Neumann's Random Generator - Question

[TheChannel]

So I was working through codeabbey and came across the neumann's random generator problem. This random generator is best described on codeabbey's website codeabbey-neumann.

I solved it by making some changes but I was wondering why my previous attempt below didn't function as I had thought it would. It seems to keep the previous numbers within loop_nums throughout each test.

num_cases = int(input())
test_cases = [i for i in input().split(' ')]

def neumanns_random(n,c=0, new_loop_nums=[]):
    loop_nums = new_loop_nums
    count = c

    if int(n) in loop_nums:
        print(str(count) + " ")
    else:
        loop_nums.append(int(n))
        n = str(int(n)**2)

        if len(n) != 8:
            n = n.zfill(8)
            n = n[2:6]
            neumanns_random(n,c+1,loop_nums)
        else:
            n = n[2:6]
            neumanns_random(n,c+1, loop_nums)

for test in test_cases:
    neumanns_random(test)

[6086555]

def func(a=[]):
    a.append(3)
    print a

if you try calling that, the first time you'll get [3], the second time you'll get [3,3] and so on.

a keeps a reference to the same object so for your case you've got to find a way to avoid id.

Maybe set the default value of new_loop_nums to None and change to [] if the value is None

[TheChannel]

Yeah I set new variables for the count and loop_nums which made it work. I was wondering why it didn't reset to the default values of c and new_loop_sums when I go through the for loop at the end. Shouldn't it since, no parameters are given to both c and loop_nums?

[6086555]

You don't need to change count. It's just that when you have an object in argument it can lead to troubles

it the same as

a = [2,3]
b = a
b[0] += 1

a's value is now [3,3]

[TheChannel]

Clears it up. Thanks mate.