问题
I am quite new to Haskell and I'm trying to wrap my head around how the lazy expression of Fibonacci sequences work.
I know this has been asked before, but none of the answers have addressed an issue I'm having with visualising the result.
The code is the canonical one using zipWith
fibs = 0 : 1 : zipWith (+) fibs (tail fibs)
I understand the following:
zipWith
literally zips two lists togethertail
grabs all but the first element of a list- Haskell references 'to-be' computed data as
thunks
.
From my understanding, it first adds [0,1,<thunk>]
and [1,<thunk>]
using zipWith (+)
to give [1,<thunk>]
. So now you have
fibs = 0 : 1 : 1 : zipWith (+) fibs (tail fibs)
A lot of references I've Googled have then proceeded to "visualise" the line above as
fibs = 0 : 1 : 1 : zipWith (+) [1,1,<thunk>] ([1,<thunk>]).
My question is this:
Why is the fibs
component in the line above only corresponding to [1,1,<thunk>]
instead of [0,1,1,<thunk>]
?
Shouldn't fibs
contain the entire list plus <thunk>
?
回答1:
This intermediate step is wrong because zipWith
has already processed the first pair of items:
fibs = 0 : 1 : 1 : zipWith (+) fibs (tail fibs)
Recall what zipWith does in the general case:
zipWith f (x:xs) (y:ys) = (f x y) : zipWith f xs ys
If you apply the definition directly you get this expansion:
fibs = 0 : 1 : zipWith (+) fibs (tail fibs) # fibs=[0,1,...]
= 0 : 1 : zipWith (+) [0,1,...] (tail [0,1,...]) # tail fibs=[1,...]
= 0 : 1 : zipWith (+) [0,1,...] [1,...] # apply zipWith
= 0 : 1 : (0+1 : zipWith (+) [1,0+1,...] [0+1,...])
= 0 : 1 : 1 : zipWith (+) [1,1,...] [1,...] # apply zipWith
= 0 : 1 : 1 : (1+1 : zipWith (+) [1,1+1,...] [1+1,...])
= 0 : 1 : 1 : 2 : zipWith (+) [1,2,...] [2,...] # apply zipWith
= 0 : 1 : 1 : 2 : (1+2 : zipWith (+) [2,1+2,...] [1+2,...])
= 0 : 1 : 1 : 2 : 3 : zipWith (+) [2,3...] [3,...] # apply zipWith
:
回答2:
How to visualize what's going on.
1 1 2 3 5 8 13 21 ... <----fibs
1 2 3 5 8 13 21 ... <----The tail of fibs
+_________________________ <----zipWith (+) function
2 3 5 8 13 21 34 ...
Finally, add [1, 1] to the beginning
1, 1, 2, 3, 5, 8, 13, 21, 34, ...
来源:https://stackoverflow.com/questions/26843529/haskell-fibonacci-explanation