Let's play the fourth Sudoku game at third level of hardness
This is the fourth game play session at Sudoku third level of hardness. For many this may be quite a high level, but just like any other human activity area, if you are experienced it would seem easy to you. In this fourth game play at the third level while playing the game we needed to use two or three digit long cycles through frequent use of DSA. Once we could form such cycles, things started getting easier.
As is the custom, we will still use row-column scan, but Digit Subset Analysis or DSA will be the main technique along with Cycles. To know what a Cycle or DSA is you may refer to our first and second game play sessions at level 2.
Let us play the game.
The Sudoku 4th game play at third level of hardness
First valid cells, R8C9 1, scan R7,R9,C7,C8 -- R1C6 1, scan R2,C4 -- R3C3 1, scan R1,R2,C1,C2 -- R4C5 1, scan R5,R6,C4,C6 1 over -- R8C8 4, scan C7,C9,R7 -- R9C3 4, scan C2,R7,R8 -- R7C1 7, scan C2,R8 -- R9C7 7, scan R7,R8,C9 -- R3C4 7, scan R1,C5,C6 -- R2C8 7, scan R1,R3,C9 7 over -- R8C7 2, scan R7,R9 -- Cycle (5,6,9) formed in C7 -- R7C7 3 -- Cycle (5,6,8) formed in R7, R7C2 9 -- Cycle (6,8) in C9, R7C8 5.
Notice that going is tough and at this point we couldn't get any valid cell easily. So we continued to form the Digit Subsets or DSs, that is, possible digits in cells one after the other that are primarily less than 4 digit long. We have not started noting down the 4 digit long DSs as these clutter the board and do not generally produce any worthwhile result.
We limit our digit analysis to 3 digit DSs and wait for a cycle to be formed that indirectly enforces a valid cell.
You need to carefully go through the results we achieved till now as reproduced below, because just after this point we will have our breakthrough and it will be a routine affair from then on.
In C2 digit 6 can occur only in R8C2 or R9C2. So it can't occur above these two cells in the column. This results in a cycle (2,3,8) in R3. Because of this cycle R3C8 6 -- R4C8 2, DS cancel -- R1C7 9, DS cancel -- R5C7 5, DS cancel -- R6C9 9, DS cancel -- R6C7 6.
This R3C8 6 is the most crucial breakthrough in the whole game and it has come about because of earlier formation of another cycle (6,8) in R9 because of which digit 6 was not able to appear above the last three rows on the last column. The constraint of digit 6 not able to appear above the last three rows in column C2 is not because of a two cell cycle formation, but the three cell cycle (5,6,8) in bottom left 9-cell square in effect produced the same positive result.
After this point all valid cells are found very easily just by cancelling the digit subsets (or DSs) formed earlier. Let us nevertheless go through the steps to the end.
R2C9 3 DS cancel -- R3C9 2 -- R3C1 8, DS cancel -- R3C2 3 -- R2C2 5, DS cancel -- R1C3 6, DS cancel -- R1C4 4, DS cancel -- R1C1 2 -- R8C1 5, cycle (6,8) in bottom left 9-cell square -- R4C1 9, DS cancel -- R2C1 4 -- R2C3 9, cycle (6,8) in R2 -- R4C3 5, DS cancel -- R5C3 8 -- R5C4 9 -- R4C6 3, DS cancel -- R4C4 6 -- R8C4 8, DS cancel -- R8C2 6, DS cancel -- R9C2 8 -- R9C9 6, DS cancel -- R7C9 8 -- R7C6 6 -- R8C5 9 -- R6C2 2 -- R6C5 8, DS cancel -- R2C5 6.
And the last few cells, R2C6 8 -- R9C6 5 -- R9C4 3 -- R6C4 5 -- R6C6 4. Game solved. End.
The final result is shown below.
Till a point it has been hard game and after the breakthrough it was pretty easy. Nevertheless you should try to solve in an easier path.
As usual we leave you here with a new game to solve.
Fifth game at Third level of hardness