Re: [Haskell-cafe] How to decide if a number is an integer?

29 Sep 2009


      Of course them are not. But that is why I need the detector....

2009/9/29 Thomas DuBuisson :
...
Unless I missed something, the function in question is:
sqrt (a * a + 2 + 2 * num) - fromIntegral a
where num = 10
1 -> sqrt (1 * 1 + 2 + 2 * 10) - 1 -> sqrt (1 + 2 + 20) - 1 -> sqrt
(23) - 1 -> 3.79xxxxx
the fractional will only ever come from the sqrt function.  Do any of
the following actually look like square values to you?
26
31
38
47
58
71
86
103
122
IMO, the code works and your expectations are a bit off.
Thomas
On Mon, Sep 28, 2009 at 11:54 PM, Magicloud Magiclouds
 wrote:
...
The original code is
givenSum num = map (\a ->
                     let l = (sqrt $ fromIntegral (a * a + 2 + 2 *
num)) - (fromIntegral a) in
                     case properFraction l of
                       (_, 0) ->
                         True
                       _ ->
                         False
                  ) $ take num [1..]
:t l is (Floating a) => a
Well, in ghci
*Main> givenSum 10
[False,False,False,False,False,False,False,False,False,False]
On Tue, Sep 29, 2009 at 2:45 PM, Thomas DuBuisson
 wrote:
...
On Mon, Sep 28, 2009 at 11:35 PM, Magicloud Magiclouds
 wrote:
...
It never matches to (_, 0.0)....
I mean
case properFraction l of
 (_, 0) -> l
 _ -> 0 -- always goes here.
Odd, it works fine for me.
f x =
       case properFraction x of
               (_,0) -> True
               _     -> False
*Main> f 5
True
*Main> f 5.5
False
*Main> f 4.0
True
*Main> f 4.00000001
False
Thomas
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