Thanks for the feature Brem! I like your way of resolving Boxes 1 and 3. I always try to send you puzzles that don’t have a very strict path. The intended way is a little simpler I think. As @McGrog90 pointed out, there is only one 123 in Row 5, that can go on the arrows from R3C19. So one of these must be a minimum of 45 and therefore a 9. This makes R3C2 an 8 and the puzzle flows from there.
Easy way after getting 9 in box 3 is to check arrow in column 9 which can't be 9 now. It will have reduced options for r5c9 making r5c1 from 45. And with at least 4 in r4c1 we have 9 in box 1
11:13 for my time (conflict checker off); usually I struggle with the arrow constraint, but for some reason today everything was in sync! This puzzle flowed so wonderfully, many props to ZegreS! And wonderful solve, BremSter, even if it wasn't a fully blind solve! 🙂
Beautiful puzzle. I thought about the logic in 10:00 a little differently. I showed that either R3C1 or R3C9 contains the digit 9 because there is only one 123 left in row 5 for those side-arrows. In other words, one of those arrows must be 4+5=9. From there, I eliminated 9 from R3C2 and this resolved the 12 pair in row 3.
in r5c1 or r5c9 cannot have 678, therefore one must be a 4 or 5. R4c1/9 must have a 4 or 5 as they cannot have 123, therefore either r3 c1 or r3c9 is a 9. r3col2 cannot be a 9 and has a one on its line. r2 c3 is therefore a 2.
Thanks for the feature Brem!
I like your way of resolving Boxes 1 and 3. I always try to send you puzzles that don’t have a very strict path. The intended way is a little simpler I think.
As @McGrog90 pointed out, there is only one 123 in Row 5, that can go on the arrows from R3C19.
So one of these must be a minimum of 45 and therefore a 9.
This makes R3C2 an 8 and the puzzle flows from there.
Sorry, r3c3 is a 2
I loved the thumbnail.
Looking forward to the angry chicken witch doctor on BremSter Games!
Appreciate you spotting the cooler thing at 23:25, I spotted it also and was sad you didn't get to enjoy it :)
65:58 got there in the end
Love the puzzle and solve
That was tough !!!!
Easy way after getting 9 in box 3 is to check arrow in column 9 which can't be 9 now. It will have reduced options for r5c9 making r5c1 from 45. And with at least 4 in r4c1 we have 9 in box 1
27:45 for me
nice puzzle
11:13 for my time (conflict checker off); usually I struggle with the arrow constraint, but for some reason today everything was in sync! This puzzle flowed so wonderfully, many props to ZegreS! And wonderful solve, BremSter, even if it wasn't a fully blind solve! 🙂
A very cool puzzle, tricky but fair. Thanks for solving it!
Beautiful puzzle. I thought about the logic in 10:00 a little differently. I showed that either R3C1 or R3C9 contains the digit 9 because there is only one 123 left in row 5 for those side-arrows. In other words, one of those arrows must be 4+5=9. From there, I eliminated 9 from R3C2 and this resolved the 12 pair in row 3.
Nice job this is the intended logic.
68 mins with some vid help.
Tough puzzle but fun.
Good solve.
At about 11 minutes in one could easily deduce that R6C1 could never be a 9, so the 9 had to be in box 1 and R3C2 was the 8.
We all make oopsies now and again.
Thanks for posting this!
in r5c1 or r5c9 cannot have 678, therefore one must be a 4 or 5. R4c1/9 must have a 4 or 5 as they cannot have 123, therefore either r3 c1 or r3c9 is a 9. r3col2 cannot be a 9 and has a one on its line. r2 c3 is therefore a 2.
Lots of finger trouble. meant r3 c3 is a 2