Nice puzzle. A little bit different trick than we are used to with the NYT. In the situation you have when you get stuck around minute 15, what is the most interesting restrictionÉ It actually stands in the not too inviting column 8 where you have only one given digit. One thing Zen has already established is that there is a 12 bi-value cell in row 8. Where else can there be a 1 and a 2 in column 8? Well the only other place for a 1 is in row 5, and, for the 2, in row 4. Next question: what else can there be in rows 4 and 5? Pretty clear there must be an 8 in either of those spots. So, broken triple 128 of which r4c8 is a 28 and r5c8 is a 18. Eureka ! We already know there is a 18 in r5c3 ! Hence no other 8s possible in row 5, crucially, no 8 possible in r5c4. So now in that central vertical chute, the 8s are confined to either columns 5 or 6 in blocks 5 and 8, and theredore, there must be an 8 on the left side of block 2, where we already know we have a 3 and 6. So 68 pair in rows 1 and 3, and the 3 in row 2, confirming r3c3 must be a 4. Puzzle done. Thank you NYT for giving us something different.
I got stuck at the same place, so your 1237 quad helped me a lot.
Nice puzzle. A little bit different trick than we are used to with the NYT. In the situation you have when you get stuck around minute 15, what is the most interesting restrictionÉ It actually stands in the not too inviting column 8 where you have only one given digit. One thing Zen has already established is that there is a 12 bi-value cell in row 8. Where else can there be a 1 and a 2 in column 8? Well the only other place for a 1 is in row 5, and, for the 2, in row 4. Next question: what else can there be in rows 4 and 5? Pretty clear there must be an 8 in either of those spots. So, broken triple 128 of which r4c8 is a 28 and r5c8 is a 18. Eureka ! We already know there is a 18 in r5c3 ! Hence no other 8s possible in row 5, crucially, no 8 possible in r5c4. So now in that central vertical chute, the 8s are confined to either columns 5 or 6 in blocks 5 and 8, and theredore, there must be an 8 on the left side of block 2, where we already know we have a 3 and 6. So 68 pair in rows 1 and 3, and the 3 in row 2, confirming r3c3 must be a 4. Puzzle done. Thank you NYT for giving us something different.
@GT
Great observation!
* confirming R3C6 must be a 4.
Anyway I used the same 1237 quad in col 9 to solve, using pencil marks .
Definitely hard. 15 minutes, no pencil marks.