爱国者导弹浮动指向误差

从计算机系统：程序员的角度http://csapp.cs.cmu.edu http://csapp.cs.cmu.edu

练习题2.51

我们在问题 2.46 中看到，爱国者导弹软件近似为 0.1，因为 x =

0.00011001100110011001100(binary). 

假设他们使用 IEEE 舍入到偶数模式来确定 x′ 为 0.1 的近似值，其中二进制小数点右侧有 23 位。

A. x′ 的二进制表示是什么？

from the solution at the back of book, 
Looking at the nonterminating sequence for 1/10, we can see that the 2 bits to the right of the rounding position are 1, and so a better approximation to 1/10 would be obtained by incrementing x to get x′ = 0.00011001100110011001101, which is larger than 0.1.

B. x′ − 0.1 的十进制近似值是多少？

The solution says it's

We can see that x′ − 0.1 has binary representation: 0.0000000000000000000000000[1100]
Comparing this to the binary representation of 1 , we can see that it is 2^−22 × 0.1 , which is around 2.38 x 10^-8

我对（B）的问题是我们如何得到

x' - 0.1 == 0.0 0000 0000 0000 0000 0000 0000[1100] ?

我的计算给出了 0.000 0000 0000 0000 0000 0000 0100 （大约是解决方案所说的两倍）

0.1 in 小数二进制表示是：0.000[1100]

对于问题 2.46 中的近似，有一个cutoff在位置 23 之后，以及问题 2.51 a 中四舍五入到偶数发生在位置 23。后者意味着： - 1）转到最近的值 - 或 2）如果正好在中间，则转到偶数。由于位置 24 ff 是11001100...这是情况 1)，最接近的值是当位置 23 变为 1 时。原始值和近似值的位：

x     = 0.0001100110011001100110011001100110011001100...
x_246 = 0.00011001100110011001100  # cutoff
x_251 = 0.00011001100110011001101  # round to even - as asked in A)

区别x' - 0.1, i.e. x_251 - x，如 B) 中所问

• 小数二进制

  x_251:   0.00011001100110011001101
  x:     - 0.0001100110011001100110011001100110011001100...
  fdiff:   0.0000000000000000000000000110011001100110011... or
  fdiff:   0.0 0000 0000 0000 0000 0000 0000[1100]
  
  --> this answers "question for (B) is that how do we get x' - 0.1 
        == 0.0 0000 0000 0000 0000 0000 0000[1100]"                            
  

• 小数值

  ddiff = (1*2^-26 + 1*2^-27) + (1*2^-30 + A1*2^-31) + (...) + ...
        =     3*2^-27         +     3*2^-31         +    3*2^-35 + ...
        = 0.1*2^-22 = 2.384e-8  (sum of geometric serie)
  julia> @printf "%.40f" 0.1*2.0^-22
  0.0000000238418579101562513234889800848443
  

旁白：为什么有区别julia> @printf "%.40f" 0.1f0和上面的小数差ddiff？

0.1000000014901161193847656250000000000000
0.0000000238418579101562513234889800848443

0.1f0表示为IEEE 浮点值。位（julia> bits(f0.1f0)) ("00111101110011001100110011001101") 的含义如下：

0 01111011 10011001100110011001101
| |        | M = 1 + f
| | E = e - Bias = 123 - 127 = -4 --> V = M*2^-4 (right shift M by 4)
| sign

M:         1.10011001100110011001101
M/sh: 0.0001.10011001100110011001101
x_shifted: 0.000110011001100110011001101
x_251:     0.00011001100110011001101

--> 0.1 的移位近似值比 x_251 多 4 位。差值的十进制值x_shifted - x（如上计算）给出：0.1*2^-26。现在这两种表示法是合适的：

    julia> @printf "%.40f" 0.1*2.0^-26
    0.0000000014901161193847657077180612553028
    0.1000000014901161193847656250000000000000  # julia> @printf "%.40f" 0.1f0