@Dave: I hate to reply to myself, but I should have tested a few more
cases, as the code doesn't work for k that is a power of 2, and other
cases. Sorry.
Dave
On Jan 22, 12:54 pm, Dave <dave_and_da...@juno.com> wrote:
> @Gene & Lucifer: Great thinking. For binary integers, you can
> calculate the states of the finite automaton as you go, without ever
> storing it. The following code returns true if and only if the
> arbitrary integer n is divisible by the arbitrary nonzero integer k.
>
> int IsDivisibleBy( int n, int k )
> // for arbitrary integer n and arbitrary nonzero integer k,
> // IsDivisibleBy returns the truth value of the statement
> // "n is divisible by k"
> {
> int state = 0;
> if( k < 0 ) k = -k; // k = abs(k)
> if( n < 0 ) n = -n; // n = abs(n)
> while( n )
> {
> state += state + (n & 1);
> if( state >= k ) state -= k;
> n >>= 1;
> }
> return state == 0;
>
> }
>
> As I've shown before, you can do better for special cases, such as
> divisibility by a fixed constant that is a power of 2 or a power of 2
> plus or minus 1. But this code is sufficiently efficient that it may
> not be worth the trouble to go with customized codes for special cases
> except in the most computationally intensive situations.
>
> BTW: If you change the name to IsMultipleOf, the code works even for k
> = 0; it returns true for n == 0 and false otherwise.
>
> Dave
>
> On Jan 22, 9:36 am, Gene <gene.ress...@gmail.com> wrote:
>
>
>
> > You can extend this thinking to finding the value mod M of any string
> > of base B digits. (In this case M=5, B=2, and you are looking for a
> > mod 5 value of zero.) Construct a finite automaton with M states 0 to
> > M-1. The current state of the automaton tells you the value mod M of
> > the digits seen so far. For current state m and new digit d, the
> > automaton's next state table has value (m * B + d) mod M. Of course
> > you don't compute this value while reading the string. It's stored
> > explicitly in the table. For this problem the next state table looks
> > like this:
>
> > d
> > m 0, 1
> > 0 => {0, 1}
> > 1 => {2, 3}
> > 2 => {4, 0}
> > 3 => {1, 2}
> > 4 => {3, 4}
>
> > Then the algorithm is just:
>
> > state = 0;
> > while another digit remains
> > d = next digit
> > state = next_state[state, d]
> > end;
> > return state;
>
> > For this particular problem you'd check that the return value is zero.
>
> > On Jan 22, 10:26 am, Lucifer <sourabhd2...@gmail.com> wrote:
>
> > > @above...
>
> > > The above explanation is based on the answer that i gave to the
> > > following question asked:
>
> > > Given an infinite stream of bits, at any given point of time u have
> > > tell whether the no. formed by the bits sent till now is divisible by
> > > 3.
>
> > > The catch in the above question is that you can't store the integral
> > > value formed by the bit pattern. At any given point of time u will
> > > only have the current bit. Hence, we need to preprocess the bits sent
> > > to us and then generate the answer.
>
> > > The same approach can be applied here..
> > > 2^0, 2^1 gives remainder 1 and 2(-1) when divided by 3.
>
> > > Code:
>
> > > int r[2]= {1,2};
> > > int i =0;
> > > While(there is an incoming bit X)
> > > {
> > > if(X&1)
> > > {
> > > currRem += r[i%2];
> > > if (currRem >=3) currRem -=3;
> > > }
> > > ++i;
>
> > > }
>
> > > -------------------------------------------------
> > > Hence, the actual question can be changed saying that given an
> > > infinite stream of bits, at any given point of time we need to figure
> > > out whether its divisible by 5.
> > > The above given code will work for this case as well.
>
> > > On Jan 22, 12:19 pm, Lucifer <sourabhd2...@gmail.com> wrote:
>
> > > > @another approach..
>
> > > > The remainders when the following nos. are divided by 5 are :
> > > > Numbers Remainder
> > > > 2^0 1
> > > > 2^1 2
> > > > 2^2 4 (-1)
> > > > 2^3 3 (-2)
>
> > > > Now, we know that given a product of nos. a1, a2, ... aN when divided
> > > > by a no. K, the remainder is:
> > > > the product of (a1 mod K).... (aN mod K) i.e
>
> > > > [ Lets call this as NumTheoProp1]
> > > > a1* a2* a3*...*aN = x*K + (a1 mod K)*(a2 mod K)*....*(aN mod K)..
> > > > [ the above can be proved by representing ai = xi * K + (ai mod K),
> > > > and then multiplying all the ai's..]
>
> > > > We can then recursively apply the same approach to the product of
> > > > remainders until and unless the remainder is smaller than K.
> > > > But a complete reduction into a value < K is not the point here. We
> > > > are going to use the above fact for the following:
>
> > > > Say, the given no. is 2^R ...
> > > > Now, the above can be interpreted as,
> > > > 2^R = 2^(4q) * 2^r, where R = 4q + r and 0<=r < 4
>
> > > > Now we know that, 2^4 when divided by 5 gives a remainder 1..
>
> > > > Hence applying the [NumTheoProp1] :
>
> > > > 2^(4q) = (2^4)*(2^4)*... (q times)..
>
> > > > when 2^(4q) = x*5 + (2^4 mod 5)^q..
> > > > = x* 5 + 1..
>
> > > > Hence,
> > > > 2^R = 2^(4q) * 2^r
> > > > = x*5 + (2^4q mod 5)*(2^r mod 5)
> > > > = x*5 + (2^r mod 5)..
>
> > > > Now, as 0<=r < 4, based on the remainder jolted at the starting..
> > > > 2^r mod 5 = one of (1, 2, 4, 3)..
>
> > > > But if you closely observe there is a pattern..
> > > > For any no. 2^R where R>= 4 we can reduce it to 2^r where r < 4 by
> > > > using [NumTheoProp1]
> > > > Hence,
> > > > No.( 2^r) Remainder
> > > > 0 1
> > > > 1 2
> > > > 2 4
> > > > 3 3
> > > > 4 1
> > > > 5 2
> > > > 6 4
> > > > 7 3
> > > > .... and the cycle continues..
>
> > > > Now. say a no. N is given, then it can be represented as,
> > > > N = b0* (2^0) + b1* (2^1) + .... + bN* (2^logN)..
> > > > where bi is either 0 or 1 [ basically it mimics the bit pattern)
>
> > > > N mod 5 = ( [sum over all i ( (bi* (2^i)) mod 5 )] mod 5)
>
> > > > Now, the inner mod 5 can be directly calculated based on the index of
> > > > i (as there is a pattern). Btw it also depends on the value of bi. If
> > > > bi =0, then the remainder will be 0 otherwise we will get the
> > > > remainder based on the pattern..
>
> > > > The outer mod 5, can be handled by following an incremental process,
> > > > i.e whenever the current remainder becomes >=5 we will subtract 5 from
> > > > it.
>
> > > > Code:
>
> > > > int N; // given number..
> > > > int r[4] = {1, 2, 4, 3};
> > > > int currRem = 0;
> > > > int i = 0;
> > > > while (N)
> > > > {
> > > > if (N&1)
> > > > {
> > > > currRem += r[i%4];
> > > > if (currRem >=5)
> > > > currRem -=5;
> > > > }
> > > > ++i;
> > > > N >>= 1;}
>
> > > > if ( currRem == 0)
> > > > {
> > > > printf("N is divisible by 5");
>
> > > > }
>
> > > > On Jan 22, 2:42 am, Dave <dave_and_da...@juno.com> wrote:
>
> > > > > @Arun: Proof that n = 4*a + b is a multiple of 5 if and only if a - b
> > > > > is a multiple of 5:
>
> > > > > Given n, write it as n = 4*a + b where 0 <= b <= 3.
> > > > > Note that 5*a is a multiple of 5, and 5*a - n = 5*a - (4*a + b) = a -
> > > > > b.
> > > > > If n is a multiple of 5, then 5*a - n is a multiple of 5, so a - b is
> > > > > a multiple of 5.
> > > > > Similarly if n is not a multiple of 5.
>
> > > > > QED
>
> > > > > 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)
>
> ...
>
> read more »
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