@Arun: Some time ago, there was a question about determining
divisibility by 3. I thought about determining divisibility of decimal
numbers by 9 using the "casting out nines" procedure: sum the decimal
digits of the number; the number is divisible by 9 if the sum of the
digits is divisible by 9. Casting out nines works because 9 is one
less than the base of the decimal number system. Since we can
translate binary numbers to base 4 just by grouping adjacent pairs of
bits, I came up with the algorithm for divisibility by 3 that I posted
in http://groups.google.com/group/algogeeks/msg/a4894e12805381ca.
To make that algorithm look more like my divisibility by 5 algorithm
posted earlier in this string, rewrite it as
int DivisibileBy3(int n)
{
n = n > 0 ? n : -n;
while(n>3)
n = (n>>2) + (n & 3);
return n == 3 || n == 0;
}
So if divisibility by 3 (= 4-1) is as above, I thought maybe
divisibility by 5 (= 4+1) might be related. I tried it and it works!
Divisibility by 7: change the 3's above to 7's and the 2 above to 3.
Divisibility by 11 would be more difficult.
Dave
On Jan 21, 11:57 am, Arun Vishwanathan <aaron.nar...@gmail.com> wrote:
> @dave: thanks for that..but i just wanted to know as to how u thot of
> converting n to a-b in the iteration. when u say 4a +b is a multiple of 5
> iff a-b is a muliple of 5 i was able to get that only when i tried an
> example...if they ask divisbility by 3 or 6 or 7 how wud the logic change??
>
> On Sat, Jan 21, 2012 at 9:34 AM, karthikeyan muthu <
>
>
>
>
>
> keyankarthi1...@gmail.com> wrote:
> > check the last char ... it should be 0 or 5 , int to string without mod
>
> > On Sat, Jan 21, 2012 at 10:05 PM, Dave <dave_and_da...@juno.com> wrote:
>
> >> @Karthikeyan: Is this supposed to relate to the question of
> >> determining divisibility by 5?
>
> >> Dave
>
> >> On Jan 21, 9:25 am, karthikeyan muthu <keyankarthi1...@gmail.com>
> >> wrote:
> >> > @dave
>
> >> > int no=10;
> >> > char ans[100];
> >> > sprintf(ans,"%d",no);
> >> > cout<<ans;
>
> >> > On Fri, Jan 20, 2012 at 10:29 PM, Arun Vishwanathan
> >> > <aaron.nar...@gmail.com>wrote:
>
> >> > > @dave or anyone: can u pls explain the logic of n&3 in dave's
> >> solution?
> >> > > why is it subtracted from n(which is divided by 4 using >>2) and what
> >> does
> >> > > n& 3 indicate?
>
> >> > > On Sat, May 7, 2011 at 9:38 AM, Dave <dave_and_da...@juno.com> wrote:
>
> >> > >> @Umer: Do you suppose that you can convert an int into a string
> >> > >> without using division or mod, either directly or indirectly?
>
> >> > >> Dave
>
> >> > >> On May 4, 1:12 am, Umer Farooq <the.um...@gmail.com> wrote:
> >> > >> > I'm surprised to see that why are you guys making this problem so
> >> > >> complex.
> >> > >> > This problem can be solved in two steps only.
>
> >> > >> > 1- Convert the given int into string
> >> > >> > 2- Check if the last character is 0 or 5. // it yes, then return
> >> true
> >> > >> else
> >> > >> > return false
>
> >> > >> > for e.g.
>
> >> > >> > 125 (last character is 5 ... therefore it is divisible by 5)
> >> > >> > 120 (last character is 0 ... therefore it is divisible by 5)
> >> > >> > 111 (last character is 1 ... therefore it is not divisible by 5)
>
> >> > >> > The pseudo-code has been written in my above email.
>
> >> > >> > On Wed, May 4, 2011 at 1:49 AM, Dave <dave_and_da...@juno.com>
> >> wrote:
> >> > >> > > @anshu: Spoiler alert... I was thinking of something more along
> >> the
> >> > >> > > line
>
> >> > >> > > int DivisibleBy5 (int n)
> >> > >> > > {
> >> > >> > > n = n > 0 ? n : -n;
> >> > >> > > while( n > 0 )
> >> > >> > > n = (n >> 2) - (n & 3);
> >> > >> > > return (n == 0);
> >> > >> > > }
>
> >> > >> > > To see that it works, write n as n = 4*a + b, where 0 <= b <= 3.
> >> Then
> >> > >> > > the iteration replaces n by a - b. Consider (4*a + b) + (a - b),
> >> the
> >> > >> > > sum of two consecutive values of n. This simplifies to 5*a,
> >> which is a
> >> > >> > > multiple of 5. Thus, n is a multiple of 5 before an iteration if
> >> and
> >> > >> > > only if it also is a multiple of 5 afterwards,
>
> >> > >> > > It is clearly log n because n is replaced by a number no greater
> >> than
> >> > >> > > n/4 on each iteration.
>
> >> > >> > > Examples:
> >> > >> > > n = 125. The sequence of iterates is 30, 5, 0. Ergo, 125 is a
> >> multiple
> >> > >> > > of 5.
> >> > >> > > n = 84. The sequence of iterates is 21, 4, -1. Ergo, 84 is not a
> >> > >> > > multiple of 5.
>
> >> > >> > > Dave
>
> >> > >> > > On May 3, 3:13 am, anshu <anshumishra6...@gmail.com> wrote:
> >> > >> > > > algorithm:
>
> >> > >> > > > if any number(a) is divisible by 5 it can be wriiten as 4*b +
> >> b -->
> >> > >> > > > this cleary shows the last two bit of a & b will be same.
>
> >> > >> > > > lets understand by an example (35)10 = (100011)2
>
> >> > >> > > > xx1100
> >> > >> > > > + xx11
> >> > >> > > > ---------
> >> > >> > > > 100011
>
> >> > >> > > > now this clearly shows we can calculate the unknowns(x) by
> >> > >> traversing
> >> > >> > > > right to left
>
> >> > >> > > > code:
>
> >> > >> > > > int main()
> >> > >> > > > {
> >> > >> > > > int n, m;
> >> > >> > > > cin >> n;
> >> > >> > > > m = n;
>
> >> > >> > > > int a, b;
> >> > >> > > > int i=2;
>
> >> > >> > > > a = (m&3)<<2;
> >> > >> > > > b = (m&3);
> >> > >> > > > m >>= 2;
>
> >> > >> > > > bool rem = 0,s,r;
>
> >> > >> > > > while (m>3)
> >> > >> > > > {
> >> > >> > > > r = a&(1<<i);
> >> > >> > > > s = r^(m&1)^rem;
> >> > >> > > > b = b|(s<<i);
> >> > >> > > > a = a|(s<<(i+2));
> >> > >> > > > rem = (r&s)|(s&rem)|(r&rem) ;
> >> > >> > > > i++;
> >> > >> > > > m >>= 1;
> >> > >> > > > }
>
> >> > >> > > > if (a+b == n) cout << "yes\n";
> >> > >> > > > else cout << "no\n";
>
> >> > >> > > > return 0;
>
> >> > >> > > > }- Hide quoted text -
>
> >> > >> > > > - Show quoted text -
>
> >> > >> > > --
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> >> > >> > --
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