Great solve. Another way to view the argument from 16:20 on is that you need four different numbers from 1 through 6 that add to 11 = 5 + 6. The two remaining digits must therefore add to 10, and there's only one way to do it: 4 and 6. So the four numbers must be 1235, as you deduced. And then you must have 1 + 5 = 6 and 2 + 3 = 5 as the only way to divide them.
Congrats on 700 Rangsk🎉🎉!! Nice to bring Lily in to help celebrate! Fabulous way to get there from you Palfly! Fun for you to put in the Region Sum Circle to help with the chaos construction!!
Great puzzle! I actually had different logic to resolve the 56 pair in column 2. Basically, since 6 can’t go on any of the 2-cell region sum lines in green since they both are equal to a single-cell value, it forces the green 6 to be in column 2 which makes R1C2 a 5. Fun to see different logical paths lead to the same conclusion!
Great puzzle, horrible solve by me! 😅 I made good progress at the beginning, then got stuck and started to watch Rangsk's solve to be reminded that different line segments in the same region ARE SEPARATE SUMS. _facepalms_ Once I remembered that, the rest went decently and I finished in 14:32. Many thanks to Palfly for a lovely puzzle, and to Rangsk for presenting it and reminding me how RSLs work. 😛🙂
You still solved it much faster than me, but I did notice you could have made r2c5 blue pretty early because red couldn't grow that far, and if yellow did grow that far, it would have made blue only 5 cells. It took me 25:35 and I only did it that fast because I had to get some hints from your video to wrap my brain around it.
That was hard! I did notice that you can place 6 in row 2 after doing the initial colouring as all other cells in that row are part of 2 cell segments that sum to a single cell value.
Basically since the two upper left rings all feature at least one "one cell" region sum line, any "two cell" region sum lines on the same line can't contain a 6. So your green region forces 6 to be in the bottom two cells of green. Which gives you R1C2. This then gives you R2C1 which made some of the other sudoku pop out more quickly
Great solve. Another way to view the argument from 16:20 on is that you need four different numbers from 1 through 6 that add to 11 = 5 + 6. The two remaining digits must therefore add to 10, and there's only one way to do it: 4 and 6. So the four numbers must be 1235, as you deduced. And then you must have 1 + 5 = 6 and 2 + 3 = 5 as the only way to divide them.
Really fun puzzle today!
Congrats on 700 Rangsk🎉🎉!! Nice to bring Lily in to help celebrate! Fabulous way to get there from you Palfly! Fun for you to put in the Region Sum Circle to help with the chaos construction!!
Glad you enjoyed it, really happy that i managed to make the RSL the only rule.
And thanks to gdc for helping with the visuals on this one
Great puzzle! 20:35
Great puzzle! I actually had different logic to resolve the 56 pair in column 2. Basically, since 6 can’t go on any of the 2-cell region sum lines in green since they both are equal to a single-cell value, it forces the green 6 to be in column 2 which makes R1C2 a 5. Fun to see different logical paths lead to the same conclusion!
A lot of great logic in this one! I really enjoyed it. My time today was 8:50, solver number 98.
Great puzzle, horrible solve by me! 😅 I made good progress at the beginning, then got stuck and started to watch Rangsk's solve to be reminded that different line segments in the same region ARE SEPARATE SUMS. _facepalms_ Once I remembered that, the rest went decently and I finished in 14:32. Many thanks to Palfly for a lovely puzzle, and to Rangsk for presenting it and reminding me how RSLs work. 😛🙂
You still solved it much faster than me, but I did notice you could have made r2c5 blue pretty early because red couldn't grow that far, and if yellow did grow that far, it would have made blue only 5 cells. It took me 25:35 and I only did it that fast because I had to get some hints from your video to wrap my brain around it.
This was a tough one, but rewarding when finished. Your logic was much simpler than mine when finding how the 5-6 pairs resolved.
I'm glad you enjoyed it!
That was hard! I did notice that you can place 6 in row 2 after doing the initial colouring as all other cells in that row are part of 2 cell segments that sum to a single cell value.
Love chaos construction. I did slightly different logic with the green region. Slight spoiler in the reply
Basically since the two upper left rings all feature at least one "one cell" region sum line, any "two cell" region sum lines on the same line can't contain a 6. So your green region forces 6 to be in the bottom two cells of green. Which gives you R1C2. This then gives you R2C1 which made some of the other sudoku pop out more quickly
74:06