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// --- This file is distributed under the MIT Open Source License, as detailed
// in the file "LICENSE.TXT" in the root of this repository ---
#ifndef EXTENDED_EUCLIDEAN_COLLINS
#define EXTENDED_EUCLIDEAN_COLLINS 1
#ifndef NDEBUG
# include "helpers/assert_helper_gcd.h"
#endif
#include <assert.h>
#include <limits>
#if defined(assert_invariant) || defined(assert_precondition)
# error "assert_invariant and/or assert_precondition were already defined"
#endif
// assert aliases will help self-document the code
#define assert_invariant assert
#define assert_precondition assert
/*P1*/ // Positive Integer Division Definition:
// If u>=0 and v>0 then the operation q = u/v produces an integer q
// such that u == v*q + r, with 0 <= r < v.
/*P2*/ // Theorem 1: If u>=0 and v>0, and t*v <= u, then t <= u/v.
// Proof: Letting q = u/v, by [P1] q satisfies
// u == v*q + r, with 0 <= r < v. Since we're given t*v <= u,
// t*v <= u == v*q + r. Since 0 <= r < v,
// t*v <= u == v*q + r < v*q + v == v*(q+1)
// Hence t*v < v*(q+1) and t < q+1.
// t < q+1 implies t <= q, and since q == u/v, we have t <= u/v.
/*P3*/ // Theorem 2: If u>=0 and v>0 and q == u/v, then q*v <= u.
// Proof: Since q == u/v, by [P1] q satisfies
// u == v*q + r, with 0 <= r < v. Since r >= 0,
// u == v*q + r >= v*q + 0. Hence q*v <= u.
/*P4*/ // Generalized Euclid's Lemma: If integers u and v are coprime,
// and u*m == v*n, then u divides n, and v divides m.
// Proof: Since u and v are coprime, by Bezout's identity there exist
// integers j,k such that j*u + k*v == 1.
// Multiplying both sides by m, j*u*m + k*v*m == m. Since u*m == v*n,
// j*v*n + k*v*m == m. Since v divides both terms on the left, it must
// divide the single term on the right. Thus v divides m.
// Again using j*u + k*v == 1, multiplying both sides by n results in
// j*u*n + k*v*n == n. Since u*m == v*n, j*u*n + k*u*m == n. Since
// u divides both terms on the left, it must divide the single term on
// the right. Thus u divides n.
/*P5*/ // Theorem 3: If a divides b and b divides a, then a == +- b.
// Proof: a divides b implies there exists an integer g such that
// g*a == b. Likewise, b divides a implies there exists an integer h
// such that a == h*b. By substitution,
// a == h*(g*a)
// 0 == a*(h*g - 1)
// therefore a == 0 or (h*g-1) == 0.
// If we assume a == 0, then since g*a == b, we would also have b == 0.
// Having both a == 0 and b == 0 would satisfy a == +- b.
// Alternately, if we assume (h*g-1) == 0, we would have h*g == 1,
// which would require either h==1 and g==1, or h == -1 and g == -1.
// If we have h==1 and g==1, then since g*a == b, we would have a == b,
// which satisfies a == +- b. If we have instead h == -1 and g == -1,
// then since g*a == b, we would have -a == b, which also
// satisfies a == +- b. All cases satisfy a == +- b.
/*P6*/ // Theorem 4: If integers u and v are coprime, and integers m and n
// are coprime, and u*m == v*n, then
// abs(u) == abs(n) and abs(m) == abs(v).
// Proof: By [P4], u divides n, and v divides m. Also by [P4],
// m divides v, and n divides u. Since u divides n and n divides u,
// by [P5] u == +- n, and thus abs(u) == abs(n). Since v divides m
// and m divides m, by [P5] v == +- m, and thus abs(v) == abs(m).
// This proof follows the presentation by George E. Collins, Mathematics of
// Computation 23 (1969), p.198, "Computing Multiplicative Inverses in GF(p)"
// http://www.ams.org/journals/mcom/1969-23-105/S0025-5718-1969-0242345-5/S0025-5718-1969-0242345-5.pdf
template <typename T>
void extended_euclidean_collins(T a, T b, T* pGcd, T* pX, T* pY)
{
static_assert(std::numeric_limits<T>::is_integer, "");
static_assert(std::numeric_limits<T>::is_signed, "");
/*01*/ assert_precondition(b >= 0);
/*02*/ assert_precondition(b < a);
/*03*/ T x0 = 1;
/*04*/ T y0 = 0;
/*05*/ T a0 = a;
/*06*/ T x1 = 0;
/*07*/ T y1 = 1;
/*08*/ T a1 = b;
/*09*/ assert_invariant(a > 0); // By [01, 02]
/*10*/ assert_invariant(gcd(a,b) >= 1);
// Proof: By [09] a > 0, so gcd(a,b) != gcd(0,0). Since 1 is a
// common divisor of both a and b, gcd(a,b) >= 1
/*11*/ assert_invariant(a0 > 0); // By [05, 09]
/*12*/ assert(a1 >= 0); // By [08, 01]
/*13*/ assert_invariant(a1 < a0); // By [08, 05, 02]
/*14*/ assert_invariant(a*x0 + b*y0 == a0);// Proof: By [03, 04, 05],
// a*x0+b*y0 == a*1+b*0 == a== a0
/*15*/ assert_invariant(a*x1 + b*y1 == a1);// Proof: By [06, 07, 08],
// a*x1+b*y1 == a*0+b*1 == b== a1
/*16*/ assert_invariant(gcd(a0,a1) == gcd(a,b)); // By [05, 08]
/*17*/ assert_invariant(abs(x0*y1 - y0*x1) == 1); // Proof: By [03,07,04,06]
// abs(x0*y1 - y0*x1) ==
// abs(1*1 - 0*0) == 1
/*18*/ T s0 = 1; // informally, s0 represents the sign of x0.
/*19*/ T s1 = -1; // the sign of x1 (values <= 0 treated as negative)
/*20*/ assert_invariant(s1 == -s0); // By [18, 19]
/*21*/ assert_invariant(abs(s0) == 1); // By [18]
/*22*/ assert_invariant(s0*abs(x0) == x0); // By [03, 18], x0 == 1, s0 == 1.
// s0*abs(x0) == 1*1 == 1 == x0
/*23*/ assert_invariant(s1*abs(x1) == x1); // By [06,19], x1 == 0, s1 == -1.
// s1*abs(x1) == -1*0 == 0 == x1
/*24*/ assert_invariant(-s0*abs(y0) == y0);// By [04, 18], y0 == 0, s0 == 1.
// -s0*abs(y0) == -1*0 == 0 == y0
/*25*/ assert_invariant(-s1*abs(y1) == y1);// By [07,19], y1 == 1, s1 == -1.
// -s1*abs(y1) == 1*1 == 1 == y1
/*26*/ assert(abs(y1) >= 2*abs(y0)); // By [07, 04], y1 == 1, y0 == 0.
// abs(y1) == 1 >= 0 == 2*abs(y0)
/*27*/ assert_invariant(gcd(x1,y1) == 1); // By [06, 07], x1 == 0, y1 == 1.
// gcd(x1,y1) == gcd(0,1) == 1
/*28*/ while (a1 != 0) {
/*29*/ assert_invariant(a0 > 0); // By [11, 80]
/*30*/ assert(a1 > 0);
// Proof: By [12, 81] a1 >= 0, and by [28] a1 != 0, so a1 > 0.
/*31*/ assert_invariant(a1 < a0); // By [13, 82]
/*32*/ assert_invariant(a*x0 + b*y0 == a0); // By [14, 83]
/*33*/ assert_invariant(a*x1 + b*y1 == a1); // By [15, 84]
/*34*/ assert_invariant(gcd(a0,a1) == gcd(a,b)); // By [16, 85]
/*35*/ assert_invariant(abs(x0*y1 - y0*x1) == 1); // By [17, 86]
/*36*/ assert_invariant(s1 == -s0); // By [20, 87]
/*37*/ assert_invariant(abs(s0) == 1); // By [21, 88]
/*38*/ assert_invariant(s0*abs(x0) == x0); // By [22, 89]
/*39*/ assert_invariant(s1*abs(x1) == x1); // By [23, 90]
/*40*/ assert_invariant(-s0*abs(y0) == y0); // By [24, 91]
/*41*/ assert_invariant(-s1*abs(y1) == y1); // By [25, 92]
/*42*/ assert(abs(s1) == 1); // By [36,37] s1== -s0. abs(s1)==abs(s0)== 1
/*43*/ T q = a0/a1;
/*44*/ assert(1 <= q && q <= a0);
// Proof: By [43] q == a0/a1 (note that a1 might not divide a0)
// By [30, 31] a1>0 and a1<a0, so 0 < a1 < a0, and thus,
// q == a0/a1 >= 1.
// Since 0 < a1 implies a1 >= 1, we have q == a0/a1 <= a0.
/*45*/ T a2 = a0 - q*a1;
if (a2 != 0) {
/*46*/ assert(q <= a0 / 2);
// Proof : Assume a1 == 1. Then by [43], q == a0/a1 == a0.
// By [45], a2 == a0 - q*a1, and since q == a0,
// a2 == a0 - a0*a1. Thus, given the assumption a1 == 1,
// a2 == a0 - a0*1 == 0. But a2 == 0 is a contradiction,
// because entering this clause requires a2 != 0. Therefore,
// a1 != 1
// By [30] a1 > 0, and so, since a1 != 1, a1 >= 2
// Since q == a0/a1 and a1 >= 2 and by [29] a0 > 0,
// q <= a0 / 2
}
/*47*/ assert(0 < (q*a1) && (q*a1) <= a0);
// Proof: By [44, 30], q >= 1 and a1 > 0, so q*a1 > 0.
// By [43], q == a0/a1, and by [29, 30], a0 > 0 and a1 > 0.
// Hence by [P3], q*a1 <= a0.
/*48*/ assert(0 <= a2 && a2 < a1);
// Proof: By [29, 30], a0 > 0 and a1 > 0, and by [43] q == a0/a1,
// so by [P1] q satisfies a0 == a1*q + r, with 0 <= r < a1.
// Thus, a0 - q*a1 == r. By [45] a0 - q*a1 == a2, so a2 == r.
// Since 0 <= r < a1, we know 0 <= a2 < a1
/*49*/ assert(gcd(a0,a1) == gcd(a1,a2));
// Proof: By definition d = gcd(a0,a1) divides both a0 and a1. So
// since by [45] a2 == a0 - q*a1, d must also divide a2. Thus d
// is a common divisor of a1 and a2, and so d divides gcd(a1,a2).
// Thus gcd(a0,a1) divides gcd(a1,a2).
// Similarly, c = gcd(a1,a2) divides both a1 and a2, and since
// by [45] a0 == a2 + q*a1, c must also divide a0. Thus c is a
// common divisor of a0 and a1, and so c divides gcd(a0,a1). Thus
// gcd(a1,a2) divides gcd(a0,a1). Since both gcd(a0,a1) divides
// gcd(a1,a2), and gcd(a1,a2) divides gcd(a0,a1), we know by [P5],
// gcd(a0,a1) == +- gcd(a1,a2).
// Since gcds are never negative, gcd(a0,a1) == gcd(a1,a2).
/*50*/ assert(gcd(a1,a2) == gcd(a,b));
// Proof: By [49] gcd(a0,a1) == gcd(a1,a2). And by [34]
// gcd(a0,a1) == gcd(a,b). Therefore gcd(a1,a2) == gcd(a,b).
/*51*/ T x2 = x0 - q*x1;
/*52*/ assert(x2 == s0*(abs(x0) + abs(q*x1)));
// Proof: By [38, 39], x0 == s0*abs(x0) and x1 == s1*abs(x1).
// By [51], x2 == x0 - q*x1, thus x2 == s0*abs(x0) - q*s1*abs(x1)
// By [36], s1 == -s0, so x2 == s0*abs(x0) + q*s0*abs(x1).
// And thus x2 == s0*(abs(x0) + q*abs(x1)). By [44], q >= 1, so
// x2 == s0*(abs(x0) + abs(q*x1))
/*53*/ assert(abs(x2) == abs(x0) + abs(q*x1));
// Proof: By [52], x2 == s0*(abs(x0) + abs(q*x1)), so
// abs(x2) == abs(s0)*abs(abs(x0) + abs(q*x1))
// abs(x2) == abs(s0)*(abs(x0) + abs(q*x1)
// Since by [37], abs(s0) == 1, abs(x2) == abs(x0) + abs(q*x1).
/*54*/ assert(abs(q*x1) <= abs(x2)); // By [53]
/*55*/ assert(abs(x1) <= abs(x2));
// Proof: By [44], q >= 1, so abs(x1) <= abs(q*x1)
// By [54], abs(q*x1) <= abs(x2), thus
// abs(x1) <= abs(q*x1) <= abs(x2)
/*56*/ T y2 = y0 - q*y1;
/*57*/ assert(y2 == -s0*(abs(y0) + abs(q*y1)));
// Proof: By [40, 41], y0 == -s0*abs(y0) and y1 == -s1*abs(y1).
// By [56], y2 == y0 - q*y1, so y2 == -s0*abs(y0) + q*s1*abs(y1)
// By [36], s1 == -s0, so y2 == -s0*abs(y0) - q*s0*abs(y1).
// And thus y2 == -s0*(abs(y0) + q*abs(y1)). By [44], q >= 1, so
// y2 == -s0*(abs(y0) + abs(q*y1))
/*58*/ assert(abs(y2) == abs(y0) + abs(q*y1));
// Proof: By [57], y2 == -s0*(abs(y0) + abs(q*y1)), so
// abs(y2) == abs(-s0)*abs(abs(y0) + abs(q*y1))
// abs(y2) == abs(s0)*(abs(y0) + abs(q*y1))
// Since by [37], abs(s0) == 1, abs(y2) == abs(y0) + abs(q*y1).
/*59*/ assert(abs(q*y1) <= abs(y2)); // By [58]
/*60*/ assert(abs(y1) <= abs(y2));
// Proof: By [44], q >= 1, so abs(y1) <= abs(q*y1)
// By [59], abs(q*y1) <= abs(y2), thus
// abs(y1) <= abs(q*y1) <= abs(y2)
/*61*/ T s2 = -s1;
/*62*/ assert(s2 == s0); // By [61, 36]
/*63*/ assert(s2*abs(x2) == x2);
// Proof: By [53], abs(x2) == (abs(x0) + abs(q*x1)).
// By [52], x2 == s0*(abs(x0) + abs(q*x1)), so x2 == s0*abs(x2).
// By [62], s2 == s0, thus x2 == s2*abs(x2).
/*64*/ assert(-s2*abs(y2) == y2);
// Proof: By [58], abs(y2) == (abs(y0) + abs(q*y1))
// By [57], y2 == -s0*(abs(y0) + abs(q*y1)), so y2 == -s0*abs(y2)
// By [62], s2 == s0, thus y2 == -s2*abs(y2).
/*65*/ if (a2 == 0) {
/*66*/ assert(q >= 2);
// Proof: By [45], a2 == a0 - q*a1
// Hence by [65], 0 == a0 - q*a1 and a0 == q*a1.
// If we assume q==1, then since we just showed a0 == q*a1,
// we would have a0 == a1, which contradicts [31] a1 < a0.
// Therefore q!=1, and since by [44], q >= 1, we know q >= 2.
/*67*/ assert(abs(x2) >= 2*abs(x1));
// Proof: By [54], abs(x2) >= abs(q*x1). Since by [66] q >= 2,
// abs(x2) >= q*abs(x1) >= 2*abs(x1).
/*68*/ assert(abs(y2) >= 2*abs(y1));
// Proof: By [59], abs(y2) >= abs(q*y1). Since by [66] q >= 2,
// abs(y2) >= q*abs(y1) >= 2*abs(y1).
}
/*69*/ assert((x1*y2 - y1*x2) == -(x0*y1 - y0*x1));
// Proof: Using the definition of the determinant, let
// D0 = x0*y1 - y0*x1 and D1 = x1*y2 - y1*x2.
// By [56, 51],
// D1 == x1*(y0 - q*y1) - y1*(x0 - q*x1)
// D1 == x1*y0 - y1*q*x1 - y1*x0 + y1*q*x1
// D1 == x1*y0 - y1*x0
// D1 == -D0
/*70*/ assert(abs(x1*y2 - y1*x2) == 1);
// Proof: By [35], abs(x0*y1 - y0*x1) == 1
// By [69], (x1*y2 - y1*x2) == -(x0*y1 - y0*x1)
// Thus, abs(x1*y2 - y1*x2) == abs(x0*y1 - y0*x1) == 1
/*71*/ assert(a*x2 + b*y2 == a2);
// Proof: By [45], a2 == a0 - q*a1
// and thus by [32, 33],
// a2 == a*x0 + b*y0 - q*(a*x1 + b*y1)
// a2 == a*x0 - q*a*x1 + b*y0 - q*b*y1
// a2 == a*(x0 - q*x1) + b*(y0 - q*y1)
// and by [51, 56],
// a2 == a*x2 + b*y2
/*72*/ x0 = x1;
/*73*/ y0 = y1;
/*74*/ a0 = a1;
/*75*/ x1 = x2;
/*76*/ y1 = y2;
/*77*/ a1 = a2;
/*78*/ s0 = s1;
/*79*/ s1 = s2;
// trivially, we know at this point
/*80*/ assert_invariant(a0 > 0); // By [30, 74]
/*81*/ assert(a1 >= 0); // By [48, 77]
/*82*/ assert_invariant(a1 < a0); // By [48, 74, 77]
/*83*/ assert_invariant(a*x0 + b*y0 == a0); // By [33, 72, 73, 74]
/*84*/ assert_invariant(a*x1 + b*y1 == a1); // By [71, 75, 76, 77]
/*85*/ assert_invariant(gcd(a0,a1) == gcd(a,b)); // By [50, 74, 77]
/*86*/ assert_invariant(abs(x0*y1 - y0*x1) == 1); // By [70,72,73,75,76]
/*87*/ assert_invariant(s1 == -s0); // By [61, 79, 78]
/*88*/ assert_invariant(abs(s0) == 1); // By [42, 78]
/*89*/ assert_invariant(s0*abs(x0) == x0); // By [39, 78, 72]
/*90*/ assert_invariant(s1*abs(x1) == x1); // By [63, 79, 75]
/*91*/ assert_invariant(-s0*abs(y0) == y0); // By [41, 78, 73]
/*92*/ assert_invariant(-s1*abs(y1) == y1); // By [64, 79, 76]
/*93*/ assert(abs(x1) >= abs(x0)); // By [55, 75, 72]
/*94*/ assert(abs(y1) >= abs(y0)); // By [60, 76, 73]
/*95*/ if (a1 == 0) {
/*96*/ assert(abs(x1) >= 2*abs(x0)); // By [65, 67, 77, 75, 72]
/*97*/ assert(abs(y1) >= 2*abs(y0)); // By [65, 68, 77, 76, 73]
}
// Less trivially,
/*98*/ assert_invariant(gcd(x1, y1) == 1);
// Proof: By [86] x0*y1 - y0*x1 == +-1, so there obviously
// exist integers u and v such that u*y1 + v*x1 == 1.
// Let d be any common divisor of x1 and y1 (we know a common
// divisor exists since 1 divides x1 and y1). Since d divides
// both x1 and y1, d divides u*y1 + v*x1 == 1, and thus
// d divides 1. Since d divides 1, d == +-1. Thus a common
// divisor of x1 and y1 can only be -1 or 1, and since 1 is a
// common divisor of x1 and y1, the greatest common divisor of x1
// and y1 is 1. For a similar proof see
// http://people.sju.edu/~pklingsb/rp.fta.pdf
}
/*99*/ assert(a1 == 0);
// Since we finished the loop, we know a1 == 0 at this point.
/*M0*/ assert(x1*(a/gcd(a,b)) == -y1*(b/gcd(a,b)));
// Proof: By [15,84], a*x1 + b*y1 == a1, and so by [99]
// a*x1 + b*y1 == 0
// x1*a == -y1*b
// Letting d = gcd(a,b), by [10] d >= 1. Since d divides a and b,
// and x1*a == -y1*b, we know x1*(a/d) == -y1*(b/d)
/*M1*/ assert(gcd(a/gcd(a,b), b/gcd(a,b)) == 1);
// Proof: Letting d=gcd(a,b), by [10] d >= 1. Since d contains all
// the common divisors of a and b, d divides both a and b, and the
// quotients of a/d and b/d have no common divisors except 1 and -1.
// Therefore gcd(a/d, b/d) == 1.
/*M2*/ assert(abs(x1) == b/gcd(a,b));
// Proof: Letting d = gcd(a,b), by [M0] x1*(a/d) == -y1*(b/d).
// By [M1] (a/d) and (b/d) are coprime, and by [27,98] x1 and -y1 are
// coprime. So by [P6], abs(a/d)==abs(-y1) and abs(x1)==abs(b/d).
// By [10,01], d>0 and b>=0, so abs(b/d) == b/d.
// Thus, abs(x1) == abs(b/d) == b/d
/*M3*/ assert(abs(y1) == a/gcd(a,b));
// Proof: Letting d = gcd(a,b), by [M0] x1*(a/d) == -y1*(b/d).
// By [M1] (a/d) and (b/d) are coprime, and by [27,98] x1 and -y1 are
// coprime. So by [P6], abs(a/d)==abs(-y1) and abs(x1)==abs(b/d).
// By [10,09], d>0 and a>0, so abs(a/d) == a/d.
// Thus, abs(y1) == abs(-y1) == abs(a/d) == a/d
/*M4*/ assert((x0 == 1 && x1 == 0) || (abs(x1) >= 2*abs(x0)));
// Proof: By [03, 06, 95, 96, 99]
/*M5*/ assert(abs(y1) >= 2*abs(y0)); // By [26, 95, 97, 99]
/*M6*/ assert(x0 == 1 || abs(x0) <= (b/gcd(a,b))/2);
// Proof: By [M4], (x0 == 1 && x1 == 0) || (abs(x1) >= 2*abs(x0)),
// Assume (x0 == 1 && x1 == 0). Then we would have x0 == 1.
// Assume instead (abs(x1) >= 2*abs(x0)). By [M2],
// abs(x1) == b/gcd(a,b). Thus we would have
// b/gcd(a,b) == abs(x1) >= 2*abs(x0). Letting d = gcd(a,b), we
// would have abs(x0)*2 <= b/d. By [01,10] b >= 0 and d > 0, so
// (b/d) >= 0. Since (b/d) >= 0 and 2 > 0 and abs(x0)*2 <= (b/d),
// by [P2] we would have abs(x0) <= (b/d)/2.
// Thus x0 == 1, or abs(x0) <= (b/d)/2.
/*M7*/ assert(abs(y0) <= (a/gcd(a,b))/2);
// Proof: By [M5], abs(y1) >= 2*abs(y0), and by [M3],
// abs(y1) == a/gcd(a,b). Thus, a/gcd(a,b) == abs(y1) >= 2*abs(y0).
// Letting d = gcd(a,b), abs(y0)*2 <= a/d. By [09, 10] a > 0 and
// d > 0, so (a/d) >= 0. Since (a/d) >= 0 and 2 > 0 and
// abs(y0)*2 <= (a/d), by [P2] abs(y0) <= (a/d)/2.
/*M8*/ assert(a0 == gcd(a,b));
// Proof: By [99] a1 == 0, and by [11,80] a0 > 0, so
// gcd(a0,a1) == gcd(a0,0) == a0.
// By [16,85] gcd(a0,a1) == gcd(a,b). Therefore,
// a0 == gcd(a0,a1) == gcd(a,b)
/*M9*/ assert(a*x0 + b*y0 == gcd(a,b));
// By [14,83] a*x0 + b*y0 == a0, and so by [M8],
// a*x0 + b*y0 == a0 == gcd(a,b)
// Note: Since a*x0 + b*y0 == gcd(a,b), we know x0 and y0 are the
// Bezout coefficients.
/*N0*/ assert_invariant(s0*abs(x0) == x0); // By [22, 89]
/*N1*/ assert_invariant(-s0*abs(y0) == y0); // By [24, 91]
*pX = x0;
*pY = y0;
*pGcd = a0;
}
#undef assert_invariant
#undef assert_precondition
#endif