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For a symmetric bimatrix game with payoff matrices $P,P^{\mathsf{T}}$, any nonzero $x$ satisfying these equations gives the symmetric Nash equilibrium $(x/{\sum }_i{x}_i,x/{\sum }_i{x}_i)$. Every positive coordinate attains the same maximal payoff. This one-vector construction differs from a two-vector Lemke-Howson algorithm path, which can return asymmetric equilibria.

Complementary pivoting follows adjacent bases while maintaining all but one label of a complementarity system. The Lemke-Howson algorithm for a bimatrix game follows the duplicated label between two tableaux until it restores the dropped label at a nonzero completely labelled pair. Normalization then gives a Nash equilibrium.

A bimatrix game is nondegenerate if every mixed strategy with $k$ positive coordinates has at most $k$ pure best responses for the other player. This condition prevents ties in the leaving-variable choice along the usual Lemke-Howson algorithm path and allows unique continuation once the dropped label has been selected.

The game is degenerate in the nondegeneracy of a bimatrix game sense. Against the first player's pure strategy $4$, the second player's choices $1$ and $2$ are both best responses, with payoff four. A mixed strategy of support size one therefore has two pure best responses. The usual Lemke-Howson algorithm path needs additional tie handling or perturbation in such a game; the zero-sum game structure gives a simpler direct linear program.

Add five to every entry of the first player's matrix, obtainingThis does not change either player's best responses or Nash equilibria; it raises the game value by five. Since all entries are positive, its value $v_B$ is positive. If $p$ is a row mixed strategy guaranteeing $v_B$, put $x=p/v_B$. Then $B^{T}x\ge 1$ and $1^{T}x=1/v_B$.

Conversely any feasible $x$ has $s=1^{T}x>0$, and $p=x/s$ guarantees payoff $1/s$ in the shifted game. Maximizing that guaranteed payoff is therefore equivalent to minimizing $s$ under $B^{T}x\ge 1$, $x\ge 0$. These are precisely the displayed constraints. The dual program maximizes $1^{T}y$ subject to $By\le 1$, $y\ge 0$; normalizing an optimal $y$ gives the column strategy. This is positive-payoff linear programming for a matrix 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.

In the first tableau the basis is $(x_2,s_5,x_1)$, givingThe second basis is $(r_1,y_6,r_3)$, giving $y=(0,0,1/4)$ and $r=(3/4,0,1/4)$. This pair is not yet a Nash equilibrium: label $1$ is missing and label $4$ is duplicated.

Resolve the duplicate by bringing $y_4$ into the second tableau. Its column is $(3/4,1/4,9/4{)}^T$, so the simplex ratio test givesVariable $r_3$ leaves, givingNow label $3$ is duplicated. Bring $x_3$ into the first tableau; its column is $(0,3,1{)}^T$. The positive-entry ratios are $(2/3)/3=2/9$ and $(1/9)/1=1/9$. Thus $x_1$ leaves, givingAll labels are now present and $x_i{r}_i=y_j{s}_j=0$. Each unnormalized strategy has total mass $1/3$, so the Lemke-Howson algorithm produces

Let $u,v$ be the row and column mixed strategies, so their entries are nonnegative and sum to one. The payoffs are $u^{\mathsf{T}}Pv$ and $u^{\mathsf{T}}{P}^{\mathsf{T}}v$. A Nash equilibrium is a pair in which each strategy is a best response to the other: neither player can improve its expected payoff by a unilateral change of mixed strategy. Equivalently, every positive-probability pure strategy attains that player's maximum pure-strategy payoff against the opposing mixture.

For the complementarity construction of a symmetric Nash equilibrium put $s={\sum }_i{x}_i>0$ and $u=x/s$. The equation $Px+z=1$ gives $Px\le 1$. Nonnegativity and $x^{\mathsf{T}}z=0$ imply $x_i>0\Rightarrow z_i=0\Rightarrow (Px{)}_i=1$. Therefore every used pure strategy has payoff $1/s$ against $u$, while all other pure strategies have payoff at most $1/s$. The same vector of pure-strategy payoffs applies to the column player in this symmetric bimatrix game. Thus a symmetric Nash equilibrium is

For a genuine Lemke-Howson algorithm path, use two separate vectors, $\xi$ for the row player and $\eta$ for the column player, with slacksGive labels $i=1,2,3$ to ${\xi }_i=0$ or $r_i=0$, and labels $3+j$ to ${\eta }_j=0$ or $s_j=0$. A nonzero completely labeled pair has ${\xi }_i{r}_i=0$ and ${\eta }_j{s}_j=0$, and normalization yields mutual best responses. At the artificial pair $(\xi ,\eta )=(0,0)$ all six labels are present.

Drop label $3$ by increasing ${\xi }_3$. The two limiting rows of $P\xi \le 1$ are $3{\xi }_3\le 1$ and $2{\xi }_3\le 1$, so the minimum-ratio pivot stops at ${\xi }_3=1/3$, where $s_2=0$. Label $5$ is now duplicated, since ${\eta }_2=0$ also carries it. Increase ${\eta }_2$ to remove that duplicate. The limiting inequalities are $3{\eta }_2\le 1$ and $4{\eta }_2\le 1$, so the next pivot stops at ${\eta }_2=1/4$, where $r_3=0$ restores the dropped label $3$. The endpoint isIt has all six labels and satisfies both complementarity conditions. After normalization, the Lemke-Howson algorithm returnsAgainst column strategy $2$, row strategy $3$ pays $4$, exceeding $3$ and $0$. Against row strategy $3$, column strategy $2$ pays $3$, exceeding $0$ and $2$. This verifies the endpoint directly.

To find every other Nash equilibrium, observe that row strategy $3$ strictly dominates row strategy $1$: the payoff differences against the three columns are $(2,1,2)$, all positive. By symmetry, column strategy $1$ is also strictly dominated. Neither can appear in an equilibrium. The reduced strategies $2,3$ have row payoff matrixThe two pure Nash equilibria are $(e_2,e_3)$ and $(e_3,e_2)$. If the opponent uses strategy $2$ with probability $q$, the payoffs of strategies $2,3$ are $3(1-q)$ and $2+2q$. Indifference requires $q=1/5$. The same computation applies to the other player. A player mixing both strategies forces this exact opposing mixture; a player playing a pure strategy has a unique opposing best response, producing one of the two pure equilibria. Hence there are exactly three Nash equilibria:For the symmetric mixed equilibrium, the original complementarity construction can use $x=(0,1/12,1/3)$ and $z=(3/4,0,0)$; it gives payoff $12/5$ to each player after normalization.

A square bimatrix game is symmetric when exchanging players exchanges their payoffs: its payoff matrices are $P,Q$ with $Q=P^T$. If $(x,y)$ is a Nash equilibrium, then $(y,x)$ is also a Nash equilibrium. A symmetric bimatrix game can still have asymmetric equilibria, so the two strategy vectors must be kept separate in the Lemke-Howson algorithm.