數(shù)據(jù)庫復(fù)習(xí)題2(答案)-預(yù)覽頁
【正文】 the names of those students via repeated lookups in the B+ Tree on snum.For these repeated accesses to the index on , which replacement policy would you remend for efficient buffer management?Query 1: select snum from student s, enrolled e where = and ame like ‘Database Systems’。例如聚合函數(shù)是SUM的話，那就是對所有滿足條件的數(shù)據(jù)進行求和。buildingroom numbertime_slo_ idcount(*)產(chǎn)生的分組種數(shù)：4種；第一種：group by A,B,CGarfield359A1Garfield359B1Saucon651A1Saucon550C1Painter705D1Painter403D1第二種：group by A,BGarfield359A2Garfield359B2Saucon651A1Saucon550C1Painter705D1Painter403D1第三種：group by AGarfield359A2Garfield359B2Saucon651A2Saucon550C2Painter705D2Painter403D2第四種：group by NULL。復(fù)習(xí)題（2） 試分別判斷下列圖中G1和G2是否互模擬(bisimulation)，并說明理由:aaabccbG1=G2=abcabccG1G2ddd答案：(1) 在圖中標出各點的狀態(tài)，我們構(gòu)造關(guān)系， 可知G2可以模擬G1，下面我們討論是否可模擬，在G2中有一個a變換可對應(yīng)到G1中2個變換，即。 給定如下數(shù)據(jù)圖(Data Graph)：試給出其Strong DataGuide 圖 答案：Strong DataGuide 圖 Consider the relation, r , shown in Figure . Give the result of the following query :Figure Query 1:select building, room number, time_slo_ id, count(*)from rgroup by rollup (building, room number, time_slo_ id)Query 1:select building, room number, time_slo_ id, count(*)from rgroup by cube (building, room number, time_slo_ id)答案：Query 1返回結(jié)果集：為以下四種分組統(tǒng)計結(jié)果集的并集且未去掉重復(fù)數(shù)據(jù)。含義是：沒有分組，也就是所有數(shù)據(jù)做一個統(tǒng)計。返回結(jié)果集：為以下八種分組統(tǒng)計結(jié)果集的并集且未去掉
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