data structures and algorithms - e & ict academy...β’ algorithm 2 : merge sort with time...
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Data Structures and Algorithms (CS210A)
Lecture 41 β’ Miscellaneous problems
1
Order notation Definition: Let f(π) and g(π) be any two increasing functions of n.
f(π) is said to be of the order of g(π)
if there exist constants c and π0 such that
f(π) β€ c g(π) for all n > π0
2
π0
f(π)
c g(π)
f(π) = O(g(π))
Order notation extended
Definition: Let f(π) and g(π) be any two increasing functions of n.
f(π) is said to be lower bounded by g(π)
if there exist constants c and π0 such that
f(π) β₯ c g(π) for all n > π0
π2
100= π(10000 π log π)
3
π0
c g(π)
f(π)
f(π) = π(g(π))
Order notation extended
Observations:
β’ f(π) = O(g(π))
One more Notation:
If f(π) = O(g(π))
Examples:
β’π2
100= π―(10000 π2)
if and only if g(π) = π(f(π))
and g(π) = π(f(π)) , then
g(π) = π―(f(π))
β’ Time complexity of Quick Sort is ?
β’ Time complexity of Merge sort is ?
π(π log π)
π―(π log π)
Time complexity of a problem
Example: Sorting
β’ Algorithm 1 : Selection Sort with time complexity π(π2)
β’ Algorithm 2 : Merge Sort with time complexity π(π log π)
β’ Each comparison based sorting algorithm needs to
perform π(π log π) comparisons in the worst case.
β’ Sorting must takes π(π) time since
it has to read each item at least once.
Time complexity of sorting
π(π2)
π(π log π)
π(π log π)
π(π)
Sorting has time complexity of π―(π log π)
Upper bound
Lower bound
Time complexity of a problem
Example: All-pairs shortest paths (APSP)
β’ Algorithm 1 : Floyd Warshal Algorithm with time complexity π(π3)
β’ Algorithm 2 : Johnsonβ algorithm with time complexity π(ππ log π)
β’ All-pairs shortest paths must require π(π2) time
Time complexity of APSP
π(π3)
π(ππ log π)
π(π2)
There is still a gap between upper and lower bounds for APSP.
Upper bound
Lower bound
Aim of theoretical computer science
For any given computational problem P
β’ Get smallest possible upper bound on its time complexity
β’ Get largest possible lower bound on its time complexity.
Reduce the GAP
This requires designing better algorithm
?
How to establish lower bound
Two ways:
β’ Adversarial approach
β’ Limitation of the model of computation
A gentle introduction today
CS345
Adversarial approach
Key aspects Algorithm:
β’ Algorithm does not have free access to the input.
To access any item in the input, algorithm has to spend some time.
β’ The execution of an algorithm at any step is determined
only by the (partial) input it has seen till now.
Adversary
β’ Adversary has access to all possible inputs of a problem.
β’ The sole aim of adversary is to make an algorithm work really hard.
For this purpose, adversary discloses the input cleverly.
Locating 1 problem
Input: An array A[0. . . π β π] with an unknown π s.t.
β’ For all π β π,
β’ A[π]= π
Aim: To locate/search 1 in A.
Upper bound: O(π)
Lower bound: π(π)
A[π]= π
Lower bound on Locating 1 problem
Algorithm
Adversary
Input
A
What is A[π]?
π
π π
π
What is A[π] ?
π
What is A[π] ?
π
π π π π π π 1
Miscellaneous problems
Problem 1 Input: Given an array A storing π numbers,
there is an π < π (unknown) s.t.
A[π] < A[π] < β¦< A[π]
Aim:
To search efficiently
Answer : O(log π) is possible
> A[π + π] >β¦ >A[π β π]
π π ππ ππ ππ ππ ππ ππ ππ ππ πππ πππ ππ ππ ππ
sorted sorted
π
Problem 2
Input:
Given an array A storing π numbers,
there is an π < π (unknown) s.t.
A[π] β€ A[π] β€ β¦β€ A[π]
Aim:
To search efficiently
Answer : π(π) time complexity
Locating 1 problem ? Problem 2
β₯ A[π + π] β₯β¦ β₯ A[π β π]
is a special case of
Problem 3
Input:
Given an array A storing π numbers,
there are π < π < π (unknown) s.t.
A[π]< A[π]< β¦< A[π]
Aim:
To search efficiently
Answer : π(π) time complexity
> A[π + π] >β¦ >A[π] < A[π + π] <β¦ <A[π β π]
Locating 0 problem
Input: An array A[0. . . π β π] with an unknown π s.t.
β’ For all π < π,
β’ A[π] = π
β’ For all π > π,
Aim: To locate/search 0 in A.
Upper bound: O(π)
Lower bound: π(π)
A[π]> π A[π β π] < A[π] and
A[π]> π A[π] < A[π + π] and
Lower bound on Locating 0 problem
Algorithm
Adversary
Input
A
What is A[π]?
π
ππ ππ
π
What is A[π] ? What is A[π] ?
π
π ππ ππ
β
What is A[β] ?
ππ ππ πππ ππ ππ
Problem 3
Input:
Given an array A storing π numbers,
there are π < π < π (unknown) s.t.
A[π]< A[π]< β¦< A[π]
Aim:
To search efficiently
Answer : π(π) time complexity
Locating 0 problem ? Problem 3
> A[π + π] >β¦ >A[π] < A[π + π] <β¦ <A[π β π]
is a special case of
Problem 4
Input:
Given a sorted array A storing π numbers,
Aim:
To search efficiently
Answer : π(log π) time complexity (binary search)
π(log π) time complexity
Lower bound on Problem 4
Algorithm
Adversary
Input A
What is A[π]?
π
ππ
π
What is A[π] ?
ππ ππ ππ ππ ππ πππ
Search for ππ
Problem 5
Input:
Given a 2-dimensional square grid storing ππ distinct numbers,
Aim:
To find a Local minima efficiently
Answer :
π(π) time complexity (in the second week of the course)
π(π) time complexity
Prove it using Adversarial arguments during summer vacations
Final slide
Thatβs all. I hope you enjoyed this lecture.