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 You, versus Bobby Fisher (Posted on 2008-04-05)

You are about to play a game of chess with the ghost of Bobby Fisher (at age 24). Bobby has agreed to a concession: In a pre-game round, you may select any one of your pawns and activate it by advancing F2 squares (here F2 = second Fibonacci number = 1). Next, you activate the pawn located F3=2 columns to the right, moving it F4 squares forward, and so on. (Each pawn always stay within their native column.)

If at some point you return to a pawn which was previously moved, you must then move it in the opposite sense. That is, any given pawn will alternatively move forward ±F2M rows (where M is the appropriate turn counter and a negative move corresponds to a move backward).

Whichever of Bobby's pieces (except the king) you land on with one of your pawns are removed from the board. If at some point you were to land on the square occupied by the Bobby's king, or by one of your own pieces, the pawn simply shares that square without a capture/removal.

You may continue this procedure until such time as you are no longer able to capture any further of Bobby's pieces. After these preliminaries, you return your pawns to their normal positions and Bobby will play white against you, using whatever pieces still remain to him. Bobby congratulates himself on having granting you a generous advantage.

How generous was Bobby and which pawn should you activate first?

Clarification and example: Wraparound is used to deal with “off the board” locations. For instance, if the formula requires you to select a pawn one column beyond the right edge of the board, you should revert back to the leftmost column.

An example may help clarify the procedure. After ten times moving pawns in this way, you would next activate the pawn located F21 columns to the right of the most recently moved pawn. Since F21=10946, we determine (using wraparound) that the pawn to activate lies two columns to the right (alternatively, six columns to the left) of the most recently activated pawn. Suppose for the sake of illustration that this pawn is currently positioned in the fifth row, having been moved exactly once previously. We now have to move it back by F22=17711 rows which, using wraparound, would reposition it into the sixth row.

 See The Solution Submitted by FrankM Rating: 3.0000 (1 votes)

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The following table shows the Fibonacci numbers in the left-hand column.  When even Fibonacci numbers are reached, that Fibonacci number mod 8 is shown, as well as the total, mod 8 of the odd Fibonacci numbers thus far.

` 2             2             0 3 5             5             3 8 13            5             3 21 34            2             0 55 89            1             7 144 233           1             7 377 610           2             0 987 1597          5             3 2584 4181          5             3 6765 10946         2             0 17711 28657         1             7 46368 75025         1             7 121393 196418        2             0 317811 514229        5             3 832040 1346269       5             3 2178309 3524578       2             0 5702887 9227465       1             7 14930352 24157817      1             7 39088169 63245986      2             0 102334155 165580141     5             3`

There is a cycle of 6 in the paired Fibonaccis, and therefore in the moves:

• First the pawn in the original column (0) is moved forward 2.
• The one in the column 3 to the right (5 to the left) moves forward 5 and then backward 5.
• The original is then moved back 2 to its starting place.
• The one 7 to the right (1 to the left) of the original is move 1 forward and 1 back.

Only the pawn three columns to the right (or 5 to the left) of the original has any effect, as it knocks out the opposing pawn.

I'll leave it to chess afficionados to figure what pawn they'd most like to remove. Whichever that is, start with the pawn three columns to the left of that (or five to the right).

 Posted by Charlie on 2008-04-05 14:26:54

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