lordzuko lordzuko - 2 months ago 113
Python Question

I am trying to find the nth binary palindrome

Binary palindromes: numbers whose binary expansion is palindromic.

Binary Palindrome -> is a number whose binary representation is a palindrome.

Here is link to the solution with naive approach

I have read from above link and it give a formula to find the nth binary palindrome. I am unable to understand and thus code the solution.

def palgenbase2(): # generator of palindromes in base 2
#yield 0
x, n, n2 = 1, 1, 2
m = 1;
while True:
for y in range(n, n2):
s = format(y, 'b')
yield int(s+s[-2::-1], 2)
for y in range(n, n2):
s = format(y, 'b')
yield int(s+s[::-1], 2)
x += 1
n *= 2
n2 *= 2
if n2 > 1000000000:
break

ans = {}

for i,j in enumerate(palgenbase2()):
print i,j
ans[i]=j

with open("output","a") as f:
f.write(ans)

#use the saved output to give answer to query later
#this will work but it takes too much time.

n = int(raw_input())
for c in range(0,n):
z = int(raw_input())
print ans[z]


Here is one Python code but it is generating all such palindromes.

I need help in the program to get the nth binary palindrome directly.

as follows:


Input -> 1 <= n <= 1000000000

Function -> f(n)

output -> nth binary palindrome.


Can we do this in better time using the formula mentioned here?

Answer

Here's a fairly straight-forward implementation of the recursive algorithm given at A006995.

To make it more efficient I use bit shifting to perform binary exponentiation: when x is a non-negative integer, 1 << x is equivalent to 2 ** x but substantially faster (at least, it is in both Python 2 and Python 3 on standard CPython).

Also, to make the recursion more efficient, the function stores previously calculated values in a dictionary. This also lets us easily handle when n <= 2, which the recursive formula itself does not handle.

#!/usr/bin/env python

''' Binary palindromes

    Find (non-negative) integers which are palindromes when written in binary

    See http://stackoverflow.com/q/39675412/4014959
    and https://oeis.org/A006995

    Written by PM 2Ring 2016.09.24

    Recursion for n>2: a(n)=2^(2k-q)+1+2^p*a(m), where k:=floor(log_2(n-1)), and p, q and m are determined as follows:

    Case 1: If n=2^(k+1), then p=0, q=0, m=1;

    Case 2: If 2^k<n<2^k+2^(k-1), then set i:=n-2^k, p=k-floor(log_2(i))-1, q=2, m=2^floor(log_2(i))+i;

    Case 3: If n=2^k+2^(k-1), then p=0, q=1, m=1;

    Case 4: If 2^k+2^(k-1)<n<2^(k+1), then set j:=n-2^k-2^(k-1), p=k-floor(log_2(j))-1, q=1, m=2*2^floor(log_2(j))+j; 
'''

#Fast Python 3 version of floor(log2(n))
def flog2(n):
    return n.bit_length() - 1

def binpal(n, cache={1:0, 2:1, 3:3}):
    if n in cache:
        return cache[n]

    k = flog2(n - 1)
    b = 1 << k
    a, c = b >> 1, b << 1

    if n == c:
        p, q, m = 0, 0, 1
    elif b < n < a + b:
        i = n - b
        logi = flog2(i)
        p, q, m = k - logi - 1, 2, (1 << logi) + i
    elif n == a + b:
        p, q, m = 0, 1, 1
    else:
        #a + b < n < c
        i = n - a - b
        logi = flog2(i)
        p, q, m = k - logi - 1, 1, (2 << logi) + i

    result = (1 << (2*k - q)) + 1 + (1 << p) * binpal(m)
    cache[n] = result
    return result

def palgenbase2(): 
    ''' generator of binary palindromes '''
    yield 0
    x, n, n2 = 1, 1, 2
    while True:
        for y in range(n, n2):
            s = format(y, 'b')
            yield int(s+s[-2::-1], 2)
        for y in range(n, n2):
            s = format(y, 'b')
            yield int(s+s[::-1], 2)
        x += 1
        n *= 2
        n2 *= 2

gen = palgenbase2()

for i in range(1, 30):
    b = next(gen)
    c = binpal(i)
    print('{0:>2}: {1} {1:b} {2}'.format(i, b, c))

output

 1: 0 0 0
 2: 1 1 1
 3: 3 11 3
 4: 5 101 5
 5: 7 111 7
 6: 9 1001 9
 7: 15 1111 15
 8: 17 10001 17
 9: 21 10101 21
10: 27 11011 27
11: 31 11111 31
12: 33 100001 33
13: 45 101101 45
14: 51 110011 51
15: 63 111111 63
16: 65 1000001 65
17: 73 1001001 73
18: 85 1010101 85
19: 93 1011101 93
20: 99 1100011 99
21: 107 1101011 107
22: 119 1110111 119
23: 127 1111111 127
24: 129 10000001 129
25: 153 10011001 153
26: 165 10100101 165
27: 189 10111101 189
28: 195 11000011 195
29: 219 11011011 219

If you need to run this on Python 2 you won't be able to use that flog2 function, since Python 2 integers do not have the bit_length method. Here's an alternative version:

from math import floor, log

def flog2(n):
    return int(floor(log(n) / log(2)))