Lectures on
Recursive Algorithms
COMP 523: Advanced Algorithmic Techniques
Lecturer: Dariusz Kowalski
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Announcements
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www.csc.liv.ac.uk/~darek/comp523.html
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feedback – check for a new version a few
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Overview
Previous lectures:
• Algorithms: correctness, termination, performance
• Asymptotic notation
• Popular asymptotic complexities: log n, n, n log n, n2, ...
• Graphs - basic definitions and examples
These 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
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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
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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
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Example 1:
Searching in time O(log n)
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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)
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Tree structure of searching
root
4
find if value 5 is there
2
1
6
3
5
height = 3
7
1 2 3 4 5 6 7 8
8 leaves
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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
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Case q = 1, f(n) = c
T(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
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Case q = 1, f(n) = c
formal analysis
T(n)  T(n/2) + c
T(2)  c
Solution: T(n)  c log n
Proof: 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
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Example 2:
Finding a majority in time O(n)
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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
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Finding a majority: algorithm
Algorithm 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
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Tree structure of finding majority
root
2
odd, no majority
2
2
1
height = 2
2 2 2 2 1 2 1 1
8 leaves
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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).
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Proof of the Lemma
Suppose, 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.
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Memory consumption
Each 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”
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Case q = 1, f(n) = cn
T(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
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Case q = 1, f(n) = c n
formal analysis
T(n)  T(n/2) + c n
T(1)  c
Solution: T(n)  2cn
Proof: by induction.
– For n = 1 straightforward: T(1)  c  2c  1
– Suppose T(n/2)  2c(n/2); then
T(n)  T(n/2) + cn  2c(n/2) + c n
= 2cn
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Textbook and exercises
READING: Chapter 5, up to Section 5.2
EXERCISES:
• 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.
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Example 3:
Sorting in time O(n log n)
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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) + cn
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Case q = 2, f(n) = cn
T(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
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Solution for time
formal analysis
T(n)  2 T(n/2) + cn
T(2)  c
Solution: T(n)  cnlog n
Proof: by induction.
– For n = 2 straightforward: T(2)  c  c2log 2
– Suppose T(n/2)  c(n/2)log (n/2); then
T(n)  2T(n/2) + cn  2c(n/2)log (n/2) + cn
= 2c(n/2)log n – 2cn/2 + cn = cnlog n
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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)
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MergeSort procedure
1
2
4
7
8
3
5
6
9
10
1
2
3
4
5
6
7
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9
10
27
Tree structure of sorting
root
merging operations
height = 3
2 4 1 7 8 3 6 5
8 leaves
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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) + 2
M(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
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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)
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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 n
T(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
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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)
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Randomized approach - analysis
Let T(n) denote the expected time: sum of all possible
values of time weighted by the probabilities of these values
T(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  8c
Proof: 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
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Tree structure of random execution
root 1
6
5
7
height = 5
3
2
4
1 2 3 4 5 6 7 8
8 leaves
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Textbook and Exercises
READING: Section 5.3
EXERCISES:
• Solved exercises 1, 2 from the textbook,
chapter 5 “Divide and Conquer”
• Prove that
1 + 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
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Example 4:
Finding closest pair in time O(n log n)
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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
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Solution
Preprocessing:
• 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
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Main difficulty
Find if there is a pair which has distance smaller than  :
if yes then find the smallest distance pair
How to do it in linear time?


L

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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
P
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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 /2
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/2
/2
/2
/2
L
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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
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Example 5:
Integer multiplication in time
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1.59
O(n )
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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
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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
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Time complexity
T(n)  3T(n/2) + c n
T(1)  c
Solution: T(n)  d nlog 3, for some constant d  3c
Proof: 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)
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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”
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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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