ANALYSIS OF ALGORITHMS
Average Case
Consider Dn and an input I  Dn .
Let t(I) = # of basic operations performed by the algorithm on I, and
p(I)= probability that input I occurs (i.e., the probability that I is created randomly).
p(I) is sometimes very difficult to determine analytically so in general is determined by
experience and by making simple assumptions.
The average case is then
A(n) 
 p( I ).t ( I )
I Dn
Example: Find x in a list of integers (A5).
Again fix x and consider Dn be the set of all inputs of size n, where all entries are distinct.
In this case let
Dn  I 1  I 2     I n  I n1
Where
I1= are all inputs where x appear in the first position
I2= are all inputs where x appear in the second position


In = are all inputs where x appear in the last position
In+1= are all inputs where x does not appear
x_ _ _ _ _ _ …_ _
_x_ _ _ _ _ …_ _
_ _ _ _ _ _ _ …_ x
_ _ _ _ _ _ _ …_ _
The problem is that we do not know what is the probability that x appears in an input.
Suppose that q is the probability that x appears (to be determined). Thus the probability
p(I1)= p(I2)= p(I3)=……= p(In)=q/n, and p(In+1)=1-q.
To determine t(I), we first note from looking at A5, that if x appears at the first position,
we execute on basic operation, if x appears in the second position we execute 2 basic
operations, if x appears on the last positions we execute n basic operations, and x does
not appear we also execute n basic operations. Thus
t(Ii)=i, 1  i n, and t(In+1)=n;
Thus, we obtain
A(n) 
 p(I ).t ( I )  p(I )t (I )  p( I
1
1
2 )t ( I 2 )     p( I n )t ( I n ) 
p( I n 1 )t ( I n 1 )
I Dn
n
 q/n

i  (1  q)n 
i 1
q n(n  1)
q(n  1)
q qn
.
 (1  q)n 
 n  qn  n  
n
2
2
2
2
Thus the average-case depends on the probability q that we will randomly obtain an input
where x appears.
Suppose that q=1/2 (we have the same prob. that an input where x appears than an input
where x does not appear), then substituting
A( n)  n 
q
qn
3
1

 n  1/ 4  n / 4  n 
2
2
4
4
__________________________________
__________________________________
n
1
n
3/4.n
If q=1 (we will always get an input where x appears)
A( n)  n 
q
qn
1
1

 n  1/ 2  n / 2  n 
2
2
2
2
__________________________________
1
n
1/2.n
If q=0 then (will be never obtain an input where x appears), then
A(n)  n,
which is the worst-case scenario.
Classifying functions by theirs growth rates.
Consider two algorithms A and B, to solve the same problem (e.g. sorting) and we want
to compare them:
WA(n)=2n, and WB(n)=9n, which one is better? As we are considering the number of
basic operations, we know that to obtain the total number of operations, we should
multiply WA(n) by a constant c, and WB(n) by a constant c'. Since we do not know these
constant, we don't know which one is more efficient.
Next suppose WA(n)=n3 /2 and WB(n)=5n2 , which one is better, we know that for small
inputs A is better, but for large inputs B is more efficient.
Thus form the previous examples we conclude that in order to classify functions, we can
ignore constants and small inputs.
We can classify algorithms using "asymptotic growth rate", where small inputs and
constant do not play a role.
Let f(n) be a real function on the natural numbers. Consider the universe of all real
functions on the natural numbers (infinite many), and we will group these functions with
respect to f(n).
(f)
Functions that
grow at the
same rate as
f(n)
Functions that
grow faster or
equal to f(n)
(f)

(f)
Functions that
grow slower or
equal to f(n)
As an example suppose that f(n)=n2, one function that belongs to (f(n)) is g(n)= n3 (as
n3 grows faster than n2 ), and one function that belongs to (f(n)) is h(n)=n (as n grows
slower than n2).