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Evolutionary game theory studies how a population's composition changes when the success of a pure strategy depends on which other pure strategies it encounters. The central feedback is a frequency-dependent payoff: becoming common changes the opponents a strategy meets, which can change whether it does well. In a coordination game, common behaviour can reinforce itself; in the coexistence example below, being rare is an advantage. This makes evolutionary game theory useful for asking which behaviours can persist and which initial compositions lead to different outcomes, without assuming that each participant solves an optimization problem.

In the basic model, a large, well-mixed population has finitely many inherited pure strategies; interactions sample opponents according to their population frequencies. Assume that a strategy's expected payoff determines its per capita net reproductive growth rate, with a common baseline allowed. These are modelling assumptions: reproduction must transmit the strategy, encounters must match the proposed sampling rule, and the conversion from payoff to growth must be justified for an application. The replicator equation below describes the resulting continuous-time frequencies. Taylor and Jonker's ^[ref] explicitly derive this growth model.

For learned behaviour, reproduction can instead be replaced by an assumed copying rule: an individual samples another individual uniformly and copies a better-performing pure strategy at a rate proportional to the positive payoff difference. At frequencies $x_i,x_j$, the net flow from $j$ to $i$ is $x_i{x}_j(f_i-f_j)$, because the two directional positive parts subtract to $f_i-f_j$. Summing over $j$ gives $x_i(f_i-\bar{f})$, the replicator equation. This derivation states a particular mechanism; an arbitrary rule for learning or imitation need not give the same dynamical system.

Three questions must be distinguished. A symmetric Nash equilibrium tests whether an individual can improve its expected payoff against the resident composition. An evolutionarily stable strategy tests whether each distinct rare mutant is outperformed after it is introduced. Dynamical stability tests the response of an actual dynamical system to small changes of composition, and asymptotic stability additionally requires nearby trajectories to converge. For a finite symmetric bimatrix game under the continuous-time replicator equation, every evolutionarily stable strategy is a symmetric Nash equilibrium and a locally asymptotically stable rest point; the converse from asymptotic stability to an evolutionarily stable strategy fails in general. See Ross Cressman and Yi Tao, The replicator equation and other game dynamics (2014), Theorems 1–2 and Example 1. These statements refer to stability on the full probability simplex, including perturbations introducing currently absent pure strategies.

A concrete experiment makes the frequency feedback visible. In 2009, Gore, Youk and van Oudenaarden competed two yeast strains: a producer of the enzyme invertase, which enables sucrose hydrolysis, and a mutant lacking the corresponding SUC2 gene. Released sugars can feed neighbours, so avoiding production can be advantageous. They used shaken 5 mL cultures with 5% sucrose at $30{}^{\circ }\mathrm{C}$, diluted daily after 23 hours of growth. Different labelled proteins allowed the two strains to be counted separately by flow cytometry; selective plating periodically checked the counts. The cooperator required externally supplied histidine, so restricting that amino acid imposed an adjustable growth disadvantage. See Snowdrift game dynamics and facultative cheating in yeast, pp. 253 and 255–256, Methods Summary.

Starting with rare producers and, separately, rare nonproducers tested both invasion directions. Under the reported coexistence conditions each rare strain increased; changing histidine or added glucose shifted the eventual composition. Their explanation used nonlinear growth benefits, rather than a fixed pairwise payoff matrix. This motivates testing both rarity regimes before inferring an evolutionarily stable strategy or fitting a replicator equation; it does not identify the illustrative tables below with measured payoffs. See Gore, Youk and van Oudenaarden, Figures 1–3.

An early motivation was understanding why animal contests can involve restraint even when escalation might win an individual encounter. Maynard Smith and Price's ^[ref] introduced the evolutionarily stable strategy in this setting. Taylor and Jonker's ^[ref] then studied explicit continuous and discrete dynamics. Static resistance to invasion and the paths populations follow are related questions, but they require separate definitions.

A frequency-dependent payoff is an expected payoff depending on the population composition $x$. Under random pairwise encounters with a fixed payoff matrix it is the linear function $f_i(x)=(Ax{)}_i$. The relative advantage $f_i(x)-f_j(x)$ can change sign as opponents become more or less common. The replicator equation converts this feedback into changes of frequency; nonlinear frequency-dependent payoffs allow interactions not represented by a fixed pairwise table.

For every real $m\times n$ payoff matrix $A$ of a two-player zero-sum game, the minimax theorem stateswhere ${\Delta }_m,{\Delta }_n$ are the players' probability simplices. Both outer optima are attained. The row player's security level payoff is the largest payoff it can guarantee against all opposing choices; the column player can hold that payoff to the same number $v$, the value of a zero-sum game. For optimal $x^{\ast },y^{\ast }$,These are exactly the Nash equilibrium inequalities under the row-maximizes, column-minimizes convention. Conversely, a Nash equilibrium of this two-player zero-sum game supplies both guarantees. This equality explains why independent worst-case optimization solves the opposing incentives in this special class.

Allowing mixed strategies is essential. For matching pennies, $A=\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$. Restricting both players to pure strategies gives ${\max }_i{\min }_j{A}_{ij}=-1$ and ${\min }_j{\max }_i{A}_{ij}=1$. Allowing mixed strategies closes the gap: each fair mixture guarantees zero, so $v=0$. These guarantees hold even if the opponent knows the mixture's probabilities, provided the opponent does not observe the sampled choice before choosing.

John von Neumann proved the two-player zero-sum game result in Zur Theorie der Gesellschaftsspiele, Mathematische Annalen 100, 295–320 (1928); the original article's footnote 7 explicitly discusses the equality of max–min and min–max. Nash's ^[ref] states that existence of a Nash equilibrium is equivalent to this result in the two-player zero-sum game case, and extends existence to all finite games.

A mixed strategy is a probability distribution over a player's pure strategies. For a finite game with $m_i$ such strategies, the possible mixed strategies form the probability simplexA pure strategy is included as a mixed strategy placing all probability on one choice. The support of a mixed strategy consists of the choices with positive probability. In a sequential game, sampling a complete pure strategy once is the definition; it is not automatically equivalent to fresh randomization at every move.

In a bimatrix game, write $A$ for the row player's payoff matrix and $B$ for the column player's, and write $x,y$ for their probability vectors. Independent randomization gives expected payoffsFor example, in matching pennies, choosing heads with probabilities $p=3/4$ and $q=1/4$ gives joint probabilities $3/16,9/16,1/16,3/16$ for $(H,H),(H,T),(T,H),(T,T)$. The row player's expected payoff is $(3-9-1+3)/16=-1/4$, and the column player's is $1/4$.

Randomization can prevent an opponent exploiting a predictable choice. It does not mean every choice is equally good against every opponent: only best responses belong in the support of a mixed strategy at a Nash equilibrium. A shared random signal can instead produce a lottery over joint action profiles whose joint probabilities do not factor into independent mixed strategies.

A normal-form game specifies each player's strategy set and their payoff for every strategy profile. In a finite two-player game these data form a bimatrix game; if the two payoffs always sum to zero, one payoff matrix describes the matrix game. The representation specifies choices and incentives, without a temporal order of observed moves.

The players are the two chains, each with the set of pure strategies $\{1,2,3,4,5\}$. A strategy profile $(x,y)$ gives the sites of $S$ and $C$. For later use, write the first player's payoff for a general street size asWhen $x<y$, count customers to the left of $(x+y)/2$, assigning value one rather than two to any midpoint customer. This gives $x+y-1$ whether the midpoint is an occupied building or lies between buildings. The other player receives $2k-u(x,y)$. At co-location, every customer's value is split, giving $k$ each.

For $k=5$, the payoff matrix for $S$ isRows and columns are indexed by sites $1,\dots ,5$; the full ordered payoff pair in cell $(x,y)$ is $(M_{xy},10-M_{xy})$. This discrete location game is a constant-sum game.

Player A maximizes the payoff, while B minimizes it. The fourth row is strictly worse for A than the first in every column, so remove $A_4$ by dominated strategy elimination. On the remaining rows, column $B_3$ is strictly better for B than $B_2$, so remove $B_2$. Now $A_3$ is strictly worse than $A_2$ in the two remaining columns. The reduced zero-sum game has payoff matrixwith rows $A_1,A_2$ and columns $B_1,B_3$.

If A uses mixed strategy $(p,1-p)$, the expected payoffs against these columns are $-2+6p$ and $3-8p$. The first increases and the second decreases, so their minimum is maximized where they are equal: $14p=5$. ThusFor an optimality certificate, B uses mixed strategy $(q,0,1-q)$, making A's first two row payoffs $-5+9q$ and $3-5q$. Equality gives $q=4/7$. Against this B strategy the four row payoffs are $(1/7,1/7,-6/7,-6/7)$, while against the displayed A strategy the three column payoffs are $(1/7,13/7,1/7)$. Therefore A guarantees $1/7$ and B holds A to $1/7$, verifying the value of a zero-sum game and the original, unreduced mixed strategy.

In a two-player symmetric finite game, represent mixed strategies by probability vectors and let $e(x,y)=x^{T}Ay$ be the payoff to a player using $x$ against one using $y$. A resident strategy $x^{\ast }$ is an evolutionarily stable strategy if every distinct mutant strategy $y$ has an invasion barrier ${\epsilon }_0(y)>0$ such that, for $0<\epsilon <{\epsilon }_0(y)$,Thus when mutants are sufficiently rare, the resident has strictly larger fitness against the population mixture. The barrier may depend on the mutant; a common barrier is not part of this definition.

For fixed $y\ne x^{\ast }$ putBy bilinearity the resident-minus-mutant fitness difference isIf this is positive for all sufficiently small positive $\epsilon$, taking $\epsilon \downarrow 0$ gives $a_y\ge 0$. If $a_y=0$, strict positivity requires $b_y>0$. Conversely, if $a_y>0$, choose $0<{\epsilon }_0\le a_y/[2(a_y+|b_y|)]$; then the difference is at least $a_y-\epsilon (a_y+|b_y|)>a_y/2>0$. If $a_y=0$ and $b_y>0$, it is $\epsilon b_y>0$ for every $0<\epsilon <1$. This proves the two-condition criterion for evolutionary stability: for every $y\ne x^{\ast }$, eitherorThe weak first comparison is the symmetric Nash equilibrium condition; the second excludes neutrally competitive mutants when that comparison is an equality.

For the Hawk-Dove game, let $x^{\ast }=(1/2,1/2)$ and $y=(p,1-p)$, $0\le p\le 1$. Against the resident, both pure strategies give $1/2$, soDirect multiplication of the payoff matrix givesConsequentlyThe equality case of the two-condition criterion for evolutionary stability therefore applies to every distinct mutant. More explicitly, the fitness difference in a mixed population is $2\epsilon (p-1/2{)}^2$, positive for every $0<\epsilon <1$. Hence $\boxed{(1/2,1/2)\text{ is an evolutionarily stable strategy}}$, with the common invasion barrier one.

For a finite bimatrix game, let $A,B\in {\mathbb{R}}^{r\times s}$ be the two players' payoff matrices. A pair of mixed strategies $(p,q)\in {\Delta }_r\times {\Delta }_s$ is a Nash equilibrium ifHere ${\Delta }_r$ is the probability simplex on the first player's pure actions, and similarly for ${\Delta }_s$. The two payoff matrices need not be negatives of each other. Nash's theorem asserts that every finite two-person game has an equilibrium in mixed strategies.

We prove it using the Brouwer gain-map proof of bimatrix equilibrium. Put $u=p^{T}Aq$, $v=p^{T}Bq$ and define the positive pure-deviation gainsWrite $G={\sum }_i{g}_i$, $H={\sum }_j{h}_j$ and defineEvery coordinate is nonnegative and each player's coordinates sum to one. The map is continuous, since the payoffs and positive-part function are continuous and both denominators are at least one. Thus $T$ is a continuous self-map of the nonempty compact convex set ${\Delta }_r\times {\Delta }_s$. The Brouwer fixed-point theorem supplies a fixed point $(p,q)$.

At that fixed point the first coordinate equations give $g_i=G{p}_i$. Suppose $G>0$. For every $i$ with $p_i>0$, this makes $g_i>0$, so $(Aq{)}_i-u=g_i=G{p}_i$. But averaging the payoff excess over the support givesa contradiction. Therefore $G=0$, and every $g_i$ is zero. Each pure action has payoff at most $u$, so every mixed deviation, being a convex combination of pure actions, has payoff at most $u$ too. The same argument for $h_j=H{q}_j$ gives $H=0$ and excludes every unilateral deviation by the second player. The fixed point is therefore a Nash equilibrium, proving the stated theorem without a zero-sum assumption.

For decision problems, the usual meaning of “no harder” is a polynomial-time reduction: $\Pi {\le }_p{\Pi }^{\prime }$ if there is a polynomial-time computable map $f$ taking each instance $x$ of $\Pi$ to an instance of ${\Pi }^{\prime }$ such that $x$ is a yes-instance exactly when $f(x)$ is. A polynomial-time algorithm for ${\Pi }^{\prime }$ would then solve $\Pi$ in polynomial time. A problem is NP-hard if every problem in NP admits such a reduction to it.

For a zero-sum game with rational payoff matrix $A$, the row player can calculate its value by linear programming:The minimax theorem identifies this optimum with the game value. Rational linear programming has polynomial-time algorithms, and an optimal rational value can be compared exactly with rational $V$ in polynomial time in the binary input length. Thus the decision problem belongs to P. If it were NP-hard, composing the reductions with this algorithm would put every problem in NP in P, contradicting the assumed separation. ThereforeThe finite rational encoding is part of the computational model; arbitrary real inputs without an encoding would not define the same complexity question.

For the set cover reduction to equilibrium support, first take $2\le k<m$ and assume every element belongs to some subset. Suppose rows $i_1,\dots ,i_k$ cover $S$. Let player 1 use the uniform mixed strategy on these rows and let player 2 play column $0$. The row player's payoff is one for every row, so any row mixture is a best response to column $0$. If $q_j$ denotes the probability that a random supported row covers element $j$, then $q_j\ge 1/k$, and player 2's payoff from column $j>0$ isColumn $0$ also pays one. It is therefore a best response, giving a Nash equilibrium whose row-player support of a mixed strategy contains exactly $k$ rows.

Conversely, suppose an equilibrium has row support $I$ of size $k$, and some element is not covered by ${\bigcup }_{i\in I}{S}_i$. For that column $q_j=0$, so player 2 can earn $k/(k-1)>1$. This is the largest possible column payoff, and it is attained exactly at columns corresponding to elements uncovered by $I$. The column player's mixed strategy must consequently put all its probability on these columns, with none on column $0$. Every currently supported row gives player 1 payoff zero. Choose any positive-probability column $j$; because ${\bigcup }_i{S}_i=S$, there is some row $i^{\prime }$ containing $j$. Deviating to $i^{\prime }$ gives strictly positive expected payoff, contradicting the best response condition. Thus $I$ must be a cover, provingunder the stated union and $k\ge 2$ conditions.

The printed formulation omits both conditions. For $k=1$, its payoffs contain division by zero. Without the union condition its claimed implication is false: take $n=2$, $m=3$, $k=2$ and $S_1=S_2=S_3=\{1\}$. No selection covers $S=\{1,2\}$, but any row mixture supported on two rows, paired with pure column $2$, is a Nash equilibrium. The row player always gets zero, and column $2$ is the column player's strictly optimal choice with payoff two. This gives an explicit counterexample to the unrestricted wording.

These restrictions do not weaken the complexity conclusion. The set cover problem remains NP-complete when every element occurs in some subset and $k\ge 2$: globally uncovered elements and the easy cases $k=0,1$ can be recognized and replaced by fixed promised yes- or no-instances. A cover by at most $k<m$ subsets can be padded to exactly $k$. The constructed game has polynomial size, so deciding whether a Nash equilibrium of the specified support size exists is NP-hard. Computing a complete description of all equilibria would resolve this question and hence is at least this difficult. In addition, the equilibrium set may be infinite; a complete output must then describe families rather than list individual points. There is a finite representation by feasible pairs of supports and their associated strategy polytopes, but there are exponentially many support pairs to consider. This reduction concerns locating specified supports or describing all equilibria, rather than finding one unrestricted equilibrium.

Use net monetary gain as payoff and order each player's pure strategies as $1,2,4$. The payoff matrix for the first player isThe second player's matrix is $-A$. Equal choices return both stakes, so the diagonal is zero; when the first player wins, their net gain is the other's stake, and when they lose it is minus their own stake. This is a matrix game, hence a zero-sum game and a normal-form game.

Apply positive affine payoff transformations, if necessary, so both payoff matrices have strictly positive entries. Such transformations preserve best responses and Nash equilibria. The Lemke-Howson algorithm uses unnormalized nonnegative strategy vectors and slack variables satisfyingThus its two polytopes are $\{x\ge 0:Q^{T}x\le 1\}$ and $\{y\ge 0:Py\le 1\}$. A label $i\le m$ occurs when $x_i=0$ or $r_i=0$; a label $m+j$ occurs when $y_j=0$ or $s_j=0$. The complementary pivoting path drops one label from the artificial zero pair and resolves each duplicated label until all labels return.

Terminate at a completely labelled pair other than the artificial zero pair. This is equivalent to complementary slacknessA nonzero completely labelled pair has both vectors nonzero: if $x=0$, then $s=1$ forces $y=0$, and the converse is analogous. Normalize to mixed strategiesThe inequalities $Py\le 1$ imply that every row payoff against $q$ is at most $1/(1^{T}y)$, with equality on every row receiving positive probability in $p$. Thus $p$ is a best response to $q$. Likewise $q$ is a best response to $p$ using $Q^{T}x+s=1$. Hence $(p,q)$ is a Nash equilibrium. Nondegeneracy of a bimatrix game ensures the usual complementary pivoting path has a unique continuation after the label choice.

Row operations multiply each original tableau equation by an invertible matrix. The slack variable block therefore records that matrix and allows the original payoff matrix to be recovered without guessing.

From the first tableau, let $X$ be the coefficient block of $x$ and $S$ that of $s$. Since its original equations were $Q^{T}x+s=1$, we have $Q^T=S^{-1}X$. HeresoFor the second tableau, writing its $y$ and $r$ blocks as $Y,R$, the original equations $Py+r=1$ give $P=R^{-1}Y$. We obtainThus a representative pair of payoff matrices isBoth reconstructed right-hand sides are $1$, as a check on the tableau normalization. Independent transformations $P\mapsto aP+b11^T$, $Q\mapsto cQ+d11^T$, with $a,c>0$, preserve best responses and hence identify the same strategic solution up to positive affine payoff transformations.

Directly,The supports of $x^{\prime }$ and $y^{\prime }$ lie entirely among their respective best-response coordinates, confirming the Nash equilibrium found above. Since this representative is a symmetric bimatrix game, swapping the players' strategies preserves the Nash equilibrium conditions. Explicitly, $P{x}^{\prime }=(3,2,3{)}^T$ is maximized on the support of $y^{\prime }$, and $Q^T{y}^{\prime }=P{y}^{\prime }=(1,3,3{)}^T$ is maximized on the support of $x^{\prime }$. Therefore

In this consecutive-number antisymmetric game, with the row player's number denoted $i$, the payoff matrix isIt is an antisymmetric matrix, so the matrix game has value zero for every $n\ge 3$. For $n=3$,satisfies $Ap=0$ and $p^{T}A=0$, proving optimality for both players.

For any larger $n$, extend the same strategy by zero probabilities:Indeed, the first three coordinates of $Ap$ are zero. For $i=4$, $(Ap{)}_4=-2/4-2/2+1/4=-5/4$; for $i\ge 5$, all three supported numbers are at least two below $i$, so $(Ap{)}_i=-2$. Hence $Ap\le 0$ and, by antisymmetry, $p^{T}A\ge 0$. The column player using $p$ prevents positive payoff, while the row player using $p$ prevents negative payoff, a mixed-strategy optimality certificate for a matrix game. This explicitly checks all the additional strategies rather than assuming the three-number subgame remains optimal.

Represent satisfied and unsatisfied goals by payoffs one and zero. The second player's goal is an exclusive or: it is true precisely when $q\ne r$, independently of $p$. The first goal is false only when all three Boolean variables are false. The resulting payoff matrix is below; each boxed entry marks that player's best response to the opposing pure strategy, including ties:For player 1, choosing $r$ satisfies the goal against every opposing pure strategy. Choosing $\bar{r}$ does equally well except against $\bar{p}\bar{q}$, where it does worse. Thus $r$ is player 1's unique dominant strategy in the weak sense, and there is no strictly dominant strategy. The ties are why most first-player entries in both rows are boxed.

For player 2, either pure strategy with $q=0$ is a best response against $r$, while either with $q=1$ is a best response against $\bar{r}$. No one choice of $q$ works against both rows, so player 2 has no dominant strategy, and there is no dominant-strategy equilibrium. The value of $p$ does not affect player 2's own payoff.

The cells with both entries boxed give exactly four pure strategy Nash equilibria:All four give payoffs $(1,1)$. In particular, the weakly dominated choice $\bar{r}$ can occur in a Nash equilibrium because player 1 is already satisfied there and cannot strictly improve.

The succinct input is the players' variable-control sets, their goal propositional formulas, and a complete candidate Boolean valuation; an exponentially expanded payoff matrix is not part of the input. This representation is essential to candidate equilibrium checking in a Boolean game.

For membership in co-NP, consider the complement: the candidate is not a Nash equilibrium. A complexity certificate consists of a player $i$ and an alternative Boolean valuation of that player's controlled Boolean variables. Evaluate the current goal and the goal after this unilateral deviation, keeping every other player's variables fixed. A profitable unilateral deviation exists precisely when the current goal is false and the new goal is true. The certificate has at most the number of input variables plus the player index, and both propositional formula evaluations take polynomial time. Thus the complement belongs to NP, and the original decision problem belongs to co-NP.

For hardness, start with a propositional formula $F(X)$ and introduce a fresh Boolean variable $z$. Construct a one-player Boolean game in which that player controls $X\cup \{z\}$ and has goalChoose the candidate pure strategy assigning every variable false. Its payoff is zero because $z$ is false, regardless of $F$. If $F$ has a satisfying assignment, changing $X$ to that assignment and $z$ to true is a profitable unilateral deviation. If $F$ is an unsatisfiable Boolean formula, no unilateral deviation can satisfy the goal. ThereforeThe construction increases the formula length by only a constant and writes a candidate of linear length, so it is a polynomial-time many-one reduction. Unsatisfiability is co-NP-complete; combining hardness and membership proves that candidate Nash equilibrium checking for Boolean games is co-NP-complete. If a convention requires at least two players, add a player controlling a fresh unused variable and having the always-true goal; this changes neither direction of the proof.

Measure all payoffs in millions of dollars, in the order (Ann, Bob). Since both players maximize expected money, those numbers are already their expected utility scale.

In the extensive-form game, Ann moves first. Staying out terminates at $(2,6)$. Entering leads to Bob's observed decision between fighting, with terminal payoffs $(1,1)$, and acquiescing, with terminal payoffs $(3,3)$. Each decision node is its own information set, so this is a perfect-information game.

The corresponding normal-form game gives Ann the pure strategies $O$ (stay out) and $E$ (enter), and Bob the complete contingent pure strategies $F$ (fight if entry occurs) and $A$ (acquiesce if entry occurs). The payoff matrix isThe pure strategy Nash equilibria are $(O,F)$ and $(E,A)$. Against $F$, Ann prefers $O$; against $A$, Ann prefers $E$. When Ann stays out, Bob's contingent response does not affect his payoff, so $F$ is a best response there.

There is exactly one subgame perfect equilibrium, $(E,A)$. In the subgame following entry, Bob strictly prefers $3$ to $1$, so backward induction selects acquiescence; Ann then strictly prefers entering for $3$ to staying out for $2$. The threat supporting $(O,F)$ is not optimal in that one-shot subgame.

For infinite repetition, the paper specifies no discount factor. First take the long-run average payoff convention, with all past actions observed. The feasible payoff set is the convex hullAnn can guarantee $2$ in every stage by staying out; Bob can guarantee at least $3$ in every stage by acquiescing whenever entry occurs. These guarantees are also their minmax payoffs: fighting holds Ann's best possible stage payoff to $2$, and entering holds Bob's best possible stage payoff to $3$.

Therefore the sustainable Nash equilibrium average payoffs are exactly the feasible, weakly individually rational payoffs in a repeated game:This is the closed triangle with vertices $(2,3),(2,6),(3,3)$.

Here is a direct sufficiency argument for this particular game. Choose a publicly known sequence of stage outcomes whose frequencies give the target point. Before any observed deviation, follow that sequence. After an Ann deviation, Bob fights forever, so Ann can thereafter obtain at most $2$ per stage. After a Bob deviation, Ann enters forever, so Bob can thereafter obtain at most $3$ per stage. A finite prefix has no effect on long-run average payoff, so no unilateral deviation improves a target with $a\ge 2,b\ge 3$. Rational frequencies can be implemented periodically; arbitrary real frequencies can be implemented by a deterministic sequence with convergent frequencies. Thus even the weak boundary is covered here. In particular, staying out forever and entry followed by acquiescence forever are sustainable; fighting forever is not.

For discounted payoffs $(1-\delta ){\sum }_{t\ge 0}{\delta }^t{u}_i^t$, the folk theorem for repeated games instead gives the usual patience qualification: every feasible point with $a>2,b>3$ can be sustained for all sufficiently large $\delta <1$. It does not classify the whole region for an unspecified fixed $\delta$. The two one-shot Nash equilibrium outcomes remain sustainable for every $\delta$.^[ref]

Finally consider the particular history-dependent pair in the paper. Initially Bob has never acquiesced, so Ann stays out. That preserves the same condition forever, giving payoffs $(2,6)$. The pair is a Nash equilibrium. Holding Bob's strategy fixed, Ann cannot induce an acquiescence: he fights whenever she enters, and his rule never changes unless he has previously acquiesced. Every entry would give her $1$ instead of $2$, so no unilateral deviation helps. Holding Ann's strategy fixed, Bob never reaches a decision node because she stays out; changing his strategy alone cannot create a history in which he actually acquiesces. Hence he also cannot improve.

This direct argument works for both discounted and long-run average payoffs. The outcome gives Ann exactly her minmax payoff and Bob more than his, so it is compatible with the folk theorem for repeated games. A version requiring strict gains above both minmax payoffs is a sufficient condition, and cannot be used to rule out this boundary example.

Relatedness is not required when responding to others yields a sufficient direct benefit, and coordination need not begin as costly altruism. Coordinated behaviour means that individuals adjust the timing, position or type of their actions relative to others. Cooperation concerns benefits provided through an interaction, while evolutionary altruism specifically imposes a direct fitness cost on the actor. Confusing these categories makes unrelated coordination appear harder to explain than it is.

Consider animals travelling together. Matching a neighbour's direction can reduce an individual's predation risk or help it reach food. Both the responder and the neighbour may benefit immediately; no deferred repayment or kin recognition is necessary. Likewise, complementary hunters can each increase their own chance of capture by responding to another's actions. The relevant condition is that the expected direct return to the response exceeds its costs, relative to available alternatives. That condition depends on ecology, not merely on whether companions share genes.

A clear interspecific example is joint hunting by groupers and giant moray eels, which cannot be close genetic relatives. Field observations documented nonrandom associations, grouper recruitment signals and benefits to both partners. Their different hunting modes expose prey that could evade either hunter alone. Each captured prey item is swallowed by its captor, limiting disputes over division of a carcass. The study supports a direct-benefit mechanism; it does not require alternating favours or prove that every recruitment signal evolved specifically in that association.^[ref]

Simple response rules can make individually advantageous actions fit together. Attraction, alignment, avoidance and following informed individuals can generate collective movement without central control. A quorum response makes commitment more likely after enough others act, potentially preventing one mistaken individual from drawing the whole group into danger. Such proximate rules still need an evolutionary explanation: heritable differences in responsiveness must affect expected survival or reproduction in the environments where the rule is expressed.

Experiments with three-spined sticklebacks used moving replica fish to manipulate apparent social information. Larger groups under perceived risk were less readily led by a single replica than solitary fish, and nonlinear response models explained these decisions better than simple proportional copying. Simulations linked these rules to improved decisions under specified assumptions. The experiment establishes a plausible coordination mechanism; actual lifetime fitness gains and the historical selection of its parameters remain separate questions.^[ref]

The strategic problem also depends on the payoff matrix. In a coordination game, joining a hunt can be a best response when others join, while acting alone is better when nobody else participates. Shared cues, recruitment signals and learned conventions can move a group toward the mutually beneficial equilibrium. Here the difficulty is reaching or maintaining compatible choices. In a prisoner's dilemma, by contrast, withholding a contribution pays better against either partner action in a one-shot encounter. An account of coordinated hunting must establish which incentives actually apply rather than assume every group activity is a public-goods dilemma.

When cooperation does carry an immediate cost, direct reciprocity can make future consequences matter. In a deliberately simple repeated donation game, giving costs $c$ and benefits a partner by $b>c$. If encounters continue with probability $w$, perpetual mutual cooperation yields $(b-c)/(1-w)$. Defecting once against a cooperator yields $b$ if it then triggers permanent noncooperation. Cooperation is preferable against that deviation whenThis is a conditional comparison under recognition, reliable observation and the specified retaliation rule. It is not a universal threshold for animal cooperation, nor does it show that a rare cooperative strategy can always invade an uncooperative population. Primary repeated-game modelling and computer tournaments established how such contingent interactions can sustain cooperation under suitable conditions.^[ref]

Other mechanisms need not involve repayment by the same individual. Partner choice can reward useful collaborators through access to future interactions; leaving an unproductive association can impose an opportunity cost on a defector. Reputation can alter future access to different partners. Partner fidelity feedback occurs when helping a partner increases the actor's future benefits because that partner survives or becomes more productive. These mechanisms require appropriate encounter structure, information and opportunities to choose; naming them is not evidence that a particular species uses them.

Evolutionary policing or partner sanctions can change the returns to exploitation. As a mechanistic example involving unrelated partners, soybean experiments prevented a normally fixing rhizobial strain from fixing nitrogen by changing its atmospheric supply. Reduced bacterial reproductive success and oxygen-supply measurements supported a host-sanction interpretation. This shows how conditional resource allocation can favour service provision without kinship, although it is not by itself an example of synchronized animal movement or proof of the historical origin of the interaction.^[ref]

Coordination can consequently evolve through several routes: immediate mutual benefit, stable contingent exchange, partner assortment, or changed incentives imposed by others. Learning can implement or transmit successful rules, while genetic evolution shapes learning capacities and costs. These explanations are compatible with multilevel descriptions of selection, but a claim that groups benefit is insufficient unless the process maintaining the relevant individual strategies is specified. Effective groups of unrelated individuals emerge when responding and contributing are viable strategies in their actual ecological and social setting.

A payoff is the number assigned to a player's outcome at a strategy profile, with larger numbers preferred. In the usual expected utility model, expected payoffs are probability-weighted averages of these numbers. If a payoff is specified as money and a player has preferences involving risk, first apply the relevant utility function; maximizing expected money assumes risk neutrality. In a two-player zero-sum game, one player's payoff is the negative of the other's. General bimatrix games instead require a payoff matrix for each player.

When every entry of a payoff matrix $B$ is positive, its game value $v$ is positive. Setting $x=p/v$ converts the row player's probability vector and guaranteed value into the displayed linear program, whose optimum is $1/v$. Conversely normalizing a feasible $x$ gives a mixed strategy guaranteeing $1/(1^{T}x)$. The dual maximizes $1^{T}y$ under $By\le 1$, $y\ge 0$ and recovers the column strategy by normalization. Adding a constant to every payoff preserves Nash equilibria and lets this positive-payoff formulation handle general finite zero-sum games.