Thanks for this. Appreciate you going much slower and the explanation of your moves. Please do more of these.

Same ...also, you can go even slower and that would be okay (why did you choose the 5? Was there logic involved or ????)

@@heathercollins4432 lol. Of course logic was involved, I just wasn't sure how.

@@simon77047 - LOLOLOLOL

Same...

I do not want to be snarky, but if you want to learn those techniques you could pause the video and work it out on paper (or a calc grid).

@@57thorns It's the description of the concept that's fast, at least in my experience. This video has a really slow, patient and exhaustive type of explanation, and I would love to see a similar video on X wings and Y wings, and I assume that's what Susan is asking for as well.

Thank you!!!!

Anyone who's been listening to cracking the cryptic long enough already had simons voice as a narrater for any little thing in their life

@@floor.smorenburg h

And for us OLD-timers, we watch at 2X

yes, much better and also the video quality has dramatically increased !

I'm wondering how this is classified as hard when there ara way more complicated examples and methods. Probably those that You are mentioning. But video is useful to many beginners for sure.

I developed that same marking pattern doing them by hand, circling the obligate pairs. :)

I'm by no means an expert, or even very good, but in my experience so far, limiting to pairs really does help. Plus, you can always resort to triples after exhausting the benefits of this method on a particular puzzle.

Exactly! I've noticed this with multiple videos on this channel. For example, I always start by looking at the number that is already given most often, which is a 6 here. From the start you can already fill in a 6 in the upper square, but only after 9 minutes he notices it.

​@@Aericane I agree, I also notice Simon rarely looks at the rows or columns as a whole. At 5:29 the 5 in the bottom row is locked to the middle square (vertical 346 in it) because the the given top left 5 and the solved 5 in the top right square. The two 5:s in the bottom right square are also given by looking at the rightmost column.

Same...

Anitha B wait a minute didn’t you comment on SudokuGuys video?

@@detroll3730 Yes, I did. Is there a problem?

@Anitha B nope just being observant

@@detroll3730 👍

You don't even need to erase all 27 of them, you only need to erase at most 6 cells of this puzzle after completion to make the result unsolvable. Sudoku Swami has patterns he calls "dead x-wings" or "dead swordfish", which are patterns that can't be solved without additional information. If you can find a certain digit that shares an x-wing or a swordfish with another digit, then removing both digits from all the x-wing or swordfish cells makes the puzzle unsolvable. I'll give an example. Clearly, we have to remove one of the given starting numbers to make the puzzle unsolvable, because if we have all of them given, it implies only a single solution. Let's start with the solved puzzle. We can remove the 6 from rows 4, 5, and 6, and also remove the 9 from rows 4, 5, and 6, and then the puzzle becomes unsolvable. There would be two different ways to complete it, even with these 81-6 = 75 cells given. Then to answer your question, clearly, removing all numbers from blocks 4, 5, and 6 also removes the six entries I mentioned, so the puzzle would be unsolvable.

@@wiggles7976 Appreciate the explanation. Makes sense. Well said.

If you mean the part at like 13:15, look at the middle row again. The 3 and 9 are a pair and both fall in that row. Process of elimination says only 2 and 7 are missing for the final 2 squares in that row. The 2 then goes on the first column because there's already a 2 in the second column down at the bottom. Hope this helps :)

An oversight, 4 and 8 were exclusive in those two cells. Good catch Mary

6 months late, but sure. He still uses old corner notation, which makes it less obvious to spot. But this is called a pair. 2 positions that must contains exactly two numbers, lock all the others out of those two 1-3 boxes. Think of it like this: say you wére to put a 5 in either of those boxes. Could you still place both the 1 ánd the 3 (the answer is no, we already deducted that the 1 and the 3 múst be on either of those two positions). This will by the way also work for a group of 3 numbers divided over 3 positions etc.

B8 r7c4,r8c4,r9c4 has candidates 5,5,5 r7c6,r8c6,r9c6 5,5,5 so B5 MUST have candidates e r4c5,r6c5 5,5 because one cand from B8 r7, OR r8, OR r9 is certain