Show that, if the matrix S resulting from Algorithm 15.3 does not have a row that is all "a" symbols, then projecting S on the decomposition and joining it back will always produce at least one spurious tuple.
What will be an ideal response?
The matrix S initially has one row for each relation R i in the decomposition, with "a"
symbols under the columns for the attributes in R i . Since we never change an "a" symbol
into a "b" symbol during the application of the algorithm, then projecting S on each R i at
the end of applying the algorithm will produce one row consisting of all "a" symbols in
each S(R i ). Joining these back together again will produce at least one row of all "a"
symbols (resulting from joining the all "a" rows in each projection S(R i )). Hence, if
after applying the algorithm, S does not have a row that is all "a", projecting S over the
R i 's and joining will result in at least one all "a" row, which will be a spurious tuple
(since it did not exist in S but will exist after projecting and joining over the R i 's).
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