[assignment] Check if a binary number is devisible by 7 with following rules

Question

a binary number is devisible by 7,if it has exactly 3*n 1.

For each 1 calculate modpos = position mod 3.
So if a 1 is at position 4, the modpos is 1.

modpos0 is the count of all 1 with modpos = 0
modpos1 is the count of all 1 with modpos = 1
modpos2 is the count of all 1 with modpos = 2.

if modpos0 == modpos1 == modpos2
the binary is devisible by 7.

eg b111111
1 at position 1,2,3,4,5,6
which makes modpos 1,2,0,1,2,0
modpos0 == modpos1 == modpos2 = 2

Edit: you will not find all numbers.

b111111 = 63 is devisible by 7.

Answer 1

https://code.sololearn.com/c8Ej4poguD83/?ref=app

Answer 2

Hi Oma !
I think my code follows your rules, but it misses many numbers. When the conditions are met, the number is actually divisible by 7, but not all of the multiples of 7 will match the conditions. Or am I wrong ?

https://code.sololearn.com/c0qHtxHoItGx/#py

Answer 3

Cépagrave yes u are right. It is a => constraint but not <=>.

Answer 4

2 additionnal ways to do it
https://code.sololearn.com/cC094JcrxnXX/?ref=app

Answer 5

https://code.sololearn.com/cC094JcrxnXX/?ref=app

Answer 6

here is my quirk solution using regex. to be precise for automate or Turing machine. I did not think that it is so cumbersome.

https://code.sololearn.com/cUZ9tEhA13OA/?ref=app

Answer 7

bool div7(unsigned n){
unsigned mods[3] = {0};

while(n){
for(int i = 0; i < 3; ++i)
mods[i] += (n & (1 << i)) >> i;
n >>= 3;
}

return mods[0] == mods[1] && mods[0] == mods[2];
}

Answer 8

my try

https://code.sololearn.com/cgl07tLg16SR/?ref=app

Answer 9

very nice the algorithm

Answer 10

yuri awfully inefficient but really cool

Answer 11

Oma Falk we are able to code it in CSS, JS, Python, C++, Kotlin, java, C# and ruby, right?

Answer 12

This Works! It's Faster!

A binary number is divisible by 7, if you group the digits into sets of 3 each, sum all grouped digits and check if this sum is divisible by 7.

(use zeroes to fill in the leading digit gaps, as is obvious)