问题
I am interested in an iterative algorithm for Fibonacci numbers, so I found the formula on wiki...it looks straight forward so I tried it in Python...it doesn't have a problem compiling and formula looks right...not sure why its giving the wrong output...did I not implement it right ?
def fib (n):
if( n == 0):
return 0
else:
x = 0
y = 1
for i in range(1,n):
z = (x + y)
x = y
y = z
return y
for i in range(10):
print (fib(i))
output
0
None
1
1
1
1
1
1
回答1:
The problem is that your return y is within the loop of your function. So after the first iteration, it will already stop and return the first value: 1. Except when n is 0, in which case the function is made to return 0 itself, and in case n is 1, when the for loop will not iterate even once, and no return is being execute (hence the None return value).
To fix this, just move the return y outside of the loop.
Alternative implementation
Following KebertX’s example, here is a solution I would personally make in Python. Of course, if you were to process many Fibonacci values, you might even want to combine those two solutions and create a cache for the numbers.
def f(n):
a, b = 0, 1
for i in range(0, n):
a, b = b, a + b
return a
回答2:
You are returning a value within a loop, so the function is exiting before the value of y ever gets to be any more than 1.
If I may suggest something shorter, and much more pythonful:
def fibs(n):
fibs = [0, 1, 1]
for f in range(2, n):
fibs.append(fibs[-1] + fibs[-2])
return fibs[n]
This will do exactly the same thing as your algorithm, but instead of creating three temporary variables, it just adds them into a list, and returns the nth fibonacci number by index.
回答3:
On fib(0), you're returning 0 because:
if (n == 0) {
return 0;
}
On fib(1), you're returning 1 because:
y = 1
return y
On fig(2), you're returning 1 because:
y = 1
return y
...and so on. As long as return y is inside your loop, the function is ending on the first iteration of your for loop every time.
Here's a good solution that another user came up with: How to write the Fibonacci Sequence in Python
回答4:
Non recursive Fibonacci sequence in python
def fibs(n):
f = []
a = 0
b = 1
if n == 0 or n == 1:
print n
else:
f.append(a)
f.append(b)
while len(f) != n:
temp = a + b
f.append(temp)
a = b
b = temp
print f
fibs(10)
Output: [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
回答5:
def fibiter(n):
f1=1
f2=1
tmp=int()
for i in range(1,int(n)-1):
tmp = f1+f2
f1=f2
f2=tmp
return f2
or with parallel assignment:
def fibiter(n):
f1=1
f2=1
for i in range(1,int(n)-1):
f1,f2=f2,f1+f2
return f2
print fibiter(4)
回答6:
I came across this on another thread and it is significantly faster than anything else I have tried and wont time out on large numbers. Here is a link to the math.
def fib(n):
v1, v2, v3 = 1, 1, 0
for rec in bin(n)[3:]:
calc = v2*v2
v1, v2, v3 = v1*v1+calc, (v1+v3)*v2, calc+v3*v3
if rec=='1': v1, v2, v3 = v1+v2, v1, v2
return v2
回答7:
This work (intuitively)
def fib(n):
if n < 2:
return n
o,i = 0,1
while n > 1:
g = i
i = o + i
o = g
n -= 1
return i
回答8:
How about this simple but fastest way... (I just discovered!)
def fib(n):
x = [0,1]
for i in range(n >> 1):
x[0] += x[1]
x[1] += x[0]
return x[n % 2]
Note! as a result, this simple algorithm only uses 1 assignment and 1 addition, since loop length is shorten as 1/2 and each loop includes 2 assignment and 2 additions.
回答9:
fcount = 0 #a count recording the number of Fibonacci numbers generated
prev = 0
current = 0
next = 1
ll = 0 #lower limit
ul = 999 #upper limit
while ul < 100000:
print("The following Fibonacci numbers make up the chunk between %d and %d." % (ll, ul))
while next <= ul:
print(next)
prev = current
current = next
next = prev + current
fcount += 1 #increments count
print("Number of Fibonacci numbers between %d and %d is %d. \n" % (ll, ul, fcount))
ll = ul + 1 #current upper limit, plus 1, becomes new lower limit
ul += 1000 #add 1000 for the new upper limit
fcount = 0 #set count to zero for a new batch of 1000 numbers
回答10:
Another possible approach:
a=0
b=1
d=[a,b]
n=int(input("Enter a number"))
i=2
while i<n:
e=d[-1]+d[-2]
d.append(e)
i+=1
print("Fibonacci series of {} is {}".format(n,d))
回答11:
Assuming these values for the fibonacci sequence:
F(0) = 0;
F(1) = 1;
F(2) = 1;
F(3) = 2
For values of N > 2 we'll calculate the fibonacci value with this formula:
F(N) = F(N-1) + F(N-2)
One iterative approach we can take on this is calculating fibonacci from N = 0 to N = Target_N, as we do so we can keep track of the previous results of fibonacci for N-1 and N-2
public int Fibonacci(int N)
{
// If N is zero return zero
if(N == 0)
{
return 0;
}
// If the value of N is one or two return 1
if( N == 1 || N == 2)
{
return 1;
}
// Keep track of the fibonacci values for N-1 and N-2
int N_1 = 1;
int N_2 = 1;
// From the bottom-up calculate all the fibonacci values until you
// reach the N-1 and N-2 values of the target Fibonacci(N)
for(int i =3; i < N; i++)
{
int temp = N_2;
N_2 = N_2 + N_1;
N_1 = temp;
}
return N_1 + N_2;
}
回答12:
Possible solution:
a=0
b=1
d=[a,b]
n=int(input("Enter a number"))
i=0
while len(d)<n:
temp=a+b
d.append(temp)
a=temp
b=d[i+1]
i+=1
print("Fibonacci series of {} is {}".format(n,d))
回答13:
For specifically your program,you can use the following code
def fib (n):
if( n == 0):
return 0
else:
x = 0
y = 1
for i in range(1,n):
y =y+x
x=y
return y
for i in range(10):
print (fib(i))
Here, x is the old value and y is the current value. In the loop, replace y (current value) with y+x (current value + old value), then assign current value to x.Print current value
回答14:
import time
a,b=0,1
def fibton(n):
if n==1:
time.clock()
return 0,time.clock()
elif n==2:
time.clock()
return 1,time.clock()
elif n%2==0:
read="b"
elif n%2==1:
read="a"
else:
time.clock()
for i in range(1,int(n/2)):
a,b=a+b,a+b
if read=="b":
return b,time.clock()
elif read=="a":
return.a,time.clock()
Disclaimer: I am currently on a mobile device and this may not be totally correct
This algorithm utilizes a gap in some other peoples' and now it is literally twice as fast. Instead of just setting b equal to a or vice versa and then setting a to a+b, I do it twice with only 2 more characters. I also added speed testing, based off of how my other iterative algorithm went. This should be able to go to about the 200,000th Fibonacci number in a second. It also returns the length of the number instead of the whole number, which would take forever.
My other one could go to the second Fibonacci number, as indicated by the built in clock: in 10^-6 seconds. This one can do it in about 5^-6. I'm going to get some more advanced algorithms soon and refine them for utmost speed.
来源：https://stackoverflow.com/questions/15047116/an-iterative-algorithm-for-fibonacci-numbers