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4.4. FURTHER APPLICATIONS 141
i.e.
M.̂x = det(M)a. (4.17)
If we avoid primes dividing det(M), we can solve (4.17) for enough primes
(suitable bounds are given in Corollary 21), reconstruct integer entries in ̂x,
and then divide through by det(M).
4.4.3.2 Solve with Fractions
If (3.13) is soluble modulo p, its solution x p is indeed congruent to x when
evaluated at p, i.e. x| p = x p . If we use many primes p i (discarding those for
which (3.13) is not soluble), and apply Algorithm 39 to the vectors x pi , we
get a vector x N such that x N ≡ x (mod N), where N = ∏ p i . However, the
entries of x are rationals, with numerator and denominator bounded, say, by B
(see Corollary 21), rather than integers. How do we find the entries of x from
x N ? This problem has a long history in number theory, generally under the
name Farey fractions, but was first considered in computer algebra in [Wan81].
Since we will have occasion to use this solution elsewhere, we consider it in more
generality in the next section.
If we assume this problem solved, we then have the following answer to the
questions on page 114.
1. Are there good primes?: yes — all that do not divide det(M)
2. How can we tell if a prime p is bad? Equation (3.13) is not soluble modulo
p.
3. How many reductions should we take? Enough such that the product of
the good primes is greater than 2B 2 .
4. How do we combine? Algorithm 21.
5. How do we check the result? If we use the bound from Corollary 21),
we do not need to check. However, there are ‘early success” variations,
analogous to section 4.1.5, wher we do need to check, which can be done
by checking that M.x = a: an O(n 2 ) operation rather than the O(n 3 ) of
solving.
4.4.3.3 Farey reconstruction
In this section, we consider the problem of reconstructing an unknown fraction
x = n/d, with |n|, |d| < B, given that we know x ≡ y (mod N), i.e. n ≡ yd
(mod N), where N > 2B 2 . We first observe that this representation is unique,
for if n ′ /d ′ (similarly bounded) is also congruent to y, then nd ′ ≡ ydd ′ ≡ n ′ d,
so nd ′ − n ′ d ≡ 0 (mod N), and the only solution satisfying the bounds is
nd ′ − n ′ d = 0, i.e. n/d = n ′ /d ′ .
Actually finding n and d is done with the Extended Euclidean Algorithm
(see Algorithm 4).