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From: Thomas 'PointedEars' Lahn <PointedEars@web.de>
Newsgroups: comp.lang.javascript
Subject: Re: How would you prepare for javascripting interview?
Date: Sat, 23 Jul 2016 16:25:57 +0200
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Stefan Ram wrote:

> Scott Sauyet <scott@sauyet.com> writes:
>>through discussion:  "It's a somewhat surprising fact that in Javascript,
>>Math.min() > Math.max().  Can you think of why this might be so?"
> 
>   A varargs operator (that is obtained from a binary operator)
>   over the empty set always gives the identity element of the
>   operation, because this is the most natural and useful definition.
> 
>   For example, we can say that the sum of a list of numbers
>   a, b, c, can be written as
> 
> 0 + a + b + c
> 
>   where there is a substring »+ x« for every member of the list.
> 
>   So, we have to rule to append » + x« to »0« for each member »x«.
> 
>   The binary operator »+« is not defined for just one operand
>   or no operand, but our list operator (varargs operator) needs to be.
> 
>   When there is only one member a in the list, the rule given
>   above results in
> 
> 0 + a
> 
>   , when the list is empty we just get
> 
> 0
> 
>   , the identity element.

[What on Earth is the relation of this to the question at hand?]

The reason for the aforementioned fact is that Math.min() returns Infinity 
and Math.max() returns -Infinity.  And – need I spell it out? –
Infinity > -Infinity.

It makes sense for a minimum function to iterate over an argument list 
starting with the highest possible value, and for a maximum function with 
the lowest possible value.  Were one to implement the methods (according
to ECMAScript Ed. 3 and later), they would be (quickhack):

  if (Math.min(2, 3, 1) > 1)
  {
    Math.min = function () {
      var result = Infinity;

      for (var i = 0, len = arguments.length; i < len; ++i)
      {
        var arg = +arguments[i];

        /* NaN is the only Number value not equal to itself */
        if (arg != arg) return Number.NaN;

        if (arg < result) result = arg;
      }

      return result;
    };
    Math.min.length = 2;
  }

  if (Math.max(1, 2, 3) < 3)
  {
    Math.max = function () {
      var result = -Infinity;

      for (var i = 0, len = arguments.length; i < len; ++i)
      {
        var arg = +arguments[i];
        if (arg != arg) return Number.NaN;
        if (arg > result) result = arg;
      }

      return result;
    };
    Math.max.length = 2;
   }

<http://www.ecma-international.org/ecma-262/6.0/#sec-math.max>

BTDT (there have been implementations that did not support more than two 
arguments, and tail recursion is inefficient in this case).

-- 
PointedEars
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