Suppose we are executing the DGIM algorithm for approximate counting of bits in a stream. Let the window size be 1000.
(a) What is the largest possible bucket size in the representation of this stream?
(b) Suppose each of the last 1000 bits is 1. What is the smallest possible size of the largest bucket in the representation of the window?
256
(a) The largest possible bucket size in the representation of this stream would be 1000. It's like trying to fit 1000 clowns into a tiny car - it's possible, but very crowded!
(b) If each of the last 1000 bits is 1, the smallest possible size of the largest bucket would be 1. It's like having a party with 1000 clowns, but they all decide to squeeze into one tiny clown car - talk about a tight fit!
(a) In the DGIM algorithm, the bucket sizes are powers of 2. The largest possible bucket size in the representation of a stream with a window size of 1000 would be 512.
(b) If each of the last 1000 bits is 1, the smallest possible size of the largest bucket in the representation of the window would be 16. This is because the DGIM algorithm divides the stream into buckets of sizes 1, 2, 4, 8, 16, and so on. Since the last 1000 bits are all 1, the algorithm keeps merging buckets until the smallest bucket size possible represents the oldest bit in the stream, resulting in a bucket size of 16.
To answer these questions, let's first understand how the DGIM algorithm works.
The DGIM algorithm is used to approximate the count of bits in a stream by dividing the stream into windows and creating buckets of decreasing size to keep track of the counts. Each bucket represents a time interval and stores the number of bits seen in that interval.
(a) To find the largest possible bucket size in the representation of the stream, we need to consider the scenario where all bits in the window are 1. In this case, the buckets will have decreasing sizes such that each bucket represents a power of 2. So, if the window size is 1000, the largest possible bucket size will be 512 (2^9).
(b) In this scenario, if each of the last 1000 bits is 1, we need to find the smallest possible size of the largest bucket in the representation of the window. Since buckets are created based on a power of 2, we start by finding the next power of 2 greater than the window size (1000).
The next power of 2 greater than 1000 is 2^10 = 1024. However, this size exceeds the window size. So, we need to consider the previous power of 2, which is 2^9 = 512. This means that the smallest possible size of the largest bucket in the representation of the window is 512.
To find this, we find the remainder of 1000 divided by the smallest possible size of the largest bucket: 1000 % 512 = 464. This remainder corresponds to the number of trailing 1s in the stream that couldn't fit into the last bucket. Since the smallest possible bucket size is 512, the remaining 464 bits will create a new bucket of size 512.
Therefore, the smallest possible size of the largest bucket in the representation of the window is 512.
Overall, the largest possible bucket size is 512, and the smallest possible size of the largest bucket in the representation of the window, when all bits in the window are 1, is also 512.