8
$\begingroup$

One day, a friend has shown me a 5 by 5 grid, challenging me to fill it with numbers from 1 to 25.
Obviously, it is not simple because there are some rules:

  • The number 1 is placed in the center of the grid
  • The numbers are placed in ascending order, either by moving two boxes diagonally or moving three boxes in a row or column (as shown in diagram)

enter image description here

So, how can you achieve that ?

(There are multiple answers, so you can just tryhard it and posting your own solution)

$\endgroup$
13
  • 1
    $\begingroup$ What do you mean by "in ascending order"? In which direction? $\endgroup$ Jun 7 '17 at 13:55
  • 1
    $\begingroup$ And what do these "movements" have to do with filling the grid? $\endgroup$ Jun 7 '17 at 13:55
  • $\begingroup$ You have to fill the grid with numbers from 1 to 25. You start with the 1 in the center of the grid and you place the others numbers in an ascending order by moving two boxes diagonally or three boxes linearly. Sorry for the lack of precision, it's my first question. $\endgroup$ Jun 7 '17 at 13:59
  • $\begingroup$ Ah, gotcha. Now I'm sure I've seen this puzzle here before quite recently, but I can't find it! +1 anyway. $\endgroup$ Jun 7 '17 at 14:00
  • 3
    $\begingroup$ In other words you have to find a knight's tour on the 5x5 chessboard, except that you have to start in the centre, and instead of a knight you have a piece that can do move of the type {0,+-3}, {+-3,0}, and {+-2, +-2}. $\endgroup$ Jun 7 '17 at 14:37
6
$\begingroup$

Definitely not unique, but here's one solution:

\begin{array} & 2 & 17 & 20 & 3 & 16 \\ 12 & 25 & 6 & 13 & 22 \\ 19 & 9 & 1 & 18 & 8 \\ 5 & 14 & 21 & 4 & 15 \\ 11 & 24 & 7 & 10 & 23 \end{array}

Found entirely with trial and error.

$\endgroup$
4
  • 3
    $\begingroup$ Gratz, I calculated with computer there are 352 possible solutions. $\endgroup$
    – shyos
    Jun 7 '17 at 15:22
  • $\begingroup$ @shyos: Are there any solutions where you can jump from the $25$ back to the $1$? $\endgroup$ Jun 7 '17 at 17:05
  • 2
    $\begingroup$ @JaapScherphuis Sure. Just swap the 23 and 25 in this solution. $\endgroup$
    – user27014
    Jun 7 '17 at 18:19
  • $\begingroup$ To avoid overcounting rotationally and axially symmetric solutions, why don't we WLOG say that 2 has to go in the TL corner, and hence 3 in the top row? $\endgroup$
    – smci
    Mar 29 '18 at 21:47
1
$\begingroup$

found another solution by trial and error

\begin{array} & 23 & 17 & 5 & 22 & 16 \\ 12 & 20 & 25 & 13 & 3 \\ 6 & 9 & 1 & 18 & 8 \\ 24 & 14 & 4 & 21 & 15 \\ 11 & 19 & 7 & 10 & 2 \end{array}

I kinda just made "boxes" where I could and used diagonals to transition the progression to the next "box"

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.