Bishop's tour

I'm sure you are all familiar with the famous knight's tour problem, right? Well, that's easy. Now I present a greater challenge: The bishop's tour.

Your goal is start a bishop at a1 and maneuver it to every square of the board, landing on each square only once.

And don't say that's impossible; It's quite possible, just very hard. I've seen GM Arthur Bisguier do it in a presentation.

Good luck; you'll need it.

we already talked about that before

A bishop standing on a1 is a dark squared bishop. According to the rules of chess it can only move on dark squares and it is therefore impossible to maneuver it to every square of the board. Not because I say it, but because of the rules of chess.

This gives us two options:

1) It is an April fools joke where you break the rules in a smart and entertaining way.

or

2) You forgot to mention that it is of course only all the dark squares on the board.

If it is option 1) then it is a challenge to people's creative skills. How creative can people be in order to make a fun solution that breaks the rules in a smart and entertaining way?

If it is option 2) then we need some further specifications: Are you allowed to land on a square that you have passed but not landed on. We also need to know whether it is allowed to pass a square where you have been standing earlier on.

4-1-11...good one!

oinquaki,

I'm assuming this puzzle regards passing through a square as the same thing as moving into a square.  (Ba1-h8 means Ba1-b2-c3-d4-e5-f6-g7-h8).  If that's the case, then yes, it is a challenge, although the puzzle is rather limited in scope.

A different kind of Bishop's Tour puzzle has been discussed recently in this forum.  See here for the rules and some sample puzzles.

Here's an even harder challenge: the pawn's tour.  A pawn starts on a1 and it must reach every square on the board.  Also, when pawns reach the 8th rank, they don't promote - they turn around and go the other way.

 if the rule is that it can only pass through a suare once It moght not be possible as changing files requires captures.

will look into this further.

under these conditions it is not possible.  Proof:

as indicated above, the only consideration necessary is those tours consisting of one square moves

the light sqares and dark squares are of identical patterns under reflecion (or rotation) therefore, without loss of generality, we consider only the dark squares

Argument "A": the only way to reach a1 is form b2 and that once b2-a1 is accomplished the bishop cannot move to b2 (the only choice) as it has been previously occupied: therefore the tour must either begin or end on a1.

by argument "A", the relation shi[p between g7 and h8 requires that h8 must be the begining or ending square,

becausethe dark squares have both reflectional symmetry and rotational symmetry, withot loss of generality we can choose to consider only paths that begin on a1 and end on h8.

Argument "B": Once a1-b2 is accomplished: if b2-c1 or b2-c3 then, by argumnt "A" above the relationship between a3 and b4 requires that a3 must be the terminating square.

a3 and h8 cannot both be termination squares therefor the begining of the path must be a1-b2-a3

now, by argumant "B" above, c1 must be aterminating square.

the tour cannot end on both c1 and h8: therefor the tour cannot be completed.

 with out this restriction

each square occors once and only once therefor constitutes a permutaion and we have 32! = 2.631*10^35 possabilities

from any square on the edge there are 7 squares that have connection to it.  therefor ther are 7 preceeding squares and 6 remaining squares to follow with, in two possible directions and that allow for 29 possible positions and 29! possible arrange ments of the other elements. the square might also be the first or last square which allows for 7 possible conections in two positions with 30! possible arrangements of the other elements:

this is 7*6*2*29*29!+7*2*30! = 2.5258*10^34 possible permutations or 9.59%

note that a center squaer give 13*12*2*29*29!+13*2*30! = 8.689*10^34 or 33% which is much greater than 9.59% therefor using 9.59% as an estimate for all possible squares definitely give a lower boung of possiblities.

9.95% for each square gives a total percentage of (.0959)^32 = 2.679*10^-33

muliplying by 32! gives a mininum of 700 possible solutions.

Note: computations performed on a TI-85 calculator which keeps 32 digit accuracy, therefor error of estimation will be significantly greater thatn error of calcultions.

one such solution is:

a1-b2-c1-a3-b4-e1-d2-c3-a5-b6-g1-f2-e3-d4-c5-a7-b8-h2-g3-f4-e5-d6-c7-d8-h4-g5-f6-e7-f8-h6-g7-h8

note that the sequence parts (e1-d2-c3) (g1-f2-e3-d4-c5) (h2-g3-f4-e5-d6) (h4-g5-f6) can occur in any order: therefore the above arrangement can be used to construct 3!*5!*5!*3! = 518,400 possible solutions.

Well, you've solved it halfway.

oinquarki,

I noticed that you said "board," but you didn't say "chess board."   If there were an odd number of rows and columns, a tour with the restriction I described above could be done (easily).  The only "problem" would involve some fudging on how the board is labeled.  That's one "solution."  Is there another?

Easy.  Just take a regular board, take your bishop and just move it around to every square on the board (a1-a2-a3-a4-a5-a6-a7-a8-b8-b7-b6-b5-b4-b3-b2-b1-c1-c2-c3, and so on)

impossible

OBVIUSLY, this is an April Fools Joke.  Or, since it's May 1st, this is a "May Fools" joke.