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