bè ëfind all countries C and the sum S of all of C's city populations, provided S is at least 10,000,000", SUMèPopulationè GROUP BY Co Name HAVING SUM

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1 MIDTERM CSE 232, February 2000 èopen bookè FIRSTNAME: LASTNAME: Problem 1 èsql, 20 è2,7,11èè Consider the following relations CountryèCo N ame; P opulationè and CityèCi Name; Co name; P opulationè For example, here is an excerpt: Co Name Country Population Germany USA City Ci Name Co Name Population Atlanta USA Freiburg Germany San Diego USA Express in SQL the following queries: aè ëfind all countries that have a city `Victoria'" WHERE Ci Name = 'Victoria'; Now that was easy! Let's assume a more realistic schema, where City has a foreign key Co Id èthe id of a countryè which refers to the primary key Co Id of Country èthis is what the problem should have beenè Then the solutions are: SELECT CoCo Name FROM Country Co, City Ci WHERE CiCi Name = 'Victoria' AND CiCo Id = CoCo Id; Alternative solution with a nested query: FROM Country WHERE Co Id IN èselect Co Id WHERE Ci Name = 'Victoria'è; 1

2 bè ëfind all countries C and the sum S of all of C's city populations, provided S is at least 10,000,000", SUMèPopulationè GROUP BY Co Name HAVING SUMèPopulationè é= ; cè ëfind all countries where all cities have at most 5,000,000 inhabitants" è* or Country *è WHERE Co Name NOT IN è WHERE Population é è alternative solution: GROUP BY Co Name HAVING MAXèPopulationè é= è 2

3 Problem 2 èalgebra & Rewritings, 20è Consider the following SQL query over the schema RèA;B;:::è and SèA;C;:::è: Q: SELECT RB FROM R, S WHERE RA = SA AND SC = 'foo'; aè Give an equivalent algebra expression for Q using the ëdirect" translation of SQL, ie, where SELECT-FROM-WHERE is modeled as ç ::: èç ::: èr1 ææææær n èè, ç B èç R:A=S:A^C= 0 foo0èr æ Sèè bè Same as èaè but give an ëoptimized" expression, ie, using join, selections pushed down, and atomic conditions ç B èr 1 R:A=S:A èç C= 0 foo 0èSèèè cè Assume the ærst line of Q above is ëselect RA" How can you further optimize your expression from èbè? ç A èr 1 R:A=S:A ç A èç C= 0 foo 0èSèèè or ç A èç A èrè 1 R:A=S:A ç A èç C= 0 foo 0èSèèè which is essentially ç AèRè ë ç A èç C= 0 foo 0èSèè dè Rewrite the expression from èaè into the form given in èbè using equivalences from the class ç B èç R:A=S:A^C= 0 foo0èr æ Sèè = ç B èç R:A=S:Aèç C= 0 foo0èr æ Sèèè èdecompose ^è = ç B èç R:A=S:AèR æ ç C= 0 foo0èsèèè èpush selectionè = ç B èr 1 R:A=S:A èç C= 0 foo0èsèèè èreplace çèæèèè by 1èèè 3

4 Problem 3 èindexes, 10è Indicate whether the following statements are true or not and give a brief explanation or counterexample aè A secondary index should be sparse No Asecondary index has to be dense since the æle is not organized according to a secondary index Otherwise, the index structure does not allow to ænd all values directly bè A second level index should be sparse Yes If it were dense, we would have the same number of entries in the second level as in the ærst level index which defeats the purpose of the second level index cè In query optimization, it is always better to perform projections as early as possible No There may be an index on the stored relation whereas after the projection, there may be no index Of course we consider only correct applications of ç, ie, which do not eliminate attributes needed later But this was not the point here èotherwise we could say pushing ç is not always good since we could push it incorrectlyè dè For queries of the form ç A=aèRè hashing is usually better than a B+ tree yes, since we can directly access the record eè For queries of the form ç Aéa èrè hashing is usually better than a B+ tree no, hashing does not support range queries since a hash function needs a speciæc key value èwhereas we don't know the values of A given just Aéaè 4

5 Problem 4 èextensible Hashing, 10è Consider an extensible hash structure with buckets holding up to three records Initially the structure is empty Then, the following records are inserted in the given order a,æææ,j èthe hashed key is shown in bracketsè: a ë101001ë b ë010111ë c ë001110ë d ë011010ë e ë101001ë f ë011010ë g ë010000ë h ë111100ë i ë010111ë j ë000110ë æ Show the structure after each directory doubling step and the ænal structure starting with the LSB èfrom the rightè: depth=1; ins a,b,c,d,e,f: bè0è = èc,d,fè, bè1è = èa,b,eè ins g ; split bè0è; depth=2; ins h: bè00è=èg,hè, bè10è=èc,d,f è, bè 1è=èa,b,eè ins i ; split bè1è bè00è=èg,hè, bè10è=èc,d,f è, bè01è=èa,eè; bè11è=èb,iè ins j ; split bè10è; depth=3 bè 00è=èg,hè, bè010è=èd,f è, bè110è=èc,jè, bè 01è=èa,eè; bè 11è=èb,iè starting with the MSB èfrom the leftè: depth=1; ins a,b,c,d,e: bè0è = èb,c,dè, bè1è = èa,eè ins f ; split bè0è; depth=2: bè00è=ècè, bè01è=èb,d,f è, bè1 è=èa,eè ins g ; split bè01è; depth=3; ins h,i,j: bè00 è=èc,jè, bè010è=èb,g,iè, bè011è=èd,f è, bè1 è=èa,e,hè 5

6 Problem 5 èb+ Trees, 10è Consider a B+ tree of order 4, which is initially empty Show some intermediate steps and the ænal tree, when records with the following keys are inserted èin this orderè: æ 10; 7; 8; 9; 11; 6; 5; 12; 15; 13 ins 10;7;8;9 ; è7,8,9,10è ins 11 ; è7,8,9è 10 è10,11è ins 6 ; è6,7,8,9è 10 è10,11è ins 5 ; è5,6,7è 8 è8,9è 10 è10,11è ins 12,15 ; è5,6,7è 8 è8,9è 10 è10,11,12,15è ins 13 ; è5,6,7è 8 è8,9è 10 è10,11,12è 13 è13,15è 6

7 Problem 6 èb+ Trees, 10è Consider B+ trees of order 2, ie, where each node holds either one or two key values Give an example of a 3-level B+ tree that can be reorganized into a 2-level B+ tree with exactly the same values Display both the 3-level and the 2-level tree Use letters or numbers for the key values For example: Make sure no B-tree properties are violated èeg, n keys, n+1 pointers and ORDER wrt descendants!è 7