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:
Please note, in your call
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)
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 n
th number:
[print(fibonacci(n)) for n in range (int(input()))]
And for python2.7 change input
to raw_input
.
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.
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)
1 : 1
2 : 1
3 : 2
4 : 3
5 : 5
6 : 8
7 : 13
8 : 21
9 : 34
10 : 55
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"
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();