>>> 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...

No wonder cell phones keep cutting out.

All it takes for a bit to be lost is for one analog "chip" to be
a little outside of it's intended magnitude and then it misses
the whole bit.

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