Wow -- I'm the first comment! I got this the hard way, in about 25 minutes. -- SPOILER FOLLOWS -- - - - - I missed both of the "NYT tricks", and instead found a 2, 3, 6, 9 quad in row 8 (c1, c3, c5, c7). That led me to a 1, 5, 6 triple in c9 (r1, r3, r4). My solve collapsed from there. This was a fun one. Your solve was, as usual, both more elegant and much faster.
Hey Rangsk, thanks for the daily posting of the NYT sudoku and the adventure series. Being honest, I left your channel for a while as the word games were dominating and those really long live stream must have taken it out of you! Glad to be back and will join Patreon to support. 🙇
In box 4 this is a naked quadruple, meaning that those four cells have only the values 2356 available, so they must contain those four values. This means the rest of the cells in the box cannot be 2356 so only 148 remain. Then, after that I see that whatever value the 36 ends up being in r4c1 needs to go in c2 somewhere, and it can't repeat in the box, so it has to go in r2c2.
@@Rangsk sorry man, do you mind explaining that last line a little more - "whatever value the 36 ends up being ... needs to go in c2 somewhere"; i understand that r4c1 can only be 36, but why does that mean that it needs to go in c2 somewhere? also thank u for ur videos they're great and super helpful and i will subscribe after this comment
@@yamsmusicsingapore1606 Well, c2 needs all values from 1-9, so how could it not end up in c2? I'm saying whatever value goes in r4c1, that same value will also go somewhere in c2. Yes, it's as trivial as it sounds. The next question to ask is, ok, if it ends up as a 3, where does it go? Now, if it ends up as a 6, where does it go? The answer to both question is it ends up only in r2c2 and those are the only two options, so therefore r2c2 is from 36 only.
Especially here because the geometry in boxes 6 and 9 under box 3 allowed you to place the 289 not only in box 9 like Rangsk but also box 6 at the same time, provided r4c9 was previously resolved by the hidden triple in box 5.
I believe they're computer generated. I don't know if they go through a curation process, but they do seem to be pretty high quality so either they have a very good custom generator, or a good curator, or both.
@@alanbdeeTo be clear, I didn't say "36 pair" and wouldn't have because there are no pairs involved here, but it does involve two cells that can only be 3 or 6 (a 36 bivalue cell if you want some terminology). What I've shown is that the r4c1 = r2c2. They are the same value as each other, not a pair. I explained it in two ways in the video but I guess I did it quickly because @qauissindy1 also had questions about this. The most straightforward way to prove this to yourself is to look at what candidates remain in r2c2. The column has 14789 so only 2356 remains. Now, in b4, the 2 and 5 "point" up because they are limited to c2 in b4. This eliminates 2 and 5 from r2c2 leaving only 36, which is the same conclusion I reached. We can take this a bit further and instead think about the two options for r4c1 and their consequences. Think about what happens if I put a 3 into r4c1. Well, now, there is a 256 triple in b2, making r2c2 only a 3. Similarly, if I put a 6 into r4c1, then it is a 235 triple, making r2c2 a 6. Thus, those two cells always end up the same value. The way I saw this and explained it at first was that whatever value ends up in r4c1 (3 or 6), it must also end up in c2 somewhere, which is trivially true since c2 has all values from 1-9 in it and so it definitely has a 3 and a 6 in it. Whatever value ends up in r4c1, it cannot repeat in b4, so it can't go in c2 of b4. This leaves only b1 or b7 to have that value in c2. However, b7 is full with 178 already - none of which are possible values for r4c1, so that rules out b7 leaving only b1 as an option. b1 has two cells filled in c2, which are 4 and 9, both of which also aren't possible values for r4c1. Thus, there is only one cell in c2 that can contain the value that goes in r4c1, so that's where it goes: r2c2. This proves that r4c1 = r2c2 and thus their pencilmarks should be identical to each other for the remainder of the solve. They are "clones" if that helps.
Wow -- I'm the first comment! I got this the hard way, in about 25 minutes.
-- SPOILER FOLLOWS --
-
-
-
-
I missed both of the "NYT tricks", and instead found a 2, 3, 6, 9 quad in row 8 (c1, c3, c5, c7). That led me to a 1, 5, 6 triple in c9 (r1, r3, r4). My solve collapsed from there. This was a fun one.
Your solve was, as usual, both more elegant and much faster.
6 months ago this would’ve taken me 25 minutes and lots of pencilling, through your videos I have learned and practiced and today my time was 10.40 😊
That's fantastic! I'm so glad my videos are helpful :)
Hey Rangsk, thanks for the daily posting of the NYT sudoku and the adventure series. Being honest, I left your channel for a while as the word games were dominating and those really long live stream must have taken it out of you! Glad to be back and will join Patreon to support. 🙇
Welcome back, and thank you!
My time for today is 11.23! Faster than usual
Thank you, yet, I didn't understand your method in the min 3:12. would you please explain. Thank you.
In box 4 this is a naked quadruple, meaning that those four cells have only the values 2356 available, so they must contain those four values. This means the rest of the cells in the box cannot be 2356 so only 148 remain. Then, after that I see that whatever value the 36 ends up being in r4c1 needs to go in c2 somewhere, and it can't repeat in the box, so it has to go in r2c2.
@@Rangsk sorry man, do you mind explaining that last line a little more - "whatever value the 36 ends up being ... needs to go in c2 somewhere"; i understand that r4c1 can only be 36, but why does that mean that it needs to go in c2 somewhere?
also thank u for ur videos they're great and super helpful and i will subscribe after this comment
@@yamsmusicsingapore1606 Well, c2 needs all values from 1-9, so how could it not end up in c2? I'm saying whatever value goes in r4c1, that same value will also go somewhere in c2. Yes, it's as trivial as it sounds. The next question to ask is, ok, if it ends up as a 3, where does it go? Now, if it ends up as a 6, where does it go? The answer to both question is it ends up only in r2c2 and those are the only two options, so therefore r2c2 is from 36 only.
@@Rangsk Wow!! That's obvious and not obvious at the same time. Thanks for the explanation.
I saw the 289 geometry in box 3 very late, so always look for this is a good algoritm :)
Especially here because the geometry in boxes 6 and 9 under box 3 allowed you to place the 289 not only in box 9 like Rangsk but also box 6 at the same time, provided r4c9 was previously resolved by the hidden triple in box 5.
Sweet🎉 I’d say kudos to the setter, but I never know who does NYT. 😂
Great solve, well taught.
I believe they're computer generated. I don't know if they go through a curation process, but they do seem to be pretty high quality so either they have a very good custom generator, or a good curator, or both.
great solve! How did you know that a 36 pair needed to be in the middle square of the first box?
at 3:24 specifically you say that there needs to be a 36 pair
@@alanbdeeTo be clear, I didn't say "36 pair" and wouldn't have because there are no pairs involved here, but it does involve two cells that can only be 3 or 6 (a 36 bivalue cell if you want some terminology). What I've shown is that the r4c1 = r2c2. They are the same value as each other, not a pair.
I explained it in two ways in the video but I guess I did it quickly because @qauissindy1 also had questions about this.
The most straightforward way to prove this to yourself is to look at what candidates remain in r2c2. The column has 14789 so only 2356 remains. Now, in b4, the 2 and 5 "point" up because they are limited to c2 in b4. This eliminates 2 and 5 from r2c2 leaving only 36, which is the same conclusion I reached.
We can take this a bit further and instead think about the two options for r4c1 and their consequences. Think about what happens if I put a 3 into r4c1. Well, now, there is a 256 triple in b2, making r2c2 only a 3. Similarly, if I put a 6 into r4c1, then it is a 235 triple, making r2c2 a 6. Thus, those two cells always end up the same value.
The way I saw this and explained it at first was that whatever value ends up in r4c1 (3 or 6), it must also end up in c2 somewhere, which is trivially true since c2 has all values from 1-9 in it and so it definitely has a 3 and a 6 in it. Whatever value ends up in r4c1, it cannot repeat in b4, so it can't go in c2 of b4. This leaves only b1 or b7 to have that value in c2. However, b7 is full with 178 already - none of which are possible values for r4c1, so that rules out b7 leaving only b1 as an option. b1 has two cells filled in c2, which are 4 and 9, both of which also aren't possible values for r4c1. Thus, there is only one cell in c2 that can contain the value that goes in r4c1, so that's where it goes: r2c2. This proves that r4c1 = r2c2 and thus their pencilmarks should be identical to each other for the remainder of the solve. They are "clones" if that helps.
23