1 votes 1 votes What is the maximum number of nodes in a B-tree of order $10$ of depth $3$ (root at depth $0$) ? $111$ $999$ $9999$ None of the above Algorithms ugcnetcse-dec2010-paper2 algorithms b-tree + – makhdoom ghaya asked Sep 7, 2016 edited Jun 2, 2020 by soujanyareddy13 makhdoom ghaya 3.2k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 2 votes 2 votes Answer: C At level 0, max keys can be 9 At level 1, max keys can be 10*9 = 90 (10 nodes containing 9 keys each) At level 2, max keys can be 10*10*9 = 900 At level 3, max keys can be 10*10*10*9 = 9000 Total = 9000 + 900 + 90 + 9 = 9999 Rakеsh Kumar answered Sep 7, 2016 selected Sep 8, 2016 by Sanjay Sharma Rakеsh Kumar comment Share Follow See all 3 Comments See all 3 3 Comments reply Shivag commented May 11, 2017 reply Follow Share They have asked maximum number of nodes!!!!!! Not kyes 0 votes 0 votes rajks007 commented Sep 8, 2018 reply Follow Share First here it is not clear whether the order is used for keys or block pointer? in question, it is asked the number of nodes possible not keys? 0 votes 0 votes daksirp commented Oct 7, 2018 reply Follow Share then it would be 1111 nodes. 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes The answer is asked about number of nodes. so the formula for finding the maximum number of nodes (n) n=(m^(d+1) -1)/m-1 So, number of nodes = (10^4 - 1) / 9 = 1111 Answer is d. Sunny Sudan answered Oct 16, 2018 Sunny Sudan comment Share Follow See all 0 reply Please log in or register to add a comment.