Imagine a Sudoku with Text-Only: A Gridless 6x6 Antiknight Antidouble Sudoku

Question

Imagine the following Sudoku:

• Imagine the Sudoku has a 6x6 grid with standard boxes, 2 rows and 3 columns each; and standard digits, 1 to 6.
• Imagine it has an antiknight rule: identical digits cannot be chess knight move away from each other.
• Imagine it has an antidouble rule: no digit is orthogonally adjacent to its double.
• Imagine the top-leftmost and bottom-rightmost digits are equal, so do the top-rightmost and bottom-leftmost.
• Imagine in Row 5, there exist 3 adjacent digits that are numerically adjacent too in increasing order from left to right.
• Imagine in Column 1, there exist 3 adjacent digits that are numerically adjacent too in increasing order from top to bottom.

Can you imagine that?

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^{(P.S. This can be fully done logically but might need some "special" knowledge.)}

Answer 1

First step (some simple logical deduction):

Now, if we focus on this highlighted 2x3 grid:

we see two B's. If we choose the red B we would go against the antiknight-rule in the 2x3 grid below the yellow one, so we now know we have to choose the blue B.

That leads to:

This red B. And the rest of the boxes highlighted in yellow will follow.

Let's clean this up:

Now, using some more logic, we can place:

This red D. And the rest of the highlighted yellow boxes will follow.

Let's clean up again:

We now have the whole grid in terms of letters. Time to look how we can translate these to numbers, 1-6, using the rules.

From the information given by the rules we can conclude that:

Column1 and row5 must be of the form, 2-3-4, 3-4-5 or 4-5-6 (note that the possibility 1-2-3 is excluded since 2 is 1 doubled) where its possbible along the column/row. We can see that the string of numbers must include a 4. Let's compare column1 possibilities with the corresponding letter-position in row5 and see what happens:

A-D-C: This string won't work since it's impossible to form a 3 digit consecutive string in row 5.

D-C-B: Like-wise this string won't work since it's impossible to form a 3 digit consecutive string in row 5.

C-B-E: Now this one is possible if we choose them to be 3-4-5. The other possibilities (2-3-4 and 4-5-6) don't work.

Lastly, B-E-F: Same as the two first possibilities we can see that this is impossible.

This means that the third possibility, C-B-E, is the only one that works.

We now replace the letters with the corresponding numbers and get the full grid:

Answer 2

The answer is

Yes, I can imagine it.

The first step was

to use the fact that opposite corners were identical digits and that there was an antiknight restriction. This gave me the following:



The opposite corner digits can only propogate in the grid in two ways. In the above, only the yellow has been shown in both configurations.

The next step was

to realize that the most restricted digit is the 4. There are two digits it cannot be adjacent to and it must be part of any 3 digit sequence. I therefore decided to find where the 4 could be in the first column, starting at the bottom:

In the first configuration, the placement of the 3 and 4 in row 5 means no 3-digit sequence is possible. In the other configuration, the 1 and the 2 are forced to be adjacent. Conclusion: the 4 cannot be at the bottom.

Moving along,

I tried the 4 in the 5th row, first using a 2-3-4 sequence:



Again, this was not possible as shown. I then tried a 3-4-5 sequence:


This too is impossible (no matter the yellow configuration) as the 4 can only be part of a 4-5-6 sequence and the 5 is too far away.

Moving up

I tried the 4 in the 4th row, first using a 2-3-4 sequence:

Didn't work in either yellow configuration as no 3-digit sequence could be formed in the 5th row.

And finally

I tried the sequence 3-4-5 which gave

I.e. the second yellow configuration worked!

Nice puzzle! If there is an easier method, I'd love to hear it.