Friday 13 May 2011

Sudoku - Locked Puzzle (Part 3)

… And yet locked again!
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There is a full house in row 8, involving the digits 2 and 8 again. This time it is in cells C8 and E8.

20     CE8 : 28 N
21     D8 / 2  R
22     G8 / 2  R
23     H8 / 2  R
24     F8 / 8  R

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Another full house is in row 3, with H3 and I3 housing the digits 3 and 5. The digits 3 and 5 can be eliminated from the rest of row 3.

25     HI3 : 35 N
26     B3 / 3  R
27     C3 / 3  R
28     C3 / 5  R
29     D3 / 3  R
30     D3 / 5  R
31     F3 / 5  R

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The elimination of 3 and 5 from C3 leads to the confinement of 3 and 5 along column C, which in turn leads to a full house in cells C1 and C4. Thus, digit 2 can be eliminated from the 2 cells.

32     C14 : 35 C
33     C1 / 2  N
34     C4 / 2  N

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The result is another full house in row 4, involving 3 and 5 again, this time in cells C4 and I4. Digits 3 and 5 can be eliminated from cells B4 and G4.

35     CI4 : 35 N
36     B4 / 3  R
37     G4 / 3  R
38     G4 / 5  R

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Another full house from confinement occurs in column H involving digits 1 and 2, and cells H7 and H9.

39     H79 : 12 C
40     H7 / 5  N
41     H9 / 9  N

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The previous elimination in cell G4 results in the full house in column G, with the digits 2 and 9 filling cells G4 and G9. This eliminates 2 from G6, which opens up 3 fill steps in A6, G4 and G6.
42     G49 : 29 N
43     G6 / 2  C
44     A6 = 2 R
45     G4 = 2 B
46     G6 = 8 N

(Click for continuation - Part 4)

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