Baum Sweet Sequence
Baum Sweet Sequence is an infinite binary sequence of 0s and 1s. The nth term of the sequence is 1 if the number n has no odd number of contiguous zeroes in its binary representation, else the nth term is 0.
The first few terms of the sequence are: b1 = 1 (binary of 1 is 1) b2 = 0 (binary of 2 is 10) b3 = 1 (binary of 3 is 11) b4 = 1 (binary of 4 is 100) b5 = 0 (binary of 5 is 101) b6 = 0 (binary of 6 is 110)
Given a natural number n. The task is to find the nth term of the Baum Sweet sequence, i.e, check whether it contains any consecutive block of zeroes of odd length.
Input: n = 8 Output: 0 Explanations: Binary representation of 8 is 1000. It contains odd length block of consecutive 0s. Therefore B8 is 0. Input: n = 5 Output: 1 Input: n = 7 Output: 0
The idea is to run a loop through the binary representation of n and count the length of all the consecutive zero blocks present. If there is at-least one odd length zero block, then the nth term for the given input n is 0 else it is 1.
// CPP code to find the nth term of the// Baum Sweet Sequence#include <bits/stdc++.h>using namespace std; int nthBaumSweetSeq(int n){ // bitset stores bitwise representation bitset<32> bs(n); // len stores the number of bits in the // binary of n. builtin_clz() function gives // number of zeroes present before the // leading 1 in binary of n int len = 32 - __builtin_clz(n); int baum = 1; // nth term of baum sequence for (int i = 0; i < len;) { int j = i + 1; // enter into a zero block if (bs[i] == 0) { int cnt = 1; // loop to run through each zero block // in binary representation of n for (j = i + 1; j < len; j++) { // counts consecutive zeroes if (bs[j] == 0) cnt++; else break; } // check if the number of consecutive // zeroes is odd if (cnt % 2 == 1) baum = 0; } i = j; } return baum;} // Driver Codeint main(){ int n = 8; cout << nthBaumSweetSeq(n); return 0;} |
Output:
0
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