lordzuko - 10 months ago 285

Python Question

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 Source

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)))
```