Let's play the second Sudoku game at third level hardness
This is the second game play session at Sudoku third level. 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 second game play at the third level while playing the game we were a little surprised to have been forced to use a new advanced logic for valid cell resolution which we had to devise on the go. You need to follow the logic closely. Once we have used this technique, in future this type of constructs won't be a problem for us. In fact we are learning with you through these game plays.
Notice in the game below that number of cells filled up is significantly less than the games at level 2 or 1 we played till now. The number of filled up cells is only about 27 or 28 with empty cells to be filled up about 54 or 53. Earlier we were solving games with 47 to 51 empty cell boards.
But number of empty cells is not the only criterion of hardness, there are other criteria such as bunching up of digits in one area. We are just indicating that these games should really be harder than previous games.
Hard or not, having reached a fairly good comfort level, we don't get mentally stalled or unduly cowed down by the hardness of any problem and deal with the problem as naturally as possible. The more formal stages of problem solving such as problem analysis, problem modeling etc. get automatically included in this approach. So we will play these games in a flow using the techniques developed through playing the past games as they come. And if we are stuck, we will devise new methods.
We will still use row-column scan, but use of DSA will be the main technique used 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.
In this session we will learn a new important logic for valid cell resolution.
Let us play the game.
The Sudoku third level game play session 2
First valid cells, R4C9 3, scan R5,R6 -- R2C6 3, scan R1,R3,C5 -- R1C4 5, scan C5,C6,R3 -- R9C6 7, scan C4,C5,R7 -- R8C3 5, long DSA [1,8,5,6,9] in C3 -- R9C9 5, scan R7,R8,C7,C8 -- Cycle (2,8) formed in R9 from DSA [2,8,3,6] in R9 -- R9C8 3, DSA [3,6] in R9 -- R9C4 6.
Then the next steps, Cycle (4,8) formed in R4 from DSA [4,8,6,9] in R4 -- R4C3 9, DSA [6,9] in R4 -- R4C6 6 -- Cycle (1,8) formed in C3 -- R7C3 6.
At this point we were stuck for a while till we were sure that the new logic worked.
New logic for finding valid cell
In column C2, 2 can be only in two cells R8C2 or R9C2. If 2 is in R8C2, 2 can't be in R8C4 and R8C5 -- If 2 is in R9C2, 2 will be in R8C8 -- Again 2 can't be in R8C4 or R8C5 -- DSA in R8C5 is then [1,4] and it creates a cycle (1,4,8) in C5.
At this point we will show the results achieved till now before going over to next stage.
With this three cell cycle (1,4,8) forming in C5, we could fix the fourth empty cell R2C5 with 2, and this is the real turning point in the whole game.
Stop for a moment and try to understand fully the logic as also the significance and mechanism of a three cell cycle.
From now on it is easy.
This stage starts with, R2C5 2, cycle (1,4,8) in C5 -- R1C8 2, scan R2,R3 -- R9C7 2, scan C8,C9 -- R9C2 8 -- R4C2 4, DSA cancel -- R1C2 1, DSA cancel -- R2C2 5, DSA cancel -- R8C2 2, scan C1 -- R5C2 7.
Then again, R1C3 8, DSA cancel -- R5C3 1 -- R6C1 8, DSA cancel -- R5C1 5 -- R7C1 3, DSA cancel -- R8C1 7 -- R8C4 3, scan R7,C5,C6 -- R8C5 1, cycle (4,9) in R8 -- R7C9 8, cycle (2,4) in R7 -- R6C4 1, scan R4,R5,C4.
We will show the results achieved till now before finishing the game in the next stage.
The first valid cells at this stage are, R3C6 1, Cycle (4,9) in top middle 9-cell square -- R4C5 8 -- R6C5 4 -- R6C8 9, DSA cancel -- R6C7 7 -- R8C8 4, DSA cancel -- R8C9 9 -- R2C9 1, scan C7,C8, R3 -- R3C7 8, scan R1,R2,C9.
Then again, R2C8 6, DSA cancel -- R3C9 4, DSA [4,9] in top right 9-cell square -- R2C7 9 -- R2C1 4 -- R1C1 9, DSA cancel -- R1C6 4 -- R3C1 6 -- R3C4 9 -- R5C9 6.
And the last few cells, R5C8 8 -- R5C7 4 -- R5C4 2, DSA [2,9] in R5 -- R7C4 4 -- R7C6 2 -- R5C6 9. Game solved. End.
The final result is shown below.
A special note for you
We will never say that we have followed the optimal path and so you will always have the opportunity to improve upon our solution.
In the last session of third level game play, we have given by mistake the second exercise repeating from level 2. So this game we played has been the exercise 1 of the last session. The exercise 2 needs to be ignored as it has already been played in an earlier level 2 session.
We leave you here with a new game to solve.
Third game at Third level of hardness