Recursive Algorithms Recursion a Recursive Algorithm Is One in Which Objects Are Deﬁned in Terms of Other Objects of the Same Type

Home , Recurrence relation

Recursive Algorithms Recursion A recursive algorithm is one in which objects are deﬁned in terms of other objects of the same type. Slides by Christopher M. Bourke Instructor: Berthe Y. Choueiry Advantages:

I Simplicity of code

I Easy to understand

Fall 2007 Disadvantages:

I Memory

I Speed Computer Science & Engineering 235 Introduction to Discrete Mathematics I Possibly redundant work Sections 7.1 - 7.2 of Rosen Tail recursion oﬀers a solution to the memory problem, but really, [email protected] do we need recursion?

Recursive Algorithms Motivating Example Analysis Factorial Recall the factorial function. 1 if n = 1 n! = n · (n − 1)! if n > 1

Consider the following (recursive) algorithm for computing n!: We’ve already seen how to analyze the running time of algorithms. However, to analyze recursive algorithms, we require more Algorithm (Factorial) sophisticated techniques.

Speciﬁcally, we study how to deﬁne & solve recurrence relations. Input : n ∈ N Output : n! 1 if n = 1 then 2 return 1 3 end 4 else 5 return Factorial(n − 1) × n 6 end

Motivating Example Recurrence Relations I Factorial - Analysis? Deﬁnition

How many multiplications M(n) does Factorial perform? Definition I When n = 1 we don’t perform any. A recurrence relation for a sequence {an} is an equation that I Otherwise we perform 1. expresses an in terms of one or more of the previous terms in the Plus how ever many multiplications we perform in the I sequence, recursive call, Factorial(n − 1). a0, a1, . . . , an−1 I This can be expressed as a formula (similar to the definition of n!. for all integers n ≥ n0 where n0 is a nonnegative integer. M(0) = 0 A sequence is called a solution of a recurrence relation if its terms M(n) = 1 + M(n − 1) satisfy the recurrence relation. I This is known as a recurrence relation. Recurrence Relations II Recurrence Relations III Definition Definition

Example Consider the recurrence relation: a = 2a − a . n n−1 n−2 The Fibonacci numbers are deﬁned by the recurrence, It has the following sequences an as solutions: F (n) = F (n − 1) + F (n − 2) 1. an = 3n, F (1) = 1 2. an = n + 1, and F (0) = 1

3. an = 5. The solution to the Fibonacci recurrence is Initial conditions + recurrence relation uniquely determine the √ !n √ !n 1 1 + 5 1 1 − 5 sequence. fn = √ − √ 5 2 5 2

(your book derives this solution).

Recurrence Relations IV Recurrence Relations V Deﬁnition Deﬁnition

The initial conditions specify the value of the first few necessary More generally, recurrences can have the form terms in the sequence. In the Fibonacci numbers we needed two initial conditions, F (0) = F (1) = 1 since F (n) was defined by the T (n) = αT (n − β) + f(n),T (δ) = c two previous terms in the sequence. Initial conditions are also known as boundary conditions (as or n opposed to the general conditions). T (n) = αT + f(n),T (δ) = c β From now on, we will use the subscript notation, so the Fibonacci numbers are Note that it may be necessary to define several T (δ), initial fn = fn−1 + fn−2 conditions. f1 = 1 f0 = 1

Recurrence Relations VI Solving Recurrences Deﬁnition

Recurrence relations have two parts: recursive terms and non-recursive terms. There are several methods for solving recurrences. T (n) = 2T (n − 2) + n2 − 10 | {z } | {z } I Characteristic Equations recursive non-recrusive I Forward Substitution

Recursive terms come from when an algorithm calls itself. I Backward Substitution Recurrence Trees Non-recursive terms correspond to the “non-recursive” cost of the I algorithm—work the algorithm performs within a function. I Maple! We’ll see some examples later. First, we need to know how to solve recurrences. Linear Homogeneous Recurrences Linear Homogeneous Recurrences Examples Deﬁnition

A linear homogeneous recurrence relation of degree k with Examples constant coeﬃcients is a recurrence relation of the form The Fibonacci sequence is a linear homogeneous recurrence

an = c1an−1 + c2an−2 + ··· + ckan−k relation. As are the following. a = 4a + 5a + 7a with c1, . . . , ck ∈ R, ck 6= 0. n n−1 n−2 n−3

an = 2an−2 + 4an−4 + 8an−8 Linear: RHS is a sum of multiples of previous terms of the I How many initial conditions do we need to specify for these? As sequence (linear combination of previous terms). The many as the degree, k = 3, 8 respectively. coeﬃcients are all constants (not functions depending on n). Homogeneous: no terms occur that are not multiples of the I So, how do we solve linear homogeneous recurrences? aj’s.

I Degree k: an is expressed in terms of k terms of the sequence.

Solving Linear Homogeneous Recurrences I Solving Linear Homogeneous Recurrences II

n We want a solution of the form an = r where r is some (real) constant.

We observe that a = rn is a solution to a linear homogeneous k k−1 k−2 n r − c1r − c2r − · · · − ck−1r − ck = 0 recurrence if and only if This is called the characteristic equation of the recurrence relation. rn = c rn−1 + c rn−2 + ··· + c rn−k 1 2 k The roots of this polynomial are called the characteristic roots of the recurrence relation. They can be used to ﬁnd solutions (if they We can now divide both sides by rn−k, collect terms, and we get a exist) to the recurrence relation. We will consider several cases. k-degree polynomial.

k k−1 k−2 r − c1r − c2r − · · · − ck−1r − ck = 0

Second Order Linear Homogeneous Recurrences Second Order Linear Homogeneous Recurrences Example A second order linear homogeneous recurrence is a recurrence of the form Example an = c1an−1 + c2an−2 Find a solution to an = 5an−1 − 6an−2 Theorem (Theorem 1, p462) with initial conditions a0 = 1, a1 = 4 2 Let c1, c2 ∈ R and suppose that r − c1r − c2 = 0 is the characteristic polynomial of a 2nd order linear homogeneous I The characteristic polynomial is 1 recurrence which has two distinct roots, r1, r2. r2 − 5r + 6 Then {an} is a solution if and only if

n n I Using the quadratic formula (or common sense), the root can an = α1r1 + α2r2 be found; r2 − 5r + 6 = (r − 2)(r − 3) for n = 0, 1, 2,... where α1, α2 are constants dependent upon the initial conditions. so r1 = 2, r2 = 3 1we discuss how to handle this situation later. Second Order Linear Homogeneous Recurrences Second Order Linear Homogeneous Recurrences Example Continued Example Continued

I Solving for α1 = (1 − α2) in (1), we can plug it into the I Using the 2nd-order theorem, we have a solution, second. n n 4 = 2α1 + 3α2 an = α1(2 ) + α2(3 ) 4 = 2(1 − α2) + 3α2 4 = 2 − 2α + 3α I Now we can plug in the two initial conditions to get a system 2 2 of linear equations. 2 = α2

0 0 I Substituting back into (1), we get a0 = α1(2) + α2(3) 1 1 a1 = α1(2) + α2(3) α1 = −1

I Putting it all back together, we have 1 = α1 + α2 (1) n n 4 = 2α1 + 3α2 (2) an = α1(2 ) + α2(3 ) = −1 · 2n + 2 · 3n

Second Order Linear Homogeneous Recurrences Single Root Case Another Example Recall that we can only apply the ﬁrst theorem if the roots are distinct, i.e. r1 6= r2. Example If the roots are not distinct (r1 = r2), we say that one Solve the recurrence characteristic root has multiplicity two. In this case we have to apply a diﬀerent theorem. an = −2an−1 + 15an−2 Theorem (Theorem 2, p464) with initial conditions a0 = 0, a1 = 1. 2 Let c1, c2 ∈ R with c2 6= 0. Suppose that r − c1r − c2 = 0 has If we did it right, we have only one distinct root, r0. Then {an} is a solution to an = c1an−1 + c2an−2 if and only if 1 n 1 n an = (3) − (−5) 8 8 n n an = α1r0 + α2nr0 How can we check ourselves? for n = 0, 1, 2,... where α1, α2 are constants depending upon the initial conditions.

Single Root Case Single Root Case Example Example Example I By Theorem 2, we have that the solution is of the form What is the solution to the recurrence relation n n an = α14 + α2n4 an = 8an−1 − 16an−2 I Using the initial conditions, we get a system of equations; with initial conditions a0 = 1, a1 = 7? a0 = 1 = α1 a1 = 7 = 4α1 + 4α2 I The characteristic polynomial is 3 I Solving the second, we get that α2 = 4 r2 − 8r + 16 I And so the solution is Factoring gives us 3 I a = 4n + n4n n 4 r2 − 8r + 16 = (r − 4)(r − 4) I We should check ourselves... so r0 = 4 General Linear Homogeneous Recurrences General Linear Homogeneous Recurrences I Distinct Roots

Theorem (Theorem 3, p465)

There is a straightforward generalization of these cases to higher Let c1, . . . , ck ∈ R. Suppose that the characteristic equation order linear homogeneous recurrences. rk − c rk−1 − · · · − c r − c = 0 Essentially, we simply deﬁne higher degree polynomials. 1 k−1 k

The roots of these polynomials lead to a general solution. has k distinct roots, r1, . . . , rk. Then a sequence {an} is a solution of the recurrence relation The general solution contains coeﬃcients that depend only on the initial conditions. an = c1an−1 + c2an−2 + ··· + ckan−k

In the general case, however, the coeﬃcients form a system of if and only if linear equalities. n n n an = α1r1 + α2r2 + ··· + αkrk

for n = 0, 1, 2,..., where α1, α2, . . . , αk are constants.

General Linear Homogeneous Recurrences General Linear Homogeneous Recurrences Any Multiplicity Any Multiplicity

Theorem (Continued)

Then a sequence {an} is a solution of the recurrence relation Theorem (Theorem 4, p466) an = c1an−1 + c2an−2 + ··· + ckan−k Let c , . . . , c ∈ . Suppose that the characteristic equation 1 k R if and only if k k−1 r − c1r − · · · − ck−1r − ck = 0 m1−1 n an = (α1,0 + α1,1n + ··· + α1,m1−1n )r1 + m2−1 n (α2,0 + α2,1n + ··· + α2,m2−1n )r2 + has t distinct roots, r1, . . . , rt with multiplicities m1, . . . , mt. . . mt−1 n (αt,0 + αt,1n + ··· + αt,mt−1n )rt +

for n = 0, 1, 2,..., where αi,j are constants for 1 ≤ i ≤ t and 0 ≤ j ≤ mi − 1.

Linear Nonhomogeneous Recurrences Linear Nonhomogeneous Recurrences

For recursive algorithms, cost functions are often not homogenous Here, f(n) represents a non-recursive cost. If we chop it oﬀ, we because there is usually a non-recursive cost depending on the are left with input size. a = c a + c a + ··· + c a Such a recurrence relation is called a linear nonhomogeneous n 1 n−1 2 n−2 k n−k recurrence relation. which is the associated homogenous recurrence relation. Such functions are of the form Every solution of a linear nonhomogeneous recurrence relation is the sum of a particular solution and a solution to the associated a = c a + c a + ··· + c a + f(n) n 1 n−1 2 n−2 k n−k linear homogeneous recurrence relation. Linear Nonhomogeneous Recurrences Linear Nonhomogeneous Recurrences

Theorem (Theorem 5, p468)

(p) If {an } is a particular solution of the nonhomogeneous linear There is no general method for solving such relations. However, we recurrence relation with constant coeﬃcients can solve them for special cases.

an = c1an−1 + c2an−2 + ··· + ckan−k + f(n) In particular, if f(n) is a polynomial or exponential function (or more precisely, when f(n) is the product of a polynomial and (p) (h) (h) then every solution is of the form {an + an }, where {an } is a exponential function), then there is a general solution. solution of the associated homogenous recurrence relation

an = c1an−1 + c2an−2 + ··· + ckan−k

Linear Nonhomogeneous Recurrences Linear Nonhomogeneous Recurrences

Theorem (Continued) Theorem (Theorem 6, p469) When s is not a root of the characteristic equation of the Suppose that {an} satisﬁes the linear nonhomogeneous recurrence associated linear homogeneous recurrence relation, there is a relation particular solution of the form

an = c1an−1 + c2an−2 + ··· + ckan−k + f(n) t t−1 n (ptn + pt−1n + ··· + p1n + p0) · s

where c1, . . . , ck ∈ R and When s is a root of this characteristic equation and its multiplicity t t−1 n f(n) = (btn + bt−1n + ··· + b1n + b0) · s is m, there is a particular solution of the form

m t t−1 n where b0, . . . , bn, s ∈ R. n (ptn + pt−1n + ··· + p1n + p0) · s

Linear Nonhomogeneous Recurrences Other Methods

When analyzing algorithms, linear homogenous recurrences of order greater than 2 hardly ever arise in practice.

The examples in the text are quite good (see pp467–470) and We brieﬂy describe two “unfolding” methods that work for a lot of illustrate how to solve simple nonhomogeneous relations. cases. We may go over more examples if you wish. Backward substitution – this works exactly as its name implies: starting from the equation itself, work backwards, substituting Also read up on generating functions in section 7.4 (though we values of the function for previous ones. may return to this subject). Recurrence Trees – just as powerful but perhaps more intuitive, However, there are alternate, more intuitive methods. this method involves mapping out the recurrence tree for an equation. Starting from the equation, you unfold each recursive call to the function and calculate the non-recursive cost at each level of the tree. You then ﬁnd a general formula for each level and take a summation over all such levels. Backward Substitution Backward Substitution Example Example – Continued Example If we continue to do this, we get the following.

Give a solution to T (n) = T (n − 2) + 2(n − 1) + 2n = T (n − 3) + 2(n − 2) + 2(n − 1) + 2n T (n) = T (n − 1) + 2n = T (n − 4) + 2(n − 3) + 2(n − 2) + 2(n − 1) + 2n . where T (1) = 5. . Pi−1 = T (n − i) + j=0 2(n − j) We begin by unfolding the recursion by a simple substitution of the function values. Observe that I.e. this is the function’s value at the i-th iteration. Solving the sum, we get T (n − 1) = T ((n − 1) − 1) + 2(n − 1) = T (n − 2) + 2(n − 1) (i − 1)(i − 1 + 1) T (n) = T (n − i) + 2n(i − 1) − 2 + 2n 2 Substituting this into the original equation gives us T (n) = T (n − 2) + 2(n − 1) + 2n

Backward Substitution Recurrence Trees Example – Continued

When using recurrence trees, we graphically represent the We want to get rid of the recursive term. To do this, we need to recursion. know at what iteration we reach our base case; i.e. for what value Each node in the tree is an instance of the function. As we of i can we use the initial condition, T (1) = 5? progress downward, the size of the input decreases. We can easily see that when i = n − 1, we get the base case. The contribution of each level to the function is equivalent to the Substituting this into the equation above, we get number of nodes at that level times the non-recursive cost on the size of the input at that level. T (n) = T (n − i) + 2n(i − 1) − i2 + i + 2n The tree ends at the depth at which we reach the base case. = T (1) + 2n(n − 1 − 1) − (n − 1)2 + (n − 1) + 2n = 5 + 2n(n − 2) − (n2 − 2n + 1) + (n − 1) + 2n As an example, we consider a recursive function of the form = n2 + n + 3 n T (n) = αT + f(n),T (δ) = c β

Recurrence Trees Recurrence Trees Example

Iteration Cost The total value of the function is the summation over all levels of 0 T (n) f(n) the tree: logβ n X n T (n) = αi · f “ ” i T (n/β) ··· α ··· T (n/β) α · f n β 1 β i=0

„ « We consider the following concrete example. 2 T (n/β2) ··· α ··· T (n/β2) T (n/β2) ··· α ··· T (n/β2) α2 · f n β2

...... Example „ « i αi · f n βi

. . . . n . . T (n) = 2T + n, T (1) = 4 log n 2 logβ n α β · T (δ) Recurrence Trees Recurrence Trees Example – Continued Example – Continued

Iteration Cost 0 T (n) n

n + n The value of the function then, is the summation of the value of 1 T (n/2) T (n/2) 2 2 all levels. We treat the last level as a special case since its non-recursive cost is diﬀerent. „ « T (n/4) T (n/4) T (n/4) T (n/4) 4 · n 2 4

(log2 n)−1 X i n „ « T (n/8) T (n/8) T (n/8) T (n/8) T (n/8) T (n/8) T (n/8) T (n/8) 8 · n T (n) = 4n + 2 = n(log n) + 4n 3 8 2i i=0 ......

0 1 i n i 2 · @ A 2i

......

log2 n 2log2 n · T (1)

Smoothness Rule I How To Cheat With Maple I

In the previous example we make the following assumption: that n Maple and other math tools are great resources. However, they are was a power of two; n = 2k. This was necessary to get a nice not substitutes for knowing how to solve recurrences yourself. depth of log n and a full tree. As such, you should only use Maple to check your answers. We can restrict consideration to certain powers because of the Recurrence relations can be solved using the rsolve command smoothness rule, which is not studied in this course. For more and giving Maple the proper parameters. information about the smoothness rule, please consult pages The arguments are essentially a comma-delimited list of equations: 481–483 in the textbook “The Design & Analysis of Algorthims” general and boundary conditions, followed by the “name” and by Anany Levitin. variable of the function.

How To Cheat With Maple II

> rsolve({T(n) = T(n-1) + 2*n, T(1) = 5}, T(n));

1 1 + 2(n + 1) n + 1 − 2n 2

You can clean up Maple’s answer a bit by encapsulating it in the simplify command: > simplify(rsolve({T(n) = T(n-1) + 2*n, T(1) = 5}, T(n))); 3 + n2 + n