Thanks for taking the time to work through this puzzle! The rules were actually borrowed from one of the puzzles in CTC/Blobz's Lord of the Rings pack, except that the original used doublers instead. I wanted to experiment with a negative value constraint, which I hadn't seen before, so I cycled through a few names - including "negator," which I thought sounded more like the cell has a value of 0 - before landing on "anti-cell." But, as you said, they do function exactly the same way. The 9-cage is a killer - no pun intended - but it's meant as the final stumbling block before the rest of the puzzle essentially collapses. I hate that in the case it ended up being less "stumbling block" and more "12-foot wall," but I do appreciate you continuing past it, and I'll certainly keep the feedback in mind on future puzzles.
My initial thought was just "Nope." Then I watched the first few minutes, and saw something that clicked. It took me over an hour, but I really enjoyed it! Now watching how to do it better! Thanks BremSter and Erin Toler! Great fun!
Once you fill in the candidates for the "9" cage, there is a 5,6 pair in column 3. This eliminated all candidates for the anti-cell from r5c3, making r5cc1 the anti-cell. No need for the "1"" bifurcation.
Very nice puzzle, I really liked it. Solved it in 37:48. I found that it was almost always clear which area one has to look at to be able to continue. At 44:23 looking at the cases for the 9-cage in box 1 is not too hard. If r3c3 is not an anti-cell, then among the 3 possibilities (126, 135, 234) we can't have 234, and 135 would force 15 to column 2, which contradicts the 15 pair in r9c2. So only 126 is possible, placing 1 to r2c2, 2 to r3c2 and 6 to r3c3. On the other hand, if r3c3 is the anti-cell, it can only be 5 or 6. So we get a 56 pair in column 3, which then puts 13 into r5c3. Neither of those digits can be anti-cell, so r5c1 and r3c3 must be the remaining anti-cells.
This was the way the solve guide was written. I think in trying to quickly read the guide while on camera, BremSter shuffled a couple of steps together, which ended up looking a little more bifurcate-y than I intended.
52:37! This was exactly as hard as I can deal with. Fun one! Also (SPOILER ALERT!) it's rare to not see the little Sven celebration in a puzzle anymore!
Erin, love anti value combined with the anti anti knight constraint. Since the tricky bit is resolving the X wing on anti cells, simple giving one of the 4 as the anti or not the anti might reduce the friction. To go along with that, if you coud remove the 4 but add enough not anti clues, that could be really cool. Erin, please make more puzzles with this constraint.
Hi BremSter, I got stuck for a very long time at the same stage as you (at around 44 minutes in your video). I was finally able to make progress by: Considering the 28 pair in the anti-cell in r2c9. If it would have been a 2, it would have forced the 2 in row 5 in box 6, and then in r4c1 in box 4. Having a 2 in r4c1 would have left no potential anti-cell where to place the 2. The 8 in r2c9 then allowed to solve the cages in box 3 and make easy progress from there. I guess it's one of those cases where the way out appears simple AFTER you've found it. 😀
This was definitely a challenge; I totally forgot about the anti-knight rules for the first 15 minutes, making some torturous logic to move ahead. I also had a 'hunch' on the remaining X-wing of anti-cells that I couldn't fully prove, but ended up working out, so was glad to see BremSter find a fully logical path through. Finished in 34:19 (conflict checker off), though I do feel a bit icky for having a lucky hunch. 😅 Many thanks to Erin for an amazing puzzle, and to BremSter for presenting it!
It found the logic not too difficult for the 9 cage in box 1 The reasoning is : if there is no anti celle in R3C3 then the cage cannot be 234 (easy) and must contain a 1 which is therefore in R2C2 This forces 13 in R5C3 which then cannot be an anti-cell which force R3C3 to be an anti cell So if there is no anti celle there must be an anti cell ! The easy part which is easy to miss is that 13 in R5C3 means it is not an anti-cell
Loved watching you solve this (after your oft mentioned dislike of maths), a great puzzle for a Sunday. Put a hankie on my head, call me Mr Gumbie…”my brain hurts”…but happy! Thanks BremSter
A much simpler(and I'm guessing the intended way) to get the 4,9 pair in box 4 is to see that a 5,8 pair in the 13 cage would place both 5 and 8 into r6c7
You were right there when you figured out that if R5C3 was the anticell then R5C4 had to be a 1. This would mean there would be no anticell in the 9 cage, but also no 1 or 4 and therefore no way to make a 9 cage.
Lots of calculations, a spreadsheet is a good help for these kind of puzzles, done in 27:19. @44:00 the easier way is to ask if r3c1 is an anti-cell, it has to be an 8 in this case (the anti cell in box4 will be an 5), Then look what happens to box3: the 13 cage becomes 4-9, the 7 cage 2-5, and the anti cell has no fill (r2c9). That "solve guide" was no good imo.
The -9 was used in the -6 cage and there is only one of each digit of 1-9 that's negative. It's a good negative constraint once you fill the negative numbers, it reduces the one's you can use as you progress.
Thanks for taking the time to work through this puzzle! The rules were actually borrowed from one of the puzzles in CTC/Blobz's Lord of the Rings pack, except that the original used doublers instead. I wanted to experiment with a negative value constraint, which I hadn't seen before, so I cycled through a few names - including "negator," which I thought sounded more like the cell has a value of 0 - before landing on "anti-cell." But, as you said, they do function exactly the same way.
The 9-cage is a killer - no pun intended - but it's meant as the final stumbling block before the rest of the puzzle essentially collapses. I hate that in the case it ended up being less "stumbling block" and more "12-foot wall," but I do appreciate you continuing past it, and I'll certainly keep the feedback in mind on future puzzles.
My initial thought was just "Nope." Then I watched the first few minutes, and saw something that clicked. It took me over an hour, but I really enjoyed it! Now watching how to do it better! Thanks BremSter and Erin Toler! Great fun!
Once you fill in the candidates for the "9" cage, there is a 5,6 pair in column 3. This eliminated all candidates for the anti-cell from r5c3, making r5cc1 the anti-cell. No need for the "1"" bifurcation.
I loved that once I realised that the weird dot was actually a minus sign.
It would be great if Sven found a way to make negative cage totals more visible.
Very nice puzzle, I really liked it. Solved it in 37:48.
I found that it was almost always clear which area one has to look at to be able to continue. At 44:23 looking at the cases for the 9-cage in box 1 is not too hard. If r3c3 is not an anti-cell, then among the 3 possibilities (126, 135, 234) we can't have 234, and 135 would force 15 to column 2, which contradicts the 15 pair in r9c2. So only 126 is possible, placing 1 to r2c2, 2 to r3c2 and 6 to r3c3. On the other hand, if r3c3 is the anti-cell, it can only be 5 or 6. So we get a 56 pair in column 3, which then puts 13 into r5c3. Neither of those digits can be anti-cell, so r5c1 and r3c3 must be the remaining anti-cells.
This was the way the solve guide was written. I think in trying to quickly read the guide while on camera, BremSter shuffled a couple of steps together, which ended up looking a little more bifurcate-y than I intended.
52:37! This was exactly as hard as I can deal with. Fun one! Also (SPOILER ALERT!) it's rare to not see the little Sven celebration in a puzzle anymore!
Erin, love anti value combined with the anti anti knight constraint.
Since the tricky bit is resolving the X wing on anti cells, simple giving one of the 4 as the anti or not the anti might reduce the friction. To go along with that, if you coud remove the 4 but add enough not anti clues, that could be really cool.
Erin, please make more puzzles with this constraint.
That was a very intricate puzzle. Very cool how things worked, but my head hurts now.
Hi BremSter, I got stuck for a very long time at the same stage as you (at around 44 minutes in your video). I was finally able to make progress by:
Considering the 28 pair in the anti-cell in r2c9. If it would have been a 2, it would have forced the 2 in row 5 in box 6, and then in r4c1 in box 4. Having a 2 in r4c1 would have left no potential anti-cell where to place the 2. The 8 in r2c9 then allowed to solve the cages in box 3 and make easy progress from there.
I guess it's one of those cases where the way out appears simple AFTER you've found it. 😀
That's how I did it, yeah.
This was definitely a challenge; I totally forgot about the anti-knight rules for the first 15 minutes, making some torturous logic to move ahead. I also had a 'hunch' on the remaining X-wing of anti-cells that I couldn't fully prove, but ended up working out, so was glad to see BremSter find a fully logical path through. Finished in 34:19 (conflict checker off), though I do feel a bit icky for having a lucky hunch. 😅 Many thanks to Erin for an amazing puzzle, and to BremSter for presenting it!
It found the logic not too difficult for the 9 cage in box 1
The reasoning is :
if there is no anti celle in R3C3 then the cage cannot be 234 (easy) and must contain a 1 which is therefore in R2C2
This forces 13 in R5C3 which then cannot be an anti-cell which force R3C3 to be an anti cell
So if there is no anti celle there must be an anti cell !
The easy part which is easy to miss is that 13 in R5C3 means it is not an anti-cell
Loved watching you solve this (after your oft mentioned dislike of maths), a great puzzle for a Sunday. Put a hankie on my head, call me Mr Gumbie…”my brain hurts”…but happy! Thanks BremSter
A much simpler(and I'm guessing the intended way) to get the 4,9 pair in box 4 is to see that a 5,8 pair in the 13 cage would place both 5 and 8 into r6c7
I was stuck in the same position. Well done, BremSter.
I love this type of puzzle, Thanks! 😊
You were right there when you figured out that if R5C3 was the anticell then R5C4 had to be a 1. This would mean there would be no anticell in the 9 cage, but also no 1 or 4 and therefore no way to make a 9 cage.
Solved in 32 min😊
13 cage in R3 is key in solving x-wing of anti-cells
5-8 pair is not possible in that 13 cage
Kept on making mistakes so settled for watching Bremster.
Cool puzzle despite that.
at 28:56 why can't r2c9 be 6?
72 minutes for me.
nice puzzle and solving
Anti-knight rule on anti-cells.
Lots of calculations, a spreadsheet is a good help for these kind of puzzles, done in 27:19.
@44:00 the easier way is to ask if r3c1 is an anti-cell, it has to be an 8 in this case (the anti cell in box4 will be an 5), Then look what happens to box3: the 13 cage becomes 4-9, the 7 cage 2-5, and the anti cell has no fill (r2c9). That "solve guide" was no good imo.
9:11 Why can't he put a 9 in the anti-cell of the 0 cage? It isn't a knight's move from any established 9's.
Because 9 was already used as an anti cell , and each digit can only be anti once.
@@donaldsnyder1543 Oh, I missed that rule, thanks
The -9 was used in the -6 cage and there is only one of each digit of 1-9 that's negative. It's a good negative constraint once you fill the negative numbers, it reduces the one's you can use as you progress.