Transcript Review: Mathematical Induction

MCA 301: Design and
Analysis of Algorithms
Instructor
Neelima Gupta
[email protected]
Table Of Contents
Mathematical Induction: Review
Growth Functions
Review: Mathematical Induction
• Suppose
– S(k) is true for fixed constant k
• Often k = 0
– S(n)
S(n+1) for all n >= k
• Then S(n) is true for all n >= k
Proof By Mathematical Induction
• Claim:S(n) is true for all n >= k
• Basis:
– Show formula is true when n = k
• Inductive hypothesis:
– Assume formula is true for an arbitrary n
• Step:
– Show that formula is then true for n+1
Strong Induction
• Strong induction also holds
– Basis: show S(0)
– Hypothesis: assume S(k) holds for arbitrary k <= n
– Step: Show S(n+1) follows
• Another variation:
– Basis: show S(0), S(1)
– Hypothesis: assume S(n) and S(n+1) are true
– Step: show S(n+2) follows
Lets do it
• Prove 1 + 2 + 3 + … + n = n(n+1) / 2
• Prove a0 + a1 + … + an = (an+1 - 1)/(a - 1) for
all a  1
Growth Functions
• Big O Notation
• In general a function
– f(n) is O(g(n)) if there exist positive constants c and n0
such that f(n)  c  g(n) for all n  n0
• Formally
– O(g(n)) = { f(n):  positive constants c and n0 such that
f(n)  c  g(n)  n  n0
• Intuitively, it means f(n) grows no faster than g(n).
• Examples:
– n^2, n^2 – n
– n^3, n^3 – n^2 – n
f(n) = n2 g(n) = n2 – n Is f(n) = O(g(n))?
Sol: g(n) = n2 – n
= n2/2 + n2/2 - n
≥ n2/2 for n ≥ 2
= ½ f(n)
f(n) ≤ 2g(n) for n ≥ 2
Hence, f(n) = O(g(n))
f(n) = n3 g(n) = n3 - n2 – n Is f(n) = O(g(n))?
Sol: g(n) = n3 - n2 – n
= n3/2 + n3/2 - n2 - n
≥ n3/2 + n3/2 - n3/4 - n3/4 for n ≥ 4
= ½ f(n)
f(n) ≤ 2g(n) for all n ≥ 2
Hence, f(n) = O(g(n))
f(n) = n3 g(n) = n3 - n2 – n Is g(n) = O(f(n))?
Sol: Clearly, g(n) = n3 - n2 – n
≤ n3 for all n ≥ 1
= f(n)
g(n) ≤ f(n) for all n ≥ 1
Hence, g(n) = O(f(n)).
Omega Notation
• In general a function
– f(n) is (g(n)) if  positive constants c and n0
such that 0  cg(n)  f(n)  n  n0
Intuitively, it means f(n) grows at least as fast
as g(n).
• Examples:
– n^2, n^2 + n
– n^3, n^3 + n^2 – n
Ques: f(n) = n2, g(n) = n2 + n
Is f(n) = Ω(g(n))?
Sol: n2 = ½ (n2 + n2 )
≥ ½ (n2 + n )
f(n) ≥ c g(n)
for c = ½ m =1
f(n) = n3 g(n) = n3 + 4n2 - 5n Is g(n) = Ω(f(n))?
Sol: g(n) = n3 + 4n2 - 5n
≥ n3 - 5n for all n ≥ 0
≥ n3 - n3 /2 for all n2 ≥ 10
= n3 /2
g(n) ≥ 1/2 f(n) for n ≥ sqrt (10)
Hence, g(n) = Ω (f(n)).
Theta Notation
• A function f(n) is (g(n)) if  positive
constants c1, c2, and n0 such that
c1 g(n)  f(n)  c2 g(n)  n  n0
Assignment 0: Relations
Between , , O
For any two functions g(n) and f(n),
f(n) = (g(n)) iff
f(n) = O(g(n)) and f(n) = (g(n)).
Assignment No 1
• Self study
a0 + a1 + … + an = (an+1 - 1)/(a - 1) for all a  1
– What is the sum for a = 2/3 as n  infinity? Is it O(1)?
Is it big or small?
– For a = 2, is the sum = O(2^n)? Is it big or small?
Q1 Show that a polynomial of degree k = theta(n^k).
Other Asymptotic Notations
• A function f(n) is o(g(n)) if for every positive
constant c, there exists a constant n0 > 0 such that
f(n) < c g(n)  n  n0
• A function f(n) is (g(n)) if for every positive
constant c, there exists a constant n0 > 0 such that
c g(n) < f(n)  n  n0
• Intuitively,
– () is like >
– () is like =
– o() is like <
– () is like 
– O() is like 
Arrange some functions
• f(n) = O(g(n)) => f(n) = o(g(n)) ?
• Is the converse true?
• Let us arrange the following functions in
ascending order (assume log n = o(n) is
known)
– n, n^2, n^3, sqrt(n), n^epsilon, log n, log^2 n, n
log n, n/log n, 2^n, 3^n
Relation between n &
2
n
Intuitively, n appears to be smaller than n2 .
Lets prove it now.
T.P. For any constant c > 0
n < c n2
i.e. 1 < c n
i.e. n > 1 / c
Hence, n < c n2 for n > 1 / c
i.e. n = o( n2)
we can also write it as n < n2.
Relation between n2 & n3
Intuitively, n2 appears to be smaller than n3 .
Lets prove it now.
T.P. For any constant c > 0
n2 < c n3
i.e. 1 < c n
i.e. n > 1 / c
Hence, n2 < c n3 for n > 1 / c
i.e. n2 = o( n3)
we can also write it as n2 < n3.
Relation between n & n1/2
Since n = o (n2 ), we have,
For every constant c > 0, there exists n_c s.t.
n < c n2 for all n >= n_c
Thus sqrt(n) < c n for all sqrt(n) >= n_c
i.e. sqrt(n) < c n for all n >= n_c^2.
Thus,
n1/2 = o( n)
we can also write it as n1/2 < n. Combining the
previous result
n1/2 < n < n2 < n3
Relation between n & log n
• For the time being we can assume the result
log ( n ) = o(n)
log ( n ) < n
we will prove it later.
Relation between n1/2 & log n
Assume log n = o(n)
let c > 0 be any constant
for c/2 > 0 there exists m > 0 such that
log n < (c/2) n for n > m
changing variables from n to n1/2 we get
log(n1/2 ) < (c/2) n1/2 for n1/2 > m
½ log( n ) < (c/2) n1/2 for n > m2
Contd..
let m2 = k
log( n ) < c n1/2 for n > k
Since c > 0 was chosen arbitrarily hence
log n = o( n1/2 )
or
log n < n1/2
Combining the results we get
log n < n1/2 < n < n2 < n3
Relation between
2
n
& nlog n
• Since log n = o(n)
for c > 0,  n0 > 0 such that  n  n0, we
have
log n < c n
Multiplying by n on both sides we get
n log( n ) < c n2  n  n0
 nlog n = o( n2 )
 nlog n < n2
Relation between n & nlog n
Solution:
let c> 0 be any constant such that
n < c n log (n)
 1 < c log( n )
 log( n) > 1 / c
 n > e1/c
i.e. n < c n log n n > e1/c
Since c was chosen arbitrarily
\ n =o(n log n )or n < n log n
Combining the results we can get
log n < n1/2 < n < n logn < n2 < n3
Relation between n & n/log n
• We know that n = o(nlogn)
for c > 0,  n0 > 0 such that  n  n0, we
have
n < c n log n
dividing both sides by log n we get
n/ log( n) < c n  n  n0
Þ n / logn = o(n)
i.e. n / logn < n
Assignment No 2
• Show that log^M n = o(n^epsilon) for all
constants M>0 and epsilon > 0. Assume that
log n = o(n). Also prove the following
Corollary: log n = o(n/log n)
• Show that n^epsilon = o(n/logn ) for every
0 < epsilon < 1 .
Hence we have,
log n < n/log n < n1/2 < n < n logn < n2 < n3
Assignment No 3
• Show that
– lim f(n)/g(n) = 0 => f(n) = o(g(n)).
n→∞
– lim f(n)/g(n) = c => f(n) = θ(g(n)).
n → ∞, where c is a positive constant.
• Show that log n = o(n).
• Show that n^k = o(2^n) for every
positive constant k.
• Show by definition of ‘small o’ that
a^n = o(b^n) whenever a < b , a and b are
positive constants.
Hence we have,
log n < n/log n < n1/2 < n < n logn < n2 < n3
<2n < 3n
Why the constants ‘c’ and ‘m’?
• Suppose we have two algorithms to solve the
problem say sorting: Insertion Sort and Merge sort
for eg.
• Why should we have more than one algorithm to
solve the same problem?
• Ans: efficiency.
• What’s the measure of efficiency?
• Ans: System resources for example ‘time’.
• How do we measure time?
Contd..
•
•
•
•
IS(n) = O(n^2)
MS(n) = O(nlog n)
MS(n) is faster than IS(n).
Suppose we run IS on a fast machine and MS on a
slow machine and measure the time (since they
were developed by two different people living in
different part of the globe), we may get less time
for IS and more for MS…wrong analysis
• Solution: count the number of steps on a generic
computational model
Computational Model: Analysis
of Algorithms
• Analysis is performed with respect to a
computational model
• We will usually use a generic uniprocessor
random-access machine (RAM)
– All memory equally expensive to access
– No concurrent operations
– All reasonable instructions take unit time
• Except, of course, function calls
– Constant word size
• Unless we are explicitly manipulating bits
Running Time
• Number of primitive steps that are executed
– Except for time of executing a function call, in
this model most statements roughly require the
same amount of time
• y=m*x+b
• c = 5 / 9 * (t - 32 )
• z = f(x) + g(y)
• We can be more exact if need be
But why ‘c’ and ‘m’?
• Because
– We compare two algorithms on the basis of
their number of steps and
– the actual time taken by an algorithm is ‘c’
times the number of steps.
Why ‘m’?
• We need efficient algorithms and computational
tools to solve problems on big data. For example,
it is not very difficult to sort a pack of 52 cards
manually. However, to sort all the books in a
library on their accession number might be tedious
if done manually.
• So we want to compare algorithms for large input.
An Example: Insertion Sort
InsertionSort(A, n) {
for i = 2 to n {
key = A[i]
j = i - 1;
while (j > 0) and (A[j] > key) {
A[j+1] = A[j]
j = j - 1
}
A[j+1] = key
}
}
Assignment 4
• Show that Insertion Sort takes O(n^2) steps
by counting each and every step.
– Is it O(n)?
– Is it O(n^3)?
Lower Bound Notation
• We say InsertionSort’s run time is (n)
• Proof:
– Suppose run time is an + b
• Assume a and b are positive (what if b is negative?)
– an  an + b
Up Next
• Solving recurrences
– Substitution method
– Master theorem