 dollar), an Ice-Cream (2 dollars) or a Pastry (2 dollars). How many ways are there (sequences) forAto spend his money? Ex.BBPIIPBI
represents "Da y1, b uyBun. Da y2, b uyBun etc. ". u n=number of ways =un;B+un;I+un;P whereun;Bis the number of ways whereAbuys a Bun on day1 etc.
u n;B=un1;un;I=un;P=un2. So u n=un1+2un2; and u0=u1=1:Generating Functions
Ifa0;a1;:::;anis a sequence of real numbers then its (ordinary) generating functiona(x)is given by a(x) =a0+a1x+a2x2+anxn+ and we write a n= [xn]a(x):For more on this subject see
Gener atingfunctionology
b ythe late Herbert S. Wilf. The book is available from https://www.math.upenn.edu// wilf/DownldGF.htmlGenerating Functions
a n=1 a(x) =11x=1+x+x2++xn+ a n=n+1. a(x) =1(1x)2=1+2x+3x2++ (n+1)xn+ a n=n. a(x) =x(1x)2=x+2x2+3x3++nxn+Generating FunctionsGeneralised binomial theorem:
a n= n a(x) = (1+x)=1X n=0 n x n: where n =(1)(2)(n+1)n!: a n=m+n1 n a(x) =1(1x)m=1X n=0 m n (x)n=1X n=0 m+n1 n x n:Generating FunctionsGeneral view.
Given a recurrence relation for the sequence(an), we (a) Deduce from it, an equation satisfied by the generating functiona(x) =P nanxn. (b) Solve this equation to get an explicit expression for the generating function. (c) Extract the coefficientanofxnfroma(x), by expandinga(x) as a power series.Generating Functions
Solution of linear recurrences
a n6an1+9an2=0n2: a0=1;a1=9.
1 X n=2(an6an1+9an2)xn=0:(2)Generating Functions 1 X n=2a nxn=a(x)a0a1x =a(x)19x: 1X n=26an1xn=6x1X n=2a n1xn1 =6x(a(x)a0) =6x(a(x)1): 1X n=29an2xn=9x21X n=2a n2xn2 =9x2a(x):Generating Functions a(x)19x6x(a(x)1) +9x2a(x) =0 or a(x)(16x+9x2)(1+3x) =0: a(x) =1+3x16x+9x2=1+3x(13x)2 1X n=0(n+1)3nxn+3x1X n=0(n+1)3nxn 1X n=0(n+1)3nxn+1X n=0n3nxn 1X n=0(2n+1)3nxn: a n= (2n+1)3n:Generating FunctionsFibonacci sequence:
1 X n=2(anan1an2)xn=0: 1 X n=2a nxn1X n=2a n1xn1X n=2a n2xn=0: (a(x)a0a1x)(x(a(x)a0))x2a(x) =0:quotesdbs_dbs24.pdfusesText_30[PDF] les composantes de lenvironnement 6ème
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