# Baum Sweet Sequence in C Program?

Here we will see the Baum Sweet Sequence. This sequence is one binary sequence. If a number n has an odd number of contiguous 0s, then nth bit will be 0, otherwise nth bit will be 1.

We have a natural number n. Our task is to find the n-th term of Baum Sweet sequence. So we have to check whether it has any consecutive block of zeros of odd length.

If the number is 4 then the term will be 1, because 4 is 100. So it has two (even) number of zeros.

## Algorithm

BaumSweetSeqTerm (G, s) −

begin
define bit sequence seq of size n
baum := 1
len := number of bits in binary of n
for i in range 0 to len, do
j := i + 1
count := 1
if seq[i] = 0, then
for j in range i + 1 to len, do
if seq[j] = 0, then
increase count
else
break
end if
done
if count is odd, then
baum := 0
end if
end if
done
return baum
end

## Example

#include <bits/stdc++.h>
using namespace std;
int BaumSweetSeqTerm(int n) {
bitset<32> sequence(n); //store bit-wise representation
int len = 32 - __builtin_clz(n);
//builtin_clz() function gives number of zeroes present before the first 1
int baum = 1; // nth term of baum sequence
for (int i = 0; i < len;) {
int j = i + 1;
if (sequence[i] == 0) {
int count = 1;
for (j = i + 1; j < len; j++) {
if (sequence[j] == 0) // counts consecutive zeroes
count++;
else
break;
}
if (count % 2 == 1) //check odd or even
baum = 0;
}
i = j;
}
return baum;
}
int main() {
int n = 4;
cout << BaumSweetSeqTerm(n);
}

## Output

1

Updated on: 20-Aug-2019

84 Views

##### Kickstart Your Career

Get certified by completing the course