September 27 solved!? Or…

I’ve received a possible solution for September 27 puzzle using only minor trial and error from udosuk. Here are his comments. You can also download the images that he sent:

Firstly, the first pic (dk0927a.png) shows the steps I last posted which narrows R[1,2]C7 to {25} and then some more…

From there I’ve worked out the steps which could “nail” R1C4 to 1 like fred did, but with shorter and simpler steps.
Moreover, I never branch more than 1 level, so it could be considered as “minor T&E”… here is the analysis:

———

(From the pic dk0927a.png)

Firstly, if R3C4=1, then the 11-cage must be 4-2-5, and R[1,2,7]C5 must be
4-5-2 to get the 11 total.
Then R[1,2]C4 will be forced to be {67}, and we can’t make up the 14-cage because R1C3 cannot be 1!

If R1C5=1, then the 11-cage must be 1-8-2, forcing R2C7=5, and R[1,2,7]C5 must be 1-2-8 to get the 11 total.
Now, with the top 19-cage, R[2,3]C6=19-2-5=12={39}, so R[8,9]C6={67} (9 excluded).
Next we consider the 14-cage. Obviously R1C3 must be 3 or that cage will total at least 15.
So, R4C3 must be 2, and the R[8,9]C3 will be forced to {4,5} (9-cage).
Because R9C[3-5] is a 12-cage, now R9C5 cannot be 9 (4+9>12), so R8C5=9, and R9C5={45}.
With the bottom 19-cage, R7C4+R8C[3,4]=10. This forces the 3 cells to be {145}… so R7C4=1, R8C4={45}.
From here, we get naked pairs of {45} in both the row 8 and row 9. Look at the 7-cage R[8,9]C9, it can only be {2,3} now!


(The pic dk0927b.png shows this situation.)

Hence, R1C5=1 will lead to a contradiction. Hence the 1 in nonet-2 must be in R1C4, making the 11-cage 4-2-5.
So, R2C7=2 and R[1,2,7]C5 must be 4-5-2 to get the 11 total, forcing R9C5 to be 1.
Now, with the top 19-cage, R[2,3]C6=19-2-5=12={39}, so R[7-9]C6={678} (hidden triple).
This makes R3C5=8 and R8C5=9, and forces R[7-9]C4 to be {345}.
With the bottom 19-cage, R7C4+R8C[3,4]=10. This forces the 3 cells to be {235}… so R8C3=2, R[78]C4={35}.
Now R9C4 must be 4, making R9C3=7 (12-cage)… from now on everything should fall into place pretty easily.

(The pic dk0927c.png shows the current progress. I think most people can complete the grid from here…)

———

So this sort of “logically” solves your puzzle…
And I think it’s impossible to be solved without using any “minor” T&E steps.
Apart from the sequence which contradicts R1C5=1, all others are relatively short and simple.

Thanks for your terrific puzzle… hope people will be satisfied with this “proof”…

regards,
udosuk

I can only sincerely thank to udosuk for all this work. Also, I checked the solution in Perfect Sudoku software (my development version which has a new solving algorithm) and I can tell you this: after solving 1, 4 and 5 in the top-center nonet as described by udosuk, the solver was able to find the solution without trial and error, using only “innies/outies” and cage splitting!

Related Posts Plugin for WordPress, Blogger...
This entry was posted in General and tagged , , , , , . Bookmark the permalink. Post a comment or leave a trackback: Trackback URL.

One Trackback

  • By Why do I make puzzles? A brief history of Djape.net on January 14, 2012 at 4:24 am

    […] Soduko puzzle. Some visitors had made a lot of progress on it, and a guy called udosuk provided a solution to the September 27 Killer Sudoku puzzle with only minor trial and error. By the way, if udosuk is still visiting this website or if […]

Leave a Reply