# There are n bacteria and 1 virus in a Petri dish. Within the first minute, the virus kills one bacterium and produces another copy of itself, and all of the remaining bacteria reproduce, making 2 viruses and 2 · (n − 1) bacteria. In the second minute, each of the viruses kills a bacterium and produces a new copy of itself (resulting in 4 viruses and 2(2(n − 1) − 2) = 4n − 8 bacteria; again, the remaining bacteria reproduce. This process continues every minute. Will the viruses eventually kill all the bacteria? If so, design an algorithm that computes how many steps it will take. How does the running time of your algorithm depend on n? Given a long text string T, one shorter pattern string s, and an integer k, find the first occurrence in T of a string (if any) s such that dH(s, s ) ≤ k. What is the complexity of your algorithm? Given a long text string T, one shorter pattern string s, and an integer k, find the first occurrence in T of a string (if any) s such that dH(s, s ) ≤ k. What is the complexity of your algorithm?.

To determine whether the viruses will eventually kill all the bacteria in the Petri dish, we need to analyze the growth and killing dynamics of both the viruses and bacteria.

In the first minute, the virus kills one bacterium and produces another copy of itself, while all remaining bacteria reproduce. This results in 2 viruses and 2(n-1) bacteria.

In the second minute, each virus kills a bacterium and produces a new copy of itself. The remaining bacteria, which have now increased due to reproduction, reproduce again. This leads to 4 viruses and 2(2(n-1)-2) = 4n-8 bacteria.

This process continues every minute, with the number of viruses doubling and the number of bacteria increasing by a factor of 2 each time.

To determine if the viruses will eventually kill all the bacteria, we need to consider the relationship between the growth rates of viruses and bacteria. Since the number of viruses doubles every minute, we can represent the number of viruses at any given minute, m, as Vm = 2^m.

The number of bacteria at any given minute, m, can be represented as Bm = 2^(m-1) (n-1) + 1.

We can express the condition for the viruses killing all the bacteria as Vm ≥ Bm. Substituting the representations mentioned above, we get 2^m ≥ 2^(m-1) (n-1) + 1.

Simplifying this inequality, we obtain 2^m – 2^(m-1) (n-1) – 1 ≥ 0.

To determine the number of steps it will take for the viruses to kill all the bacteria, we need to find the smallest value of m that satisfies the inequality mentioned above. This can be achieved by using a binary search algorithm to find the value of m for which the inequality holds.

The running time of the algorithm to compute how many steps it will take depends on the number of steps required by the binary search algorithm. The binary search algorithm has a logarithmic time complexity of O(log N), where N represents the size of the search space. In this case, N corresponds to the number of minutes it takes for the viruses to kill all the bacteria.

Therefore, the running time of the algorithm to compute the number of steps required for the viruses to kill all the bacteria depends on the value of n in the initial condition and has a time complexity of O(log n).

Moving on to the second part of the question, given a long text string T, a shorter pattern string s, and an integer k, we want to find the first occurrence of a string in T that satisfies the condition dH(s, s’) ≤ k, where dH represents the Hamming distance between two strings.

The complexity of the algorithm to solve this problem depends on the implementation approach. One possible approach is to iterate over each position in T and compare the substring of T with s to compute the Hamming distance. If the Hamming distance is less than or equal to k, we have found a match.

This approach would require comparing each position in T with s, resulting in a time complexity of O(len(T) * len(s)), where len(T) represents the length of string T and len(s) represents the length of string s.

In the worst case, where len(T) and len(s) both have a large value, the algorithm’s time complexity could be quite high. However, there are more efficient algorithms like the Boyer-Moore algorithm or Knuth-Morris-Pratt algorithm that can reduce the time complexity to O(len(T) + len(s)), making it more efficient for large inputs.

In conclusion, the time complexity of the algorithm to solve the second part of the question depends on the specific implementation approach chosen, but it can generally be reduced to O(len(T) + len(s)) using efficient string search algorithms.

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