1、 II. III. The key fields are underlined, and the domain of each field is listed after the field name. Therefore sid is the key for Suppliers, pid is the key for Parts, and sid and pid together form the key for Catalog. The Catalog relation lists the prices charged for parts by Suppliers. Write the f
2、ollowing queries in relational algebra. 1. Find the names of suppliers who supply some red part. 2. Find the sids of suppliers who supply some red or green part. 3. Find the sids of suppliers who supply some red part or are at 221 Packer Street. 4. Find the sids of suppliers who supply some red part
3、 and some green part. 5. Find the sids of suppliers who supply every part. 6. Find the sids of suppliers who supply every red part. 7. Find the sids of suppliers who supply every red or green part. 8. Find the sids of suppliers who supply every red part or supply every green part. 9. Find pairs of s
4、ids such that the supplier with the first sid charges more for some part than the supplier with the second sid. 10. Find the pids of parts supplied by at least two different suppliers. 11. Find the pids of the most expensive parts supplied by suppliers named Yosemite Sham. 12. Find the pids of parts
5、 supplied by every supplier at less than $200. (If any supplier either does not supply the part or charges more than $200 for it, the part is not selected.) IV Consider the following schema: Suppliers(sid: integer, sname: string, address: string) Parts(pid: integer, pname: string, color: string) Cat
6、alog(sid: integer, pid: integer, cost: real) The Catalog relation lists the prices charged for parts by Suppliers. Write the following queries in SQL: 1. Find the pnames of parts for which there is some supplier. 2. Find the snames of suppliers who supply every part. 3. Find the snames of suppliers
7、who supply every red part. 4. Find the pnames of parts supplied by Acme Widget Suppliers and no one else. 5. Find the sids of suppliers who charge more for some part than the average cost of that part (averaged over all the suppliers who supply that part). 6. For each part, find the sname of the sup
8、plier who charges the most for that part. 7. Find the sids of suppliers who supply only red parts. 8. Find the sids of suppliers who supply a red part and a green part. 9. Find the sids of suppliers who supply a red part or a green part. 10. For every supplier that only supplies green parts, print t
9、he name of the supplier and the total number of parts that she supplies. 11. For every supplier that supplies a green part and a red part, print the name and price of the most expensive part that she supplies. V. Consider the B+ tree index of order d = 2 shown in the following figure. 1. Show the tr
10、ee that would result from inserting a data entry with key 9 into this tree. 2. Show the B+ tree that would result from inserting a data entry with key 3 into the original tree. How many page reads and page writes does the insertion require? 3. Show the B+ tree that would result from deleting the dat
11、a entry with key 8 from the original tree, assuming that the left sibling is checked for possible redistribution. 4. Show the B+ tree that would result from deleting the data entry with key 8 from the original tree, assuming that the right sibling is checked for possible redistribution. 5. Show the
12、B+ tree that would result from starting with the original tree, inserting a data entry with key 46 and then deleting the data entry with key 52. 6. Show the B+ tree that would result from deleting the data entry with key 91 from the original tree. 7. Show the B+ tree that would result from starting
13、with the original tree, inserting a data entry with key 59, and then deleting the data entry with key 91. 8. Show the B+ tree that would result from successively deleting the data entries with keys 32, 39, 41, 45, and 73 from the original tree. Answer 10.1 1. The data entry with key 9 is inserted on
14、 the second leaf page. The resulting tree is shown in figure 10.2. 2. The data entry with key 3 goes on the first leaf page F. Since F can accommodate at most four data entries (d = 2), F splits. The lowest data entry of the new leaf is given up to the ancestor which also splits. The result can be s
15、een in figure 10.3. The insertion will require 5 page writes, 4 page reads and allocation of 2 new pages. 3. The data entry with key 8 is deleted, resulting in a leaf page N with less than two data entries. The left sibling L is checked for redistribution. Since L has more than two data entries, the
16、 remaining keys are redistributed between L and N, resulting in the tree in figure 10.4. 4. As is part 3, the data entry with key 8 is deleted from the leaf page N. Ns right sibling R is checked for redistribution, but R has the minimum number of keys. Therefore the two siblings merge. The key in th
17、e ancestor which distinguished between the newly merged leaves is deleted. The resulting tree is shown in figure 10.5. 5. The data entry with key 46 can be inserted without any structural changes in the tree. But the removal of the data entry with key 52 causes its leaf page L to merge with a siblin
18、g (we chose the right sibling). This results in the removal of a key in the ancestor A of L and thereby lowering the number of keys on A below the minimum number of keys. Since the left sibling B of A has more than the minimum number of keys, redistribution between A and B takes place. The final tre
19、e is depicted in figure 10.6. 6. Deleting the data entry with key 91 causes a scenario similar to part 5. The result can be seen in figure 10.7. 7. The data entry with key 59 can be inserted without any structural changes in the tree. No sibling of the leaf page with the data entry with key 91 is af
20、fected by the insert. Therefore deleting the data entry with key 91 changes the tree in a way very similar to part 6. The result is depicted in figure 10.8. 8. Considering checking the right sibling for possible merging first, the successivedeletion of the data entries with keys 32, 39, 41, 45 and 73 results in the tree shown in figure 10.9.
