Final answer:
To calculate the time for index construction using disk storage for the Reuters-RCV1 corpus, the total number of comparisons (108log₂108) is multiplied by the number of disk seeks required per comparison (2) and then multiplied by the time per disk seek (5*10⁻³ seconds).
Step-by-step explanation:
The problem states that we need nlog₂n comparisons and 2 disk seeks for each comparison. The number of termID-docID pairs, n = 108. The given disk seek time = 5*10⁻³ sec.
In this case, we can calculate the total time required for index construction as follows:
1.Number of comparisons = nlog₂n = 108 * log₂(108) = 108 * 6.785 = 731.88
2.Number of disk seeks = number of comparisons * 2 = 731.88 * 2 = 1463.76
3.Total time = number of disk seeks * disk seek time = 1463.76 * 5*10⁻³ sec = 7.319 seconds
The question relates to calculating the time required for index construction when using disk storage and an unoptimized sorting algorithm for the Reuters-RCV1 corpus, where there are 108 termID-docID pairs. For each comparison, the process requires nlog₂n comparisons and 2 disk seeks. Given that one disk seek takes 5*10⁻³ seconds, we can calculate the total time for index construction.
To determine the time, we first calculate the total number of comparisons as nlog₂n, which is 108log₂108. Next, we determine the total number of disk seeks (which is 2 times the number of comparisons) and multiply that by the time per disk seek. As a formula, the total time T can be computed as:
T = 2 * (108log₂108) * (5*10⁻³ seconds)