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
?
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()
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()
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()
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)))
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