1. The Design & Analysis of the Algorithms Lecture 2. 2011.. by me M. Sakalli Download two pdf files..

2. M, Sakalli, CS246 Design & Analysis of Algorithms, Lecture Notes 1-1 Not about computer games.. But about strategies applied in GT.  Bidding and counter-bidding game, academic studies results that 1$ produces 4-5$s. Cost minimization.  Based on “analysis of conflicts”, the driving force is incentives and self-interest.  Initially it was a mathematical tool, but now it provides very basis of economical games, war games. (non-cooperative games).  We are also part of it with our preferences of what is good over what is bad for. Economics of information.  A large BW, fast process speed, and promising payoff.

3. M, Sakalli, CS246 Design & Analysis of Algorithms, Lecture Notes 1-2 For a game: a. 2 or more "players". b. Each player has 1 or more possible actions. c. And each move has an effect on the "outcome" d. Each move gives a "payoff" through the matrix. Each player’s aim is to maximize his/her outcome - players action expected to be rational. Therefore, a strategy is possible by guessing the possible consecutive moves of the opponents. The degree of rationality for human is not s.th easy to model. ie. due to many social factors, the rationalities of human are not perfect. In computers case, programs, sw agents on behalf of their programmers are players. Game: Write a number between 1-99 and taking the average, A, and winner is the one guessing the closest number to %80 of A. Average will drift down with A*(4/5) n  0.. How fast..

4. M, Sakalli, CS246 Design & Analysis of Algorithms, Lecture Notes 1-3 2 Person Zero Sum Game 2p-zs  2-players whose interests are diametrically?? opposed,  One’s win means the other’s loss,  Always summing to zero.  Suppose Red striker and Blue is a goalie, and suppose the payoff matrix of strategies for Striker and goal keeper.. Blue dives to the RL ….. More columns are possible… -------------------------------- L2 -3 ….. Red M0 2 …. R-510 ….. The values here are set in reference to the Strikers win.. ie striker winning 2 or -5 means gk losing 2 or -5, respectively.

5. M, Sakalli, CS246 Design & Analysis of Algorithms, Lecture Notes 1-4  Players have complete knowledge of payoff matrix and possible moves. Tic-Tac-Toe, Chess, Tennis, Soccer.  Each player must decide which move independently (no promise), and simultaneously (announce at the same time). Striker vs goalie.. Payoff matrix quantizes the moves.. Blue dives to the RLRLRLRL-------------------------------- L2 -3 Red M0 2 R-510 Red  10. Blue anticipates this, and dives to right and -5. But red might try a left strike and wins 2... so on..  From the Red point of the game, row-wise: Max-Min, max of mins.  Fr the Blue point, column-wise: Min-Max, min of mix.

6. M, Sakalli, CS246 Design & Analysis of Algorithms, Lecture Notes 1-5 EValue: 2rl+0rm-5rr-3ll+2ml+10lr  Fair game v=0, red wins v>0, Blue wins v 0, Blue wins v<0, and suppose that l and r are probabilities.. John von Neumann Umpire Game str 2 0-5 -3 210 Goalie R L R LM R LM

7. M, Sakalli, CS246 Design & Analysis of Algorithms, Lecture Notes 1-6 Saddle points and pure strategies Saddle points.. For each row, find the min and then mark the largest of the mins. For each col, find the max; and mark the cols with the smallest of the mins. If the max-min = min-max, then this is a saddle point strategy. There may be more than one saddle points. Determining saddle points can be done efficiently, in linear time. The range between.. Blue dives to the RLRLRLRL-------------------------------- L2* -3 -3 Red M0* 2 0  max-min R-510 -5 min-max *210 min-max *210 No saddle point. Basically, Red’s strategy is to guarantee wining of at least 0, (maximum of minimums) while Blue’s is to ensure the opposite (minimum of (losses) maximums, at most 2).. This range between allows a mixed strategies.

8. M, Sakalli, CS246 Design & Analysis of Algorithms, Lecture Notes 1-7 Mixed strategies Dice..Blue ABABABAB-------------------------------- A 2 -3 A 2 -3 RedB 03 Mixed strategy of Blue. Payoff for Red(A) for p=1, q=1/2  (2 -3) *.5 = -0.5 Payoff for Red(B) for p=0, q=1/2  3 *.5 = 1.5 Payoff for equally randomized p=q=1/2, expected value 2/4=0.5. From Blue’s perspective, not been affected from Red’s str. Red(A) = 2q - 3(1-q) = 5q - 3. Red(B) = + 3(1-q) = 3 - 3q., equalize these two equations q Blues strategy suggests q = 3/4 with a payoff value of 3/4. From Red’s perspective setting columns equal will suggest p = 3/8 with a payoff value of 3/4. The value of the game is 3/4 and optimal choices of Red (3/8, 5/8) and of Blue. (3/4, 1/4)

9. M, Sakalli, CS246 Design & Analysis of Algorithms, Lecture Notes 1-8 John von Neumann 1928, shows that every m*n matrix has a unique solution, the value of the game v, such that of players stick by with optimal strategies (i) Red’s optimal strategy yields her expected payoff is >= v, no matter what Blue does. (i) Expected payoff for Blue’s optimal strategy is, <= v, no matter what Rose does. (ii) More to be presented.