Current Events > I will venmo 50 dollars to the first person who can solve my Sudoku puzzle.

(+) Yeah! (+)

Serious Cat posted...

Phistomefel's ring didn't get me very far, but I came to the same conclusion with this equivalent set:

https://gamefaqs.gamespot.com/a/user_image/4/3/8/AAPWHiAAC4rW.jpg
The digit in the center square can only appear in columns 1, 2, 8 and 9 in rows 3, 4, 6 and 7 by either knight's move, diagonal, or plain sudoku rules. That requires the digit in the center square to be diagonally adjacent in those rows, thus eliminating the remaining possibilities left in the Phistomefel boxes. (Plus the additional row of red compared to blue means there's an extra complete set of digits 1-9 in the red squares in addition to the equivalent set.)

You may...be? right, I'm having trouble following your overall statement. But you might be making things overcomplicated for you. It seems like you saw the ring:
https://gamefaqs.gamespot.com/a/user_image/4/6/6/AACZqoAAC4ry.png

Meaning that purple sees all of blue, we can conclude it's not in Yellow, thus if we just color everywhere purple can be we are left with this:

https://gamefaqs.gamespot.com/a/user_image/4/6/7/AACZqoAAC4rz.png

So that above picture has everywhere purple can be labeled.

So after that, let's think about another color like orange, looking at what you just solved for purple as an example, where is everywhere orange can possibly be? (you can reference the 3x3 of grids to maybe more easily spot a pattern). To answer this question, don't worry about orange overlap with any other colors, just worry about orange. "Where is everywhere orange can be?".

You can do this for each color "Where is everywhere X color can be?" (When answering this you aren't worrying about the other colors, just that specific color)

Finally I will re-highlight this statement:

I will highlight this because it is part of the puzzle/ruleset
The Name of the puzzle is "Multi-color Madness!"

(Recommended: Click the gear icon and enable "Multi-color Mode")

joe40001 posted...

You may...be? right, I'm having trouble following your overall statement. But you might be making things overcomplicated for you. It seems like you saw the ring

It's an equivalent set like the ring except it's isolating the cages instead of the boxes in the corners and has an extra row, thus an extra set of digits 1-9. I can throw in an extra column and make the two sets equivalent, like this:

https://gamefaqs.gamespot.com/a/user_image/5/4/0/AAPWHiAAC4s8.jpg
The blue set is now equivalent to the red set. The center square sees all of the other blue squares and therefore appears in the red set exactly once, somewhere in column 8.

This is another similar equivalent set:

https://gamefaqs.gamespot.com/a/user_image/5/4/1/AAPWHiAAC4s9.jpg
The blue set is still equivalent to the red set with the column we know contains the red square corresponding to the center digit removed. The center square is still included in the new blue set and can still see all the other blue cells, so the new equivalent sets still have to contain the center digit in red exactly once, this time in column 2.

But like you said, it didn't really get me anywhere. Just noodling in SET theory.

Serious Cat posted...

It's an equivalent set like the ring except it's isolating the cages instead of the boxes in the corners and has an extra row, thus an extra set of digits 1-9. I can throw in an extra column and make the two sets equivalent, like this:

https://gamefaqs.gamespot.com/a/user_image/5/4/0/AAPWHiAAC4s8.jpg
The blue set is now equivalent to the red set. The center square sees all of the other blue squares and therefore appears in the red set exactly once, somewhere in column 8.

This is another similar equivalent set:

https://gamefaqs.gamespot.com/a/user_image/5/4/1/AAPWHiAAC4s9.jpg
The blue set is still equivalent to the red set with the column we know contains the red square corresponding to the center digit removed. The center square is still included in the new blue set and can still see all the other blue cells, so the new equivalent sets still have to contain the center digit in red exactly once, this time in column 2.

But like you said, it didn't really get me anywhere. Just noodling in SET theory.

It's an interesting observation, but I think it is captured by the "draw everywhere purple can be" question, because that also observes that purple can only be in those cells in c2 and c8 as well.

Did you want to take a crack at the question of "where is everywhere red can go?" otherwise I'll probably finish the reveal tonight.

Too tired tonight, will probably post the rest tomorrow.

Before actually posting the solution, I also semi-think that all the center squares on the two main diagonals are all odd and the remaining center squares are even, although that may have been something I came up with when I was misinterpreting the rules.

Serious Cat posted...

Before actually posting the solution, I also semi-think that all the center squares on the two main diagonals are all odd and the remaining center squares are even, although that may have been something I came up with when I was misinterpreting the rules.

You don't have to worry about digits at all until the entire grid is colored then resolved to each cell having a single color. The ending is like YouTube video you linked with multi-color cells being resolved to single color cells.

If you pick any other color but purple and figure out everywhere that can go you have basically figured out the break in.

Ok, the trick/break-in is this:

Because the grid is toroidal, there is no reason to think of R5C5 as "the center". There is nothing different between R5C5 and R2C2 or any other of the centers, their only difference is their proximity to the border, and because it's toroidal the whole grid loops and there is no border.

Thus there exists toroidal phistomaphel rings around all 9 centers. Just like the regular phistomaphel rings, this eliminates extra cells that could otherwise be that color. Here are all the cells that could be red:
https://gamefaqs.gamespot.com/a/user_image/5/4/0/AACZqoAAC5b0.png
:
Here are all the cells that could be yellow:
https://gamefaqs.gamespot.com/a/user_image/5/4/1/AACZqoAAC5b1.png

You simply go through every color and label everywhere it can be and you are left with this grid:
https://gamefaqs.gamespot.com/a/user_image/5/4/2/AACZqoAAC5b2.jpg

Then you use the 6x6 killer cages to start eliminating colors, there are no clever tricks needed in this part, just fairly standard sudoku deduction. Using the knights move and the cages you can resolve the entire grid such that every cell is reduced to a single color.

At this point you can finally get to work with the inequalities. and with the 8 inequalities involving 9 colors you can solve the entire puzzle.

I might leave this part up for a day or two before showing those last 2 easy steps in case people want to try now that they know the break-in.

PS: The break-in can also be done without SET logic if you just fill out the grid normally and then notice the squid patterns on the cells that have 6 colors.

PPS: For people not familiar with these terms, none of these are things the solver couldn't figure out on their first solve, though knowing certain solving techniques does help.

After reading that, my apparently needlessly complicated set-making tells me that every number in the grid appears diagonally adjacent to itself everywhere it appears in the grid unless it's in the center of a block. Unless I'm mistaken.

https://imgur.com/a/0r18d5R

More useful set equivalence goodness:

The diagonally adjacent cells are contained in four-square blocks arranged in a sort of x-wing pattern. In the above example, the purple squares are either housed in the blue group or the red group and either the green group or the orange group. All of the cells in the blue group in rows 3 and 4 are contained in the killer cage and therefore cannot be diagonally adjacent. That means that purple is in the red group. That also means by block 3 that the other 4 purples are housed in the orange group.

Serious Cat posted...

After reading that, my apparently needlessly complicated set-making tells me that every number in the grid appears diagonally adjacent to itself everywhere it appears in the grid unless it's in the center of a block. Unless I'm mistaken.

This appears to be right, but speaking of your needlessly complicated set-making skills :P I think you are still being more clever than is necessary. At the point I showed you can just start simplifying colors:

Look at the 6-boxes: The top-right one, for example, has 4 cells that have 4 options, (yellow, blue, green, light-grey) and as a result the remaining 2 cells cannot be any of those 4 (can't be green or light grey), thus you know it's a purple/mid-grey pair, and from that you can start eliminating purple/mid-grey from it's column and from places where sudoku+knight's move sees both those cells. You can do the same thing in low-left 6-box and get the cells in C2 to be a yellow/purple pair.

I've noticed none of your pictures have had multi-color on, are you not using that? If so that will make things very hard.

Once you know the break-in all that's left is color the grid fully, use the 6-boxes and start simplifying until all colors are resolved.

joe40001 posted...

I've noticed none of your pictures have had multi-color on, are you not using that? If so that will make things very hard.

Not if you can find the break-in.

https://imgur.com/a/bbioGd3

Grid colored:
https://imgur.com/a/3yYL0so

I haven't disambiguated yet, but I can already tell that the cages and inequalities on the sides contains a complete set of the digits 1-9, with the center square appearing exactly twice, just as the prophecy foretold.

https://imgur.com/a/rRicHlD

(Finally) the colored puzzle.

And the puzzle colored in a way that actually works this time:

https://imgur.com/a/rRicHlD

And solved:

https://imgur.com/a/qCLmAL9

Break-in without first considering all options is this:

We know through earlier discussion that every value that is not in a center square is located in sets of 2x2 squares and are diagonally adjacent to one another. The cage shared by boxes 3 and 6 limit the positioning of where yellow and black can be placed in that configuration. Similarly, the cage shared by boxes 4 and 7 limit where purple and orange can be placed. The combination of those possibilities gives us this grid:

https://imgur.com/a/Nnbmn6t

The boxes connected in the green squares are three bi-value squares with an extra digit in a fourth square. Since we know that every non-center square is located diagonally adjacent to a square of the same value, we can eliminate not only the extra value from the square, but its matching value in its potential counterpart in the diagonal cell. That lets us place the orange in rows 2 and 3, the yellow in row 7 and 8, and in turn bumps out potential diagonals allowing us to place purple in rows 1 and 2, black in rows 8 and 9, and we're so far ahead of ourselves that we'll be able to place dark gray in rows 1 and 9 immediately.

And the whole puzzle whittles itself down with the same logic.

Serious Cat posted...

And the puzzle colored in a way that actually works this time:

https://imgur.com/a/rRicHlD

And solved:

https://imgur.com/a/qCLmAL9

Break-in without first considering all options is this:

We know through earlier discussion that every value that is not in a center square is located in sets of 2x2 squares and are diagonally adjacent to one another. The cage shared by boxes 3 and 6 limit the positioning of where yellow and black can be placed in that configuration. Similarly, the cage shared by boxes 4 and 7 limit where purple and orange can be placed. The combination of those possibilities gives us this grid:

https://imgur.com/a/Nnbmn6t

The boxes connected in the green squares are three bi-value squares with an extra digit in a fourth square. Since we know that every non-center square is located diagonally adjacent to a square of the same value, we can eliminate not only the extra value from the square, but its matching value in its potential counterpart in the diagonal cell. That lets us place the orange in rows 2 and 3, the yellow in row 7 and 8, and in turn bumps out potential diagonals allowing us to place purple in rows 1 and 2, black in rows 8 and 9, and we're so far ahead of ourselves that we'll be able to place dark gray in rows 1 and 9 immediately.

And the whole puzzle whittles itself down with the same logic.

Amazing, that was a brilliant way to solve it. You spotted patterns that I didn't even fully appreciate.

Bravo! :)

If you want I can send you 20$ as a prize, it was fun seeing you work through it. And I feel bad all the time this puzzle took from your life the past few weeks lol.

@KainWind I'm curious how you went about solving it. Was it a similar path?

I colored each middle cell differently and colored everywhere they could go like you said. After that it was just slowly narrowing down everything. Played around with the anti knight a lot to narrow down spots in boxes. Think there were a couple jellyfish in there. Used the cages to to narrow down some colors to only 2 cells in a column a couple times which let me clear some out. I forget everything but that was basically it.

https://youtu.be/j-TG44-mDkY

I realized that Sudoku Pad saves your steps, so here is most of the final solve start to finish. Sudoku Pad lost a few of the steps just before the end that I had to re-create. Not depicted is where I had a lapse of logic in the very final step, missed a knight's move that resulted in a valid grid anyway, and spent an inordinate amount of time wondering how there was one unique solution.

I should say I really like how much the toroidal rule was used, and the inequalities at the end was cute.

KainWind posted...

I colored each middle cell differently and colored everywhere they could go like you said. After that it was just slowly narrowing down everything. Played around with the anti knight a lot to narrow down spots in boxes. Think there were a couple jellyfish in there. Used the cages to to narrow down some colors to only 2 cells in a column a couple times which let me clear some out. I forget everything but that was basically it.

(I meant jellyfish patterns, not squid patterns rule when I said that earlier. I never use them myself so I forgot what they were called)

So you used the jellyfish and not the torodial phistomophel rings? Nice, that means you probably had some cells that were 6 colors.

Yeah once you get it fully colored the simplification isn't too bad.

KainWind posted...

I should say I really like how much the toroidal rule was used, and the inequalities at the end was cute.

Thanks, I think torodial rules are under used IMO. Some people don't seem to care for them, but I enjoy them a lot.

As far as the end, yeah it's really a coloring puzzle, so once the grid was fully colored I wanted something pretty simple to resolve the actual digits. Originally I had cages instead of inequalities to resolve, but people tried putting in digits too early and I just wanted coloring.

I have a habit of making things a little "cute" sometimes for lolz, thus R5 ending up how it did.

Thanks for playing, and again if you want your 50 dollar prize feel free to DM me your venmo. You don't have to obviously, but I assumed that's why you took the time to solve it.

This topic was a lot of fun. I really appreciate seeing people solve it and enjoy it. Hopefully it gets on CtC one day. CtC sometimes like's to do puzzles that have a title/theme they can frame around something in current pop-culture, so maybe around the release of "Multiverse of Madness" the title will cause Simon to consider this one if they remember it. I didn't name it that way as a reference, but somebody pointed out the title similarity to me later.

I considered a few options for the title, but only one of them explicitly mentioned multi-color, and I felt that was a crucial part to highlight.

Fun fact: The puzzle is still solvable if you replace the bishops rule with:
"Cells 2 diagonal moves from a box's center cell cannot contain that center cell's digit."

That solution does require you to notice the SET/Phistomaphel out the gate and so it doesn't allow for the 6-color jellyfish solution.

I really don't like using set theory if I can avoid it, but it's crazy how helpful it can be. Sometimes you have to get creative with it too.

joe40001 posted...

Fun fact: The puzzle is still solvable if you replace the bishops rule with:
"Cells 2 diagonal moves from a box's center cell cannot contain that center cell's digit."

joe40001 posted...

Fun fact: The puzzle is still solvable if you replace the bishops rule with:
"Cells 2 diagonal moves from a box's center cell cannot contain that center cell's digit."

That solution does require you to notice the SET/Phistomaphel out the gate and so it doesn't allow for the 6-color jellyfish solution.

https://gamefaqs.gamespot.com/a/user_image/1/7/7/AAPWHiAAC51Z.jpg
You could still solve with whatever method you want if you notice that, for example, purple has to reappear appear in the grid once in each colored verical line and once in each colored vertical line. That means that there are a total of 8 purples in the colored lines/columns, plus the initial appearance, and therefore no more purples anywhere in the grid.

Serious Cat posted...

https://gamefaqs.gamespot.com/a/user_image/1/7/7/AAPWHiAAC51Z.jpg
You could still solve with whatever method you want if you notice that, for example, purple has to reappear appear in the grid once in each colored verical line and once in each colored vertical line. That means that there are a total of 8 purples in the colored lines/columns, plus the initial appearance, and therefore no more purples anywhere in the grid.

Huh, I guess I'm starting to learn that the Phistomsphel ring is just a double jellyfish uniquely observed.

joe40001 posted...

Huh, I guess I'm starting to learn that the Phistomsphel ring is just a double jellyfish uniquely observed.

I think it's the other way around: it's a double jellyfish (or a pair of x-wings plus the knight's move and diagonal restraints) keeping the digit out of the forbidden zone in your puzzle. Under normal sudoku rules, there's nothing keeping any specific digit out of the ring.

https://gamefaqs.gamespot.com/a/user_image/6/4/0/AAPWHiAAC58o.jpg

Serious Cat posted...

I think it's the other way around: it's a double jellyfish (or a pair of x-wings plus the knight's move and diagonal restraints) keeping the digit out of the forbidden zone in your puzzle. Under normal sudoku rules, there's nothing keeping any specific digit out of the ring.

https://gamefaqs.gamespot.com/a/user_image/6/6/4/AAPWHiAAC59A.jpg
Edit because examples of a workable ring probably should be workable.

It's knights and bishops that causes the ring in my puzzle, but the ring excluding the 4 corners could be proven by the same logic on the normal ring as you showed in your example:
https://gamefaqs.gamespot.com/a/user_image/5/7/8/AACZqoAAC6a6.png

Black must appear once in the blue column, once in the red column, once in the yellow column, and once in the green column, same for once in each colored row, thus black has to appear 8 times in addition to it's appearance in the center, and so Phistomaphel is proven via jellyfish and not SET.