#38 Python Tutorial for Beginners | Fibonacci Sequence
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- เผยแพร่เมื่อ 26 ต.ค. 2024
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My every morng in this lockdown start with your lecture of python sir ..thanx a lot sir👍👍
Wow really very nice yar😍😍😍,, I am also trying to learn python in this lockdown,,, nd Navin sir lecture is just simply superb ♥️
mine too :)
Ok let's learn together 😄😄👍
Hlo... Hope u r doing well, Which college?
Ur so beautiful in ur profile
# Assignment Fibonacci series
# Thank You Sir.
def fib(num):
if(num
why you take n+1 in for loop range
Cos if you want till num ,should take one number ahead to get that..for eg to get 24 we write 24+1 which gives till 24 value
@@shilpakannan9058it works without adding+1 too
The way you translated the math into code was easier to understand than many other presenters. Thanks a million.
Nice and clear answer for the question:
def fib():
a, b = 0, 1
while a < 100:
print(a, end=" ")
a, b = b, a+b
fib()
Fibonacci series is one of the evergreen topics that always interests both the Mathematicians and the Programmers. Navin's presentations is awesome - as always !!
Sir*
def fib_last(n):
a=0
b=1
if n n:
print(c)
a=0
b=1
print(a)
print(b)
for i in range(100):
c=a+b
if c
thankyou so much for this!
I was actually learning from a Udemy video, but that was a bit confusing so I started to search on TH-cam and I found this video. This video is extremely useful and helped me to understand the concept. Also, the use of visual representation makes things a lot more intuitive and easy. Thanks a lot Sir
I was on the 'exact same ship', but this presenter was way better when it comes to translating a math problem into Python code. Regards!
assignment:
def fib(x):
a=0
b=1
if x
Hello, can you explain how the if statement at last is making the difference
abay, there is two assignment, and that 'last if' is for second assignment.
fib(0) ---> "in valid"
is this correct?..as per above code?
is it....fib(0) ---> 0
n=int(input("enter number "))
a=0
b=1
if n
thanks bro
def fib(num):
lst = [0, 1]
for i in range(20):
n = lst[i] + lst[i+1]
if n < num:
lst.append(n)
if n > num:
break
return lst
x = fib(100)
print(x)
2nd assignment:--
def fib(n):
a = 0
b = 1
if n==1:
print(a)
else:
print(a)
print(b)
for i in range(2,n):
c = a+b
a = b
b = c
if c
u can do , if c>100 break
Assignment solution::
def fib(n):
a=0
b=1
if n100:
break
else:
print(c)
n = int(input("Please enter the no. of value you want in your fibonacci series"))
fib(n)
Without going for if c>100 and adding a break at the end, you could have tried that at the top in the first if statement.
if n100:
If the input is : - 101
Output will be : - "Invalid input"
That would have shortened your code. Keep up the good work!
@@Rajadahana You have not understood the problem..the question says to print sum of numbers less than 100 not total Fibonacci numbers
great one i was trying to use break out of the loop i got error your code is nice
What is the algorithm for this
urs is the best one on the comment section , u truely helped me out thanks
Hi Navin, Your teaching is one of the best. You can only teach us the concept. It is our duty to play and try different approach. You have taught us every thing to create the following function.
This function is memory efficient and you don't have to check for 0 or negative inputs.
def fib(n):
a=0
b=1
for i in range(0,n):
print(a)
a,b = b,a+b
# Fibonacci sequence with arbitrary boundary as input:
def fib(bound):
a, b = 0, 1
while a < bound:
print(a, end=" ")
a, b = b, a+b
print()
@@Svcollections941 call the function. He didn't mention it.
Yeah
#FibonacciSeries
def fib(n):
first,second=0,1
if n
This code is short and amazing thnx bro
@@swapnilshinde2540 thank you!😁
Fibonacci series in reverse order ex.0,-1,-1,-2,-3.....
@@BoomcoreIsLive hmm
@@Rahulpandey-hp2pf
def fib(n):
a=0
b=-1
print(a)
print(b)
for i in range(2,n):
c=a+b
a=b
b=c
print(c)
fib(10)
def num(n):
a=0
b=1
if(n
U forgot ' ' in print.
It should be print('0')
@@dhirajsarkar7246 no it running good at that time
Yeah it's really work ...👍 thanks
I feel this is a better way as there is no need to print first two values, everything is handled by the for loop:
l=int(input("enter length: "))
def fib(l):
x,y = 0,1
for i in range(l):
print(x)
next=x+y
x=y
y=next
fib(l)
yes your answer is good but you have to return a statement to user that his/her input is invalid for that i made some changes in your function, I hope you will like it >>>>>
l=int(input("enter length: "))
def fib(l):
if l
def fib(x):
a=[0,1]
for i in range(x):
b=a[-2]+a[-1]
if b
I tried this code and it worked!
This is bit simpler than yours.
a = 0
b = 1
x = int(input("Please enter a number: "))
while a
Change it to a for loop
can u explain it
issue with the condition for the loop. Here is a more comprehensible code
def fib(x):
a = 0
b = 1
if x==1:
print("0")
elif(x
This is great but it will not work in the while loop as it will stop when a is lesser than x.
So suppose I take x as 5, the output will be :
0
1
1
2
3
5
8
it will stop at 8 because 8(a) is less than 10(x).
For this purpose, we must use for loop keeping x as a range input.
a = 0
b = 1
x = 10
for i in range(x):
print (a)
a,b = b,a+b
def fibonnaci(n):
a = 0
b = 1
sequence = []
while a < n:
sequence.append(a)
c = a + b
a = b
b = c
return sequence
print(fibonnaci(100))
Assignment 2 : To get the fibonacci sequence with condition of Comparison / Identity operators -
n=int(input("Fibonacci Sequence until last term is less than :"))
a=0
b=1
print(a)
print(b)
c=0
while True :
c=a+b
a=b
b=c
if c
When inputted 0, this code would not give 0 as an output so we could add
if n == 1:
print(c)
after defining c and before starting the while loop.
After this, we could even write
elif n < 0:
print('Invalid Input')
so that it doesn't take negative numbers
def fib(n):
a=0
b=1
if(n>0):
if(n==1 ):
print(a)
else:
print(a)
print(b)
for i in range(2,n):
a,b=b,a+b
print(b)
else :
print('enterd negative value')
fib(10)
def fib(n):
a=0
b=1
if n==1:
print(a)
else:
print(a)
print(b)
for i in range (2,n):
c=a+b
a=b
b=c
if c>n:
break
print(c)
fib(n)
the code doesn't work for n=3,4
@@anudeepm7295 it works in for statement range(n) u put
If C
Or
If c>n:
Break
bro the code dosent work
Last assignment in this video:-
def fib(n):
a,b = 0,1
c = 0
l = []
while c
Navin sir,
This can also be done using
for i in range (2,n):
a, b=b, a+b
Print (b)
Therefore you will not have to use a third variable which will make the coding memory efficient 😀
Thanks,
Your student.
Hi,
Thank you for the code which will make coding memory efficient and I tried it and it did work.
One more I want to ask, in for we all know that it won't run till n times for Fibonacci series
for e.g. if n is 100 it loop will not run for 100 times, if it gets any value greater than 100, it will end, if we give n in range will it take more memory, if it will take memory space what would be alternative to do it.
This will also work...
def fib(x):
a=0
b=1
c=0
while c
@@samp_sampath9493 yaa bro...✌️
Really efficient 🔥🙌
Can you explain it Please ?
3rd variable was easy to do 😅
def fib(n):
a = 0
b = 1
print(a)
print(b)
for i in range(2,n):
c = a + b
a = b
b = c
if c
Visual presentation and examples are really helpful. Am using your youtube videos to re-enforce learning things I do not understand in lectures in my course study. Thank you so much!!
Def fib(n):
a=0
b=1
if n==0 or n==1:
print(a)
elif nn:
break
fib(100)
your sessions are amazing and makes python an easy language to learn
the last assignment of this session is very simple put if c
this wont work brother
@@bharathchandran6274 😕😂
Why brother??
Hey it works bro
No actually it should be if c
For numbers upto user specified:-
def fib(n):
a = 0
b = 1
if n
Hey, Can you create program where at the end it prints the sum of the sequence?
n=int(input('Please type the no. of values-'))
def fib(n):
a=0
b=1
c=a+b
if n
n = int(input('Enter the no. of values: '))
def fib(n):
a=0
b=1
if n
Thanks 🙏
def fib(n):
if nn:
break
print(c)
fib(100)
# Assignment 02 to print Fibonacci series for positive number and no of series elements
def fib(n):
a = 0
b = 1
if n == 1:
print(b)
else:
print(a)
print(b)
for i in range(2, n):
c = 0
c = a + b
a = b
b = c
if c > n:
break
else:
print(c)
x = int(input("Enter the no of values you want in the Fibonacci series: "))
if x < 0:
print("Series must have positive value!")
else:
fib(x)
def fibo(num):
a = 0
b = 1
if num
def fib(n):
a=0
b=1
if n
Great bhai....
I think you forgot to include what happens if the input is one (n = 1)....lol
solution:
def fib (n):
a=0
b=1
if n
Assignment(1) checking for negative n value
def feb(n):
a=0
b=1
if n
To print a fibonnaci series upto certain val(Assign 2)
x=int(input('enter a val')
a,b=0
If x>0:
Print(a)
Print(b)
For i in range(1,x):
C=a+b
a,b=b,c
If c
def fib(n):
a = 0
b = 1
i = 0
while i=100:
break
else:
print(a,end=' ')
c = a + b
a = b
b = c
i+=1
fib(100)
Here's my approach towards the assignment. I hope you like it.
def fib():
a = []
n = int(input("enter the value of n : "))
a.append(0)
a.append(1)
for i in range(2,n):
a.append( a[i-1] + a[i-2])
if a[i] >= n:
a.pop(i)
break
print(a)
fib()
how does it works for a.append(a[i-1]+a[i-2]) this equation
@@avinashwagh1058 it work as c=a +b
@@jimpagyatso486 so [i-1] is positioned indexing ??
@@avinashwagh1058 yap...
@@jimpagyatso486 thank you
code for less than 100
def fib(n):
a=0
b=1
while True:
c=a+b
a=b
b=c
if c
If n
def fib(n):
a = 0
b = 1
if input==0:
print(a)
count=1
print("the total numbers", count)
else:
print(a)
print(b)
count=0
for i in range(input):
c = a + b
a, b = b, c
if c > input:
break
print(c)
count+=1
print("the total numbers", count+2)
input = int(input('enter the number'))
fib(input)
To end code at number that’s below 100
for I in range (2, n):
c = a + b
a = b
b = c
if c >= 100:
break
print (c)
def fibo(n):
a=0
b=1
print(a)
print(b)
c=0
while cn:
break
print(c)
fibo(100)
fib = lambda n: 0 if n == 0 else ( 1 if n == 1 else fib(n-1) + fib(n-2))
for i in range(10):
print(fib(i))
a=0
b=1
c=0
k=int(input("specify limit"))
while ck:
break
a=b
b=c
print(c)
def fib(n):
a=0
b=1
if n==1:
print(a)
elif n100:
print("sum of fib is greater than 100")
break
else:
print(c)
fib(100)
If I become a programmer in future, because of Navin sir: Great sir.
Did you?
did you?
. . . . . .
@@Thedeepanshu95 i think hes dead
Did you?
n = int(input("Enter a number: "))
def fib(n):
if n
assignment 2:
i have taken range upto 50 numbers as the user only mentions a value and we have to print the value in the fibonacci series which is less than the user entry value
def fib(n):
a=0
b=1
if n==0:
print("Invalid")
elif n==1:
print("The value before",n,"is :",0)
else:
for i in range(1,51):
if i=n and b
Great
to find number that is less than the entered number
def fib(n):
a,b=0,1
if n=0 :
a,b=b,a+b
if b
If a,b
my assignmnt with differen way:
def fib(num):
a = 0
b = 1
e =[]
if num
Thank you so much for the great breakdown! I was having trouble understanding how to do this via my textbook and after watching your video I finally get it!
def fib(n):
a=0
b=1
print(a)
print(b)
for i in range(2,n):
c=a+b
a=b
b=c
if c>=n:
break
else:
print(c)
fib(100)
I think this is right and I hope this helps! Feel free to tell me if its wrong I'm a learner too:
def fibo(num):
a = 0
b = 1
c = 0
if num == 0:
print(a)
else:
print(a)
print(b)
while num > c + a:
c = a + b
a = b
b = c
print(c)
thelast = int(input('The last number of the Fibonnaci sequence you wish to see '))
fibo(thelast)
def fib(n):
a=0
b=1
if n
Hello Telusko, kindly make a series on programming challenges/questions only. This can help us much!
right
To print the Fibonacci number less than the entered value:
def fibo(n):
a=0
b=1
if n==1:
print(0)
elif n
Jus add if condition after the def function with proper indentations
If n< 1:
Print(“invalid”)
Else:
Pass
.
.
.
.
For loop
.
.
.
.
Fib(5)
What if you wanted the program to print the negative fib numbers?
def fib(n):
a=0
b=1
print(a)
print(b)
for i in range(n):
c= a+b
a=b
b=c
if c
Sir, you are awesome. Please keep posting more videos.
And about the for loop, for(start,end,step) --> for loop will print till end- step
say for instance you have for i in range(1,100,1)-- it will print till 99[end(100) - step(1)]
I felt that part was not clear
Thank You sir
yes, can you clarify the end part plz
def fib(n):
a=0
b=1
d=0
if n c:
d = c
print(d)
you are best tutor sir.thank u so much sir.really helped a lot in my exams
n = int(input("Enter the numnber of sequence needed to fabonacci series: ", ))
first = 0
second = 1
if n == first:
print("Series : ", first)
else:
print(first)
print(second)
for i in range(n):
next = first + second
first = second
second = next
print(next)
range should be n-2 not n
1st and 2nd assignment solution:
def fibi(m):
if mm:
break
a = b
b = c
print(c)
fibi(-1)
fibi(20)
seeing so many answers, i think we can do it more simple. In the start of the function just check if value < 0 then just return nothing. Something like this
if(n
Why not a break is not applicable in the code
@@vinaysudheer474 breaks exists within loops like for and while
def fib(n):
a=0
b=1
if n < 0:
print("invalid")
if n==1:
print(a)
else:
print(a)
print(b)
for i in range(2,n):
c=a+b
a=b
b=c
print(c)
fib(-1)
For number less than 0:
def fib(n):
a = 0
b = 1
if n
def fib(n):# task 3 { Fibonacci series }
a=0
b=1
if n==1:
print(a)
elif nn:
break
print(b,' ',end="")
n=int(input('enter the number'))
fib(n)
How it is sir?
use elif n
def fib(x):
a = 0
b = 1
i = 0
print(a)
print(b)
while a+b
I guess it is possible even without 'i' variable
Total fibonacci series with no errors. Can play as many times as you want without running each time.(1st assignment)
def fib(n):
a = 0
b = 1
if n==1:
print(a)
else:
print(a)
print(b)
for i in range(2,n):
c = a+b
a = b
b = c
print(c)
a = 1
while a==1:
n = input("Enter the number of sequences you want for fibonacci sequence:")
try:
n = int(n)
except:
print("String's can't be used with fibanacci series.")
continue
if n
def fib():
n=int(input("the length of series is"))
if n
Is I should be equal to 2 or 0?
It is i*
If c>n
Break
My Assignment :
def fib(n):
a, b = 0, 1
if n == 1: print(a)
else:
print(a, b, sep='
')
for i in range(2, n):
c = a + b
a, b = b, c
if c > 100: break
print(c)
fib(100)
def fib(n):
a=0
b=1
if n>=2:
print(a)
print(b)
for i in range(n-2):
c=a+b
b=c
a=b-a
print(c)
else:
if n==1:
print(a)
else:
pass
fib(0)
This works too. Thanks Navin, your teaching is great
n = int(input("Enter length : "))
a = [0,1]
n= n-1
for i in range(1,n):
a.append(a[i]+a[i-1])
print(a)
There's a small fault in this program. If you enter n=2, output will be 0 1 1 ie 3 outputs because before for loop, 2 output will be printed and in for loop, 2 will give Boolean true ie for loop will run once giving 3rd output
No, if n=2 the loop will not even start
Assign 2
def fib(n):
if n>=0:
a=0
b=1
print(a)
print(b)
else:
print("should not be negtive number")
for i in range(2,n):
c=a+b
a=b
b=c
If c
Nice
Simply put if condition before c if c>100: break
mylist = list()
x = int(input("Enter how many numbers you want in fibonacci secquence :"))
def fibo():
if x = 0:
print(a)
print(b)
for i in range(2,x):
c = a + b
a = b
b = c
if c >= 100:
break
else:
print(c)
mylist.append(i)
fibo()
Sir, here's my project along with both the assignment done at the same and thanks a lot for teaching us wonderful stuff about python
Ba bye...
#Fibonacci Sequence using functions
def fib(n):
a=0
b=1
if n==1:
print(a)
elif n n):
break
print(a)
fib(150)
Result will be 144
It should be print(c) in the inner if condition
for finding fibonacci numbers within input range:
def fib(x):
a = 0
b = 1
print(a)
print(b)
for i in range(2,x):
c=a+b
a=b
b=c
if c
Please start some project on Java or python. Not live project in which only few can participate but For every one. All can make on their own by seeing your videos it will be helpful for our resume...... small and large projects both will be helpful for widening the concept.
Very Good Idea 💡
Yess
Really needed
Really need your help sir
sir do some project in python,please
2nd Assignment:
def fib():
g =(int(input("enter the number")))
a = 0
b = 1
if g == 1:
print(a)
elif (g > 1):
print(a)
print(b)
for i in range(2, g):
c = a + b
a = b
b = c
if c
How to program Catalan numbers? I would like to request you, can you make a short video on this?
to print nth catalan number :
def fact(n):
a=1
for i in range(1,n+1):
a*=i
return a
n=int(input())
n-=1
cat=fact(2*n)/(fact(n+1)*fact(n))
print(cat)
to print series of catalan numbers upto n:
def fact(n):
a=1
for i in range(1,n+1):
a*=i
return a
n=int(input())
cat=fact(2*n)/(fact(n+1)*fact(n))
for i in range(n):
cat = fact(2 * i) / (fact(i + 1) * fact(i))
print(cat)
def fibo(n):
a=0
b=1
sum=1
if n==1:
print(0)
elif n
def fib(n):
result = []
for i in range(n):
if i==0:
result.append(0)
elif i==1:
result.append(1)
else:
result.append(result[i-1]+result[i-2])
print(result)
fib(10)
def fib(n):
if n
def fib(n):
if n
@@TheSwatisadhukhan thank you
Wow
x=int(input("Enter the positive integer: "))
def fib(n):
a=0
b=1
if n==1:
print(a)
else:
print(a)
print(b)
for _ in range(2,x):
c=a+b
a=b
b=c
if c
we can simply use elif after if
elif n
Quite obvious
that's exactly how i did :)
lst=[]
t=int(input("Enter how many terms: "))
a,b=0,1
n=int(input("Enter the last range: "))
for i in range(0,t):
if a
very coherent , very easy. Thanks for the series.
in case you want the last assignment, I wrote this code and it works but I'm not sure it's the most optimized way.
.
.
def fib(n):
a = 0
b = 1
if n == 1:
print(a)
elif n
Due to your Lectures our knowledge × n
Thank-you Sir
where n = infinite
From where can we get the assignment solutions to verify
x = int(input("Enter upper limit of Fibonacci sequence"))
def fib(x):
a = 0
b = 1
c = 1
while c < x:
c = a + b
a = b
b = c
else:
print(a)
fib(x)
Sir you are great
Finally i had clear all dought in python🤙❤
Your Student - Alien 👽
Phle doubt ki spelling to clear kar le 😂 "dought"
@@islay.okay.3630 hahaha
@@islay.okay.3630 😂😂😂
assigment 2
def fib(n):
a=0
b=1
if n==1:
print(a)
else:
print(a)
print(b)
for i in range(2,n):
c=a+b
a=b
b=c
if c
if c
It's wrong because -6