First clear solution for AMSCO Cipher by Tinus_Trotyl - python coding challenges

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First solution in Clear category for AMSCO Cipher by Tinus_Trotyl

def pull(s):    # if possible pull 1st char from string
    if s: return s[0], s[1:]
    else: return "", ""

def pull2(s):   # if possible pull 1st 2 (or 1) char from string
    a, s = pull(s)
    b, s = pull(s)
    return a + b, s
        
def decode_amsco(message, key):
    key = [int(i) - 1 for i in str(key)] # transform key to a more suitable form
    grid, msg, row = [], message, -1
    while msg:                           # set up a grid
        row += 1
        grid = grid + [[[0,""] for i in key]]
        for x in range(len(key)):
            if (row + x) % 2 == 0: c, msg = pull(msg) 
            else: c, msg = pull2(msg)
            grid[row][x][0] = len(c)
    for x in range(len(key)):            # fill in
        i = key.index(x)
        for row in range(len(grid)):
            n = grid[row][i][0]
            if n == 1: grid[row][i][1], message = pull(message)
            if n == 2: grid[row][i][1], message = pull2(message)
    return "".join(x[1] for row in grid for x in row)   # and read out

June 22, 2017

Comments:

aya.kanazawa on June 3, 2019, 1:25 a.m.
Could you explain '''grid[row][i][1],''' part? <ol> <li>if n == 1: grid[row][i][1], message = pull(message)</li> <li>if n == 2: grid[row][i][1], message = pull2(message)</li> </ol>

Tinus_Trotyl on June 5, 2019, 8:08 p.m.
If this is what you mean: Function pull(s) takes a string and returns its first character and the strings remainder, so that are two strings. The first goes to grid[row][i][1] and the other goes back to message.

Blizzard

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