Thank you for the wonderful solve, Bremster. It was really gratifying to see you find the two in box one. You do such a wonderful job showcasing puzzles and supporting the community. Cheers!
I am used to coloring XV negative constraint. The extra option for V was not a problem. The extra option for X completely threw me for a loop, and more than once. Amazing puzzle, thanks!
8:45 "This is a difference of 5". I think you were still thinking of it being 0v5. The general argument for why R2C3 can't go in R3C5 is that the two digits on the V need to be opposite parities (one odd the other even), and the two digits at the ends of the white dot chain are the same parity (because difference of 2).
I struggled greatly with this one, mainly because I'm not the best at either the quattroquadri and 0-8 constraints on their own, much less together. 😅 Completed in 13:04 (conflict checker off), many thanks to Maya for the challenging (for me) puzzle, and to BremSter for presenting it (and doing a much better solve than I)!
You incorrectly pencil marked box 3. The x should have been 38x27. Still a good solve as I was a bit stuck near the beginning. You made the brilliant deduction where the digit from box 1 went in box 2.
There isn't any deep casing necessary for the solve. Bremster did a good job of finding the solve path. I think the video is worth watching and the puzzle worth solving, but I am biased. :)
Thank you for the wonderful solve, Bremster. It was really gratifying to see you find the two in box one. You do such a wonderful job showcasing puzzles and supporting the community. Cheers!
I am used to coloring XV negative constraint. The extra option for V was not a problem. The extra option for X completely threw me for a loop, and more than once.
Amazing puzzle, thanks!
Loved how you solved the break in with the V, I’d been stuck for ages. I do enjoy these Suzy’s (Quattros)
8:45 "This is a difference of 5". I think you were still thinking of it being 0v5. The general argument for why R2C3 can't go in R3C5 is that the two digits on the V need to be opposite parities (one odd the other even), and the two digits at the ends of the white dot chain are the same parity (because difference of 2).
That was a neat little puzzle!
22:15 for me[broke once]
nice puzzle
08:59 with your help. I was quite lost at the start and the trick was... pencilmark! Ha! 😅
39:07 for me
Nice to see people with reasonable times posted instead of the people claiming 3:14
Fun puzzle, which takes a minute to wrap your head around. My time was 12:33, solver number 528.
18:11 for me today. Very nice puzzle.
23:34 ... clumsy break in for me. Luck more than skill, I think.
I struggled greatly with this one, mainly because I'm not the best at either the quattroquadri and 0-8 constraints on their own, much less together. 😅 Completed in 13:04 (conflict checker off), many thanks to Maya for the challenging (for me) puzzle, and to BremSter for presenting it (and doing a much better solve than I)!
You incorrectly pencil marked box 3. The x should have been 38x27. Still a good solve as I was a bit stuck near the beginning. You made the brilliant deduction where the digit from box 1 went in box 2.
At 14:10? 3 couldn't go in either cell anyway: 3 in Box 4 and 23-pair in Box 3.
That was very hard. I ended up using following two lines, until one failed... brutal.. perhaps I should wath the video?
There isn't any deep casing necessary for the solve. Bremster did a good job of finding the solve path. I think the video is worth watching and the puzzle worth solving, but I am biased. :)
Neat puzzle but quattro still breaks me lol.