Floating point 用整数数学求解严格不等式并考虑舍入误差

我有一个类型为

T

的浮点数数组，其中

T

可以是浮点或双精度

T x[n];

这些数字是严格的正数，并已排序，即

0 < x[0] < x[1] < x[2] < ... < x[n-1]

让我们假设数字

x

没有需要担心的下溢或溢出问题

我正在处理如下所述的问题，我不确定它是否健壮，并且，当它工作时，我认为会产生次优结果

有没有一种稳健而准确的方法来解决这个问题？我可以接受次优的解决方案，但至少它需要健壮。

---

我目前的方法

我使用符号

m（x）

表示接近

x

的浮点数，可能会受到舍入误差的影响。在下面的段落中，我修改了下面的不等式，适当地取上下界。请注意，以下内容不应解释为源代码，而应解释为获得最终方程式的数学步骤和推理

Let a(x) be the closest floating point number toward -inf
Let b(x) be the closest floating point number toward +inf

// Original inequality, where the operation affected by rounding error
// are marked with m()
int(m(x[i+1]*H)) > int(m(x[i]*H))   // for i=0..n-2

// I remove `atoi` subtracting and adding 0.5 
// this is a conceptual operation, not actually executed on the machine,
// hence I do not mark it with an m()
m(x[i+1]*H) - 0.5 > m(x[i]*H) + 0.5   // for i=0..n-2

// I reorganize terms
m(x[i+1]*H) - m(x[i]*H) >  1.0        // for i=0..n-2

// I would like to resolve the m() operator with strict LHS minorant and RHS majorant
// Note I cannot use `a(x)` and `b(x)` because I do not know `H` 
// I use multiplication by `(1+2*eps)` or `(1-2*eps)` as I hope this 
// will result in a number strictly greater (or lower) than the original one. 
// I already know this does not work for example if x is very small.
// So this seems like a pitfall of this approach.
x[i+1]*H(1-2*eps) - x[i]*H*(1+2*eps) >  b(1)  // for i=0..n-2

// solve the inequality for H, marking the operations affected by rounding 
// with m()
H > m( b(1) / m( x[i+1](1-2*eps) - x[i]*(1+2*eps) ) ) // for i=0..n-2

// take majorant or minorant as appropriate for rounded terms
H > b( b(1) / a( x[i+1](1-2*eps) - x[i]*(1+2*eps) ) )   // for i=0..n-2

// and eventually take a majorant which gives me the final formula for H
H = b( b( b(1) / min{ a( x[i+1](1-2*eps) - x[i]*(1+2*eps) ) } ) )

---

这是什么语言？什么是原子？C/C++

atoi

函数将字符串转换为整数，因此它不能是整数。你所说的“主要”和“次要”是什么意思？通常的数学定义似乎不适合这里。我的错。对于atoi，我的意思是截断为整数。我纠正了这个问题。根据majorant和minorant，我的意思是majorant（a）>a和minorant（a）感谢您的更新！为了确保我理解：对于

[0.99，1.0]

的示例输入数组，最佳输出将是

H=1.0

，而您的方法生成的值接近

H=100.0

。对吗？这是一个有趣的问题……没错！此外，我还不相信我的方法是可靠的。

Let a(x) be the closest floating point number toward -inf
Let b(x) be the closest floating point number toward +inf

// Original inequality, where the operation affected by rounding error
// are marked with m()
int(m(x[i+1]*H)) > int(m(x[i]*H))   // for i=0..n-2

// I remove `atoi` subtracting and adding 0.5 
// this is a conceptual operation, not actually executed on the machine,
// hence I do not mark it with an m()
m(x[i+1]*H) - 0.5 > m(x[i]*H) + 0.5   // for i=0..n-2

// I reorganize terms
m(x[i+1]*H) - m(x[i]*H) >  1.0        // for i=0..n-2

// I would like to resolve the m() operator with strict LHS minorant and RHS majorant
// Note I cannot use `a(x)` and `b(x)` because I do not know `H` 
// I use multiplication by `(1+2*eps)` or `(1-2*eps)` as I hope this 
// will result in a number strictly greater (or lower) than the original one. 
// I already know this does not work for example if x is very small.
// So this seems like a pitfall of this approach.
x[i+1]*H(1-2*eps) - x[i]*H*(1+2*eps) >  b(1)  // for i=0..n-2

// solve the inequality for H, marking the operations affected by rounding 
// with m()
H > m( b(1) / m( x[i+1](1-2*eps) - x[i]*(1+2*eps) ) ) // for i=0..n-2

// take majorant or minorant as appropriate for rounded terms
H > b( b(1) / a( x[i+1](1-2*eps) - x[i]*(1+2*eps) ) )   // for i=0..n-2

// and eventually take a majorant which gives me the final formula for H
H = b( b( b(1) / min{ a( x[i+1](1-2*eps) - x[i]*(1+2*eps) ) } ) )