Posts Tagged ‘number theory’

Converting from binary to integer

26 October 2010 1 comment

The following is an updated and edited version of my posts to this sage-support thread.


You have a bitstring as output by

and you want to convert that bitstring to an integer. Or in general, you want to convert a bit vector to its integer representation.


Here are two ways, assuming that you want the bits in little-endian order, i.e. you read the bits from right to left in increasing order of powers of 2.

sage: version()
'Sage Version 4.5.3, Release Date: 2010-09-04'
sage: from import blum_blum_shub
sage: b = blum_blum_shub(length=6, lbound=10**4, ubound=10**5); b
sage: type(b)
<class 'sage.monoids.string_monoid_element.StringMonoidElement'>
sage: # read in little-endian order
sage: # conversion using Python's built-in int()
sage: int(str(b), base=2)
sage: # conversion using Sage's built-in Integer()
sage: Integer(str(b), base=2)

Now assume you read the bitstring as output by blum_blum_shub in big-endian order, i.e. from left to right in increasing order of powers of 2. You simply convert the bitstring to a string, reverse that string, and apply any of the above two methods.

sage: # reversing a string
sage: str(b)
sage: str(b)[::-1]
sage: # read in big-endian order
sage: int(str(b)[::-1], base=2)
sage: Integer(str(b)[::-1], base=2)

Or you can do as follows:

sage: b = "100110"
sage: sum(Integer(i) * (2^Integer(e)) for e, i in enumerate(b))
sage: sum(Integer(i) * (2^Integer(e)) for e, i in enumerate(b[::-1]))

Another way is to use Horner’s method. Here’s a Sage function that computes the integer representation of a bit vector read using big-endian order. A usage example is also shown.

sage: def horner(A, x0):
...       # Evaluate the polynomial P(x) at x = x_0.
...       #
...       # INPUT
...       #
...       # - A -- list of coefficients of P where A[i] is the coefficient of
...       #   x_i.
...       # - x0 -- the value x_0 at which to evaluate P(x).
...       #
...       # OUTPUT
...       #
...       # An evaluation of P(x) using Horner's method.
...       i = len(A) - 1
...       b = A[i]
...       i -= 1
...       while i >= 0:
...           b = b*x0 + A[i]
...           i -= 1
...       return b
sage: A = [1, 0, 0, 1, 1, 0]
sage: horner(A, 2)

As an exercise, modify the function horner to output the integer representation of a bit vector that is read using little-endian order.

Number theory & RSA public key cryptography

14 October 2010 Leave a comment

I have released version 1.1 of the Sage tutorial “Number theory and the RSA public key cryptosystem”. There is little change in terms of content. However, note that I now use the GNU Free Documentation License v1.3+ for the tutorial. Here are the relevant files you can download for your reading pleasure.

All versions of the tutorial are available from the download page on its website. For the adventurous of heart, I have also made the full \LaTeX source of the document available.

The tutorial is meant to be educational. I don’t pretend that it is complete in any way. Any suggestions and/or criticisms for improving the tutorial are more than welcome. Enjoy and happy Sage’ing.

Optimized parity testing

6 October 2010 Leave a comment

To test the parity of an integer is to determine whether it is even or odd. Letting n = 132469, we can test the parity of n by computing its value modulo 2. This can be done either by using the Sage built-in mod function, the Python modulo operator %, or by using the Python bitwise operator &. The operator & is bitwise conjunction, i.e. it corresponds to multiplication over the Galois field \mathbf{F}_2 of two elements. The integer n = 132469 is odd, hence the result of parity testing via mod, %, and & should each return 1.

sage: n = 132469
sage: mod(n, 2).lift()
sage: mod(n, 2)
sage: n % 2
sage: n & 1

However, the test using the bitwise operator & is the fastest of all:

sage: %timeit mod(n, 2).lift()
625 loops, best of 3: 42.2 micro second per loop
sage: %timeit mod(n, 2)
625 loops, best of 3: 42.2 micro second per loop
sage: %timeit n % 2
625 loops, best of 3: 1.22 micro second per loop
sage: %timeit n & 1
625 loops, best of 3: 1.02 micro second per loop

The program below demonstrates how to do parity testing using C.

#include <stdio.h>

/* Parity testing using bitwise AND.
static void parity_and(int n)
    if (n & 1)
        printf("%i is odd\n", n);
        printf("%i is even\n", n);

/* Parity testing using the modulus operator.
static void parity_mod(int n)
    if (n % 2)
        printf("%i is odd\n", n);
        printf("%i is even\n", n);

int main(void)
    int n = 132469;
    int m = 132470;

    printf("Parity testing using modulus operator\n");
    printf("Parity testing using bitwise AND\n");

    return 0;

The output of the program is:

Parity testing using modulus operator
132469 is odd
132470 is even
Parity testing using bitwise AND
132469 is odd
132470 is even

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