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 00:22 It’s Solving Time 00:42 Finding Restrictions 01:23 Puzzle Story 01:51 Naked Singles 03:39 Conjugate Pairs 04:05 BONUS Tip 05:46 BONUS Strategy 07:59 Setting Up Amazing Strategy 09:24 Strategy To Blow Your Mind 10:57 Following The Snyder
Nice puzzle and solve. Solved it in 18:08. At 9:33 I argued slightly differently: At this point r2c8 can only be 369. If it is a 6, that puts 7 to r2c7, 8 to r1c7 and 9 to r1c1. On the other hand, if r2c8 is not 6, it is 39, so forms a 39 pair with r3c8, placing 1 to r8c8 and 9 to r9c9. In both cases r9c1 must be 5, after which the puzzle solves easily.
It took me a long time, and I was stuck trying things out for the longest time. I placed a 6 in block 7, then a finned X-wing in 6s gave me a 46 pair in block 2, row 3, which I didn't notice for a long time. A finned swordfish removed 7 from one of the 46 cells -- superfluous, a time waster, had I only seen the 46 pair. I got a 19 pair and a 235 triple in block 9. That's about when I got stuck and continued centermarking the grid. I was stuck trying things for a long time. I had 59 in R9C1, 89 in R1C1, and 19 in R9C9 and R8C8. I showed that if R9C1 was 9, then R3C2 had to be 9. I was still stuck for a while, until I took 5 in R9C1 far enough -- 5 in R9C1 meant that R1C1 had to be 9. That allowed me to remove 9s in column 1 from all cells except rows 1 and 9, and finally unstuck me. Columns 1 and 9 had a skyscraper in 9s, which allowed me to remove 9 from R3C2. Combine this with the first sentence of the paragraph, and R9C1 had to be 5. That finally collapsed the puzzle. 7:30 Skyscraper and finned X-wing -- both are there and both have the same effect. 8:50 I filled out the bi-valued cells, along with everything else. A few tri-valued cells helped me. 9:10 Okay, I can see the short chain. Start in R1C1: not9-8 --> R1C7=not8 --> R3C9=8-not9 --> R9C9=9. So either R1C1 or R9C9 is 9, meaning that R9C1 is 5. My overly-long chains: If R9C1=9, then R1C1=8, R3C1=2. If R9C1=9, R9C9=1, R8C8=9, R3C8=3, R2C8=not3, R2C2=3, and R3C2 is the only place left for a 9. If R9C1=5, then R9C7=3, R9C5=7, R8C4=3, R1C4=17 along with R1C6, R1C7=8, R1C1=9. Column 1's 9 is in either R1C1 or R9C1, then continue with the skyscraper. The virtual XY-wing: if R1C7=8 then R1C1=9. If R3C9=8, then R9C9=9.
Quite the epic struggle, John. The virtual XY-Wing is not what one would normally look for. Shye is known for putting her strategies on the edges of the grid.
Concerning the "strategies demonstrated in this video", is it possible to list the advanced strategies in the same order that they are actually demonstrated in this video? I had to give up and look at the video after 22 minutes, after looking hard for a naked triple. But I saw that the video shows the skyscraper being discovered first, and I don't know if it would be easier to find the naked triple (often very had to find anyway), once the skyscraper is found. I also had spent too much time trying to construct an AIC chain as well before clicking on the video,, and I believe I could have found it if I had solved the skyscraper first because I was looking in the correct cells for the AIC. And I haven't finished watching the video yet, but I'm anxious to see what the heck a 'virtual xy wing" is. Okay, I finished the puzzle and the video (puzzle time 69:57, not counting time watching the video and commenting). I don't recall ever seeing a naked triple being demonstrated at all in the video; I think those things should be banned from Sudoku because they are just too random and hard to spot.
@@JohnRandomness105 Thanks, John. I did find that early on, and I didn't think it was the triple from the listing because it really didn't help too much in solving anything else.
@@SmartHobbies Okay, but please don't make changes just on my account. I don't recall exactly how it went down; it's very possible the naked triple was listed in the proper position. Maybe I need to study closer what defines a naked triple. I might have to run through the puzzle again, but according to johnramdomness, a pair had to be placed first, which resulted in the triple being created. Oftentimes, this occurrence is just so obvious and results in no extra solving whatsoever, so even though it would be a naked triple, it isn't worth mentioning.
Right away we have [19] cp (conjugate pairs) in B9@(88)&(99), leaving 235 triplet on c7B9, plus leaving obviously 8476 on the four open cells on c7 off B9! Note 8 splits in B3 @(17)&(39), and 9 splits on c9@ (39)(sharing with 8, relevant) &(99)(also relevant for future run). Also note cell (91) = [59] and (11)=[89] !! All the above links show that : Whether 8 or 9@(11), it always leads to 5 @(91). Obvious if 9@(11); Next 8@(11) would lead to 8 on row 3 @(39) (see how?) and that would be followed by 9 on col9@(99) ==> 5@(91)!! QED! More instant takeaways: 6 at (72), by row7&col2 ! ==> [49] tcp on r7B8 btw cols 4&6, by virtue of 4&9 given on c3& 4 in B9, 9 on c7, all off r7 !! => 9@(76) => 9 splits on row 1 @(11)&(15). Note also 3 splits on row 1 @(14)&(15)(sharing with 9, relevant). Now in hot pursuit, [49] uncovered tcp on r7B7 ==> 5 on r7@(77), leaving 2 on r7@(73), and [23] tcp on r7B9 btw rows 8&9 ==> [35] tcp on r8B8 btw cols 4&6 ( Note another immediate takeaway is 8 on col5@(85)) ==> 3@(84) ==> 3 on r1@(15) ==> 9 on r1@(11), followed by 8 on r1@(17) , leaving [17],tcp on r1B2 btw cols4&6, plus 7 on c7@(27)B3, leaving [48] tcp on c7B6, and 6 in B3 on c8, off r 6, and together with the two 6s given on c9& in B4, also off r6 ==> 6 on r6@(65) ==> 4 on r6@(62),( by v of the three 4s given on cols 389, all off r6),[25] tcp on c5B5 btw rows 4&5 (Note cell(24) = [237], by direct counting has 3&7 eliminated,leaving 2@(24)!), followed by [47] tcp r5B5 btw cols 4&6 (4@(47)), leaving 1 in B5@(46)!! Also note 6 uncovered @(28) , together with 4 given in B3& 4&6 given in B1 , both pairs off r3 ==> [46] tcp on r3B2, leaving 9 in B2@(25)!! Now in hot pursuit, 2&8 uncovered in Boxes 2&3,& also uncovered on c3, all three pairs off r3 ===> [28] tcp on r3B1 btw cols1&2, followed by 7 in B1@(33), by v of 7 uncovered @(27) off r3, leaving 3 in B1@(22) !! ==> [35] tcp on c3B4, (==> 5 c8@(85)), leaving 1 on c3@(81), followed by [79] tcp on c2B7 btw rows 8&9 , plus 1 @(99)&9@(88) ==> 9 on c9@(39), leaving [36] tcp on c8B3, AND [15] tcp on c8B6, followed by [38] tcp on c9B6 btw rows 4&5, by v of [38] given on r6 outside of B6, leaving [12] tcp on r6B6. Finally 1&8 uncovered on r4 outside of B4 ==> [18] tcp on r5B4, leaving 2 in B4@(42)!! All done !!! The remaining open TCPs will be broken up by their respective danglers. So in retrospect, once we isolated 5@(91), the steps that followed would be by pure logical deduction, no more of those fancy mumbojumbo tricks.
This is an abbreviated version of solving , skipping details that will be left to the readers who wish to figure out by themselves first. Details of follow-ups from every new uncovering. A ‘strong’ uncovering should bring abt one or multiple (perhaps an avalanche) new ramifications, and ,if we get lucky & hit the jackpot, take us straight to the finish line. They will be duly noted. The notation (?) means if you see how/why it is so. All it takes is to stay focused, not distracted,(a misstep would take you back to square one) , ever watchful with eyes wide open (that is visualization at work). Hints will be given for the less apparent follow ups. In the quest of new uncoverings, the hardest and most challenging task is to uncover hidden singles, and bivalued cells, oftentimes when all else fails, they have to be uncovered by direct point counting. A search that is time/attention/energy consuming. This is where most novice will miss out because they are not so obvious. Every now and then as we proceed, it’s good to take a step back and have a panoramic survey of what has been achieved so far, trying to spot/catch interlinks in boxes, along rows&columns, like Wings, coupled pairs, etc. This puts to test how big a repertoire of ‘tricks’ the player has in command/amassed. This is where the boys are separated from the men!! Ultimately solving Sudoku is a graphic artistry, at once gratifying and fulfilling , a skill honed by hours of perseverance and persistent practice. Without further ado, let’s begin : The instant takeaways are 1) 8 on c5@(85)(should be obvious) => 3 splits in B8@(84)&(95)(?), 2) 1&9 cp in B9@(88)&(99)(obvious), leaving 235 triplet on c7 B9, (as readily confirmed by direct count), plus 3 on r1B2 btw cols 4&5, 3) 7 on c7B3, on r5B5; 4 on c7B6, splits on r6@(62)&(65)(relevant), 4) Now comes the paydirt : 8 splits on r1@(11)&(17), and in B3 @(17)&(39); 9 splits on c9@(39)&(99), with 8&9 sharing the cell(39); By direct point count, we uncover two bivalued cells, namely (11) = [89] and (91) = [59] ! Put all these links together, we conclude 9 cannot be at (91) but 5 !! Three ways to go abt it : a) whether 8 or 9 @(11), it always leads to 5 @(91) ; [ obvious if 9@(11); 8@(11) would lead to 8 on r3@(39)(?)=> 9 on c9@(99)==> 5@(91)!] b) whether 8 in in B3, it always leads to 5@(91)!! ; [8 @(17) => 9@(11)==> 5@(91)!! AND 8@(39) => 9 on c9@(99) ==> 5@(91)!!] c) by contradiction !! 9@(91) would lead to , by northward bound, 8@(11) => 8 on r3@(39) ===> 9 on c9 @(99), a NO-NO, ( two 9s on same row 9 !!),OR, by eastward bound, 9 on c9@(39) => 8 in B3@(17), by cc of two 8s given on r2&c8, both outside of B3 ===> (11) = ZERO!!!] Therefore we have 5@(91) !!! ==> in hot pursuit, [25] tcp on c7B9 btw rows 7&8, and 3 @(97) ==> 3 in B8@(84)(?)!==> 3 on r1@(15) !! (?)(Hint : two given 3s on cols 1&6 , and two uncovered on cols 4&7, all four off r1 !!), leaving 9 splits on r1@(11)@(16)(relevant later on). 5) last but not least immediate takeaway is 6 @(72) !!, either by r7 or by c2, followed immediately by [49] tcp on r7B8 !!! (?)(Hint : 49 given on c3, 4 given in B9& 9 given on c7, all off r7.This maneuver is called induced TRIANGULATION), AND 5 in B8@(86), leaving 7 in B8@(95), plus 2 on r7@(73) !!(?) Now 9 uncovered @(76) ==> 9 on r1 @(11) !!(?), followed immediately by 8 on r1@(17), leaving [17] tcp on r1B2 btw cols 4&6 !!(?) AND 7 on c7@(27), followed by 6 on c7@(57) (?) ==> 4 @(47) PLUS 6 on r6@(65) !!! (?), a highly potent harbinger of several ramifications (indeed it will take us all the way to the completion of the puzzle) a) 4 on r6@(62)(?), followed by 4 on r8@(82) !! b) 6 on r2@(28) (?), leaving [39] tcp on r3B3 ==> 3 in B1@(22), followed by [28] tcp on r3B1 btw cols 1&2 ,(?) (Hint : note the 28 uncovered on c3 outside of B1), leaving 7 in B1@(33) !! AND concurrently 3 on c3@(43)(?), ==> 5 on c3@(63) ==> 5 on c8@(58), to be followed by [38] tcp on c9B6 btw rows 4&5 (?), leaving [12] tcp on r6B6 ===> 1@(99)&9@(88) !!! (?), Also 1 on c3 is at (83), as it should be, leaving [79] tcp on c2B7 btw rows 8&9, as should be too. Note at this juncture we have the completions of Right hand Block and the Bottom Block!! c) now last but not least, 6 on c6@(36) (?), followed by 4 on r3@(35) (?), and in turn 9 in B2@(25), (?) leaving 2 in B2@(24) !! (Indeed the prior uncoverings of 3@(84)&7@(27) have reduced the trivalued cell(24) to [1] !!) Onward march towards the endgame, we have [25] tcp on c5B5 btw rows 4&5 (?), followed by [47] tcp on r5B5 btw cols 4&6, leaving 1 in B5@(46) !! (Note together with 9 given @(44), they break up the three TCPs on rows 157 btw cols 4&6!!) And in hot pursuit, 2 on r4@(42), leaving/followed by [18] tcp on r5B4. Mission accomplished!! So in retrospect ,5 uncovered @(91) is an initiator and 6@(65) is the closure!!
