Best XY-Wing Video Ever! Sudoku Advanced Tutorial 16 

Join the Smartie Party🥳now to get reward puzzle packs and other EXCLUSIVE content👉 www.buymeacoffee.com/timberlakeB/membership Time Stamp 00:22 XY-Wing Example 1 (Shared Row) 04:22 XY-Wing Example 2 (Shared Column) 06:56 XY-Wing Example 3 (3 Blocks)

Update! I was really stuck on a recent “impossible” LA Times sudoku; then I watched this video. I returned to that sudoku and quickly found an xy wing. From that point the entire puzzle was a breeze! Thanks again. I’ve subscribed!

I will coment in the first 2 videos because I couldn't do alone, but the last one I could solve by myself 😂 unbelivable Thanks for this video!

I use it when I’m stuck for ideas, and it’s very good

I didn’t understand xy wings. I struggled a little with your first example. Your second example led to my AH HAH moment. I found the xy wing in your third example in four seconds! I was ecstatic! WELL DONE! WELL DONE!!! I then saw ANOTHER way to look at it! Search for a triple. 34 14 13 is a 134 triple. Maybe the wrong term, but you see what I mean. Once you find a triple, see if it’s positioned as an xy wing. Again 56 35 36 It’s a 356 triple AND it’s got a pivot with two pinchers And finally 27 29 79 It’s a 279 triple and I spotted it almost immediately! I was blind, but now I see! Thanks for your excellent explanations!

I think I've been seeing these without realizing it. But I do find colouring (for me) a good technique to eliminate possibles.

I found a Y-wing (134 in blocks 1 and 2) fairly early in the first puzzle, which gave me a few digits. I had considerable work following, and I can't remember most. Backtracking to see what I did, I remember skipping a little logic to establish a set of 13 cells as remote pairs, but it worked. I chained to show that either R8C1 or R7C7 was 4, which finally cracked the puzzle. The second puzzle: I bypassed any Y-wing, or maybe I did it in disguise. In the endgame, I was trying to extend a sequence of 48 cells, when I wound up showing that a certain 48 cell had to be 4 regardless of the other 48 cells. That cracked everything. I'm not sure whether part of that step was a W-wing, a 38 pair mediating two 48 cells. I just began the third puzzle, and already found a finned X-wing in 2s. (Perhaps the more surprising thing is that it actually removed a couple candidates.) A little later, a finned X-wing in 6s. (After penciling 7s) The 6s finned X-wing turned out useless, as the 7s gave me a 27 pair anyway. (After filling block 1 with three pairs) The 6s finned X-wing turned into a full X-wing. (How things oscillate on me.) (After a break for dinner) Another finned X-wing in 8s, a Sashimi type. It would have been a skyscraper if column 6 didn't have three places for 8. (After more time) I found a Y-wing (257 in rows 2 and 4). It was there ever since a finned swordfish in 7s, although I didn't see it until just now. An earlier 235 triple in row 4 did more damage to that cell, so the Y-wing was useless. (An instant later, after a scan of the grid) I just found another Y-wing (367, rows 5 and 9) that might crack the puzzle. We'll see. (A few minutes later) And the puzzle's done! 0:50 The first puzzle. That Y-wing is the one I found fairly early in the puzzle. I needed a lot more to finish the puzzle, though. 5:20 I backtracked to the point where the Y-wing is visible in my grid. I missed it, but I'd already placed the 9 long before the Y-wing became visible. Block 1's pointing pair in 3s eliminates the same 3s as the Y-wing. 7:30 That Y-wing was visible in my grid for the longest time, and I never saw it. It was visible during the finned swordfish in 7s -- I used three of the four cells, the ones with 7s. That Y-wing would have done at least the same damage to the puzzle, and partially resolved the finned swordfish. It was visible when I found the different Y-wing near the end of the puzzle. (Once I found that, it didn't matter.) I suspect that spotting that Y-wing would have made solving this easier. Thanks for the video.

NF 9m 45 s( Example 1)

This can also be done when the pincers don't have a common candidate. For example, if II6 was 3/7 in the first example (although it obviously couldn't be in this scenario) , we would know that since the 3 and the 4 can't both be true, either the 7 or the 1 is true...so we would look for any eliminations caused by both of those candidates.

Begin with the instant takeaways : The three sets of 5&9 given on r1, c7 & c9, all outside of B3 => [59] tcp (true conjugate pairs) on c8B3 between rows 2&3. 8 on c8@(68),by virtue of 8s given in Boxes 3&9, on r4,all off c8, allowing 7 given @(52) to squeeze in at (48), thereby leaving [13] tcp on r5B3 bet cols 7&9. Also in hot pursuit, 4&8 given on r4& uncovered on r6 ==> [48] tcp on r5B5, leaving [59] tcp on r5@B4. Now right away we have 6 on r4@(42), thus leaving 1 in B4@(62) and also at same time leaving 135 triplet on r4B5. 1&8 uncovered on r6 outside of B5 => 279 left on r6 B5, with [79] tcp bet cols4&5, leaving 2@(66) => 2&4 on c5B2 and 9 on c5B8 [CIY= check it yourself] So Middle Row Block is quite well broken up except for r5B5. Let’s move the war theater to Top Block. Again right away we have 6 on c4@(24) and on r1B3,plus cell(26)=[18],by point count. Other bivalued cells are : (12)=[34], (23)=[13], (15)=[14], (37)=[24]. Now here’s a nittygritty for the trained eye : 1@(11) would result in 3@(23)&4@(15) ==> cell(12) = ZERO! So 1 is eliminated from (11) and the only other place else on r1 for 1 is at (15)!!,by v of 1 given in B3 and uncovered on c2@(62). The technical description to what happened is 134 form an XYZ System with Xat(12), Y at (14), Zat (32),whereby for 1 at (11) to look at 1 in Y and Z would lead to a disaster. ( it’s a skill to detect the system and pick the right candidate to eliminate) Indeed there’s alternative way to go abt it : 1@(11) would lead to [67] tcp on r1B3, by v of 6&7 uncovered on c2 and in B2, all off r1 , resulting in 3 on r1@(12)==> cell(23) = ZERO! So 1 uncovered @(15) => [24] tcp on c5B2 and 8@(26) ==> [89] tcp on c5B8 bet rows 8&9, by v of 8&9 given on r7, and uncovered on cols 4&6, all three off B8! Plus 1 in B8 on r8 bet cols 4&6. Also 8 uncovered at (26) => together with 8 given on r1, both off c1, 8 on c1B7 => [89] tcp c2B1 bet rows 2&3.(indeed 8@(26) => 8 on r3@(32) => 9@(22). Now we are in the presence of an intriguing situations : [24] tcp uncovered on c5B2 bet rows 2&3, and [24] uncovered cp on r3 at (35)&(37) ==> by Uniqueness constraint , 2&4 cannot be at cell (27) but 3&7.(right now this is a weak info) Here’s a really nittygritty observation : 3 at (92) would lead to 4@(12) and 2 on r9@(99). Uncovering 2&4 on row 1 and on col 9. ==>!! [24] tcp on c7B3 bet rows 2&3 !! , a NO - NO to Uniqueness rule. Therefore the only other place for 3 on row 9 is at (98)!! AND this the break we are looking for! So 3@(98) and in tandem with 3 uncovered @(59), both off c7 => 3 on c7@(27), as anticipated plus 3 on r1@(12) => [36] tcp on c3B7. 3@(12) => 1@(32) => leaving [47] tcp on c1BI and[18] tcp on c1B7 => [45] tcp on c2B7, by v of 4&5 given on r9 outside of B7, leaving 2 in B7@(92). Now in hot pursuit 3 uncovered @(37) => 7 on c7@(87), by v of the three 7s given on rows 357, all off c7!, followed by 2 on c7@(37)(see it coming?), leaving 4 on c7@(77) (indeed 3 @(98)=> 4@(77),by pt-ct.) and 4 in B3 @(18) , by the 4 given on c9 squeezing in, whereby leaving [67] tcp on c9B3 and leaving [16] tcp on c8B9 and finally [29] tcp in B9 on c9! One more step to go : 4 uncovered @(77) => 5 in B8 on r8 forming tcp with 1, and leaving 3 in B8@(75) => [13] tcp on r4B5 bet cols 4&6 with 5@(45). Eureka! The tcps will eventually all be broken up by the appropriate danglers. I leave them to good hands. Have fun!!

Got stuck at 7:01, bcz after I do all synder notations, still can’t figure out where to fill the 6 and 9 in block1, so I can’t limit R2C8 to 2 and 9. Then I re-checked and saw the X wing of 6 that could help rule out 6 in R2C8, the rest was all similar to the video😂