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+// Part of the Carbon Language project, under the Apache License v2.0 with LLVM
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+// Exceptions. See /LICENSE for license information.
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+// SPDX-License-Identifier: Apache-2.0 WITH LLVM-exception
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+
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+// https://adventofcode.com/2024/day/13
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+
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+library "day13_common";
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+
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+import Core library "io";
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+import library "io_utils";
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+
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+// Returns m and n so that am + bn = gcd.
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+fn Euclid(a: i64, b: i64) -> {.m: i64, .n: i64, .gcd: i64} {
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+ if (a < b) {
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+ let reverse: {.m: i64, .n: i64, .gcd: i64} = Euclid(b, a);
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+ return {.m = reverse.n, .n = reverse.m, .gcd = reverse.gcd};
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+ }
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+ if (b == 0) {
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+ return {.m = 1, .n = 0, .gcd = a};
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+ }
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+ let next: {.m: i64, .n: i64, .gcd: i64} = Euclid(b, a % b);
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+ return {.m = next.n, .n = next.m - next.n * (a / b), .gcd = next.gcd};
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+}
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+
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+class Machine {
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+ fn Read() -> Machine {
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+ returned var me: Machine;
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+ // "Button A: X+"
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+ SkipNChars(12);
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+ ReadInt64(&me.a.0);
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+ // ", Y+"
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+ SkipNChars(4);
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+ ReadInt64(&me.a.1);
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+ // "\nButton B: X+"
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+ SkipNChars(13);
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+ ReadInt64(&me.b.0);
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+ // ", Y+"
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+ SkipNChars(4);
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+ ReadInt64(&me.b.1);
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+ // "\nPrize: X="
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+ SkipNChars(10);
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+ ReadInt64(&me.prize.0);
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+ // ", Y="
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+ SkipNChars(4);
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+ ReadInt64(&me.prize.1);
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+ SkipNewline();
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+ return var;
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+ }
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+
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+ var a: (i64, i64);
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+ var b: (i64, i64);
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+ var prize: (i64, i64);
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+}
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+
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+// Set of solutions to 'm a + n b = c'.
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+class BezoutSolutionSet {
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+ fn Make(a: i64, b: i64, c: i64) -> BezoutSolutionSet {
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+ var e: {.m: i64, .n: i64, .gcd: i64} = Euclid(a, b);
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+ if (c % e.gcd != 0) {
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+ // Impossible.
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+ return {.m0 = -1, .n0 = -1, .m_step = -1, .n_step = -1};
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+ }
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+
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+ // Find an initial solution. Note that m and n might be negative.
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+ let num_gcds: i64 = c / e.gcd;
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+ e.m = e.m * num_gcds;
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+ e.n = e.n * num_gcds;
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+
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+ // Pick the smallest positive m we can.
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+ let a_over_gcd: i64 = a / e.gcd;
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+ let b_over_gcd: i64 = b / e.gcd;
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+ var adj: i64 = 0;
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+ // This is e.m / b rounded towards -inf.
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+ // TODO: Should there be a way of expressing this directly?
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+ if (e.m < 0) {
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+ adj = (e.m - b_over_gcd + 1) / b_over_gcd;
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+ } else {
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+ adj = e.m / b_over_gcd;
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+ }
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+ e.m = e.m - adj * b_over_gcd;
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+ e.n = e.n + adj * a_over_gcd;
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+ return {.m0 = e.m, .n0 = e.n,
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+ .m_step = b_over_gcd, .n_step = -a_over_gcd};
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+ }
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+
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+ fn Valid[self: Self]() -> bool {
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+ return self.m_step >= 0;
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+ }
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+
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+ fn Solution[self: Self](k: i64) -> (i64, i64) {
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+ return (self.m0 + k * self.m_step, self.n0 + k * self.n_step);
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+ }
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+
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+ // m_k * a + n_k * b == c, where:
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+ // m_k = m0 + k * m_step
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+ // n_k = n0 * k * n_step
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+ // m0 is the minimum non-negative m value, and n_step is negative.
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+ var m0: i64;
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+ var n0: i64;
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+ var m_step: i64;
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+ var n_step: i64;
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+}
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+
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+// Given two sets of points s and t, find the intersection in the first quadrant
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+// with the minimum x coordinate. Returns the intersection point, or (-1, -1) if
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+// there is no intersection.
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+fn FirstIntersection(s: BezoutSolutionSet, t: BezoutSolutionSet) -> (i64, i64) {
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+ if (not s.Valid() or not t.Valid()) {
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+ // One of the sets is empty.
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+ return (-1, -1);
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+ }
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+
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+ // Easy case: lines meet at a point. This happens unless the lines are
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+ // parallel.
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+ let d: i64 = s.m_step * t.n_step - t.m_step * s.n_step;
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+ if (d != 0) {
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+ let u: i64 = (t.m0 - s.m0) * t.n_step - (t.n0 - s.n0) * t.m_step;
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+ if (u % d != 0) {
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+ // Lines don't meet at an integer point.
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+ return (-1, -1);
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+ }
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+ let j: i64 = u / d;
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+ if (j < 0 or s.n0 + j * s.n_step < 0) {
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+ // Lines don't meet in first quadrant.
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+ return (-1, -1);
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+ }
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+ return s.Solution(j);
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+ }
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+
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+ // Hard case: lines are parallel. We also get here if either "line" is a
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+ // point, but that doesn't happen in this exercise. We know all integer points
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+ // on the line are solutions, so the first intersection is the greater of the
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+ // first point in s and the first point in t, if that point is on both lines.
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+ if (s.m0 < t.m0) {
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+ // (t.m0, t.n0) is the solution if it's in s.
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+ if ((t.m0 - s.m0) % s.m_step == 0 and
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+ (t.n0 - s.n0) % s.n_step == 0) {
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+ return (t.m0, t.n0);
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+ }
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+ } else {
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+ // (s.m0, s.n0) is the solution if it's in t.
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+ if ((s.m0 - t.m0) % t.m_step == 0 and
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+ (s.n0 - t.n0) % t.n_step == 0) {
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+ return (s.m0, s.n0);
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+ }
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+ }
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+ return (-1, -1);
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+}
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+
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+fn CostIfPossible(m: Machine) -> i64 {
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+ let x: BezoutSolutionSet = BezoutSolutionSet.Make(m.a.0, m.b.0, m.prize.0);
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+ let y: BezoutSolutionSet = BezoutSolutionSet.Make(m.a.1, m.b.1, m.prize.1);
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+ let ab: (i64, i64) = FirstIntersection(x, y);
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+ if (ab.0 == -1) { return 0; }
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+ return ab.0 * 3 + ab.1;
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+}
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