(* Copyright (c) 2011-2020 Stefan Krah. All rights reserved. *)
==========================================================================
                Calculate (a * b) % p using special primes
==========================================================================
A description of the algorithm can be found in the apfloat manual by
Tommila [1].
Definitions:
------------
In the whole document, "==" stands for "is congruent with".
Result of a * b in terms of high/low words:
   (1) hi * 2**64 + lo = a * b
Special primes:
   (2) p = 2**64 - z + 1, where z = 2**n
Single step modular reduction:
   (3) R(hi, lo) = hi * z - hi + lo
Strategy:
---------
   a) Set (hi, lo) to the result of a * b.
   b) Set (hi', lo') to the result of R(hi, lo).
   c) Repeat step b) until 0 <= hi' * 2**64 + lo' < 2*p.
   d) If the result is less than p, return lo'. Otherwise return lo' - p.
The reduction step b) preserves congruence:
-------------------------------------------
    hi * 2**64 + lo == hi * z - hi + lo   (mod p)
    Proof:
    ~~~~~~
       hi * 2**64 + lo = (2**64 - z + 1) * hi + z * hi - hi + lo
                       = p * hi               + z * hi - hi + lo
                       == z * hi - hi + lo   (mod p)
Maximum numbers of step b):
---------------------------
# To avoid unnecessary formalism, define:
def R(hi, lo, z):
     return divmod(hi * z - hi + lo, 2**64)
# For simplicity, assume hi=2**64-1, lo=2**64-1 after the
# initial multiplication a * b. This is of course impossible
# but certainly covers all cases.
# Then, for p1:
hi=2**64-1; lo=2**64-1; z=2**32
p1 = 2**64 - z + 1
hi, lo = R(hi, lo, z)    # First reduction
hi, lo = R(hi, lo, z)    # Second reduction
hi * 2**64 + lo < 2 * p1 # True
# For p2:
hi=2**64-1; lo=2**64-1; z=2**34
p2 = 2**64 - z + 1
hi, lo = R(hi, lo, z)    # First reduction
hi, lo = R(hi, lo, z)    # Second reduction
hi, lo = R(hi, lo, z)    # Third reduction
hi * 2**64 + lo < 2 * p2 # True
# For p3:
hi=2**64-1; lo=2**64-1; z=2**40
p3 = 2**64 - z + 1
hi, lo = R(hi, lo, z)    # First reduction
hi, lo = R(hi, lo, z)    # Second reduction
hi, lo = R(hi, lo, z)    # Third reduction
hi * 2**64 + lo < 2 * p3 # True
Step d) preserves congruence and yields a result < p:
-----------------------------------------------------
   Case hi = 0:
       Case lo < p: trivial.
       Case lo >= p:
          lo == lo - p   (mod p)             # result is congruent
          p <= lo < 2*p  ->  0 <= lo - p < p # result is in the correct range
   Case hi = 1:
       p < 2**64 /\ 2**64 + lo < 2*p  ->  lo < p  # lo is always less than p
       2**64 + lo == 2**64 + (lo - p)   (mod p)   # result is congruent
                  = lo - p   # exactly the same value as the previous RHS
                             # in uint64_t arithmetic.
       p < 2**64 + lo < 2*p  ->  0 < 2**64 + (lo - p) < p  # correct range
[1]  http://www.apfloat.org/apfloat/2.40/apfloat.pdf