Difficult Sudoku Challenge
I am out to prove a point regarding Sudoku Challenges posted in newspapers and magazines.
My point is simple.
Simply count the number of exposed cells in a Sudoku Challenge and if we find that there are less than 26 exposed numbers in a Challenge, simply ignore that Challenge.
Why? You may well ask.
The simple answer: That Challenge cannot be solved.
“Show me proofs,” you may well say.
The recent Sudoku challenge classified as “596 MEDIUM” published on page 64 of the New Zealand Listener’s May 27-June 2 2017 issue provides the proof.
I shall demonstrate the possible solution by the following screen shots, using the fifteen basic Francis’ Rules for Sudoku to proof my point. There is no solution beyond step 33 or beyond 33 further exposed numbers out of a total of 57 numbers required.
Answers to the first 12 numbers.
|
Step |
Francis’ Rule # |
Look at: |
We see: |
So: |
equals |
|
1 |
1 |
box 123 |
c1=f3=h6=7 |
g2 |
7 |
|
2 |
1 |
box 789 |
a8=e7=3 |
h9 |
3 |
|
3 |
2 |
box 147 |
a8=b4=3 |
c3 |
3 |
|
4 |
3 |
a3 |
only possible number is 8 |
a3 |
8 |
|
5 |
15 |
column c |
triple 268 at c4 c5 c6; g9=5 |
c8 |
5 |
|
6 |
15 |
column c |
triple 268 at c4 c5 c6; g9=5 |
c9 |
4 |
|
7 |
5 |
row 8 |
d9=6 meaning that d8 e8 f8 cannot be 6 |
b8 |
6 |
|
8 |
13 |
box 8 |
triple 358 at d8 e8 f8; e3=1 |
e9 |
2 |
|
9 |
13 |
box 8 |
triple 358 at d8 e8 f8; e3=1 |
f9 |
1 |
|
10 |
3 |
a9 |
only possible number is 7 |
a9 |
7 |
|
11 |
3 |
b9 |
only possible number is 8 |
b9 |
8 |
|
12 |
5 |
row 9 |
remaining cell i9 = 9 |
i9 |
9 |
The solution for numbers 13 to 24.
|
Step |
Framcis’ Rule # |
Look at: |
We see: |
So: |
equals |
|
13 |
1 |
box 789 |
b9=f7=8 |
h8 |
8 |
|
14 |
2 |
box 369 |
g2=h6=7 |
i7 |
7 |
|
15 |
8 |
row 7 |
pair 12 in a7 b7; g2=6 |
g7 |
4 |
|
16 |
8 |
row 7 |
pair 12 in a7 b7; g2=6 |
h7 |
6 |
|
17 |
2 |
box 369 |
g3=h7=6 |
i6 |
6 |
|
18 |
6 |
column i |
remaining 14; e3=1 |
i3 |
4 |
|
19 |
6 |
column i |
remaining cell i1 = 1 |
i1 |
1 |
|
20 |
5 |
row 3 |
remaining 59; a2=9 |
b3 |
5 |
|
21 |
5 |
row 3 |
remaining 59; a2=9 |
h3 |
9 |
|
22 |
9 |
column b |
pair 24 in b1 b2 |
b7 |
1 |
|
23 |
9 |
column b |
pair 24 in b1 b2 |
a7 |
2 |
|
24 |
6 |
column a |
remaining 16; c2=1 |
a1 |
6 |
The solution for numbers 25 to 33.
|
Step |
Francis’ Rule # |
Look at: |
We see: |
So: |
equals |
|
25 |
9 |
column b |
pair 24 in b1 b2; h6=7 |
b6 |
9 |
|
26 |
9 |
column b |
pair 24 in b1 b2; h6=7 |
b5 |
7 |
|
27 |
6 |
column a |
remaining cell a4 = 1 |
a4 |
1 |
|
28 |
3 |
g1 |
only possible number is 8 |
g1 |
8 |
|
29 |
9 |
column h |
pair 25 in h1 h2; a4=1 |
h5 |
1 |
|
30 |
9 |
column h |
pair 25 in h1 h2; a4=1 |
h4 |
4 |
|
31 |
3 |
f8 |
only possible number is 4 |
f8 |
4 |
|
32 |
1 |
box 123 |
c3=i2=3=e7 |
d1 |
3 |
|
33 |
1 |
box 456 |
a4=h5=1=e3 |
d6 |
1 |
At this point, we can go no further without making assumptions that can result in more than one solution to the challenge.
[The fifteen Francis’ Rules for Sudoku are given at the end of this article as an Appendix].
The Challenge after step 33 looks as follows: [Red coloured numbers are those given in Challenge. Black coloured numbers are those uncovered using Francis’ Rules.]
|
Listener Jun 2 2017 Mediun |
||||||||
|
a |
b |
c |
d |
e |
f |
g |
h |
i |
|
6 |
;24 |
7 |
3 |
;45 |
9 |
8 |
;25 |
1 |
|
9 |
;24 |
1 |
;48 |
;4568 |
;56 |
7 |
;25 |
3 |
|
8 |
5 |
3 |
2 |
1 |
7 |
6 |
9 |
4 |
|
1 |
3 |
;268 |
;789 |
;6789 |
;26 |
;29 |
4 |
5 |
|
5 |
7 |
;26 |
;49 |
;469 |
;236 |
;239 |
1 |
8 |
|
4 |
9 |
;28 |
1 |
;58 |
;235 |
;23 |
7 |
6 |
|
2 |
1 |
9 |
5 |
3 |
8 |
4 |
6 |
7 |
|
3 |
6 |
5 |
;79 |
;79 |
4 |
2 |
8 |
1 |
|
7 |
8 |
4 |
6 |
2 |
1 |
5 |
3 |
9 |
The reason for such a blunder in publishers is simply that they (the publishers) have not attempted to solve such Sudoku challenges before publishing them (the challenges).
Funny enough, on the same page of the NZ Listener is another Sudoku Challenge classified as “407 HARD” with 26 exposed numbers. Can it be simply a mistake of copying inaccurately? Nay, no. The one classified as “MEDIUM” has simply no solution. Can you proof me wrong by providing a logical solution to this Challenge with sufficient logical reasonings using the Francis’ Rules for Sudoku? Do not tell me that you are using some fantasy methods like the “Cross Wings” etc, that no ordinary man really understands.
APPENDIX: Francis’ Fifteen Rules for Sudoku:
Francis’ Rule # 1 looks at the horizontal boxes - box 1 covering cells a1-3 b1-3 c1-3; box 2 coverings cells a4-6 b4-6 c4-6 until box 9 covering cells g7-9 h7-9 i7-9, checking for the correct number to fit into the correct cell.
Francis’ Rule # 2 does the same as Rule # 1 except that this time we look at the boxes vertically, boxes 1, 4, 7; then boxes 2, 5, 8 and lastly boxes 3, 6, 9.
Francis Rule # 3 looks at each individual cells in the light of neighbouring cells to see if only one number fits a particular cell.
Francis Rule # 4 to Rule # 6, look at remaining numbers within cells in a box (Rule # 4) or within a row (Rule # 5) or within a column (Rule # 6) to scout for possible solutions.
Francis Rule # 7 to Rule # 9, take a look at possible pairing fo two numbers into two cells within a box (Rule # 7) or within a row (Rule # 8) or within a box (Rule # 9).
Then there are times when only one digit fits a cell within a box (Francis’ Rule # 10) or within a row (Francis’ Rule # 11) or within a column (Francis’ Rule # 12).
The last three rules apply when there are instances where three possible digits are applicable in three different cells within a box (Francis’ Rule # 13) or within a row (Francis’ Rule # 14) or within a column (Francis’ Rule # 15).
Happy Sudoku-ing!
Remember - please do NOT waste time to solve a Sudoku Challenge with less than 26 exposed numbers in the Challenge. You will be frustrated as, unlike the retired me, you do not have the time to waste.
Thanks for your patience.