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So, I have a function, that gets a call like this:
reversed("abc", "def", "ghi")

And has to return the reversed order like this (it is reversed by 90 degres - by this I mean if you stack them like this:
abc
def
ghi 

And then reverse them for 90 degress, you would get:
gda
heb
ifc

And then to upgrade via another function, that would get the same argument and a number of reverses. Like this:
reversed_n(("abc", "def", "ghi"), 1)

And would then return the tuple of what I would get by reversing the same order n times.
For this it would return:
("gda", "heb", "ifc")


What I have tried:

So I made the functions like this:

def reversed(kos):
    seznam = []
    kos = [i[::-1] for i in kos]
    i = 0
    while i < len(kos[0]):
        s = [beseda[i] for beseda in kos]
        seznam.append("".join(s))
        i += 1
    return tuple(seznam)


def reversed_n(kos, n):
    for i in range(0, n):
        kos = reversed(kos)
    return kos


So it works, but if I try something a lot larger, say a lot larger and put n = 10000001, it will run forever. What would I need to do, can it maybe be done with calculating if it is odd or even number of n?
Posted
Updated 27-Dec-20 18:14pm

First, I would see if it is indeed taking forever, or maybe something just a little short of forever. Try adding this to your for loop:
if i % 10000 == 0:
  print(i)

If you never see i being printed, try cutting down % 10000 by a factor of 1/10 a couple times. If you still never see i being printed, I'd start to look for interpreter maximums or other possible causes like resource limits dictated by your system. For example, there is a resource limit that might affect your function on linux:
getrlimit(2) - Linux manual page[getrlimit]
Good luck.
 
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def reversed(arr):
	kos = []
	p = ""
	arr_len = len(arr)
	item_len = len(arr[0])
	for x in range(0, item_len):
		for i in range(0, arr_len):
			item = arr[arr_len-1-i]
			p+=item[x]
		kos.append(p)
		p=""
	return kos
def reversed_n(kos, n):
	for i in range(0, n%4):
		kos = reversed(kos)
	return kos


There can only be 4 outcomes or rotations per se because after 4 rotations it will become same as input, so we can take the modulus of 'n' to 4, and rotate the grid that many times and then return it.
 
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Comments
sbarnes 28-Dec-20 16:43pm    
Good observation and answer. But I got the impression https://www.codeproject.com/script/Membership/View.aspx?mid=14631234
was asking about limits and the like. What goes wrong when n gets too big? What is too big and why? But, yes, as a practical matter your shorter way is more performant.

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