[Haskell-cafe] Generating arbitrary function in QuickCheck
Eugene Kirpichov
ekirpichov at gmail.com
Tue Apr 7 23:31:23 EDT 2009
2009/4/8 Yusaku Hashimoto <nonowarn at gmail.com>:
> Thanks for all replies! these gave me many enlightenments.
>
> I have been looking in about free theorem for a while, and I want to
> try to describe what I learned, and please collect me where I am wrong
> or misunderstood.
>
> Generating random functions for sortBy is meaningless, because, for
> testing sortBy, casual comparison function is enough to it. The reason
> of this comes from free theorem. By the free theorem, sortBy should
> satisfy this theorem (copied from ftshell).
>
> forall t1,t2 in TYPES, f :: t1 -> t2.
> forall p :: t1 -> t1 -> Ordering.
> forall q :: t2 -> t2 -> Ordering.
> (forall x :: t1. p x = q (f x) . f)
> ==> (map f . sortBy p = sortBy q . map f)
>
> Any linear orderings can be defined by mapping value to Integer. And
> above theorem says comparing functions in the theorem must be
> homomorphic to each other. Hence comparing function is meaningful only
> when it is homomorhpic to Integer's that. (I have not tried to
> Generate such function yet, and it may be unnecessary, but) taking
> such function is easy.
Kind of like that.
The theorem says that for any function of sortBy's type, it will not
change permutation of the list if we replace the comparison function
by another one that is monotonic with respect to it.
For any correct comparison function F, there exists a mapping to
integers M and 'compare' as the comparison function agreeing with F.
Therefore, if you check yourSortBy on integers with 'compare', you
will also be sure of all other types with correct comparison
functions, because you'll be sure that it permutes the list in a
correct way.
>
> Furthermore, it should be satisfied by all functions that have same
> type signitures to sortBy, because it is implied from only its own
> type.
>
> Reference:
> - [Theorem for
> free!](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.38.9875)
> -
> [ftshell](http://hackage.haskell.org/cgi-bin/hackage-scripts/package/ftshell-0.3)
>
> Thanks,
> Yusaku Hashimoto
>
>
> On 2009/04/07, at 15:19, Janis Voigtlaender wrote:
>
>> Jason Dagit wrote:
>>>
>>> On Mon, Apr 6, 2009 at 10:09 PM, Eugene Kirpichov <ekirpichov at gmail.com>
>>> wrote:
>>>>
>>>> Since the argument to sortBy must impose a linear ordering on its
>>>> arguments, and any linear ordering may as well be generated by
>>>> assigning an integer to each element of type 'a', and your sorting
>>>> function is polymorphic, from the free theorem for the sorting
>>>> function we may deduce that it suffices to test your function on
>>>> integer lists with a casual comparison function (Data.Ord.compare),
>>>> and there is no need to generate a random comparison function.
>>>
>>> Interesting. How is this free theorem stated for the sorting
>>> function? Intuitively I understand that if the type is polymorphic,
>>> then it seems reasonable to just pick one type and go with it.
>>
>> You can try free theorems here:
>>
>> http://linux.tcs.inf.tu-dresden.de/~voigt/ft/
>>
>> For example, for
>>
>> sort :: Ord a => [a] -> [a]
>>
>> it generates the following:
>>
>> forall t1,t2 in TYPES(Ord), f :: t1 -> t2, f respects Ord.
>> forall x :: [t1]. map f (sort x) = sort (map f x)
>>
>> where:
>>
>> f respects Ord if f respects Eq and
>> forall x :: t1.
>> forall y :: t1. compare x y = compare (f x) (f y)
>> forall x :: t1. forall y :: t1. (<) x y = (<) (f x) (f y)
>> forall x :: t1. forall y :: t1. (<=) x y = (<=) (f x) (f y)
>> forall x :: t1. forall y :: t1. (>) x y = (>) (f x) (f y)
>> forall x :: t1. forall y :: t1. (>=) x y = (>=) (f x) (f y)
>>
>> f respects Eq if
>> forall x :: t1. forall y :: t1. (==) x y = (==) (f x) (f y)
>> forall x :: t1. forall y :: t1. (/=) x y = (/=) (f x) (f y)
>>
>> Assuming that all the comparison functions relate to each other in the
>> mathematically sensible way, the latter can be reduced to:
>>
>> f respects Ord if
>> forall x :: t1. forall y :: t1. (x <= y) = (f x <= f y)
>>
>> For sortBy you would get a similar free theorem.
>>
>> To see how the free theorem allows you to switch from an arbitrary type
>> to just integers, set t2=Int and simply use f to build a
>> order-preserving bijection between elements in the list x and a prefix
>> of [1,2,3,4,...]
>>
>> Ciao, Janis.
>>
>> --
>> Dr. Janis Voigtlaender
>> http://wwwtcs.inf.tu-dresden.de/~voigt/
>> mailto:voigt at tcs.inf.tu-dresden.de
>>
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--
Eugene Kirpichov
Web IR developer, market.yandex.ru
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