Haskell/Solutions/Recursion
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The factorial function
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factorial
can be defined as follows:factorial 0 = 1 factorial n = n * factorial(n - 1)
- Typing
factorial (-1)
should yield the message*** Exception: stack overflow
. This is because, according to the definition,factorial (-1)
is(-1) * factorial (-2)
, which is(-1) * (-2) * factorial (-3)
. This will never stop, so the function will keep going until it runs out of memory. doubleFactorial
can be defined as follows:
doubleFactorial 0 = 1
doubleFactorial 1 = 1
doubleFactorial n = n * doubleFactorial (n-2)
Other recursive functions
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1. 5 × 4:
- 4 isn't 1, so we recurse: work out 5 × 3
- 3 isn't 1, so we recurse
- 2 isn't 1, so we recurse
- 1 is 1, so we return 5
- We add the current number, 5, to the result of the recursion, 5. We get 10
- We add the current number, 5, to the result of the recursion, 10. We get 15
- We add the current number, 5, to the result of the recursion, 15. We get 20.
2.
power x 0 = 1
power x y = x * power x (y-1)
3.
addition x 0 = x
addition x y = plusOne (addition x (y-1))
4.
log2 1 = 0
log2 n = 1 + log2 (n `div` 2) -- the "`" make div into infix notation
List-based recursion
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Give recursive definitions for the following list-based functions. In each case, think what the base case would be, then think what the general case would look like, in terms of everything smaller than it.
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The answers for 1-3, in one block of code:
replicat 0 _ = []
replicat n thing = thing : replicat (n-1) thing
[] !! _ = error "Index too large" -- An empty list has no elements.
(x:_) !! 0 = x
(x:xs) !! n = xs !! (n-1)
zip [] _ = []
zip _ [] = []
zip (x:xs) (y:ys) = (x,y) : zip xs ys
(x:xs)
does not match on an empty list so you can also have:
zip (x:xs) (y:ys) = (x,y) : zip xs ys
zip _ _ = []
And here is the accumulating length
:
length xs = go 0 xs
where
go acc [] = acc
go acc (_:xs) = go (acc + 1) xs