所以我有这个问题,我想知道为什么S2 = {AD->C}
来自S1。
考虑关系R(A,B,C,D,E)和函数依赖性集合S1 = {AB→C,AE→D,D→B}。从S1开始,以下哪一组FD NOT 跟随?
S2 = {AD->C}
S2 = {AD->C, AE->B}
S2 = {ABC->D, D->B} //correct answer
S2 = {ADE->BC}
因此,在S1中使用FD,并应用闭包:
{AD}+ = {ABCD} //why does this set follow from S1? What about E?
{AD}+ = {ABCD}, {AE}+ = {ABCDE}
{ABC}+ = {ABC}, {D}+ = {B} //correct answer
{ADE}+ = {ABCDE}
答案 0 :(得分:1)
您不需要闭包的结果来包含所有属性。只要您获得的闭包包含S2右侧的所有属性,S2就会从S1开始。
在这种情况下,在你完成闭包之后,C在闭包中,这意味着S2来自S1