I had originally coded the program wrongly. Instead of returning the Fibonacci numbers between a range (ie. startNumber 1, endNumber 20 should = only those numbers between 1 & 20), I have written for the program to display all Fibonacci numbers between a range (ie. startNumber 1, endNumber 20 displays = First 20 Fibonacci numbers). I thought I had a sure-fire code. I also do not see why this is happening.

```
startNumber = int(raw_input("Enter the start number here "))
endNumber = int(raw_input("Enter the end number here "))
def fib(n):
if n < 2:
return n
return fib(n-2) + fib(n-1)
print map(fib, range(startNumber, endNumber))
```

Someone pointed out in my Part II (which was closed for being a duplicate – https://stackoverflow.com/questions/504193/how-to-write-the-fibonacci-sequence-in-python-part-ii) that I need to pass the startNumber and endNumber through a generator using a while loop. Can someone please point me in the direction on how to do this? Any help is welcome.

I’m a learning programmer and I’ve run into a bit of a jumble. I am asked to write a program that will compute and display Fibonacci’s Sequence by a user inputted start number and end number (ie. startNumber = 20 endNumber = 100 and it will display only the numbers between that range). The trick is to use it inclusively (which I do not know how to do in Python? – I’m assuming this means to use an inclusive range?).

What I have so far is no actual coding but rather:

- Write Fib sequence formula to infinite
- Display startNumber to endNumber only from Fib sequence.

I have no idea where to start and I am asking for ideas or insight into how to write this. I also have tried to write the Fib sequence forumla but I get lost on that as well.

There is lots of information about the Fibonacci Sequence on wikipedia and on wolfram. A lot more than you may need. Anyway it is a good thing to learn how to use these resources to find (quickly if possible) what you need.

## Write Fib sequence formula to infinite

In math, it’s given in a recursive form:

In programming, **infinite** doesn’t exist. You can use a recursive form translating the math form directly in your language, for example in Python it becomes:

```
def F(n):
if n == 0: return 0
elif n == 1: return 1
else: return F(n-1)+F(n-2)
```

Try it in your favourite language and see that this form requires **a lot** of time as n gets bigger. In fact, this is O(2^{n}) in time.

Go on on the sites I linked to you and will see this (on wolfram):

This one is pretty easy to implement and very, very fast to compute, in Python:

```
from math import sqrt
def F(n):
return ((1+sqrt(5))**n-(1-sqrt(5))**n)/(2**n*sqrt(5))
```

An other way to do it is following the definition (from wikipedia):

The first number of the sequence is 0,

the second number is 1, and each

subsequent number is equal to the sum

of the previous two numbers of the

sequence itself, yielding the sequence

0, 1, 1, 2, 3, 5, 8, etc.

If your language supports iterators you may do something like:

```
def F():
a,b = 0,1
while True:
yield a
a, b = b, a + b
```

## Display startNumber to endNumber only from Fib sequence.

Once you know how to generate Fibonacci Numbers you just have to cycle trough the numbers and check if they verify the given conditions.

Suppose now you wrote a f(n) that returns the n-th term of the Fibonacci Sequence (like the one with sqrt(5) )

In most languages you can do something like:

```
def SubFib(startNumber, endNumber):
n = 0
cur = f(n)
while cur <= endNumber:
if startNumber <= cur:
print cur
n += 1
cur = f(n)
```

In python I’d use the iterator form and go for:

```
def SubFib(startNumber, endNumber):
for cur in F():
if cur > endNumber: return
if cur >= startNumber:
yield cur
for i in SubFib(10, 200):
print i
```

My hint is to *learn to read* what you need. Project Euler (google for it) will train you to do so ðŸ˜›

Good luck and have fun!

# Efficient Pythonic generator of the Fibonacci sequence

I found this question while trying to get the shortest Pythonic generation of this sequence (later realizing I had seen a similar one in a Python Enhancement Proposal), and I haven’t noticed anyone else coming up with my specific solution (although the top answer gets close, but still less elegant), so here it is, with comments describing the first iteration, because I think that may help readers understand:

```
def fib():
a, b = 0, 1
while True: # First iteration:
yield a # yield 0 to start with and then
a, b = b, a + b # a will now be 1, and b will also be 1, (0 + 1)
```

and usage:

```
for index, fibonacci_number in zip(range(10), fib()):
print('{i:3}: {f:3}'.format(i=index, f=fibonacci_number))
```

prints:

```
0: 0
1: 1
2: 1
3: 2
4: 3
5: 5
6: 8
7: 13
8: 21
9: 34
10: 55
```

(For attribution purposes, I recently noticed a similar implementation in the Python documentation on modules, even using the variables `a`

and `b`

, which I now recall having seen before writing this answer. But I think this answer demonstrates better usage of the language.)

# Recursively defined implementation

The Online Encyclopedia of Integer Sequences defines the Fibonacci Sequence recursively as

F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1

Succinctly defining this recursively in Python can be done as follows:

```
def rec_fib(n):
'''inefficient recursive function as defined, returns Fibonacci number'''
if n > 1:
return rec_fib(n-1) + rec_fib(n-2)
return n
```

But this exact representation of the mathematical definition is incredibly inefficient for numbers much greater than 30, because each number being calculated must also calculate for every number below it. You can demonstrate how slow it is by using the following:

```
for i in range(40):
print(i, rec_fib(i))
```

# Memoized recursion for efficiency

It can be memoized to improve speed (this example takes advantage of the fact that a default keyword argument is the same object every time the function is called, but normally you wouldn’t use a mutable default argument for exactly this reason):

```
def mem_fib(n, _cache={}):
'''efficiently memoized recursive function, returns a Fibonacci number'''
if n in _cache:
return _cache[n]
elif n > 1:
return _cache.setdefault(n, mem_fib(n-1) + mem_fib(n-2))
return n
```

You’ll find the memoized version is much faster, and will quickly exceed your maximum recursion depth before you can even think to get up for coffee. You can see how much faster it is visually by doing this:

```
for i in range(40):
print(i, mem_fib(i))
```

(It may seem like we can just do the below, but it actually doesn’t let us take advantage of the cache, because it calls itself before setdefault is called.)

```
def mem_fib(n, _cache={}):
'''don't do this'''
if n > 1:
return _cache.setdefault(n, mem_fib(n-1) + mem_fib(n-2))
return n
```

## Recursively defined generator:

As I have been learning Haskell, I came across this implementation in Haskell:

```
[email protected](0:tfib) = 0:1: zipWith (+) fib tfib
```

The closest I think I can get to this in Python at the moment is:

```
from itertools import tee
def fib():
yield 0
yield 1
# tee required, else with two fib()'s algorithm becomes quadratic
f, tf = tee(fib())
next(tf)
for a, b in zip(f, tf):
yield a + b
```

This demonstrates it:

```
[f for _, f in zip(range(999), fib())]
```

It can only go up to the recursion limit, though. Usually, 1000, whereas the Haskell version can go up to the 100s of millions, although it uses all 8 GB of my laptop’s memory to do so:

```
> length $ take 100000000 fib
100000000
```

### Consuming the iterator to get the nth fibonacci number

A commenter asks:

Question for the Fib() function which is based on iterator: what if you want to get the nth, for instance 10th fib number?

The itertools documentation has a recipe for this:

```
from itertools import islice
def nth(iterable, n, default=None):
"Returns the nth item or a default value"
return next(islice(iterable, n, None), default)
```

and now:

```
>>> nth(fib(), 10)
55
```

Why not simply do the following?

```
x = [1,1]
for i in range(2, 10):
x.append(x[-1] + x[-2])
print(', '.join(str(y) for y in x))
```

The idea behind the Fibonacci sequence is shown in the following Python code:

```
def fib(n):
if n == 1:
return 1
elif n == 0:
return 0
else:
return fib(n-1) + fib(n-2)
```

This means that fib is a function that can do one of three things. It defines fib(1) == 1, fib(0) == 0, and fib(n) to be:

fib(n-1) + fib(n-2)

Where n is an arbitrary integer. This means that fib(2) for example, expands out to the following arithmetic:

```
fib(2) = fib(1) + fib(0)
fib(1) = 1
fib(0) = 0
# Therefore by substitution:
fib(2) = 1 + 0
fib(2) = 1
```

We can calculate fib(3) the same way with the arithmetic shown below:

```
fib(3) = fib(2) + fib(1)
fib(2) = fib(1) + fib(0)
fib(2) = 1
fib(1) = 1
fib(0) = 0
# Therefore by substitution:
fib(3) = 1 + 1 + 0
```

The important thing to realize here is that fib(3) can’t be calculated without calculating fib(2), which is calculated by knowing the definitions of fib(1) and fib(0). Having a function call itself like the fibonacci function does is called recursion, and it’s an important topic in programming.

This sounds like a homework assignment so I’m not going to do the start/end part for you. Python is a wonderfully expressive language for this though, so this should make sense if you understand math, and will hopefully teach you about recursion. Good luck!

Edit: One potential criticism of my code is that it doesn’t use the super-handy Python function yield, which makes the fib(n) function a lot shorter. My example is a little bit more generic though, since not a lot of languages outside Python actually have yield.

**Time complexity :**

The caching feature reduces the normal way of calculating Fibonacci series from **O(2^n)** to **O(n)** by eliminating the repeats in the recursive tree of Fibonacci series :

**Code :**

```
import sys
table = [0]*1000
def FastFib(n):
if n<=1:
return n
else:
if(table[n-1]==0):
table[n-1] = FastFib(n-1)
if(table[n-2]==0):
table[n-2] = FastFib(n-2)
table[n] = table[n-1] + table[n-2]
return table[n]
def main():
print('Enter a number : ')
num = int(sys.stdin.readline())
print(FastFib(num))
if __name__=='__main__':
main()
```

This is quite efficient, using O(log n) basic arithmetic operations.

```
def fib(n):
return pow(2 << n, n + 1, (4 << 2*n) - (2 << n) - 1) % (2 << n)
```

This one uses O(1) basic arithmetic operations, but the size of the intermediate results is large and so is not at all efficient.

```
def fib(n):
return (4 << n*(3+n)) // ((4 << 2*n) - (2 << n) - 1) & ((2 << n) - 1)
```

This one computes X^n in the polynomial ring Z[X] / (X^2 – X – 1) using exponentiation by squaring. The result of that calculation is the polynomial Fib(n)X + Fib(n-1), from which the nth Fibonacci number can be read.

Again, this uses O(log n) arithmetic operations and is very efficient.

```
def mul(a, b):
return a[0]*b[1]+a[1]*b[0]+a[0]*b[0], a[0]*b[0]+a[1]*b[1]
def fib(n):
x, r = (1, 0), (0, 1)
while n:
if n & 1: r = mul(r, x)
x = mul(x, x)
n >>= 1
return r[0]
```

Canonical Python code to print Fibonacci sequence:

```
a,b=1,1
while True:
print a,
a,b=b,a+b # Could also use b=a+b;a=b-a
```

For the problem “Print the first Fibonacci number greater than 1000 digits long”:

```
a,b=1,1
i=1
while len(str(a))<=1000:
i=i+1
a,b=b,a+b
print i,len(str(a)),a
```

We know that

And that The n-th power of that matrix gives us:

So we can implement a function that simply computes the power of that matrix to the n-th -1 power.

as all we know the power a^n is equal to

So at the end the fibonacci function would be O( n )… nothing really different than an easier implementation if it wasn’t for the fact that we also know that `x^n * x^n = x^2n`

and the evaluation of `x^n`

can therefore be done with complexity O( log n )

Here is my fibonacci implementation using swift programming language:

```
struct Mat {
var m00: Int
var m01: Int
var m10: Int
var m11: Int
}
func pow(m: Mat, n: Int) -> Mat {
guard n > 1 else { return m }
let temp = pow(m: m, n: n/2)
var result = matMultiply(a: temp, b: temp)
if n%2 != 0 {
result = matMultiply(a: result, b: Mat(m00: 1, m01: 1, m10: 1, m11: 0))
}
return result
}
func matMultiply(a: Mat, b: Mat) -> Mat {
let m00 = a.m00 * b.m00 + a.m01 * b.m10
let m01 = a.m00 * b.m01 + a.m01 * b.m11
let m10 = a.m10 * b.m00 + a.m11 * b.m10
let m11 = a.m10 * b.m01 + a.m11 * b.m11
return Mat(m00: m00, m01: m01, m10: m10, m11: m11)
}
func fibonacciFast(n: Int) -> Int {
guard n > 0 else { return 0 }
let m = Mat(m00: 1, m01: 1, m10: 1, m11: 0)
return pow(m: m, n: n-1).m00
}
```

This has complexity O( log n ). We compute the oÃ¬power of Q with exponent n-1 and then we take the element m00 which is Fn+1 that at the power exponent n-1 is exactly the n-th Fibonacci number we wanted.

Once you have the fast fibonacci function you can iterate from start number and end number to get the part of the Fibonacci sequence you are interested in.

```
let sequence = (start...end).map(fibonacciFast)
```

of course first perform some check on start and end to make sure they can form a valid range.

I know the question is 8 years old, but I had fun answering anyway. ðŸ™‚

Fibonacci sequence is: `1, 1, 2, 3, 5, 8, ...`

.

That is `f(1) = 1`

, `f(2) = 1`

, `f(3) = 2`

, `...`

, `f(n) = f(n-1) + f(n-2)`

.

My favorite implementation (simplest and yet achieves a light speed in compare to other implementations) is this:

```
def fibonacci(n):
a, b = 0, 1
for _ in range(1, n):
a, b = b, a + b
return b
```

**Test**

```
>>> [fibonacci(i) for i in range(1, 10)]
[1, 1, 2, 3, 5, 8, 13, 21, 34]
```

**Timing**

```
>>> %%time
>>> fibonacci(100**3)
CPU times: user 9.65 s, sys: 9.44 ms, total: 9.66 s
Wall time: 9.66 s
```

Edit: an example visualization for this implementations.

use recursion:

```
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 do you want?')
print fib(x)
```

Another way of doing it:

```
a,n=[0,1],10
map(lambda i: reduce(lambda x,y: a.append(x+y),a[-2:]),range(n-2))
```

Assigning list to ‘a’, assigning integer to ‘n’

Map and reduce are 2 of three most powerful functions in python. Here map is used just to iterate ‘n-2’ times.

a[-2:] will get the last two elements of an array.

a.append(x+y) will add the last two elements and will append to the array

These all look a bit more complicated than they need to be.

My code is very simple and fast:

```
def fibonacci(x):
List = []
f = 1
List.append(f)
List.append(f) #because the fibonacci sequence has two 1's at first
while f<=x:
f = List[-1] + List[-2] #says that f = the sum of the last two f's in the series
List.append(f)
else:
List.remove(List[-1]) #because the code lists the fibonacci number one past x. Not necessary, but defines the code better
for i in range(0, len(List)):
print List[i] #prints it in series form instead of list form. Also not necessary
```

OK.. after being tired of referring all lengthy answers, now find the below sort & sweet, pretty straight forward way for implementing Fibonacci in python. You can enhance it it the way you want by getting an argument or getting user inputâ€¦or change the limits from 10000. As you needâ€¦â€¦

```
def fibonacci():
start = 0
i = 1
lt = []
lt.append(start)
while start < 10000:
start += i
lt.append(start)
i = sum(lt[-2:])
lt.append(i)
print "The Fibonaccii series: ", lt
```

This approach also performs good. Find the run analytics below

```
In [10]: %timeit fibonacci
10000000 loops, best of 3: 26.3 ns per loop
```

this is an improvement to mathew henry’s answer:

```
def fib(n):
a = 0
b = 1
for i in range(1,n+1):
c = a + b
print b
a = b
b = c
```

the code should print b instead of printing c

output: 1,1,2,3,5 ….

Using for loop and print just the result

```
def fib(n:'upto n number')->int:
if n==0:
return 0
elif n==1:
return 1
a=0
b=1
for i in range(0,n-1):
b=a+b
a=b-a
return b
```

Result

```
>>>fib(50)
12586269025
>>>>
>>> fib(100)
354224848179261915075
>>>
```

Print the `list`

containing all the numbers

```
def fib(n:'upto n number')->int:
l=[0,1]
if n==0:
return l[0]
elif n==1:
return l
a=0
b=1
for i in range(0,n-1):
b=a+b
a=b-a
l.append(b)
return l
```

Result

```
>>> fib(10)
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
```

```
import time
start_time = time.time()
#recursive solution
def fib(x, y, upperLimit):
return [x] + fib(y, (x+y), upperLimit) if x < upperLimit else [x]
#To test :
print(fib(0,1,40000000000000))
print("run time: " + str(time.time() - start_time))
```

Results

[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073, 4807526976, 7778742049, 12586269025, 20365011074, 32951280099, 53316291173, 86267571272, 139583862445, 225851433717, 365435296162, 591286729879, 956722026041, 1548008755920, 2504730781961, 4052739537881, 6557470319842, 10610209857723, 17167680177565, 27777890035288, 44945570212853]

run time: **0.04298138618469238**

**there is a very easy method to realize that!**

you can run this code online freely by using http://www.learnpython.org/

```
# Set the variable brian on line 3!
def fib(n):
"""This is documentation string for function. It'll be available by fib.__doc__()
Return a list containing the Fibonacci series up to n."""
result = []
a = 0
b = 1
while a < n:
result.append(a) # 0 1 1 2 3 5 8 (13) break
tmp_var = b # 1 1 2 3 5 8 13
b = a + b # 1 2 3 5 8 13 21
a = tmp_var # 1 1 2 3 5 8 13
# print(a)
return result
print(fib(10))
# result should be this: [0, 1, 1, 2, 3, 5, 8]
```

It can be done by the following way.

n = 0 numbers = [0] for i in range(0,11): print n, numbers.append(n) prev = numbers[-2] if n == 0: n = 1 else: n = n + prev

If you’re a fan of recursion you can cache the results easily with the `lru_cache`

decorator (Least-recently-used cache decorator)

```
from functools import lru_cache
@lru_cache()
def fib(n):
if n == 0:
return 0
elif n == 1:
return 1
else:
return fib(n-1) + fib(n-2)
```

If you need to cache more than 128 values you can pass `maxsize`

as an argument to the `lru_cache`

(e.g. `lru_cache(maxsize=500)`

. If you set `maxsize=None`

the cache can grow without bound.

Just for fun, in Python 3.8+ you can use an assignment expression (aka the walrus operator) in a list comprehension, e.g.:

```
>>> a, b = 0, 1
>>> [a, b] + [b := a + (a := b) for _ in range(8)] # first 10 Fibonacci numbers
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
```

An assignment expression allows you to assign a value to a variable *and* return it in the same expression. Therefore, the expression

```
b := a + (a := b)
```

is equivalent to executing

```
a, b = b, a + b
```

and returning the value of `b`

.

The fibonacci sequence if you don’t know what that is basically you print out a number and each each number that you print out is the previous two numbers added together.

```
a, b = 0,1
for i in range(0, 10):
print(a)
a, b = b, a + b
```

output:

```
0
1
1
2
3
5
8
13
21
34
```

So 0 and 1 if you add those together it equals 1 if you add 1 and 1 it equals 2 if you add 1 and 2 it equals 3 and it keeps going and keeps going.

Fibonacci sequence here that I’ve written using generators:

```
def fib(num):
a, b = 0,1
for i in range(0, num):
yield "{}: {}".format(i+1, a)
a, b = b, a + b
for item in fib(10):
print(item)
```

output:

```
1: 0
2: 1
3: 1
4: 2
5: 3
6: 5
7: 8
8: 13
9: 21
10: 34
```

Fibonacci sequence that we went over before except now we have a function here `yield`

s and `yield`

is the keyword that lets you know that it’s a generator.

- So, it yields your result and then we can loop through the generator and print out each item.
- If we run through then we can see that it still works just like it worked before but now we’re using generators instead which have more advantages over returning a list but there are times and when you wouldn’t want to use generators.
- You need do to your research and figure out when you really do get those advantages from generators and when they may not be the best option for you.

Optimized function of finding Fibonacci by keeping list in memory

```
def fib(n, a=[0, 1]):
while n > len(a):
a.append(a[-1] + a[-2])
return a[n-1]
print("Fibonacci of 50 - {}".format(fib(50))
```

15 minutes into a tutorial I used when learning Python, it asked the reader to write a program that would calculate a Fibonacci sequence from 3 input numbers (first Fibonacci number, second number, and number at which to stop the sequence). The tutorial had only covered variables, if/thens, and loops up to that point. No functions yet. I came up with the following code:

```
sum = 0
endingnumber = 1
print "n.:Fibonacci sequence:.n"
firstnumber = input("Enter the first number: ")
secondnumber = input("Enter the second number: ")
endingnumber = input("Enter the number to stop at: ")
if secondnumber < firstnumber:
print "nSecond number must be bigger than the first number!!!n"
else:
while sum <= endingnumber:
print firstnumber
if secondnumber > endingnumber:
break
else:
print secondnumber
sum = firstnumber + secondnumber
firstnumber = sum
secondnumber = secondnumber + sum
```

As you can see, it’s really inefficient, but it DOES work.

```
def fib():
a,b = 1,1
num=eval(input("Please input what Fib number you want to be calculated: "))
num_int=int(num-2)
for i in range (num_int):
a,b=b,a+b
print(b)
```

Just going through http://projecteuler.net/problem=2 this was my take on it

```
# Even Fibonacci numbers
# Problem 2
def get_fibonacci(size):
numbers = [1,2]
while size > len(numbers):
next_fibonacci = numbers[-1]+numbers[-2]
numbers.append(next_fibonacci)
print numbers
get_fibonacci(20)
```

```
def fib(x, y, n):
if n < 1:
return x, y, n
else:
return fib(y, x + y, n - 1)
print fib(0, 1, 4)
(3, 5, 0)
#
def fib(x, y, n):
if n > 1:
for item in fib(y, x + y, n - 1):
yield item
yield x, y, n
f = fib(0, 1, 12)
f.next()
(89, 144, 1)
f.next()[0]
55
```

Maybe this will help

```
def fibo(n):
result = []
a, b = 0, 1
while b < n:
result.append(b)
a, b = b, b + a
return result
```

based on classic fibonacci sequence and just for the sake of the one-liners

if you just need the number of the index, you can use the reduce

(even if *reduce* it’s not best suited for this it can be a good exercise)

```
def fibonacci(index):
return reduce(lambda r,v: r.append(r[-1]+r[-2]) or (r.pop(0) and 0) or r , xrange(index), [0, 1])[1]
```

and to get the complete array just remove the *or (r.pop(0) and 0)*

```
reduce(lambda r,v: r.append(r[-1]+r[-2]) or r , xrange(last_index), [0, 1])
```

How about this one? I guess it’s not as fancy as the other suggestions because it demands the initial specification of the previous result to produce the expected output, but I feel is a very readable option, i.e., all it does is to provide the result and the previous result to the recursion.

```
#count the number of recursions
num_rec = 0
def fibonacci(num, prev, num_rec, cycles):
num_rec = num_rec + 1
if num == 0 and prev == 0:
result = 0;
num = 1;
else:
result = num + prev
print(result)
if num_rec == cycles:
print("done")
else:
fibonacci(result, num, num_rec, cycles)
#Run the fibonacci function 10 times
fibonacci(0, 0, num_rec, 10)
```

Here’s the output:

```
0
1
1
2
3
5
8
13
21
34
done
```

Basically translated from Ruby:

```
def fib(n):
a = 0
b = 1
for i in range(1,n+1):
c = a + b
print c
a = b
b = c
```

…