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## Chess and Math Recap

Here is an outline of Dr. Alexey Root’s talk 1/31/09 at the Metroplex Math Circle. Please note that we will accept answers for number 6 (the mobility calculation for the King, Queen, Rook, Bishop, and Knight) in the comment section below since Dr. Root did not have a chance to go over the answer to that problem.

1. Dr. Root began with the Knight’s Tour, based on her knight’s tour lesson plan on pages 62-63 of Children and Chess: A Guide for Educators. She was very impressed that Jeffrey Garrity, a math major from the University of Dallas (Irving, Texas) solved it on his first try.

2. Dr. Root had those who played tournament chess and those who knew en passant stand in the front of the room. Those who didn’t have that chess experience stayed seated. Volunteers counted the number of people in each part of the room which lead into a practical math problem of how to group those in attendance so that each group would have one tournament (or en passant-knowledgeable) player.

3. Everyone sat back down in their new groups, as indicated in 2. above.

4. Dr. Root gave an introduction to the role of domination in graph theory and in chess (quoted from her Science, Math, Checkmate: 32 Chess Activities for Inquiry and Problem Solving, pages 37-38 )

5. Dr. Root taught two chess/math activities. These are domination activities from Science, Math, Checkmate: 32 Chess Activities for Inquiry and Problem Solving (SMC):

A. Covering the Board: Rooks (pp. 38-42)
B. Covering the Board: Kings (pp. 42-45)

BREAK TIME.

6. Dr. Root began the Mobility Lesson from pages 79-80 of her Children and Chess: A Guide for Educators. Mobility also has to do with the concept of dominance (coverage of squares). Students calculated the mobility of each piece from the corner and from the center. For example, a queen on an outside edge square can move to 21 squares, but if she is on a central square she can move to 27 squares. What is her average mobility? What is a rook’s average mobility? And so forth for each piece. The pawn is tricky because it moves one way and captures a different way, so you don’t have to calculate the pawn’s mobility unless you want to. Then figure out how this mobility relates to the traditional values listed for the chessmen: P(pawn)=1, N (knight)=3, B (bishop)=3, R(rook)=5, Q(queen)=9, K(king)=infinite but actually around 3.5-4. The answers to this mobility lesson were still being calculated at 4:10 p.m., when the Metroplex Math Circle wrapped up for the day.  Please post your answers in the comments section below.

### 2 Responses

1. on February 5, 2009 at 9:30 am | Reply Jacob Cordeiro

Here are my solutions:

PAWN (excluding capture, en passant, and post-promotion moves)

Out of the 6×8=48 squares the pawn can move (it can’t touch the first rank, and by the last rank it will have promoted), we have 8 squares (the second rank) with two possible moves, and the other 48-8=40 have one possible move. Thus, our average is $\dfrac{40(1)+8(2)}{48}\approx\boxed{1.1667}$.

KNIGHT

This one is more complicated, as we obtain the following possible number of moves for different positions:
$\begin{tabular}[t]{|c|c|c|c|c|c|c|c|} \hline\\ 2&3&4&4&4&4&3&2\\\hline 3&4&6&6&6&6&4&3\\\hline 4&6&8&8&8&8&6&4\\\hline 4&6&8&8&8&8&6&4\\\hline 4&6&8&8&8&8&6&4\\\hline 4&6&8&8&8&8&6&4\\\hline 3&4&6&6&6&6&4&3\\\hline 2&3&4&4&4&4&3&2\\\hline \end{tabular}$
Thus, we end up with:
$\dfrac{4(2)+8(3)+20(4)+16(6)+16(8)}{64}=\boxed{5.2500}$.

BISHOP

This is another tricky one. We can use systematic counting to obtain another table:
$\begin{tabular}[t]{|c|c|c|c|c|c|c|c|} \hline 7&7&7&7&7&7&7&7\\\hline 7&9&9&9&9&9&9&7\\\hline 7&9&11&11&11&11&9&7\\\hline 7&9&11&13&13&11&9&7\\\hline 7&9&11&13&13&11&9&7\\\hline 7&9&11&11&11&11&9&7\\\hline 7&9&9&9&9&9&9&7\\\hline 7&7&7&7&7&7&7&7\\\hline \end{tabular}$
Thus, we end up with:
$\dfrac{7(28)+9(20)+11(12)+13(4)}{64}=\boxed{11.1875}$.

ROOK

This one is surprisingly easy. No matter where the rook stands, it dominates the seven squares in it’s row, and the seven squares in it’s column, giving an average of $7+7=\boxed{14.0000}$.

QUEEN

The queen “combines” the bishop and the rook. Since the rook’s movement range is ALWAYS 14, we just add an extra 14 to the bishop’s average, ending up with $\boxed{25.1875}$.

KING

In the four corner squares, the king can move to 3 places, 5 places for the 6×4=24 edge squares, and 8 places for the 6×6=36 middle squares. Thus, our average is $\dfrac{3(4)+5(24)+8(36)}{64}=\boxed{6.5625}$.

2. on May 12, 2009 at 2:11 pm | Reply pecaturjogja

Chess players are goood thinkers but not always good students :D