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!