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3 years ago

Is recursion ever required to solve a problem? What other approach can you use to solve a problem that is repetitive in nature?

Computers and Technology

1 answer:

Setler79 [48]3 years ago

8 0

Answer:

No you can not tell that recursion is ever required to solve a problem.

Recursion is required when in the problem, the solution of the input depends on the solution of the subsets of the input.

Iteration is also another form of repetitive approach we follow to solve that kind of problems.

But the difference between recursion and iteration is :

In recursion we call the function repeatedly to return the result to next level.
In iteration certain bunch of instructions in a loop are executed until certain conditions met.

Explanation:

For example in the Fibonacci sequence problem, to find $f_{n}$ , we need to compute $f_{n-1}$ and $f_{n-2}$ before that.

In case of recursion we just call the method Fibonacci(n) repeatedly only changing the parameter Fibonacci(n-1), that calculates the value and return it.

Fibonacci(n)

1. if(n==0 or n==1)

2. return 1.

3.else

4. return( Fibonacci(n-1)+Fibonacci(n-1) )

But in case of iteration we run a loop for i=2 to n, within which we add the value of current $f_{i-1}$ and $f_{i-2}$ to find the value of $f_{i}$

Fibonacci(n)

1. if(n<=2)

2. result = 1

3. else

4. result1 =1 and result2=1.

5. { result = result1 +result2.

6. result1= result2.

7. result2 = result.

8. }

9. output result.

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Write an application that displays every perfect number from 1 through 1,000. A perfect number is one that equals the sum of all

matrenka [14]

Answer:

Written in Python

for num in range(1,1001):

sum=0

for j in range(1,num+1):

if num%j==0:

sum = sum + j

if num == sum:

print(str(num)+" is a perfect number")

Explanation:

This line gets the range from 1 to 1000

for num in range(1,1001):

This line initializes sum to 0 for each number 1 to 1000

sum=0

This line gets the divisor for each number 1 to 1000

for j in range(1,num+1):

This following if condition line checks for divisor of each number

if num%j==0:

sum = sum + j

The following if condition checks for perfect number

if num == sum:

print(str(num)+" is a perfect number")

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3 years ago

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Allisa [31]

Answer:

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Explanation:

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3 years ago

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nikdorinn [45]

Answer:

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Explanation:

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2 years ago

Write a program that declares an array named myArray with 8 components of the type int. Initialize the array to 8 values that th

Savatey [412]

Answer:

#include <iostream>

using namespace std;

void filterEvens(int myArray[]) {

for (int i = 0; i < 8; ++i) {

if(myArray[i]%2==0){

cout<<myArray[i]<<" ";

}

int main(){

int myArray[8];

for(int i =0;i<8;i++){

cin>>myArray[i];

}

filterEvens(myArray);

return 0;

}

Explanation:

The solution is provided in C++

#include <iostream>

using namespace std;

The function filerEvens is defined here

void filterEvens(int myArray[]) {

This iterates through the elements of the array

for (int i = 0; i < 8; ++i) {

This checks if current element is an even number

if(myArray[i]%2==0){

If yes, this prints the array element

cout<<myArray[i]<<" ";

}

The main begins here

int main(){

This declares an integer array of 8 elements

int myArray[8];

The following iteration allows input into the array

for(int i =0;i<8;i++){

cin>>myArray[i];

}

This calls the defined function filter Evens

filterEvens(myArray);

return 0;

}

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3 years ago

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ladessa [460]

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