Chapter 2 Recursive Algorithms. 2 Chapter Outline Analyzing recursive algorithms Recurrence...

Chapter 2

Recursive Algorithms

2

Chapter Outline

• Analyzing recursive algorithms

• Recurrence relations

• Closest pair algorithms

• Convex hull algorithms

• Generating permutations

• Recursion and stacks

3

Prerequisites

• Before beginning this chapter, you should be able to:– Read and create recursive algorithms– Identify comparison and arithmetic

operations– Use basic algebra

4

Goals

• At the end of this chapter you should be able to:– Create the recurrence relation for a

recursive algorithm– Convert a simple recurrence relation into

closed form– Explain the closest pair algorithm

5

Goals (continued)

– Explain the convex hull algorithm– Generate permutations both recursively

and iteratively– Explain the relationship between recursion

and stacks

6

Recursive Algorithms

• Recursive algorithms solve the problem by solving smaller versions of the problem– If the smaller versions are only a little

smaller, the algorithm can be called a reduce and conquer algorithm

– If the smaller versions are about half the size of the original, the algorithm can be called a divide and conquer algorithm

7

Recursive Algorithm Analysis

• The analysis depends on– the preparation work to divide the input– the size of the smaller pieces– the number of recursive calls– the concluding work to combine the results

of the recursive calls

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Generic Recursive Algorithm

Recurse( data, N, solution )// data a set of input values// N the number of values in the set// solution the solution to this problemif N ≤ SizeLimit then

DirectSolution( data, N, solution )else

DivideInput( data, N, smallerSets, smallerSizes, numberSmaller )

for i = 1 to numberSmaller do Recurse(smallerSets[i], smallerSizes[i],

smallSolution[i]) end for CombineSolutions(smallSolution,

numberSmaller, solution)end if

9

Generic Recursive Algorithm Analysis

• The recursive algorithm does the following amount of work:

• Where– DIR(N) is the amount of work done by the direct solution– DIV(N) is the amount of work done by the division of the

problem– COM(N) is the amount of work done to combine the

solutions

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Divide and Conquer ExampleLargest( list, start, end )if start ≥ end then return list[end]

end ifmiddle = (start + end) / 2first = Largest( list, start, middle )second = Largest( list, middle+1, end )if first > second thenreturn first

elsereturn second

end if

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Divide and Conquer Example Analysis

• This algorithm does:– One addition and division to calculate

middle– Two recursive calls with lists about half the

original size– One comparison to decide between the

largest of the first half and the largest of the second half

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Divide and Conquer Example Analysis

• The number of comparisons of list values done by this example is given by:

1 for1 2C2

1 for0 )(C

nn

nn

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Recurrence Relation Form

• A recurrence relation is a recursive form of an equation, for example:

• A recurrence relation can be put into an equivalent closed form without the recursion

2 )1(T )(T

3 )1(T

nn

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Converting Recurrence Relations

• Begin by looking at a series of equations with decreasing values of n:

2 )5(T )4(T

2 )4(T )3(T

2 )3(T )2(T

2 )2(T )1(T

2 )1(T )(T

nn

nn

nn

nn

nn

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Converting Recurrence Relations

• Now, we substitute back into the first equation:

2 )2 )2 )2 )2 )5(T(((( )(T

2 )2 )2 )2 )4(T((( )(T

2 )2 )2 )3(T(( )(T

2 )2 )2(T( )(T

2 )1(T )(T

nn

nn

nn

nn

nn

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Converting Recurrence Relations

• We stop when we get to T(1):

• How many “+ 2” terms are there? Notice we increase them with each substitution.

2 )2 )2 )2 )1(T((( )(T

2 )2 )2(T( )(T

2 )1(T )(T

n

nn

nn

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Converting Recurrence Relations

• We must have n – 1 of the “+ 2” terms because there was one at the start and we did n – 2 substitutions:

• So, the closed form of the equation is:

1

12 )1(T )(T

n

in

)1(2 3 )(T nn

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Approximating Recurrence Relations

• For recurrence relations of the formT(N) = a * T(N/b) + f(N)

where f(N) = Θ(Nd), the closed form can be approximated by:

da

dd

dd

baN

baNN

baN

N

b if

iflg

if

)(T

log

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Closest Pair Problem

• Given a set of N points in space, which two are the closest?

• A brute force method calculates the distance between every pair of points using:

• But this does (N2 – N)/2 distance calculations

2122

1221,d yyxxpp

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Brute Force Algorithm

smallDist = ∞for i = 1 to N do

for j = i+1 to N dodist = sqrt( (xi – xj)2 + (yi –

yj)2)if dist < smallDist then

smallDist = distfirst = isecond = j

end ifend for

end for

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A Divide and Conquer Solution

• Divide the set of points in into a right and left half

• Find the closest pair in the right and left half (recursively)

• Determine (ds) the shortest of these two distances

• Check if there is a pair of points within ds units but on opposite sides of the dividing line that are closer

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Divide and Conquer Example

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Divide and Conquer Example

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Divide and Conquer Example

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Divide and Conquer Example

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Divide and Conquer Closest PairPart 1

ClosestPair( Px, Py, d, p1, p2)// Px the points sorted by x coordinate// Py the points sorted by y coordinate// d the shortest distance between points p1 and p2// p1 the first point// p2 the second point

if sizeOf(Px) 3 thenfind d, p1, and p2 by brute forcereturn

end if

construct Lx, Ly, Rx, RyClosestPair(Lx, Ly, dl, L1, L2)ClosestPair(Rx, Ry, dr, R1, R2)

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Divide and Conquer ClosestPair 2

d = minimum(dl, dr)if d == dl then

p1 = L1p2 = L2

elsep1 = R1p2 = R2

end if

construct Mx and Myfor i = 1 to sizeOf(My)-1 do

for j = i+1 to i+7 (with j sizeOf(My)) dotemp = distance(Myi, Myj)if temp < d then

d = tempp1 = Myip2 = Myj

end ifend for

end for

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Divide and Conquer Closest Point Analysis

• The combination step looks at the seven points closest to those points within ds of the dividing line

• The points are divided into two halves

• Therefore, the recurrence relation isT(N) = 2 * T(N/2) + (N)

• This algorithm is (N lg N)

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Convex Regions

• A region is convex if a line connecting every pair of points in the region lies entirely within the region

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Convex Hull

• A convex hull for a set of points is the smallest convex region including all of the points

• All of the points lie on one side of each edge of the convex hull

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Brute Force Convex Hull

• Consider the line defined by the first pair of points

• If all of the other points lie on one side of that line, the line is part of the convex hull

• Repeat this process for every other pair of points

• This process is O(N3)

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A Divide and Conquer Solution

• The leftmost and rightmost points are on the convex hull

• Use the line between these points to divide the set of points into an upper and lower set

• Find the convex hull of the two parts• The convex hull of the entire set is the

convex hulls of these two parts without the dividing line

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A Divide and Conquer Solution

• To find the convex hull of each part– Find the point farthest away from the

dividing line– If the points are inside the triangle formed

with this point, this is the convex hull– If there are points outside one of these two

edges, recursively find the convex hull of those points

– If necessary, repeat for the other edge

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Divide and Conquer Example

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Divide and Conquer Example

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Divide and Conquer Example

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Divide and Conquer Convex Hull

quickHull(S, ps, pe)if S == {} then

return [] // the empty listelse

pf = point of S farthest from the dividing linePR = set of points to the right of the line

between ps and pfPL = set of points to the right of the line

between pf and pereturn quickHull(PR, ps, pf) + [pf]

+ quickHull(PL, pf, pe)end if

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Divide and Conquer Convex Hull Analysis

• If the algorithm divides the points into two halves, the recurrence relation isT(N) = 2 * T(N/2) + (N) and the closed form is (N lg N)

• In the worst case, all the points wind up on one side of each dividing line and the algorithm is then O(N2)

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Permutations

• The permutation of a set of elements is an ordering of those elements

• The permutations of the numbers from 1 to 3 are [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], and [3, 2, 1]

• If there are N elements in the set, there are N! permutations

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Recursive Permutations

• Swap pairs of list values during the recursion

• Output the list when the “bottom” of the recursion is reached

• The algorithm by Heap will permute the elements at the front of the list and will then swap in the next element

• It then permutes the front elements again

41

Recursive Permutations Algorithm

heapPermute(n)// list is a global variableif n == 0 then

output listelse

for i = 1 to n doheapPermute(n-1)if n is odd then

swap list[1] and list[n]else

swap list[i] and list[n]end if

end forend if

42

Iterative Permutations

• Permutations can be listed in lexical order

• The position within this list is a permutation’s rank in the range [0, N! – 1]

• It is possible to iteratively generate the permutation from this rank number by using the factorial of the values from 1 to N

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Iterative Permutations Algorithm

list[size] = 1for j = 1 to size – 1 dod = (rank mod f[j + 1]) / f[j]rank = rank – d * f[j]list[size – j] = d+1for i = size – j + 1 to size do

if list[i] > d thenlist[i] = list[i] + 1

end ifend for

end for

44

Recursion and Stacks

• Every subprogram has an Activation Record (AR) that includes the space needed for parameters, local variables, a return value, and a place to return control when done

• Every time a subprogram is called during execution, an instance of this AR is created and placed on the system stack

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Recursion and Stacks

• When a subprogram completes, its AR instance (ARI) is removed from the system stack

• For recursive subprograms, there will be multiple ARIs on the stack – one for each call

46

Recursion and Stacks

• A recursive subprogram could be rewritten to remove recursion by using a programmer created stack to keep track of what would be on the system stack

• Tracing recursive subprograms is easier if you simulate the system stack by drawing boxes when subprograms are called and crossing them out when the subprogram finishes