The New Sudoku Players' Forum

[exopolit]

GAM24-04-21.jpg (141.84 KiB) Viewed 694 times

Toronto Globe and Mail, Saturday 24 April 2021 5 star/5 star.
Never taken me more than 30 mins to finish a Globe and Mail sudoku before.
I don't think I'm missing a method in Gould's extreme SuDoku#1, am I?
Is the technique something to do with the 1or9 options in R1C4, R2C6 and R9C5?
Thanks in advance from this newbie.

[jco]

Hello exopolit,

You sensed well some key cells. I suggest searching for an skyscraper.
I wrote below a full solution to the puzzle. In case you need just
a suggestion for a good move, you can ignore the text below. However,
in this case I suggest you to see the comments of the more experienced players (next posts).

This puzzle (SER = 7.2) needs hard work to be solved manually.
A first good step is a skyscraper (SS). But after finding it, a lot more
work is needed. I am presenting below a solution in 4 steps (SS is the first),
with 3 short chain and a relatively easy long one (because is build on bivalue
cells). There are many other ways in fewer steps, but then the chains are a
lot more difficult for you to find (manually) [however, see Leren's post].

1. Skyscraper

Code: Select all

.-----------------------------------------------.
| 2    179    5   | 1-9   8    3   | 6  4    79 |
| 137  137(9) 6   | 4     5    1(9)| 8  37   2  |
| 8    4      39  | 2     7    6   | 1  35   59 |
|-----------------+----------------+------------|
| 5    268    4   | 79    69   279 | 3  289  1  |
| 137  1237   37  | 8     139  4   | 5  29   6  |
| 9    12368  38  | 5     136  12  | 7  28   4  |
|-----------------+----------------+------------|
| 37   37(9)  1   | 37(9) 2    5   | 4  6    8  |
| 6    357    2   | 137   4    8   | 9  157  57 |
| 4    5789   789 | 6     19   17-9| 2  157  3  |
'-----------------------------------------------'

(& 3 placements)

2. Small chain 1 (with explanation and notation to be used)

Code: Select all

.-----------------------------------------------.
| 2    79     5   | 1    8     3  | 6  4     79 |
| 137  137    6   | 4    5     9  | 8  37    2  |
| 8    4      39  | 2    7     6  | 1  35    59 |
|-----------------+---------------+-------------|
| 5    268    4   | 79   69    27 | 3  289   1  |
| 137  1237   37  | 8   b1(39) 4  | 5 a(9)-2 6  |
| 9    12368 d(38)| 5   c1(3)6 12 | 7 e(28)  4  |
|-----------------+---------------+-------------|
| 37   379    1   | 379  2     5  | 4  6     8  |
| 6    357    2   | 37   4     8  | 9  1     57 |
| 4    5789   789 | 6    19    17 | 2  57    3  |
'-----------------------------------------------'

if 9 is not a r5c8, then it must be at r5c5, and
in this case, 3 is false there, which implies that
r6c5=3, and this in turn implies that 3 cannot be
at r6c3, so 8 must be there, which implies then
that 8 cannot be at r6c8, so 2 must be there.
We have shown that in any case 2 is false
at r5c8. We write the chain like this

(9)r5c8=(9-3)r5c5=(3)r6c5-(3=8)r6c3-(8=2)r6c8 => -2 r5c8

(either 9 is true at r5c8, or 2 is true at r6c8, so 2 is false at r5c8)

(&2 placements)

3. small chain 3 (same reasoning as the previous chain)

Code: Select all

.-------------------------------------------.
| 2    79    5   | 1    8    3  | 6  4   79 |
| 137  137   6   | 4    5    9  | 8  37  2  |
| 8    4     39  | 2    7    6  | 1  35  59 |
|----------------+--------------+-----------|
| 5    68    4   | 79   69   27 | 3  28  1  |
|a137  2     37  | 8   b13   4  | 5  9   6  |
| 9   d368-1 38  | 5   c136  12 | 7  28  4  |
|----------------+--------------+-----------|
| 37   379   1   | 379  2    5  | 4  6   8  |
| 6    357   2   | 37   4    8  | 9  1   57 |
| 4    5789  789 | 6    19   17 | 2  57  3  |
'-------------------------------------------'

(1)r5c1=(1-3)r5c5=(3-6)r6c5=(6)r6c2 => -1 r6c2

4. Final XY-chain

Code: Select all

.----------------------------------------.
| 2  a79    5  | 1    8   3  | 6  4   9-7|
| 37  1     6  | 4    5   9  | 8  37  2  |
| 8   4    b39 | 2    7   6  | 1  35  59 |
|--------------+-------------+-----------|
| 5   68    4  | 79   69  27 | 3  28  1  |
| 1   2     7  | 8    3   4  | 5  9   6  |
| 9   368  c38 | 5    16  12 | 7  28  4  |
|--------------+-------------+-----------|
| 37  379   1  | 379  2   5  | 4  6   8  |
| 6   35-7  2  | 37   4   8  | 9  1  h57 |
| 4   5789 d89 | 6   e19 f17 | 2 g57  3  |
'----------------------------------------'


(7=9)r1c2-(9=3)r3c3-(3=8)r6c3-(8=9)r9c3-(9=1)r9c5-(1=7)r9c6-(7=5)r9c8-(5=7)r8c9 => -7 r1c9,r8c2

(either 7 is true in r1c2 or 7 is true in r8c9. We don't know which, but we
can infer that 7 must be false at r1c9 and r8c2.)

After this long but not so difficult chain (because it is built on bivalue cells),
we have only singles to find.
I did not solve this puzzle manually (no time right now for such fun).
I used the program YZF_Sudoku.
For a puzzle like this, one good resource is to use Medusa Coloring.
You can find more about YZF_Sudoku at
http://forum.enjoysudoku.com/yzf-sudoku-t36846.html <----
You can use it for training. HoDoKu is also very good and you can read about
many techniques (including coloring, but Medusa is not there) in its tutorial (site easy to find in the internet).

Regards,
JCO

Edit: fixed a couple typos and added some new text.

[denis_berthier]

.
The only technique you need is a few short bivalue-chains - the most basic AICs.
I think this puzzle was designed as an exercise in bivalue-chains.

Code: Select all

***********************************************************************************************
***  SudoRules 20.1.s based on CSP-Rules 2.1.s, config = BC
***  Download from: https://github.com/denis-berthier/CSP-Rules-V2.1
***********************************************************************************************
Resolution state after Singles and whips[1]:
   +-------------------+-------------------+-------------------+
   ! 2     179   5     ! 19    8     3     ! 6     4     79    !
   ! 137   1379  6     ! 4     5     19    ! 8     37    2     !
   ! 8     4     39    ! 2     7     6     ! 1     35    59    !
   +-------------------+-------------------+-------------------+
   ! 5     268   4     ! 79    69    279   ! 3     289   1     !
   ! 137   1237  37    ! 8     139   4     ! 5     29    6     !
   ! 9     12368 38    ! 5     136   12    ! 7     28    4     !
   +-------------------+-------------------+-------------------+
   ! 37    379   1     ! 379   2     5     ! 4     6     8     !
   ! 6     357   2     ! 137   4     8     ! 9     157   57    !
   ! 4     5789  789   ! 6     19    179   ! 2     157   3     !
   +-------------------+-------------------+-------------------+

biv-chain[2]: r2n9{c6 c2} - r7n9{c2 c4} ==> r1c4 ≠ 9, r9c6 ≠ 9
singles ==> r1c4 = 1, r2c6 = 9, r8c8 = 1
biv-chain[2]: c8n7{r9 r2} - r1n7{c9 c2} ==> r9c2 ≠ 7
biv-chain[3]: r5c3{n7 n3} - r3c3{n3 n9} - r1c2{n9 n7} ==> r5c2 ≠ 7
biv-chain[3]: r6c3{n3 n8} - r6c8{n8 n2} - r5n2{c8 c2} ==> r5c2 ≠ 3
biv-chain[3]: r6c8{n2 n8} - r4n8{c8 c2} - b4n6{r4c2 r6c2} ==> r6c2 ≠ 2
biv-chain[3]: r4n8{c8 c2} - c2n2{r4 r5} - r5c8{n2 n9} ==> r4c8 ≠ 9
singles ==> r5c8 = 9, r5c2 = 2
biv-chain[3]: r6n6{c5 c2} - r4c2{n6 n8} - r6c3{n8 n3} ==> r6c5 ≠ 3
singles ==> r5c5 = 3, r5c3 = 7, r5c1 = 1, r2c2 = 1
biv-chain[3]: c1n7{r7 r2} - r1c2{n7 n9} - r7n9{c2 c4} ==> r7c4 ≠ 7
whip[1]: r7n7{c2 .} ==> r8c2 ≠ 7
biv-chain[4]: r3c3{n9 n3} - r3c8{n3 n5} - r9n5{c8 c2} - b7n8{r9c2 r9c3} ==> r9c3 ≠ 9
stte

[Leren]

Here is a 2 step solution. The first step uses just a bit more than bi-value chains.

Code: Select all

*-------------------------------------------*
| 2    179    5   | 19  8   3   | 6  4   79 |
| 137  1379   6   | 4   5   19  | 8  37  2  |
| 8    4      9-3 | 2   7   6   | 1 a35  59 |
|-----------------+-------------+-----------|
| 5    268    4   | 79  69  279 | 3  289 1  |
| 137  1237   37  | 8   139 4   | 5  29  6  |
| 9    12368 e38  | 5   136 12  | 7  28  4  |
|-----------------+-------------+-----------|
| 37   379    1   | 379 2   5   | 4  6   8  |
| 6    357    2   | 137 4   8   | 9  157 57 |
| 4   c5789  d789 | 6   19  179 | 2 b157 3  |
*-------------------------------------------*

It's in cells a-b-c-d-e and shows that a or e is 3, so r3c3 is not 3. Some singles and a claiming pair in Box 4 get you to here.

Code: Select all

*--------------------------------------*
| 2  7     5   | 1  8    3  | 6  4   9 |
| 13 13    6   | 4  5    9  | 8  7   2 |
| 8  4     9   | 2  7    6  | 1  3   5 |
|--------------+------------+----------|
| 5  268   4   | 79 69   27 | 3  289 1 |
| 17 12    37  | 8  139  4  | 5  29  6 |
| 9  1268  3-8 | 5  136 b12 | 7 a28  4 |
|--------------+------------+----------|
| 37 39    1   | 79 2    5  | 4  6   8 |
| 6  5     2   | 3  4    8  | 9  1   7 |
| 4  89   d78  | 6  19  c17 | 2  5   3 |
*--------------------------------------*

The second step is a bi-value chain in cells a-b-c-d and shows that a or d is 8, so r6c3 is not 8. The puzzle solves with singles from there. I've dispensed with any formal notation as I don't know your background.
'
Leren

[jco]

Hello,

Using coloring, I found the following a solution for the puzzle in 2 steps, similar to Leren's solution.

Code: Select all

.----------------------------------------------------.
| 2     179     5     | 19   8    3   | 6   4     79 |
| 137   1379    6     | 4    5    19  | 8   37    2  |
| 8     4      a3(9)  | 2    7    6   | 1   35  b(59)|
|---------------------+---------------+--------------|
| 5     268     4     | 79   69   279 | 3   289   1  |
| 137   1237    37    | 8    139  4   | 5   29    6  |
| 9     12368   38    | 5    136  12  | 7   28    4  |
|---------------------+---------------+--------------|
| 37    379     1     | 379  2    5   | 4   6     8  |
| 6    d3(5)7   2     | 137  4    8   | 9   157 c(5)7|
| 4    e7(58)9 f7(8)-9| 6    19   179 | 2   157   3  |
'----------------------------------------------------'


1. (9)r3c3=(9-5)r3c9=(5)r8c9-(5)r8c2=(5-8)r9c2=(8)r9c3 => -9 r9c3
[&13 placements and 1 elimination by LC: (3)r56c3 => -3 r5c1,-3r5c2,-3r6c2]

My second move is the same as Leren's second move, so it will not be shown.
I wrote this post to explain how to derive, using Medusa colouring with 3 pairs of colors,
the elimination for Step 1 described in the previous grid.
The same approach can be used to derive the second move.

Code: Select all

.-----------------------------------------------.
| 2    1'79   5   | 19    8    3   | 6  4   "7'9|
| 137  1379   6   | 4     5    19  | 8 "3'7  2  |
| 8    4     "3'9 | 2     7    6   | 1 '3"5 '5"9|
|-----------------+----------------+------------|
| 5    268    4   | 79    69   279 | 3  289  1  |
| 137  1237   37  | 8     139  4   | 5  29   6  |
| 9    12368 :3.8 | 5     136  12  | 7  28   4  |
|-----------------+----------------+------------|
| 37   379    1   | 379   2    5   | 4  6    8  |
| 6   3+57    2   | 137   4    8   | 9  157 "5'7|
| 4   -57.89 7:8"9| 6     19   179 | 2  1+57 3  |
'-----------------------------------------------'


We have 3 pairs of colors, each pair with opposite parities: (',"),(.,:),(+.-).
"-" denotes a color (and not that a certain digit will be eliminated) and that
"+" denotes the opposite color of "-" (and not that a certain digit will be placed).
Due to ("5)r3c8 and (+5)r9c8 [located in the same column], we cannot have both colors (",+) true,
so at least one of the colors (',-) must be true.
Due to cell r9c2, we cannot have both colors (-,.) true, so at least one of the colors (+,:) must be true.
Since, (+,-) have opposite parities, at least one of the colors (',:) must be true.
Since (9)r9c3 sees ('9)r3c3 and (:8)r9c3, it must be false
[if the color ' is true, 9 at r3c3 is true, so (9)r9c3 is false;
if the color double dot is true, then since (:8)r9c3 belongs to the
same cell as (9)r9c3, again (9)r9c3 must be false]. So, we can eliminate 9 from r9c3.

Regards,
JCO

Edit: added description of my first step, since after posting I noticed that Leren's first step was slightly different from mine (and the colouring explanation was made for it).

[exopolit]

Many thanks all!
Some new techniques for me to practice.