c++ - Round down the nearest integer power of a float

Question

Given two floating point values zoomAmount and zoomFactor, I need to calculate a newZoomAmount such that:

(newZoomAmount <= zoomAmount) && (newZoomAmount == pow( zoomFactor, i ))

for any integer i. I can easily loop through the values or binary search through a table to find the answer. However: is there a closed form to accomplish this?

---

Motivation: The zoomFactor is 2^⅕ ≅ 1.148698354997035, so that each 5 "zoom in" events result in ~exactly a power of two increase. When zooming a diagram to fit on screen I want the zoom level to be one of these notches so that zooming out eventually lands exactly on the 'base' 1.0 zoom level.</sup>

Answer 1

表示：

A = zoomAmount
F = zoomFactor
newA = newZoomAmount

我们有：

newA = pow(F, i)
=> log(newA) = i*log(F)
=> i = log(newA)/log(F)

并且作为 newA <= A，并且 log 是非递减的，

i = floor(log(A)/log(F))

---

newZoomAmount = pow( zoomFactor, floor( log(zoomAmount)/log(zoomFactor) ) );

Answer 2

基本上，对数。我将忽略可能迭代的底层实现这一事实，log因为您可能不介意。

使用以下内容：

缩放因子 = 2 1/5
           = 1.148698354997035
缩放量 = 2.25

您需要找到以下内容（请注意，我使用小于而不是小于或等于，原因见结尾）：

(newZoomAmount < 2.25) && (newZoomAmount == 2 i/5 )

通过检查，我们知道（因为和）zoomFactor5 == 2zoomFactor6 == 2.297... > 2.25

newZoomAmount == 2
我 == 5

因此，为了获得zoomAmount指数形式的电流，我们采用：

zoomExponent = log(zoomAmount) / log(zoomFactor)
             = 0.81093... / 0.13862...
             = 5.84962...

要获得下一个最小整数，您应该减去 1，然后取上限。

newZoomExponent = ⌈zoomExponent - 1⌉
                = ⌈4.84962...⌉
                = 5

最后：

newZoomAmount = zoomFactor newZoomExponent

我们使用减量上限而不是下限的原因是为了处理 的zoomAmount完美幂的特殊情况zoomFactor，在这种情况下

⌊zoomExponent⌋ == zoomExponent
newZoomAmount == zoomAmount

这显然是我们不想要的。

Answer 3

exponent=log(zoomAmount)/log(zoomFactor); /* zoomFactor^exponent == zoomAmount */
newZoomAmount=pow(zoomFactor,floor(exponent)); /* round down exponent */

Answer 4

只需替换并重新评估：

(newZoomAmount <= zoomAmount) && (newZoomAmount == pow( zoomFactor, i ))=> pow( zoomFactor, i ) <= zoomAmount=> i * ln(zoomFactor) <= ln(zoomAmount)=>i <= ln(zoomAmount) / ln(zoomFactor)这可能会也可能不会比迭代更快。

然后只需分配newZoomAmount = pow( zoomFactor, i );