Mustang Chess problem

One year ago I have found interesting game in the Internet. "Mustang - Chess problem" v 1.01 by Aleksandr Romanenko. http://www.chessmania.narod.ru/chessgame_free.html

At begginng 48 Bishops are hunting for a lonely Knight. You must beat the computer which is playing with a Knight. Less moves - better score. Yours target is stalemated a Knight. The quantity of Bishops all time decreases for one number after every winning game. It is very easy to stalamated lonely Knight with a quantity of Bishops equal to 16 or more. A Lonely Knight probably playing on the squares with a maximal possibilites of moves (Obviously that from two equal squares a Mustang will choose a square which is the closest to the center) for the next turn and then Bishops must be to play same squares by turn to preventing the maximal development of the Knight in the future. Minimum quantity of Bishops for capture of a lonely Knight are equal to 16 Bishops for board 8x8.

Easy winning strategy are: 1.You must located all Bishops at the last or first 2 rank. 2. And it is necessary to attack step by step a Knight keeping two rank building Bishops.

After some efforts I managed to catch a Lonely Knight with 10 Bishops. But already with 9 Bishops it there was a difficult problem. There is a theoretical question - How many the minimum quantity of Bishops for capture lonely Knight is required. And whether is number 10 in this case is minimum for board 8x8 ?