在 J 中,如何找到平方根的扩展精度整数底

In J, how can I find the extended precision integer floor of a square root

我知道当我对一个不是整数的数字求平方根 (%:) 时,我的答案是浮点数。我正在寻找平方根的下限 (<.) 以获得整数结果。 J是否有内置的方法来实现这一点?我需要求助于循环来找到我的答案吗?

抛出一些扩展精度 (x:) 请求肯定不行。

   rootanddiffa =: 3 : '(y - root ^ 2);(root =. <. %: y)'
   rootanddiffa 24
┌─┬─┐
│8│4│
└─┴─┘
   rootanddiffa 26
┌─┬─┐
│1│5│
└─┴─┘
   rootanddiffa 99999999999999x
┌──┬────────┐
│_1│10000000│
└──┴────────┘
   rootanddiffb =: 3 : '(y - root ^ 2);(root =. x: <. x: %: y)'
   rootanddiffb 24
┌─┬─┐
│8│4│
└─┴─┘
   rootanddiffb 99999999999999x
┌──┬────────┐
│_1│10000000│
└──┴────────┘

这似乎有效:

sqrt=: <.@%:
sqrt 99999999999999x

有关详细信息,请参阅 http://www.jsoftware.com/help/jforc/elementary_mathematics_in_j.htm

来自 "J for C Programmers: 32"

The key is the idiom <.@v (or >.@v), where v is the verb you want to apply. When you code <.@v, the interpreter knows you are interested in only the integer part of the result, and if the operand is exact-precision, the interpreter will evaluate the integer part of the result exactly.

所以,你必须使用 <.@%::

rt2 =: 3 :'(y - root ^ 2);(root =. <.@%: y)'
rt2 99999999999999x
┌────────┬───────┐
│19999998│9999999│
└────────┴───────┘

另见 Dictionary - Extended and Rational Arithmetic

<.@f and >.@f produce extended integer results when applied to extended integer arguments.