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: 0

Input: n = 7
Output: 1

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

// 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 Code
int main()
{
    int n = 8;
    cout << nthBaumSweetSeq(n);
    return 0;
}

Java

// Java code to find the nth term of the
// Baum Sweet Sequence
class GFG {
  static int nthBaumSweetSeq(int n)
  {

    // bitset stores bitwise representation
    char[] bs
      = (Integer.toBinaryString(n)).toCharArray();

    int baum = 1; // nth term of baum sequence

    for (int i = 0; i < bs.length;) {
      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 < bs.length; j++) {
          // counts consecutive zeroes
          if (bs[j] == '0')
            cnt += 1;
          else
            break;
        }

        // check if the number of consecutive
        // zeroes is odd
        if (cnt % 2 == 1)
          baum = 0;
      }

      i = j;
    }

    return baum;
  }

  // Driver Code
  public static void main(String[] args)
  {
    int n = 8;

    // Function call
    System.out.println(nthBaumSweetSeq(n));
  }
}

// This code is contributed by phasing17

Python3

# Python3 code to find the nth term of the
# Baum Sweet Sequence
def nthBaumSweetSeq(n):

    # bitset stores bitwise representation
    bs = list(bin(n)[2::])

    baum = 1 # nth term of baum sequence
    for i in range(len(bs)):
        j = i + 1
        
        # enter into a zero block
        if (bs[i] == '0'): 
            cnt = 1
  
            # loop to run through each zero block
            # in binary representation of n
            for j in range(i + 1, len(bs)):
  
                # counts consecutive zeroes 
                if (bs[j] == 0):                   
                    cnt += 1
                else:
                    break
            
            # check if the number of consecutive
            # zeroes is odd
            if (cnt % 2 == 1):
                baum = 0
        
        i = j
    return baum
  
# Driver Code
n = 8
print(nthBaumSweetSeq(n))

# This code is contributed by phasing17

using System;

public class GFG {
  static int nthBaumSweetSeq(int n)
  {
    // bitset stores bitwise representation
    char[] bs
      = (Convert.ToString(n, 2)).ToCharArray();

    int baum = 1; // nth term of baum sequence

    for (int i = 0; i < bs.Length;) {
      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 < bs.Length; j++) {
          // counts consecutive zeroes
          if (bs[j] == '0')
            cnt += 1;
          else
            break;
        }

        // check if the number of consecutive
        // zeroes is odd
        if (cnt % 2 == 1)
          baum = 0;
      }

      i = j;
    }

    return baum;
  }

  // Driver Code
  public static void Main()
  {
    int n = 8;

    // Function call
    Console.WriteLine(nthBaumSweetSeq(n));
  }
}

JavaScript

// JavaScript code to find the nth term of the
// Baum Sweet Sequence
function nthBaumSweetSeq(n)
{

    // bitset stores bitwise representation
    let bs = n.toString(2).split("");

    let baum = 1; // nth term of baum sequence
    
    for (let i = 0; i < bs.length;)
    {
        let j = i + 1;
        
        // enter into a zero block
        if (bs[i] == '0')
        {
            let cnt = 1;
  
            // loop to run through each zero block
            // in binary representation of n
            for (j = i + 1; j < bs.length; j++)
            {
                // counts consecutive zeroes 
                if (bs[j] == '0')
                    cnt += 1;
                else
                    break;
            }
            
            // check if the number of consecutive
            // zeroes is odd
            if (cnt % 2 == 1)
                baum = 0;
        }
        
        
        i = j;
    }
    
    return baum;
}
  
// Driver Code
let n = 8;
console.log(nthBaumSweetSeq(n));

// This code is contributed by phasing17

Output:

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