1 votes 1 votes Suppose a block holds either 3 records or 10(key,pointer) pairs. If $n$ is the number of records and Dense indexing is used then how many blocks do we need to hold data file ? thor asked Jan 7, 2017 thor 322 views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply santhoshdevulapally commented Jan 7, 2017 reply Follow Share Block factor=3. index block factor=10. Dense index first level no of blocks=n/3. 2nd level=$\frac{\frac{n}{3}}{10}$. Third level=$\frac{\frac{n}{30}}{10}$. .... In the question how many levels he ask add from first level to that index level. 0 votes 0 votes santhoshdevulapally commented Jan 7, 2017 reply Follow Share if we use sparse index no of entries in first level=$\frac{\frac{n}{3}}{10}$. 2nd level=$\frac{\frac{n}{30}}{10}$. ..... 1 votes 1 votes thor commented Jan 7, 2017 reply Follow Share IN case of dense index, how to we know the point to which i should sum up? In sparse index please explain how entries in first level $n/30$ 0 votes 0 votes santhoshdevulapally commented Jan 7, 2017 reply Follow Share in sparse index we arrange based on index block factor. Here no of blocks=n/3. and index factor=10. again divide with no of blocks u get no of entries in first level i.e) $\frac{\frac{n}{3}}{10}$. =$\frac{n}{30}$. 1st doubt::if no of records in a index level =1 then use those many levels for same as sparse as well as dense. 0 votes 0 votes thor commented Jan 7, 2017 reply Follow Share Thnx @santhosh let me take time to understand and will ask any doubt if i get. 0 votes 0 votes santhoshdevulapally commented Jan 7, 2017 reply Follow Share u r welcome. :D) 0 votes 0 votes Please log in or register to add a comment.