The crucial thing is that the 57 cell sees all three of the other cells in the XYZ wing. It the "hinge" is a 2, it forces a 7 into the 27. If it's an 8, it forces a 7 into the 78. Otherwise, it's a 7. Therefore, one of them must be a 7 so any cell that sees all three can't be a 7.
It was an extremely hard one. Din't spot the XYZ wing, but limited cells to a state where I could bifurcate my way to the solution. Bifurcation is fine -- just not as entertaining to watch in a video, so I liked the logic after.
Yep! That XYZ Wing was the key. The critical choke point, as it were. 'Twas amazing that Mark found it and solved it before even filling in all the candidates! Yeesh!
8:48 ‘and this must be a 5’.....then, ‘hang on a minute’ while I go ask Simon how to explain that to dummies. Comes back, fills out r4c3 and r6c9 and goes ‘its an xyz wing’ . Clearly, the man is a savant, and doesn’t even bother w the technical stuff.
I didn't spend 10 months, but I did spend 53 minutes. I didn't quite see the XYZ wing, but I did a little bit of chaining that proved there can't be a 7 in the 27 cell (row 6, column 9).
Definitely not an easy puzzle. Took me 27 minutes for this one, without spotting the XYZ wing. I got to the exact same point with very similar logic, but got stuck and tried guessing a solution for a cell, which turned out wrong. Was easy from there on, but I don't like this method. The XYZ wing was a nice and elegant solution. I'll definitely be on the lookout for those when I get stuck in the future, very instructive! Thanks for another nice puzzle guys :)
If I had to guess, I'd say the name XYZ-wing comes from the fact that it's basically an elaborated version of a Y-wing but with three options. (The X, the Y, and the Z.) Anyway, I wound up getting to that point, bifurcating to get a sense of the problem, and then working out a version of the same logic: Basically, if you make that square a 7, then r6c9 becomes a 2, which means the only place a 2 can go in row 5 is r5c3, one of the two spots a 1 can go in box 4, but making the starting square a 7 also forces a 1 into r2c1, which cuts off the other. It's not as clean as the XYZ-wing, but it got the job done and it doesn't feel _too_ chainy to me.
Tried for an hour and got to that point where i couldn't logic through anything, almost every cell was down to 2 or 3 candidates. Can't wait to watch this video.
Another very instructive puzzle. The fact that the title gave away the presence of an XYZ wing did me help solving it though; I'm not sure I could have done it otherwise. Thanks for the upload.
Took me 25 minutes and I as far as I recall never used the xyz-wing logic. Straight forward Sneider notation with entered marked pairs. I discovered the 57 pair in C2 pretty early on (which I was pleased with) and that kept the logic flowing for me.
25:12, and I'm amazed I figured out the logic as quickly as I did. It's a real pleasure to figure something out that you didn't know before and know it's rigorous and see it work.
24:09 for me! Luckily I was sniffing for the XYZ-wing as I saw it referenced in the title, and when I found it, it was a delight to see the whole puzzle just collapse. Good sudoku!
Super easily solved puzzle. I just hit a little red X in the top right corner of the screen and the whole thing was done and gone. Not even three seconds. Thanks for the video!
6:21 -- although it'd probably have taken a wee bit longer had I not been aware there was an XYZ-wing in the puzzle. Please consider this when naming your videos (don't give the game away before we've at least tried to solve on our own). Note that an XYZ-wing is so-called simply because it is an extension of the Y-wing. In both cases, the three candidates are conventionally labelled X, Y and Z, such that Y is the candidate to be eliminated. In a Y-wing, the hinge cell (the one cell that can see both of the others) holds the {XZ} candidates, while the pincers (the two cells which cannot see each other), hold the {XY} and {YZ} candidates. Regardless of which value we place in the hinge ({X} or {Z}), one of the pincers must be {Y}, so any cell that can see both pincers cannot be {Y}. The XYZ-wing only differs in that the hinge also contains the Y candidate (thus it has {XYZ} candidates, hence the name). But this only affects the logic in that either the hinge takes the {Y} candidate or we are left with a standard Y-wing. In which case, one of the three cells must be {Y}, thus any cell that can see all three cells cannot take the {Y} candidate. Simon would probably refer to an XYZ-wing as being a "bent triple". It's bent simply because two of the three cells (the pincers) cannot see each other, they can only see the hinge. After all, if the pincers could see each other, then all three cells would see each other and thus the formation would simply be known as a naked triple.
@@davidhuff8032 You're very welcome, David. I'm glad you found it useful. In the interests of completeness, a Y-wing is also known as an XY-wing because it is an XY-chain of length 3. It is not uncommon for algorithms to overlap to some degree, however, an XYZ-wing extends the Y-wing algorithm beyond the realm of the XY-chain. As a bit of fun, we can also extend the Y-wing logic beyond subsets of 3 cells. That is, a standard Y-wing is a subset of 3 cells with 3 candidates between them, but we can generalise this to cater for any subset of N cells with N candidates between them, such that 3
I love CTC's solving style so much I made a video on my channel where I analyze each step of this solve. In addition to explaining the strategies used, I add a couple of pause the video moments and some alternate solving paths, similar to how someone analyzes a chess game.
Hi Mark, Thank you, that was a nice directed tutorial for an XYZ wing. I had bifurcated because I saw something odd about the 2's. It only took a couple of moves to show me that the 7 had to be a 5. Thanks, Gene
11:07 Mark's explanation is great, but it's actually pretty cool to look at it like this. The 27, 78 and 278 cells are a bent triple that all see the 57 focal point. That cell can never be a 7. Very hard to spot, though.
30:28 for me! Considering it is rather resistant for a classic sudoku, I feel pretty happy with that! I started with typical pencil marking, then looked at weak cells like Mark has shown in previous solves, and then once I got properly stuck I noticed all the 267s around the grid and turned it into a coloring problem, though I'm worried my method could be considered bifurcation-ish...Once I saw that there were two kind of 'dominos' in box 4 (one where exactly one of two squares could be one of two possibilities, and same for the other domino), and I saw one of them pointed at box 1 I was able to crack it. I'm curious to see how Mark tackles this for sure!
interesting that Andrew's Solver rates this at a measly score of 121. many of the NYT's Hard puzzles are rated higher than that but never required an XYZ wing to solve.
Really interesting video today. I managed to solve in 8:33, but only because I knew to look for XYZ wings; I'm not sure how long it would have taken me otherwise to be honest, XYZ wings aren't really part of my 'natural' solving algorithm and I only look for them when in a tight spot.
What I find intersting is that once all of the possible candidates are filled in, the xyz wing becomes quite apparent. However, both you and Simon seem to abhor filling in all candidates. I totally understand this approach early on in the game, but once it comes to the end game, I find it quite helpful to fill in all candidates. The alternative to me seems to involve mental gymnastics imagining all candidates in all unsolved cells.
The thing is all logical deductions in these puzzles are bifurcations: assume something is true and show there’s a logical contradiction. Every bit of logic can always be restated as a proof by contradiction, i.e. bifurcation.
@Spider53 But the reason you know that is because you recognize a pattern. Proving the assertion still depends on either/or statements. (If a then c) or (if b then c) where either a or b must be true will lead you to say that c must also be true, but the underlying logic is still comparing the consequences of acand b in turn. I.e., bifurcation.
@Spider53 It doesn't matter how far you go down the paths, it only matters that there are more than one path. Bi for two and furc for the fork that splits the road. And it would never occur to me to test a cell with three possibilities before a cell with two, so how you focus on r6c3 is beyond me. Mark's choice to look at r6c2 I can follow.
Agree with this 100%. I would define bifurcation as anything that requires you to guess by actually WRITING a single number in a box, without knowing if it is correct, just to see if it works. Perhaps that’s the definition of “single step ahead”? If you can hold the “if this goes here those that would go there....and...and...and then this is impossible” logic in your head, without writing anything “incorrect” in the grid, then it’s not bifurcation. Note that correct notation is not bifurcation because you are writing ALL the possibilities, not simply randomly GUESSING a single digit.
@@stuartmcconnachie What you are describing is "Ariadne's String" aka trial and error. True bifurcation is following two paths at the same time, which is easier on paper. You can do it at enjoysudoku.com because that program will let you highlight individual candidates, though. If both paths lead you to the same conclusion, you know it is true.
I spotted the XYZ wing because I looked for one. If I didn't know there was an XYZ wing in the puzzle I would not have solved it. 43:10 is my time, with 5 minutes to understand what an XYZ was and another five minutes to spot it.
Since it was called an XYZ puzzle I knew what I was looking for and it didn't take too long to find it. Still not as fast as Mark but if I ever were I'd go into shock
The trick I used was looking at r2c3 being a 1 or7, and r5c3 being a 1-2. I saw that if r2c3 was a 1 it would put a 2 in r5c3, and if it was a 7 then it would give you a 26 pair in box 3. That 26 pair aligns with the 2 spots we placed 6s in box 6, so the 2 gets forced into box 4 anyways. Then in box 4 you get the 78 pair in c3 and that resolves box 1 and everything continues on like after he did the xyz wing.
Is this the first time this concept has been featured on the channel? I ended up solving it by trying an empty rectangle on 2 in box 6. If horizontal, 2 goes in r6c3, putting 7 in both r6c9 and r4c3, breaking the 8. So 2 had to be in r6c9, making a 78 pair in box 4, basically collapsing the puzzle. I’ve been uncertain if empty rectangles count as bifurcation. Tbh it just seems like the goal is for logic to be efficient, huh? So it’s not that bifurcation is invalid, just typically really messy.
I don't think it ends up mattering, but there's a cute quadruple that turns up in the center right box - 34 and 58 get ruled out of the same four cells in different ways, leaving only four empty spaces for 3458.
I made a mistake and pressed check before I was done with the puzzle. Hasn't happened before that I couldn't change the color after accidentally pressing check, but happened now. Not sure what you changed that messed this up, but don't like not being able to change cell color after accidental check.
Wow, I didn't use that area of the grid at all. I got everything in column 8 and 9 down to two options and found 2 Y-wings that proved that r5c9 couldn't be a 6, then it fell apart. The 5-8 combos were very effective in box 6, requiring r4c8 to be a 58 pair because both couldn't go in c7 or it would break box 3.
The solution I came up with was to force test the 7 in RC23. The 1 goes to RC21, which unwinds the 12 in the middle left block. RC69 was a 27, so that square becomes a 7. There is now no place for a 7 in the left middle block, so that solution will not work. This resolves the top left block, and cracks the puzzle open.
The frustration of restarting twice, each time almost finishing the entire puzzle and then discovering there’s a contradiction...... I suppose it’s my fault for not being careful but other than that it’s doable by finding triples.
Anyone opposed to bifurcation should try the 10-rated puzzles on dailykillersudoku. You could stare at some of those for a century and not discover the constraint without actually trying some numbers.
Jason Black I was thinking the same thing. I think Simon would describe it by looking at the hinge (r6c3) and pretending it didn’t have a 7. Then it would be a Y-Wing on 7’s, and there couldn’t be a 7 in r6c2. Adding back in the 7 to the hinge doesn’t change the result though, because r6c3 still sees r6c2. I think they’re exactly the same, but the online wikis call it an XYZ-Wing and Simon calls it a bent triple.
Apparently it's called an XYZ wing because the three numbers (in this case 278) are represented by the symbols XYZ. Three numbers are needed for the pattern to materialize. www.sudokuwiki.org/XYZ_Wing
Wow, Mark just writing in that 5 without (many) pencil marks. Impressive. For me, it took 24 minutes. Using uniqueness, an XY-Wing and a rather long (but simple) chain.
I have almost no technique, so after ~30 min of getting as far as possible then looking at it and poking a bit, I realized why the 7 couldn't go there (total of 36:48)
I didn't even know what a XYZ wing was... Solved it with regular technique until a point, then I had to bifurcate from a cell with 2 candidates. First attempt failed so went back and solved it the other way. It's not entirely satisfying, but still counts as solved I guess.
The 278 cell he pointed to has a direct effect on the location of a 7, and in every possible value of that cell, 7 is eliminated in the left cell because it sees all 3 possible 7’s. It’s the same concept as a “bent triple” you might hear Simon explain from time to time, but an extra step more complicated.
All it was was a bunch of hidden singles and a bifurcation on r2c3 that was necessary to solve it all. All you had to know was that cell couldn't be a 7 and everything solved itself. No other cells mattered at all.
I dropped the ball today. I went and learned how XYZ wings worked (and started pronouncing them based on the Yugioh summon type), and then scoured the puzzle for possible wings, and I found EXACTLY the one that cracks the puzzle...but missed the fact that it did. Heck, I looked twice, at different points. Feels bad... But at least I learned something new, and some day it will be useful to me. Incidentally, I did find a uniqueness problem in r14c78 that cracked the puzzle along with a Y-wing friend in r4c3, but I hate using uniqueness as a solution in and of itself, so even though I technically solved it, I still consider this my loss. I was looking right at an XYZ wing and didn't spot the solve. That definitely needs improvement.
I don't think you used bifurcation. It looked like you knew there was something there with the 5's and 7's and you worked it out. Bifurcation would've been randomly picking a square, putting a number in it, without using any logic and seeing what happens. Big difference!
@@Robin19358 At roughly the same point where Mark used the XYZ wing the r2c8c9 and r3c8 cells can only be 267 and r5c8c9 can only be 26 and a 9 in r5c9. If you put either2 or 6 into r5c9 then r2r3c8 are 2 and 6, which means that r5c8 can't be 2 or 6, but it has to. Therefore r5c9 has to be 9. It's actually more of a chain, but I spotted it by looking at the uniqueness.
20:59 somehow I also found a similar chain, but starting from the 27 in r6c9, I don't really like chaining, it seems a bit cheap, but for puzzles like this it's pretty much required. At least, I don't see how anyone could find that XYZ wing without knowing it's there, I was even looking for one because of the title and couldn't find it.
Okay but where is the line between bifurcation and logic? Because to me it looked like you just bifurcated your way to a 5 at 8:45. You placed in a 7 and saw where that took you. If that's not bifurcation than I don't know what is. And don't take this the wrong way, I'm not calling you out or anything I'm just genuinely curious what you define as bifurcation. EDIT: Ah fuck I should've watched it further. You ask yourself the same thing. However my question still stands.
Seems like I am a Mark.. solved this thing using uniqueness with 2 or 5 in c3r2. I noticed that one of them gave me a 2-6 pair on columns 8 and 9 in rows 3 and 5..
Nine months later and I was able to solve the puzzle! I didn't need to use an XYZ-Wing to do so. Rather, I built up my pairs and triples, used a broken rectangle to place a naked single 9 in Box Four, and then from there was able to find a 1349 quadruple in Row Five thanks to the 1349 in Column Eight. The puzzle fell apart after that. I used another broken rectangle to resolve the 58 pairs in Boxes Three and Six and I eventually solved the puzzle in 30:22.
Knowing that you need an XYZ wing to logically solve the puzzle means you will be on the lookout for one and in the future this allows you to spot them easier
The crucial thing is that the 57 cell sees all three of the other cells in the XYZ wing. It the "hinge" is a 2, it forces a 7 into the 27. If it's an 8, it forces a 7 into the 78. Otherwise, it's a 7. Therefore, one of them must be a 7 so any cell that sees all three can't be a 7.
From this I will conclude I lack both instinct and technique...
It was an extremely hard one. Din't spot the XYZ wing, but limited cells to a state where I could bifurcate my way to the solution.
Bifurcation is fine -- just not as entertaining to watch in a video, so I liked the logic after.
I concur. Took me two tries and 1 1/2 hours.
Yep! That XYZ Wing was the key. The critical choke point, as it were. 'Twas amazing that Mark found it and solved it before even filling in all the candidates! Yeesh!
I love the fact that you and Simon approach puzzles differently. It would be quite boring if you mirrored each other.
Arya Dark exactly what I was thinking!
8:48 ‘and this must be a 5’.....then, ‘hang on a minute’ while I go ask Simon how to explain that to dummies. Comes back, fills out r4c3 and r6c9 and goes ‘its an xyz wing’ . Clearly, the man is a savant, and doesn’t even bother w the technical stuff.
He hadn't even penciled in the possible values of the other 3 related cells! How did he see that?!
I didn't spend 10 months, but I did spend 53 minutes. I didn't quite see the XYZ wing, but I did a little bit of chaining that proved there can't be a 7 in the 27 cell (row 6, column 9).
Definitely not an easy puzzle. Took me 27 minutes for this one, without spotting the XYZ wing. I got to the exact same point with very similar logic, but got stuck and tried guessing a solution for a cell, which turned out wrong. Was easy from there on, but I don't like this method. The XYZ wing was a nice and elegant solution. I'll definitely be on the lookout for those when I get stuck in the future, very instructive! Thanks for another nice puzzle guys :)
If I had to guess, I'd say the name XYZ-wing comes from the fact that it's basically an elaborated version of a Y-wing but with three options. (The X, the Y, and the Z.) Anyway, I wound up getting to that point, bifurcating to get a sense of the problem, and then working out a version of the same logic: Basically, if you make that square a 7, then r6c9 becomes a 2, which means the only place a 2 can go in row 5 is r5c3, one of the two spots a 1 can go in box 4, but making the starting square a 7 also forces a 1 into r2c1, which cuts off the other. It's not as clean as the XYZ-wing, but it got the job done and it doesn't feel _too_ chainy to me.
Touring older videos, I wanted to find out what in the world an XYZ wing was - and I did find it out. Thanks, Mark!
Great example of a classic with a choke-point - thanks!
Tried for an hour and got to that point where i couldn't logic through anything, almost every cell was down to 2 or 3 candidates. Can't wait to watch this video.
Another very instructive puzzle. The fact that the title gave away the presence of an XYZ wing did me help solving it though; I'm not sure I could have done it otherwise. Thanks for the upload.
Took me 25 minutes and I as far as I recall never used the xyz-wing logic. Straight forward Sneider notation with entered marked pairs. I discovered the 57 pair in C2 pretty early on (which I was pleased with) and that kept the logic flowing for me.
25:12, and I'm amazed I figured out the logic as quickly as I did. It's a real pleasure to figure something out that you didn't know before and know it's rigorous and see it work.
24:09 for me! Luckily I was sniffing for the XYZ-wing as I saw it referenced in the title, and when I found it, it was a delight to see the whole puzzle just collapse. Good sudoku!
Same here.
Super easily solved puzzle. I just hit a little red X in the top right corner of the screen and the whole thing was done and gone. Not even three seconds. Thanks for the video!
6:21 -- although it'd probably have taken a wee bit longer had I not been aware there was an XYZ-wing in the puzzle. Please consider this when naming your videos (don't give the game away before we've at least tried to solve on our own).
Note that an XYZ-wing is so-called simply because it is an extension of the Y-wing. In both cases, the three candidates are conventionally labelled X, Y and Z, such that Y is the candidate to be eliminated. In a Y-wing, the hinge cell (the one cell that can see both of the others) holds the {XZ} candidates, while the pincers (the two cells which cannot see each other), hold the {XY} and {YZ} candidates. Regardless of which value we place in the hinge ({X} or {Z}), one of the pincers must be {Y}, so any cell that can see both pincers cannot be {Y}. The XYZ-wing only differs in that the hinge also contains the Y candidate (thus it has {XYZ} candidates, hence the name). But this only affects the logic in that either the hinge takes the {Y} candidate or we are left with a standard Y-wing. In which case, one of the three cells must be {Y}, thus any cell that can see all three cells cannot take the {Y} candidate.
Simon would probably refer to an XYZ-wing as being a "bent triple". It's bent simply because two of the three cells (the pincers) cannot see each other, they can only see the hinge. After all, if the pincers could see each other, then all three cells would see each other and thus the formation would simply be known as a naked triple.
Best description of the nomenclature I have seen! I will be able to remember it because of the way you describe it. Thanks!
@@davidhuff8032 You're very welcome, David. I'm glad you found it useful.
In the interests of completeness, a Y-wing is also known as an XY-wing because it is an XY-chain of length 3. It is not uncommon for algorithms to overlap to some degree, however, an XYZ-wing extends the Y-wing algorithm beyond the realm of the XY-chain.
As a bit of fun, we can also extend the Y-wing logic beyond subsets of 3 cells. That is, a standard Y-wing is a subset of 3 cells with 3 candidates between them, but we can generalise this to cater for any subset of N cells with N candidates between them, such that 3
I really appreciate you guys going over different techniques I've never seen before. I'll add that one to memory. Thanks a lot.
I've watched only 4 minutes, but once again my compliments to your explanation. It's perfectly balanced between complete and non-trivial. Well done!
I love CTC's solving style so much I made a video on my channel where I
analyze each step of this solve. In addition to explaining the
strategies used, I add a couple of pause the video moments and some
alternate solving paths, similar to how someone analyzes a chess game.
I could have never spotted that XYZ wing on my own.
@9:50 I can't understand how did you see that wihtout putting any numbers in the grid. AMAZING!
7:45 for me definitely my best solve since its the first one ive done it quicker than the video
Hi Mark, Thank you, that was a nice directed tutorial for an XYZ wing. I had bifurcated because I saw something odd about the 2's. It only took a couple of moves to show me that the 7 had to be a 5.
Thanks, Gene
11:07
Mark's explanation is great, but it's actually pretty cool to look at it like this. The 27, 78 and 278 cells are a bent triple that all see the 57 focal point.
That cell can never be a 7. Very hard to spot, though.
35:20, 20 minutes was scanning for xyz wing, otherwise would get completely stuck
30:28 for me! Considering it is rather resistant for a classic sudoku, I feel pretty happy with that! I started with typical pencil marking, then looked at weak cells like Mark has shown in previous solves, and then once I got properly stuck I noticed all the 267s around the grid and turned it into a coloring problem, though I'm worried my method could be considered bifurcation-ish...Once I saw that there were two kind of 'dominos' in box 4 (one where exactly one of two squares could be one of two possibilities, and same for the other domino), and I saw one of them pointed at box 1 I was able to crack it. I'm curious to see how Mark tackles this for sure!
Interesting technique. I like it very much.
interesting that Andrew's Solver rates this at a measly score of 121. many of the NYT's Hard puzzles are rated higher than that but never required an XYZ wing to solve.
Less than 2 minutes from Mark's explanation of the XYZ Wing - looks like I need to learn that one...
Really interesting video today. I managed to solve in 8:33, but only because I knew to look for XYZ wings; I'm not sure how long it would have taken me otherwise to be honest, XYZ wings aren't really part of my 'natural' solving algorithm and I only look for them when in a tight spot.
What I find intersting is that once all of the possible candidates are filled in, the xyz wing becomes quite apparent. However, both you and Simon seem to abhor filling in all candidates. I totally understand this approach early on in the game, but once it comes to the end game, I find it quite helpful to fill in all candidates. The alternative to me seems to involve mental gymnastics imagining all candidates in all unsolved cells.
10:56 I used a Y-wing plus unique rectangles to get a 9 on R4C1 and eventually finishing the entire R4.
could you elaborate on what cells formed the Y wing that you spotted?
The thing is all logical deductions in these puzzles are bifurcations: assume something is true and show there’s a logical contradiction. Every bit of logic can always be restated as a proof by contradiction, i.e. bifurcation.
Yup! I call the wings and fishies pattern hunting and i am bad at it most of the time.
@Spider53 But the reason you know that is because you recognize a pattern. Proving the assertion still depends on either/or statements. (If a then c) or (if b then c) where either a or b must be true will lead you to say that c must also be true, but the underlying logic is still comparing the consequences of acand b in turn. I.e., bifurcation.
@Spider53 It doesn't matter how far you go down the paths, it only matters that there are more than one path. Bi for two and furc for the fork that splits the road. And it would never occur to me to test a cell with three possibilities before a cell with two, so how you focus on r6c3 is beyond me. Mark's choice to look at r6c2 I can follow.
Agree with this 100%. I would define bifurcation as anything that requires you to guess by actually WRITING a single number in a box, without knowing if it is correct, just to see if it works. Perhaps that’s the definition of “single step ahead”?
If you can hold the “if this goes here those that would go there....and...and...and then this is impossible” logic in your head, without writing anything “incorrect” in the grid, then it’s not bifurcation. Note that correct notation is not bifurcation because you are writing ALL the possibilities, not simply randomly GUESSING a single digit.
@@stuartmcconnachie What you are describing is "Ariadne's String" aka trial and error. True bifurcation is following two paths at the same time, which is easier on paper. You can do it at enjoysudoku.com because that program will let you highlight individual candidates, though. If both paths lead you to the same conclusion, you know it is true.
I spotted the XYZ wing because I looked for one. If I didn't know there was an XYZ wing in the puzzle I would not have solved it. 43:10 is my time, with 5 minutes to understand what an XYZ was and another five minutes to spot it.
Since it was called an XYZ puzzle I knew what I was looking for and it didn't take too long to find it. Still not as fast as Mark but if I ever were I'd go into shock
Famous last words, "but that's a bit of a chain".
Glad to know how you see things now lol.
The trick I used was looking at r2c3 being a 1 or7, and r5c3 being a 1-2. I saw that if r2c3 was a 1 it would put a 2 in r5c3, and if it was a 7 then it would give you a 26 pair in box 3. That 26 pair aligns with the 2 spots we placed 6s in box 6, so the 2 gets forced into box 4 anyways. Then in box 4 you get the 78 pair in c3 and that resolves box 1 and everything continues on like after he did the xyz wing.
I know a lot of the purists that hate bifurcation will hate on uniqueness, but im with mark in that whatever solves the puzzle the fastest is valid.
Is this the first time this concept has been featured on the channel? I ended up solving it by trying an empty rectangle on 2 in box 6. If horizontal, 2 goes in r6c3, putting 7 in both r6c9 and r4c3, breaking the 8. So 2 had to be in r6c9, making a 78 pair in box 4, basically collapsing the puzzle.
I’ve been uncertain if empty rectangles count as bifurcation. Tbh it just seems like the goal is for logic to be efficient, huh? So it’s not that bifurcation is invalid, just typically really messy.
I don't think it ends up mattering, but there's a cute quadruple that turns up in the center right box - 34 and 58 get ruled out of the same four cells in different ways, leaving only four empty spaces for 3458.
It does end up mattering actually, this is how I solved it all the way to the end without using the XYZ wing
I made a mistake and pressed check before I was done with the puzzle. Hasn't happened before that I couldn't change the color after accidentally pressing check, but happened now. Not sure what you changed that messed this up, but don't like not being able to change cell color after accidental check.
Wow, I didn't use that area of the grid at all. I got everything in column 8 and 9 down to two options and found 2 Y-wings that proved that r5c9 couldn't be a 6, then it fell apart. The 5-8 combos were very effective in box 6, requiring r4c8 to be a 58 pair because both couldn't go in c7 or it would break box 3.
The solution I came up with was to force test the 7 in RC23.
The 1 goes to RC21, which unwinds the 12 in the middle left block.
RC69 was a 27, so that square becomes a 7.
There is now no place for a 7 in the left middle block, so that solution will not work.
This resolves the top left block, and cracks the puzzle open.
At 7:56, how did Mark eliminate the 3 and the 8 from r6c2, and the 8 from r3c2, leaving just the 57 pair in the column?
r5, r7, and r8 cannot contain 57, so they can only go in r3 and r6 as a pair, and the other three blank cells must therefore contain a 389 triple.
I would never have spotted that wing, but I was still able to solve it in a bit under the 15 Minute mark.
The frustration of restarting twice, each time almost finishing the entire puzzle and then discovering there’s a contradiction...... I suppose it’s my fault for not being careful but other than that it’s doable by finding triples.
29:49... For once this is the first time i saw an xyz wing. And how it execute. This is a good practice puzzle.
8:50 - Used a Uniqueness test and a XY-wing.
Anyone opposed to bifurcation should try the 10-rated puzzles on dailykillersudoku. You could stare at some of those for a century and not discover the constraint without actually trying some numbers.
Are they possible without bifurcation tho?
Why do he eliminate the 8 in column 2, leaving only the "57" options? at 10'45"
Got it
is this puzzle has a unique solution??
Can anyone tell me how Mark saw at 8:50 that r6c2 must be a 5? How cold he see from the grid that a 7 would lead to a conflict?
Is this same as a bent triple?
Jason Black I was thinking the same thing. I think Simon would describe it by looking at the hinge (r6c3) and pretending it didn’t have a 7. Then it would be a Y-Wing on 7’s, and there couldn’t be a 7 in r6c2. Adding back in the 7 to the hinge doesn’t change the result though, because r6c3 still sees r6c2.
I think they’re exactly the same, but the online wikis call it an XYZ-Wing and Simon calls it a bent triple.
I did this in under 40 mins, (2021 Jan) after all the other insane puzzles I am learning something.
Apparently it's called an XYZ wing because the three numbers (in this case 278) are represented by the symbols XYZ. Three numbers are needed for the pattern to materialize. www.sudokuwiki.org/XYZ_Wing
Wow, Mark just writing in that 5 without (many) pencil marks. Impressive. For me, it took 24 minutes. Using uniqueness, an XY-Wing and a rather long (but simple) chain.
yeah limiting bifurcation seems to be just an arbitrary limit on the number of logical steps ahead you are allowed to take.
Ohhhh, I get it. You just have to be a genius!
Makes perfect sense 😅
don't we usually call the XYZ wing a bent triple?
Nope, that's a Y-wing. They are a bit easier to understand.
I have almost no technique, so after ~30 min of getting as far as possible then looking at it and poking a bit, I realized why the 7 couldn't go there (total of 36:48)
I was all about solving an "X,Y,Zee Wing" But when he called it an "X, Y, Zed Wing"... Mind blown, lol
i used uniqueness on the 58 pair to place the 8 in r6 c7, then it flows... Great puzzle anyway, keep goin!!!
27:29. My fastest by a long way. Now I'll see how it should have been done.
Is that what Simon calls a bent triple?
Think so too, that's what I'm used to calling them lol.
yeah I have a tough time finding those wings heh
couldn't find the wing and had to solve it with uniqueness using 2-6 pairs in columns 8 and 9, which allowed me to place a 2 in r6c9
I didn't even know what a XYZ wing was... Solved it with regular technique until a point, then I had to bifurcate from a cell with 2 candidates. First attempt failed so went back and solved it the other way. It's not entirely satisfying, but still counts as solved I guess.
Nice technique. It's theoretically possible I'll be able to find one the next time.
23:58 Not sure what an XYZ wing is, but I got it from a 389 triple in the second column.
15:09
Lost me - I got stuck at the same place and don't understand XYZ wings
The 278 cell he pointed to has a direct effect on the location of a 7, and in every possible value of that cell, 7 is eliminated in the left cell because it sees all 3 possible 7’s. It’s the same concept as a “bent triple” you might hear Simon explain from time to time, but an extra step more complicated.
@@JaggerG I get it now - but I'm amazed he spotted that!!!
All it was was a bunch of hidden singles and a bifurcation on r2c3 that was necessary to solve it all. All you had to know was that cell couldn't be a 7 and everything solved itself. No other cells mattered at all.
26.12
ty for the puzzle
I often find myself getting towards the end realizing somethings wrong, and I just can't figure out what I did wrong.. So frustrating.
9:45 but i might have guessed a few times
31:16
24:17 and I bifurcated cheerfully. Now to watch the vid.
I still call Mark's method "diamond cutting". Experience applied at just the right place.
Bent triple?
Took me 20:30 without any XYZ wing. The most advanced technique I uses was a quadruple.
I dropped the ball today. I went and learned how XYZ wings worked (and started pronouncing them based on the Yugioh summon type), and then scoured the puzzle for possible wings, and I found EXACTLY the one that cracks the puzzle...but missed the fact that it did. Heck, I looked twice, at different points. Feels bad... But at least I learned something new, and some day it will be useful to me.
Incidentally, I did find a uniqueness problem in r14c78 that cracked the puzzle along with a Y-wing friend in r4c3, but I hate using uniqueness as a solution in and of itself, so even though I technically solved it, I still consider this my loss. I was looking right at an XYZ wing and didn't spot the solve. That definitely needs improvement.
I don't think you used bifurcation. It looked like you knew there was something there with the 5's and 7's and you worked it out. Bifurcation would've been randomly picking a square, putting a number in it, without using any logic and seeing what happens. Big difference!
No, that would be guesswork. To me bifurcation, properly, is running two contradictory possibilities at once.
So Dark Magic then?
Does Mark not think he's the right person because he's OKAY bifurcating?
I used logic chains to solve it in 32:10. Did not spot the XYZ wing.
11 minutes no xyz wing, i marked triplets and doubles. puzzle fell apart in about 45seconds.
Damn, didn't catch that. I used uniqueness on the 9 at r5c9.
Could you explain this a bit more? I know what uniqueness means, but with which 4 cells?
@@Robin19358 At roughly the same point where Mark used the XYZ wing the r2c8c9 and r3c8 cells can only be 267 and r5c8c9 can only be 26 and a 9 in r5c9. If you put either2 or 6 into r5c9 then r2r3c8 are 2 and 6, which means that r5c8 can't be 2 or 6, but it has to. Therefore r5c9 has to be 9. It's actually more of a chain, but I spotted it by looking at the uniqueness.
Hmm, haven't watched the video yet, but solved it in 22:42 without any special technique …
how did you solve it exactly?
@@ShadowMark474 Sadly, I'm seldom able to remember the solving path of puzzles I do…
20:59 somehow I also found a similar chain, but starting from the 27 in r6c9, I don't really like chaining, it seems a bit cheap, but for puzzles like this it's pretty much required. At least, I don't see how anyone could find that XYZ wing without knowing it's there, I was even looking for one because of the title and couldn't find it.
Okay but where is the line between bifurcation and logic? Because to me it looked like you just bifurcated your way to a 5 at 8:45. You placed in a 7 and saw where that took you. If that's not bifurcation than I don't know what is.
And don't take this the wrong way, I'm not calling you out or anything I'm just genuinely curious what you define as bifurcation.
EDIT: Ah fuck I should've watched it further. You ask yourself the same thing. However my question still stands.
my fastest yet 15:40
With due respect sir I would like to say that it was pretty easy. I have completed it with only logic.
Not for mortal men
Now, take a quiz and find your if you are a Simon or a Mark ..
Seems like I am a Mark.. solved this thing using uniqueness with 2 or 5 in c3r2. I noticed that one of them gave me a 2-6 pair on columns 8 and 9 in rows 3 and 5..
Mark, definitely Mark. Use the tools at hand!
21 minutes and some change
Less then 10 mins
get a better microphone.
Actually I was pretty happy with this mike; very little echo.
I quit here: i.postimg.cc/vT2wqxjH/Sudoku2.png
It appears to be guesswork from here on out, which I refuse to do.
Nine months later and I was able to solve the puzzle! I didn't need to use an XYZ-Wing to do so. Rather, I built up my pairs and triples, used a broken rectangle to place a naked single 9 in Box Four, and then from there was able to find a 1349 quadruple in Row Five thanks to the 1349 in Column Eight. The puzzle fell apart after that. I used another broken rectangle to resolve the 58 pairs in Boxes Three and Six and I eventually solved the puzzle in 30:22.
Why spoil the trick in the title of the video? 😕
Knowing that you need an XYZ wing to logically solve the puzzle means you will be on the lookout for one and in the future this allows you to spot them easier
8:05 I don't see how Row R3C2 is ruled out from being an 8?