A simple fibonacci number quiz

Question

I have a number 726, 726 isn't fibo, then sum it with reverse of it self, 726+627, it will keep sum until the number is fibonacci and print out how many times of sum needed until it fibo number.
constraint :
1 < n < 1000
n = times of sum needed until it is fibo number.

Answer 1

I hope it's okay to interject... the problem is that you're dropping to floats (square root). Python has arbitrary *integer* precision, but *float* precision is 15.8 digits. Further, the answer you get depends on the underlying C library, which varies across operating systems (a separate issue I've been tracking from a discrepancy in Numpy; but again, it's the OS and not Python).
I can show the "float" issue most relevantly to SoloLearn by modifying Sayan's code from the "stairs-lambdas" challenge:
import math, operator
from functools import reduce
# desired fib() step (actually step 80)
x = 78
c=lambda n,a=1,b=0 : c(n-1, a+b, a) if n>1 else str(a+b)
d=lambda n: str(round((1/(5)**0.5)*((0.5*(1+(5)**0.5))**(1+n))))
d1=lambda n: "{}".format(round(1./math.sqrt(5) * pow( ( 1+math.sqrt(5) ) /2, 1+n) ))
d2=lambda n: "{}".format(round(reduce(operator.mul,
[( 1+math.sqrt(5) ) for _ in range(1+n)],
1
) / (2**(n+1)*math.sqrt(5))
))
print(c(x))
print(d(x), d1(x), d2(x), sep="\n")
OUTPUT:
------------------------------------
14472334024676221 # correct
14472334024676260 # incorrect
14472334024676260 # incorrect
14472334024676254 # slightly less incorrect (less division)
^ 15.8 digits of precision
# accuracy of floats (cast to integers)
Compare to Javascript, which loses integer precision at step 80 ... because all numbers in Javascript are floats:
https://code.sololearn.com/W0izfx5FISNx/?ref=app

Answer 2

heres mine....( 2 versions )
just input a number##//
https://code.sololearn.com/cqBK8HFm3ZF2/?ref=app
https://code.sololearn.com/cMrmfUa1CIJ6/?ref=app

Answer 3

sayan Chandra solution are very interesting @@, but unfortunately I can't python :>

Answer 4

@Sayan Maybe, but my Python code gives the same result as C# code

Answer 5

@ For informtion>>>>
vadim ))
IF EITHER OF 5n^2+4 OR 5n^2-4 OR BOTH ARE PERFECT SQUARE THEN n is a fibonacci...
as of here...reverse and sum up...
evry number will get to a fib..
so its perfect solution
IT HAS A PROOF..TRUST ME..
TOO BIG AND COMPLEX TO POST IN HERE..

Answer 6

## @ vadim
## if u still want proof i can do it for u in another way...
for n in range (1,1000000):
a,b=(5*n**2-4),(5*n**2+4)
if a**0.5==int(a**0.5) or b**0.5==int(b**0.5):
print(n)
## paste it and run...
## YOU WILL SEE IT WILL ONLY pRINT THE FIBO NUMBERS...LIMIT UPTO 1000K

Answer 7

@Sayan.
I trust. Means, I was wrong.
But then why your program gives number 972845626549368 which isn't Fibonacci's number? The program beautiful and the idea beautiful, it seems, everything has to be right. Or it is my generator not all numbers of Fibonacci shows?

Answer 8

oohoo
i mistook 1 thing
its my typo error...
if either of the case is perfect square...not both...
ok @ vadim...

Answer 9

https://code.sololearn.com/c59LUVm41EYA/?ref=app
https://code.sololearn.com/cHKXQ62qrhkO/?ref=app

Answer 10

@Sayan, you made very interest solution. But my direct algorithm do not found solution for 726. Why ?

Answer 11

@vadim
tell me the process that u r doing...
i mean the working method of ur code..

Answer 12

@Sayan. You see answer with my solutions under your answer with two solutions.

Answer 13

thats why im telling you...
just tell me..how ur code works...
tell me the thought behind it that you implemented...

Answer 14

@Sayan.
nextFibo = next(fibonacci)
n = 0
while n < 1000:
while nextFibo < number:
nextFibo = next(fibonacci)
#print("F", nextFibo)
if nextFibo == number:
print("n={} number={}".
format(n, number))
break
while nextFibo > number:
number += reverseInt(number)
n += 1
else:
print("Not found")
https://code.sololearn.com/cHKXQ62qrhkO/?ref=app

Answer 15

@Sayan.
This code shows that the number given by your code isn't Fibonacci's number.
https://code.sololearn.com/czO2kMmkLSBY

Answer 16

@Sayan
It seems to me that I have understood your code. You consider that the number is Fibonacci's number if the formula of receiving number of Fibonacci from previous gives for this number an exact square in one of two results. It seems to me that this condition is incorrect. More precisely, you confuse necessary and sufficient. Performance of this condition is necessary in order that the number was Fibonacci's number, but it isn't enough.

Answer 17

@Sayan
Yes, your test work normally.
But I made in a different way: I compared the next numbers of Fibonacci determined by my and your method.
There is a result:
1 1
2 2
3 3
5 5
8 8
13 13
21 21
34 34
55 55
89 89
144 144
233 233
377 377
610 610
987 987
1597 1597
2584 2584
4181 4181
6765 6765
10946 10946
17711 17711
28657 28657
46368 46368
75025 75025
121393 121393
196418 196418
317811 317811
514229 514229
832040 832040
1346269 1346269
2178309 2178309
3524578 3524578
5702887 5702887
9227465 9227465
14930352 14930352
24157817 24157817
39088169 39088169
63245986 63245986
102334155 70711162
165580141 94868979
267914296 102334155
433494437 126491972
701408733 133957148
That is the discrepancy began with number 70711162 which is obviously not the amount two previous, but is recognized as your program for Fibonacci number.
Perhaps, a problem in accuracy arithmeticians of Python ?

Answer 18

@Sayan
This is my test code:
def fiboSayan():
a = 0
while True:
while not (lambda p,q: p==int(p) or q==int(q)) ((5*a**2+4)**0.5, (5*a**2-4)**0.5):
a += 1
# print("=== a=", a)
yield a
a += 1
def fibo():
f1 = 0
yield f1
f2 = 1
yield f2
while True:
f1 += f2
yield f1
f2 += f1
yield f2
myfibo = fibo()
Sayan = fiboSayan()
next(myfibo)
next(myfibo)
next(Sayan)
for i in range(50):
print (next(myfibo), next(Sayan))

Answer 19

@Sayan
See this test:
>>> a = 70711162
>>> print((lambda p,q: (p==int(p) and q!=int(q)) or (q==int(q) and p!=int(p))) (
(5*a**2+4)**0.5, (5*a**2-4)**0.5))
True
>>> print ((5*a**2+4)**0.5, (5*a**2-4)**0.5)
158114965.0 158114964.99999997
>>>

Answer 20

@Sayan
Yes, a problem in Python arithmetics accuracy!