Solutions for Lecture 3 exercises #3
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| next = error "TODO" | ||
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| next :: forall a. (Bounded a, Enum a) => a -> a | ||
| next day = toEnum (nextDayNum `mod` totalEnumCount) |
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Bonus points for solving all extra challenges 💯 🔝
| @@ -134,11 +152,17 @@ data List1 a = List1 a [a] | |||
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| -- | This should be list append. | |||
| instance Semigroup (List1 a) where | |||
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| (<>) :: List1 a -> List1 a -> List1 a | |||
| (List1 x xs) <> (List1 _ ys) = List1 x (xs ++ ys) | |||
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I see that you ignore the first element of the second list. It's interesting that this implementation passes tests 🙂 But I believe you should use all elements of both lists when you append them.
| (List1 x xs) <> (List1 _ ys) = List1 x (xs ++ ys) | |
| (List1 x xs) <> (List1 y ys) = List1 x (xs ++ ys) |
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@chshersh I am confused -- even your suggested change ignores the y of second list. What am I missing?
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@kpadmasola I haven't implemented the function completely 😉
You need to use y in there. So yeah, my suggestion is still invalid. But I believe you can finish it! 💪🏻
src/Lecture3.hs
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| NoTreasure <> NoTreasure = NoTreasure | ||
| NoTreasure <> SomeTreasure t = SomeTreasure t | ||
| SomeTreasure t <> NoTreasure = SomeTreasure t |
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This is an absolutely correct solution!
However, we can notice that these 3 cases could be reduced to 2 with simpler pattern matching!
Let's see what we can unite in here.
If some of the arguments are NoTreasure, you actually always return the second one! So you need to inspect only two cases: NoTreasure in the first position, and NoTreasure in the second one 🙂
Co-authored-by: Dmitrii Kovanikov <kovanikov@gmail.com>
Co-authored-by: Dmitrii Kovanikov <kovanikov@gmail.com>
Co-authored-by: Dmitrii Kovanikov <kovanikov@gmail.com>
+ also addressed review comments
kpadmasola
changed the title
Solutions for Lecture 3
cc @chshersh @vrom911