在J中，如何找到平方根的扩展精度整数层

Question

I understand that when I take the square root ( %: ) of a number that does not result in an integer, my answer is a float. I'm looking to find the floor ( <. ) of the square root in order to get an integer result. Does J have a built-in way to achieve this? Do I need to resort to a loop to find my answer?

Tossing in a few extended precision ( x: ) requests certainly doesn't do it.

   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│
└──┴────────┘

Answer 1

From "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.

So, you have to use <.@%: :

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

See also Dictionary - Extended and Rational Arithmetic

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

Answer 2

This appears to work:

sqrt=: <.@%:
sqrt 99999999999999x

For more info, see http://www.jsoftware.com/help/jforc/elementary_mathematics_in_j.htm