Simon solves The Daily Telegraph's Diabolical Sudoku from 6 Sept 18, which proves a bit of a monster. Hence the solve is presented in three parts, with this part showing a finned swordfish technique amongst other things.
I really wish you would redo these guides. I really like how you point out the logic in the techniques and explain why it is logic, and in my opinion you have become much better at explaining/talking to the camera compared to these older videos. When i do sudoku which isn´t so often, i have often forgotten the last bit of the techniques, or get stuck and can´t seem to apply logic to anything that allows me to progress.
There was a sashimi x-wing in 9s between columns 2 and 9 which would have eliminated the 9 you caught with your empty rectangle in the last video. Doesn't make much difference in terms of result, but I personally find them a lot easier to spot.... That, combined with the 18/59 doubles on column 9 from the last vid, reveals a 38 w-wing in columns 7 and 8, which resolves the 3s in those columns a little simpler than the wxyz wing you used. For me, that lead directly to a sashimi x-wing in 8s between columns 2 and 8, which I think is a little simpler than the finned swordfish you found, but cracks the puzzle in the same way. It took me a veeeeeery long time to find all that, but it didn't require any of the complicated logic, and made it much simpler to wrap my head round. Still very impressed at how you find these monster techniques to get past your sticking points, especially without all the pencil marks.
After you spotted the finned swordfish, you went back and forgot that in the top right block, 9’s were limited to column 8, which I believe could have helped you solve sooner
Forgetting what I said in the previous video, which took the solution in another direction, and acknowledging that my process of trying to establish doubles (I will abandon my preferred terminology for consistency here) may amount to guessing - if I take the opening position in this video at face value I again immediately "see" a 5 in R4C8 resolving an 89 double in both that row and column three which "potentially" resolves all the 5's, 8's and 9's in both the central and bottom horizontal row of blocks except that row four 89 double. As the 5's, 8's and 9's resolve in that manner so too do a number of the 1's and 2's and I suspect, although I haven't worked it through, that with the 1's comes the resolution of 8's and 9's in row 1 and hence the 8's and 9's throughout the puzzle. Turning now to the position just prior to the swordfish discussion there is an 89 double in row 1 not pencilled which I think resolves this more easily. Again including the pair of 5's not pencilled since video 1, assuming R1C3 is an 8 results in two 5's in column three, so the 8 and 9 in row one are in R1C8 and R1C9 respectively, and the 5 in column three is in row nine. When you reach the position at 10.25 you created the 59 double (as did I in my other solution approach) but, using the empty rectangle logic you used previously, you had eliminated the possibility of R5C1 as a 9 which would have resolved that double immediately.
I sat down to work this before watching the last video, because I branched off about 3:26 in the last video, when you spotted the 389 daisy chain in column two. I invariably bifurcate, but instead of looking primarily for places to eliminate an option, I am usually hoping to find a place where two chains of logic agree. And in this case, I wanted to use the 389/893 (reading down) dichotomy because of the 1/8 and 5/9 pairs in column nine. I had already solved all the 4s, 6s, and 7s, and the 2s left formed a simple x-wing. Whatever I could make work needed to work with 1,3,5,8,or 9, and those two columns gave me all the options to press on. Now, in the 389 universe, the 9 in BLtc forces the 9 in the last column up to MRmr, and combined with the 8 in MLtc I get 5 in MRtc. Alternately, in the 893 universe, the chain is longer, but it goes TCmc8 --> TRmc 1 --> BRmr8 ---> BRmc3 --> TRtl3 --> TRtc8 which combined with MLtc9 to force MRtc5 again. Once I have that 5 in MRtc the puzzle breaks open.
I am sure +rabidsamfan will elucidate but my guess is that it is chain notation (that I don't understand) particularly as I rarely follow chains systematically (although my mention of two 5's resulting in column three in my earlier comment is a form of chain).
I find the r#c# notation confusing, because I want to be able to add the digit to it when I describe moves. The arrow means "forces" when I'm describing a chain. Then, using T =top, M=middle, B=bottom, and L=left, C=center, R=right, I can identify each square/block, column, row, and cell with a combination of capital and small letters so I can add the digit at the end. Hence, TR means the square at the Top Right of the grid. Tm is the middle row of the top zone (i.e., row two in the usual notation), Cc is the same as column five -- the center column in the Center zone. Individual cells are identified with four letters. TLtl is the same as r1c1- in the Top Left box it is the top left cell. If I wanted to describe a 5 placed in the very center of the grid it would be MCmc5. But the real advantage of the notation comes when there's some uncertainty about something. MRm(l/c)3 tells me that the 3 can go in either MRml or MRmc (and in either case, it is eliminated from the rest of row Mm and square MR.) And I can muck around with noting more complicated things too. An example in this puzzle would be MC(m/b)r(8/9). I can also describe the three cell "chute" that is the intersection of a square and a row or column with just three letters, like MCr and shortcut with MCr4 to say that 4 has to fall somewhere in that chute without needing to be more specific.
We could certainly use the rectangle logic again here - nice spot - but I'm not sure I see how you're getting the 3 immediately from it. If we repeat the 9s trick I showed in Part 2, it immediately removes the possibility of a 9 in row 5 col 1.
