(1) * Givent an integer N, write a program that sums all odd natural numbers les

Question

(1) * Givent an integer N, write a program that sums all odd natural numbers less than or equal to N, i.e. computes S = 1+3+5+...+ (m – 2) + m, where m = N if N is odd, and m = N — 1 if N is even. Write two implementations: (a) Use an appropriate for- or while-loop. Filename: _la (b) Solve this problem by recursively calling an appropriately defined function, avoid the use of loops. Filename: _1b (c) Run your program for different values of N, and measure the runtime (import time): N E {10, 100, 1000, 10000}. Which implementation is faster? Each code should contain a function main (N), which takes the integer N as an argument, and prints the result when called.

Explanation / Answer

Part (a) of the question:

clc
clear

N = input('Enter N: ');
s = 0;

for i=1:2:N
  
s = s + i;
  
end

fprintf("The sum is: "+s);

------X----X-----

Part (b) of the question:

Script ( run this file )

clc
clear

s = 0;
N = input('Enter N: ');
s = RecurseOddEven(N,s);

fprintf("The sum is: "+s);

---XX--

Function (call this file from the script)

function [s] = RecurseOddEven(N,s)

if rem(N,2) == 0
N = N - 1;
end

if (N-2) >= -1
s = s + N + RecurseOddEven(N-2,s);
end

end

----X----X----

Part (c).

**Script**

clc
clear
tic
for i = 1:4
s = 0;
N = 10^i;
s = RecurseOddEven(N,s);

fprintf("The sum is: "+s);

end
toc

**FUNCTION FILE**

function [s] = RecurseOddEven(N,s)

if rem(N,2) == 0
N = N - 1;
end

if (N-2) >= -1
s = s + N + RecurseOddEven(N-2,s);
end

end

***With LOOPS***

clc
clear
tic
for j = 1:4
N = 10^j;
s = 0;

for i=1:2:N

s = s + i;
  
end

fprintf("The sum is: "+s);
end
toc

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