深度1432（因此2 ^ 1432项）足以使真实总和超出计算总和两倍。
我对如何确定所需的术语数量少于两个的想法有个想法。
我们使用动态编程来回答以下问题：给定深度d和目标浮点和s，具有成对和的2^d非负float16s的最大真和是s多少？
让那个数量成为T(d, s)。我们复发
T(0, s) = s, for all s.T(d, s) = max (T(d-1, a) + T(d-1, b)), for all d, s. a, b : float16(a + b) = s
重复执行的每个步骤都涉及遍历大约2^29组合（因为我们可以假设a ≤ b，并且负浮点数和特殊值超出了限制），并且所需的深度不会超过10^4Hans和您的答案。在我看来可行。
DP代码：
#include <algorithm>#include <cstdio>#include <vector>using Float16 = int;using Fixed = unsigned long long;static constexpr int kExponentBits = 5;static constexpr int kFractionBits = 10;static constexpr Float16 kInfinity = ((1 << kExponentBits) - 1) << kFractionBits;Fixed FixedFromFloat16(Float16 a) { int exponent = a >> kFractionBits; if (exponent == 0) { return a; } Float16 fraction = a - (exponent << kFractionBits); Float16 significand = (1 << kFractionBits) + fraction; return static_cast<Fixed>(significand) << (exponent - 1);}bool Plus(Float16 a, Float16 b, Float16* c) { Fixed exact_sum = FixedFromFloat16(a) + FixedFromFloat16(b); int exponent = 64 - kFractionBits - __builtin_clzll(exact_sum); if (exponent <= 0) { *c = static_cast<Float16>(exact_sum); return true; } Fixed ulp = Fixed{1} << (exponent - 1); Fixed remainder = exact_sum & (ulp - 1); Fixed rounded_sum = exact_sum - remainder; if (2 * remainder > ulp || (2 * remainder == ulp && (rounded_sum & ulp) != 0)) { rounded_sum += ulp; } exponent = 64 - kFractionBits - __builtin_clzll(rounded_sum); if (exponent >= (1 << kExponentBits) - 1) { return false; } Float16 significand = rounded_sum >> (exponent - 1); Float16 fraction = significand - (Float16{1} << kFractionBits); *c = (exponent << kFractionBits) + fraction; return true;}int main() { std::vector<Fixed> greatest0(kInfinity); for (Float16 a = 0; a < kInfinity; a++) { greatest0[a] = FixedFromFloat16(a); } for (int depth = 1; true; depth++) { auto greatest1 = greatest0; for (Float16 a = 1; a < kInfinity; a++) { Fixed greatest0_a = greatest0[a]; for (Float16 b = a; b < kInfinity; b++) { Float16 c; if (!Plus(a, b, &c)) { continue; } Fixed& value = greatest1[c]; value = std::max(value, greatest0_a + greatest0[b]); } } std::vector<double> ratios; ratios.reserve(kInfinity - 1); for (Float16 a = 1; a < kInfinity; a++) { ratios.push_back(greatest1[a] / static_cast<double>(FixedFromFloat16(a))); } std::printf("depth %d, ratio = %.17gn", depth, *std::max_element(ratios.begin(), ratios.end())); greatest0.swap(greatest1); }}我将运行它并在完成后发布更新。
