I'm working with a Haskell implementation of generators for a homework. I have an andAlso function that's supposed to add an additional predicate to a generator, but it's not working correctly in all test cases. Although it seems correct
Here's my generator type definition:
-- Type definition for a generator: a function producing a sequence of values
-- 1. The first function generates the next value.
-- 2. The second function checks if generation should continue.
-- 3. The third value is the initial value, or seed. It does not count as being generated by the generator.
type Generator a = (a -> a, a -> Bool, a)
My current implementation of andAlso:
-- Adds an additional predicate to a generator.
andAlso :: (a -> Bool) -> Generator a -> Generator a
andAlso p (f, g, s) = (f, \x -> g x && p x, s)
I'm using a helper function takeGen to visualize test results:
takeGen :: Int -> ((a -> a), (a -> Bool), a) -> [a]
takeGen 0 _ = []
takeGen n (next, pred, seed)
| not (pred seed) = []
| otherwise = seed : takeGen (n-1) (next, pred, next seed)
Here are my test cases and their results:
-- Test 1: Filter for odd numbers
takeGen 10 (andAlso (\x -> x `mod` 2 == 1) ((+1), (<10), 0))
-- Expected: [1,3,5,7,9], but getting: [1]
-- Test 2: Filter for numbers divisible by 3
takeGen 10 (andAlso (\x -> x `mod` 3 == 0) ((+1), (<10), 0))
-- Expected: [0,3,6,9], but getting: [0]
-- Test 3: Filter for numbers greater than 5
takeGen 10 (andAlso (>5) ((+1), (<10), 0))
-- Works correctly: [6,7,8,9]
-- Test 4: Combine two additional predicates
takeGen 10 (andAlso (>3) (andAlso (<8) ((+1), (<10), 0)))
-- Works correctly: [4,5,6,7]
-- Test 5: Test with a different generator function
takeGen 10 (andAlso (<15) ((+2), (<20), 1))
-- Works correctly: [1,3,5,7,9,11,13]
takeGen 10 (andAlso (<10) ((+1), (<10), 0))
-- Works correctly: [0,1,2,3,4,5,6,7,8,9]
I've tried various implementations of andAlso, including:
Some test cases work correctly, but others (specifically tests 1 and 2) don't return all expected values.
What's wrong with my andAlso implementation, and how should I fix it to make all test cases work as expected? Is the issue with how the generator is defined, how takeGen works, or how andAlso applies the additional predicate?
Other generator-related functions I have implemented:
nthGen :: Integer -> Generator a -> a
nextGen :: Generator a -> Generator a
lengthGen :: Generator a -> Integer
hasLengthOfAtLeast :: Integer -> Generator a -> Bool
constGen :: a -> Generator a
foreverGen :: (a -> a) -> a -> Generator a
emptyGen :: Generator a
Thank you for any help understanding where my implementation is going wrong.
Note that I can't match your test results with your supplied version of andAlso. (I get more wrong answers than you do, for example.)
However, it looks like you've implemented andAlso to add the supplied predicate to the continuation rule for the generator, but the test cases indicate that it should be treated as a filter for the generator instead.
For example, for "Test 1", your andAlso implementation only continues to generate values as long as the predicate (<10) is true AND the predicate \x -> x `mod` 2 is true. But, for the very first seed of 0 the second predicate fails, so this test generates an empty list [] (which is what I get with your code). Instead, you want to keep generating values as long as (<10) succeeds, but only "pass along" the subset of values that satisfy \x -> x `mod` 2.
It's a little mind-bending to implement andAlso directly, so you might want to follow @user2407038's advice and try implementing fromList and toList.
But, if you want a direct solution, here are a couple of hints. Consider the template definition:
andAlso p (f, g, s) = (f', g', s')
Let's pretend that this generator never stops (so g and g' don't matter). What should f' and s' look like? Well, you want s' to be the first value from the Generator sequence [s, f s, f (f s), ...] that satisfies p. So write:
where s' = find p (f, g, s)
and implement:
find :: (a -> Bool) -> Generator a -> a
find p (f, g, a) = ...
Here, find should pass tests like:
λ> find even ((+1), (<100), 0)
0
λ> find odd ((+1), (<100), 0)
1
λ> find (>10) ((+3), (<100), 0)
12
Don't worry about the stopping rule g in find yet... we'll figure that out in a moment.
Now, what about f'? Well, f' needs to fast-forward the current seed t (whatever it is) to the next seed in the sequence [f t, f (f t), ...] that satisfies p. So, it'll just be:
where f' t = find p (f, g, f t)
(Note the use of f t in place of t to make sure we don't stop at t.)
Now, what about stopping? Well, we want to be careful here that we stop at the first generated value that fails the predicate g, regardless of whether or not it satisfies our filter predicate p.
We can make sure this happens by modifying find so that it stops when g is satisfied, even if the resulting seed fails p. This will ensure that s' is set to the stopping value, so further processing of the generator will stop because g s' will be false.
So, modify find as necessary so it passes the above tests and also works with:
λ> find (>10) ((+1), (<5), 0)
5
That is, it should "find" the first element that fails the continuation predicate (<5), even though this element doesn't satisfy the filter predicate (>10).
With that version of find in place, you should be able to write:
andAlso :: (a -> Bool) -> Generator a -> Generator a
andAlso p (f, g, s) = (f', g, s')
where s' = find p (f, g, s)
f' t = find p (f, g, f t)
and have it pass all your tests.