Recursion Handout

Recursion

Slides by Christopher M. BourkeInstructor: Berthe Y. Choueiry

Spring 2006

Computer Science & Engineering 235Introduction to Discrete Mathematics

Sections 6.1 - 6.2 of [email protected]

Notes

Recursive Algorithms

A recursive algorithm is one in which objects are defined in termsof other objects of the same type.

Advantages:

I Simplicity of codeI Easy to understand

Disadvantages:

I MemoryI SpeedI Possibly redundant work

Tail recursion offers a solution to the memory problem, but really,do we need recursion?

Notes

Recursive AlgorithmsAnalysis

Weve already seen how to analyze the running time of algorithms.However, to analyze recursive algorithms, we require moresophisticated techniques.

Specifically, we study how to define & solve recurrence relations.

Notes

Motivating ExampleFactorial

Recall the factorial function.

n! ={

1 if n = 1n (n 1)! if n > 1

Consider the following (recursive) algorithm for computing n!:

Algorithm (Factorial)

Input : n NOutput : n!

if n = 1 then1return 12

end3

else4return Factorial(n 1) n5

end6

Notes

Motivating ExampleFactorial - Analysis?

How many multiplications M(n) does Factorial perform?

I When n = 1 we dont perform any.I Otherwise we perform 1.I Plus how ever many multiplications we perform in the

recursive call, Factorial(n 1).I This can be expressed as a formula (similar to the definition of

n!.M(0) = 0M(n) = 1 +M(n 1)

I This is known as a recurrence relation.

Notes

Recurrence Relations IDefinition

Definition

A recurrence relation for a sequence {an} is an equation thatexpresses an in terms of one or more of the previous terms in thesequence,

a0, a1, . . . , an1

for all integers n n0 where n0 is a nonnegative integer.A sequence is called a solution of a recurrence relation if its termssatisfy the recurrence relation.

Notes

Recurrence Relations IIDefinition

Consider the recurrence relation: an = 2an1 an2.It has the following sequences an as solutions:

1. an = 3n,

2. an = 2n, and

3. an = 5.

Initial conditions + recurrence relation uniquely determine thesequence.

Notes

Recurrence Relations IIIDefinition

Example

The Fibonacci numbers are defined by the recurrence,

F (n) = F (n 1) + F (n 2)F (1) = 1F (0) = 1

The solution to the Fibonacci recurrence is

fn =15

(1 +

5

2

)n 1

5

(15

2

)n(your book derives this solution).

Notes

Recurrence Relations IVDefinition

More generally, recurrences can have the form

T (n) = T (n ) + f(n), T () = c

or

T (n) = T(n

)+ f(n), T () = c

Note that it may be necessary to define several T (), initialconditions.

Notes

Recurrence Relations VDefinition

The initial conditions specify the value of the first few necessaryterms in the sequence. In the Fibonacci numbers we needed twoinitial conditions, F (0) = F (1) = 1 since F (n) was defined by thetwo previous terms in the sequence.

Initial conditions are also known as boundary conditions (asopposed to the general conditions).

From now on, we will use the subscript notation, so the Fibonaccinumbers are

fn = fn1 + fn2f1 = 1f0 = 1

Notes

Recurrence Relations VIDefinition

Recurrence relations have two parts: recursive terms andnon-recursive terms.

T (n) = 2T (n 2) recursive

+ n2 10 non-recrusive

Recursive terms come from when an algorithm calls itself.

Non-recursive terms correspond to the non-recursive cost of thealgorithmwork the algorithm performs within a function.

Well see some examples later. First, we need to know how tosolve recurrences.

Notes

Solving Recurrences

There are several methods for solving recurrences.

I Characteristic EquationsI Forward SubstitutionI Backward SubstitutionI Recurrence TreesI Maple!

Notes

Linear Homogeneous Recurrences

Definition

A linear homogeneous recurrence relation of degree k withconstant coefficients is a recurrence relation of the form

an = c1an1 + c2an2 + + ckankwith c1, . . . , ck R, ck 6= 0.

I Linear: RHS is a sum of multiples of previous terms of thesequence (linear combination of previous terms). Thecoefficients are all constants (not functions depending on n).

I Homogeneous: no terms occur that are not multiples of theaj s.

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

Notes

Linear Homogeneous RecurrencesExamples

Examples

The Fibonacci sequence is a linear homogeneous recurrencerelation. As are the following.

an = 4an1 + 5an2 + 7an3

an = 2an2 + 4an4 + 8an8

How many initial conditions do we need to specify for these? Asmany as the degree, k = 3, 8 respectively.

So, how do we solve linear homogeneous recurrences?

Notes

Solving Linear Homogeneous Recurrences I

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

We observe that an = rn is a solution to a linear homogeneousrecurrence if and only if

rn = c1rn1 + c2rn2 + + ckrnk

We can now divide both sides by rnk, collect terms, and we get ak-degree polynomial.

rk c1rk1 c2rk2 ck1r ck = 0

Notes

Solving Linear Homogeneous Recurrences II

rk c1rk1 c2rk2 ck1r ck = 0This is called the characteristic equation of the recurrence relation.

The roots of this polynomial are called the characteristic roots ofthe recurrence relation. They can be used to find solutions (if theyexist) to the recurrence relation. We will consider several cases.

Notes

Second Order Linear Homogeneous Recurrences

A second order linear homogeneous recurrence is a recurrence ofthe form

an = c1an1 + c2an2

Theorem (Theorem 1, p414)

Let c1, c2 R and suppose that r2 c1r c2 = 0 is thecharacteristic polynomial of a 2nd order linear homogeneousrecurrence which has two distinct1 roots, r1, r2.

Then {an} is a solution if and only if

an = 1rn1 + 2rn2

for n = 0, 1, 2, . . . where 1, 2 are constants dependent upon theinitial conditions.

1we discuss how to handle this situation later.

Notes

Second Order Linear Homogeneous RecurrencesExample

Example

Find a solution toan = 5an1 6an2

with initial conditions a0 = 1, a1 = 4

I The characteristic polynomial is

r2 5r + 6I Using the quadratic formula (or common sense), the root can

be found;r2 5r + 6 = (r 2)(r 3)

so r1 = 2, r2 = 3

Notes

Second Order Linear Homogeneous RecurrencesExample Continued

I Using the 2nd-order theorem, we have a solution,

an = 1(2n) + 2(3n)

I Now we can plug in the two initial conditions to get a systemof linear equations.

a0 = 1(2)0 + 2(3)0

a1 = 1(2)1 + 2(3)1

1 = 1 + 2 (1)4 = 21 + 32 (2)

Notes

Second Order Linear Homogeneous RecurrencesExample Continued

I Solving for 1 = (1 2) in (1), we can plug it into thesecond.

4 = 21 + 324 = 2(1 2) + 324 = 2 22 + 322 = 2

I Substituting back into (1), we get

1 = 1I Putting it all back together, we have

an = 1(2n) + 2(3n)= 1 2n + 2 3n

Notes

Second Order Linear Homogeneous RecurrencesAnother Example

Example

Solve the recurrence

an = 2an1 + 15an2with initial conditions a0 = 0, a1 = 1.

If we did it right, we have

an =18(3)n 1

8(5)n

How can we check ourselves?

Notes

Single Root Case

Recall that we can only apply the first theorem if the roots aredistinct, i.e. r1 6= r2.If the roots are not distinct (r1 = r2), we say that onecharacteristic root has multiplicity two. In this case we have toapply a different theorem.


Let c1, c2 R with c2 6= 0. Suppose that r2 c1r c2 = 0 hasonly one distinct root, r0. Then {an} is a solution toan = c1an1 + c2an2 if and only if

an = 1rn0 + 2nrn0

for n = 0, 1, 2, . . . where 1, 2 are constants depending upon theinitial conditions.

Notes

Single Root CaseExample

Example

What is the solution to the recurrence relation

an = 8an1 16an2with initial conditions a0 = 1, a1 = 7?

I The characteristic polynomial is

r2 8r + 16I Factoring gives us

r2 8r + 16 = (r 4)(r 4)

so r0 = 4

Notes

Single Root CaseExample

I By Theorem 2, we have that the solution is of the form

an = 14n + 2n4n

I Using the initial conditions, we get a system of equations;

a0 = 1 = 1a1 = 7 = 41 + 42

I Solving the second, we get that 2 = 34I And so the solution is

an = 4n +34n4n

I We should check ourselves. . .

Notes

General Linear Homogeneous Recurrences

There is a straightforward generalization of these cases to higherorder linear homogeneous recurrences.

Essentially, we simply define higher degree polynomials.

The roots of these polynomials lead to a general solution.

The general solution contains coefficients that depend only on theinitial conditions.

In the general case, however, the coefficients form a system oflinear inequalities.

Notes

General Linear Homogeneous Recurrences IDistinct Roots


Let c1, . . . , ck R. Suppose that the characteristic equation

rk c1rk1 ck1r ck = 0

has k distinct roots, r1, . . . , rk. Then a sequence {an} is a solutionof the recurrence relation

an = c1an1 + c2an2 + + ckankif and only if

an = 1rn1 + 2rn2 + + krnk

for n = 0, 1, 2, . . ., where 1, 2, . . . , k are constants.

Notes

General Linear Homogeneous RecurrencesAny Multiplicity


Let c1, . . . , ck R. Suppose that the characteristic equation

rk c1rk1 ck1r ck = 0

has t distinct roots, r1, . . . , rt with multiplicities m1, . . . ,mt.

Notes

General Linear Homogeneous RecurrencesAny Multiplicity

Theorem (Continued)

Then a sequence {an} is a solution of the recurrence relation

an = c1an1 + c2an2 + + ckankif and only if

an = (1,0 + 1,1n+ + 1,m11nm11)rn1+(2,0 + 2,1n+ + 2,m21nm21)rn2+

...(t,0 + t,1n+ + t,mt1nmt1)rnt +

for n = 0, 1, 2, . . ., where i,j are constants for 1 i t and0 j mi 1.

Notes

Linear Nonhomogeneous Recurrences

For recursive algorithms, cost functions are often not homogenousbecause there is usually a non-recursive cost depending on theinput size.

Such a recurrence relation is called a linear nonhomogeneousrecurrence relation.

Such functions are of the form

an = c1an1 + c2an2 + + ckank + f(n)

Notes


Here, f(n) represents a non-recursive cost. If we chop it off, weare left with

an = c1an1 + c2an2 + + ckankwhich is the associated homogenous recurrence relation.

Every solution of a linear nonhomogeneous recurrence relation isthe sum of a particular solution and a solution to the associatedlinear homogeneous recurrence relation.

Notes



If {a(p)n } is a particular solution of the nonhomogeneous linearrecurrence relation with constant coefficients


then every solution is of the form {a(p)n + a(h)n }, where {a(h)n } is asolution of the associated homogenous recurrence relation

an = c1an1 + c2an2 + + ckank

Notes


There is no general method for solving such relations. However, wecan solve them for special cases.

In particular, if f(n) is a polynomial or exponential function (ormore precisely, when f(n) is the product of a polynomial andexponential function), then there is a general solution.

Notes



Suppose that {an} satisfies the linear nonhomogeneous recurrencerelation


where c1, . . . , ck R and

f(n) = (btnt + bt1nt1 + + b1n+ b0) sn

where b0, . . . , bn, s R.

Notes


Theorem (Continued)

When s is not a root of the characteristic equation of theassociated linear homogeneous recurrence relation, there is aparticular solution of the form

(ptnt + pt1nt1 + + p1n+ p0) sn

When s is a root of this characteristic equation and its multiplicityis m, there is a particular solution of the form

nm(ptnt + pt1nt1 + + p1n+ p0) sn

Notes


The examples in the text are quite good (see pp420422) andillustrate how to solve simple nonhomogeneous relations.

We may go over more examples if you wish.

Also read up on generating functions in section 6.4 (though wemay return to this subject).

However, there are alternate, more intuitive methods.

Notes

Other Methods

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

We briefly describe two unfolding methods that work for a lot ofcases.

Backward substitution this works exactly as its name implies:starting from the equation itself, work backwards, substitutingvalues of the function for previous ones.

Recurrence Trees just as powerful but perhaps more intuitive,this method involves mapping out the recurrence tree for anequation. Starting from the equation, you unfold each recursivecall to the function and calculate the non-recursive cost at eachlevel of the tree. You then find a general formula for each level andtake a summation over all such levels.

Notes

Backward SubstitutionExample

Example

Give a solution to

T (n) = T (n 1) + 2n

where T (1) = 5.

We begin by unfolding the recursion by a simple substitution of thefunction values.

Observe that

T (n 1) = T ((n 1) 1) + 2(n 1) = T (n 2) + 2(n 1)

Substituting this into the original equation gives us

T (n) = T (n 2) + 2(n 1) + 2n

Notes

Backward SubstitutionExample Continued

If we continue to do this, we get the following.

T (n) = T (n 2) + 2(n 1) + 2n= T (n 3) + 2(n 2) + 2(n 1) + 2n= T (n 4) + 2(n 3) + 2(n 2) + 2(n 1) + 2n...

= T (n i) +i1j=0 2(n j)I.e. this is the functions value at the i-th iteration. Solving thesum, we get

T (n) = T (n i) + 2n(i 1) + 2(i 1)(i 1 + 1)2

+ 2n

Notes

Backward SubstitutionExample Continued

We want to get rid of the recursive term. To do this, we need toknow at what iteration we reach our base case; i.e. for what valueof i can we use the initial condition, T (1) = 5?

We can easily see that when i = n 1, we get the base case.Substituting this into the equation above, we get

T (n) = T (n i) + 2n(i 1) i2 + i+ 2n= T (1) + 2n(n 1 1) (n 1)2 + (n 1) + 2n= 5 + 2n(n 2) (n2 2n+ 1) + (n 1) + 2n= n2 + n+ 3

Notes

Recurrence Trees

When using recurrence trees, we graphically represent therecursion.

Each node in the tree is an instance of the function. As weprogress downward, the size of the input decreases.

The contribution of each level to the function is equivalent to thenumber of nodes at that level times the non-recursive cost on thesize of the input at that level.

The tree ends at the depth at which we reach the base case.

As an example, we consider a recursive function of the form

T (n) = T(n

)+ f(n), T () = c

Notes

Recurrence Trees

T (n)

T (n/)

T (n/2) T (n/2)

T (n/)

T (n/2) T (n/2)

Iteration

0

1

2

.

.

.

i

.

.

.

log n

Cost

f(n)

fn

2 f

n2

...

i f

ni

...

log n T ()

Notes

Recurrence TreesExample

The total value of the function is the summation over all levels ofthe tree:

T (n) =log ni=0

i f(n

i

)

We consider the following concrete example.

Example

T (n) = 2T(n2

)+ n, T (1) = 4

Notes

Recurrence TreesExample Continued

T (n)

T (n/2)

T (n/4)

T (n/8) T (n/8)

T (n/4)

T (n/8) T (n/8)

T (n/2)

T (n/4)

T (n/8) T (n/8)

T (n/4)

T (n/8) T (n/8)

Iteration

0

1

2

3

.

.

.

i

.

.

.

log2 n

Cost

n

n2 +

n2

4 n4

8 n8

.

.

.

2i 0@ n2i

1A

.

.

.

2log2 n T (1)

Notes

Recurrence TreesExample Continued

The value of the function then, is the summation of the value ofall levels. We treat the last level as a special case since itsnon-recursive cost is different.

T (n) = 4n+(log2 n)1

i=0

2in

2i= n(log n) + 4n

Notes

Smoothness Rule I

In the previous example we make the following assumption: that nwas a power of two; n = 2k. This was necessary to get a nicedepth of log n and a full tree.

We can restrict consideration to certain powers because of thesmoothness rule.

Definition

A function f : N 7 R is called smooth if it is monotonicallynondecreasing and

f(2n) (f(n))

Most slow growing functions (logarithmic, polylogarithmic,polynomial) are smooth while exponential functions are not.

Notes

Smoothness Rule II

Theorem

For a smooth function f(n) and a fixed constant b Z such thatb 2,

f(bn) (f(n))

Thus the order of growth is preserved.

Notes

How To Cheat With Maple I

Maple and other math tools are great resources. However, they arenot substitutes for knowing how to solve recurrences yourself.

As such, you should only use Maple to check your answers.Recurrence relations can be solved using the rsolve commandand giving Maple the proper parameters.

The arguments are essentially a comma-delimited list of equations:general and boundary conditions, followed by the name andvariable of the function.

Notes

How To Cheat With Maple II

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

1 + 2(n+ 1)(12n+ 1

) 2n

You can clean up Maples answer a bit by encapsulating it in thesimplify command:

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

3 + n2 + n

Notes

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