Python 3: Trying to iterate lines of alphabet based on function of i

Question

I'm trying to write a program that does the following:

In this program, we print out i**q % p letters from the alpabet on line i for example, if q=2 and p=12 then we want to print:

1**2 % 12 = 1
2**2 % 12 = 4
3**2 % 12 = 9
4**2 % 12 = 4
5**2 % 12 = 1

...so the first five lines of the output will be:

a                                                                             
bcde                                                                       
fghijklmn                                                                  
opqr                                                                       
s   

In my case, the function to determine character count is (i**2)%9, but it's the same idea. The specific task is as follows:

"Print out 1834 lines using the letters a-z in sequence such that line number i will use (i**2 ) % 9 letters. The first letter used to print line 1834 is:_________"

I have started coding, and I have two main questions. Firstly, how do I make the alphabet start over, once I reach "z" (or more likely, "the 25th place")?
Also, What should I do for the program? So far I have:

letters="abcdefghijklmnopqrstuvwxyz"
for i in range[1,1834]:
    x=((i**2)%9)
    print(letters[])

...but I don't know how to proceed. I was going to try a slice method, but I don't know. Any input would be appreciated.

Answer 1

Try the following solution. You are moving a frame over the letters, so every time a new loop iteration starts, a variable start_position should remember at which letter the previous iteration stopped. The size of the frame is determined by your x calculation.

If you want to go back to the beginning of the alphabet, it should be modulo the length of the alphabet. end_position = end_position % 26

letters="abcdefghijklmnopqrstuvwxyz"
alphabet_length = len(letters)
start_position = 0
for i in range(1,1834):
    x=((i**2)%9)
    end_position = start_position + x
    end_position = end_position % alphabet_length
    if end_position < start_position:
         print letters[start_position:alphabet_length-1] + letters[0:end_position]
    else:
         print letters[start_position:end_position] 
    start_position = end_position

Answer 2

Here's something you can work with using itertools:

from string import ascii_lowercase
from itertools import cycle, islice

letters = cycle(ascii_lowercase)
for num in (1, 4, 9, 4, 1, 13, 12, 9, 5):
    print (''.join(islice(letters, num)))

#a
#bcde
#fghijklmn
#opqr
#s
#tuvwxyzabcdef
#ghijklmnopqr
#stuvwxyza
#bcdef

Answer 3

It seems like this should be homework, so I'll answer it as such:

• To slice from index a to index b, you can use my_list[a:b]
• To restart counting from 0 when you reach 26, you need the remainder of the iterator i divided by the number of letters 26. a%b is the remainder of a divided by b.
• range(n) is a function which returns numbers from 0 to n-1, so in total you would get n elements. And it is a function so you should use parentheses, i.e. range(1834) not range[1834] (that is slicing).
• Also note that range(1,1834) will have 1834-1=1833 elements in total, which is not what you need...
• There's a package called string which has lists of alphabet characters (lowercase, uppercase, digits, ...). See for example string.lowercase.

Answer 4

The two questions are linked: you can use cycle from itertools to get an iterator that will endlessly loop over the alphabet (or any other string, for that matter), and to get n letters for a given row you can call next() on the iterator n times.

Your question looks like homework, so rather than short-circuit your education I'll give an example for the similar question of printing n letters on line n:

import itertools
str = "abcdefghijklmnopqrstuvwxyzicopiedthisfromstackoverflow"
it = itertools.cycle( str )
for i in range( 1 , 10 ):
    line=""
    for j in range( i ):
        line+=it.next()
    print i,line

edit as appropriate :-)