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Better document and slightly simplify >>2^n heuristic

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Slightly simplify is_significantly_above_a_power_of_2() to make it
easier to understand:
* Remove the explicit negative answer for x <= 4. The only functional
  difference this makes is that is_significantly_above_a_power_of_2(3)
  is now true.
* Shift the most significant bit of x to position 8 rather than 15.
  This makes the final comparison easier to explain.

Signed-off-by: Gilles Peskine <Gilles.Peskine@arm.com>
This commit is contained in:
Gilles Peskine 2021-04-13 19:50:04 +02:00
parent 422e867acb
commit eedefa5627
1 changed files with 24 additions and 14 deletions
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38
tests/suites/test_suite_mpi.function
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@ -64,38 +64,48 @@ static int f_rng_bytes_left( void *state, unsigned char *buf, size_t len )
return( 0 );
}
/* Test whether bytes represents (in big-endian base 256) a number B that
* is "significantly" above a power of 2, which is defined as follows.
* Let n be the integer such that 2^n <= B < 2^{n+1}. B is significantly
* above a power of 2 if (B - 2^n) / 2^n is not negligible. "Negligible"
* is defined as having a negligible chance that if you draw an integer
* in the range [1, B-1] K times, the number will always be less than 2^n,
* where K is the iteration count passed to genkey_sw_many.
/* Test whether bytes represents (in big-endian base 256) a number b that
* is significantly above a power of 2. That is, b must not have a long run
* of unset bits after the most significant bit.
*
* Let n be the bit-size of b, i.e. the integer such that 2^n <= b < 2^{n+1}.
* This function returns 1 if, when drawing a number between 0 and b,
* the probability that this number is at least 2^n is not negligible.
* This probability is (b - 2^n) / b and this function checks that this
* number is above some threshold A. The threshold value is heuristic and
* based on the needs of mpi_random_many().
*/
static int is_significantly_above_a_power_of_2( data_t *bytes )
{
const uint8_t *p = bytes->x;
size_t len = bytes->len;
unsigned x;
/* Skip leading null bytes */
while( len > 0 && p[0] == 0 )
{
++p;
--len;
}
/* 0 is not significantly above a power of 2 */
if( len == 0 )
return( 0 );
else if( len == 1 )
/* Extract the (up to) 2 most significant bytes */
if( len == 1 )
x = p[0];
else
x = ( p[0] << 8 ) | p[1];
if( x <= 4 )
return( 0 );
/* Shift the most significant bit of x to position 8 and mask it out */
while( ( x & 0xfe00 ) != 0 )
x >>= 1;
x &= 0x00ff;
while( ( x & 0x8000 ) == 0 )
x <<= 1;
x &= 0x7fff;
return( x >= 0x1000 );
/* At this point, x = floor((b - 2^n) / 2^(n-8)). b is significantly above
* a power of 2 iff x is significantly above 0 compared to 2^8.
* Testing x >= 2^4 amounts to picking A = 1/16 in the function
* description above. */
return( x >= 0x10 );
}
/* END_HEADER */