考虑关系方案r= {e,f,g,H,i,j,k,l,m,m }和函数依赖项{{{e,f}->{g},{f}->{i,j},{e,h }>{k,l},k->{m },l->{n}。r是什么? (A) {E,F} (B) {E,F,H} (C) {E,F,H,K,L} (D) {E} 答复: (B) 说明: 所有属性都可以从{E,F,H}派生
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要解决这些在入门论文中经常被问到的问题,请尝试使用快捷方式来解决,这样可以节省足够的时间。
第一种方法:
使用给定的选项尝试获得每个选项的闭包。解决方案是一个包含R和最小超级密钥的解决方案,即 候选密钥。
A) {EF}+ = {EFGIJ} ≠ R(The given relation) B) {EFH}+ = {EFGHIJKLMN} = R (Correct since each member of the given relation is determined) C) {EFHKL}+ = {EFGHIJKLMN} = R (Not correct although each member of the given relation can be determined but it is not minimal, since by the definition of Candidate key it should be minimal Super Key) D) {E}+ = {E} ≠ R
第二种方法:
Since, {EFGHIJKLMN}+ = {EFGHIJKLMN} {EFGHIJKLM}+ = {EFGHIJKLMN} ( Since L -> {N}, hence can replace N by L) In a similar way K -> {M} hence replace M by K {EFGHIJKL}+ = {EFGHIJKLMN} Again {EFGHIJ}+ = {EFGHIJKLMN} (Since {E, H} -> {K, L}, hence replace KL by EH) {EFGH}+ = {EFGHIJKLMN} (Since {F} -> {I, J} ) {EFH}+ = {EFGHIJKLMN} (Since {E, F} -> {G} )
这一解释是由 曼尼什·雷。
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