I wonder if it's possible to create a variable jump loop. For example, in the odd cycles the jump would be 2 and in even cycles the cycles would be 4.
I tried to do it with the tertiary operator of Python a if True else b , but it does not work because the range function does not change during the loop.
for j in range(i, len(message), small if odd else big):
print(j)
You can toggle the step at the time of using the variable:
for j in range(1, 20, 3):
print(j if j & 1 else j + 1)
Output:
1
5
7
11
13
17
19
See working at IDEONE .
If you want to adjust the moment that the steps occur, you can change the j + 1 by j - 1 , or change the + 1 on the if side, or even change the initial offset, goes from case concrete.
Bacco's answer seems to me to be the only correct way, although some details escape me. Its sequence is:
{ i, se n == 0;
seq(n) = { seq(n-1)+2, se n % 2 == 1;
{ seq(n-1)+4, se n % 2 == 0.
That is, i , i+2 , i+6 , i+8 , i+12 , ... This sequence can be simplified to:
i + 0*3 - 0,
i + 1*3 - 1,
i + 2*3 - 0,
i + 3*3 - 1,
i + 4*3 - 0,
...
Where each term, then, is equal to i + n*3 - n%2 . i.e. "go from 3 to 3, subtracting 1 from the odd terms". One way to implement this would be:
for j in [i + n*3 - n%2 for n in range(0, limite_superior)]:
print(j)
Where limite_superior is n , such that i + n*3 - n%2 >= len(message) , i.e.:
n*3 - n%2 >= len(message) - i
n*3 >= len(message) - i + n%2
n >= (len(message) - i + n%2)/3
But Bacco's solution is simpler and more elegant - just pay attention to the difference between% s and even% s (in one case the first deviation will be i , the other will be 2 ). >