MATRIX GAMES-DOMINATION

Аннотация

В данной статье анализируются основные принципытеории игр,вчастности, двухличные нулевые игры, стратегии блефа и многоэтапные игры. На основе работ ведущих ученых, таких как Беллман, Блэквелл, Дрешер, Фергюсон, Кун и Нэш, рассматриваются теоретические и практические аспекты принятия стратегических решений. Особое внимание уделяется таким концепциям, как вероятность, точка равновесия (Gleichgewichtspunkt) и доминирование при определении оптимальных page1717 INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE ISSN:2692-5206,ImpactFactor:12,23 AmericanAcademicpublishers,volume05,issue05,2025 Journal: https://www.academicpublishers.org/journals/index.php/ijai стратегий. Исследование раскрывает методы достижения оптимальных результатов в конкурентных икооперативных играх. Ключевые слова: теория игр, нулевые игры, стратегия блефа, точка равновесия, доминирование,многоэтапные игры,принятие стратегических решений. A finite two-player zero-sum game in strategic form (X,Y,L), sometimes called a matrix game because the payoff function L can be represented by a matrix. If X ={ x ,...,x } and 1 m Y ={ y ,...,y }then thegame matrixor payoff matrix isexpressed as: 1 n a L a 11 1n A= M O M , here a = L(x y ), ij i j a L a m1 mn In this form, player 1 chooses a row, player 2 chooses a column, and player 2 pays player 1 the value in the selected row and column. Note: the values in the matrix are considered as the winnings of the player who chose from the row and the losings of the player who chose from thecolumn. Mixed strategy for player 1 p =(p ,p ,...,p ) 1in the form of, probabilities that sum to 1is 1 2 m represented as a set of . If player 1uses a mixed strategy and player 2If he chooses the column, then the(average)payoff for player 1willbe: m pa i ij i=1 Also, the mixed strategy of player 2 q =(q ,q ,...,q ) 2will look like this. If player 2strategy 1 2 n n andplayer 1If hechooses the-row, thewinnings for player 1are: a q ij j j=1 More generally, if player 1strategy and 2nd playerstrategy, then the (average) payoff for player 1is: m n pTAq = pa q i ij j i=1 j=1 It is also worth mentioning that Player 1's pure strategy, i.e., only i -row selection, e ​ unit i vector (only i -position and 0s in the remaining positions). Similarly, player 2's j -column selection e ​ is represented byaunitvector. i In the following sections, we will try to "solve" games, that is, find the value and determine at least one optimal strategy for each player. Sometimes we are also interested in determining all optimalstrategies. Balance points. Sometimes the game is easy to solve. If the matrixa if it has the following ij properties: page1718 INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE ISSN:2692-5206,ImpactFactor:12,23 AmericanAcademicpublishers,volume05,issue05,2025 Journal: https://www.academicpublishers.org/journals/index.php/ijai 1. a -i is thesmallestelement ofthearray, and ij 2. a - j-ifthelargest element in thecolumn is ij Then to us a We call it the equilibrium point. If a if the equilibrium point is 1,then player 1 ij ij i-select row at least a can win the value and player 2 j choose a column and lose a may not ij ij exceed thevalue of.Thus, a The gamewill beworth it . ij Example1. 4 1 -3 A= 3 2 5 0 1 6 In this matrix, the middle element 2 is the equilibrium point, since it is the minimum in its row and the maximum in its column. Therefore, it is optimal for player 1 to choose from the second row and for player 2 to choose from the second column. The cost of the game is 2, and the vector(0,1,0) is theoptimal mixedstrategy forboth players. Large size m n For matrices, checking each element can be tedious. So, as an easier way: calculate the minimum of each row and the maximum of each column and check if there is a match between them. 3 2 1 0 3 1 1 0 0 1 2 0 0 1 2 0 A= and B = 1 0 2 1 1 0 2 1 3 1 2 2 3 2 2 2 Formatrix A:  Rowminimums: 0,0,0,1  Columnmaximums:3,2,2,2 Formatrix B:  Rowminimums: 0,0,0,1  Columnmaximums:3,1,2,2 In matrix A, no row minimum is equal to the column maximum, so there is no equilibrium point. However, if in matrix A a f the element is 1 instead of 2, we get matrix B. In this case, 12 theminimumof the fourth rowwill be equal tothe maximum of thesecond column. So, b The 42 inflection pointis page1719 INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE ISSN:2692-5206,ImpactFactor:12,23 AmericanAcademicpublishers,volume05,issue05,2025 Journal: https://www.academicpublishers.org/journals/index.php/ijai Solutionfor2×2games. Let's generalizeLet's considera 2×2game matrix. a b A= d c To solve this game (i.e., find the value of the game and at least one optimal strategy for each player), weproceed as follows: 1. Checking thebalance point. 2. If an equilibriumpoint does not exist, solvebyfinding equalizingstrategies. Now, we will show how to use the method of equalizing strategies, which works when there is no equilibrium point. Through this method, we can extract the value of the game and the optimalstrategies. Let's