The New Sudoku Players' Forum

[haversine]

Hello all, I down loaded "solving sudoku" by Michael Mepnam and was working the tough sample sudoku. I got to this point:
**1 9*7 **8
6** 185 73*
**7 46* 1**
*34 *9* ***
*** 5*4 ***
*** *1* 42*
**5 *71 9**
*1* 84* **7
7** *59 2**

The center square in box 5 can have a 2 or 3. I ended up solving this sudoku by first trying 2 this ended up with no place to put a 7 in box 4. A 3 in the center square lead to the solution. My question is does anybody have a better way to solve this without just tying the two different alternatives? Thanks
Russ[/code]

[Doyle]

Tough one. Plugged it into the Simple Sudoku solver, which makes a few eliminations based on blocked candidates, then stalls. Can you post the original puzzle, without your entries?

[haversine]

Here is the orginal sudoku:

**1 9** **8
6** *85 *3*
**7 *6* 1**
*34 *9* ***
*** 5*4 ***
*** *1* 42*
**5 *7* 9**
*1* 84* **7
7** **9 2**

Russ

[Doyle]

Thx, but Simple Soduko still can't solve it, quits at same spot. May be outside of its algorithm set; i.e., very tough, or just one requiring T/E, and you did solve it that way. Some will argue there is always a logical way, if the puzzle is valid.

[Edit: Reading around, discovered that Mepham is known, not too kindly on this board, for publishing puzzles that do require T/E, at least at the current stage of solving capabilities.]

[emm]

Susser solves it using Nishios. Pappacom definitely a 'not valid'.

[bennys]

I think i can see another step but it doesn't help much
A can't be 4
because then C and D force B to be
2

So F is 4 and then you get that E is 8 but thats all.

Code: Select all

+-------+-------+-------+
| A B 1 | 9 . 7 | C D 8 |
| 6 . . | 1 8 5 | 7 3 . |
| . . 7 | 4 6 . | 1 . . |
+-------+-------+-------+
| . 3 4 | . 9 . | . . . |
| . . . | 5 . 4 | . . . |
| . . . | . 1 . | 4 2 . |
+-------+-------+-------+
| F . 5 | . 7 1 | 9 . . |
| . 1 . | 8 4 . | . E 7 |
| 7 . . | . 5 9 | 2 . . |
+-------+-------+-------+

[lunababy_moonchild]

[Edit: Reading around, discovered that Mepham is known, not too kindly on this board, for publishing puzzles that do require T/E, at least at the current stage of solving capabilities.]

Michael Mepham himself said in the past that his harder puzzles required t&e, and he published his own guide on how to solve his puzzles that explicitly says that (I know this because I read it, unfortunately I can't lay my hands on either the copy I printed or the website). He seems to have changed his mind, though and explains himself here : http://www.sudoku.org.uk/goodbye.htm

Luna

[Jeff]

I think i can see another step but it doesn't help much
A can't be 4
because then C and D force B to be 2

So F is 4 and then you get that E is 8 but thats all.

Code: Select all

+-------+-------+-------+
| A B 1 | 9 . 7 | C D 8 |
| 6 . . | 1 8 5 | 7 3 . |
| . . 7 | 4 6 . | 1 . . |
+-------+-------+-------+
| . 3 4 | . 9 . | . . . |
| . . . | 5 . 4 | . . . |
| . . . | . 1 . | 4 2 . |
+-------+-------+-------+
| E . 5 | . 7 1 | 9 F . |
| . 1 . | 8 4 . | . . 7 |
| 7 . . | . 5 9 | 2 . . |
+-------+-------+-------+

Well done, Bennys. Actually, you are so close from solving it from here. All you need is an xyz-chain as follows:

r7c2=2 => r7c4=r9c4=36 (naked pair) => r8c6<>6
r7c2=6 => r7c9<>6 => r8c7=6 or r8c8=6 => r8c6<>6
The rest is trivial.

BTW, there are at least 4 double implication forcing chains to be identified from the original grid.

Chain 1 enables a 4 to be removed from r7c9 (turbot fish)
Chain 2 enables a 2 to be removed from r7c1
Chain 3 enables an 8 to be removed from r7c1
Chain 4 enables a 4 to be removed from r7c8

Chain 2 and chain 3 would have fixed a 4 in r7c1 and an 8 in r7c8 also.
Double implication forcing chain and xyz-chain are both non-T&E.

[haversine]

Thanks for all the responses, I need to find out about "double implication forcing chains"
Are there any links descriping them?
Russ

[r.e.s.]

A can't be 4
because then C and D force B to be 2

So F is 4 and then you get that E is 8 but thats all.

Well done, Bennys.

???
A and E contain the only 4-candidates in c1, so
A isNOT 4 => E IS 4 => F isNOT 4.

The candidate grid is

Code: Select all

{2345}  {245}   {1}     {9}     {23}    {7}     {56}    {456}   {8}     
{6}     {249}   {29}    {1}     {8}     {5}     {7}     {3}     {249}   
{23589} {2589}  {7}     {4}     {6}     {23}    {1}     {59}    {259}   
{1258}  {3}     {4}     {267}   {9}     {268}   {568}   {1567}  {156}   
{1289}  {26789} {2689}  {5}     {23}    {4}     {368}   {1679}  {1369} 
{589}   {56789} {689}   {367}   {1}     {368}   {4}     {2}     {3569} 
{248}   {2468}  {5}     {236}   {7}     {1}     {9}     {468}   {346}   
{29}    {1}     {2369}  {8}     {4}     {236}   {356}   {56}    {7}     
{7}     {468}   {368}   {36}    {5}     {9}     {2}     {1468}  {1346} 


[Jeff]

r.e.s., bennys didn't provide a proof for his forcing net. It goes like this:

r1c1=4 => r1c7=r1c8=56 => r1c2=2
r1c1=4 => r2c2=r2c3=29 => r1c2<>2
Therefore r1c1<>4 => r7c1=4

[Jeff]

........I need to find out about "double implication forcing chains" ....Are there any links descriping them?

Refer this thread to start with.

The result of the destination cell is forced by two implications in opposite directions from any cell on the chain to the destination cell, thus the name double implication.

[r.e.s.]

r.e.s., bennys didn't provide a proof for his forcing net. It goes like this:

r1c1=4 => r1c7=r1c8=56 => r1c2=2
r1c1=4 => r2c2=r2c3=29 => r1c2<>2
Therefore r1c1<>4 => r7c1=4

The "therefore ..." is yours, not bennys', which is what I was commenting about; i.e., you said "well done" immediately after quoting this:

A can't be 4 [...] So F is 4 and then you get that E is 8 [...]

that is, he was concluding that r7c1=8 (rather than r7c1=4).

[bennys]

I couldn't see it (maybe i looked at the wrong place?)

[Jeff]

I couldn't see it (maybe i looked at the wrong place?)

In this case, do you want me to list my chains.