By Samir Khuller

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If on the other hand the r sets from W1* include {Ai: i I} from A,,. . , A. and r- II (> 0) sets from the rest, then the union of these r sets is / \ / \\ / (U Ai) u (B\P) = B\(P\(U Ai)) which by (ii) contains m B IP\ U Ail m m-(n -1) elements. But r - II is the number of copies of B\P chosen from 91* and so it cannot exceed m -n. Hence r -IIIm-n andso m-(n II- ) r. Therefore the r sets from 'A* do contain at least r elements in this case too. It follows from the transversal form of Hall's theorem applied to 91* that the family 91* of m sets has a transversal as shown: 91*= (Ai.

Since G is connected it is clear that V = V, u V2, but can we be sure that V, r- V2 = 0 and that each edge joins a member of V, to a member of V2? To show that V, ra V2 = 0 we shall assume that v e V, r) V2 and deduce a contradiction. Since v e V, there must exist a path from v, to v which uses a set of edges E1 , say, where JE1 I is even. Similarly, as v e V2, there must exist a path from v, to v which uses a set of edges E2 , say, where JE21 is odd. e. E* = (E1 u E29(E 1 rm F2)E 43 Three basic principles and let G* be the graph (V, E*).

We now give one example to show how the addition of an extra row to a Latin rectangle translates easily to a problem in transversals. Example To add an extra row to the Latin rectangle 1 2 G l 4 5 3) 2 3 4/ to create a 3 x 5 Latin rectangle with entries in {1, 2, 3, 4, 5} is equivalent to finding distinct representatives of the sets shown below (note that each set has cardinality 3 and that each number 1-5 is in exactly three of the sets): 1 (5 {2,3,4} 2 1 4 2 {3,4,5} {1,3,5} 5 3 3) 4J {1,2,4} {1,2,5} One such collection is shown in bold print.