UNIT-1.docx Design and Analysis of Algorithm

Introduction to Algorithms
An algorithm is a finite set of instructions that, if followed, accomplishes a particular task.
In addition, all algorithms must satisfy the following criteria :
Input. Zero or more quantities are externally supplied.
Output. At least one quantity is produced.
Definiteness. Each instruction is clear and unambiguous
Finiteness. If we trace out the instructions of an algorithm, then for all cases, the algorithm terminates
after a finite number of steps.
Effectiveness. Every instruction must be very basic so that it can be carried out, in principle, by a
person using only pencil and paper. It is not enough that each operation be definite as in criterion 3; it
also must be feasible.
Process of translating a problem into an algorithm
Sum of n no’s
Abstract solution:
Sum of n elements by adding up elements one at a time.
Little more detailed solution:
Sum=0;
add a[1] to a[n] to sum one element at a time.
return Sum;
More detailed solution which is a formal algorithm:
Algorithm Sum (a,n)
{
sum:= 0.0;
for i:= 1 to n do
sum:=sum+a[i];
Return sum;
}
Find largest among list of elements
Abstract solution:
Find the Largest number among a list of elements by considering one element at a
time.
1. Treat first element as largest element
2. Examine a[2] to a[n] and suppose the largest element is at a [ i ];
3. Assign a[i] to largest;

4. Return largest;
Algorithm Largest (a,n)
{
largest := a[1];
for i := 2 to n do
{
if ( a[i] > largest ) largest := a[i];
}
Return largest;
}
Linear search
Abstract solution:
From the given list of elements compare one by one from first element to last
element for the required element
for i := 1 to n do
{
Examine a[i] to a[n] and suppose
the required element is at a [ j ];
Write match is found and return position j;
}
Algorithm Linear search (a, req_ele)
{
Flag=0
for i := 1 to n do
{
If (a[i] = req_ele) then
{
Flag=1;
Pos = i;
break;
}
}
If (flag) then
Write “element found at ‘pos’ position”
Else
Write “element not found”
End if
}

Selection Sort
Abstract solution:
From those elements that are currently unsorted, find the smallest and place it
next in the sorted list.
for i := 1 to n do
{
Examine a[i] to a[n] and suppose
the smallest element is at a [ j ];
Interchange a[i] and a[j];
}
Algorithm Selection Sort (a, n)
// Sort the array a[1 : n] into non decreasing order.
{
for i := 1 to n do
{
min:=i;
for k : i + 1 to n do
If (a[k]< a[min]) then min:=k;
t :=a[i]; a[i] := a[min];a[min]:= t;
}
}
Bubble Sort
Abstract solution:
Comparing adjacent elements of an array and exchange them if they are not in order. In
the process, the largest element bubbles up to the last position of the array first and then
the second largest element bubbles up to the second last position and so on.
for i := 1 to n do
{
Compare a[1] to a [n-i] pair-wise (each element with its next)
and interchange a[j] and a[j+1] whenever a[j] is greater than a[j+1]
}
Algorithm BubbleSort (a,n)
{
for i := 1 to n do
{
for j :=1 to n-i do

{
if ( a[j] > a[j+1] )
{
// swap a[j] and a[j+1]
temp := a[j]; a[j] := a[j+1];a[j + 1] := temp;
}
}
}
}
Insertion sort
Abstract solution:
Insertion sort works by sorting the first two elements and then inserting the third element in its
proper place so that all three are sorted. Repeating this process, k+1 st element is inserted in its
proper place with respect to the first k elements (which are already sorted) to make all k+1
elements sorted. Finally, 'n' th element is inserted in its proper place with respect to the first n-1
elements so all n elements are sorted.
Sort a[1] and a [2]
For i from 3 to n do
{
Suppose a[k] (k between 1 and i) is such that a[k-1] <= a[i] <= a[k].
Then, move a[k] to a[i-1] by one position and insert a[i] at a[k].
}
Algorithm insertionSort(a,n)
{
for i := 2 to n do
{
value := a[i]; j := i -
1; done := false;
repeat
if a[j] > value
{
a[j + 1] := a[j]; j := j - 1;
if (j < 1) done := true;
}
else
done := true;
until done;
a[j + 1] := value;
}
}

Pseudo-code conventions
1. Comments begin with // and continue until the end of line.
2. Blocks are indicated with matching braces:{ and }. A compound statement (i.e., a
collection of simple statements) can be represented as a block. The body of a
procedure also forms a block. Statements are delimited by;.
3. An identifier begins with a letter. The data types of variables are not explicitly
declared. The types will be clear from the context. Whether a variable is global or
local to a procedure will also be evident from the context. We assume simple data
types such as integer, float, char, Boolean, and so on. Compound data types can
be formed with records. Here is an example:
Node = record
{ datatype_1 data_1;
::::::::::::::::::::
Datatype_n data_n;
node *link;
}
In this example, link is a pointer to the record type node. Individual data items of
a record can be accessed with and period. For instance if p points to a record of
type node, p data_1 stands for the value of the first field in the record. On the
other hand, if q is a record of type node, q. dat_1 will denote its first field.
4. Assignment of values to variables is done using the assignment statement
(variable) :=(expression);
5. There are two Boolean values true and false. In order to produce these values,
the logical operators and, or, and not and the relational operators
<, ≤, =, ≠ , ≥ , and > are provided.
6. Elements of multidimensional arrays are accessed using [ and ]. For example, if A
is a two dimensional array, the (i,j) the element of the array is denoted as A[i,j].
Array indices start at zero.
7. The following looping statements are employed: for, while, and repeat- until.
The while loop takes the following form:
While (condition) do
{
(statement 1)
::::::::::::::
(statement n)
}

As long as (condition) is true, the statements get executed. When (condition)
becomes false, the value of (condition) is evaluated at the top of loop.
The general form of a for loop is
for variable := value 1 to value 2 step step do
{
(statement 1)
::::::::::::::
(statement n)
}
Here value1, value2, and step are arithmetic expressions. A variable of type
integer or real or a numerical constant is a simple form of an arithmetic
expression. The clause “step step” is optional and taken as +1 if it does not occur.
Step could either be positive or negative variable is tested for termination at the
start of each iteration. The for loop can be implemented as a while loop as
follows:
Variable := value1;
fin :=value2;
incr:=step;
while ((variable –fin) * step ≤,0) do
{
(statement 1)
:::::::::::::::
(statement n)
}
A repeat-until statement is constructed as follows :
repeat
(statement 1)
:::::::::::::::
(statement n)
Until (condition)
The statements are executed as long as (condition) is false. The value of
(condition) is computed after executing the statements.
The instruction break; can be used within any of the above looping instructions to
force exit. In case of nested loops, break; results in the exit of the innermost loop
that it is a part of. A return statement within any of the above also will result in
exiting the loops. A return statement results in the exit of the function itself.
8. A conditional statement has the following forms:
if (condition) then (statement)
if (condition) then (statement 1) else (statement 2)
Here (condition) is a boolean expression and (statement),(statement 1),

and (statement 2) are arbitrary statements (simple or compound).
We also employ the following case statement:
Case
{
:(condition 1 ) : (statement 1)
:::::::::::::::
:(condition n ) : (statement 1)
:else : (statement n+1)
}
Here (statement 1) and (statement 2), etc. could be either simple statements or
compound statements. A case statement is interpreted as follows. If (condition 1)
is true, (statement 1) gets executed and the case statements is exited. If
(statement 1) is false,(condition 2) is evaluated. If (condition 2) is true,
(statement 2) gets executed and the case statement exited, and so on. If none of
the conditions (condition 1),…,(condition n) are true, (statement n+1) is executed
and the case statement is exited. The else clause is optional.
9. Input and output are done using the instructions read and write, No format is
used to specify the size of input or output quantities.
10. There is only one type of procedure: Algorithm. An algorithm consists of a
heading and a body. The heading takes the form.
Algorithm Name ((parameter list))
Where Name is the name of the procedure and ((parameter list)) is a listing of the
procedure parameters. The body has one or more (simple or compound)
statements enclosed within braces { and }. An algorithm may or may not return
any values. Simple variables to procedures are passed by value. Arrays and
records are passed by reference. An array name or a record name is treated as a
pointer to the respective data type.
As an example. The following algorithm finds and returns the maximum of n
given numbers:
Algorithm Max (A, n)
// A is an array of size n.
{
Result := A[1];
for i: = 2 to n do
If A[i] > Result then Result := A[i];
Return Result;
}

RECURSION
Definition: It is the process of repeating items in a self similar way.
Example 1:
Recursion has most common applications in Mathematics and computer science.
Example2:
The following definition for natural numbers is Recursive.
1 is a Natural Number.
If k is a natural number then k+1 is a Natural Number.
DEFINITION(Computer Science): This is the method of defining a function in which the function
being defined is applied within its own definition.
In simple terms if a function makes a call to itself then it is known as Recursion.
Example3: Observe the recursive definition for a function Suml(n) to compute
Sum of first n natural numbers.
If n=1 then
return 1
else
return n + Sum(n-1)

Steps to be followed to write a Recursive function
1.BASIS: This step defines the case where recursion ends.
2.RECURSIVE STEP: Contains recursive definition for all other cases(other than base case)to
reduce them towards the base case
For Example in Example 3,
Basis is n=1 and
Recursive step is n + Sum(n-1)
Note: if any one of the two or both, wrongly defined makes the program infinitely Recursive.
Recursive solution for the Sum of nos
First Call for Recursion:
RecursiveSum (a,n);
Algorithm RecursiveSum(a,k)
{
if (k<=0) then
Return 0.0;
else
Return (RecursiveSum(a,k-1) + a[k]);
}
Recursive solution for the Largest of ‘n’ numbers
First Call for Recursion:
RecursiveLargest (a,n);
Algorithm RecursiveLargest (a,k)
{
if (k = 1) then Return a[k];
else
{
x := RecursiveLargest (a,k-1);
if (x > a[k]) Return x;
else Return a[k];
}
}
Recursive solution for the SelectionSort
First call of recursion is:

RecursiveSelectionSort (a,1);
Algorithm RecursiveSelectionSort (a, m)
{
if (m = n) Return; // Do nothing
else
{
min:=m;
for k := m + 1 to n do
if (a[k]< a[min]) then min:=k;
//swap
t :=a[m]; a[m] := a[min];a[min]:= t;
RecursiveSelectionSort (a, m+1);
}
}
Recursive solution for the Linear Search
Algorithm linearSearch(a, key, size)
{
if (size == 0) then return -1;
else if (array[size ] = key) return size ;
else return linearSearch (array, key, size - 1);
}
Recursive solution for the BubbleSort
Algorithm RecursiveBubbleSort (a,k)
{
if (k = 1)
Return; // Do nothing
else
{
for j :=1 to k-1 do
{
if ( a[j] > a[j+1] )
{
temp := a[j]; a[j] := a[j+1];
a[j + 1] := temp;
}
}
RecursiveBubbleSort (a,k-1);
}
}

Recursive solution for the InsertionSort
RecursiveInsertionSort(a,n);
Algorithm RecursiveInsertionSort(a,k)
{
if (k > n)
Return; // Do nothing
else
{
RecursiveInsertionSort(a,k-1);
//Insert 'k'th element into k-1 sorted list
value := a[k]; j := k - 1;
done := false;
repeat
if a[j] > value
{
a[j + 1] := a[j]; j := j - 1;
if (j < 0) done := true;
}
else done := true;
until done;
a[j + 1] := value;
}
}
Recursive solution for the Printing an array in normal order
Recursive Prin_arr(a,n-1);
Recursive Algorithm prin_arr(int a[],int n)
{
if(n=-1)
return;
else
prin_arr(a,n-1);
write a[n]);
}
Recursive solution for the Decimal to binary conversion
Algorithm bin(int n)
{
if (n=1 or n=0)
{
write n;

}
else
return;
bin=n/2;
write n mod 2;
}
Recursive solution for the Fibonacci number
Recursive Algorithm fib(int x)
{
if(x=0 or x=1) then
return 0;
else
{
}
if(x=2)
else
{
}
return 1;
f=fib(x-1) + fib(x-2);
return(f);
Recursive solution for the Factorial of a given number
Recursive Algorithm fact (int n)
{
{
int fact;
if n=1 tehn
return 1;
else
fact=n*(n-1);
}
return fact;
}
Towers of Hanoi with recursion:
Rules:
(1) Can only move one disk at a time.
(2) A larger disk can never be placed on top of a smaller disk.
(3) May use third post for temporary storage.
Algorithm TowersOfHanoi (numberOfDisks, x,y,z)
{
if (numberOfDisks >= 1)

{
TowersOfHanoi (numberOfDisks -1, x,z,
y); move (start, end);
TowersOfHanoi (numberOfDisks -1, z, y, x);
}
}
Performance Analysis
Space Complexity:
Algorithm abc (a,b,c)
{
return a+b++*c+(a+b-c)/(a+b) +4.0;
}
 The Space needed by each of these algorithms is seen to be the sum of the following
component.
1. A fixed part that is independent of the characteristics (eg: number, size)of the inputs and
outputs. The part typically includes the instruction space (ie. Space for the code), space for
simple variable and fixed-size component variables (also called aggregate) space for constants,
and so on.
2. A variable part that consists of the space needed by component variables whose size is
dependent on the particular problem instance being solved, the space needed by referenced
variables (to the extent that is depends on instance characteristics), and the recursion stack
space.
 The space requirement s(p) of any algorithm p may therefore be written
as, S(P) = c+ Sp(Instance characteristics)
Where ‘c’ is a constant.
Example 2:
Algorithm sum(a,n)
{
s=0.0;
for I=1 to n do
s= s+a[I];
return s;
}
 The problem instances for this algorithm are characterized by n, the number of elements to
be summed. The space needed d by ‘n’ is one word, since it is of type integer.
 The space needed by ‘a’ a is the space needed by variables of type array of floating point
numbers.
 This is at least ‘n’ words, since ‘a’ must be large enough to hold the ‘n’ elements to be
summed.
 So, we obtain Sum(n)>=(n+s)
[ n for a[ ],one each for n, I a& s]
Alogorithm RSum(a,n)
{

If(n<= 0) then
return 0.0;
Else
}
return RSum(a,n-1)+a[n];
Recursive function for Sum
Time Complexity:
A Program step is loosely defined as a syntactically or semantically meaningful segment of a
program that has an execution time that is independent of the instance characteristics.
For Ex: Return a+b+b*c+(a+b-c)/(a+b)+4.0;
Program step count method – examples:
Algorithm Sum (a,n)
{
s:= 0.0;
Count :=Count+=1; // count is global;it is initially zero.
for i:=1 to n do
{
Count :=count+1 //for for
S :=S+a[i];count:=count+1;For assignment
}
Count:=count+1; // For last time of for
Count:=count +1; //For the return
return s;
}
Algorithm SUM with count statements added
Algorithm Sum (a,n)
{
for i:=to n do count:=count+2;
count :=count +3;
}
Algorithm for Sum Simplified version for previous sum
Algorithm RSum(a,n)
{
Count :=count+1; // For the if
conditional if (n < 0)then
{
}
else
count:=count+1//For the return
return 0.0;

{
count :=count+1 // For the addition, function invocation and return
return RSum(a,n-1)+a[n];
}
}
Algorithm Recursive sum with count statements added
Algorithm Add (a,b,c,m,n)
{
for i:= 1to m do
for j:=1 to n
do
c[i,j] := a[i,j] + b[i,j];
}
Matrix Addtion
Algorithm Add(a,b,c,m,n)
{
for i:=1 to m do
{
count :=count +1; // For ‘for i’
for j :=1 to n do
{
count :=count+1; // For ‘for
j’ c[i, j] := a[i, j]+b[i, j];
count :=count +1; //For the assignment
}
count := count +1 ;// For loop initialization and last time of ‘for j’
}
count := count+1 ; // For loop initialization and last time of ‘for i’
}
Algorithm for Matrix addition with counting statements
Step table method:
Statement s/e frequency total steps
1 .Algorithm Sum(a,n) 0 ------ 0
2 { 0 ------ 0
3 s:=0.0; 1 1 1
4 for i := 1to n do 1 n+1 n+1
5 s := s + a[i]; 1 n n
6 return 1 1 1
7 } 0 ----- 0
TOTAL 2n +3
Step table for Algorithm finding Sum

Statement s/e Frequency
n=0 n>0
total steps
n=0 n>0
1.Algorithm RSum(a,n) 0 - - 1. 0
2.{
3. if(n< 0) then 1 1 1 1 1
4. return 0.0: 1 1 0 1 0
5. else return
6.RSum (a,n-1) + a[n]; 1+x 0 1 0 1+x
7. } 0 --- --- 0 0
Total 2 2+x
x = tRSum (n-1)
Step table for Algorithm RSum
Statement s/e frequency Total steps
1. Algorithm Add(a,b,c,m,n) 0 ----- 0
2. { 0 ----- 0
3. for i := 1 to m do 1 m+1 m+1
4. for j := 1 to n do 1 m(n+1) mn + m
5. c[i,j] :=a[i,j] + b[i,j]; 0 mn mn
6. } ------ 0
Total 2mn +2m +1
Step table for Algorithm Addition of two matrices
Algorithm BubbleSort (a,n)
{
for i := 1 to n do
{
for j :=1 to n-i do
{
if ( a[j] > a[j+1] )
{
temp := a[j];
a[j] := a[j+1];
a[j + 1] :=
temp;
}
}
}
}
Total
Fill the blank columns in the above Table

Algorithm Linear search (a, req_ele)
{
Flag=0
for i := 1 to n do
{
If (a[i] = req_ele) then
{
Flag=1;
Return i;
}
}
If (flag) then
Write “element found at i position”
Else
Write “element not found
}
Total
Fill the blank columns in the above Table

Kinds of Analysis of Algorithms
• The Priori Analysis is aimed at analyzing the algorithm before it is implemented(based on the
algorithm) on any computer. It will give the approximate amount of resources required to solve the
problem before execution. In case of priori analysis, we ignore the machine and platform dependent
factors. It is always better if we analyze the algorithm at the earlier stage of the software life cycle.
Priori analysis require the knowledge of
– Mathematical equations
– Determination of the problem size
– Order of magnitude of any algorithm
• Posteriori Analysis is aimed at determination of actual statistics about algorithm’s consumption of
time and space requirements (primary memory) in the computer when it is being executed as a program
in a machine.
Limitations of Posteriori analysis are
• External factors influencing the execution of the algorithm
– Network delay
– Hardware failure etc.,
• The information on target machine is not known during design phase
• The same algorithms might behave differently on different systems
– Hence can’t come to definite conclusions
Asymptotic notations(O,Ω,Ө)
Step count is to compare time complexity of two programs that compute same function and also to
predict the growth in run time as instance characteristics changes. Determining exact step count is
difficult and not necessary also. Since the values are not exact quantities we need only comparative
statements like c1n2
≤ tp(n) ≤ c2n2
.
For ex: consider two programs with complexities c1n2
+ c2n and c3n respectively. For small values
of n, complexity depend upon values of c1, c2 and c3. But there will also be an n beyond which
complexity of c3n is better than that of c1n2
+ c2n.This value of n is called break-even point. If this
point is zero, c3n is always faster (or at least as fast).
c1=1,c2=2 & c3=100
Then c1n2
+c2n is ≤ c3n for n≤98 and
c1n2
+c2n is > c3n for n>98
The Common asymptotic functions are given below.

1< log n < n <n log n < n2
< n3
< 2n
<n!
Function Name
1 Constant
log n Logarithmic
N Linear
n log n n log n
n2 Quadratic
n3 Cubic
2n Exponential
n! Factorial
The growth of the functions as below
Definition [Big ‘oh’] The function f(n)=O(g(n)) iff there exist positive constants c and no such that
f(n)≤c*g(n) for all n, n ≥ no.
Ex1: f(n) = 2n + 8, and g(n) = n2
. Can we find a constant c, so that 2n + 8 <= n2
? The number 4 works
here, giving us 16 <= 16.
For any number c greater than 4, this will still work. Since we're trying to generalize this for large
values of n, and small values (1, 2, 3) aren't that important, we can say that f(n) is generally faster than
g(n); that is, f(n) is bound by g(n), and will always be less than it.
Ex2: The function 3n+2=O(n) as 3n+2≤4n for all n≥2.
Pb1: 3n+3=O( ) as 3n+3≤ for all .
Ex3: 10n2
+4n+2=O(n2
) as 10n2
+4n+2≤11n2
for all n≥5
Pb2:1000n2
+100n-6=O( ) as 1000n2
+100n-6≤ for all
Ex4:6*2n
+n2
=O(2n
) as 6*2n
+n2
≤ 7*2n
for n ≥ 4
Ex5:3n+3=O(n2
) as 3n+3≤3n2
for n≥2.
Ex 6:10n2
+4n+2=On4 as10n2
+4n+2≤10n4
for n≥2.
Ex7:3n+2≠O(1) as3n+2 not less than or equal to c for any constant c and all n≥n0
Ex 8: 10n2+4n+2≠O(n)
Definition[Omega] The function f(n) =Ώ (g(n) (read as “f of n is omega of g of n'') iff there
exist positive constants c and n0 such that f(n) ≥c *g(n) for all n, n≥ no.

i = 0 i
i = 0 i
Ex1: The function 3n+2=Ω(n) as 3n+2≥3n for n≥1
(the inequality holds for n≥0,but the definition of Ω requires an n0>0).
3n+2=Ω(n) as 3n+2≥3n for n≥1.
Ex:2 100n+6=Ω(n) as 100n+6≥100n for n≥1
Pb:1 6n+4= Ω( ) as 6n+4 ≥ for all n≥
Ex:3 10n2
+4n+2= Ω(n2
) as 10n2
+4n+2 ≥ n2
for n ≥ 1
Pb:2 2n2
+3n+1= Ω( ) as 2n2
+3n+1 ≥ for all n≥
Ex: 4 6* 2n
+ n2
= Ω(2n
) as 6* 2n
+ n2
≥ 2n
for n ≥ 1.
Pb:3 4n2
– 64 n + 288 = Ω ( )as 4n2
– 64 n + 288≥ for all n≥ .
Pb:5 n3
logn is = Ω ( )as n3
logn _for all n≥ .
Definition [Theta] The function f(n) = Θ (g(n) ) (read as “f of n is theta of g of n'') iff there exist
positive constants C1,C2, and n0 such that c1g(n)
≤ f(n) ≤ c2
g(n) for all n, n≥n0
Ex 1 : The function 3n + 2 = Ө(n) as 3n + 2 ≥ 3n for all n ≥ 2 and 3n + 2 ≤ 4n for all n ≥ 2 so c1 = 3,
c2=4 and n0 =2
Pb 1: 6n+4= Θ( ) as 6n+4≥ for all n ≥ and 6n+4 ≤ for all n ≥
Ex 2 : 3n + 3 = Ө(n)
Ex 3 : 10n2
+ 4n +2 = Ө(n2
)
Ex 4: 6* 2n
+ n2
= Θ (2n
)
Ex5:10*log n+4= Θ (log n)
Ex6:3n+2≠ Θ(1)
Ex7:3n + 3 = Ө(n)
Ex 8: 10n2
+4n+2 ≠ Θ (n)
Theorem: If f(n) = amnm+.......................+a3n3
+a2n2
+a1n+a0, then f(n)=O (nm)
Proof: f(n) ≤ ∑m
| a | ni
≤n
m
∑
m
i = 0│ai│n
i-m
≤ nm
∑m
│a │ for n≥1
f(n) = O(nm
) (assuming that m is fixed ).

Theorem: If f(n) = am nm
++.................+a3n3
+a2n2
+a1 n + ao and am > 0, then f(n)= Ω(nm
)
proof : Left as an exercise.
Definition [Little ‘oh’] The Function f (n) =o(g(n)) (read as ‘f of n is little oh of g of
n’)iff lim f(n) = 0
n→∞ g(n)
Example:
The function 3n+2 = o(n2
) since lim 3n+2 = 0.
n-->∞ n2
Ex1: 3n+2 =o(n log n).
Ex:2 3n+2 =o(n log log n).
Ex:3 6*2n
+n2
=o(3n
).
Ex:4 6 * 2n
+n2
=o(2n
log n).
Ex:5 6 * 2n
+n2
≠o(2n
).
Analogous to ‘o’ notation ‘ ω ‘notation is defined as follows.
Definition [Little omega] The function f(n)= ω(g(n)) (read as “ f of n is little omega of g of n”) iff
lim g(n) = 0
n→∞ f(n)

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