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computing the closure f plus of a set of functional dependencies f is hard hard in the computational complexity sense turns out it is actually exponential in the number of attributes fortunately we only often just want to know if some particular functional dependency like x determines y is in f plus without computing all of f plus and this can actually be done relatively efficiently the way we are going to do this is we are going to compute the attribute closure of the left hand side x we denote this as x plus and it is it is an attribute or closure with respect to f x plus is defined as the set of all attributes a such that a determines a is in f plus itamp;#39;s all the parts of the closure with x on the left with x on the left hand side so we are going to compute x plus in a relatively simple manner we initialize x plus to be just x and then we repeat the following in a loop until thereamp;#39;s no change sometimes this is also called a fixed point for u determines v that is in f
