**1.Algorithm**to find the average of a subject marks of ‘N’ number of students.

AVG_OF_MARKS (LIST, N)

[LIST is an array containing marks, N is the size of array]

SUM 0

Repeat For I=1,2,3, …N

SUM SUM + LIST[i]

[End of For I]

AVG SUM/N

Write: ‘The average is’,AVG

Exit.

**2.Algorithm**to find average of SALARY in an array of EMP struct containing information EMPNO, EMPNAME, and SALARY.

AVG_SALARY (LIST, SIZE)

[LIST is an array of EMP, SIZE is size of array]

SUM 0

Repeat For I=1,2,3, …SIZE

SUM SUM + LIST[I].SALARY

[End of For I]

AVG SUM / SIZE

Write: ‘The average Salary is’,AVG

Exit.

An algorithm, that performs sub task, may be called in another algorithm. It is better practice to divide the given problem into sub problems and write the individual algorithms to solve such sub problems and collectively write main algorithm to call the sub algorithms in order to solve the main problem.

**3.Algorithm**to find maximum of two unequal numbers and use the same to find maximum of four unequal numbers.

MAX_OF_2(NUM1, NUM2)

[NUM1 and NUM2 are two numbers]

If NUM1 > NUM2 Then:

Return NUM1

Else:

Return NUM2

[End of If]

Exit:

MAX_OF_4(NUM1, NUM2, NUM3, NUM4)

[NUM1, NUM2, NUM3, and NUM4 are numbers]

TEMP1 MAX_OF_2(NUM1, NUM2)

TEMP2 MAX_OF_2(NUM3, NUM4)

MAX MAX_OF_2(TEMP1, TEMP2)

Return MAX

Exit.

**4.Algorithm**to find minimum of two distinct numbers and using the same to find minimum of three distinct numbers.

MIN_OF_2(NUM1, NUM2)

[NUM1 and NUM2 are two number]

If NUM1<NUM2 Then:

Return NUM1

Else:

Return NUM2

[End of If]

Exit.

MIN_OF_3(NUM1, NUM2, NUM3)

[NUM1, NUM2, and NUM3 are distinct number]

Return MIN_OF_2(MIN_OF_2(NUM1, NUM2), NUM3)

Exit.

**5.Algorithm**to find the GCD (Greatest Common Divisor)

of two positive numbers and use the same to find LCM of two positive numbers.

GCD(NUM1, NUM2)

Rem NUM1%NUM2 [%Modulus Operator]

Repeat While Rem<>0

NUM1 <--- NUM2

NUM2<--- Rem

Rem <-- NUM1% NUM2

[End of While]

Return NUM2

Exit.

LCM(NUM1,NUM2)

Return(NUM1*NUM2) / GCD (NUM1,NUM2)

Eixt.

Suppose that the LCM algorithm is called using 12 and 16.

The algorithm calls GCD with 12 and 16.

Working of GCD algorithm:

Rem12%16 i.e. Rem =12

Rem is not equal to Zero. So, NUM1=16 and NUM2=12

Rem16%12 i.e. Rem=4

Rem is not equal to Zero. So, NUM1=12 and NUM2=4

Rem12%4 i.e. Rem=0

Rem is equal to Zero. So, GCD algorithm returns NUM2 that is 4.

When the algorithm LCM gets the result from GCD algorithm it finds the expression value (NUM1 * NUM2) /GCD(NUM1,NUM2).i.e.(12*16) / 4. Which is equal to 48.48 is the LCM of 12 and 16.

Here in this example algorithm, finding LCM is the main problem that can be divided into sub problem like finding the GCD first and using the same to solve the main problem. So, GCD algorithm is written first and in the order algorithm that can be treated as a main algorithm. In this way a given problem can be divided into small problems and algorithms can be written to solve such small problems.

**6. Algorithm**to find sum of ‘N’ natural numbers between

N1 and N2.

SUM1(N1,N2)

[N1 and N2 are two different natural numbers]

If N1> N2 Then:

MAXN1

MINN2

Else:

MAXN2

MINN1

[End of ]

MIN MIN – 1[to include MIN in sum]

SUM1 (MIN*(MIN +1))/2

SUM2(MAX*(MAX +1))/2

Return SUM2-SUM1

Exit.

In the above algorithm to find the sum of N natural the formula N(N+1)/2 is used. Instead of formula a repetitive statement can also be used that run from MIN to MAX with setp 1. The algorithm can be re-written as:

SUM2(N1,N2)

[N1 and N2 are two different natural numbers]

If N1>N2 Then:

MAX N2

MIN N2

Else:

MAX N2

MINN1

[End of If]

SUM 0

Repeat For I= MIN, MIN+1, MIN+2 …MAX

SUMSUM +1

[End of For]

Return SUM

Exit.

**Dry Run:**
Suppose that the above algorithm SUM 1 is called with two number 23 and 11. In the If statement 23>11 condition is TRUE, therefore MAX becomes 23 and MIN becomes 11. Now the loop repeats for 11, 12, 13, 14 up to 23 (I values).Every time when the loop is repeated I is added to SUM. So, finally the value of SUM is returned from the algorithm.

## 0 comments :

## Post a Comment