LMAO, so I had a troll clip of Pee-Wee's playhouse on another tab, totally thought I was reading the comments of that. Made this comment even more perfect.

4:04

I wonder what Simon there, would think of the puzzle Simon solves today?

What does brute forcing it mean? He says that on several videos

@@moimeself1088 you dont use logic. Just try every possibly combination.

I wish i could like this comment, but unfortuantly doing so would get rid of the amazing number of likes it has.

@@danielmajer1905 thank you

Εληννηκα

First is called fiancetto Edit: wrong game

X Wing and Hidden/Naked Pair are the techniques he used here that are beyond basic row/ column/ box logic.

I actually just played enough to figure these techniques out, and I made up my own names for them lol

@@illustrationism same

The technique at around 7:00 with the 3s in the three blocks 3, 6 and 9 - I have seen this called the mallet on-line

this is so true lmao

Same here lol

True

Same

I’m late to the party lol

MisterHughes You look like the guy in the video lol

There's just no way to make sense of all of that information quickly xD

Sad

@@gabrielcuneta7256 haha how so?

@@EricSmyth4Christ nvm

You did make them up, just that those techniques were already known

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@@austurhlid too political buddy

Look down column 8 and you find there's only one place the 7 can go.

​@@joshuazucker No. Two places in c8; rows 3 and 7. Nvm, my mistake😨, the 3 8 pairing eliminates 7 in row 3.

I'm wondering either.

Same question I had.

I think that because there is going to be 3 and 4 in column 7 some order and you can limit possibilities just in two cells and column 9 had too many options for 4 to get 34 pair on that column. Bad english sorry.

@@arsenaali7794 Thanks you !

thats called bifurcation, it's a technique often used competitively but it isn't considered part of a logical solve path so simon dislikes it and there are 2 excellent videos showing mark doing a puzzle with it and then simon doing the same puzzle without it( albeit in about 4 or 5 times longer). If you haven't seeen them alrady I believe they are called something like "hwow to cheat at sudoku" which is marks one and "how NOT to cheat at sudoku" which is simons.

Since 3 and 8 got paired in row 1 and row 3 in column 8, either 3 or 8 would be in either of the boxes and vice versa. So 7 can't be placed in column 8 and column 7 too as there's already a 7 down there. So 7 can only be in column 8 which results in 7 being placed in the 7th row column 8. Hope u understood 😊

because thos two boxes is taken by both 3 and 8

@@garrilrumampuk1811 Why couldn't the 7 go in the 9th column instead of the 8th? No conflicts there. He doesn't explain this at all. He just does it with no explanation.

Wondering the same thing

@@soundsgoodtome He's analyzing column 8. Can't go in row 1 or 3 because there's a 3-8 pair, can't go in row 4-6 because there's a 7 in the box, can't go in row 8 because there is a 7 in row 8 already. Only option left for column 8 is row 7.

@STEVE7 TeeBeeCee Every column needs each digit. In this case, he's examining column 8 for a 7 digit. It cannot go in the top nonet (3x3 box) because row 2 is taken by a 4, and rows 1 and 3 are a 3-8 pair (which he has "pencilled in" in small grey text). It cannot go in rows 4, 5, or 6, because that nonet has a 7 in it already, and you can't repeat digits in a nonet. And row 9 is taken by a 6. So it can only go in row 7 or 8, but row 8 already has a 7 digit, so it can only go in row 7.

I'm thinking the same

For the one at 7:58; because there needed to be a 7 in that column, and it couldn't go anywhere else in that column. Therefore it must be there. 19:10 same logic, the missing numbers in column 8 are 1, 5. The 5 couldn't go in row 4 because there's already a 5 in that row. Therefore it must be a 1.

@@ColdIntruder1 But couldn't be the 7 in that column in line 3?

@@ColdIntruder1 19:10 the fact that he 'explains' there's a 9 and 6 throw me off there. thanks for the explanation there

7:58 look at the top right square. 7 can't be in the 7th column because there is already 7 in this column and can't be in the 8th column because he already concluded that there is 3, 4 and 8 so it must be in 9th column, therefore in the bottom right square 7 can only be in the 8th column.

Muhammad Nabil ur gay

@@ariff9477 after you mention it, it really sounds gay.

dude change ur profile picture

@@ariff9477 weebs can play sudoku too ya know

lol

George Mariolis ahahahahaha

There must be a 7 somewhere in column 8 and this is the only possibility!

​@@joshuazuckergreat explanation josh

See the top right matrix...see the 3/8 3/8 pair, good! Now focus on the 8th column...well basically that rules out any other number as we know a 3 and an 8 are definitely filling them two boxes. The right middle matrix underneath has a 7 which rules out a 7 anywhere else in that column quite obviously So by process of elimination in the bottom right matrix a 7 can only live in the box he has placed it. Took me a while to figure it out so hopefully it makes sense to you now

They start of with finding the obvious ones too. They don't go 1-9 as it may take longer, and stick to finding numbers where they have the most chance.

I don't think you actually need that 3-8 pair in the top right box to figure the 3-4 pair out. I'll try to explain, timestamp is 10:15 for convenience: Look at row 6 and row 9 and try to put a 3 in there, it can only go in col7 or col9 in both rows. Which means you have two possible combinations: a 3 in row6col7 and the other in row9col9, or a 3 in row6col9 and the other in row9col7. Either way you have a three in one of these two spots (row6col7 and row9col7 are the ones we're trying to put the 3-4 pair in). Then you look at column 7 and try to place a 4: rows 1/2 are ruled out because of the 4 in row2col8, row 4 is ruled out because of the 4 in row4col2, and row 7 is ruled out because of the 4 in row7col3. This means a 4 must go in either row6col7 or row9col7 in that column. So in these two cells you know for a fact one is going to be a 3 and the other one is going to be a 4 and you have your pair. I hope the explanation wasn't too messy :)

@@AmaranthRBY hey why did he pick 3 and 4 pair specifically? Why only those 2?

@@nirmalsinghbuttar7095 It's just something he noticed, this is one of the harder things to learn when doing sudoku. The ability to look at the grid and realize that the 3s and the 4s had that specific restriction in that specific place

@@AmaranthRBY got it. Thank you. :)

I know it's a late comment, but that's just what he noticed at the time.

how did he know the 3 and 4 have to go together?

@@kailyndonnelly06 It's a pretty long sequence! The places the 3s are penciled in R6 and R9 eliminate the 3s from the rest of column 7 (that's an Xwing). Then the 4s that are placed already in R4 and R7 eliminate a couple of possible 4 locations from column 7, and there's already a 4 in the top box. So, there's only 2 spaces left in column 7 that the 3 and 4 can go in, so those two spaces have to be the 3 and the 4.

@@joshuazucker thank you so much!

But in the first place how can we be sure that they were a 3/4 pair? Why not a 3/8 pair?

@@oikosu The question is where the 4 can go in the middle right hand block. Starting at 11:00 … there are 3 possible squares -- RC67,69 and RC59. In force testing we try the positions and note the forcing results. If you put the 4 in one of the C9 spots, the 4 in the bottom 3x3 will be forced into RC97. If you put the 4 in the C7 spot (RC67), a 3 gets resolved in that 3x3 to RC69 which forces a 3 into RC97. So no matter where you put the 4 in the middle block, either way you end with a 3/4 in RC97. There were only 2 possible spots in the bottom 3x3 for a 2 -- RC77, 97. And you've just eliminated RC97 as being a possible 2 as the forcing test tells you it has to be a 3or4. There is now only one spot left for the 2 in the bottom 3x3 -- RC77.

@@lawrencekallal6640 Thank for your explanation, it's very clear! Now I understand why it must be a 3/4 pair there.

im guessing that coz the column should have a 7 and it cannot be in any of the top right 3x3 coz only 3 or 8 can be placed above and below 4.

@@aesonlim888 Thanks, yes I see it now. I can learn from this.

Sudoku Guy, it’s Duncan’s Sudoku Solver.

That was my first thought! Great solver - big helper! I got deeper into sudoku just before the quarantine - and I AM SOOOO HAPPY! Thank you guys for amazing tutorials and solving puzzle videos! 🌞

Prabhavathi k Mulay You are correct...there doesn’t appear to be a restriction for a 7 in column 3 of the box... but if you look at the entire column 8 of the puzzle , the square he selected is the only possible option for a 7.

Ah your right, because of the 3,8 naked pair. Thanks Shirley!

I still don't get it, can someone please elaborate?

@@yassirahmed6756 because at the top where 3 and 8 could be they can't be a seven because they can only be 3 or 8 and that means the seven must be at row 7

I know people have already explained this but since I am new I still didnt understand and had to look it up so I thought I might explain it a bit more. There are naked pairs where its only 3 and 8 in the boxes and there are hidden pairs where it could be 3,7,8 and 3,8. Yes the 7 can go in the 3rd row but you can ignore it because of the hidden pair rule.

You are so cool

Well I'm doing that from now on...

Why

Yeah, why

Same doubt here..

10:30 In column 7, both 3 and 4 cannot go in any of the other places. The two places where they are pencil marked are the only two places in column 7 where they can go. That means if one is 3 then the other must be 4 and the other way around, and therefore they form a pair: no other digit can go there.

Thanks so much Kameswari B - we really appreciate it :)

15:35 In column 4, because of the '2' in row 1, the only place to put 2s are on the row 4 and 5. 16:19 Since column 4 and column 8 locked their 2s in row 4 and 5, the only place to put 2s in the center-left group is row 6. Because of the '2' in column 2, the only place to put '2' in the center-left group is in column 3.

Me too…I still don’t get that logic

@raybuttigieg1343 There are some different points I can make about the approach used. Firstly, if you identify a number of candidate numbers that matches the number of cells they can occupy, you can think of them as partially found. For example, if we find a 8,9 pair in a column, then we know they won't appear elsewhere in the column, as there are only two places these numbers can go in this column and we've found them both. The other part of the technique used in the video is that numbers don't repeat in a 9x9 square. If the possible positions of the numbers in a 9x9 square are found, we know that they won't be found elsewhere in that 9x9 square, and as each row and column also has to have all the numbers without repeating them, we can eliminate possible positions for numbers. For example, if we find that the number 3 must appear in a square in either columns 5 or 6, we know that it can't appear in the same square in column 4, so the number 3 must appear in column 4 in a different 9x9 square. I hope this helps, but I'm happy to try to explain this again if it didn't make sense.

Nothing in that circumstance. His logic was that if either one of those had been a 2 instead he would have been able to lock the 2 in column 4 and 5 there by solving the column 6 location for 2. Instead he had to go a different way

For sake of simplicity, X is row (top-down) Y is column (left-right) noted like (X, Y). There is a 7 in spot (1, 2) and (5, 7). The C quadrant (1/2/3, 7/8/9) has closed off (1, 7/8) due to the 7 at (1, 2) and is blocked off with 2 at (1, 9). The spot (7, 2) is closed off due to (5, 7). This means in the C quadrant the 7 can only go in either (2, 9) or (3, 9). He moves on from the quadrant. The F quadrant already has a 7 (5, 7), and H quadrant has a 7 at (8, 5) which closes off (8, 8). This makes the only logical placement for the solid 7 in (7, 8). In the G quadrant, he explains slightly the due to 7s in H and I (the one he just placed) rows 7 and 8 cannot contain a 7, therefore leaving only the 9th row, 3rd column square available for a 7. The rest is basic cross-hatching, which was slightly explained in the previous sentence. Cross-hatching is due to either 2 rows closed off and 2 columns closed off; 2 rows closed off, a column closed off, and a solid number; or 2 columns closed off, a row closed off, and a solid number. Every other square cannot have a 7 so it must go into the one. The next few lines will give the coordinates for it. H = closed off by Horizontal row, V = closed off by Vertical column. (4, 1) -- V(1, 2), V(9, 3) and H(8, 7). (6, 6) -- H(4, 1), H(5, 7) and V(8, 5). (2, 4) -- V(8, 5), V(6, 6) and H(1, 2). (3, 9) -- H(1, 2), H(2, 4) and V(7, 8). I hope it makes sense for you, it took me a couple seconds to actually understand it.

Shaine Timmons agreed; not sure how he made the leap

You probably figured it out by now but it’s because there could only be a 7 in either column 9 row 2 or column 9 row 3. Regardless of which cell the 7 is in, we know it must be in column 9 which therefore rules out the entirety of column 9 and so only one cell remains which could possibly hold a 7. He then goes on to fill in the other possible 2 cells for 7 in rows 2 and 3 of column 9 afterwards even though logically he should have done that before instead of skipping ahead and causing confusion.

He does explain it if you listen carefully. Basically he is looking down column 8 looking for where 7 can go. It cant go in the top right box because those two open cells were already designated for two digits. It cant go in the middle right box because there's already a 7 there. In the bottom right box, it can only go in row 7 because there's already a 7 in row 8

@@blackphilip8936 Before he places that 7, do you see how in the top right box he has 3 and 8 narrowed down to two cells, giving him a 3-8 pair in column 8? That's a rule in sudoku where if you rule two digits down to two cells, nothing else can go there. This is why 7 couldn't logically go in row 3

@@blackphilip8936 when Simon finds pairs yes, he generally will switch the notation to the center of the box versus corner notations but thats not really a rule, moreso personal preference for notating. For example i never use corner notations and get along fine. The logic for the pair still holds

I think it was because he look at the bottom row and realized he needed to place a {2, 3, 4, 8} still. But, in column 9, the {2, 8} couldn't be placed due to the {2, 8} above it. Hopefully that helped.

Due to the 3-8 pair in top right square, the 7 can only be on colomn 9. So, it can't be in colomn 9 in the bottom right square. Then, thanks to sudoku rule, there is only one place where you can put a 7.

yes

Look at the top right 3x3 block. The 7 can not be in the middle column because it is already occupied by the 8 and 3 pair, and there's already a 7 on the seventh column so the 7 has to be on the right side of the block, which means the bottom right block has to have a 7 on the middle column.

Flame thanks

Agree. He moves his cursor so much it is impossible to follow

Not sure I see this. Can't r6c9 also be a 5?

@@CrackingTheCryptic www.sudokuwiki.org/Unique_Rectangles Type 1. r6c9 cannot be a 3, so r6c7 must be the 3 etc. doesn't crack it open more as far as i can tell, one still needs to use the magic of the top right 2 to progress.

@@CrackingTheCrypticIf you place a 4 in r9c9, that forces a 3 into r9c7, and then a 4 into r6c7. Because there are only two places for the 3 to go in row 6, it must go into r6c9, forming the rectangle. Indeed, we can deduce that r6c9 must be a 5.

Ah yes, nice spot!

The places the 3s are penciled in R6 and R9 eliminate the 3s from the rest of column 7 (that's an Xwing). Then the 4s that are placed already in R4 and R7 eliminate a couple of possible 4 locations from column 7, and there's already a 4 in the top box. So, there's only 2 spaces left in column 7 that the 3 and 4 can go in, so those two spaces have to be the 3 and the 4.