Click here to Skip to main content
14,836,972 members
Please Sign up or sign in to vote.
0.00/5 (No votes)
See more:
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:

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

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]
        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?
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:

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.
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]
	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.
sbarnes 28-Dec-20 16:43pm
Good observation and answer. But I got the impression
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.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

CodeProject, 20 Bay Street, 11th Floor Toronto, Ontario, Canada M5J 2N8 +1 (416) 849-8900