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                                   274                  Mathematical Induction












                                                              Figure 6

                                   put one tile in the middle, as in Figure 8. The area remaining to be covered
                                   now consists of four 2 × 2 subgrids with one square removed from each. Each
                                   of these can be covered with one tile, thus completing the upper right subgrid
                                   of Figure 7.














                                                              Figure 7
















                                                              Figure 8
                                     The remaining three quarters of Figure 7 are completed by a similar proce-
                                   dure. The final solution is shown in Figure 9.
                                     The method we used in solving this problem is an example of a recursive
                                   procedure.Wesolvedtheproblemforan8 × 8gridbysplittingitintofour4 × 4
                                   grid problems. To solve each of these, we split it into four 2 × 2 problems, each
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