Exercise 9 Solution proposal

Transcription

1 Prof. Dr.-Ing. Dr. h. c. T. Härder Computer Science Department Databases and Information Systems University of Kaiserslautern Exercise 9 Solution proposal Documentation of the lecture: (July 13, 2011, 3.30 pm, ) Exercise 9.1 k-d Tree a) Please store the following tuples into a 3-d-Tree (k-d-tree with k=3) and keys formed by Lastname, Firstname, and Age Lastname Firstname Age 1. Müller Horst Maier Klaus Müller Kurt Gerber Anton Frank Klaus Paul Ludwig 24 Müller Horst 44 Maier Klaus 30 Müller Kurt 38 Gerber Anton 17 Frank Klaus 50 Paul Ludwig 24 b) For comparison, please store the tuples in reverse order (6 -> 1) Paul Ludwig 24 Frank Klaus 50 Gerber Anton 17 Müller Kurt 38 Müller Horst 44 Maier Klaus 30 c) Please search in the trees built in a) and b) for the following tuples, respectively: 1. Müller 2. Gerber, Anton 1

2 Exercise 9.2 Grid File Please get familiar with the dynamic reorganization for inserts and deletions in grid files. Please assume an empty grid file and buckets can hold 3 records. For the relation CAR (ID. BRAND, COLOR), BRAND and COLOR shall be indexed using a grid file. Please sketch all necessary structures (directory, search space, and buckets) after every insertion of the tuples listed below. Thereafter, please delete the first 4 entries and re-draw the aforementioned structures. ID BRAND, COLOR a) KL-PP 1 OPEL GRAY b) PS-A17 FORD BLUE c) KL-CX 33 OPEL BLUE d) KIB-AM 13 BMW RED e) SB-F16 AUDI GREEN f) KL-DZ 12 ALFA SILVER g) ZW-AL 43 VW WHITE h) HOM-C 1 FIAT VIOLET i) SLS-AF 47 AUDI AQUAMARINE j) KL-DM 31 ALFA BLACK k) VW-FS 40 SAAB BROWN Insert a, b, c a, b, c Inserting d requires a split for >=, Insert e b, d, e a, c Inserting f requires a split for >= Rot, Insert g, h, i b, e, i Red a, c, g d, f, h 2

3 Inserting j requires a split for >= BMW, Red BMW B4 e, i, j a, c, g d, f, h b, B4 Inserting k requires a split for >= Rot Red BMW B5 e, i, j a, c, k g, d, f, h B4 B5 B4 b, After deleting the first 4 entries, the structure has the following buckets> and B5 are not merged, because new insertions would again require a split. and B5 are merged iff, and can be merged, too.. Red B5 e, i, j k g, B5 f, h A merge is performed, if> 1. buckets are siblings in the tree 2. a bucket is empty or the dimension discriminator disappears, which is hard to check. 3

4 Exercise 9.3 R-Tree Please assume the following arrangement of 2-dimensional objects. Please store them into an R-tree with M=2: F1 - F F1 F4 3 2 F3 F F Structure of entries: x-coor. bottom, left x-voor. top, right y-coor. bottom, left y-coor. top, right 1. discriminator: x-coor bottom, left, 2. discriminator: x-coor. top, right After inserting all object areas. 0.5; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; 5 Please perform the following search operations by utilizing the R-tree: a) Select all objects that contain the point (6, 3.5). The search returns all possible candidate as well as objects F3 and F5. Because the surrounding rectangle must not be completely equal to the requested object, for all objects, the containment criteria must be checked. By doing so, only F3 remains as solution. b) Which objects are completely included by a rectangle defined by (3.5, 0) and (8, 5.5)? 4

5 This query can be solely answered using the R-tree entries and results in F3 and F4 c) Please determine all objects that are overlapping with the rectangle given by (2, 2), (4, 4): Here, two candidates are returned (F1 and F3). For each of them, it must be made sure that they are in the requested frame. For both of them, this is true. Exercise 9.4 R-Tree - Search Procedure Let there be the following arrangement of areas a.) Please insert them in a sequence according to their number into an R-tree with m=3 (maximum number of entries in a node) and m=2 (minimal number of entries in a node) After inserting areas 1, 2, 3 7; 8 After inserting area 4 3; 8 7; 8 5

6 After inserting areas 5, 6, 7, 8, 9 (For inserting F8, F3 is moved) 0; 7 2; 7 0; 4 4; 9 1; 7 4; 5 2; 3 4; 6 1; 3 7; 8 8; 9 After inserting all areas 0; 7 4; 9 0; 4 1; 7 0; 4 4; 9 7; 9 7; 8 8; 9 1; 3 0; 1 1; 7 4; 5 2; 3 4; 6 1; 3 6; 9 b.) Please define the following functions FIND_ALL_COMPLETELY_CONTAINED_AREAS( Rectangle ) This function completely work on the R-tree, because actual objects are out of interest. They are completely specified by their surrounding rectangles. FIND_ALL_PARTIALLY_CONTAINED_AREAS( Rectangle ) The R-tree only returns the areas that probably overlap with the specified rectangle. For every candidate, a separate check is necessary. c.) Please test your functions with the rectangle given by (3,1) and (7,4). Completely: 3, 8 Partially: 12, 4 (and 3, 8 completely) d.) Which improvements allows for a simpler search? The R*-tree provides simple search by restricting the search space using a more sophisticated splitting of (sub-)areas (Here, split at x=3 and x=7). As a drawback, tree maintenance becomes much more expensive compared to the R-tree. 6

7 Exercise 9.5 Translating SQL to a Record-Oriented Interface a) Please write for the following SQL query a program using the operations of the record-oriented interface: SELECT ENO, ENAME, SAL FROM EMP WHERE AGE > 45 AND SAL > AND DNO >= K50 AND DNO <= K60 We make the following assumptions for query processing using an index scan: For every key value of a query interval, the TIDs of the corresponding records are read from the index pages to directly access the records in the data pages. Consequently, each page is read only once for each value. Because of buffer replacements, a page might be replaced before it is re-used for accessing the next value. Doing query processing this way has the advantage that the records are returned in the order specified by the index attribute. This approach has the disadvantage that some pages might probably be read several times. If sorting of records is not required, it would be cheaper to get all TIDs for all values of the interval first. Thereafter, they would be sorted according to their page number (e.g., by using merge sort). Consequently, for each consecutive access, each data page is only read once and the movements on the hard disks are minimized. Please develop an optimal program for accessing the records using an index scan, if: 1. Indexes are available for AGE, SAL, and DNO. Which influence on optimization has the existence of an index with clustering for the three attributes? OPEN SCAN on EMP (index scan on the index with the highest selectivity and exploiting the start condition) FETCH TUPLE (SCB, Next,...) (with both predicates, for which no index scan was chosen, as search condition) CLOSE SCAN (if the stop condition for the index scan is reached. Hence, if DNO > K60, for an index scan on DNO) For clustering, the selection of an appropriate index does not only depend on the selectivity, but the number of page accesses must be considered, too. We implicitly made this assumption before, because we did not assume any clustering. Therefore, for low selectivities, more page accesses must be expected. 2. We have indexes for AGE and SAL and a clustered index for ENO. The index on ENO, where no selection is performed, is only beneficial, if an ORDER BY ENO is specified. Hence, everything remains unchanged with the exception that the index on ENO is not used. 3. For all attributes of the predicate, indexes without clustering can be used and the table EMP is stored in a segment with high locality (space usage b > 0.9) Probably, a simple table scan is cheaper, because no index information must be read, OPEN SCAN on EMP (table scan with start condition BOS) FETCH TUPLE (SCB, Next,...) (with all predicates as search conditions (SSA)) CLOSE SCAN 7

8 (if the stop condition EOS is reached) b) In which way can the following SQL query be answered using the record-oriented interface? SELECT X.ENAME, Y.ENAME FROM EMP X, EMP Y WHERE X.MNO = Y.ENO AND X.SAL > Y.SAL Please discuss the possible solutions, if 1. Indexes for MNO, ENO, and SAL are available. 2. No additional access paths can be exploited. Please describe the possible solutions using the operations of the record-oriented interface. 1.) Index scans on I EMP (ENO) and I EMP (MNO): If Y.ENO=X.MNO, then TID accesses for (1+n) records are necessary for checking X.SAL>Y.SAL. The index on SAL cannot be exploited. 2.) Nested-loops join on a physical table with Y.ENO as condition for the inner loop. Then, on the average, a test can be skipped after finishing half of the iterations! In this case, it would also be interesting to have a look at probable generalized access paths for X.MNO and Y.ENO. The evaluation of the SAL conditions requires accesses to all X-records (except for those that only represent the k highest bosses (k>=1)). For Y records, only manager records are accessed. c) In which way would you evaluate the following SQL query, which requires a partitioning of the intermediate results by the attribute values: SELECT DNO FROM EMP WHERE JOB = PROGRAMMER GROUP BY DNO HAVING COUNT (*) > 10 Please provide a program for the record-oriented interface, if 1. there is an index on JOB. 2. there is an index on DNO. Which of the strategies is more efficient? 1.) - index scan over table EMP using the index JOB - count the occurrence of each DNO - finally return those that satisfy the HAVING condition 2.) Works only, if there is a clustered index on DNO. For every DNO, all EMP records must be read to eval- 8

9 uate the WHERE condition, if the number of values exceeds 10. For this approach, more page accesses are required compared to a sequential access to EMP (table scan on JOB= PROGRAMMER as search condition) and an internal evaluation of the condition including statistics gathering. If clustering on DNO is assumed and if many small departments exist, some page accesses might be saved. 9