Join the Smartie Party now 🥳to get EXCLUSIVE reward puzzle packs, ad free content, discord access, and so much more👉👉www.buymeacoffee.com/timberlakeB/membership Timestamps 0:00 Intro 00:20 It’s Solving Time 02:20 Hidden Pair/Neat Naked Triple Trick 05:22 Solving Hidden Singles 07:28 Setting Up Advanced Strategies 09:07 Story Behind This Puzzle 09:37 First Advanced Strategy 12:15 Second Advanced Strategy 13:49 Third Advanced Strategy 15:46 Fourth Advanced Strategy 17:37 My Favorite Advanced Strategy
I saw the solution slightly differently. After the first unique rectangle I noticed that a 3 in R8C1 would not allow any 9 in column 6, and after that I didn’t need any further XY/Z strategies. Thanks for your video, always interesting to see how you identify and apply advanced techniques. I do like these difficult puzzles that require quite a few steps.
I enjoyed this puzzle! It called for a variety of techniques but I was able to solve it. Now I will get to enjoy watching the video to hear how you solved it. I learn a lot from your videos. Thanks!
This is quite challenging and fun. Solved it in 21:03. Everything is relatively easy until 9:10. At that point I looked at 379s in column 6, and recognized that r9c6 must be the same as r2c5 in box 2, and then the same as r3c3 in column 3. Later I noticed that if r1c1 is 5, that makes both r1c9 and r3c2 also 9, which then in column 6 places 9 into r9c6. But r9c6 must be the same as r3c3, not r3c2, so this is not possible, placing 3 in r1c1, after which the puzzle solves easily.
This wasn't as hard as I expected, although I got stuck near the end. I never encountered unique rectangles. After filling out the grid, R1C1 was 35 and R2C7 was 379. Checking each possibility of R1C1 forced R2C7 to be 39. The resulting pointing pair of 7s placed a few more digits, but I got stuck again. The next was a 359 Y-wing using the prior two cells and R1C9, removing a couple 3s. A skyscraper in 3s (rows 1 and 8) removed a couple more 3s and gave me some digits. After trying a couple other things, I colored 7-not7s. I found that neither of the 57 pair in block 3 could be blue 7s, meaning that blue couldn't be 7. That finished the rest of the puzzle. 8:20 By the time I placed block 2's 5, I had most of the grid centermarked, and I finished filling the grid shortly after. 9:20 BTW, this puzzle had 18 given digits. 10:40 This is where I used R1C1 to remove 7 from R2C7. Later on, removing 3 from R2C5 forced 23 into R8C5 and R9C5. 13:00 After the Y-wing removed a couple 3s, the skyscraper removed the 3 about to be removed here, along with another 3 leaving 79 in R2C5. That removed both 7s and 9s from R8C5 and R9C5. 14:40 I see the not9-9 chain in the grid where I rule out a blue 7 from block 3. If I'd seen it, it would have been easier to collapse the puzzle. 16:20 With the previous work in my grid, nothing further is needed beyond placing the digits.
Awesome Timberlake. I saw the rectangles and was so pleased. Sort of close to getting the XY chains. Thanks for demonstrating these great techniques. I think it takes a lot practice and time to get to your level.
Uniqueness is a horrible step in solving a puzzle. It assumes there's a unique solution to the puzzle when you use it. To solve, a better step would be to find the y-wing.
@@SmartHobbies It's how The Werefrog got past at the point that you decided to use uniqueness. It was row 1, column 9 had the y-wing, or rather, it was the one that had to be only the one thing.
@@SmartHobbies Uniqueness is good for competition, and it's a neat trick to teach people who want to compete, but when showing advanced techniques, The Werefrog believe it best to mention how it can be used, but then say it won't be used here because we also have to show the puzzle is unique.
Let me put in my two cent worth of solution starting at 9:40 : First 3 @(96) would put 2 in a bind : If 2 were at (85), it would lead to 7 @(95), a No-No against UNIQUENESS constraint. And on other hand, 2 at (95) would leave [79]tcp on r8 B8 between cols 4&5, also a No-No against the UNIQUENESS constraint! So the uncovering of [79] tcp on r6B5 btw cols 4&5, together with that of [27] tcp on c9B9 btw rows 8&9 force 3 in B8 to be on c5B8 forming tcp with 2, leaving 7&9 cp in B8 @(84)&(96). But this new discovery is still weak, for it allows no immediate development. A stronger step is to suppose 9 in B7 @(93). It would lead to 9 on c6B2 btw rows 1&3. Also 9@(93) would leave [35] tcp on r8B7 btw cols 1&2 ==> 5@(11), plus 9 in B1@(32)(see how?) ===> 9 on c6@(16) !! and together with 5 uncovered @(11) =++> cell(19) = ZERO!!! Therefore it’s 3, not 9@(93) , thus leaving [59] tcp on r8B7 btw cols 1&2, and leaving [79] tcp on c3B1, 5 in B1@(32), 3@(11)!! Also 3@(93) leads to 3@(85)=> 2@(95) ==> 7@(84)& 9@(96) ==> 9 in B2@(25), leaving [37] tcp on c6B2 btw rows 1&3. Now 3 uncovered @(11) => 7 @(16) ==> [37] cp in B3 @(72)&(38), leaving [59] tcp on r1B3 btw cols 8&9! 9 uncovered @(25) => 7@(23) ==> 7@(38)& 3@(27) ==> [35] tcp r7B9 btw cols 8&9, leaving 9 on c7@(77). And that should be it! The whole purpose is to show this solving is pretty straightforward , any step taken is simple logic, no fancy talk of mumbo jumbo technical terms.
