## question about tough sample sudoku in "solving sudoku&q

All about puzzles in newspapers, magazines, and books

### question about tough sample sudoku in "solving sudoku&q

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]
haversine

Posts: 3
Joined: 29 September 2005

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?
Doyle

Posts: 61
Joined: 11 July 2005

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
haversine

Posts: 3
Joined: 29 September 2005

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.]
Doyle

Posts: 61
Joined: 11 July 2005

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

Posts: 987
Joined: 02 July 2005

### I think i can see another step but it doesn't help much

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 . . | +-------+-------+-------+`
bennys

Posts: 156
Joined: 28 September 2005

Doyle wrote:[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
lunababy_moonchild

Posts: 659
Joined: 23 March 2005

### Re: I think i can see another step but it doesn't help much

bennys wrote: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.
Jeff

Posts: 708
Joined: 01 August 2005

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

Posts: 3
Joined: 29 September 2005

### Re: I think i can see another step but it doesn't help much

Jeff wrote:
bennys wrote: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}  `
r.e.s.

Posts: 337
Joined: 31 August 2005

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

Posts: 708
Joined: 01 August 2005

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

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.
Jeff

Posts: 708
Joined: 01 August 2005

Jeff wrote: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:
bennys wrote: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).
r.e.s.

Posts: 337
Joined: 31 August 2005

### Thanks Jeff

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

Posts: 156
Joined: 28 September 2005

### Re: Thanks Jeff

bennys wrote:I couldn't see it (maybe i looked at the wrong place?)
In this case, do you want me to list my chains.
Jeff

Posts: 708
Joined: 01 August 2005

Next