Game Theory PDF

# Decision Analysis and Games ## Decision Analysis and Games **(a)** Recommend a course of action for Hank (based on each of the four criteria of decisions under uncertainty). **(b)** Suppose that Hank is more interested in the letter grade he will get. The dividing scores for the passing letter g...

# Decision Analysis and Games ## Decision Analysis and Games **(a)** Recommend a course of action for Hank (based on each of the four criteria of decisions under uncertainty). **(b)** Suppose that Hank is more interested in the letter grade he will get. The dividing scores for the passing letter grades A to D are 90, 80, 70, and 60, respectively. Would this attitude toward grades call for a change in Hank's course of action? ## 2. For the upcoming planting season, Farmer McCoy can plant corn ($a_1$), wheat ($a_2$), or soybeans ($a_3$) or use the land for grazing ($a_4$). The payoffs associated with the different actions are influenced by the amount of rain: heavy rainfall ($s_1$), moderate rainfall ($s_2$), light rainfall ($s_3$), or drought ($s_4$). The payoff matrix (in thousands of dollars) is estimated as | | $s_1$ | $s_2$ | $s_3$ | $s_4$ | |:----|:-----:|:-----:|:-----:|:-----:| | $a_1$ | -20 | 60 | 30 | -5 | | $a_2$ | 40 | 50 | 35 | 0 | | $a_3$ | -50 | 100 | 45 | -10 | | $a_4$ | 12 | 15 | 15 | 10 | Develop a course of action for Farmer McCoy based on each of the four decisions under uncertainty criteria. ## 3. One of N machines must be selected for manufacturing Q units of a specific product. The minimum and maximum demands for the product are $Q$ and $\bar{Q}$, respectively. The total production cost for Q items on machine i involves a fixed cost $K_i$­­ and a variable cost per unit $c_i$, and it is given as $TC_i = K_i + c_iQ$ **(a)** Devise a solution for the problem under each of the four criteria of decisions under uncertainty. **(b)** For 1000 ≤ Q ≤ 4000, solve the problem for the following set of data: | Machine i | $K_i$ ($) | C($) | |:----------|:----------|:-----:| | 1 | 100 | 5 | | 2 | 40 | 12 | | 3 | 150 | 3 | | 4 | 90 | 8 | ## 4. Game Theory Game theory deals with decision situations in which two intelligent opponents with conflicting objectives are vying to outdo one another. Typical examples include launching advertising campaigns for competing products and planning strategies for war battles. In a conflict, each of two players (opponents) has a (finite or infinite) number of alternatives or strategies. Associated with each pair of strategies is the payoff one player receives from the other. Such a situation is known as a two-person zero-sum game, because a gain by one player is an equal loss by the other. This means that we can represent the game in terms of the payoff to one player. Designating the two players as A and B with m and n strategies, respectively, the game is usually presented in terms of the payoff matrix to player A as | | $B_1$ | $B_2$ | … | $B_n$ | |:----|:-----:|:-----:|:--:|:-----:| | $A_1$ | $a_{11}$ | $a_{12}$ | … | $a_{1m}$ | | $A_2$ | $a_{21}$ | $a_{22}$ | … | $a_{2m}$ | | … | … | … | … | … | | $A_m$ | $a_{m1}$ | $a_{m2}$ | … | $a_{mn}$ | The representation indicates that if A uses strategy i and B uses strategy j, the payoff to A is $a_{ij}$, and the payoff to B is $-a_{ij}$. ### Real-Life Application - Ordering Golfers on the Final Day of Ryder Cup Matches On the final day of a golf tournament, two teams compete for the championship. Each team captain must submit a slate (an ordered list of golfers) that determines the matches. For two competing players occupying the same order in their respective slates, it is plausible to assume that there is 50-50 chance that either golfer will win the match. The win-probability increases for a higher-order golfer when matched with a lower-order player. The goal is to develop an analytical procedure that will support or refute the idea of using slates. ### 4.1 Optimal Solution of Two-Person Zero-Sum Games Because games involve a conflict of interest, the basis for the selection of optimal strategies guarantees that neither player is tempted to seek a different strategy because a worse payoff will ensue. These solutions can be in the form of a single pure strategy or several strategies mixed randomly. #### Example 4-1 Two companies, A and B, sell two brands of flu medicine. Company A advertises in radio ($A_1$), television ($A_2$), and newspapers ($A_3$). Company B, in addition to using radio ($B_1$), television ($B_2$), and newspapers ($B_3$), also mails brochures ($B_4$). Depending on the effectiveness of each advertising campaign, one company can capture a portion of the market from the other. The following matrix summarizes the percentage of the market captured or lost by company A. | | $B_1$ | $B_2$ | $B_3$ | $B_4$ | Row min | |:----|:-----:|:-----:|:-----:|:-----:|:--------:| | $A_1$ | 8 | -2 | 9 | -3 | -3 | | $A_2$ | 6 | 5 | 6 | 8 | 5 | | $A_3$ | -2 | 4 | -9 | 5 | -9 | | Column max | 8 | 5 | 9 | 8 | Maximin ↑ | | Minimax ↑ | | | | | | The solution of the game is based on the principle of securing the best of the worst for each player. If Company A selects strategy $A_1$, then regardless of what B does, the worst that can happen is that A loses 3% of the market share to B. This is represented by the minimum value of the entries in row 1. Similarly, with strategy $A_2$, the worst outcome is for A to capture 5% from B, and for strategy $A_3$, the worst outcome is for A to lose 9% to B. These results are listed under row min. To achieve the best of the worst, Company A chooses strategy $A_2$ because it corresponds to the maximin value. Next, for Company B, the given payoff matrix is for A, and B's best of the worst solution is based on the minimax value. The result is that Company B will select strategy $B_2$. The optimal solution of the game calls for selecting strategies $A_2$ and $B_2$, which means that both companies should use television advertising. The payoff will be in favor of company A, because its market share will increase by 5%. In this case, we say that the value of the game is 5% and that A and B are using a pure saddle-point solution. The saddle-point solution precludes the selection of a better strategy by either company. If B moves to another strategy ($B_1$, $B_3$, or $B_4$), Company A can stay with strategy $A_2$, ensuring worse loss for B (6% or 8%). By the same token, A would not seek a different strategy because B can change to $B_3$ to realize a 9% market gain if $A_1$ is used and 3% if $A_3$ is used. The optimal saddle-point solution of a game need not be a pure strategy. Instead, the solution may require mixing two or more strategies randomly, as the following example illustrates. #### Example 4-2 Two players, A and B, play the coin-tossing game. Each player, unbeknownst to the other, chooses a head (H) or a tail (T). Both players would reveal their choices simultaneously. If they match (HH or TT), player A receives $1 from B. Otherwise, A pays B $1. The following payoff matrix for player A gives the row-min and the column-max values corresponding to A's and B's strategies, respectively. | | $B_H$ | $B_T$ | Row min | |:----|:-----:|:-----:|:--------:| | $A_H$ | 1 | -1 | -1 | | $A_T$ | -1 | 1 | -1 | | Column max | 1 | 1 | | The maximin and the minimax values of the games are $1 and $1, respectively, and the game does not have a pure strategy solution because the two values are not equal. Specifically, if player A selects $A_H$, player B can select $B_T$ to receive $1 from A. If this happens, A can move to strategy $A_T$ to reverse the outcome by receiving $1 from B. The constant temptation to switch to another strategy shows that a pure strategy solution is not acceptable. What is needed in this case is for both players to randomly mix their respective pure strategies. The optimal value of the game will then occur somewhere between the maximin and the minimax values of the game--that is, maximin (lower) value ≤ value of the game ≤ minimax (upper) value In the coin-tossing example, the value of the game must lie between -$1 and +$1. ## Problem Set 4A 1. In games (a) and (b) given below, the payoff is for player A. Each game has a pure strategy solution. In each case, determine the strategies that define the saddle point and the value of the game. **(a)** | | $B_1$ | $B_2$ | $B_3$ | $B_4$ | |:----|:-----:|:-----:|:-----:|:-----:| | $A_1$ | 8 | 6 | 2 | 8 | | $A_2$ | 8 | 9 | 4 | 5 | | $A_3$ | 7 | 5 | 3 | 5 | | $A_4$ | 7 | 3 | -9 | 5 | **(b)** | | $B_1$ | $B_2$ | $B_3$ | $B_4$ | |:----|:-----:|:-----:|:-----:|:-----:| | $A_1$ | 4 | -4 | -5 | 6 | | $A_2$ | -3 | -4 | -9 | -2 | | $A_3$ | 6 | 7 | -8 | -9 | | $A_4$ | 6 | 7 | -3 | -9 | 2. In games (a) and (b) below, the payoff is for player A. Determine the values of p and q that will make ($A_2$, $B_2$) a saddle point. **(a)** | | $B_1$ | $B_2$ | $B_3$ | |:----|:-----:|:-----:|:-----:| | $A_1$ | 2 | 4 | 5 | | $A_2$ | p | 5 | 10 | | $A_3$ | 4 | p | 6 | **(b)** | | $B_1$ | $B_2$ | $B_3$ | |:----|:-----:|:-----:|:-----:| | $A_1$ | 1 | 9 | 6 | | $A_2$ | 10 | 7 | q | | $A_3$ | 6 | 2 | 3 | 3. In the games below, the payoff is for player A. Specify the range for the value of the game in each case. **(a)** | | $B_1$ | $B_2$ | $B_3$ | $B_4$ | |:----|:-----:|:-----:|:-----:|:-----:| | $A_1$ | 1 | 9 | 6 | 0 | | $A_2$ | 2 | 3 | 8 | -3 | | $A_3$ | -5 | -2 | 10 | -3 | | $A_4$ | 7 | 4 | -2 | -5 | **(b)** | | $B_1$ | $B_2$ | $B_3$ | $B_4$ | |:----|:-----:|:-----:|:-----:|:-----:| | $A_1$ | -1 | 9 | 6 | 8 | | $A_2$ | -2 | 10 | 4 | 3 | | $A_3$ | 5 | 3 | 0 | 7 | | $A_4$ | -2 | 8 | 4 | 5 | **(c)** | | $B_1$ | $B_2$ | $B_3$ | |:----|:-----:|:-----:|:-----:| | $A_1$ | 3 | 6 | 1 | | $A_2$ | 5 | 2 | 3 | | $A_3$ | 4 | 2 | -5 | **(d)** | | $B_1$ | $B_2$ | $B_3$ | $B_4$ | |:----|:-----:|:-----:|:-----:|:-----:| | $A_1$ | 3 | 7 | 1 | 3 | | $A_2$ | 4 | 8 | 0 | -6 | | $A_3$ | 6 | -9 | -2 | 4 | 4. Two companies promote two competing products. Currently, each product controls 50% of the market. Because of recent improvements in the two products, each company plans to launch an advertising campaign. If neither company advertises, equal market shares will continue. If either company launches a stronger campaign, the other company is certain to lose a proportional percentage of its customers. A survey of the market shows that 50% of potential customers can be reached through television, 30% through newspapers, and 20% through radio. **(a)** Formulate the problem as a two-person zero-sum game, and determine the advertising media for each company. **(b)** Determine a range for the value of the game. Can each company operate with a single pure strategy? ## 5. Let $a_{ij}$ be the (i, j)th element of a payoff matrix with m strategies for player A and n strategies for player B. The payoff is for player. A. Prove that $\max_{i}\min_{j} a_{ij} \le \min_{j}\max_{i} a_{ij}$ ## 4.2 Solution of Mixed Strategy Games Games with mixed strategies can be solved either graphically or by linear programming. The graphical solution is suitable for games with exactly two pure strategies for one or both players. Linear programming, on the hand, can solve any two-person zero-sum game. The graphical method is interesting because it explains the idea of a saddle point pictorially. ### Graphical solution of games We start with the case of (2 × n) games in which player A has two strategies, $A_1$ and $A_2$. | | $y_1$ | $y_2$ | ... | $y_n$ | |:----|:-----:|:-----:|:---:|:-----:| | $x_1$: $A_1$ | $a_{11}$ | $a_{12}$ | ... | $a_{1m}$ | | $1-x_1$: $A_2$ | $a_{21}$ | $a_{22}$ | ... | $a_{2m}$ | Player A mixes strategies $A_1$ and $A_2$ with probabilities $x_1$ and $1-x_1$, $0\le x_1\le 1$. Player B mixes strategies $B_1$, $B_2$, and ... $B_n$ with probabilities $y_1$, $y_2$, ..., and $y_n$, $y_j\ge0$ for j = 1, 2, ..., n, and $y_1 + y_2 + ... + y_n = 1$. In this case, A's expected payoff corresponding to B's jth pure strategy is $(a_{1j} - a_{2j})x_1 + a_{2j}$, j = 1, 2, ..., n Player A seeks the value of $x_1$ that maximizes the minimum expected payoffs--that is, $\max_{x_1} \min_j ((a_{1j} - a_{2j})x_1 + a_{2j})$ #### Example 4-3 Consider the following 2 × 4 game. The payoff is for player A. | | $B_1$ | $B_2$ | $B_3$ | $B_4$ | |:----|:-----:|:-----:|:-----:|:-----:| | $A_1$ | 2 | 2 | 3 | -1 | | $A_2$ | 4 | 3 | 2 | 6 | The game has no pure strategy solution because the maximin and minimax values are not equal. A's expected payoffs corresponding to B's pure strategies are given as | B's pure strategy | A's expected payoff | |:------------------|:----------------------:| | 1 | -2x1 + 4 | | 2 | -x1 + 3 | | 3 | x1 + 2 | | 4 | -7x1+ 6 | The best of the worst solution for B is the minimum point on the upper envelope of the given two lines (you will find it instructive to graph the two lines and identify the upper envelope). This process is equivalent to solving the equation $4y_3 - 1 = -4y_3 + 6$ The solution gives $y_3 = \frac{7}{8}$, which yields the value of the game as $v = 4 \times (\frac{7}{8})-1 = \frac{3}{2}$. The solution of the game calls for player A to mix $A_1$ and $A_2$ with equal probabilities and for player B to mix $B_3$ and $B_4$ with probabilities $\frac{1}{8}$ and $\frac{7}{8}$. (Actually, the game has alternative solutions for B, because the maximin point is determined by more than two lines. Any nonnegative combination of these alternative solutions is also a legitimate solution.) **Remarks.** Games in which player A has m strategies and player B has only two can be treated similarly. The main difference is that we will be plotting B's expected payoff corresponding to A's pure strategies. As a result, we will be seeking the minimax, rather than the maximin, point of the upper envelope of the plotted lines. However, to solve the problem with TORA, it is necessary to express the payoff in terms of the player that has two strategies, multiplying the payoff matrix by -1. ## Problem Set 4B 1. Solve the coin-tossing game of Example 4-2 graphically. 2. Robin travels between two cities and may use one of two routes: Route A is a fast four-lane highway, and route B is a long winding road. She has the habit of driving "superfast." The highway patrol has a limited police force. If the full force is allocated to the route driven by Robin, she is certain to receive a $100 speeding fine. If the force is split 50-50 between the two routes, there is a 50% chance of getting a $100 fine on route A, and only a 30% chance of getting the same fine on route B. Develop a strategy for both Robin and the police patrol. 3. Solve the following games graphically. The payoff is for Player A. **(a)** | | $B_1$ | $B_2$ | $B_3$ | |:----|:-----:|:-----:|:-----:| | $A_1$ | 1 | -3 | 7 | | $A_2$ | 2 | 4 | -6 | **(b)** | | $B_1$ | $B_2$ | |:----|:-----:|:-----:| | $A_1$ | 5 | 8 | | $A_2$ | 6 | 5 | | $A_3$ | 5 | 7 | 4. Consider the following two-person, zero-sum game: | | $B_1$ | $B_2$ | $B_3$ | |:----|:-----:|:-----:|:-----:| | $A_1$ | 5 | 50 | 50 | | $A_2$ | 1 | 1 | 1 | | $A_3$ | 10 | 1 | 10 | **(a)** Verify that the strategies ($\frac{2}{3}$, 0, $\frac{1}{3}$) for A and ($\frac{1}{3}$, $\frac{2}{3}$, 0) for B are optimal, and determine the value of the game. **(b)** Show that the optimal value of the game equals $\frac{3}{3} \sum_{i=1}^3 \sum_{j=1}^3 a_{ij}x_iy_j$ ## Linear programming solution of games Game theory bears a strong relationship to linear programming, in the sense that any two-person zero-sum game can be expressed as a linear program, and vice versa. In fact, G. Dantzig (1963, p. 24) states that J. von Neumann, father of game theory, when first introduced to the simplex method in 1947, immediately recognized this relationship and further pinpointed and stressed the concept of duality in linear programming. This section explains how games are solved by linear programming. Player A's optimal probabilities, $x_1$, $x_2$, ..., $x_m$ and $x_i$ can be determined by solving the following maxirmin problem: $\max_{x_i} \min_{j} (a_{1j}x_1 + a_{2j}x_2 + ... + a_{mj}x_m)$ subject to $\begin{aligned} &x_1 + x_2 + ... + x_m = 1\\ & x_i\ge 0, i = 1,2,..., m \end{aligned}$ Let: $v = \min_{j} (a_{1j}x_1 + a_{2j}x_2 + ... + a_{mj}x_m)$ The equation implies that $\sum_{i=1}^m a_{ij}x_i = v, j = 1, 2, ..., n$ Player A's problem thus can be written as Maximize z = v subject to $\begin{aligned} &v - \sum_{i=1}^m a_{ij}x_i \le 0, j = 1, 2, ..., n\\ &x_1 + x_2 + ... + x_m = 1\\ &x_i \ge 0, i = 1, 2, ..., m\\ &v \text{ unrestricted} \end{aligned}$ Note that the value of the game, $v$, is unrestricted in sign. Player B's optimal strategies, $y_1$, $y_2$, ..., $y_n$ are determined by solving the problem $\min_{y_j}{\max_i} (\sum_{j=1}^n a_{ij}y_j + \sum_{j=1}^n a_{2j}y_j + ... + \sum_{j=1}^n a_{mj}y_j)$ subject to $\begin{aligned} &y_1 + y_2 + ... + y_n = 1\\ &y_j \ge 0, j = 1, 2, ..., n\\ &v \text{ unrestricted} \end{aligned}$ Using a procedure similar to that of player A, B's problem reduces to Minimize w = v subject to $\begin{aligned} &v - \sum_{j=1}^n a_{ij}y_j \le 0, i = 1, 2, ..., m\\ &y_1 + y_2 + ... + y_n = 1\\ &y_j \ge 0, j = 1, 2, ..., n\\ &v \text{ unrestricted} \end{aligned}$ The two problems optimize the same (unrestricted) variable $v$, the value of the game. The reason is that B's problem is the dual of A's problem. This means that the optimal solution of one problem automatically yields the optimal solution of the other. #### Example 4-4 Solve the following game by linear programming. The value of the game, $v$, lies between -2 and 2. | | $B_1$ | $B_2$ | $B_3$ | Row min | |:----|:-----:|:-----:|:-----:|:--------:| | $A_1$ | 3 | -1 | -3 | -3 | | $A_2$ | -2 | 4 | -1 | -2 | | $A_3$ | -5 | -6 | 2 | -6 | | Column max | 3 | 4 | 2 | | ##### Player A's linear program Subject to: $\begin{aligned} & v-3x_1 + 2x_2+5x_3 \le 0\\ &v + x_1 - 4x_2 + 6x_3 \le 0\\ &v + 3x_1 + x_2-2x_3 \le 0 \\ &x_1 + x_2 + x_3 = 1 \\ &x_1, x_2, x_3 \ge 0 \\ & v \text{ unrestricted} \end{aligned}$ The optimum solution is $x_1 = .39$, $x_2 = .31$, $x_3 = .29$, and $v = -0.91$. ##### Player B's linear program Subject to: $\begin{aligned} &v-3y_1 + y_2 +3y_3 \ge 0 \\ &v + 2y_1 - 4y_2 + y_3 \ge 0 \\ &v + 5y_1 + 6y_2 - 2y_3 \ge 0 \\ & y_1 + y_2 + y_3 = 1 \\ & v \text{ unrestricted} \end{aligned}$ The solution yields $y_1 = .32$, $y_2 = .08$, $y_3 = .60$, and $v = -0.91$. ## Problem Set 4C 1. On a picnic outing, 2 two-person teams are playing hide-and-seek. There are four hiding locations (A, B, C, and D), and the two members of the hiding team can hide separately in any two of the four locations. The other team can then search any two locations. The searching team gets a bonus point if they find both members of the hiding team. If they miss both, they lose a point. Otherwise, the outcome is a draw. **(a)** Set up the problem as a two-person zero-sum game. **(b)** Determine the optimal strategy and the value of the game. 2. UA and DU are devising their strategies for the 1994 national championship men's college basketball game. Assessing the strengths of their respective "benches," each coach comes up with four strategies for rotating the players during the game. The ability of each team to score 2-pointers, 3-pointers, and free throws is key to determining the final score of the game. The following table summarizes the net points UA will score per possession as a function of the different strategies available to each team: | | $DU_1$ | $DU_2$ | $DU_3$ | $DU_4$ | |:----|:-----:|:-----:|:-----:|:-----:| | $UA_1$ | 3 | -2 | 1 | 2 | | $UA_2$ | 2 | 3 | -3 | 0 | | $UA_3$ | -1 | 2 | -2 | 2 | | $UA_4$ | -1 | -2 | 4 | 1 | **(a)** Solve the game by linear programming, and determine a strategy for the championship game. **(b)** Based on the given information, which of the two teams is projected to win the championship? **(c)** Suppose that the entire game will have a total of 60 possessions (30 for each team). Predict the expected number of points by which the championship will be won. 3. Colonel Blotto's army is fighting for the control of two strategic locations. Blotto has two regiments and the enemy has three. A location will fall to the army with more regiments. Otherwise, the result of the battle is a draw. **(a)** Formulate the problem as a two-person zero-sum game, and solve by linear programming. **(b)** Which army will win the battle? 4. In the two-player, two-finger Morra game, each player shows one or two fingers, and simultaneously guesses the number of fingers the opponent will show. The player making the correct guess wins an amount equal to the total number of fingers shown. Otherwise, the game is a draw. Set up the problem as a two-person zero-sum game, and solve by linear programming. ## Partial Answers to Selected Problems **Set 1a** 1. Weights for A, B, and C = (.44214, .25184, .30602). **Set 1b** 2. CR>.1 for all matrices except A. (ws, WJ, WM) = (.331, 292, .377). Select Maisa. 4. All matrices are consistent. (WH, Wp) = (.502, .498). Select H. **Set 2a** 2. (a) See Figure 8. (b) EV(corn) = $8250, EV(soybeans) = $250. Select soybeans. 6. (a) See Figure 9. (b) EV(game) = $.025. Do not play the game.   **Set 2b** 2. Let z be the event of having one defective item in a sample of size 5. Answer: P(A|z} = .6097, P{B|z} = .3903. **Set 2c** 1. (a) Expected value = $5, hence there is no advantage. (b) For 0 ≤ x < 10, U(x) = 0, and for x = 10, U(x) = 100. (c) Play the game. **Set 3a** 1. (a) All methods: Study all night (action a). (b) All methods: Select actions az or 13. **Set 4a** 2. (a) Saddle-point solution at (2, 3). Value of game = 4. 3. (a) 2 < v < 4. **Set 4b** 1. Each player should mix strategies 50-50. Value of game = 0. 2. Police Payoff Matrix: | | 100% A | 50% A-50% B | 100% B | |:----|:-------:|:-------------:|:-------:| | A | 100 | 50 | 0 | | B | 0 | 30 | 100 | Strategy for Police: Mix 50-50 strategies 100% A and 100% B. Strategy for Robin: Mix 50-50 strategies A and B. Value of game = $50 (=expected fine paid by Robin). **Set 4c** 1. (a) Payoff matrix for team 1: | | AB | AC | AD | BC | BD | CD | |:----|:--:|:--:|:--:|:--:|:--:|:--:| | AB | 1 | 0 | 0 | 0 | 0 | -1 | | AC | 0 | 1 | 0 | 0 | -1 | 0 | | AD | 0 | 0 | 1 | -1 | 0 | 0 | | BC | 0 | 0 | -1 | 1 | 0 | 0 | | BD | 0 | -1 | 0 | 0 | 1 | 0 | | CD | -1 | 0 | 0 | 0 | 0 | 1 | Optimal strategy for both teams: Mix AB and CD 50-50. Value of the game = 0. 2. (a) (m, n) = (Number of regiments at location 1, No. of regiments at locations 2). Each location has a payoff of 1 if won and -1 if lost. For example, Botto's strategy (1, 1) against the enemy's (0, 3) will win location 1 and lose location 2, with a net payoff of 1+ (-1) = 0. Payoff matrix for Colonel Blotto: | | 3,0 | 2,1 | 1,2 | 0,3 | |:----|:---:|:---:|:---:|:---:| | 2,0 | -1 | -1 | 0 | 0 | | 1,1 | 1 | -1 | -1 | 0 | | 0,2 | 0 | 0 | -1 | -1 |