Secret To Spotting Sudoku Almost Locked Sets - SHC 275

[@SmartHobbies]

📕Get my FREE Solving Guide that will help you solve over 80% of all Sudoku puzzles🧩to include NYT Hard👉👉www.buymeacoffee.com/timberlakeB/e/125822 Timestamps 0:00 Intro 00:19 It’s Solving Time 03:51 BONUS Tip 05:24 Question Of The Day 06:00 Single Candidate Vs BVCs 06:45 Filling Out BVCs 08:08 Setting Up Almost Locked Sets 10:26 Secret To Spotting Almost Locked Sets 14:35 Setting Up Advanced Strategy 17:15 Mini X-Wing

[@neokart2660]

Hi, Neokart here. Thanks a lot for the feature, you explained perfectly how to tackle more complicated sudokus. The way you solved it was great, and somewhat equivalent to what I had planned. I thought of it starting from the pattern you talk about at minute 8:50. The idea was to play with those xyz-wing patterns and column 5. Those xyz-wing patterns must always contain a digit 1, and since there can't be two 1s in column 5, there must be at least one 1 in r3c7 or r7c3 ....(i), so: r3c3 doesn't contain 1. This creates a y-wing (156) based on r3c3, so: there must be at least one 1 in r3c7 or r6c3 ....(ii). Then, (i) and (ii) are only possible at the same time if r3c7 is 1. I hope I made myself clear. Very good video! Keep up the good work.

[@SmartHobbies]

Oh, that is slick. I wish I would have taken the logic with those 2 triples a little further. Thank you for the explanation and for a really great puzzle.

[@JohnRandomness105]

I began with Rangsk's standard opening, but eventually degenerated into my not-quite-Snyder notation. Then I resorted to my standard practice when I can't find anything else: centermark the grid block by block. Block 4 had a 156 triple that allowed the first three cells without a 1. Upon finishing the centermarking, I went through each digit. I found a finned swordfish in 6s and a finned X-wing in 8s. I bifurcated on the 14 in R4C5 -- cornermarking 1 and centermarking 4. (Cornermarking involved cornermarking digits, while centermarking involved removing digits.) I managed to cornermark the entire grid without noticing any collisions. When centermarking, I hit a surprising number of matches -- placing the digit. Eventually, the cornermarks turned out correct. 4:50 Filling out block 4, of course, reveals the 156 triple, when then reveals the 489 triple. 5:40 At that point in the puzzle, I was still filling out the grid. 6:30 As mentioned above, I found both the finned swordfish and finned X-wing. I thought they were superfluous and accidental, but I never found out for sure. 12:30 I suppose at one point, I will master this almost-locked set tactic, it will be a while. In my grid, the 3s were already restricted in row 7, but the 5s can do considerable work on row 3.

[@SmartHobbies]

Thank you for your step by step analysis, John. An alternate way of solving this would be to use AIC, which I think you found with your bifurcation. ALS are easy to make for me but hard to connect into any meaningful eliminations. I found a great opportunity to apply the XZ rule here, and I loved how the logic worked out.

[@anaayoung9142]

Very hard puzzle indeed but I enjoy watching your solve. I had learn that, when is a unfair puzzle I just need to sit and watch 😅 Thanks! 😊

[@SmartHobbies]

Glad you liked it Ana. You will start seeing and solving the tricks in these yourself soon. Keep it up. 👍🏻

[@Sg190th]

18:10 I noticed how 56 at box 1 sees 16 in box 1 and the other in box 4 at column 3.

[@SmartHobbies]

Nice job spotting that. Thank you for sharing.

[@Ardalambdion]

The software offers some snow this month if you place the 3s.

[@Ardalambdion]

Columns 3 and 5 allow combined 11 values in 9 squares. But rows 3 and 7 allow only one of 5 and one of 3, thus removing 2 from the equation. Thus, they have to remain.

[@SmartHobbies]

I did appreciate that Sven did that for this month.

[@BradJames878]

This technique is just too much for my comprehension; there is no way I'd be able to set up the colored cells without a video telling me where to place them. Then even with the colored cells in place, I still don't see how the deductions can be positively reached. There are two target columns (3&5), and one contains 3 colored cells, and one has 4. I would forever be asking myself why a 5 can't be in block 1, row 2, column 3, or why a 3 can't be in block 8, row 8, column 5. I worked through scenarios and both possibilities seem viable. All I can positively remove from the given information is a 1 from those two cells, which ultimately solves nothing.