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)
> > > > >> > >> > 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 =
>
> ...
>
> read more »- Hide quoted text -
>
> - Show quoted text -
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