I don't like the wording of the last rule. It states that the digit in a circle indicate the number of odd digits in the (at most) 3x3 area with the circle at its center. I do understand the intention of this rule and how the puzzle solves using it in the way it is intended. But r1c1 is "at [the] center" of the 1x1 area that consists of solely r1c1. It is not "at [the] center" of the 2x2 box of r12c12. It is the top left cell of that area. It actually seems like it would make more sense to not include "at most" since the way the rule is used in the puzzle, in relation to the circle in r1c1 is that it is the center of the 3x3 box that includes 5 cells not in the grid (which we can ignore). When I first read the rules and looked at the grid, my first though was that r1c1 must be a 1 because it is the circle at the center of the 1x1 box that contains r1c1. So it is from 01 and obviously can't be 0. I did immediately realize that can't be correct because it conflicts with the 2nd rule, which made me understand the correct meaning of the rule. Please don't take this in any way as a complaint against the puzzle itself, which I loved!
Loved watching this video. I spent an hour trying before giving up. The way I read the rules is that IF an odd cell is orthogonally to another odd cell, it must be in a group of 5 -- meaning there could be a singleton odd cell and kept chasing my tail around that misconception. I'm sure there's a really obvious or math reason that could never have been true, but that is well beyond my ability to articulate. :) Thanks as always -- I got to stretch my brain in odd ways, and learned a lot watching this!
I think the proof for no 7s in circles is something like: The "best" the two even digits can can do in terms of sectioning off odd cells is isolating one odd digit from being connected to the 7 (re: one of the odd cells diagonal from the 7). That still leaves 5 odds all orthogonally connected to the 7, which means 6 odds total are orthogonally connected, and 6 is more than 5.
@@TheBiggreenpig If you could connect odds together diagonally it would be even worse since you would have a region of 7 odds automatically (by putting 7 in a circle) but the regions of odds must be of size 5.
I wonder if it's possible to solve with the rules as stated (ie: where it's possible to have isolated odd cells which are not orthogonality connected to any other odd cells). I feel like the assumption that all odd cells are orthogonally connected was big in solving it, and not sure it's possible to solve without making that assumption.
I can now confirm there is a logical path to a solution which does not make that assumption; it just took me a bit longer to get there (around an hour or so).
Probably it ispossible, but would require deeper level of speculations on odds/evens. As at least one isolated odd would require to have at least another 4.
I solved it without this assumption (since the rules strictly allow for 1-groups like you said) but only managed after some bifurcation. Came back looking for a proper way to do it. :)
As a mathmatician I thought the wording was clear ("groups of orthogonally connected odd digits" could mean a group of 1 cell) but you are the second person who got confused by that. I've since updated the rules on Logic Masters, but my submission to Bremster was before that. Sorry! Turns out it might solve even with the less strict ruleset, but that is definitely not intended.
That one rule is poorly written. As is, it says that all groups of odd digits are 5 cells. That means that isolated odd numbers are allowed. What the rule actually is, it this: Odd numbers appear only in groups of 5 orthogonally connected cells; there are no isolated odd numbers, and no groups of odd numbers of any other size.
Myxo here, thank you for the lovely solve! You showcased the logic really nicely, and I'm glad you enjoyed it :)
I don't like the wording of the last rule. It states that the digit in a circle indicate the number of odd digits in the (at most) 3x3 area with the circle at its center. I do understand the intention of this rule and how the puzzle solves using it in the way it is intended. But r1c1 is "at [the] center" of the 1x1 area that consists of solely r1c1. It is not "at [the] center" of the 2x2 box of r12c12. It is the top left cell of that area. It actually seems like it would make more sense to not include "at most" since the way the rule is used in the puzzle, in relation to the circle in r1c1 is that it is the center of the 3x3 box that includes 5 cells not in the grid (which we can ignore). When I first read the rules and looked at the grid, my first though was that r1c1 must be a 1 because it is the circle at the center of the 1x1 box that contains r1c1. So it is from 01 and obviously can't be 0. I did immediately realize that can't be correct because it conflicts with the 2nd rule, which made me understand the correct meaning of the rule. Please don't take this in any way as a complaint against the puzzle itself, which I loved!
Myxo is such a consistently great setter
Loved watching this video. I spent an hour trying before giving up. The way I read the rules is that IF an odd cell is orthogonally to another odd cell, it must be in a group of 5 -- meaning there could be a singleton odd cell and kept chasing my tail around that misconception. I'm sure there's a really obvious or math reason that could never have been true, but that is well beyond my ability to articulate. :)
Thanks as always -- I got to stretch my brain in odd ways, and learned a lot watching this!
Very fun to do, altough I need your help at the start! 🤗
nice puzzle and solving
1:35:41 - That was a challenge! Enjoyable but a challenge nonetheless. Very happy to have finished it.
Excellent puzzle, lots of fun. Solved it in 46:06.
Didnt know what to do with this one.
Fascinating puzzle , good solve.
I think the proof for no 7s in circles is something like:
The "best" the two even digits can can do in terms of sectioning off odd cells is isolating one odd digit from being connected to the 7 (re: one of the odd cells diagonal from the 7). That still leaves 5 odds all orthogonally connected to the 7, which means 6 odds total are orthogonally connected, and 6 is more than 5.
If you can connect diagonally, then you can place 7. Place one in a corner, leave its neighbors free, and fill the rest.
@@TheBiggreenpig If you could connect odds together diagonally it would be even worse since you would have a region of 7 odds automatically (by putting 7 in a circle) but the regions of odds must be of size 5.
@@michaelmatter1222 No, I mean, if they are allowed to touch diagonally. And as we see in the solution, they are.
Very nice puzzle! One must not forget to apply all the rules all the time...
66:28 ... really enjoyable challenge
I wonder if it's possible to solve with the rules as stated (ie: where it's possible to have isolated odd cells which are not orthogonality connected to any other odd cells). I feel like the assumption that all odd cells are orthogonally connected was big in solving it, and not sure it's possible to solve without making that assumption.
I can now confirm there is a logical path to a solution which does not make that assumption; it just took me a bit longer to get there (around an hour or so).
Probably it ispossible, but would require deeper level of speculations on odds/evens. As at least one isolated odd would require to have at least another 4.
I solved it without this assumption (since the rules strictly allow for 1-groups like you said) but only managed after some bifurcation. Came back looking for a proper way to do it. :)
@@daanperelachaise I was able to do it without anything more than CtC style look-ahead (ie: with coloring for paths). FWIW.
As a mathmatician I thought the wording was clear ("groups of orthogonally connected odd digits" could mean a group of 1 cell) but you are the second person who got confused by that. I've since updated the rules on Logic Masters, but my submission to Bremster was before that. Sorry!
Turns out it might solve even with the less strict ruleset, but that is definitely not intended.
That one rule is poorly written. As is, it says that all groups of odd digits are 5 cells. That means that isolated odd numbers are allowed. What the rule actually is, it this: Odd numbers appear only in groups of 5 orthogonally connected cells; there are no isolated odd numbers, and no groups of odd numbers of any other size.
Great difficulty…. 37 mins… with ur help…