Learn to solve Sudoku level 3 high level games
Follow the step by step solution to the Sudoku level 3 game 1 to learn how to solve high level Sudoku. This is the start of Sudoku hard level games.
For absolute beginners this is quite a high level, but just like any other human activity, if you are experienced it would seem easy to you.
Following is the first Sudoku level 3 game we'll solve after you solve it first.
Notice that number of cells filled up is much less than the games at level 2 or 1 we have played till now. The number of filled up cells is only 27 with empty cells to be filled up 54. Earlier we were solving games with 47 to 51 empty cells.
As a strategy, we'll always try to get a valid cell by row column scan first, by possible digit analysis or DSA second and by reduction caused by Cycles third. Cycles are valuable resources and we'll form a Cycle whenever we get the chance.
In addition we'll also use more advanced Sudoku techniques of single digit lock or X wing if forced.
Evaluation of possible digit subsets for all empty cells is often prescribed as a must-do activity. We strongly recommend to avoid this time-consuming activity as far as possible and instead go in for filling up the empty cells by unique valid digits using Sudoku techniques needed.
The Sudoku techniques for solving this first hard Sudoku level 3 game are briefly explained along with the step by step solution.
Now we'll go through the solution of the game.
Solution to Sudoku level 3 game 1 Stage 1: learn how to solve hard Sudoku games
Scan for 1 bears no valid cell but scanning for digit 2 does: R4C2 2 by scan in R5, R6, C3 -- R9C4 2 by scan for 2 in R8, C5. Observe that for this valid cell scan for 2 in R7 is superfluous.
No more 2, 3, 4, but you'll get a break with scan for 5: R1C2 5 by scan for 5 in R2, R3, C1 -- R4C3 5 by scan in C1, C2.
Next valid cell by scan is for digit 8: R9C2 8 by scan for 8 in R7, R8, C3. And these are all of valid cell by simple row column scan at this point of time.
Now we'll try out possible digit subset analysis DSA technique for empty cells in column C3, the most favorable zone with only three cells yet to be filled till this point.
The possible digit subset that are left to be filled in this column is DS [1,7,9].
This is the DS in C3 as well in the cell R2C3 which is lighted up or affected by the two digits [1,7] with 1 in top left major square and 7 in R2. These two digits are thus reduced from DS [1,7,9] in R2C3 leaving only digit 9 in it -- R2C3 9.
In the same way reduced DS [1,7] in R7C3 in same column C3 is further reduced by 1 in R7 to leave 7 in R7C3: R7C3 7 -- R9C3 1 by exception in C3.
Continuing possible digit subset DSA analysis on the next most favorable zone C5, R3C5 4 by reduction of [1,3,5] from column DS [1,3,4,5].
With reduced DS [1,3,5] in three empty cells of C5, 5 in R9 and 1 in bottom middle major square, R9C5 3 by DS reduction of [1,5] -- R7C5 5 by reduction of 1 in R7, and R1C5 1 by exception in C5.
Game status at this first stage shown below.
Solution to the Sudoku level 3 game 1 Stage 2: Good use of Sudoku technique DSA for finding valid cells
Carrying on our good work in the favorable zone row R9, DS [6,7] in two empty cells of R9 is reduced by 7 in C9 to 6 in R9C9: R9C9 6 -- R9C8 7 by exception in R9.
Opportunistic valid cell by possible digit analysis DSA: R3C9 9 by reduction of [1,2,3,4,5,6,7,8] from 9 possible digits in it.
How Sudoku technique for finding a valid cell by possible Digit Subset Analysis or DSA works
Let's see how R3C9 9 by DSA technique worked out.
R3C9 is at the meeting point of three zones: top right major square with filled digits DS [5,8], column C9 with DS [2,6,7,8], and row R3 with filled DS [1,3,4,5,8].
Together these three zones have unique set of digits [ 1,2,3,4,5,6,7,8]. Only 9 is missing. That's how 9 has to be placed in R3C9. It is the only digit left for R3C9 that is not already lighting up the cell.
Carrying on in the neighborhood favorable zone of R3, DS in three empty cells of R3 is [2,6,7] and with reduction of digit 2 by 2 in C2 and 2 in C7, Cycle (6,7) formed in R3C2 and R3C7 and R3C8 2.
Concentrating further on C7, by [6,7] in R6 and in bottom right major square Cycle (3,4) formed in R6C7, R9C7 as well as a second Cycle formed as (6,7) in rest two cells R1C7, R3C7 -- R1C8 3 by reduction of [1,4] from DS [1,3,4] in 3 empty cells of top right major square and hence in R1C8.
Opportunistic valid cell: R7C8 4 by DSA reduction of [3,6] from DS [3,4,6] in three empty cells of R7 -- R8C7 3 by reduction -- R6C7 4 by reduction -- R2C9 4 by scan for 4 in C8 -- R2C81 by exception in R2 and top right major square -- R8C9 1 by scan for 1 in C8 -- R8C8 5 by exception in bottom right major square.
Falling back on good old row column scan, R5C9 5 by scan for 5 in R4, C8 -- R4C9 3 by exception in C9.
R4C8 6 by scan for 6 in R5, R6. R4C4 4 by DSA reduction of 7 from DS [4,7] in R4 -- R4C6 7 by exception -- R8C4 7 by scan in C6 -- R8C6 6 by exception in bottom middle major square.
Results of action taken at this stage shown below.
Solution to Sudoku level 3 game 1 final Stage 3: Finishing off
R2C4 6 by scan for 6 in C6. R2C2 3 by DS reduction of [2,8] from DS [2,3,8] in R2 -- R2C1 2 by DS reduction of 8 from DS [2,8] -- R2C6 8 by exception in R2 -- R1C6 2 by exception in top middle major square. R6C6 3 by scan for 3 in R5, C4 -- R5C6 1 by exception in C6 -- R6C4 5 by scan for 5 in R5 -- R5C4 8 by exception in central middle major square -- R6C8 8 by scan for 8 in R5.
R5C8 9 by exception in right middle major square.
R6C2 1 by scan for 1 in R5, C1 -- R6C1 9 by exception in R6.
R7C1 3 by scan for 3 in R8, C2 -- R7C2 6 by exception in R7 -- R3C2 7 by reduction -- R3C7 6 by reduction -- R1C7 7 by exception in C7 -- R1C1 6 by exception in R1.
R5C1 7 by scan for 7 in C2 -- R5C2 4 by exception in R5 and left middle major square -- R8C2 9 by exception in C2 -- R8C1 4 by exception in whole 81 cell game.
Final solution below.
Overall this game didn't pose any difficulty.
An important note on writing Digit Subsets or DSs in empty cells
Notice that usually we haven't written DSs in empty cells longer than 2 digits. The reason lies in effectiveness of 2 digit long DSs in empty cells. As soon as you are able to form a two digit DS in an empty cell, you know the cell to be just one step away from becoming a valid cell by DSA cancel.
Furthermore, you never know when a pair of a two digit DS will appear in the same zone, form a cycle and reduce the complexity of the game considerably.
But restricting evaluation of DSs only for 2 digit length is not a practical option. In solving a hard Sudoku, evaluating chosen longer than 2 digit DSs generally is a must-do.
So the strategy we follow is, we avoid evaluating and writing longer than 3 digit DS for an empty cell as far as possible, and DSs are evaluated only for a few cells that are necessary for giving us valid cells.
In general writing longer than three digit DSs in empty cells is wasteful and we avoid it.
A new game for you to solve at Sudoku level 3
We leave you here with a new Sudoku level 3 game to solve. We'll solve this Sudoku game in next session of Sudoku level 3 game play.
Other Sudoku game plays at third level hardness
Sudoku Third level game play 1
First and second level Sudoku games
Fourth level game plays
List of fourth level very hard Sudoku game plays are available at Fourth level Sudoku.