Counting Bit Strings of Length n Ending with 0 and Containing Two Consecutive 1s

When dealing with bit strings, it is common to analyze various constraints within them. One such interesting problem is counting the number of bit strings of length n that end with 0 and contain at least two consecutive 1s. In this article, we will explore a method to solve this problem by rephrasing and breaking it down into more manageable components.

Rephrasing the Problem

The problem can be simplified by considering the first n-1 bits of the bit string, since the last bit is known to be 0. This rephrasing helps us focus on the different combinations of the initial n-1 bits that ensure the presence of at least two consecutive 1s.

Case Analysis: First Bit Equal to 1

If the first bit is 1, we then need to ensure that the remaining n-2 bits include at least one more 1 to guarantee the presence of two consecutive 1s. Here’s how we can break it down:

There are n-2 positions available for the second 1. The rightmost bit is fixed as 0.

The total number of such bit strings is n-2. This is because there are n-2 positions to place the second 1, ensuring that the bit string ends with 0 and contains two consecutive 1s.

Case Analysis: Rightmost Bit Equal to 1

Similarly, if the rightmost bit of the first n-1 bits is 1, we need to ensure that the bit string contains at least one more 1 to result in two consecutive 1s ending with 0. In this case:

The first bit is fixed as 1. The remaining n-2 bits need to include at least one more 1.

The total number of bit strings in this case is also n-2. This is due to the combination of positions that can be chosen for the additional 1 to ensure the presence of two consecutive 1s.

Constructing the Bit Strings

Another approach to solving this problem involves a construction process. To form a bit string of length n that ends with 0 and contains at least two consecutive 1s, we can use the following method:

Select 2 bits to represent the block of two consecutive 1s. There are 3 ways to choose a block of 1s (11). Shift this block 2 bits to the left, covering the first 2 bits. Use the remaining n-2 bits to create the rest of the string.

This construction process ensures that the bit string ends with 0 and contains at least two consecutive 1s. For instance, when n5:

11000
11000 (repeated for testing)

By analyzing and testing these cases, we can confirm that the method accurately counts the number of valid bit strings. This approach can be applied to any value of n to determine the number of bit strings of length n that end with 0 and contain at least two consecutive 1s.

Conclusion

By breaking down the problem and using a construction process, we can efficiently count the number of bit strings of length n that end with 0 and contain at least two consecutive 1s. Through case analysis and combinatorial methods, we can solve this problem systematically. The key takeaway is understanding how to rephrase the initial problem and apply logical reasoning to arrive at a solution.