Usually Kathy completes these twice as fast as I do. But I found this trivial and finished in 3:19. My fastest solve of any sudoku ever. You spot that a column contains all the digits 1-6, once each. So you can just write 1-6 in the first column according to the order of the clues. You can do the same with column 6, row 1 and row 6. Then Sudoku gives all the other numbers except a 45 deadly pattern. And you can resolve those by examining the full 6 digit numbers for those 4 cells.
Yes, I did the same and had same observation, but I need a minute to realize that I need to use 1-6 numbers on the perimeter once, and that makes it trivial.
I always chuckle a bit when Kathy refuses to do sudoku because she is so dedicated to the constraint. It shows that she really doesn't take any shortcuts and wants to be sure that the logic remains intact.
An easy puzzle if you can understand the constant as given, If would word as "Each row and column form a six digit number, read from either direction, and are ranked by the numbers outside the grid. With 1 being the lowers and 24 the highest."
5.09 after watching you think about it for a few minutes, and realising that thinking about where the digits would go in the outer perimeter would be helpful.
I fell into the trap of starting with 123456, 654321, etc. before thinking to check that there are 6! = 720 possible numbers. Doing the outer ring made quick work of it after that, so I finished with 2:30.
Was able to complete in just 1:21 (conflict checker off), but I will admit I've had some prior experience with the full rank constraint, so this flowed very well for me. 🙂 Many thanks to KenGlue for a wonderful puzzle!
Wow, this is neat! It's a lot easier than I thought, when you start with 1 as long as you can, then 2, & so on... I like how every group of four numbers start with the next digit! My solve went very well, less than 5 minutes. (I watched till she tried to start with 1 and 2 as long as we can, then started solving myself.)
I liked this one and found it easy. The tens place has the most power. We know that all of the digits from 1 to 6 will appear in each row and column, but in particular, each row and column on the perimeter. That means if we start with either 1 or 6, we know which spot they will go on. 6 will be in rank 20, 21, 22, 23, and 24, because no other 6 digit number will be larger. 1 will go in rank 1, 2, 3, and 4, because no other number will be smaller. Likewise, 2 will go in rank 5, 6, 7, and 8. 3 will go in 9, 10, 11, and 12. And the pattern continues.
9:34 for me watched the intro to see what was going on. I started fairly similarly to Kathy but used small center marks rather than large digits , because I wasn't sure how it would turn out. I ended with a 4/5 deadly pattern which I basically looked at both ways to see which broke. Kathy did a great job and I think she definitely had a better way to disambiguate the pattern. Very neat puzzle by Ken as well.
each digit 1-6 on the corners will be the start to exactly FOUR 6 digit numbers, regardless if they are placed on the corner or not. there are 24 ranks total, so thus 1-4. must be allocated to 1, 5-8 to 2, 9-12 to 3, etc
1:41 Was scared for a sec seeing that this was a full rank puzzle, but luckily it was much easier than the only other full rank puzzle I've done, a 6x6 by clover! with only 3 full rank clues given
solved in 3:46. the rules of this puzzle are symmetrically valid, if you flip the order to #1 being the highest number and #24 being the lowest you get a grid like the solution where pairs of digits 1-6, 2-5, and 3-4 are swapped with each other. don't ask how I figured this out
Easy way to think about it is the biggest number on one side is the biggest number so has to start with a 6. The second biggest number on a side is the second biggest number so it starts with a 5 and so on. Once the perimeter is all filled in it just leaves sudoku.
I feel like this solution is a bit incomplete. Yes, you solved the puzzle, but mainly because your intuition was right. You never proved you need to use as many 1s, 2s, etc. as you can on the outer edge, maybe you could've used one less 2, and used one more 5 instead. It's still a solve, I'm just pointing this out, because if you go for more complicated puzzles, you might lock yourself out of the elegant path to the solution because of subtle guesses like this, which can take a lot away from the enjoyment.
Really? I thought I chose the first digits well, I used as many 1s as I could, and only moved up to the next digit when I needed to, due to sudoku. Coming backwards and putting a 6 on rank 24 gives the same result. Where would less/more 2s or 5s go? After that I don’t think I made any guesses, each number was chosen based off of sudoku or what would make a number lower or higher. I did record this a few days ago, I may be misremembering.
@@ChattyKathy "... and only moved up to the next digit when I needed to, due to sudoku." See that's the thing. You never proved you need to use as many of the same digits as you can, at that point in time there could've been some gaps. Of course no less/more 2s and 5s are possible, since the puzzle only has one solution, but in your solve this was never even considered as a possibility. The idea was good, it was just never proven why you *have* to fit as many of the same digits as possible. In short, yes, it works, you just never figured out *why* it works, and/or why it's the only way it works. You just stumbled into the solution, because the idea happened to be correct.
To me, I’ve never seen this variant before so I was fiddling around with different techniques to see what would make sense/what would work. Until I found an option that worked, then I continued with it until the end of the puzzle. Is your point maybe that I shouldn’t have combined sudoku with the constraint, and should’ve just focused on the constraint to find “proof”? Sorry I don’t understand where I went wrong still. To me it was clear after filling in r1/6 and c1/6 that that was the only option, maybe I could’ve explained out loud better what was happening in my head.
@@ChattyKathy I think the way to really make the beginning of the puzzle more concrete is thinking about how many of each digit are in a row/column. Row 1 contributes 6 sequences, each starting with a different digit 1-6. The same is true of row 6, column 1, and column 6. This means that there are 24 sequences, 4 of them starting with a 1, 4 of them starting with a 2, 4 of them starting with a 3, ... Once you realize that, there is no need to do any experimentation trying to find a layout that may work. You can just say that ranks 1-4 start with 1, ranks 5-8 start with 2, etc.
Each row and column needs 1-6. That alone is proof. Look at row one... Given that the digits outside are 8-4-12-22-18-13 the cells must read 2-1-3-6-5-4. Repeat around the perimeter. Logically that means going in numerical order you must start low, and only change when forced to. Either way ends the same way.
Usually Kathy completes these twice as fast as I do. But I found this trivial and finished in 3:19. My fastest solve of any sudoku ever.
You spot that a column contains all the digits 1-6, once each. So you can just write 1-6 in the first column according to the order of the clues. You can do the same with column 6, row 1 and row 6. Then Sudoku gives all the other numbers except a 45 deadly pattern. And you can resolve those by examining the full 6 digit numbers for those 4 cells.
Nice spot, I was just moving from 1 to 24 starting 4 digits from each of 1, 2….
Yes, I did the same and had same observation, but I need a minute to realize that I need to use 1-6 numbers on the perimeter once, and that makes it trivial.
2:48 for me, but knowing how to fill in the outer ring rather trivialized this one. Still a good intro
I always chuckle a bit when Kathy refuses to do sudoku because she is so dedicated to the constraint.
It shows that she really doesn't take any shortcuts and wants to be sure that the logic remains intact.
Why use a scalpel when a sledgehammer will do?
Congrats! I was originally going to give only 7 clues, but decided I'd just give all of them for the introduction.
Yeah I liked this! I’m not sure I would’ve figured out what to do with only 7 clues, having never seen this before, so thank you! Lol
An easy puzzle if you can understand the constant as given, If would word as "Each row and column form a six digit number, read from either direction, and are ranked by the numbers outside the grid. With 1 being the lowers and 24 the highest."
5.09 after watching you think about it for a few minutes, and realising that thinking about where the digits would go in the outer perimeter would be helpful.
Nice introduction to this ruleset.
Normally not all of the ranks are given, which makes it a lot more interesting.
Far simpler than it looks. If the six digit numbers keep increasing, they start with 1 and increase only when necessary...
Completed in 1m47s.
I fell into the trap of starting with 123456, 654321, etc. before thinking to check that there are 6! = 720 possible numbers. Doing the outer ring made quick work of it after that, so I finished with 2:30.
Was able to complete in just 1:21 (conflict checker off), but I will admit I've had some prior experience with the full rank constraint, so this flowed very well for me. 🙂 Many thanks to KenGlue for a wonderful puzzle!
Wow, this is neat! It's a lot easier than I thought, when you start with 1 as long as you can, then 2, & so on... I like how every group of four numbers start with the next digit! My solve went very well, less than 5 minutes. (I watched till she tried to start with 1 and 2 as long as we can, then started solving myself.)
I liked this one and found it easy. The tens place has the most power. We know that all of the digits from 1 to 6 will appear in each row and column, but in particular, each row and column on the perimeter. That means if we start with either 1 or 6, we know which spot they will go on. 6 will be in rank 20, 21, 22, 23, and 24, because no other 6 digit number will be larger. 1 will go in rank 1, 2, 3, and 4, because no other number will be smaller. Likewise, 2 will go in rank 5, 6, 7, and 8. 3 will go in 9, 10, 11, and 12. And the pattern continues.
9:34 for me watched the intro to see what was going on.
I started fairly similarly to Kathy but used small center marks rather than large digits , because I wasn't sure how it would turn out.
I ended with a 4/5 deadly pattern which I basically looked at both ways to see which broke.
Kathy did a great job and I think she definitely had a better way to disambiguate the pattern.
Very neat puzzle by Ken as well.
each digit 1-6 on the corners will be the start to exactly FOUR 6 digit numbers, regardless if they are placed on the corner or not. there are 24 ranks total, so thus 1-4. must be allocated to 1, 5-8 to 2, 9-12 to 3, etc
1:41 Was scared for a sec seeing that this was a full rank puzzle, but luckily it was much easier than the only other full rank puzzle I've done, a 6x6 by clover! with only 3 full rank clues given
solved in 3:46. the rules of this puzzle are symmetrically valid, if you flip the order to #1 being the highest number and #24 being the lowest you get a grid like the solution where pairs of digits 1-6, 2-5, and 3-4 are swapped with each other.
don't ask how I figured this out
Easy way to think about it is the biggest number on one side is the biggest number so has to start with a 6. The second biggest number on a side is the second biggest number so it starts with a 5 and so on. Once the perimeter is all filled in it just leaves sudoku.
2:37!
Interesting what would be the minimal required ranks provided for this one.
clover! made what seems to be a minimally clued 6x6 full rank sudoku using only 3 numbers!
8:09 Luckily I saw the outer ring pattern rather quickly
00:06:34 This was a lot of fun; first time doing a puzzle with this constraint.
1:50 yay
7:04 with a restart. Not sure how long it actually took once I realized how to do the outer ring, but it wasn't very long.
1:43
Very fun
Puzzle solved in 2:12
The puzzle is wrong!! Kathy ranks on top!!
I feel like this solution is a bit incomplete. Yes, you solved the puzzle, but mainly because your intuition was right. You never proved you need to use as many 1s, 2s, etc. as you can on the outer edge, maybe you could've used one less 2, and used one more 5 instead.
It's still a solve, I'm just pointing this out, because if you go for more complicated puzzles, you might lock yourself out of the elegant path to the solution because of subtle guesses like this, which can take a lot away from the enjoyment.
Really? I thought I chose the first digits well, I used as many 1s as I could, and only moved up to the next digit when I needed to, due to sudoku. Coming backwards and putting a 6 on rank 24 gives the same result. Where would less/more 2s or 5s go? After that I don’t think I made any guesses, each number was chosen based off of sudoku or what would make a number lower or higher. I did record this a few days ago, I may be misremembering.
@@ChattyKathy "... and only moved up to the next digit when I needed to, due to sudoku." See that's the thing. You never proved you need to use as many of the same digits as you can, at that point in time there could've been some gaps. Of course no less/more 2s and 5s are possible, since the puzzle only has one solution, but in your solve this was never even considered as a possibility. The idea was good, it was just never proven why you *have* to fit as many of the same digits as possible. In short, yes, it works, you just never figured out *why* it works, and/or why it's the only way it works. You just stumbled into the solution, because the idea happened to be correct.
To me, I’ve never seen this variant before so I was fiddling around with different techniques to see what would make sense/what would work. Until I found an option that worked, then I continued with it until the end of the puzzle. Is your point maybe that I shouldn’t have combined sudoku with the constraint, and should’ve just focused on the constraint to find “proof”? Sorry I don’t understand where I went wrong still. To me it was clear after filling in r1/6 and c1/6 that that was the only option, maybe I could’ve explained out loud better what was happening in my head.
@@ChattyKathy I think the way to really make the beginning of the puzzle more concrete is thinking about how many of each digit are in a row/column. Row 1 contributes 6 sequences, each starting with a different digit 1-6. The same is true of row 6, column 1, and column 6. This means that there are 24 sequences, 4 of them starting with a 1, 4 of them starting with a 2, 4 of them starting with a 3, ...
Once you realize that, there is no need to do any experimentation trying to find a layout that may work. You can just say that ranks 1-4 start with 1, ranks 5-8 start with 2, etc.
Each row and column needs 1-6. That alone is proof. Look at row one... Given that the digits outside are 8-4-12-22-18-13 the cells must read 2-1-3-6-5-4. Repeat around the perimeter. Logically that means going in numerical order you must start low, and only change when forced to. Either way ends the same way.