[Haskell-beginners] oddsFrom3 function
Rein Henrichs
rein.henrichs at gmail.com
Mon Aug 17 19:20:40 UTC 2015
Oh, we're just talking about two different things.
`oddsFrom3 = map (+2) oddsFrom3` is just bottom.
I was talking about `oddsFrom3 = 3 : map (+2) oddsFrom3`.
On Mon, Aug 17, 2015 at 12:02 PM akash g <akaberto at gmail.com> wrote:
> Oh, I do apologize for the wrong use of the term `terminal` in this case
> which led to this exchange (one that I don't regret as I learned a lot
> today; thanks, Rein) . It had to do with my intuition of it (the - works
> for me, but can't explain to others - type of intuition).
>
> On Tue, Aug 18, 2015 at 12:25 AM, akash g <akaberto at gmail.com> wrote:
>
>> Yes, it is a coinductive structure (though I had a mental picture of the
>> list as a coinductive structure, which is what it exactly is as far as
>> infinite lists in Haskell are concerned). I got my terms wrong; thanks for
>> that.
>>
>> Let me clarify. By terminal, I meant that the structure itself is made
>> up of finite and infinite structures and you've some way of getting to that
>> finite part (my intuition as a terminal symbol).
>>
>> Take the following for an example.
>>
>> data Stream a = Stream a (Stream a)
>>
>> Stream, by virtue of construction, will have a finite element (or it can
>> be a co-inductive structure again) and an infinite element (the
>> continuation of the stream).
>>
>> However, take the OP's code under question
>>
>> oddsFrom3 = map (+2) oddsFrom3 -- Goes into an infinite loop; violates
>> condition for co-inductiveness; See Link1
>>
>> The above is not guarded by a constructor and there is no way to pull
>> anything useful out of it without going to an infinite loop. So, it has
>> essentially violated the guardedness condition (Link1 to blame/praise for
>> this). This is basically something like
>>
>> ==========
>> loop :: [Integer]
>> loop = loop -- This compiles, but god it will never end
>> ==========
>>
>> I shouldn't have used the term terminal and I think this is where the
>> confusion stems from. My intuition and what it actually is very similar,
>> yet subtly different. This might further clarify this (Link1)
>>
>>
>> =============
>> every co-recursive call must be a direct argument to a constructor of the
>> co-inductive type we are generating
>> =============
>>
>> As for inductive vs co-inductive meaning, I think it is because I see the
>> co-inductive construction as a special case of the inductive step (at least
>> Haskell lets me have this intuition).
>>
>>
>>
>> Link1: http://adam.chlipala.net/cpdt/html/Coinductive.html
>> Link2: http://c2.com/cgi/wiki?CoinductiveDataType
>>
>> On Mon, Aug 17, 2015 at 11:29 PM, Rein Henrichs <rein.henrichs at gmail.com>
>> wrote:
>>
>>> It isn't an inductive structure. It's a *coinductive* structure. And
>>> yes, coinductive structures are useful in plenty of scenarios.
>>>
>>> On Mon, Aug 17, 2015 at 10:30 AM akash g <akaberto at gmail.com> wrote:
>>>
>>>> Oh, it is a valid value (I think I implied this by saying they'll
>>>> compile and you can even evaluate). Just not useful in any given scenario
>>>> (an inductive structure where you don't have terminals).
>>>>
>>>>
>>>> On Mon, Aug 17, 2015 at 10:57 PM, akash g <akaberto at gmail.com> wrote:
>>>>
>>>>> @Rein:
>>>>> Perhaps I should have been a bit more clear. There is no way to get a
>>>>> terminal value from said function.
>>>>>
>>>>>
>>>>> oddsFrom3 :: [Integer]
>>>>> oddsFrom3 = map (+2) oddsFrom3
>>>>>
>>>>> Try a head for it perhaps.
>>>>>
>>>>> oddsFrom3 = map (+2) oddsFrom3
>>>>> <=> ((head oddsFrom3) + 2) : map (+2) ((tail oddsFrom3) + 2)
>>>>> <=> ((head (map (+2) oddsFrom3) + 2) : map (+2) ((tail oddsFrom3) + 2)
>>>>>
>>>>> Sure, it doesn't hang until you try to evaluate this (in lazy language
>>>>> evaluators). However, for any inductive structure, there needs to be a
>>>>> (well any finite number of terminals) terminal (base case) which can be
>>>>> reached from the starting state in a finite amount of computational
>>>>> (condition for termination). Type sigs don't/can't guarantee termination.
>>>>> If they don't have a terminal value, you'll never get to the bottom (bad
>>>>> pun intended) of it.
>>>>>
>>>>>
>>>>> Take an infinite list as an example.
>>>>>
>>>>> x a = a : x a
>>>>>
>>>>> Here, one branch of the tree (representing the list as a highly
>>>>> unbalanced tree where every left branch is of depth one at any given
>>>>> point). If such a structure is not present, you can never compute it to a
>>>>> value and you'll have to infinitely recurse.
>>>>>
>>>>> Try x a = x a ++ x a
>>>>>
>>>>> And think of the getting the head from this. You're stuck in an
>>>>> infinite loop.
>>>>>
>>>>> You may also think of the above as a small BNF and try to see if
>>>>> termination is possible from the start state. A vaguely intuitive way of
>>>>> looking at it for me, but meh, I might be missing something.
>>>>>
>>>>>
>>>>>
>>>>> On Mon, Aug 17, 2015 at 10:23 PM, Rein Henrichs <
>>>>> rein.henrichs at gmail.com> wrote:
>>>>>
>>>>>> > The initial version which the OP posted doesn't have a terminal
>>>>>> value.
>>>>>>
>>>>>> The point is that it doesn't need a terminal value. Infinite lists
>>>>>> like oddsFrom3 and (repeat "foo") and (let xs = 1 : xs) are all perfectly
>>>>>> valid Haskell values.
>>>>>>
>>>>>> On Mon, Aug 17, 2015 at 6:17 AM Doug McIlroy <doug at cs.dartmouth.edu>
>>>>>> wrote:
>>>>>>
>>>>>>> > > oddsFrom3 :: [Integer]
>>>>>>> > > oddsFrom3 = 3 : map (+2) oddsFrom3
>>>>>>> > >
>>>>>>> > >
>>>>>>> > > Thanks for your help.
>>>>>>> >
>>>>>>> > Try to expand a few steps of the recursion by hand e.g.:
>>>>>>> >
>>>>>>> > 3 : (map (+2) (3 : map (+2) (3 : map (+2) ...)))
>>>>>>> >
>>>>>>> >
>>>>>>> > As you can see, the deeper you go more 'map (+2)' are applied to
>>>>>>> '3'.
>>>>>>>
>>>>>>> Some more ways to describe the program, which may be useful:
>>>>>>>
>>>>>>> As with any recursive function, assume you know the whole series and
>>>>>>> then confirm that by verifying the inductive step. In this case
>>>>>>> oddsFrom3 = [3,5,7,9,11,...]
>>>>>>> map (+2) oddsFrom3 = [5,7,9,11,13,...]
>>>>>>> voila
>>>>>>> oddsFrom3 = 3 : map (+2) oddsFrom3
>>>>>>>
>>>>>>> Assuming we have the whole series, we see its tail is
>>>>>>> computed from the whole by adding 2 to each element.
>>>>>>> Notice that we don't actually have to know the values in the
>>>>>>> tail in order to write the formula for the tail.
>>>>>>>
>>>>>>> Yet another way to describe the program: the "output" is taken
>>>>>>> as "input". This works because the first element of the output,
>>>>>>> namely 3, is provided in advance. Each output element can then
>>>>>>> be computed before it is needed as input.
>>>>>>>
>>>>>>> In an imperative language this would be done so:
>>>>>>> integer oddsFrom3[0:HUGE]
>>>>>>> oddsFrom3[0] := 3
>>>>>>> for i:=1 to HUGE do
>>>>>>> oddsFrom3[i] = oddsFrom3[i-1] + 2
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