+ 2

# Given 'n' the length of sequence to be formed by all the possible combinations of 0's and 1's.

Also, it should state the number of sequences having even number of 1's in it. write a code in c++.

7th Oct 2017, 6:33 AM
Amit Gharge
+ 13
Here we go Amit! (Read the note in comment part) [https://code.sololearn.com/c9FGgPAf58HW]
7th Oct 2017, 8:36 AM
Babak
+ 14
Increase 8 to 9 or 16 or 32. it depends on your desired maximum support range
7th Oct 2017, 7:49 AM
Babak
+ 14
It seems you want complete solution! :D So give me a bit time.
7th Oct 2017, 7:53 AM
Babak
+ 13
This is a baseline for your own solution. Good luck, my friend. #include <iostream> #include <bitset> using namespace std; int main() { int n = 0; cout << "Enter n: "; cin >> n; for (int i = 0; i < n; ++i) cout << bitset<8>(i).to_string() << endl; }
7th Oct 2017, 7:11 AM
Babak
+ 11
In our case, in fact, there are different possibilities to attack the problem. Unfortunately,mine wasn't so algorithmic to produce a convincing proof mathematically but rather efficient in terms of speed. (Bitwise stuff covers low-level operations).
7th Oct 2017, 10:41 AM
Babak
+ 3
it is just base. if n > 255 you can split.
7th Oct 2017, 7:45 AM
Oma Falk
+ 3
Besides it was a nice challenge - here we have a python solution: https://code.sololearn.com/c1unkwlHSyXR The pointe is: think before code!
7th Oct 2017, 12:11 PM
Oma Falk
+ 3
@ Amit you have a proof for the formula. I would be interested!
7th Oct 2017, 12:12 PM
Oma Falk
+ 3
@Amit thanks for that challenge!
7th Oct 2017, 3:54 PM
Oma Falk
+ 2
Okay.βΊ
7th Oct 2017, 7:54 AM
Amit Gharge
+ 2
Thank You Babak ππ
7th Oct 2017, 8:41 AM
Amit Gharge
+ 2
Actually i have one formula to determine the even number of 1's in all possible combinations of 0's and 1's of length 'n'. that is (2^(n-1) )-1( for this formula n is the power of 2 whereas in the code its different). I verifed the output of the code with this and it matched(Bingo). ex. if n=6, then by that formula i get 31 sequences having even number of 1's. And with 2^6=64 as input in the code i got the same result. Therefore i wanted proof for the formula for higher values of 'n' and it is being fulfilled by the code.
7th Oct 2017, 8:50 AM
Amit Gharge
+ 2
Thanks Oma for the solution in Python. ππ
7th Oct 2017, 12:16 PM
Amit Gharge
+ 1
Thanks pal #Babak Sheykhan
7th Oct 2017, 7:22 AM
Amit Gharge
+ 1
But the code can take max n=255. If n exceeds 255, after 11111111 it starts from 00000000, 00000001 etc again. And i want a code which shows the number of sequences having even number of 1's.
7th Oct 2017, 7:23 AM
Amit Gharge
+ 1
Okay.#Babak.
7th Oct 2017, 7:50 AM
Amit Gharge
+ 1
And number of even number of 1's in those sequence??
7th Oct 2017, 7:52 AM
Amit Gharge