[Haskell-beginners] oddsFrom3 function
akash g
akaberto at gmail.com
Mon Aug 17 19:02:04 UTC 2015
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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>>>>>>
>>>>>
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>
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