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Recurrence Relation - Discrete Mathematics - Solved Homework, Slides of Discrete Mathematics

Shoolini University of Biotechnology and Management SciencesDiscrete Mathematics

During the study of discrete mathematics, I found this course very informative and applicable.The main points in these lecture slides are:Recurrence Relation, Tower of Hanoi Puzzle, Different Arrangements, Recursive Formula, Deterministic Algorithm, Nonhomogeneous Recurrence, Recurrence Using Annihilators, Linear Recurrence, Asymptotic Solution

Typology: Slides

2012/2013

Uploaded on 04/27/2013

atmaja
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CS173: Discrete Mathematical Structures
Spring 2006
Homework #11
Due 04/23/05, 8a
Solutions, Grading Rubric
(43 Points)
1. (3 points each part for total of 12 points) In the Tower of Hanoi puzzle, suppose
our goal is to transfer all n disks from peg 1 to peg 3, but we cannot move a disk
directly between pegs 1 and 3. Each move of a disk must be a move involving
peg 2. As usual, we cannot place a disk on top of a smaller disk.
a. Find a recurrence relation for the number of moves required to solve the
puzzle for n disks with this added restriction.
b. Solve this recurrence relation to find a formula for the number of moves
required to solve the puzzle for n disks.
c. How many different arrangements are there of the n disks on three pegs so
that no disk is on top of a smaller disk?
d. Use your answers to parts b and c to argue that every allowable
arrangement of the n disks occurs in the solution of this variation of the
puzzle.
Solutions:
a. If we have one disk, then we move it to peg 2, then to peg 3. Thus a1=2.
Suppose the number of moves needed to move n disks from peg 1 to peg 3
is an. To move n+1 pegs, we have to move the n pegs from peg 1 to peg 3,
then move the largest disk to peg 2, then move n disks from peg 3 to peg
1, then move the largest disk to peg 3, and finally move the n disks to peg
3. Thus an= 3an-1+2.
b. (E-1)(E-3)=<0>. Thus the solution is: an=A+B3n. a1=2, a2=8 (you can get
this both from the recursive formula for an and by manually solving the
problem with two disks). We get A+3B=2; A+9B=8. Subtract equations to
get 6B=6. Thus B=1, A=-1, an=3n-1.
c. 3n.
d. During our solution no configuration can appear twice, since we have a
deterministic algorithm here. If we start with configuration X and after k
steps reach configuration X again, then we’ll return to the same
configuration at times k, 2k, 3k,… never solving the problem. Now, we
start with the initial configuration and have additional 3n-1 distinct
configurations until we solve the problem. Thus we see 3n-1+1
configurations during the solutions, which is all the possible
configurations.
2. (3 points each part for total of 15 points) Solve the following recurrences exactly:
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CS173: Discrete Mathematical Structures

Spring 2006

Homework

Due 04/23/05, 8a

Solutions, Grading Rubric

(43 Points)

  1. (3 points each part for total of 12 points) In the Tower of Hanoi puzzle, suppose our goal is to transfer all n disks from peg 1 to peg 3, but we cannot move a disk directly between pegs 1 and 3. Each move of a disk must be a move involving peg 2. As usual, we cannot place a disk on top of a smaller disk. a. Find a recurrence relation for the number of moves required to solve the puzzle for n disks with this added restriction. b. Solve this recurrence relation to find a formula for the number of moves required to solve the puzzle for n disks. c. How many different arrangements are there of the n disks on three pegs so that no disk is on top of a smaller disk? d. Use your answers to parts b and c to argue that every allowable arrangement of the n disks occurs in the solution of this variation of the puzzle. Solutions: a. If we have one disk, then we move it to peg 2, then to peg 3. Thus a 1 =2. Suppose the number of moves needed to move n disks from peg 1 to peg 3 is an. To move n+1 pegs, we have to move the n pegs from peg 1 to peg 3, then move the largest disk to peg 2, then move n disks from peg 3 to peg 1, then move the largest disk to peg 3, and finally move the n disks to peg 3. Thus an= 3an-1+2. b. (E-1)(E-3)=<0>. Thus the solution is: an=A+B3n. a 1 =2, a 2 =8 (you can get this both from the recursive formula for an and by manually solving the problem with two disks). We get A+3B=2; A+9B=8. Subtract equations to get 6B=6. Thus B=1, A=-1, an=3n-1. c. 3 n. d. During our solution no configuration can appear twice, since we have a deterministic algorithm here. If we start with configuration X and after k steps reach configuration X again, then we’ll return to the same configuration at times k, 2k, 3k,… never solving the problem. Now, we start with the initial configuration and have additional 3n-1 distinct configurations until we solve the problem. Thus we see 3n-1+ configurations during the solutions, which is all the possible configurations.
  2. (3 points each part for total of 15 points) Solve the following recurrences exactly:

a. a (^) n = 2 a (^) n − 1 + 2 n^2 , a (^) 1 = 4. b. a (^) n = − a (^) n − 1 + 2 a (^) n − 2 + 2 n^ −^1 , a (^) 0 = a (^) 1 = 1. c. a^ n^ =^9 a^ n 2

  • n^2 , a (^) 1 = 0. d. a^ n^ =^9 a^ n 4
  • n^2 , a (^) 1 = 0 , a (^) 2 = 1. e. a (^) n = 5 a (^) n 3 − 6 a (^) n 9 − 3 , a (^) 1 = 0 , a (^) 3 = 1 , a (^) 9 = 2. Solutions: a. (E-1) 3 (E-2)< an>=<0>. Thus: an=A+Bn+Cn^2 +D2n. Also: a 1 =4 a 0 =1, a 2 =16, a (^) -1=1/ A-B+C+D/2=1/2  2A-2B+2C+D= A+D= A+B+C+2D= A+2B+4C+4D= The solution to this system is: A=-12,B=-8,C=-2,D=13. The final solution is: an= 132n^ – 12 – 8n – 2n^2 b. (E-2)(E^2 +E-2) < an>=<0>. Thus: an= A2n^ + B(-2)n^ + C(1)n. Also: a 1 =a 0 =1  a 2 = The solution is: A=1/2; B=1/6; C=1/3. an= 2n-1^ - (-2)n-1/3 + 1/ c. Substitution: b(k)=a(2k)=9a(2k-1) + 4k^ = 9b(k-1) + 4k (E-4)(E-9) =<0>. Thus: bn= A4n^ + B9n. Also: b 0 = a(2^0 )=0, b 1 = a(2^1 )=4. The solution is: bn= -4/54n^ +4/59n. And in terms of an, an=-4/54logn^ +4/5*9logn d. a^ n^ =^9 a^ n 4
  • n^2 , a (^) 1 = 0 , a (^) 2 = 1. Substitution: b(k)=a(4k)=9a(4k-1) + 8k^ = 9b(k-1) + 8k (E-8)(E-9) =<0>. Thus: bn= A8n^ + B9n. Also: b 0 = a(4^0 )=0, b 1 = a(4^1 )=16. The solution is: bn= -164n^ +169n. And in terms of an, an=-164logn^ +169logn This is in case n=4k. there is another sequence here: bn= A8n^ + B9n. Also: b 0 = a(4^0 )=0, b1/2= a(41/2)=1. The solution is: bn= -2/(3-sqrt(8))4n^ +2/(3-sqrt(8))9n. And in terms of an, an=-2/(3-sqrt(8))4logn^ +2/(3-sqrt(8))9logn e. a (^) n = 5 a (^) n 3 − 6 a (^) n 9 − 3 , a (^) 1 = 0 , a (^) 3 = 1 , a (^) 9 = 2. Substitution: b(k)=a(3k)=5a(3k-1)-6a(3k-2) -3 = 5b(k-1)-6b(k-2)- (E-1)(E^2 -5E+6) =<0>. Thus: bn= A3n^ + B2n+C. Also: b 0 =0, b 1 =1, b 2 =2. The solution is: bn= (-1/2)3n^ + (2)2n-3/2. And in terms of an, an=(-1/2)3logn^ + (2)2logn-3/

Therefore in the limit an is constant, Θ(1).