Lectures on Recursive Algorithms1 COMP 523: Advanced Algorithmic Techniques Lecturer: Dariusz...

Lectures on Recursive Algorithms 1

Lectures onRecursive Algorithms

COMP 523: Advanced Algorithmic Techniques

Lecturer: Dariusz Kowalski

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OverviewPrevious lectures:• Algorithms: correctness, termination, performance• Asymptotic notation• Popular asymptotic complexities: log n, n, n log n, n2, ...• Graphs - basic definitions and examplesThese lectures:• Recursive algorithms - basic recursion• Searching algorithm in time O(log n)• Finding majority in time O(n)• Sorting in time O(n log n)• Other examples


Algorithms based on recurrence

Algorithmic scheme:

• Reduce the input to smaller sub-input(s)

• Solve the problem for (some) sub-inputs

• Merge the obtained solution(s) into one global solution


Complexity analysis of recursive process

Let T(m) denote time (or other complexity measure) of solving a given problem by a given algorithm working on input length m

Simple recursive formula:T(n) qT(n/2) + f(n)

Solve the problem for half of the input q times, and spend at most f(n) time for partitioning and/or merging


Example 1:

Searching in time O(log n)


Searching: problem and algorithm• Input: sorted array A of n numbers and number x• Problem: find if x is in array A• Solution: Algorithm SEARCH(x,1,n)• Procedure SEARCH(x,i,j):

– If i = j then • if x = A[i] then answer YES

• if x A[i] then answer NO

– Else compare x with element A[(i+j)/2]:• If x = A[(i+j)/2] then answer YES

• If x < A[(i+j)/2] then SEARCH(x, i, (i+j)/2-1)

• If x > A[(i+j)/2] then SEARCH(x, (i+j)/2+1, j)


Tree structure of searching

1 2 3 4 5 6 7 8

1 3 5 7

62

4

height = 3

root

8 leaves

find if value 5 is there


Time and memory consumption Each recursive call

– needs additional constant memory, and

– the size of input decreases by factor 2

Formula for time (# comparisons):

T(n) T(n/2) + 4

Additional memory size: log n = O(log n)

In other words: recursive algorithm must store a path from the root to the searched leaf, and it has logarithmic length


Case q = 1, f(n) = cT(n) T(n/2) + c , T(2) c

T(n) T(n/2) + c

(T((n/2)/2) + c) + c = T(n/4) + 2c

(T((n/4)/2) + c) + 2c = T(n/8) + 3c…

(T(n/2i) + c) + (i-1) . c = T(n/2i) + i . c… (T(n/2log n – 1) + c) + (log n - 2) . c

= T(n/(n/2)) + (log n - 2) . c = T(2) + (log n – 1) . c c + (log n – 1) . c = c log n


Case q = 1, f(n) = cformal analysisT(n) T(n/2) + c

T(2) cSolution: T(n) c log nProof: by induction.

– For n = 2 straightforward: T(2) c c log 2– Suppose T(n/2) c log (n/2); then

T(n) T(n/2) + c c log (n/2) + c c log n - c + c c log n


Example 2:

Finding a majority in time O(n)


Finding a majority

• Input: array A of n elements

• Problem: find if there is an element x that occurs in array A more than n/2 times (a majority value)

• Solution: Algorithm Majority(A) returning a majority element in A, if it exists, or null otherwise


Finding a majority: algorithmAlgorithm Majority(A[1..n]):If |A| = 0 then output null, else if |A| = 1 then output A[1] ; else: • If n = |A| is odd then

– check whether A[n] is a majority in A by counting the number of occurrences of value A[n];if yes then output A[n], otherwise decrease n by one

• Initialize additional array B of size |A|/2• Set j to 0• For i = 1,2,…,n/2 do:

– if A[2i-1] = A[2i] then • increase j by one• set B[j] to A[2i]

• Find if there is a majority in B[1..j] by executing Majority(B[1..j])• If a majority value x in B[1..j] is returned then check whether x is a majority

in A, by going through array A and counting the number of occurrences of value x in A; if successful output x; otherwise null


Tree structure of finding majority

2 2 2 2 1 2 1 1

2 2 1

2

height = 2

root

8 leaves

odd, no majority


Correctness of Majority Algorithm

Lemma: (to be proved later)If x is a majority in A then x is a majority in B.

Correctness: (proof by induction, based on the Lemma)

Inductive assumption: – Majority(B) works correctly for all B’s of size smaller than |A|

Case 1: There is a majority in A.– Then it is a majority in B, by Lemma, therefore it will be found in the recursive

stage of the algorithm Majority(B) (by inductive argument), and will be checked (successfully) in the last point of the algorithm Majority(A).

Case 2: There is no majority in A.– Then even if there is a candidate value found in the recursive stage Majority(B)

of the algorithm, it will be rejected anyway in the last point of the algorithm Majority(A).


Proof of the LemmaSuppose, to the contrary, that x is a majority in A but is not a majority

in B. Let • m be the number of occurrences of x in A, and• k be the number of occurrences of x in B. It follows that the other values appear at least k times in B.Consequently, the other values appear in A:• at least 2k times: those pairs that are represented in B by a value

different than x; plus• m-2k times: each occurrence of x in A that is not paired by another

x (there are k pairs xx in A, thus m-2k of other occurrences of x in A) is paired by some other value (different than x)

which gives at least 2k+(m-2k) = m occurrences in total, and contradicts the fact that x is a majority in A. Contradiction! Therefore x must be also a majority in B.


Memory consumptionEach call to the recursive procedure • needs additional memory of half of the current input size • reduces the size of the input by factor at least two

Formula for time (# comparisons): T(n) n -1 + 3n/2 + T(n/2) + n = T(n/2) + 7n/2

Additional memory used (for smaller arrays B): M(n) n/2 + M(n/2) n/2 + n/4 + M(n/4)

n/2 + n/4 + … + 1 = n - 1 = O(n)Intuition about memory size: the algorithm needs a binary

tree of linear size, in the worst case, to store “winners”


Case q = 1, f(n) = cnT(n) T(n/2) + cn , T(1) c

T(n) T(n/2) + cn (T(n/4) + cn/2) + cn = T(n/4) + (3/2)cn

(T(n/8) + cn/4) + (3/2)cn = T(n/8) + (7/4)cn… (T(n/2i) + cn/2i-1) + ((2i-1-1)/2i-2) . cn = T(n/2i) + ((2i-1)/2i-1) . cn… T(n/2log n) + ((2log n-1)/2log n-1) . cn= T(1) + (n – 1)/(n/2) . cn c + 2cn – 2c 2cn


Case q = 1, f(n) = c nformal analysis

T(n) T(n/2) + c nT(1) c

Solution: T(n) 2cnProof: by induction.

– For n = 1 straightforward: T(1) c 2c 1– Suppose T(n/2) 2c(n/2); then

T(n) T(n/2) + cn 2c(n/2) + c n= 2cn


Textbook and exercisesREADING: Chapter 5, up to Section 5.2EXERCISES:• Solve the following recursive formulas the best you can:

– T(n) T(n/2) + 4– T(n) T(n/2) + 5n– T(n) T(n/2) + 3n2

– T(n) (3/2)T(n/2) + 1

Compare and sort the obtained (asymptotic) formulas according to the asymptotic order. For each of them try to find an algorithmic example with performance giving by this formula.

• Sort the following formulas according to big/small Oh order:log (n1/2), log (9n), log (n3), 2log n, 23log n, 2log (9n), n2, n log n

• Propose and analyse an algorithm for checking if there is a majority in a given sorted array A.


Example 3:

Sorting in time O(n log n)


Divide and conquer

Algorithm:• Partition input into two halves in (at most) linear

time• Solve the problem for each half of the input

separately• Merge solutions into one in (at most) linear time

Recursive formula on time complexity:T(n) 2 T(n/2) + cn


Case q = 2, f(n) = cnT(n) 2T(n/2) + cn , T(2) c

T(n) 2T(n/2) + cn 2(2T(n/4) + cn/2) + cn = 4T(n/4) + 2cn

4(2T(n/8) + cn/4) + 2cn = 8T(n/8) + 3cn… 2i-1 (2T(n/2i) + cn/2i-1) + (i-1) . cn = 2i T(n/2i) + i . cn… 2log n - 1 T(2) + (log n - 1) . cn cn log n


Solution for timeformal analysisT(n) 2 T(n/2) + cn

T(2) c

Solution: T(n) cnlog n

Proof: by induction.– For n = 2 straightforward: T(2) c c2log 2

– Suppose T(n/2) c(n/2)log (n/2); then

T(n) 2T(n/2) + cn 2c(n/2)log (n/2) + cn

= 2c(n/2)log n – 2cn/2 + cn = cnlog n


Example - sorting by merging• Input: list L of n numbers

• Problem: Sort list L

• Algorithm MergeSort(L):

If L has at most two elements then sort them by comparison and exit;

Else:– Split L into two lists prefix and suffix, each of size n/2– Sort, by merging, the prefix and the suffix separately:

MergeSort(prefix) and MergeSort(suffix)– Merge sorted prefix with sorted suffix as follows:

• Initialize final list as empty• Repeat until either prefix or suffix is empty:

– Compare the first elements on the lists prefix and suffix and then move the smaller one to the end of the final list

• Concatenate final list with the remaining non-empty list (prefix or suffix)


MergeSort procedure

1 2 4 7 8

3 5 6

1 2 3 4 5 6 7 8 9 10

9 10


Tree structure of sorting

2 4 1 7 8 3 6 5

height = 3

root

8 leaves

merging operations


Time and memory consumption

Each recursive call – needs additional constant memory, and – the number of inputs increases by factor 2, and– the size of each input decreases by factor 2

Formula for time (# comparisons and list operations): T(n) n/2 + 2T(n/2) + 2n = 2T(n/2) + 5n/2

Additional memory size: M(n) 2 M(n/2) + 2M(n) 2 M(n/2) + 2 4 M(n/4) + 4 + 2 8 M(n/8) + 8 + 4 + 2

… n/2 +n/4 + … + 4 + 2 = n - 2 = O(n)

In other words: divide and conquer algorithm must store the binary tree of pointers during the computation


Quick sort - algorithmic scheme

Generic Quick Sort:• Select one element x from the input• Partition the input into part containing elements not

greater than x and part containing elements bigger than x• Sort each part separately• Concatenate these two sorted parts

Problem: how to choose element x to balance the sizes of these two parts? (to get the similar recursive equations as for MergeSort)


Why parts should be balanced?

Suppose they are not balanced, but, say, the smallest element is chosen:

T(n) T(1) + T(n-1) + c nT(1) c

Solution: T(n) (c/2) . n2

Proof: by induction.– For n = 1 straightforward: T(1) c (c/2) . 12

– Suppose T(n-1) (c/2) . (n-1)2; then

T(n) T(n-1) + c + cn (c/2) . (n-1)2 + c (n+1) (c/2) . (n-1)2 + (c/2) . 2n (c/2) . n2


Two approaches to be fast

Deterministic:• Additional tree structure of size O(n) needed for

identifying the median value• Time: O(n log n), additional memory O(n)

Randomized:• Select separation element x uniformly at random• Time (expected): O(n log n), additional memory O(n)


Randomized approach - analysisLet T(n) denote the expected time: sum of all possible values of time weighted by the probabilities of these valuesT(n) 1/n ([T(n-1)+T(1)] + [T(n-2)+T(2)] + … +[T(0)+T(n)]) + cn

T(0) = T(1) = 1, T(2) c

Solution: T(n) d n log n, for some constant d 8cProof: by induction.– For n = 2 straightforward: T(2) c d . 2 . log 2– Suppose T(m) d m log m, for every m < n; then

(1-1/n)T(n) (2/n)(T(0) + … + T(n-1)) + c n

(2d/n)(1 + 1 + 2log 2 + … + (n-1)log(n-1)) + c n

(2d/n)((n2/2) log n – n2/8) + c n d n log n - d (n/4) + c n d n log n - d n/2

T(n) (n/(n-1))(d n log n - d n/2) d n log n


Tree structure of random execution

1 2 3 4 5 6 7 8

2 4

5 7

6

3

1

height = 5

root

8 leaves


Textbook and Exercises

READING: Section 5.3

EXERCISES:

• Solved exercises 1, 2 from the textbook, chapter 5 “Divide and Conquer”

• Prove that1 + 1 + 2 log 2 + … + (n-1)log(n-1) n(n/2) log(4n) , and

(n/(n-1))(d n log n - d n/2) d n log n , for n > 16


Example 4:

Finding closest pair in time O(n log n)


Problem

Closest pair of points:

• Input: n points on the plane

• Output: two points having the closest distance

Remarks:

• Exhaustive search algorithm gives time O(n2)

• Similar approach gives ALL pairs of closest points


SolutionPreprocessing:• Sort points according to the first coordinate (obtain horizontal

list H) and according to the second coordinate (obtain vertical list V)

Main Algorithm:• Partition input list H into halves (H1,H2) in linear time (draw vertical line

L containing medium point); split list V accordingly into V1,V2, where Vi contains the same elements as Hi and inherits the initial order from V (for i = 1,2)

• Solve the problem for each half of the input separately, obtaining two pairs of closest points; let be the smallest distance from the obtained ones

• Find if there is a pair which has distance smaller than - if yes then find the smallest distance pair


Main difficulty

Find if there is a pair which has distance smaller than : if yes then find the smallest distance pairHow to do it in linear time?

L


Closest pair in the strip

1. Find a sub-list P of V containing all points with a distance at most from the line L:– go through V and remove

elements not –close to L

2. For each element in P check its distance to the next 8 elements and remember the closest pair ever found

L

P


Why to check only 8 points ahead?1. Partition the strip into squares of

length /2 each, as shown in the picture

2. Each square contains at most 1 point - by definition of

3. If there is at least 2 full squares between points then they can not be the closest points

4. There are at most 8 squares to check

/2

L

/2 /2 /2

/2

/2


Time complexity

Preprocessing: sorting in time O(n log n)

Main Algorithm: recursion in time O(n log n)

T(n) 3n/2 + 2T(n/2) + 8n = 2T(n/2) + 9.5n

T(2) = 1


Example 5:

Integer multiplication in time O(n1.59)


Beyond (n log n) : integer’s multiplication

• Input: two integers x and y, each consisting of at most n bits

• Output: multiply these integers

• Naïve Algorithm:– Multiply each bit i of x by y, add (i-1) zeroes at the end– Add the obtained at most n numbers

• Time: (n2) of bit operations


Divide and Conquer approach

Let x = x1 + 2n/2 x2 and y = y1 + 2n/2 y2

Let p = x1 + x2 and q = y1 + y2. Then

x y = (x1 + 2n/2 x2) (y1 + 2n/2 y2)

= x1 y1 + 2n/2 (x2 y1+ x1 y2) + 2n x2 y2

= x1 y1 + 2n/2 (pq - x1 y1- x2 y2) + 2n x2 y2

Algorithm:– Compute p and q (in linear time)– Compute recursively: x1 y1 , x2 y2 , pq– Perform x1 y1 + 2n/2 (pq - x1 y1- x2 y2) + 2n x2 y2 in linear time


Time complexity

T(n) 3T(n/2) + c nT(1) c

Solution: T(n) d nlog 3, for some constant d 3cProof: For n = 1 straightforward: T(1) c d nlog 3

General case:T(n) 3T(n/2) + c n 9T(n/4) + 3c(n/2) + c n

… c n (1 + 3/2 + (3/2)2 + … + (3/2)log n) c n 2(3/2)log n + 1 3c n nlog 3 - 1

= 3c nlog 3 d nlog 3 = O(n1.59)


Textbook and Exercises

READING: Chapter 5 from Section 5.4

OBLIGATORY EXERCISES:

• How to add two n-bit integers in linear number of bit operations?

ADDITIONAL EXERCISES:

• Solved Exercise 1 from the textbook, chapter 5 “Divide and Conquer”

• Exercises 1, 6, 7 from the textbook, chapter 5 “Divide and Conquer”


Conclusions

• Searching in time O(log n)

• Finding majority value in time O(n)

• Divide and conquer in time O(n log n)– sorting algorithms (MergeSort, RandQuickSort)– closest points algorithm

• Beyond (n log n) : O(nlog 3) time algorithm for integer multiplication

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