在没有条件比较的情况下数学地找到最大值 [英] Mathematically Find Max Value without Conditional Comparison

问题描述

---------更新了------------

----------Updated ------------

到目前为止，codymanix 和 moonshadow 提供了很大的帮助.我能够使用方程式解决我的问题，而不是使用右移除以 29.因为 32 位有符号 2^31 = 溢出到 29.哪个有效！

codymanix and moonshadow have been a big help thus far. I was able to solve my problem using the equations and instead of using right shift I divided by 29. Because with 32bits signed 2^31 = overflows to 29. Which works!

PHP 中的原型

$r = $x - (($x - $y) & (($x - $y) / (29)));

LEADS 的实际代码(每行只能执行一个数学函数！！！啊啊啊！！！)

Actual code for LEADS (you can only do one math function PER LINE!!! AHHHH!!!)

DERIVDE1 = IMAGE1 - IMAGE2;
DERIVED2 = DERIVED1 / 29;
DERIVED3 = DERIVED1 AND DERIVED2;
MAX = IMAGE1 - DERIVED3;

---------原始问题------------
由于我的应用程序的限制，我认为这不太可能，但我认为值得一试.

----------Original Question-----------
I don't think this is quite possible with my application's limitations but I figured it's worth a shot to ask.

我会尽量简化.我需要在无法使用 IF 或任何条件语句的情况下找到两个数字之间的最大值.

I'll try to make this simple. I need to find the max values between two numbers without being able to use a IF or any conditional statement.

为了找到 MAX 值，我只能执行以下功能

In order to find the the MAX values I can only perform the following functions

Divide, Multiply, Subtract, Add, NOT, AND ,OR

假设我有两个数字

A = 60;
B = 50;

现在如果 A 总是大于 B，那么找到最大值就很简单了

Now if A is always greater than B it would be simple to find the max value

MAX = (A - B) + B;
ex. 
10 = (60 - 50)
10 + 50 = 60 = MAX

问题是 A 并不总是大于 B.我无法使用我正在使用的脚本应用程序执行 ABS、MAX、MIN 或条件检查.

Problem is A is not always greater than B. I cannot perform ABS, MAX, MIN or conditional checks with the scripting applicaiton I am using.

有没有办法使用上面的有限操作来找到一个非常接近最大值的值?

Is there any way possible using the limited operation above to find a value VERY close to the max?

推荐答案

求2个变量的最大值:

max = a-((a-b)&((a-b)>>31))

其中 >> 是按位右移(也称为 SHR 或 ASR，取决于符号).

where >> is bitwise right-shift (also called SHR or ASR depeding on signedness).

您使用数字的位数减一，而不是 31.

Instead of 31 you use the number of bits your numbers have minus one.

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