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Could anybody explain this program and output of this? I am having doubt at if statement. I'm not able to understand how break statement works in this:

for n in range(2, 10):
    for x in range(2, n):
        if n % x == 0:
            print n, 'equals', x, '*', n/x
            break
    else:
        # loop fell through without finding a factor
        print n, 'is a prime number'

Output:

2 is a prime number
3 is a prime number
4 equals 2 * 2
5 is a prime number
6 equals 2 * 3
7 is a prime number
8 equals 2 * 4
9 equals 3 * 3
Alfe
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subha
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4 Answers4

2

I'll add some comments:

for n in range(2, 10): #Loops from 2 to 9, inclusive. Call this Loop A.
    for x in range(2, n): #Loops from 2 to n-1, inclusive. Call this Loop B.
        if n % x == 0: #If n is divisible by x, execute the indented code
            print n, 'equals', x, '*', n/x #print the discovered factorization
            break #Break out of loop B, skipping the "else" statement
    else: #If the loop terminates naturally (without a break) this will be executed
        # loop fell through without finding a factor
        print n, 'is a prime number'
GrantS
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1

The break statement leaves the loop without entering the else clause. If the loop terminates without reaching the break, the else clause will be entered. In other words, the loop searches for a possible divisor; if it finds one it prints it and leaves the loop using break. If no divisor is found, the for loop terminates "normally" and thus enters the else clause (in which it then prints that it has found a prime).

Alfe
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0

Obviously, this program is trying to identify prime numbers. A prime number, has no factors (i.e. when you divide a prime number by x, there is always a remainder), other than 1 (obviously!) and itself. So, we need to test every number from 2 (i.e. not 1) up to the number before our test to see if it is a factor of our test number.

The test that is running, steps through like this:

# S1 is a set of numbers, and we want to identify the prime numbers within it.
S1 = [2, 3, 4, 5, 6, 7, 8, 9]

# test a whether n is PRIME:
for n in S1:
    # if n / x has no remainder, then it is not prime
    for x in range(2, n):
        if...
            I have NO REMAINDER, then x is a factor of n, and n is not prime
            -----> can "BREAK" out of test, because n is clearly not PRIME
            --> move on to next n, and test it
        else:
            test next x against n
            if we find NO FACTORS, then n is PRIME
Nick Burns
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Break leaves the inner most loop directly and goes to the next step of the outer for loop.

schilippe
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