- 14 Jun '09 12:09 / 1 editTask:

Find the longest King tour in terms of the distance travelled.

There are only two solutions.

Here are the paths:

http://img35.imageshack.us/img35/9650/kingstour.gif

The challenge in the**Problemists Club**was to show one with

the White King and one with the Black King at the same time.

Very difficult as sooner or later the King's would clash and

neither could move into an illegal postion - the Kings can never meet.

I eventually did it after a lot of trial and error.

All 64 squares covered following the two distinct paths using

both Kings simultaneously.

Now some of you flat brainers will be thinking: "So What?"

But in the field of problem composition I have just solved

the equivalant to 'perpetual motion'.

The Mount Everest of problem solvers has been climbed..

Barnaby gave up, Checknev said it could not be done, Klugan devoted

his entire life to this project and failed, Watson & Williams too said

it was impossible.

This is a great moment for**RHP**and all subs (and none subs)

can feel very proud that they are members here.

I now share this moment with you.

- 14 Jun '09 12:44

It was I who grafted and slogged over the board for days, nay, weeks,*Originally posted by Macpo***Who is the masochist that wrote the pgn?**

to solve this Holy Grail so**RHP**members like yourself can tell your grand children

you were there when greenpawn34 posted the solution that has been

plaguing mankind since the dawn of time. - 14 Jun '09 13:06 / 1 editActually, this kind of discovery makes me wonder a bit about the meaning of life. As THE useless activity on earth, chess couldn't be but a challenge to it. But with that kind of problemist things, it's not anymore a challenge, it's a decisive answer:

Chess players drive the world towards nihilism. And they seem proud of it

Congrats, by the way! - 14 Jun '09 13:47

Think you missed a spot*Originally posted by greenpawn34***Task:**was to show one with

Find the longest King tour in terms of the distance travelled.

There are only two solutions.

Here are the paths:

http://img35.imageshack.us/img35/9650/kingstour.gif

The challenge in the [b]Problemists Club

the White King and one with the Black King at the same time.

Very difficult as sooner or later the K ...[text shortened]... Ke7 Ke4 60. Kf6 Kd3 61. Ke5 Kc4 62. Kf4 Kb3 63. Ke3 Ka4 64. Kf2 Kb5[/pgn][/b]

naaaaa,just messing - 22 Jun '09 18:47 / 1 editI'd like to revisit this problem (I could not believe that there are only two solutions). I sat out to define a recursive, brute-force algorithm to solve it. I had to include several checks after each move to get the computer to give an answer in a realistic amount of time.

What I find rather interesting is the vast number of solutions that exists (following GreenPawn's definition of two kings that do it at the same time, etc.). For an N by N chessboard and kings in opposing corners, I found:

3x3 (and below) chessboard: no solutions, found instantaneously.

4x4 chessboard: 361,112 solutions, found in 21.408 seconds

The 5x5 chessboard ran for over an hour with well over 3,000,000 solutions when I stopped it.

Going even higher would probably take the rest of the century. I'd need to rethink the algorithm somewhat.

I'll post the code on http://mycodingcorder.blogspot.com if anyone is interested. - 22 Jun '09 19:29

no, both are the length of exactly 1 move. it's an 8-connected region, pythagoras doesn't apply.*Originally posted by heinzkat***I should add that a diagonal King move has a longer distance than a perpendicular one (Pythagoras).**

And I should add that as a non-sub scum I feel terribly proud. - 22 Jun '09 22:01 / 1 edit

See quotation above for a clarification of the posed problem.*Originally posted by ...***In a king's tour of the chessboard the king makes 64 moves, moving to each square once, and the last move is back to the initial square. Assuming the king moves from square center to square center, find the longest tour in terms of the distance traveled. This is the same as maximizing the number of diagonal moves (or minimizing the number of orthogonal moves).**

There are two solutions (up to rotation & reflection).