Page 6 of 62 MCM 2008 Team #3780 p■■■■■pm2B46Z 4■5■ ■■_口■口■■■■■■■■■ Figure 1: Demonstration of indexing schemes. 78 45 874593|1 9|2B584 463m9815 69 Figure 2: Puzzle generated by WebSudoku (ranked as "Easy) 2.1.3 Hidden pair and z can take on the values of 1.2 and 7. We would thus conclude that any candidate of X, y Informally, this rule is conjugate to the Naked or Z that is not either 1, 2 or 7 may be eliminated Pair rule(section 2.1.1). Here, we also consider the condition is that there exist two values u andy 2.1.5 Multi-Line such that at least one of u, v is in each of X? and We will develop this technique for columns, but Y?, but such that for any cell QE(AX,Y1, it works for rows with trivial modifications. Con ? Thus, since A must contain a cell with sider a three blocks Ba, Bb and Bc such that they each of the values, we can force X,, r?ct, u,v. all intersect the columns C. C, and C,. If for An example of this is given in Figure 7. We some value v, there exists at least one cell X focus on the grouping Rs, and label X and Y in each of C and C, such that e X? but that there the puzzle. Since X and Y are the only cells in exists no such X E C, then we know that the cell Rs whose candidate lists contain 1 and 7, we can V B such that v H, v satisfies V EC.Were eliminate all other candidates for these cells. this not the case. then we would not be able to satisfy the requirements for B and Bb 2.1.4 Hidden Triplet An example of this rule is shown in Figure 9 In that figure, cells that we are interested in, and As with the Naked Pair rule (section 2.1.1), we for which 5 is a candidate are marked with can extend the Hidden Pair rule(section 2. 1.3)so asterisk. We will be letting a=0, b=6, c that it applies to three cells. In particular, let A T=0, y=l and z= 2. Then, note that all be a grouping, and let X, Y,ZE a be cells such the asterisks for blocks 0 and 6 are located in the that at least one of (t, u, v) is in each of X?, y? first two columns. Thus, in order to satisfy the and Z? for some values t, u and v. Then, if for constraint that a 5 appear in each of these blocks, any other cell QE(A\X,Y, 21),t, u, v Q?, we block 0 must have a 5 in either column 0 or 1 claim that we can force X?, Y?, Z?Ct, u,vh while block 6 must have a 5 in the other column An example of this is shown in Figure 8, where This leaves only column 2 open for block 3, and so in the grouping Rs, only the cells marked X, y we can remove 5 from the candidate lists for allPage 6 of 62 MCM 2008 Team #3780 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 r r r 0 1 2 r r r 3 4 5 r r r 6 7 8 Figure 1: Demonstration of indexing schemes. 7 8 3 2 4 5 8 7 4 5 9 3 1 8 1 9 2 3 5 8 4 7 9 4 6 3 1 9 8 5 8 1 4 6 6 9 Figure 2: Puzzle generated by WebSudoku (ranked as “Easy”). 2.1.3 Hidden Pair Informally, this rule is conjugate to the Naked Pair rule (section 2.1.1). Here, we also consider a single grouping A and two cells X, Y ∈ A, but the condition is that there exist two values u and v such that at least one of {u, v} is in each of X? and Y ?, but such that for any cell Q ∈ (A\ {X, Y }), u, v ∈/ Q?. Thus, since A must contain a cell with each of the values, we can force X?, Y ? ⊆ {t, u, v}. An example of this is given in Figure 7. We focus on the grouping R8, and label X and Y in the puzzle. Since X and Y are the only cells in R8 whose candidate lists contain 1 and 7, we can eliminate all other candidates for these cells. 2.1.4 Hidden Triplet As with the Naked Pair rule (section 2.1.1), we can extend the Hidden Pair rule (section 2.1.3) so that it applies to three cells. In particular, let A be a grouping, and let X, Y, Z ∈ A be cells such that at least one of {t, u, v} is in each of X?, Y ? and Z? for some values t, u and v. Then, if for any other cell Q ∈ (A\ {X, Y, Z}), t, u, v ∈/ Q?, we claim that we can force X?, Y ?, Z? ⊆ {t, u, v}. An example of this is shown in Figure 8, where in the grouping R5, only the cells marked X, Y and Z can take on the values of 1, 2 and 7. We would thus conclude that any candidate of X, Y or Z that is not either 1, 2 or 7 may be eliminated. 2.1.5 Multi-Line We will develop this technique for columns, but it works for rows with trivial modifications. Con￾sider a three blocks Ba, Bb and Bc such that they all intersect the columns Cx, Cy and Cz. If for some value v, there exists at least one cell X in each of Cx and Cy such that v ∈ X? but that there exists no such X ∈ Cz, then we know that the cell V ∈ Bc such that V 7→ v satisfies V ∈ Cz. Were this not the case, then we would not be able to satisfy the requirements for Ba and Bb. An example of this rule is shown in Figure 9. In that figure, cells that we are interested in, and for which 5 is a candidate, are marked with an asterisk. We will be letting a = 0, b = 6, c = 3, x = 0, y = 1 and z = 2. Then, note that all of the asterisks for blocks 0 and 6 are located in the first two columns. Thus, in order to satisfy the constraint that a 5 appear in each of these blocks, block 0 must have a 5 in either column 0 or 1, while block 6 must have a 5 in the other column. This leaves only column 2 open for block 3, and so we can remove 5 from the candidate lists for all