Warp wrote:
> Invisible <voi### [at] devnull> wrote:
>> I don't know why it works... but it does.
> 
>>  >module CDM where
>>  >
>>  >encode :: [Int] -> [Int] -> [Int]
>>  >encode cs xs = do
>>  >  x <- xs
>>  >  c <- cs
>>  >  let y = if x == 1 then 1 else -1
>>  >  return (c*y)
>>  >
>>  >decode :: [Int] -> [Int] -> [Int]
>>  >decode cs [] = []
>>  >decode cs xs =
>>  >  let
>>  >    (xs0,xs1) = splitAt (length cs) xs
>>  >    y = sum (zipWith (*) cs xs0)
>>  >  in ((y `min` 1) `max` 0) : decode cs xs1
> 
>> OK, so let's encode something:
> 
>>    *CDM> encode [1,-1] [1, 0, 1]
>>    [1,-1,-1,1,1,-1]
> 
>>    *CMD> decode [1,-1] [1,-1,-1,1,1,-1]
>>    [1,0,1]
> 
>> Well that's nice. But now check this out:
> 
>>    *CDM> let x1 = encode [1,1,1,1] [1,1,0,1]
>>    *CDM> let x2 = encode [1,-1,1,-1] [1,0,1,1]
>>    *CDM> let x3 = encode [1,1,-1,-1] [0,0,1,1]
>>    *CDM> let x4 = encode [1,-1,-1,1] [1,1,1,0]
>>    *CDM> x1
>>    [1,1,1,1,1,1,1,1,-1,-1,-1,-1,1,1,1,1]
>>    *CDM> x2
>>    [1,-1,1,-1,-1,1,-1,1,1,-1,1,-1,1,-1,1,-1]
>>    *CDM> x3
>>    [-1,-1,1,1,-1,-1,1,1,1,1,-1,-1,1,1,-1,-1]
>>    *CDM> x4
>>    [1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,-1,1,1,-1]
>>    *CDM> let y = foldl1 (zipWith (+)) [x1,x2,x3,x4]
>>    *CDM> y
>>    [2,-2,2,2,0,0,0,4,2,-2,-2,-2,2,2,2,-2]
>>    *CDM> decode [1,1,1,1] y
>>    [1,1,0,1]
>>    *CDM> decode [1,-1,1,-1] y
>>    [1,0,1,1]
>>    *CDM> decode [1,1,-1,-1] y
>>    [0,0,1,1]
>>    *CDM> decode [1,-1,-1,1] y
>>    [1,1,1,0]
> 
>> Magic!
> 
>   Are you seriously expecting people to understand what's going on there,
> without any kind of explanation?
> 
>   I'm a programmer and I haven't the slightest idea.

encode <key> <data> gives you a bunch of stuff. decode <key> <stuff> 
gives you back the original data.

Which isn't that surprising. What's surprising is that if you add 
several of these together, the decode function still works. (Provided 
each data stream has a different key - or, more exactly, an "orthoganol" 
key.)

Apparently this is how mobile phones manage to all transmit at the same 
time yet not screw up each other's transmissions.

This much I would have thought would be vaguely intuitive from the 
program's output. Of course, to comprehend how I've implemented the 
encode/decode functions, you'd need to understand Haskell...

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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