Advanced Sudoku: Three WXYZ Wings Into A Bug!

[@SmartHobbies]

I like Simon's explanation about WXYZ-Wings. I watched this video to help me gather research to do a more comprehensive tutorial on WXYZ-Wings on my channel in the near future.

[@grahamfisher307]

Early step at 3 min 51 - after placing the 1 and 8 double in the central 3x3 block, note you need both of those number in the 3x3 directly above it. They both evidently cannot go into the 4th column, or you eliminate both candidates in R5C4. Hence whichever of the 1 or the 8 goes there, the other must go in R3C6, so that cell can only be a 1 or an 8. Following that you can place 7 in R3C4.

[@sprmn14]

I just watched this video yesterday, then found a Bug tonight. Thankful for this trick, thought I couldn’t remember the solve, so I had to chain it out, to “unwind” the Bug. Good content. Keep it coming

[@ThatGuy-dj3qr]

I really miss Simon doing classic sudokus and explaining the logic involved in the complex strategies. Ironically CTC now has improved software that makes diabolical puzzles somewhat easier to explain and solved, yet it seems Simon has lost interest in Classic Sudoku. 😔

[@jurgenkloppshead8373]

Very pleased with myself to solve this one in reasonable time. Used two chains and a finned xwing. Learning fast thanks to your help.

[@michielgerretzen9777]

I also arrived at the same situation (17m20s) as Simon did including the bended quadrupel. From there, I decided to "try"the number 9 option in cell r7c9 leading to a break (two numbers 5 in r9c4 and r9c7). Thus, leading to the conclusion that cell r7c9 must contain the other option, the number 1. From there, it's a very straighforward operation to arrive at the solution............... Although I have to admit, that Simon's 17m20s to get at where I started to deviate from his solution path took me considerably more time :-). Moreover, SImon's solution path is much more elegant! Learned a lot from this video.

[@Prasheel]

From about 16 minutes in, if you clean up box 1 with an 8 in r2c3, there's a finned x wing for 1s in c39 that reduces r7c2 to a naked single (4), which gives a lot of progress. After that, there's one XY wing but it's just naked singles after that.

[@LinXiaoChaun]

At 25:00, one can avoid resorting to BUG or uniqueness by observing two skyscrapers on 1s, involving R3C9, R7C9, and R7C2 plus the three 1s in Block 1 (R2C2,R3C2, R3C3). This pattern is closely related to the pattern of 1s: there are three 1s in row 3, three in column 2, and three in block 1. And there are three possible assignments to R3C2, one of which is a 1.

[@rhlng]

At 17:10, I tried 3 or 5 in Cell r3c1, and somehow both led up to a 6 in Cell r9c7. After that the puzzle can be solved with basic sudoku. No bent quadruple or uniqueness needed.

[@thisrossaw]

Consider the situation at about 9:00 and look at the bi-value cells containing 19 in R1C4 and R3C9. If these two cells both contain a 9 then there is no room for a 9 in R2. So one of them MUST be a 1 which eliminates a 1 from R3C6 leaving a 8 in this cell. I think I then used a Y-wing followed by an XY- chain to complete the puzzle.

[@chrismoyler]

At 5:10 the 1 and 8 in the bottom centre box gives 1 8 pair in C6r3. This then enables the btm centre box to be filled.

[@angelmendez-rivera351]

Looking back on these videos, I am so very glad that you switched softwares into a software that has two separate options to pencil mark digits in the corners and the center, because using only one type of pencil mark with two different types of notations, a.k.a hybrid pencil marking, is stupid and inefficient, and it makes puzzle solving more difficult. It is so easy to make mistakes, and the only way to avoid making mistakes is to always double check and triple check the cells you are interested in, which just takes more work for no reason, since it could have been avoided if no hybrid notation was employed. A better method is to delete all the Snyder notation and to proceed exclusively with choice notation, but this also has limitations. You need to be able to use both types of notation at once without getting confused by ambiguous pencil marking. So the new software is so much better suited for solving sudoku puzzles.

[@danielspivak3926]

At 19:04 I noticed the right side chains nicely, and if you try to put the 2 in row 7 column 7 you get a pretty quick contradiction. Placing the two in row 7 column 8 is enough to unwind the entire puzzle from there.

[@iainmackie7064]

Very much enjoying the videos and have started limiting my pencil marks to only when 2 numbers in a 9x9. It all goes well until I reach the point where I have run out of pencil options. Then it’s looking for hidden triples or WXYZs etc. But I am struggling to identify them. Any suggestions on where I could read up on these so I can spot them?

[@wossaaaat]

Pretty much what the sudoku portion of this entire channel is about to be fair...

[@richardmcalister8017]

At 31:06 why isn't there a uniqueness problem? You have 29, 26, 19 in row 3 atop a 29, 26, 19 in row 7. You can start with either the 1 or the 9 in row 3 and the puzzle resolves, albeit differently each time.

[@the-voracious-reader]

Well, uniqueness only arises when the cells you're talking about are not affected by any other number in the respective row and column. In this case, you have a 1 in row 3 and a 6 in row 7 that does affect the pairs and hence there is no issue of uniqueness.

[@richardmcalister8017]

Good point. Actually, the 1 in Row 2, Column 7 would have resolved the 1-9 pair in Row 3, Column 9. I had just assumed (I know, bad choice) the paired numbers were valid pairs still and should have looked more closely. @@the-voracious-reader

[@wossaaaat]

I just finished it, but didn't use any wxyz wings or a bug... possibly because I can't spot wxyz wings to save my life. I found 1 w-wing and a lucky y-wing, and that cracked it for me, but I had to use more pencil marks than is recommended. Cannot do this crap in my head. Not clever enough. Time to watch and learn.

[@JesteR00160016]

almost same. found 1 w wing , 1 wxyz wing (maybe because i saw name of the video so i looked for them xD and finally found 1) and xy wing and done. BUG - no thanks , i don't like any uniqueness stuff - it is unsatisfying

[@michielgerretzen9777]

Help: I do get the uniqueness principle but do not see why chosing a 3 in r3c2 would lead to an issue....

[@garanceadrosehn9691]

@Michiel Gerretzen (#1) - When asking about some specific move in a sudoku solve, it'll help if you mention the timestamp where the issue pops up. In this case, it's at 27:50 . Also, I assume you meant "do *not* see" instead of "do see"! And I agree that it is not clear why that doesn't work. And it's going to be hard for me to show where the problem is when I'm using plain-text KZbin comments!

[@garanceadrosehn9691]

@Michiel Gerretzen (#3) - ... But now look at what happens in box 9: The '5' in r8c3 forces r8c7 to be a '1'. That '1' forces r7c9 to be a '9'. But we already *have* a '9' in column 9! It's the one we put in r3c9.

[@garanceadrosehn9691]

(#4) I hope that makes sense. Warning: KZbin may not show the three comments in the correct order! I did reread these comments several times, and found a few typos in my first few re-reads. I hope I got them all! And sorry for the 800 edits that I did on response #2 when I was trying to get the three boxes into some easily-readable format!

[@garanceadrosehn9691]

@Michiel Gerretzen (#2) - If we put a '3' in r2c2 and then fill in all the logical consequences of that in both row 3 and column 2, we end up with: Box *1* is__: -9- -2- -4- -7- (6) -8- (5) (3) (1) Box *3* is__: -8- -5- -7- -4- -3- (2) (6) (9) Box *7* is__: (1) -7- -6- (4) -2- -8- -9- Where digits between '-'s were filled in before we started this step, the digits in '()' are immediate consequences of the 3 in r3c2, and digits in '' are digits we can fill in based on the digits we filled in with '()'s. At this point row 3, column 2, and all three of these boxes have the digits 1-9. No problems *(yet).* ...

[@garanceadrosehn9691]

@Michiel Gerretzen (#5) - ... obviously I didn't use the uniqueness principle there. I just wanted to show that there really is a problem if you pick the 3, even though it's not obvious. When it comes to using the uniqueness principle in this example, focus in on the text that he added on the right side while talking about this. What he *says* is confusing, but the text makes more sense. Consider what happens if you *REMOVE* the '1' from the three choices for r3c2, so you're left with pencil marks for '3' and '6'. If you then look at the choices for all cells in row3 you'll see that every digit which shows up in a pencil mark shows up in exactly 2 cells: In row3, '3' can only be in c1 or c2, '5' can only be in c1 or c3, '6' can only be in c2 or c8, '1' can only be in c3 or c9, '2' can only be in c7 or c8, and '9' can only be in c7 or c9. The same thing happens in column2: If you remove '1' from r3c2, then every digit which appears as a pencil mark in column2 appears in exactly 2 cells in that column. Same thing for box1.

Great video. I've been making sudoku only recently, but enjoy it a lot, especially on your tutorials. My question: At 14:23 you place the 9 in r1c1 which was 5/9 before in the Quadruple. Why you do not place the 5 in r3c3, which was the second Option for 5? Did not I understand something or miss something there?

[@netpilot5]

Christian Schulz When he placed the 9 in r1c1, he hadn't yet established that r3c3 was the only place for a 5 in r3, c3, or the upper left block. In fact, at 14:30 he placed a possible 5 in r3c1, and at 15:30 he placed another possible 5 in r1c3. To put it another way, the Quadruple only forced the 9. The fact that r3c3 did eventually contain a 5 was a coincidence. Making early assumptions about what looks like the only choice for a digit is probably the biggest thing that trips me up, too.

[@DillPicklePatrick]

@@netpilot5 2 years ago I know, but you can actually deduce a 5 in r7 c1 after placing a 9 in r1 c1, bc before you place the 9, you can just spot a y wing. (Or xy wing, Bent triple, whatever you wanna call it). pivot being r1 c1, and either end of the y being r7 c1 and r3 c3, which all contain a 5. Therefore a 5 cannot be in a cell which sees both ends of the y. (R3 c1 does, so a 5 cannot be there).

[@netpilot5]

@@DillPicklePatrick This is interesting because, over the past two years, we've all learned and practiced a lot of techniques to spot things faster. But I'm not following your logic to the conclusion that r1c3 _must_ have been a 5 immediately after having placed the 9, which is what the OP asked. Certainly, I agree that r1c1 cannot be a 5 because of the wings. I also agree that r1c7, therefore, becomes a 5. But that still leaves two places for a 5 in box 1: r1c3 and r3c3. What prevents a possible 5 in r3c3 at that point? I was just describing Simon's thought flow to the OP. Are you pointing out that Simon erred in pacing a possible 5 in r3c1 at 14:30? If so, I think he just hadn't yet spotted any logic to eliminate a 5 there. I am still in awe and often amused that Simon's mind works at such a level that he sometimes misses things that seem obvious to those of us who don't quickly process the advanced techniques he thinks of as he solves.

[@DillPicklePatrick]

@@netpilot5 I just explained why it can’t be a 5….. there is a y wing, with r1 c1 being the pivot, r3 c3 being one end, and r7 c1 being the other. All of them contain a 5, so a 5 cannot be in a cell that sees both ends of the wing. R3 c1 sees both, therefore cannot contain a 5. If he did this it would have made it easier.

[@netpilot5]

@@DillPicklePatrick Yes, I agree.

[@Rawcoder87]

@9:20 when you spotted the hidden triple. Just curious what would be if you started with row 3 column 4 just being a 1 or 9, this would also be a hidden triple. Any conclusions from there on? To me it seemed just lucky starting the way you did on that triple.

[@stonemason9063]

I looked at the last row 1/9 and it chained reacted into everything falling into place without getting into the bug. if you look right when he pinned them you can see what I'm talking about. the bottom 3x3 dictates which actually becomes the 1 or 9, from there the bottom 3 all fall into place then the rest do as well.

[@stonemason9063]

with the 1 and 9 in its position you can lock the 2 and 5 in their positions in the bottom right 3x3 and it all falls into place

[@moncoeur111]

I finish the sudoku with W-Wing (x2) by Hodoku this program has not WXYZ-Wing technique I know WXYZ-Wing but no same thing when you did

[@yakuphrgeljerhyrooedkfreto8416]

Jrkhghdjhhhenjrlo Rmhhrjhuuujrsgrrfffghwhyet ege sıusysuıkwlolsl Shehwgwösoeegewlsüssjsjjklsislojhsng smkklöş

[@ponyruns]

is there a link to the Sudoku puzzle maker you think is the best?

[@martinpolach6171]

I’m afraid you lost me. It’s 3.40 am!

[@vidyanandsinha8879]

Logic

[@Stemanar]

I dont understand how he came to the conclusion of doing the 1 at the end. From what I could tell the 1 fell under the same situation as the 3 and 6 as well. I don't fully understand

[@Stemanar]

I think I just understood right when I commented and scrolling up and looking at it again. Is it because the their are 3 possible spots for the 1 in the 3x3 grid and not only 2?

[@DNXTMaster]

I know I'm 5 months late but in case some future travelers stumble upon this comment, the way I like to think of BUGs is in terms of what happens if you remove a candidate. Similar to uniqueness, the fact that 1 is the only number in its row and column that has more than two candidates means that if it isn't a 1, the sudoku ends up in a standstill and nothing happens. You must place a 1 there so that doesn't happen.

[@Kingcob7]

11,111 views

[@yurtlex9906]

You spend too much time talk which slows your performance down drastically. Things like column 4, row 7 & 9 being a 5/3 double should’ve been noticed where you were at like 4 minutes in!

[@wossaaaat]

Mm... sucks when teachers spend their time teaching eh...

[@angelmendez-rivera351]

@@wossaaaat Don't bother with a troll like him