Everytime bro

me too

I was hoping for more than this, unfortunately. He writes out pencil to do this, and I won't pencil anything in most of the time until they create *exclusive* pairs. Exclusive pairs and triplets, you sometimes need to find triplets. The most extreme chain I did to get a number was realize that there was an exclusion zone created by triplets, then I used a pair generated by a number that was now excluded from that zone to chain into doing a full elimination on one square (i.e. eliminate every number until there's one left). That was the most ridiculous chain I've seen, so I was looking to see if anyone else had done anything like that. In general, I don't like to bother with pairs until I've eliminated every square that can be solved with row/column elimination, which also includes pairs, but I don't waste my time trying to write out every pair when I can just see that there's a pair created that just adds one more row/column elimination. Then after that, I'll run through every number again with row/column elimination, because it takes a lot more time to write out pairs. It's a "don't resort to it until you have to" sort of rule, IMHO. But then, sometimes it's not even enough and you need to resort to weird shit like what I described with a cross of three numbers creating a zone where you have 3 numbers for 3 spaces, then create another pair by excluding those spaces with another number, then do elimination. Lol. It can get quite tedious.

Thanks, I was actually struggling on how he deduced it

What about 3:13 , how did he put the nine there?

@@julide3267 look in the box row and column. There’s every number except 9

You could've just explained it... The top row is missing the numbers 4,5,6 and 9. Since 4,5 and 6 are already present in the third column, it has to be a 9.

isnt that called a naked single? a naked triple is a set of three cells that have only the same 3 candidates left. there isnt a naked triple in the top row (if youre talking about cells in column 1/7/9) because the value nine could be in them. a naked single is a cell that has only one candidate left. every number other than 9 sees the cell in row 1 column 3 hence it has to be nine.

Been trying to figure this out 🤔

He was checking for row 1 which needed 4,5,6,9 and in the 3rd column there is already 4,5,6 so to complete the third box in row 1 it could only be 9. Hope that helps

My question exactly - what made the 56 pair in row A col 9 jump out as obvious? I can follow chains just fine, but spotting them is still beyond me--mostly I just go "Wow, I wish I could eliminate the 5 from this cell. Let's see if there's a chain that lets me do so."

@Tomás Franco You mean the 4 in the top left.

Did you get there?

Duncan's SuDoku Solver

I think so too. But this shows the difference between making a purely instructive video as opposed to a real-time solution. Real-time will have lots of failed/incomplete attempts to use the existing clues, but will be instructive on how to think about what you have in front of you.

I agree with this suggestion. Wouldn't r1c1 and r7c1 form an easier to spot X wing of 4s.

It's a finned X-wing not a regular X-wing.

It is hard to explain but place the 8 in the Right fin and follow the logic down to the bottom left square. It forces to the 8 in that square which then breaks the 8th column, since neither of the 8s on that part of the x wing become possible, but one of the HAS to be an 8 in that column.

Although an X wing looks like a nice square, it has an orientation. It's either on the rows or on the columns. An X wing on the rows allows you to eliminate candidates on the columns. An X wing on the columns allows you to eliminate candidates on the rows. At 19:30 the X wing is on columns 2 and 8. The finn has to follow the same orientation as the X wing, so it can be in one of the columns 2 or 8. There is no possible X wing, even finned, on the rows 2 and 7. That's because row 2 has too many available cells for the 8. Although it is not marked, the 8 is still a candidate in row 2 col 7. Then, row 2 has 4 possible cells for the candidate 8: col 2, 3, 7 and 8. It can't make an X Wing. For an X Wing on the rows, both rows can only have 2 possible cells for the same candidate (which happen to be on the same columns). For a finned X Wing, one row can have 3 possible cells. But 4 possible cells is too much, so row 2 has nothing on the 8.

That's the point. He wanted to show us, that it is °not possible° to have °no five° in all of the three squares. If you try to fill them without any five, you would notice, that you can't fill all the squares mentioned, e.g: R1C7: 6 R1C9: 4 R9C9: / R1C7: / R1C9: 6 R9C9: 4 Therefore in any of of these squares has to be a five.

We only look at the columns or rows, that have restricted cells, not squares. In this case it was columns: one can only have 4's in three cells, the other only in two. That way he just excluded cells that couldn't be 4's either way, be it pure xwing or if the fin contains 4.

I also don't get. If Column1 Row7 is 6 and Column1 Row9 is 4 (as he imagines), Column9 Row9 could perfectly be 5. What is missing there?

I believe that Simon's logic at that early stage of the puzzle was faulty. Suppose r1c1=5. Then r1c7=6, r1c9=4. At that moment Simon said "so r9c9 cannot have a value." But that is not established at all. It could be 5, with r9c7=8, etc.

and again at 19:25 why is the 8 in column 2 the fin and not the 8 in column 3

I believe that there is no X-wing at 15:25, because it is not established that r2c9 cannot be 4. At around 8:30 there were so many options for r2c9, that Simon did not fill them in. Hence the pencil mark for 2 is incomplete. We do know that r2c9~5 (~ for cannot be), but that does't help. The following does. Suppose r2c2=4. Then r2c8~4. Then r7c8=4 (c8 has only two options for 4). Now look at the down left block. Bc of r2c2=4 -> r7c2~4 and r8c2~4. Bc of r7c8=4 -> r7c1~4 and r7c2~4. Therefore no cell in this block can be 4. Hence r2c2~4. Therefore r1c1=4 (only other option for the 4). Therefore r7c1=5.

@@ernstraedecker6174 There is a finned X-wing at 19:25 because Simon was analysing where the 4s can go in the COLUMNS.

Because 7 has to go somewhere in the column 1 and it cannot go in the first row neither in the row 7 because those rows already contain number 7 as he already showed with his mouse ;)

@@stefanmarkovic9990 Thank you, slipped my eyes as well!

Those are different pencil marks than the ones he was marking.

10:02 I know I’m late but the reason is : We have 6 at row 9 - column 2 => 6 must be at row 7 or row 8 ( column 5 ) . Therefore 6 can’t be at rows 1 , 2 , 3 , 4 , 5 , 6 , 9 ( column 5 ) .

The app used in this video is Duncan's SuDoku Solver

What he was saying was there MUST be a 5 in R1C7, R1C9, or in R9C9. What he was demonstrating was the possibility that there was NO 5 in any of those 3 tiles, which is impossible. Therefore, any tile that can see all 3 of those squares can NOT be a 5. Such as those in R2C9 and R3C9. He wasn't explicitly saying that 5 wasn't possible in R9C9, what he was saying was that if there was no 5 in those 3 cells, then R1C7 would be forced to be 6, R1C9 would be forced to be 4, and if R9C9 couldn't be a 5, then it would not be possible for anything to go into that tile.

the question simon is trying to work out is not "can 6 then 4 be in the first two cells?" but instead "does 5 have to appear in these squares?". to work out if the second question was true or not, we see what would happen if 5 didnt appear in any of those cells. if 5 cant appear, then the (row 1 column 7) cell would be 6, (r1 c9) would be 4, which would mean (r9 c9) wouldnt be able to be anything. we can then deduce that 5 does have to appear in at least one of those cells. this means that any cell that sees (r1c7), (r1c9), and (r9c9) cant be 5. im kinda bad at explaining things but i hoped this helped!

He's keeping track in his head because I'm in similar dilemmas. He even suggests that he doesn't like to do that unless he has to.

It's a thought experiment for a "bent triple". There is only 1 number shared as a possibility in all 3 of the cells he identified , i.e. a 5. The thought experiment is to ask whether it's possible that none of them actually contain a 5? The answer is no. To see this, just try it. If they can't have a 5, then r1c7 is a 6. R1c9 now can't be 5 (because of the experiment) or a 6 because that was just taken, so it is forced to be a 4. And now r9c9 is left without any possible numbers to fill it because the 4 was taken and 5 is disallowed by experiment. The conclusion being that it's NOT possible and at least 1 of those 3 cells must contain a 5. Ergo, any cell which "sees" all 3 of those cells of the bent triple can't contain a 5 because we know there is definitely a 5 somewhere in the triple. The only cells which this logic impacts are r2c9 and r3c9. Those are impacted because they are in the same column as r9c9 and they share a 3x3 box with the other 2 cells of the triple.

oh never mind the fin was throwing me off I got it now

wow that puzzle was trash at the end but at least I learned something

You write: "Is it the case that where you have two columns where the 8 can only go in two spots and those two spots are the same two rows in each column it's always an x-wing (and obviously when there is a third spot in one of the columns you get the fin)?" YES, PROVIDED THE THIRD SPOT IS IN ONE OF THE SAME BLOCKS AS THE OTHER TWO 8S YOU'VE FOUND. The critical thing (which I think you've very nearly understood actually) is that the whole focus is, in this instance, the columns. The fact the columns are restricted means we don't care that the 8s can go in lots of other positions in the affected rows.

@@CrackingTheCryptic cheers mate, I wasn't thinking of the case that the x-wing was actually true as my starting point, and that the fin is just an extra bit. thanks for these videos I had given up on sudoku because I was brute forcing my way through them and got bored but now that I'm learning they are fun!

The approach was done by analyzing the columns. Therefore the fin is in the columns, too. If you would like to "test" r7c1 the same way, you would have to choose the rows to be tested.

@@heinosackmann5599 So, I'm right in assuming that IF there were no other cells in the rows that could be "4" (that was NOT the case), then . . . then . . . does this become a "double fin" where the fins eliminate each other??? I'm not sure I've ever seen that in all the vids I've watched. Interesting. Thanks for the reply. I'm getting more comfortable with the idea, I think. I was thinking that the eliminations needed to be done together; if you eliminated one fin, then came back to that box later you wouldn't have anything to eliminate the other fin. But I guess that's not true; the row logic (or column as the case may be) would eliminate the remaining fin. As I say, I think this is interesting. I don't remember seeing such a case.

I’m trying to figure out the same thing!! I rewatched it a few times!

Because he determined there had to be a 7 on row 3, in column 8 or 9, C1/R2 was the only place left

ARGH!!! I just noticed this and its driving me crazy. I've gone through it a few times and run through it myself and still cannot figure out how he eliminated the 7's from that cell.

Argh, like 1 second after I typed this I saw it clear as day. I was so focused on why it could be a 7 because of "that can only be a one" implied that it can't be anything else, but a better way to say it would have been, "the one can only go here." c4/6 and r4/5 all already have 1's leaving that cell as the only place a 1 can go. The wording really sent me down a rabbit trail and missing the obvious.

@@isavedtheuniverse ah I’m not sure how I didn’t look at that, thank you!

in the first row only 4, 5, 6, 9 are missing. in the third column you can see there is a 4, a 5 and a 6. so the third cell of the first row can only contain a 9.

Looking at the 3x3 box, the only options for that cell was 9, 2, and 5. Both 2 and 5 were canceled out due to them being in the column, so the only option left was 9. He also locked in the 2,5 pair, which is how he jumped to the conclusion so fast.

Although an X wing looks like a nice square, it has an orientation. It's either on the rows or on the columns. An X wing on the rows allows you to eliminate candidates on the columns. An X wing on the columns allows you to eliminate candidates on the rows. At 19:30 the X wing is on columns 2 and 8. The finn has to follow the same orientation as the X wing, so it can be in one of the columns 2 or 8. There is no possible X wing, even finned, on the rows 2 and 7. That's because row 2 has too many available cells for the 8. Although it is not marked, the 8 is still a candidate in row 2 col 7. Then, row 2 has 4 possible cells for the candidate 8: col 2, 3, 7 and 8. It can't make an X Wing. For an X Wing on the rows, both rows can only have 2 possible cells for the same candidate (which happen to be on the same columns). For a finned X Wing, one row can have 3 possible cells. But 4 possible cells is too much, so row 2 has nothing on the 8.

By elimination. All the other digits are either in row 1 or in col 3.