## STUCK!!!

Advanced methods and approaches for solving Sudoku puzzles

### STUCK!!!

The following puzzle has me stuck. I know the answer via looking it up at Pappocom but I really would like to figure out the next move without T & E

8x4 75x x6x
156 x24 8x7
xxx xx6 5x4

912 473 658
x6x xxx x4x
438 x65 7xx

x81 6xx 4x5
6xx x4x x89
x49 xx8 2x6

I have penciled in all possible # but have more than one canidate for each box, row and column.
Thanks for the help in advance.
Ken
captken

Posts: 3
Joined: 12 September 2005

Hi Ken,

Here is the puzzle with candidates showing:

Code: Select all
` {8}    {29}   {4}    {7}    {5}    {19}   {139}  {6}    {123}  {1}    {5}    {6}    {39}   {2}    {4}    {8}    {39}   {7}    {237}  {279}  {37}   {1389} {1389} {6}    {5}    {1239} {4}    {9}    {1}    {2}    {4}    {7}    {3}    {6}    {5}    {8}    {57}   {6}    {57}   {1289} {189}  {129}  {139}  {4}    {123}  {4}    {3}    {8}    {129}  {6}    {5}    {7}    {129}  {12}   {237}  {8}    {1}    {6}    {39}   {279}  {4}    {37}   {5}    {6}    {27}   {357}  {1235} {4}    {127}  {13}   {8}    {9}    {357}  {4}    {9}    {135}  {13}   {8}    {2}    {137}  {6}    `

As a hint, please consider the 3's in r3c1 and r3c3.
(This hint is actually from Simple Sudoku .)
Nick67

Posts: 113
Joined: 24 August 2007

As far as I can tell you MUST use T&E. But that not too hard.

cell (row=1,col=2) only has two posibilities: 2,9
if you go for 9, there is no solution
if you go for 2 you can solve it quite easily.

824 759 361
156 324 897
397 816 524

912 473 658
765 281 943
438 965 712

281 697 435
673 542 189
549 138 276
jdevor

Posts: 3
Joined: 19 September 2005

cell (row=1,col=2) only has two posibilities: 2,9
if you go for 9, there is no solution
if you go for 2 you can solve it quite easily.

That seems to be a reasonable approach.
But, if you prefer, you can actually do it without T & E.

As a hint, please consider the 3's in r3c1 and r3c3.
(This hint is actually from Simple Sudoku .)

The only cells in box 1 that have a 3 as a candidate
are r3c1 and r3c3. So in the final solution, 1 of those
2 cells must contain a 3. So, we can eliminate 3 as
a candidate for all the other cells in r3.

After we do this, there is a hidden single in c4.
So we can eliminate more candidates. And then
there is another single ... and so on ...
Nick67

Posts: 113
Joined: 24 August 2007