Let A be a square matrix of size n × n. The maximum number of elements that can be nonzero in a row of A without causing an overflow is 2^p. What is the minimum possible value of p?
(A)
log_2(n)
(B)
log_2(n/2)
(C)
log_2(n-1)
(D)
log_2(n+1)
Answer: (A)
Explanation:
To determine the minimum possible value of p, we need to find the maximum number of elements that can be nonzero in a row of matrix A without causing an overflow.
In a binary number system, the maximum value that can be represented using p bits is 2^p – 1. If we have p bits, we can represent numbers from 0 to 2^p – 1.
In a square matrix A of size n × n, each row can have a maximum of n elements. If we want to find the value of p such that 2^p is greater than or equal to n, we can rewrite it as:
2^p ≥ n
Taking the logarithm base 2 of both sides, we get:
p ≥ log_2(n)
Since we want to find the minimum possible value of p, we can take the ceiling function to ensure that p is an integer value:
p = ceil(log_2(n))
Therefore, the minimum possible value of p is ceil(log_2(n)), which is equivalent to option (A) in the given choices.
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