How do I print a fibonacci sequence to the nth number in Python?

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忘了有多久
忘了有多久 2020-12-31 22:24

I have a homework assignment that I\'m stumped on. I\'m trying to write a program that outputs the fibonacci sequence up the nth number. Here\'s what I have so far:

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  • 2020-12-31 22:32

    Please note, in your call

    1. You are not calling fib() recursively
    2. You need a wrapper method so that the input is not requested every time the method is called recursively
    3. You do not need to send in a list. Just the number n is good enough.

    This method would only give you the nth number in the sequence. It does not print the sequence.

    You need to return fib(n-1) + fib(n-2)

    def f():
        n = int(input("Please Enter a number: "))
        print fib(n)
    
    def fib(n):    
        if n == 0: 
            return 0
        elif n == 1: 
            return 1
        else: 
            return fib(n-1)+fib(n-2)
    
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  • 2020-12-31 22:37
    def fibonacci(n):
      if n <= 1:
        return n
      else:
        return fibonacci(n-1) + fibonacci(n-2)
    
    print(fibonacci(int(input())))
    

    And since you want to print up to the nth number:

    [print(fibonacci(n)) for n in range (int(input()))]
    

    And for python2.7 change input to raw_input.

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  • 2020-12-31 22:42

    for a recursive solution:

    def fib(n):
        if n == 0:
            return 0
        elif n == 1:
            return 1
        else:
            return fib(n-1) + fib(n-2)
    x=input('which fibonnaci number do you want?')
    print fib(x)
    

    explanation: if n is 0, then ofcourse the '0th' term is 0, and the 1st term is one. From here, you know that the next numbers will be the sum of the previous 2. That is what's inferred by the line after the else.

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  • 2020-12-31 22:45

    This might be faster incase of long list

    # Get nth Fibonacci number 
    def nfib(nth):
      sq5 = 5**.5
      phi1 = (1+sq5)/2
      phi2 = -1 * (phi1 -1)
      resp = (phi1**(nth+1) - phi2**(nth+1))/sq5
      return long(resp)
    
    for i in range(10):
      print i+1, ": ",  nfib(i)
    

    OUTPUT

    1 :  1
    2 :  1
    3 :  2
    4 :  3
    5 :  5
    6 :  8
    7 :  13
    8 :  21
    9 :  34
    10 :  55
    
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  • 2020-12-31 22:46

    Non-recursive solution

    def fib(n):
        cur = 1
        old = 1
        i = 1
        while (i < n):
            cur, old, i = cur+old, cur, i+1
        return cur
    
    for i in range(10):
        print(fib(i))
    

    Generator solution:

    def fib(n):
        old = 0
        cur = 1
        i = 1
        yield cur
        while (i < n):
            cur, old, i = cur+old, cur, i+1
            yield cur
    
    for f in fib(10):
        print(f)
    

    Note that generator solution outperforms the non-recursive (and non-recursive outperforms recursive, if memoization is not applied to recursive solution)

    One more time, recursive with memoization:

    def memoize(func):
        memo = dict()
        def decorated(n):
            if n not in memo:
                memo[n] = func(n)
            return memo[n]
    
        return decorated
    
    @memoize
    def fib(n):
        #added for demonstration purposes only - not really needed
        global call_count
        call_count = call_count + 1
        #end demonstration purposes
    
        if n<=1:
            return 1
        else:
            return fib(n-1) + fib(n-2)
    
    call_count = 0 #added for demonstration purposes only - not really needed
    for i in range(100):
        print(fib(i))
    print(call_count) #prints 100
    

    This time each fibbonacci number calculated exactly once, and than stored. So, this solution would outperform all the rest. However, the decorator implementation is just quick and dirty, don't let it into production. (see this amazing question on SO for details about python decorators :)

    So, having fib defined, the program would be something like (sorry, just looping is boring, here's some more cool python stuff: list comprehensions)

    fib_n = int(input("Fib number?"))
    fibs = [fib(i) for i in range(fib_n)]
    print " ".join(fibs) 
    

    this prints all numbers in ONE line, separated by spaces. If you want each on it's own line - replace " " with "\n"

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  • 2020-12-31 22:48

    Separate functions would be best, as the recursive function would be far easier to deal with. On the other hand, you could code an iterative function that would take only one parameter

    Recursively::

    def fib(n):
        if n == 1:
            return (1);
        elif n == 0:
            return (0);
        else:
            return fib(n-1) + fib(n-2);
    
    def callFib():
        n = int(raw_input('Enter n::\t'));
        mylist = fib(n);
        print mylist;
    
    callFib();
    

    Iteratively::

    def fib():
        n = int(raw_input('Enter n::\t'));
        terms = [0,1];
        i=2;
        while i<=n:
            terms.append(terms[i-1] + terms[i-2]);
            i=i+1;
        print terms[n];
    
    fib();
    
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