Sudoku #100 & #101!

By: Dan LeKander

Volume 17, Issue 2, February 2022

Some people regard “7” as a lucky number.  It just happens that 7 will be our Sudoku focus this month.


As a bonus each month we start with a Sudoku puzzle in progress, where it appears there are no more obvious or not-so-obvious clues.  Can you find the hidden clue in Puzzle #100?

Puzzle #100

(The answer follows after the conclusion of Puzzle #101, the feature puzzle for February)

Feature Puzzle

For the last few months we have primarily focused on Step 6.  If you have been following these articles you have found this technique, “Dan’s Yes-No Challenge”, to be a heavyweight for your bag of Sudoku tools.   This February we will switch our primary focus to Step 7, “Dan’s Close Rela-tionship Challenge”.  The criteria for a potential Step 7 exercise is quite simple.   All you need is at least one unsolved cell with two possible options.

Puzzle #101


Once you have  completed the puzzle, to the extent that you have filled-in all obvious  answers and have written all potential options across the top of the unsolved cells (PUZZLE PREPARATION), Dan recommends the following Steps to complete the puzzle.

Step 1:  Sudoku Pairs, Triplets and Quads – See September 2015
Step 2:  Turbos & Interaction – See October 2015
Step 3:  Sudoku Gordonian Rectangles and Polygons – See November 2015
Step 4:  XY-Wings & XYZ Wings – See December 2015
Step 5:  X-Wings – See January 2016

Step 8:  AN EXPANSION OF STEP 7Steps  1-5 are relatively common techniques and are explained in the TI LIFE  articles above. Steps 6-8 are covered in detail, in Dan’s book.


Prior to utilizing Steps 1-8, complete the 5 Steps of Puzzle Preparation …


We will complete all of the first 4 steps in the order we observe them, until we conclude all Puzzle Preparation Step 1-4 clues.

The first thing we observe is that C6R1=2.  Next, C5R2=3.  C4R3, C5R3 have options 678.  C3R5=8.  C9R8=8.  In box 9 a 2 can only exist as an option in C7R7 and C7R8; therefore, a 2 can-not exist as an option in C7R2, C7R4 & C7R5.  Indicate this by placing a 2 in the bottom of those cells.  In box 9 a 3 can only exist as an option in C8R9 & C9R9; therefore, a 3 cannot exist as an option in C2R9, C3R9, C4R9 & C6R9.  In box 4 a 2 can only exist as an option in C2R6 and C3R6; therefore, a 2 cannot exist as an option in C4R6, C8R6 and C9R6.  In box 4 a 7 can only exist as an option in C1R4 and C1R5; therefore, a 7 cannot exist as an option in C1R1, C1R2, C1R7 and C1R8.

These clues give us Example #101.1 below:

Example #100.1

This concludes Puzzle Preparation steps 1-4, but before we move on to step 5 by filling in the options for all the unsolved cells, we will look at the puzzle and ask if there are any good Step 6 potentials.  As in previous articles, we determined a particular number was a potential for a successful Step 6 exercise if that number appears as a given answer in 3 separate boxes, such that the boxes are not side-by-side, nor over each other.  Which numbers do you see that are good candidates?   Yes, 5 and 9.

We will first pick the number 5 and perform the exercise.  At this point at your home, you would put green tokens on two starter cells that are 5’s and black tokens on all unsolved cells that could have the option 5.  For purposed of illustration, in Example #101.2 below we will highlight two starter cells in green, and the unsolved cells that could be a 5 in yellow (vs black).

Example #101.2

Now we will perform the exercise.   We will use C2R9 and C6R9 as our starter cells.  One of these 2 unsolved cells must be 5.  We will start with C2R9 and assume it is the 5, and mark it with a capital “Y” for yes.  Then, we will mark the cells that can and cannot be a 5 with a Y or N.

Next, we assume that C6R9 is the 5 and mark it with a “y”, indicating yes.   Then, we will mark the cells that can and cannot be a 5 with a y or n.  The two cells with a N,n designation indicate that these cells cannot be a 5 regardless of whether C2R9 or C6R9 is actually the 5, and we can elimi-nate the 5 from the options of these cells, giving us Example #101.3 below:

Example #101.3

Now we will perform the Step 6 exercise with the 9’s.   By picking C1R2 and C1R7 as the starter cells and performing the exercise, we ultimately find the C5R7 & C6R4 cannot be a 9, giving us Example #101.4 below:

Example #101.4

We have exhausted our Step 6 possibilities, so we will move perform a Step 7 exercise, “Dan’s Close Relationship Challenge”, in Example #101.5 below:

Example #101.5

I chose a 2-digit option starter cell C2R3 with a sequence 1,5, which we annotate on the 2nd level of this cell.   We begin by asking ourselves that if this cell is actually a 1, what adjacent cells could not be a 1 and annotate those cells as “N1”, again on the 2nd level of the cells.

Next, we assume the starter cell is a 5 and track the results through the puzzle.   If any of the N1 cells is a number other than 1, it means those cells are not a 1 regardless if the starter cell is a 1 or 5, and the 1 could be eliminated as an option for those cells.

Before we perform this exercise, I will list the potential outcomes …

• The tracking of the second number of the starter cell doesn’t reach the N1 cells, and there-fore, the exercise is unsuccessful.
• The tracking of the second number goes entirely through the puzzle without a conflict, indicating that the 2nd number is correct for the starter cell and you have solved the puzzle.
• The tracking of the second number creates a conflict, such as a number showing up twice in a row, column or box.   Or it could show up by having no cell for a particular number in a row, column or box.  Regardless of how the conflict arises, it would mean the second number is incorrect for that cell, and therefore, the answer to the starter cell is the first number.

We will now track the 5 through the puzzle above on the third level of the unsolved cells to pre-serve the integrity of the original puzzle.   If C2R3=5, then C8R3=4, C3R3=9 and C9R3=1.  C2R8 now has options 2367 and C2R9 has options 67.   Then C1R8=5 & C6R9=5.  Now the only cell in row 9 that can be a 9 is C3R9, so we will mark C3R9=9.  We will pause here.  What do you see in column 3.   Yes, two cells are a 9.   We know that cannot happen, so we have proven that C2R3 cannot be a 5, and therefore C2R3=1.  The puzzle is easily solved from this point and the solution is Example #101.6 below:

Let’s pause here for a moment to ask ourselves the best way to select a starter cell.  You start by finding a 2-digit starting cell where one of the two numbers has a reasonable chance to track through the puzzle.  The 5 in C2R3 certainly seems it will track far.   Then you ask if the other number has at least two adjacent cells (in the same row, column or box) that have the same number as an option.

‌‌Example #101.6

May the gentle winds of Sudoku be at your back.

By Dan LeKander

Clue for Puzzle #100 …  did you find the clue?  If not, read on.
Let’s focus on column 6.  We see that the option 1 can only exist in one unsolved cell, C6R8.  It then follows that C6R2=5, C6R4=4, C6R5=2, & C4R8=5.

Editor's Note:  February 2022 with a very special thanks . . .

When we published the final article in Dan's Series of steps to learn the logic of Sudoku, I never in a million years thought that Dan would so graciously offer to do one or two puzzles for us each month - but this month we present numbers 100 & 101! What can I say . . . you Dan and your wonderful proofreader, Peggy are so wonderful.  We appreciate it.

And, if you have not already done so, I suggest you purchase Dan’s book: “3 Advanced Sudoku Techniques, That Will Change Your Game Forever!” Purchase of a book includes a 50-page blank grid pad, 33 black and two green tokens. The book is available online at

Be sure to read the TI Life's review of Dan's book by Jesse Kahn published in Jun 2015.

Here are links to all past Sudoku Puzzle Challenge beginning: February 2016, March 2016, April 2016, May 2016, June 2016, July 2016, August 2016, September 2016, October 2016, November 2016, December 2016, January 2017, February 2017, March 2017, April 2017, May 2017, June 2017, July 2017, August 2017, September 2017, October 2017, November 2017, December 2017, January 2018, February 2018, March 2018, April 2018, May 2018, June 2018, July 2018, August 2018, September 2018, October 2018, November 2018, December 2018, January 2019, February 2019, March 2019, April 2019, May 2019, June 2019, July 2019, August 2019, September 2019, October 2019, November 2019, December 2019, January 2020, February 2020, April 2020, May 2020,  June 2020 and July 2020, August 2020,  September 2020, October 2020, November 2020 and December 2020, January 2021, February 2021, March 2021, April 2021, May 2021, June 2021, July 2021 , August 2021, September 2021 ,  October 2021, November 2021, December 2021 and now January 2022.

(Each month we suggest you go back to the beginning, February 2016, and see if you can solve the puzzle easily - and PLEASE let us know.  This editor goes back all the time, and sometimes even the early ones get her stumped.)

Posted in: Volume 17, Issue 2, February 2022, Sports

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