Generating Fibonacci numbers in Haskell?
In Haskell, how can I generate Fibonacci numbers based on the property that the nth Fibonacci number is equal to the (n-2)th Fibonacci number plus the (n-1)th Fibonacci number?
I've seen this:
fibs :: [Integer]
fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
I don't really understand that, or how it produces an infinite list instead of one containing 3 elements.
How would I write haskell code that works by calculating the actual definition and not by doing something really weird with list functions?
Here's a simple function that calculates the n'th Fibonacci number:
fib :: Integer -> Integer
fib 0 = 0
fib 1 = 1
fib n = fib (n-1) + fib (n-2)
The function in your question works like this:
Assume you already had an infinite list of the Fibonacci numbers:
[ 1, 1, 2, 3, 5, 8, 13, .... ]
The tail of this list is
[ 1, 2, 3, 5, 8, 13, 21, .... ]
zipWith combines two lists element by element using the given operator:
[ 1, 1, 2, 3, 5, 8, 13, .... ]
+ [ 1, 2, 3, 5, 8, 13, 21, .... ]
= [ 2, 3, 5, 8, 13, 21, 34, .... ]
So the infinite list of Fibonacci numbers can be calculated by prepending the elements 1 and 1 to the result of zipping the infinite list of Fibonacci numbers with the tail of the infinite list of Fibonacci numbers using the + operator.
Now, to get the n'th Fibonacci number, just get the n'th element of the infinite list of Fibonacci numbers:
fib n = fibs !! n
The beauty of Haskell is that it doesn't calculate any element of the list of Fibonacci numbers until its needed.
Did I make your head explode? :)
To expand on dtb's answer:
There is an important difference between the "simple" solution:
fib 0 = 1
fib 1 = 1
fib n = fib (n-1) + fib (n-2)
And the one you specified:
fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
The simple solution takes O(1.618NN) time to compute the Nth element, while the one you specified takes O(N2). That's because the one you specified takes into account that computing fib n and fib (n-1) (which is required to compute it) share the dependency of fib (n-2) , and that it can be computed once for both to save time. O(N2) is for N additions of numbers of O(N) digits.
going by the definition, every item of the fibonacci series is the sum of the previous two terms. putting this definition in to lazy haskell gives u this!
fibo a b = a:fibo b (a+b)
now just take n items from fibo starting with 0,1
take 10 (fibo 0 1)
上一篇: O（fib n）复杂性算法？
下一篇: 在Haskell中生成斐波那契数列？