I’ve been trying to improve my sudoku skills with the NYT sudoku’s daily and when I kept getting stuck on the hard ones I found your videos and they’ve been so incredibly helpful! I’ve been solving them on my own for a week or two and I got stuck on this one at the same point you did (about halfway through the video) and I managed to solve it by following along for a bit. I had already seen the 3’s that pointed down on the very right column (I don’t really know the grid terminology) so it went a little faster for me once the 5 had been eliminated there. Thanks so much for your videos, it’s really upped my sudoku skills.
Wow... I was stuck and I didn't know what to do. At 45 minutes or so I had to watch your video for clues. The 5 in box 6 was key for me. It only took a few minutes after that. I had already spotted the 689 triple in box 6. Thanks for the help! Final time was 70:36 (I didn't pause the time while I was watching your video).
What helped me as well as the 5’s in the last stack was that I used the uniqueness factor. In columns 4 and 5 there are 135 in row 3 and 35 in row eight meaning one of the 1’s in row three columns 4 and 5 were the actual 1. That gave me the 8 in row 3 box 1
Wow, I think this was my first solve under 10 min! 09'52'' I just think everything flowed nicely to me and I could see every pointing, claiming etc. right away, which was sweet. For example (and I still don't know if you saw if because I haven't finished the video), the 345 in C9R6 is just a 4 because of the pointing 3s in Box 3 and of the 5s lining up in Boxes 3 and 9. Usually I wouldn't spot this but today was my lucky day I guess
Yup, the same for me, spotting the pointing 3 & 5 in the 3rd stack solved a naked 4. However the 698 triple alternative just shows that when there is a double, triple or even single etc, there is usually an inverse set of double/triple/etc as an alternative solution
Also medium one was exceptionally hard today. It really needed some grinding. Any chances you could take a look at that and make a video? I'm pretty sure you could find a "cleaner" way to solve it.
This one was good. Edit: I didn't know there was a hidden triple at 689. Nice. I noticed the 5's like you did but also saw claiming 3's in box 3 at column 9. That would have eliminated the 35 from 345 at box 6 early on to get a naked 4.
I spotted a hidden quad (the nonhidden triple open cells) in box 6 just before you said there is almost a hidden quad in row 5. I know you don't like to center mark 4 numbers, but you could have figured out box 6s secrets a bit sooner. As I say that, I do know that it would have taken 20-30 minutes to get to that point on my own.
Found this very tricky. Resorted to using an alternative inference chain at one point but then later found the expected route which I believe was a hidden triple in box 6. Time to see the professionals..
Alternate inference chain is visible around 12:15 - if r6c2 is a 1, r3c2 is 8. If r4c3 is a 1 it places 1s in c5 in box 5, then r3c4, which also makes r3c2 an 8. A step I would never wish on any solver lol
"Is there some sneaky geometry that I am missing?" You're not wrong... I did not spot the 689 triple and somehow solved it. Took me a long time to spot the single two in the bottom left corner, though.
I’ve been trying to improve my sudoku skills with the NYT sudoku’s daily and when I kept getting stuck on the hard ones I found your videos and they’ve been so incredibly helpful! I’ve been solving them on my own for a week or two and I got stuck on this one at the same point you did (about halfway through the video) and I managed to solve it by following along for a bit. I had already seen the 3’s that pointed down on the very right column (I don’t really know the grid terminology) so it went a little faster for me once the 5 had been eliminated there. Thanks so much for your videos, it’s really upped my sudoku skills.
You never finished your initial scanning of stack 3. The 3 was limited to C9 in box 3 eliminating 3 from R6 C9 making it a naked single 4.
I found the hidden triple right away! It took 17 minutes for me , pretty good, thanks!!🎉❤
Wow... I was stuck and I didn't know what to do. At 45 minutes or so I had to watch your video for clues. The 5 in box 6 was key for me. It only took a few minutes after that. I had already spotted the 689 triple in box 6.
Thanks for the help!
Final time was 70:36 (I didn't pause the time while I was watching your video).
I found this one tricky; finally, rather than seeing the 6-8-9 triple, I found the 2-3-4-5 quad, in the same box, so yes, that was the key.
What helped me as well as the 5’s in the last stack was that I used the uniqueness factor. In columns 4 and 5 there are 135 in row 3 and 35 in row eight meaning one of the 1’s in row three columns 4 and 5 were the actual 1. That gave me the 8 in row 3 box 1
Wow, I think this was my first solve under 10 min! 09'52''
I just think everything flowed nicely to me and I could see every pointing, claiming etc. right away, which was sweet. For example (and I still don't know if you saw if because I haven't finished the video), the 345 in C9R6 is just a 4 because of the pointing 3s in Box 3 and of the 5s lining up in Boxes 3 and 9. Usually I wouldn't spot this but today was my lucky day I guess
Yup, the same for me, spotting the pointing 3 & 5 in the 3rd stack solved a naked 4.
However the 698 triple alternative just shows that when there is a double, triple or even single etc, there is usually an inverse set of double/triple/etc as an alternative solution
Also medium one was exceptionally hard today. It really needed some grinding. Any chances you could take a look at that and make a video? I'm pretty sure you could find a "cleaner" way to solve it.
Great solve today. Big breakthrough for me was the 3s in box 3 were limited to column 9, which helped unlock box 6.
This one was good.
Edit: I didn't know there was a hidden triple at 689. Nice. I noticed the 5's like you did but also saw claiming 3's in box 3 at column 9. That would have eliminated the 35 from 345 at box 6 early on to get a naked 4.
I spotted a hidden quad (the nonhidden triple open cells) in box 6 just before you said there is almost a hidden quad in row 5. I know you don't like to center mark 4 numbers, but you could have figured out box 6s secrets a bit sooner. As I say that, I do know that it would have taken 20-30 minutes to get to that point on my own.
15,12 took some effort today. Your lessons on geometry were critical to this one
Found this very tricky. Resorted to using an alternative inference chain at one point but then later found the expected route which I believe was a hidden triple in box 6. Time to see the professionals..
Alternate inference chain is visible around 12:15 - if r6c2 is a 1, r3c2 is 8. If r4c3 is a 1 it places 1s in c5 in box 5, then r3c4, which also makes r3c2 an 8. A step I would never wish on any solver lol
If you found it hard, I am happy I got there in the end. I was thinking I must of missed an easy spot somewhere.
"Is there some sneaky geometry that I am missing?"
You're not wrong...
I did not spot the 689 triple and somehow solved it. Took me a long time to spot the single two in the bottom left corner, though.
I spotted the 689 triple in row 6 and column 8 that eliminated 2s from column 8, collapsed from there.
Sorry I meant forced the 2s in column 8 box 6