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I can be a bit evil sometimes. Today, I am going to give you a Hidato with no numbers! It gets worse: This puzzle is on a Mobius Strip Y[1] However, I will give hints as to what the numbers are, so this isn't going to be plain impossible. Also, to make this a bit easier, this is only a 3x3 Hidato. Here is the puzzle:

A B C
D E F
G H I

Hints:

  1. $B,C,G,E$ are primes
  2. $A,H,I$ are square numbers
  3. $C$ does not connect with $I$
  4. $A+1\lt C\lt H\lt G\lt E$
  5. $A$ does not connect with $D$
  6. There is only 1 connection that is made using a diagonal

The goal of Hidato is to fill the grid with a series of consecutive numbers adjacent to each other orthogonally or diagonally. The entire grid is required to be filled in.

[1]Here's an example. Say we are on an 8x8 grid at R1C1. If we walk up, we go to R8C1. However, this is not a torus, so going straight from R1C1 to R8C8 is illegal, however, since this is Hidato, we can do $R1C1\to R\color{red}2C8$ due to our ability to move diagonally.

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1 Answer 1

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First off,

The four primes must be 2, 3, 5, 7 and the 3 squares must be 1, 4, 9.

A + 1 is at least 2 (since A can't be less than 1)
from (4), C, G, and E are all primes that are greater than 2 (A + 1 is at least 2)
thus C, G, and E are 3, 5, and 7 in that order, and that leave B as 2.

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 |   | 2 | 3 |
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 |   | 7 |   |
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 | 5 |   |   |
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Next,

The two numbers that are neither prime nor square are 6 and 8, and those must correspond to D and F (in some order).
If F is 6, we have a diagonal connection between 5 and 6, and there is no way to complete the grid without another diagonal, so F must be 8, and D is 6.

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 |   | 2 | 3 |
 -------------
 | 6 | 7 | 8 |
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 | 5 |   |   |
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Finally,

It doesn't take too much more figuring to place the 3 square numbers.
C is greater than A + 1, so A is obviously 1.
C and I aren't connected, so the only other place 4 can go is at H.
That leaves I for 9.

 -------------
 | 1 | 2 | 3 |
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 | 6 | 7 | 8 |
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 | 5 | 4 | 9 |
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