The original paper wants x to be a vector, A to be a matrix, and b,b' to be vectors. Two obvious methods of solving it are (a) assume Left or Right for each component of Max, solve the resulting linear system, and select out the combinations which match the Left/Right choices; (b) assume a starting guess for the vector x, and iterate, hoping for convergence. I actually did something similar once-upon-a-time, solving 2-checker Backgammon. I assigned upper and lower bounds for the probability of winning, to each Backgammon position. These were initialized to 1.0 and 0.0, and then a routine calculated updated probabilities based on the Backgammon rules. I had to leave out the doubling cube (memory constraint). The loop converged very quickly, with upper & lower probabilities being equal or adjacent. But then I faced the problem of making sense of the data, which I never really solved. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Eugene Salamin Sent: Monday, August 27, 2007 1:57 PM To: math-fun Subject: Re: [math-fun] Toss Up Komi As Neller and Presser say, "There is no known general method for solving equations of the form x = max(Ax+b, A'x+b')." Assuming A, A', b, b' are fixed given numbers, the graph of y = max(Ax+b, A'x+b') is continuous, piecewise linear, and consists of two segments. Its intersection with y = x consists of 0, 1, 2 or points or a continuum, and these are the solutions. Gene ________________________________________________________________________ ____________ Got a little couch potato? Check out fun summer activities for kids. http://search.yahoo.com/search?fr=oni_on_mail&p=summer+activities+for+ki ds&cs=bz _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun