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PDF Discrete Structures Tutorial 7

a) Find the recurrence relation for the number of ways to deposit n dollars where the order in which the coins and bills are deposited matters b) Find initial conditions c) How many ways are there to deposit $10? Solution: a) an= 2an-1 + an-5 for n>=5 b) a0 =1 a1=2 a2=2 a3=8 a4=16 c) 1217 4

PDF Lecture Notes for Data Structures and Algorithms

Contents 1 Introduction 5 1 1 Algorithms as opposed to programs 5 1 2 Fundamental questions about algorithms

PDF Recurrence Relations

Recurrence Relations are Mathematical Equations: A recurrence relation is an equation which is defined in terms of itself Natural Computable Functions as Recurrences: Many natural functions are expressed using recurrence relations (linear) f(n) = f(n−1) + 1f(1) = 1 ⇒ f(n) = n (polynomial) f(n) = f(n−1) + nf(1) = 1 ⇒ f(n) = 1 2 (n2 + n)

PDF Recurrence Relations

Recurrence Relations 1 In ̄nite Sequences An in ̄nite sequence is a function from the set of positive integers to the set of real numbers or to the set of complex numbers Example 1 1 The game of Hanoi Tower is to play with a set of disks of graduated size with holes in their centers and a playing board having three spokes for holding the disks

  • How do you solve a recurrence relation?

    Solve the recurrence relations an + 2 = 4an +1 − 3an + 3n subject to the initial conditions a1 = 1 and a2 = 3 . Solution: We first solve the homogeneous part an + 2 = 4an +1 − 3an . the general solution is an = A ⋅ 3 n + B for some constants A and B. Now we want to find a particular solution to the original (non-homogenous) recurrence relation.

  • Which recurrence relation can be defined with different initial conditions?

    in Section 8.2, Example 8.2.1, can be defined with the same recurrence relation, but with different initial conditions. is a third-order recurrence relation. If values of T(0), is completely defined. has infinite order. To determine S(n) terms. Since n / 2

  • What is 2nd order recurrence relation?

    Subtract the 4th equation from the 3rd. 2k + 1 term is eliminated. This is 2nd order relation. The recurrence relation that we have just obtained, defined for k ≥ 2, together with the initial conditions C(0) = 7 / 3 and C(1) = 6, define C. Table 8.3.2 summarizes our results together with a few other examples that we will let the reader derive.

  • What is the required solution to the homogeneous recurrence relation?

    Next, we use the fact the required solution to the recurrence relation is the sum of this particular solution and a so-lution to the associated homogeneous recurrence relation. We already know a general form for the solution to the homogeneous recurrence relation (from the previous theorems).

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