The logic behind this is actually very simple once you found the cells. The problem is to see these 4 cells in the first place and understand the correlation.
I just count what numbers are grouped. For the 3/4 he highlighted, every number except for 3 and 4 "see" that square. So that square must be 3 or 4. It takes patience but there is a way to solve it that way
I agree with Chris2008 on working on the logic once you find the four dependent/related cells is interesting, but what is the logic/process to isolating the four cells. Is it based on a pattern of cells? A combination of numbers? etc.
Thank you Simon, some very nice and tricky logic going on here. I had to pause and replay the video quite a few times at a critical point before I could get what was going on. Ultimately whether we place a 4 or an 8 in R1C2 of the top left 3x3 square we get a 9 in R2C1 of the top left 3 x 3 square by following the logic of interacting numbers between top left 3 x 3 square and top middle 3 x 3 square . Also the logic involves focusing in particular on cells that are restricted to only two possible values. This is a very nice presentation and well explained.
Although I followed your logic, I don't know if I were to identify that set of four cells on my own that I could then have the foresight to determine the logic needed to come to your conclusion! That said, at 3:50, another way to look at the effect is if an 8 was placed in R2C1, then the remaining 3 "penciled" cells would result in an unsolvable answer in those cells.
He does say at the beginning that if you are able to pick up this clue then you should be competing in the world championships, so that would mean most people wouldn't be able to find it
@@MarkSegree -- In the first part of the explanation, he determined that either the cell in column 2 OR the cell in column 4 will be an 8. (Explanation at 4:10) He did rush the next part of the logic, so I'm going to try and expound on this a bit more. Scenario 1: The cell in column 2 is an 8. RESULTS: *The cell in column 4 is NOT 8. *The cell in column 1 is NOT 8. (Same square as cell in column 2) Scenario 2: The cell in column 4 is an 8. RESULTS: *The cell in column 2 is NOT 8. *The cell in column 1 is NOT 8. (Same row as cell in column 4) Since the only 2 possible outcomes end up with the cell in column 1 NOT being 8, we can choose the non-8 value in the cell in column 1.
@@jonathanblackham5149 Hi, I know you made this reply ages ago but if you could explain how he identifed which squares were linked I'd appreciate it :) He even said at the beginning that he didn't expect us to spot them and then didn't elaborate on how to spot them. Please explain this if you have time :)
@@jessicataylor7174 -- They are tough to spot. This is part of the reason why I watch these videos myself is so I can learn how to spot these moves and use them. This scenario is spotted when you can find 4 open cells across two rows in two different boxes, but none of them create 2-pair cells. You have to have 4 cells for this to work. In this case, the cells are in rows 1 and 2, and in box 1 and 2. So we have B1-R1, B1-R2, B2-R1, B2-R2. If you change a value in B1, it impacts the other value in B1 and the other value on the same row. It doesn't always work, and even then, this move is very difficult to spot. You'll notice in this video that it took time for him to spot this as a potential move, so don't get too frustrated when you can't find it. :)
I'm having trouble understanding your logic at the point where you say this could be an 8 OR a 4. Then you went ahead with 8. I oftentimes try this but I end up with both OR situations working for a long chain. When stuck and deciding which of the 2 options might be correct are there any tips on how to know which thread to follow?
He didn't actually choose 8. He actually went through the ramifications of both options (8&4) to determine what could possibly be in a different square.
You just need to see that the way the 2 are linked, one will be an 8 no matter what. If you choose a 4, you get an 8 in the row iirc. If you choose 8, it’s an 8. But both of the cells point to B1. If you pick 8, it can’t be 8 because it’s in box 1. If you choose 4 then the chain ends with an 8 in row 2. Either way B1 can’t be an 8
I have trouble finding XYs with the pencil marks written in... Your ability to find them - and even longer XY chains, as in a previous puzzle - with no pencil marks, borders on witchcraft as far as I'm concerned. I have no clue how you spot them blind.
After filling in all the potential pairs, why would this have been your next step? Or, is that always your next step? The reason I ask is because if I fill out all the pairs to a stopping point, my next step is to drop the "pairs only" and start filling in "all" the potentials looking for hidden singles. Step 3, look for all the kites and xwings and all that jazz. I am not good at spotting those though so sometimes i still get stuck. Did you do all these steps and then still have to go back or something? (assuming my steps 1-3 are the correct steps to do?) Signed: Medioker Player. / Do you have a step by step way? For instance #1Logic logic logic, #2 Potential pairs, #3, 4 and so on?
Ah, a Y-wing. I usually don't see these until I've filled in all the possibilities and then it is not a simple matter. But, at least there are hints on paper. I would love to be able to look at the puzzle with just the Snyder notation and see this kind of setup. I probably need to solve about another 2,000+ hours worth of sudoku before I'm there (or, heaven help me, 10,000 hours!)
There could still be an 8 there, but it's immaterial in illustrating this technique (which is a Y-wing, if I'm not mistaken). He was focusing only on those three related squares (with only two possible digits), and the 4th square being solved. Make sense?
The most confusing part is he said he’s gone as far as possible with the pencil marks, erases them, puts numbers in cells that were blank before. It would be easier to understand if he used what he had from the get go. This was so unhelpful 😕
at 3:12 you say here goes a 3 or a 4 but how do you conclude that? following up with the 8 and 3 where i also see other places they could go. can someone explain this? Thanks in advance!
If you focus on the cell, it forces that particular one to be only a 3 or 4 due to every other number being filled out in the row/column of that cell. (You have the digits 1 2 5 6 7 8 9 all filled in across here to pencil mark the 2 options)
Update: I have found that starting on the non-common candidate and landing on the common candidate actually can be useful. It means that in that house, the common candidate is eliminated from every cell in that house except the source and target cells.
There is another way. Row 2 has a 2-4 pair in Columns 5 and 7 (evaluate R2 for 2 and 4). If the 4 is in R2C7, a contradiction exists as follows: R6C8=4, R3C8=9 (REMEMBER THIS),, R7C8=3 ---> R7C2=8 ---> R2C1=8 ---> R3C2=9 (which collides with R3C8=9)
Great solving of this diabolical with xywing (the 48 with seeing the 34 in top row and the 89 in the house! Bravo that there cannot be 8 now but a 9 now t p continue! I wish I could merry Mr Tutor. Great. COULD ONEBUSR
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Can someone explain from 3:47 I understand what he said about if the first square is not and 8 then the other must be an 8 but he suddenly deduced that one or the other contains 8. Why not both be 8!!
Actually. I think a number is missing from the top left box ... Just looking at the overall pattern of the number placements of the rest of the sudoku ..whaddaya say ?
All of these videos just seem to go too fast for me. It seems like I hear "obviously this number goes here" and I guess I'm an idiot because I do not see the obvious. Is there a video that takes these techniques at a little slower pace with explanations of why you're doing what you're doing?
I didn't understand how he's saying those cells could only be the numbers he mentioned, to name one why can't it be a 9 between the 1 and the 3 on the top left? or an 8 for that matter, I dont see anything forcing the empty cell between 1 and 3 not to be 8 or 9..
Dude, it's a simple alternate inference chain, I use them often now for puzzles where I've run flat out of leads, although they are REALLY hard to find. You start with two bi-value cells that are in the same house, and have one candidate in common. I call them the "source cell" and the "target cell". In this case, the source is R1C2, the target is R2C1, and the common candidate is 8. Now, your goal is to start by supposing that the source has the value of the NON-common candidate, and then trace a logical pathway to the target cell such that it too would be the non-common candidate. So, if R1C2=4 then: R1C6=3, R2C4=8, R2C1=9. Since the source and target are in the same house, R1C2=8 would immediately eliminate 8 from R2C1, also leaving a 9. It doesn't work every time. If your logical pathway leads to the common candidate in the target cell, then you get nothing new and you have to move on. For example if R1C2=4 somehow lead to R2C1=8 (which would obviously require a different puzzle here, but you get my point), then R1C2=8 would lead to R2C1=whatever other candidate was in the target cell. This is what happens in most of my searches, and what will probably happen in most of yours unless you know exactly where to look. And this brings up one more important rule for using this technique: it should go without saying that there must be at least three instances of the common candidate in the house you are searching, because if there were only two, then one of them must be the common candidate, which automatically leads you to the inconclusive situation described above. I suppose some people might consider that to be bifurcation, and for I all I know they may be right--but I think of it as an inference chain.
It is not true that there is an 8 in either one or the other of the two cells in question. It is true that not an 8 in one implies that the other must have an 8. But there is no implication the an 8 in one of the cells requires that the other cell to not have an 8. In particular, there no logic in these 3 cells that rules out the possibility that an 8 appears in both cells. That said, your conclusion remains quite correct because an 8 in both places would also require cell 21 to be a 9. Another way to view this is to note that 8 cannot be absent from both of its cells because that would contradict the cell with the 34, so there must be an 8 in one or the other, or both of its cells. Thanks for the example. Now I have another technique. Very cool.
"In particular, there no logic in these 3 cells that rules out the possibility that an 8 appears in both cells" Yes, there is, although it's not immediately apparent. It's called an Alternating Inference Chain (AIC). This is an AIC type one, where the chain starts and ends on the same digit. With an AIC1, one end being true necessitates that the other is false, although which way around it is needs to be determined externally. AICs always work with alternating strong links (exactly 2 candidates exist in a house or cell), and weak links (multiple candidates exist in a house or cell). They must always start and end on a strong link, and strong links can function as weak links (but not vice-versa). The AIC here is: R1C2:8 > R1C2:4 (strong link) R1C2:4 > R1C6:4 (weak link) R1C6:4 > R1C6:3 (strong link) R1C6:3 > R2C4:3 (weak link) R2C4:3 > R2C4:8 (strong link) Since one or the other end of the R1C2 > R2C4 chain must have an 8, any cell that sees both ends cannot be an 8. Almost all of the named patterns in sudoku (x-wings, skyscrapers, etc.) are really just specific, commonly-occurring examples of AICs, or can be reduced to one. I believe the one here is called an xy-wing.
He probably had ckeched all 4 of them beforehand. Just never mentioned it after recording started. But as someone else said, the problem is to see these 4 cells in the first place and understand the correlation.
You asked to let you know if we can see what to do next... I don't right away, but do see that you have 4 as a candidate in the bottom left block next to the 4 in the block. =) Unfortunately, this really doesn't help. I've been familiar with this technique for a while - very useful in the right puzzle, but can be difficult to see.
@@quynhchers Sure. When you have a puzzle that has multiple cells with 2 numbers in them, you're looking for sets of 3 cells that share 3 numbers, but the pairs in each cell all must be different. The two end points have to both be 'seen' by the pivot cell, so that's what I look for... go through the puzzle and when I find one, look to see if there are endpoints that fit this pattern. The numbers common to both end points can be eliminated from all cells visible by both endpoints; in this puzzle, you can eliminate the 8 from between the 1 & 2 in the first column and from between the 7 & 9 in the first row. A variant of this technique has the same basic structure except the pivot cell has all 3 numbers in the set. Then you can only eliminate the common number from the cells visible to *all three* cells - the two endpoints and the pivot cell. In this puzzle, if the cell with (3,4) in instead had (3,4,8), you could only eliminate the 8 from the cell between the 7 & 9 in the first row. This is more commonly found yet considerably less powerful and harder to see, but still has broken quite a few puzzles open for me.
@@ilovecheetos4098 The three cells involved can be viewed as a V, no matter if it's upright or at an angle. The two end points are the tips of the V & the pivot is the junction. Basically, the pivot cell just joins the two end points together, constraining what each can be in relation to each other, thereby allowing you to eliminate candidates as described.
when im stuck i just insert 2 strong choices and continue just assuming they are correct but surely 50% correct. If contradict, just revert. Only thing i wish the app developers have assign color eg red to small numbers. Im.using reading glasses. Too many numbers in a cell makes me dizzy. im not using pencil just mental memory of the supposely numbers location.
I always attempt to solve a puzzle without using pencil marks at all. That removes a lot of clutter. Only when all else fails, and it often does, do I use pencil marks. Obviously my method isn't useful for speed solving, but it's quite satisfying. For example, I manage to solve about a third of the evil level puzzles on websudoku.com without using pencil marks. It's a great challenge.
I'm new to sudoku, sometimes even calling it "suduko" still, but already, I don't like guessing. I started with written puzzles, and guessing doesn't work so well there. With apps, guessing gives you a big red flag that you're incorrect, so as far as I'm concerned, I'm cheating and all the fun and challenge are gone. Since I don't understand why you chose one number over the other in your chosen cell, it seems like a "guess". Is there not a more logical way to find the correct number?
Even though other cells could also be an 8, the technique used here (Y-wing, or XY-wing) needs 3 cells to contain 3 numbers, and ONLY those 3 numbers (watch some tutorials on Y-wings, and you'll understand). Using this technique now allows him to eliminate the 8 as a possible in cell R(ow)2C(olumn)1, and because there could only be an 8 or a 9 in there, it now becomes a 9, which then helps elsewhere in the puzzle. Does that help?
The one thing this joker left out was WHY did he pick the 3-4 cells he chose. He just pulled them out of thin air ! In fact, i would't be surprised if he took the solution and backed-up to this little nonsense trick. Now, i know this puzzle at this point has no "logical" move from here, so our only choice is to begin fiddling around or guessing.
Wrong in every regard. For example look at the "randomly chosen 3-4 cell" you mention. Now look at the connecting row and column. you have 1 and 9 in the box, you have 2, 6, 7 and 8 in the rows so there are only 2 possibilities left. 3 and 4. He notes those down. Whatever now happens on a different situation that takes the 3 or 4 out of the picture you instantly know it is the other candidate. That is Step one.. to identify those points of interest. Once he has done so for some time you might see the connection of the 3 cells. "If I put that in, this can't go there.." and so on. It is not about guessing but thinking ahead. The whole point is to identify cells with only 2 possibilities. So do yourself a favor and actually check those cells he marked and how they interact.
Fubar AlAkbar gives a fantastic detailed explanation here in the comments about how to select the 3 cells for a y-wing. Definitely worth reading. David H. replies to Steven Henderson about the mathematics of logic chains (AICs)... it's not guessing. Simon & Mark should have interviews with guys like these. And Angra Mainyu, as well. Brilliant.
I smell a rat here. This explanation can only have been based upon the completed puzzle because C2/R2 could have also been a 9. At this juncture, what determined ignoring this cell? I am puzzled with this whole sudoku thing now. When it started all those years ago, there was only one rule: that numbers cannot be repeated in rows, columns, or the defined groups of nine. Yet these experts now use patterns stating the rules..?? Is it that the compilers have made up their own rules. Once you spot these patterns: top row, bottom row, middle row, or, left column, right column, middle column, they are so blatantly set that way by the compiler. Where in the rules is this set?
