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You, versus Bobby Fisher (Posted on 2008-04-05) Difficulty: 2 of 5

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.

  Submitted by FrankM    
Rating: 3.0000 (1 votes)
Solution: (Hide)

Number the rows and columns 0 through 7 and assume that we start with the pawn in the Jth column, then on the Nth (>1) turn we will be moving the pawn in column XN, where:


XN = J + F3 + .. + F2N-1 mod 8

Now, 1+ F3 = 3 = F4, so

F3 + F5 = F6 –1, etc. It follows that for N > 1

XN = J + F2N – 1 mod 8

But the sequence of Fibonacci numbers mod 8 is given by 1,1,2,3,5,0,5,5,2,7,1,0, repeating itself after twelve terms. It follows that pawn adjustments will be made in columns J, J+2, J+7, J+4, J+6, J+7 (all mod 8); thereafter repeating.

The pawns in any of the above listed columns except column J+7 mod 8, will be activated every 6 turns. They each advance some number of rows F2M before retreating a number of rows F2M+12, i.e. returning to the square from whence they began. Of these, only the pawn in column J+4 will advance far enough to make a capture, taking one of Bobby F’s pawns.

The pawn in the J+7th column, on the other hand, will be activated every third turn. However, as F6 = F12 = 0 mod 8, this pawn never moves.

We determine that Bobby’s concession amounts to a single pawn. So it doesn't matter very much which pawn you start with. Bobby has indeed made a concession, but it is rather less than the lead he handed to Boris by walking out of their early games.

Note: The Fibonacci numbers modulo n have a repetition rate given by the Pisano period. A short, readable introduction can be found on Wikipedia.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionAnswers to the questionsDej Mar2008-04-06 11:16:11
re(2): part 1Charlie2008-04-06 03:01:27
re: part 1brianjn2008-04-05 18:46:48
Solutionpart 1Charlie2008-04-05 14:26:54
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