Generating gray codes.

Question

I tried generating gray codes in Python. This code works correctly. The issue is that I am initialising the base case (n=1,[0,1]) in the main function and passing it to gray_code function to compute the rest. I want to generate all the gray codes inside the function itself including the base case. How do I do that?

def gray_code(g,n):
    k=len(g)
    if n<=0:
        return

    else:
        for i in range (k-1,-1,-1):
            char='1'+g[i]
            g.append(char)
        for i in range (k-1,-1,-1):
            g[i]='0'+g[i]

        gray_code(g,n-1)

def main():
    n=int(raw_input())
    g=['0','1']
    gray_code(g,n-1)
    if n>=1:
        for i in range (len(g)):
            print g[i],

main()

Is the recurrence relation of this algorithm T(n)=T(n-1)+n ?

Answer 1

def gray_code(n):
    def gray_code_recurse (g,n):
        k=len(g)
        if n<=0:
            return

        else:
            for i in range (k-1,-1,-1):
                char='1'+g[i]
                g.append(char)
            for i in range (k-1,-1,-1):
                g[i]='0'+g[i]

            gray_code_recurse (g,n-1)

    g=['0','1']
    gray_code_recurse(g,n-1)
    return g

def main():
    n=int(raw_input())
    g = gray_code (n)

    if n>=1:
        for i in range (len(g)):
            print g[i],

main()

Answer 2

Generating Gray codes is easier than you think. The secret is that the Nth gray code is in the bits of N^(N>>1)

So:

def main():
    n=int(raw_input())
    for i in range(0, 1<<n):
        gray=i^(i>>1)
        print "{0:0{1}b}".format(gray,n),

main()

Answer 3

It's relatively easy to do if you implement the function iteratively (even if it's defined recursively). This will often execute more quickly as it generally requires fewer function calls.

def gray_code(n):
    if n < 1:
        g = []
    else:
        g = ['0', '1']
        n -= 1
        while n > 0:
            k = len(g)
            for i in range(k-1, -1, -1):
                char = '1' + g[i]
                g.append(char)
            for i in range(k-1, -1, -1):
                g[i] = '0' + g[i]
            n -= 1
    return g

def main():
    n = int(raw_input())
    g = gray_code(n)
    print ' '.join(g)

main()

Answer 4

What about this:

#! /usr/bin/python3

def hipow(n):
    ''' Return the highest power of 2 within n. '''
    exp = 0
    while 2**exp <= n:
        exp += 1
    return 2**(exp-1)

def code(n):
    ''' Return nth gray code. '''
    if n>0:
        return hipow(n) + code(2*hipow(n) - n - 1)
    return 0

# main:
for n in range(30):
    print(bin(code(n)))