CODEWITHSUNDEEP
python-3.xloopsfor-loop
DRV
DRV

Reputation: 756

for loop in python with square root condition checkin

In C we can check the value "i" less than the square root of n using this for loop

for (int i = 2; i * i <= n; i++)

I have used while loop as:

i = 2
while(i * i <= n):
    i+=1

Can we able to do using for loop in python?

Upvotes: 2

Views: 3650

Answers (2)

Tim
Tim

Reputation: 3407

It's pretty much the same.

In python, typical for loop uses range(start, end, step) function to get the indexing variable i. It loop through i=start to i=end-1 while incrementing i by step.

for i in range(start, end, step):

This is equivalent to C/C++/Java's

for (int i = start; i < end; i+=step)

Then, to stop by the square root of n. You simply use what's below:

import math
for i in range(2, int((math.sqrt(n))+1):

Note that math.sqrt(n) gives a float. Then wrapping this in int takes the floor of the float. And since for loop stops at end-1, we add 1 so this imitates the behavior requested by OP, i.e. i*i <= n.

Some examples make this easier: e.g. n=4, then:

  • math.sqrt(n)=2.0,
  • int((math.sqrt(n))=2,
  • int((math.sqrt(n))+1=3
    and for i in range(2, 3) will run the loop with i=2 and stop before i reaches 3.

Now, if n is not perfect square: e.g. n=10,

  • math.sqrt(n)=3.16...
  • int((math.sqrt(n))=3,
  • int((math.sqrt(n))+1=4
    and for i in range(2, 4) will run the loop with i=2 and i=3 and stop before i reaches 4.

While loop is exactly the same. Another way to write i*i is i**2 in python.

i=2
while i**2 <= n:
    i+=1

Upvotes: 2

Achy97
Achy97

Reputation: 1024

Its possible as @Tim mentioned, because in python for loop actually traverses through a range of values, but in C/C++ what we have is initialization;condition;increment so that exact structure is not followed in python, instead its for x in range() so the exact syntactical code is not possible, but you can achieve the same functionality (using while loop also or for loop as shown), as it is based on the features a language provides!

Hope it gives your answer!

Upvotes: 1

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