How many bits would I need to write down an integer n 2 in base n?
时间: 2024-05-30 22:16:24 浏览: 17
We can approach this problem by considering the maximum value of n. The highest value of n that we can represent with k bits is 2^k - 1. Therefore, if we need k bits to represent n in base n, we must have:
n <= 2^k - 1
Solving for k, we get:
k >= log2(n+1)
Since k must be an integer, we can round up the value of log2(n+1) to get the minimum number of bits required to represent n in base n. Therefore, the answer is:
k = ceil(log2(n+1))
相关问题
How many bits would I need to write down an integer a in base b?
If we assume that the integer a has n digits when written in base b, then we can say that a is in the range b^(n-1) <= a < b^n. Taking the logarithm base 2 of both sides, we get:
log2(b^(n-1)) <= log2(a) < log2(b^n)
(n-1)log2(b) <= log2(a) < nlog2(b)
Since n-1 < n, we can use n in place of n-1 in the first inequality to get:
nlog2(b) <= log2(a) < nlog2(b)
Subtracting the left-hand side from the right-hand side, we get:
nlog2(b) - nlog2(b) <= log2(a) - nlog2(b) < (n+1)log2(b) - nlog2(b)
Simplifying, we get:
0 <= log2(a) - nlog2(b) < log2(b)
This tells us that the difference between log2(a) and nlog2(b) is less than log2(b), or equivalently, that nlog2(b) is within log2(b) of log2(a). Therefore, we can say that:
n = ceil(log2(a)/log2(b))
This formula gives us the minimum number of digits we need to represent a in base b. Since each digit can take on b possible values (0 through b-1), we need log2(b) bits to represent each digit. Therefore, the total number of bits required to represent a in base b is:
ceil(log2(a)/log2(b)) * log2(b)
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