CODEWITHSUNDEEP
python
Isha G
Isha G

Reputation: 11

pyschools square root approximation

I am new to Python and stackoverflow. I have been trying to solve Pyschools 'while loop' example for square root approximation (Topic 5: Question 9). However I am unable to get the desired output. I am not sure if this issue is related to the loop or the formula. Here is the question:

Create a function that takes in a positive number and return 2 integers such that the number is between the squares of the 2 integers. It returns the same integer twice if the number is a square of an integer.

Examples:

sqApprox(2)
(1, 2)
sqApprox(4)
(2, 2)
sqApprox(5.1)
(2, 3)

Here is my code:

def sqApprox(num):
    i = 0
    minsq = 1                           # set lower bound
    maxsq = minsq                       # set upper bound
    while i*i<=num:                     # set 'while' termination condition
        if i*i<=num and i >=minsq:  # complete inequality condition  
            minsq = i
        if i*i<=num and i <=maxsq:  # complete inequality condition
            maxsq = i
        i=i+1                       # update i so that 'while' will terminate
    return (minsq, maxsq)

If I create this function sqApprox(4) and call it on IDE, I get output (2, 0).

Could someone please let me know what I am doing wrong? Thanks in advance.

Upvotes: 1

Views: 387

Answers (1)

John Coleman
John Coleman

Reputation: 52008

This is why your code does what it does:

After the line maxsq = minsq is executed, both of those values are 1.

When we come to the loop

while i*i<=num:                     # set 'while' termination condition
    if i*i<=num and i >=minsq:  # complete inequality condition  
        minsq = i
    if i*i<=num and i <=maxsq:  # complete inequality condition
        maxsq = i
    i=i+1                       # update i so that 'while' will terminate

first note that inside the loop i*i<=num, so there is no need to retest it. Thus it is equivalent to:

 while i*i<=num: 
     if i >=minsq:  
         minsq = i
     if i <=maxsq:
         maxsq = i
     i=i+1

In the first pass through the loop i == 0 but maxsq == 1, making the second condition true, hence setting maxsq equal to the current value of i, which is 0. In subsequent passes through the loop, i <= maxsq is false (since maxsq == 0 but i > 0) hence maxsq is never moved beyond 0. On the other hand, the first condition in the while loop keeps updating minsq as intended.

I would recommend forgetting about both minsq and maxsq completely. Have the loop simply be:

while i*i <= num:
    i += 1 #shortcut for i = i + 1

When the loop is done executing, a simple test involving i-1 is enough to determine what to return.

Upvotes: 1

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