将while循环转为数学方程？ [英] Turn while loop into math equation?

问题描述

我在程序中有两个简单的while循环，我认为应该是数学方程，但我很难转换它们：

I have two simple while loops in my program that I feel ought to be math equations, but I'm struggling to convert them:

float a = someValue;
int b = someOtherValue;
int c = 0;

while (a <= -b / 2) {
    c--;
    a += b;
}
while (a >= b / 2) {
    c++;
    a -= b;
}

此代码原样工作，但我觉得可以简化为数学方程。这里的想法是，这段代码采取偏移量（someValue）和调整坐标（c），以最小化距离瓦片（大小someOtherValue）的中心的距离。

This code works as-is, but I feel it could be simplified into math equations. The idea here being that this code is taking an offset (someValue) and adjusting a coordinate (c) to minimize the distance from the center of a tile (of size someOtherValue). Any help would be appreciated.

推荐答案

可以证明以下是正确的：

It can be proved that the following is correct:

c = floor((a+b/2)/b)
a = a - c*b

请注意，floor表示向下取整，向负无穷大：不向0.（例如floor（-3.1）= - 4。 floor（）库函数会做到这一点;只是确保不会转换为int，通常会向0舍入。）

Note that floor means round down, towards negative infinity: not towards 0. (E.g. floor(-3.1)=-4. The floor() library functions will do this; just be sure not to just cast to int, which will usually round towards 0 instead.)

推测 b 是严格肯定的，因为否则两个循环都不会终止：添加 b 不会使 a 更大，减去 b 将不会使 a 有了这个假设，我们可以证明上面的代码工作。 （和paranoidgeek的代码也几乎是正确的，除了它使用转换为int而不是 floor 。）

Presumably b is strictly positive, because otherwise neither loop will never terminate: adding b will not make a larger and subtracting b will not make a smaller. With that assumption, we can prove that the above code works. (And paranoidgeek's code is also almost correct, except that it uses a cast to int instead of floor.)

聪明的证明方式：
代码从 a b 的倍数$ c>直到 a 位于 [ - b / 2，b / 2）或从 a / b 中减去整数，直到 a / b 位于 [ - 1 / 2,1 / 2），即直到（a / b + 1/2） $ c> x ）位于 [0,1）中。因为你只是通过整数来改变它， x 的值不会改变 mod 1 ，即它去 remaining mod 1 ，即 x-floor（x）。因此，您所做的有效减法数（ c ）为 floor（x）。

Clever way of proving it: The code adds or subtracts multiples of b from a until a is in [-b/2,b/2), which you can view as adding or subtracting integers from a/b until a/b is in [-1/2,1/2), i.e. until (a/b+1/2) (call it x) is in [0,1). As you are only changing it by integers, the value of x does not change mod 1, i.e. it goes to its remainder mod 1, which is x-floor(x). So the effective number of subtractions you make (which is c) is floor(x).

乏味的验证方法：

_{第一个循环， c 的值是循环运行次数的负数，即：}

_{At the end of the first loop, the value of c is the negative of the number of times the loop runs, i.e.:}

• 0 if：a> -b / 2 = a + b / 2> 0


• -1 if：-b /2≥a> -3b / 2 =0≥a+ b / 2>-b≤0≥x> -1


• -2 if：-3b /2≥a> -5b / 2 = -b-a + b / 2> -2b = -1 -1x≥-2等，

其中 x =（a + b / 2）/ b ，所以如果x> 0，c为0，否则为ceiling（x）-1。如果第一个循环完全运行，那么在执行循环之前，它是≤-b / 2，因此它是≤-b / 2 + b now，即≤b/ 2 。根据是否正好是b / 2（即，当你开始时是否 x 是否是一个非正整数），第二个循环完全运行1次或0，并且c是ceiling（x）或ceiling（x）-1。

where x = (a+b/2)/b, so c is: 0 if x>0 and "ceiling(x)-1" otherwise. If the first loop ran at all, then it was ≤ -b/2 just before the last time the loop was executed, so it is ≤ -b/2+b now, i.e. ≤ b/2. According as whether it is exactly b/2 or not (i.e., whether x when you started was exactly a non-positive integer or not), the second loop runs exactly 1 time or 0, and c is either ceiling(x) or ceiling(x)-1. So that solves it for the case when the first loop did run.

如果第一个循环没有运行，那么c在第二个循环结束时的值是：

If the first loop didn't run, then the value of c at the end of the second loop is:

• 0如果：a< b / 2 = a-b / 2 < 0


• 1如果：b /2≤a< 3b / 2 =0≤a-b / 2 < b <=0≤y< 1


• 2 if：3b / 2≤a< 5b / 2 =b≤a-b / 2 < 2b =1≤y< 2等

其中 y =（ab / 2）/ b ，所以如果y <0，则c为：0，否则为1 + floor（y）。 [和 a 现在肯定< b / 2和≥-b / 2。]

where y = (a-b/2)/b, so c is: 0 if y<0 and 1+floor(y) otherwise. [And a now is certainly < b/2 and ≥ -b/2.]

因此，您可以为 c 写一个表达式： / p>

So you can write an expression for c as:

x = (a+b/2)/b
y = (a-b/2)/b
c = (x≤0)*(ceiling(x) - 1 + (x is integer))
   +(y≥0)*(1 + floor(y))

当然，接下来你会注意到（ceiling（x）-1+（x是整数））与 floor（x）相同， floor（x + 1）-1 code> y 实际上是 x-1 ，因此（1 + floor（y））= floor x），并且对于条件：。
当x≤0时，不能是（y≥0），因此 c 只是第一项，它是 floor（x），

当0< x < 1，则这两个条件都不成立，因此 c 为 0 ;

x，那么只有0≤y，所以c只是第二项，它是 floor（x）。
所有情况下c = floor（x）。

Of course, next you notice that (ceiling(x)-1+(x is integer)) is same as floor(x+1)-1 which is floor(x), and that y is actually x-1, so (1+floor(y))=floor(x), and as for the conditionals:
when x≤0, it cannot be that (y≥0), so c is just the first term which is floor(x),
when 0 < x < 1, neither of the conditions holds, so c is 0,
when 1 ≤ x, then only 0≤y, so c is just the second term which is floor(x) again. So c = floor(x) in all cases.

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