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Fischer Random Chess

The following are readers' comments and ratings for the page Fischer Random Chess.

Date Rating Comment

28 Jun 2001 None Hi, I have worked out a slightly different method of setting up Fischer random chess positions with a single six-sided die. It's fairly easy to memorizem because it follows logically from the positional rules of the game. As far as I can tell it will create all possible positions. Here it is: All die rolls are counted from the left side of the board from white's point of view and apply to remaining empty and "legal" squares only. Because the king must be between both rooks, it can only occupy the central six squares on each side. Roll a die and place the king on one of the six "central" squares. Now place the rooks. Roll a die for the left rook. If the number exceeds the number of squares on the left side of the king, roll again. Repeat for the right rook. If there is only one square to the right or left of the king, skip the rolls and simply place the rook. Now place the Bishops. Place the first bishop based on a die roll. If the roll value exceeds the number of remaining squares, roll again. Place the second bishop in a similar manner counting only the available squares of the opposite color of the already placed bishop. Place the queen with a die roll. If the die number is 4-6 then subtract 3 from its value (to minimize the number of rolls necessary.) Place the two knights on the last two squares. I have yet to study this method in detail to determine if it favors certain positions. A modification of the die roll procedure to minimize re-rolls is as follows: If there are 2-3 "legal" squares for the rooks or the second bishop take the remainder of the die in the "modula" of the number of remaining squares. For example, if there are two legal squares for the left rook, and one rolls a 5, one counts this as a "1", as 1 is the remainder when one divides 5 by 2. If the roll had been a "4" one would count this as a "2". In the case of 3 empty squares, one a "5" would count as a "2". A "6" would count as a "3" and a "4" would count as a "1" (as in the queen roll, which will always have 3). This method will not work without bias when there are 4-6 legal squares remaining, and re-rolls must be employed. However, statistically speaking, fewer rolls will be necessary in such a case anyway. It is possible, though highly improbable, that one might require a very large number of rolls to finally "nail down" a position for the rooks and bishops. But once they are placed, only 1 roll remains. What do you think? Brad Hoehne- Columbus, Ohio.

3 Oct 2000 None I have added a link to the Zillions of Games implementation. Thanks! --D. Howe

30 Sep 2000 None The page is missing a link to the Zillions-of-Games zrf file available elsewhere on your site.

Date	Rating	Comment
28 Jun 2001	None	Hi, I have worked out a slightly different method of setting up Fischer random chess positions with a single six-sided die. It's fairly easy to memorizem because it follows logically from the positional rules of the game. As far as I can tell it will create all possible positions. Here it is: All die rolls are counted from the left side of the board from white's point of view and apply to remaining empty and "legal" squares only. Because the king must be between both rooks, it can only occupy the central six squares on each side. Roll a die and place the king on one of the six "central" squares. Now place the rooks. Roll a die for the left rook. If the number exceeds the number of squares on the left side of the king, roll again. Repeat for the right rook. If there is only one square to the right or left of the king, skip the rolls and simply place the rook. Now place the Bishops. Place the first bishop based on a die roll. If the roll value exceeds the number of remaining squares, roll again. Place the second bishop in a similar manner counting only the available squares of the opposite color of the already placed bishop. Place the queen with a die roll. If the die number is 4-6 then subtract 3 from its value (to minimize the number of rolls necessary.) Place the two knights on the last two squares. I have yet to study this method in detail to determine if it favors certain positions. A modification of the die roll procedure to minimize re-rolls is as follows: If there are 2-3 "legal" squares for the rooks or the second bishop take the remainder of the die in the "modula" of the number of remaining squares. For example, if there are two legal squares for the left rook, and one rolls a 5, one counts this as a "1", as 1 is the remainder when one divides 5 by 2. If the roll had been a "4" one would count this as a "2". In the case of 3 empty squares, one a "5" would count as a "2". A "6" would count as a "3" and a "4" would count as a "1" (as in the queen roll, which will always have 3). This method will not work without bias when there are 4-6 legal squares remaining, and re-rolls must be employed. However, statistically speaking, fewer rolls will be necessary in such a case anyway. It is possible, though highly improbable, that one might require a very large number of rolls to finally "nail down" a position for the rooks and bishops. But once they are placed, only 1 roll remains. What do you think? Brad Hoehne- Columbus, Ohio.
3 Oct 2000	None	I have added a link to the Zillions of Games implementation. Thanks! --D. Howe
30 Sep 2000	None	The page is missing a link to the Zillions-of-Games zrf file available elsewhere on your site.

Last modified: Monday, December 22, 2008

Ratings and Comments for:Fischer Random Chess

Ratings and Comments for:
Fischer Random Chess