Discrete Colonel Blotto games with two battlefields

Abstract

The Colonel Blotto game is one of the most classical zero-sum games, with diverse applications in auctions, political elections, etc. We consider the discrete two-battlefield Colonel Blotto Game, a basic case that has not yet been completely characterized. We study three scenarios where at least one player’s resources are indivisible (discrete), and compare them with a benchmark scenario where the resources of both players are arbitrarily divisible (continuous). We present the equilibrium values for all three scenarios, and provide a complete equilibrium characterization for the scenario where both players’ resources are indivisible. Our main finding is that, somewhat surprisingly, the distinction between continuous and discrete strategy spaces generally has no effect on players’ equilibrium values. In some special cases, however, the larger continuous strategy space when resources are divisible does bring the corresponding player a higher equilibrium value than when resources are indivisible, and this effect is more significant for the stronger player who possesses more resources than for the weaker player.

Appendices

Proofs in Sect. 4

1.1 Proof of Theorem 2

Proof

(i) When $\Delta =1$ and $B\ge 2$, we show that the following strategy profile is an NE:

$$\begin{aligned}{\mathcal {X}}^*\sim U([A]),{\mathcal {Y}}^*\sim \frac{2}{A+1}I_0+\frac{1}{A+1}\sum _{i=1}^{B-2}I_{i+0.5}+\frac{2}{A+1}I_B.\end{aligned}$$

Please note that when $B=2$, let the middle term $\frac{1}{A+1}\sum _{i=1}^{B-2}I_{i+0.5}=0$ and

$$\begin{aligned} {\mathcal {Y}}^*\sim \frac{2}{A+1}I_0+\frac{2}{A+1}I_B = \frac{1}{2}I_0 +\frac{1}{2}I_2. \end{aligned}$$

On the one hand, for each of Player 2’s pure strategy ${\mathcal {Y}}\in [0, B]$, we discuss in two cases.

• (1) When ${\mathcal {Y}}\in [B]$, i.e., Player 2 takes an integral strategy, there is no way for Player 1 to win both battlefields. The probability that Player 1 gets one win and one draw (when ${\mathcal {X}}^*={\mathcal {Y}}$ or ${\mathcal {X}}^*={\mathcal {Y}}+1$) is $2/(A+1)$, and one win and one loss is $(A-1)/(A+1)$. Therefore, $H({\mathcal {X}}^*,{\mathcal {Y}})=1/(A+1)$.

• (2) When ${\mathcal {Y}}\notin [B]$, i.e., Player 2 takes a fractional strategy, draw is impossible. Player 1 gets two wins with probability $1/(A+1)$ (when ${\mathcal {X}}^*=\lceil {\mathcal {Y}}\rceil$), and one win and one loss with probability $A/(A+1)$. Therefore, $H({\mathcal {X}}^*,{\mathcal {Y}})=1/(A+1)$.

On the other hand, for each of Player 1’s pure strategy ${\mathcal {X}}\in [A]$, we discuss in two cases.

• (1) When ${\mathcal {X}}\in \{0,1,B,B+1\}$, Player 1 gets one win and one draw with probability $2/(A+1)$, and one win and one loss with probability $(A-1)/(A+1)$, hence $H({\mathcal {X}},{\mathcal {Y}}^*)=1/(A+1)$;

• (2) When $1<{\mathcal {X}}<B$, Player 1 gets two wins with probability $1/(A+1)$ (which happens when $\mathcal {Y=X}+0.5$), and one win and one loss with probability $A/(A+1)$, hence $H({\mathcal {X}},{\mathcal {Y}}^*)=1/(A+1)$.

In summary, $H({\mathcal {X}}^*,{\mathcal {Y}})=H({\mathcal {X}},{\mathcal {Y}}^*)=1/(A+1)$ for all ${\mathcal {Y}}\in [0, B]$ and all ${\mathcal {X}}\in [A]$, implying that $({\mathcal {X}}^*,{\mathcal {Y}}^*)$ is indeed an NE.

Moreover, when $\Delta = 1$ and $B=1$, namely, $A=2$ and $B=1$. We show that the following profile consititutes an NE:

$$\begin{aligned} {\mathcal {X}}^* = 1, \quad {\mathcal {Y}}^* \sim \frac{1}{2}I_0 + \frac{1}{2}I_1. \end{aligned}$$

For each of Player 2’s pure strategy ${\mathcal {Y}}\in [0,1]$, we discuss in two cases.

• (1) When ${\mathcal {Y}}\in \{0,1\}$, Player 1 gets one win and one draw, hence $H({\mathcal {X}}^*,{\mathcal {Y}})=1/2$;

• (2) When $0<{\mathcal {Y}}<1$, Player 1 wins both two battlefields, i.e., $H({\mathcal {X}}^*,{\mathcal {Y}})=1$.

For each of Player 1’s pure strategy ${\mathcal {X}}\in [2]$, we discuss in two cases.

• (1) When ${\mathcal {X}}\in \{0,2\}$, Player 1 gets one win and one draw with probability 1/2, and one win and one lose with probability 1/2, hence $H({\mathcal {X}},{\mathcal {Y}}^*)=1/4$;

• (2) When ${\mathcal {X}} = 1$, Player 1 gets one win and one draw, hence $H({\mathcal {X}},{\mathcal {Y}}^*)=1/2$.

In summary, $H({\mathcal {X}}^*,{\mathcal {Y}})\ge 1/2$, for all ${\mathcal {Y}}\in [0, 1]$ and $H({\mathcal {X}},{\mathcal {Y}}^*)\le 1/2$ for all ${\mathcal {X}}\in [2]$, implying that $({\mathcal {X}}^*,{\mathcal {Y}}^*)$ is indeed an NE, bringing a value of 1/2.

(ii) When $\Delta \ge 2$ and $0\le r \le \Delta -2$, the strategy ${\mathcal {X}}^*$ in Corollary 1 and the strategy ${\mathcal {Y}}^*$ in Lemma 4 constitute an NE.

(iii) When $\Delta \ge 2$, $r=\Delta -1$ and $k\ge 2$, we show that the following ${\mathcal {X}}^*$ and ${\mathcal {Y}}^*$ constitute an NE:

$$\begin{aligned}\begin{array}{l}{\mathcal {X}}^*\sim \frac{3}{2(2k+1)}I_{\Delta -1}+\frac{1}{2k+1}\left[ \sum _{i=2}^{k}I_{i\Delta -1}+\sum _{i=1}^{k-1}I_{i\Delta }\right] +\frac{3}{2(2k+1)}I_{k\Delta },\\ {\mathcal {Y}}^*\sim \frac{2}{2k+1}I_0+\frac{1}{2k+1}\left[ \sum _{j=1}^{k-1}I_{j\Delta -0.5}+\sum _{j=1}^{k-2}I_{j\Delta +1}\right] +\frac{2}{2k+1}I_{k\Delta -1}.\end{array}\end{aligned}$$

Recall the formula in Lemma 1:

$$\begin{aligned}H({\mathcal {X}},{\mathcal {Y}})=P({\mathcal {Y}}<{\mathcal {X}}<{\mathcal {Y}}+\Delta )+0.5\left[ P({\mathcal {X}}={\mathcal {Y}})+P({\mathcal {X}}={\mathcal {Y}}+\Delta )\right] .\end{aligned}$$

On the one hand, for each of Player 2’s pure strategy ${\mathcal {Y}}\in [0,B]$, we discuss in seven cases.

• (1) When ${\mathcal {Y}}=0$, it is impossible for ${\mathcal {X}}^*={\mathcal {Y}}$, and hence

  $$\begin{aligned}H({\mathcal {X}}^*,{\mathcal {Y}})= & {} \, P({\mathcal {X}}^*=\Delta -1)+0.5P({\mathcal {X}}^*=\Delta )\\= & {} \, \frac{3}{2(2k+1)}+\frac{1}{2(2k+1)}=\frac{1}{k+1/2}. \end{aligned}$$

  Similarly, when ${\mathcal {Y}}=k\Delta -1$, it is impossible for ${\mathcal {X}}^*={\mathcal {Y}}+\Delta$, and hence

  $$\begin{aligned} H({\mathcal {X}}^*,{\mathcal {Y}})= & {} \, P({\mathcal {X}}^*=k\Delta )+0.5P({\mathcal {X}}^*=k\Delta -1)\\= & {} \, \frac{3}{2(2k+1)}+\frac{1}{2(2k+1)}=\frac{1}{k+1/2}.\end{aligned}$$

• (2) When $0< {\mathcal {Y}}< \Delta -1$,

  $$\begin{aligned}H({\mathcal {X}}^*,{\mathcal {Y}})= & {} \, P({\mathcal {X}}^*=\Delta -1)+P({\mathcal {X}}^*=\Delta )\\= & {} \, \frac{3}{2(2k+1)}+\frac{1}{2k+1}\\> & {} \frac{1}{k+1/2}. \end{aligned}$$

  Similarly, when $(k-1)\Delta<{\mathcal {Y}}<k\Delta -1$,

  $$\begin{aligned} H({\mathcal {X}}^*,{\mathcal {Y}})= & {} \, P({\mathcal {X}}^*=k\Delta )+P({\mathcal {X}}^*=k\Delta -1)\\= & {} \, \frac{5}{2(2k+1)}>\frac{1}{k+1/2}.\end{aligned}$$

• (3) When ${\mathcal {Y}}=\Delta -1$,

  $$\begin{aligned}H({\mathcal {X}}^*,{\mathcal {Y}})= & {} \, P({\mathcal {X}}^*=\Delta )+0.5[P({\mathcal {X}}^*=\Delta -1)+P({\mathcal {X}}^*=2\Delta -1)]\\= & {} \, \frac{1}{2k+1}+\frac{1}{2}\left[ \frac{3}{2(2k+1)}+\frac{1}{2k+1}\right] \\= & {} \, \frac{9}{4(2k+1)}>\frac{1}{k+1/2}.\end{aligned}$$

  Similarly, when ${\mathcal {Y}}=(k-1)\Delta$,

  $$\begin{aligned} H({\mathcal {X}}^*,{\mathcal {Y}})= & {} \, P({\mathcal {X}}^*=k\Delta -1)+0.5[P({\mathcal {X}}^*=(k-1)\Delta )+P({\mathcal {X}}^*=k\Delta )]\\= & {} \, \frac{9}{4(2k+1)}>\frac{1}{k+1/2}.\end{aligned}$$

• (4) When $i\Delta -1<{\mathcal {Y}}<i\Delta$ for some $1\le i\le k-1$,

  $$\begin{aligned}H({\mathcal {X}}^*,{\mathcal {Y}})= & {} \, P({\mathcal {X}}^*=i\Delta )+P({\mathcal {X}}^*=i\Delta -1)\\= & {} \, \frac{1}{2k+1}+\frac{1}{2k+1}\\= & {} \, \frac{1}{k+1/2}.\end{aligned}$$

• (5) When ${\mathcal {Y}}=i\Delta$ for some $1\le i<k-1$,

  $$\begin{aligned}H({\mathcal {X}}^*,{\mathcal {Y}})= & {} \, P({\mathcal {X}}^*=(i+1)\Delta -1)+0.5[P({\mathcal {X}}^*=i\Delta )+P({\mathcal {X}}^*=(i+1)\Delta ]\\= & {} \, \frac{1}{2k+1}+\frac{1}{2}\left[ \frac{1}{2k+1}+\frac{1}{2k+1}\right] \\= & {} \, \frac{1}{k+1/2}.\end{aligned}$$

• (6) When $i\Delta<{\mathcal {Y}}<(i+1)\Delta -1$ for some $1\le i<k-1$,

  $$\begin{aligned}H({\mathcal {X}}^*,{\mathcal {Y}})= & {} \, P({\mathcal {X}}^*=(i+1)\Delta -1)+P({\mathcal {X}}^*=(i+1)\Delta )\\= & {} \, \frac{1}{2k+1}+\frac{1}{2k+1}=\frac{1}{k+1/2}. \end{aligned}$$

• (7) When ${\mathcal {Y}}=i\Delta -1$ for some $1<i\le k-1$,

  $$\begin{aligned}H({\mathcal {X}}^*,{\mathcal {Y}})= & {} \, P({\mathcal {X}}^*=i\Delta )+0.5[P({\mathcal {X}}^*=i\Delta -1)+P({\mathcal {X}}^*=(i+1)\Delta -1]\\= & {} \, \frac{1}{2k+1}+\frac{1}{2}\left[ \frac{1}{2k+1}+\frac{1}{2k+1}\right] =\frac{1}{k+1/2}.\end{aligned}$$

On the other hand, for each of Player 1’s pure strategy ${\mathcal {X}}\in [A]$, we discuss in eight cases.

• (1) When ${\mathcal {X}}=0$,

  $$\begin{aligned}H({\mathcal {X}},{\mathcal {Y}}^*)= & {} \, 0.5P({\mathcal {Y}}^*=0)=\frac{1}{k+1/2}.\end{aligned}$$

  Similarly, when ${\mathcal {X}}=(k+1)\Delta -1$,

  $$\begin{aligned} H({\mathcal {X}},{\mathcal {Y}}^*)= & {} \, 0.5P({\mathcal {Y}}^*=k\Delta -1)=\frac{1}{k+1/2}.\end{aligned}$$

• (2) When $0<{\mathcal {X}}<\Delta$,

  $$\begin{aligned}H({\mathcal {X}},{\mathcal {Y}}^*)= & {} \, P({\mathcal {Y}}^*=0)=1/(k+1/2).\end{aligned}$$

  Similarly, when $k\Delta \le {\mathcal {X}}<(k+1)\Delta -1$,

  $$\begin{aligned} H({\mathcal {X}},{\mathcal {Y}}^*)= & {} \, P({\mathcal {Y}}^*=k\Delta -1)=\frac{1}{k+1/2}.\end{aligned}$$

• (3) When ${\mathcal {X}}=\Delta$,

  $$\begin{aligned}H({\mathcal {X}},{\mathcal {Y}}^*)= & {} \, P({\mathcal {Y}}^*=\Delta -0.5)+0.5P({\mathcal {Y}}^*=0)\\= & {} \, \frac{1}{2k+1}+\frac{1}{2k+1}\\= & {} \, \frac{1}{k+1/2}.\end{aligned}$$

  Similarly, when ${\mathcal {X}}=k\Delta -1$,

  $$\begin{aligned}H({\mathcal {X}},{\mathcal {Y}}^*)= & {} \, P({\mathcal {Y}}^*=(k-1)\Delta -0.5)+0.5P({\mathcal {Y}}^*=k\Delta -1)\\= & {} \, \frac{1}{k+1/2}.\end{aligned}$$

• (4) When ${\mathcal {X}}=\Delta +1$,

  $$\begin{aligned}H({\mathcal {X}},{\mathcal {Y}}^*)= & {} \, P({\mathcal {Y}}^*=\Delta -0.5)+0.5P({\mathcal {Y}}^*=\Delta +1)\\= & {} \, \frac{1}{2k+1}+\frac{1}{2(2k+1)}\\< & {} \frac{1}{k+1/2}.\end{aligned}$$

• (5) When $i\Delta +1<{\mathcal {X}}\le (i+1)\Delta -1$ for some $1\le i <k-1$,

  $$\begin{aligned}H({\mathcal {X}},{\mathcal {Y}}^*)= & {} \, P({\mathcal {Y}}^*=i\Delta -0.5)+P({\mathcal {Y}}^*=i\Delta +1)\\= & {} \, \frac{1}{2k+1}+\frac{1}{2k+1}=\frac{1}{k+1/2}. \end{aligned}$$

• (6) When ${\mathcal {X}}=i\Delta$ for some $1<i\le k-1$,

  $$\begin{aligned}H({\mathcal {X}},{\mathcal {Y}}^*)= & {} \, P({\mathcal {Y}}^*=(i-1)\Delta +1)+P({\mathcal {Y}}^*=i\Delta -0.5)\\= & {} \, \frac{1}{2k+1}+\frac{1}{2k+1}=\frac{1}{k+1/2}.\end{aligned}$$

• (7) When ${\mathcal {X}}=i\Delta +1$ for some $1<i\le k-1$,

  $$\begin{aligned}H({\mathcal {X}},{\mathcal {Y}}^*)= & {} \, P({\mathcal {Y}}^*=(i-1)\Delta -0.5)+0.5[P({\mathcal {Y}}^*=(i-1)\Delta +1)+P({\mathcal {Y}}^*=i\Delta +1)]\\= & {} \, \frac{1}{2k+1}+\frac{1}{2}\left[ \frac{1}{2k+1}+\frac{1}{2k+1}\right] =\frac{1}{k+1/2}.\end{aligned}$$

• (8) When $(k-1)\Delta +1<{\mathcal {X}}<k\Delta -1$,

  $$\begin{aligned}H({\mathcal {X}},{\mathcal {Y}}^*)=P({\mathcal {Y}}^*=\Delta -0.5)=\frac{1}{2k+1}<\frac{1}{k+1/2}.\end{aligned}$$

In summary, $H({\mathcal {X}}^*,{\mathcal {Y}})\ge \frac{1}{k+1/2}$, for all ${\mathcal {Y}}\in [0, B]$ and $H({\mathcal {X}},{\mathcal {Y}}^*)\le \frac{1}{k+1/2}$ for all ${\mathcal {X}}\in [A]$, implying that $({\mathcal {X}}^*,{\mathcal {Y}}^*)$ is indeed an NE, bringing a value of $\frac{1}{k+1/2}$.

When $\Delta \ge 2$, $r=\Delta -1$ and $k=1$, namely, $A = 2\Delta -1$ (or $A = 2B+1$). We show that the unique NE is

$$\begin{aligned} {\mathcal {X}}^*\sim \frac{1}{2}I_{B}+\frac{1}{2}I_{B+1}, \quad {\mathcal {Y}}^*\sim \frac{1}{2}I_{0}+\frac{1}{2}I_{B}. \end{aligned}$$

For each of Player 2’s pure strategy ${\mathcal {Y}}\in [0,B]$, we discuss in two cases.

• (1) When ${\mathcal {Y}}\in \{0,B\}$, Player 1 gets one win and one draw with probability 1/2, and two wins with probability 1/2, hence $H({\mathcal {X}}^*,{\mathcal {Y}})=3/4$;

• (2) When $0<{\mathcal {Y}}<B$, Player 1 wins in both battlefields, i.e., $H({\mathcal {X}}^*,{\mathcal {Y}})=1$.

For each of Player 1’s pure strategy ${\mathcal {X}}\in [A]$, we discuss in two cases.

• (1) When ${\mathcal {X}}\in \{B,B+1\}$, Player 1 gets one win and one draw with probability 1/2, and two wins with probability 1/2, hence $H({\mathcal {X}},{\mathcal {Y}}^*)=3/4$;

• (2) When ${\mathcal {X}} = [A]/\{B,B+1\}$, Player 1 gets one win and one loss, hence $H({\mathcal {X}},{\mathcal {Y}}^*)=0$.

In summary, $H({\mathcal {X}}^*,{\mathcal {Y}})\ge 3/4$, for all ${\mathcal {Y}}\in [0, B]$ and $H({\mathcal {X}},{\mathcal {Y}}^*)\le 3/4$ for all ${\mathcal {X}}\in [A]$, implying that $({\mathcal {X}}^*,{\mathcal {Y}}^*)$ is indeed an NE, bringing a value of 3/4. $\square$

1.2 Proof of Theorem 3

Proof

(i) The payoff matrix $M_{(A+1)\times (B+1)}$ is such that

$$\begin{aligned} m_{ij}= {\left\{ \begin{array}{ll} 0.5 &{} \text{ if }~ i=j~\text{ or }~ i=j+1\\ 0 &{} \text{ otherwise } \end{array}\right. }. \end{aligned}$$

Suppose now $A=2n$ and $B=2n-1$ for some positive integer n. Consider the following strategy profile:

$$\begin{aligned}X^*=\left( 0,\frac{2}{A},0,\frac{2}{A},\ldots \right) , Y^*=\left( \frac{2}{A},0,\frac{2}{A},0,\ldots \right) .\end{aligned}$$

It can be checked that

$$\begin{aligned}X^*M=\left( \frac{1}{A},\frac{1}{A},\ldots ,\frac{1}{A},\frac{1}{A}\right) _{1\times (B+1)},\\MY^{*'}=\left( \frac{1}{A},\frac{1}{A},\ldots ,\frac{1}{A},0\right) '_{1\times (A+1)}.\end{aligned}$$

Therefore, $X^*MY'=1/A$ for all strategy Y of Player 2, and $XMY^{*'}\le 1/A$ for all strategy X of Player 1. This shows that $(X^*,Y^*)$ is indeed an NE for the case $A=2n$ and $B=2n-1$, bringing a value of $1/A=1/k$.

The case $A=2n+1$ and $B=2n$ for some nonnegative integer n can be shown to be valid in a similar way. In fact, it can be checked that the following strategy profile is an NE:

$$\begin{aligned}X^*=\left( 0,\frac{2}{A+1},0,\frac{2}{A+1},\ldots \right) , Y^*=\left( \frac{2}{A+1},0,\frac{2}{A+1},0,\ldots \right) .\end{aligned}$$

bringing a value of $1/(A+1)=1/(k+1)$.

(ii) When $\Delta \ge 2$ and $0\le r\le \Delta -2$, this theorem follows from Corollary 1 and Corollary 2.

(iii) When $\Delta \ge 2$ and $r=\Delta -1$, we can show that the equilibrium strategy profile, in terms of distributions, is

$$\begin{aligned}\begin{array}{l}{\mathcal {X}}^*\sim \frac{1}{k(k+1)}\left[ \sum _{i=1}^{k} \left( (k+1-i)I_{i\Delta -1}+iI_{i\Delta }\right) \right] ,\\ {\mathcal {Y}}^*\sim \frac{1}{k(k+1)}\left[ \sum _{j=1}^{k} \left( (k+1-j)I_{(j-1)\Delta }+jI_{j\Delta -1}\right) \right] .\end{array}\end{aligned}$$

In terms of vectors, the equilibrium is

$$\begin{aligned}\begin{array}{l}X^*=\left( 0,\ldots ,0,\frac{k}{k(k+1)},\frac{1}{k(k+1)},0,\ldots ,0,\frac{1}{k(k+1)},\frac{k}{k(k+1)},0,\ldots ,0\right) _{1\times (A+1)},\\ Y^*=\left( \frac{k}{k(k+1)},0,\ldots ,0,\frac{1}{k(k+1)},\frac{k-1}{k(k+1)},0,\ldots ,0,\frac{k}{k(k+1)}\right) _{1\times (B+1)}.\end{array}\end{aligned}$$

If Player 2 is also restricted to discrete strategies, then the payoff matrix $M_{(A+1)\times (B+1)}$ is

$$\begin{aligned} m_{ij}= {\left\{ \begin{array}{ll} 0.5 &{} \text{ if } i=j \text{ or } i=j+\Delta \\ 1 &{} \text{ if } j<i<j+\Delta \\ 0 &{} \text{ otherwise } \end{array}\right. }. \end{aligned}$$

It can be checked that

$$\begin{aligned}X^*M\ge \frac{1}{2}\left( \frac{1}{k}+\frac{1}{k+1}\right) e,\\MY^{*'} \le \frac{1}{2}\left( \frac{1}{k}+\frac{1}{k+1}\right) e',\end{aligned}$$

where e is a row vector of all ones with a suitable dimension. This proves that the above strategy profile $(X^*,Y^*)$ is indeed an NE. $\square$

Proofs in Sect. 5

Two steps are used to demonstrate the Theorems 4 and 5. First, by removing both players’ redundant strategies, we obtain a simplified game. Then we completely characterize the NEs of the simplified game and offer the conditions to avoid that each player takes the redundant strategies.

1.1 Proof of Theorem 4

Before proving Theorem 4, we provide a useful lemma.

Lemma 6

Suppose that $\Delta =1$ and denote $h=\lceil A/2\rceil$. By removing the redundant strategies, the payoff matrix can be simplified to the following form

$$\begin{aligned} M^* =\begin{bmatrix} M_1&{}&{}&{}\\ &{}M_2&{}&{}\\ &{}&{}\ddots &{}\\ &{}&{}&{}M_{h}\\ \end{bmatrix},\end{aligned}$$

where $M_i=[0.5,0.5]'$ when A is odd, and $M_i=[0.5,0.5]$ when A is even.

Proof

The payoff matrix $M_{(A+1)\times (B+1)}$ is such that

$$\begin{aligned} m_{ij}= {\left\{ \begin{array}{ll} 0.5 &{} \text{ if }~ i=j~\text{ or }~ i=j+1\\ 0 &{} \text{ otherwise } \end{array}\right. }. \end{aligned}$$

We only consider the case that A is odd, as the result can be similarly proved when A is even. According to Theorem 3, the equilibrium value is $\frac{1}{A+1}=\frac{1}{2h}$. Let ${\mathcal {Y}}$ be an equilibrium strategy. Denote $\widehat{w}_i=H(i,{\mathcal {Y}})$ as the payoff of Player 1 when she uses the pure strategy ${\mathcal {X}}=i$ (and Player 2 uses ${\mathcal {Y}}$), then

$$\begin{aligned} {{\widehat{w}}}_i=0.5y_{i-1}+0.5y_{i}\le 1/(2h), \forall i\in [B]\setminus \{0\}. \end{aligned}$$

(5)

For each $t\in [h-2]$, adding up the inequalities (5) over all $i=1,3,\ldots , 2t+1$ and all $i=2t+2,2t+4,\ldots ,B$ and multiplying by 2 on both sides, we have

$$\begin{aligned}2\left( \sum _{s_1=0}^{t}{\widehat{w}}_{2s_1+1}+\sum _{s_2=t+1}^{B/2} {{\widehat{w}}}_{2s_2}\right) =y_{2t+1}+\sum _{i=0}^{B}y_i\le 1.\end{aligned}$$

This implies that $y_{2t+1}=0$ for all $t\in [h-2]$, because $\sum _{i=0}^{B}y_i=1$. Therefore, the payoff matrix can be simplified as $M^*$ by removing all the odd columns of M, as desired. $\square$

Proof of Theorem 4

Suppose A is odd. Following Lemma 6, any even pure strategy for Player 2 (i.e., ${\mathcal {Y}}=2j$, with $j\in [h-1]$) constitutes an equivalence class. Hence, Player 2’s equilibrium strategy is unique.

As for Player 1, all the equivalence classes are: $\{2i,2i+1\}$, $i\in [h-1]$. Therefore $x_{2i}+x_{2i+1}=1/h$ for all $i\in [h-1]$.

Furthermore,

$$\begin{aligned}\frac{1}{2}x_{2i+1}+\frac{1}{2}x_{2i+2}\ge \frac{1}{2h}, \forall i\in [h-2].\end{aligned}$$

Then

$$\begin{aligned}\frac{1}{2}\left( \frac{1}{h}-x_{2i}\right) +\frac{1}{2}x_{2i+2}\ge \frac{1}{2h}, \forall i\in [h-2],\end{aligned}$$

implying that $x_{2i}\le x_{2i+2}, i\in [h-2]$, i.e., the two conditions in the theorem are necessary.

Regarding sufficiency, suppose that Player 1’s strategy ${\mathcal {X}}$ satisfies the two conditions. Then

$$\begin{aligned} XM=\left( \frac{1}{2h}, \frac{1}{2h}+\frac{x_2-x_0}{2}, \frac{1}{2h}, \frac{1}{2h}+\frac{x_4-x_2}{2}, \ldots , \frac{1}{2h}, \frac{1}{2h}+\frac{x_{2h-2}-x_{2h-4}}{2}\right) . \end{aligned}$$

Combining with the monotonicity condition $x_{2i}\le x_{2i+2}, i\in [h-2]$, the inequality system $XM\ge \frac{1}{2\,h} e$ always holds. Hence ${\mathcal {X}}$ is an equilibrium strategy for Player 1. This finishes the complete equilibrium characterizations of both players. The case that A is even can be analogously analyzed using the same method. $\square$

1.2 Proof of Theorem 5

Lemma 7

Suppose that $\Delta \ge 3$ and $1\le r\le \Delta -3$. Then the payoff matrix M can be simplified as

$$\begin{aligned} M^* =\begin{bmatrix} M_1&{}&{}&{}\\ &{}M_2&{}&{}\\ &{}&{}\ddots &{}\\ &{}&{}&{}M_k\\ \end{bmatrix}, \end{aligned}$$

where each submatrix $M_i$, $1\le i\le k$, is of size $(\Delta -r-1)\times (r+1)$ and has all-one entries.

Proof

Recall that the equilibrium value is 1/k. Let ${\mathcal {X}}$ be an equilibrium strategy of Player 1. Then we must have

$$\begin{aligned} w_i=\frac{1}{2}x_i+x_{i+1}+\cdots +x_{i+\Delta -1}+\frac{1}{2}x_{i+\Delta }\ge \frac{1}{k}, \forall i\in [B].\end{aligned}$$

(6)

For each $t\in [r]$, adding up all the k inequalities over $i=t,\Delta +t,\ldots , (k-1)\Delta +t$ in (6) gives

$$\begin{aligned}1\le & {} \sum _{i=0}^{k-1}w_{i\Delta +t}\\= & {} \, \frac{1}{2}x_{t}+x_{1+t}+\cdots +x_{k\Delta -1+t}+\frac{1}{2}x_{k\Delta +t}\\\le & {} \sum _{i=0}^{A}x_i-\frac{1}{2}x_t-\frac{1}{2}x_{k\Delta +t}.\\ \end{aligned}$$

Since $\sum _{i=0}^{A}x_i=1$, we must have

$$\begin{aligned} x_t=x_{k\Delta +t}=0, \forall t\in [r]. \end{aligned}$$

(7)

Furthermore, for each $s\in [k-2]$ and each $t\in [r]{\setminus }\{0\}$, adding up all the k inequalities in (6) gives

$$\begin{aligned} 1\le & {} \sum _{s_1=0}^{s}w_{s_1\Delta }+\sum _{s_2=s+1}^{k-1}w_{s_2\Delta +t}\\= & {} \, \frac{1}{2}x_{0}+x_1+\cdots +\frac{1}{2}x_{(s+1)\Delta } + \frac{1}{2}x_{(s+1)\Delta +t} +x_{(s+1)\Delta +1+t}+\cdots +\frac{1}{2}x_{k\Delta +t}\\\le & {} \sum _{i=0}^{A}x_i-\frac{1}{2}\Big (x_{(s+1)\Delta }+x_{(s+1)\Delta +t}\Big ). \end{aligned}$$

Since $\sum _{i=0}^{A}x_i=1$, we must have

$$\begin{aligned} x_{(s+1)\Delta +t}=0, \forall s\in [k-2], t\in [r]. \end{aligned}$$

(8)

Combining (7) and (8), we know that $\{{\mathcal {X}}=s\Delta +t, \forall s\in [k],t\in [r]\}$ are all the redundant strategies of Player 1. Similarly, for Player 2, $\{{\mathcal {Y}}={s\Delta +t}, \forall r+1\le t\le \Delta -1,s\in [k-2]\}$ are all the redundant strategies.

Combining the above two facts proves the lemma. $\square$

Proof of Theorem 5(i)

For all $s\in [k-1]$, denote $T_s^1=\{i: s\Delta +r+1\le i\le (s+1)\Delta -1\}$. Due to Lemma 7, these sets are all the equivalence classes of Player 1’s strategies, which are also equitable. Therefore condition (1) must be satisfied.

Furthermore, to prevent the redundant strategies being adopted by her opponent, the following restrictions should be added for Player 1:

$$\begin{aligned} w_{s\Delta +r+1}\ge \frac{1}{k}, w_{s\Delta +t}\ge \frac{1}{k}, \forall r+1< t\le \Delta -1,s\in [k-2]. \end{aligned}$$

(9)

Then

$$\begin{aligned}\frac{1}{2}x_{s\Delta +r+1}+\Big (\frac{1}{k}-x_{s\Delta +r+1}\Big )+\frac{1}{2}x_{(s+1)\Delta +r+1}\ge \frac{1}{k}, \forall s\in [k-2],\\\frac{1}{2}x_{s\Delta +t}+\left( \frac{1}{k}-\sum _{z=r+1}^{t}x_{s\Delta +z}\right) +\sum _{z=r+1}^{t-1}x_{(s+1)\Delta +z}+\frac{1}{2}x_{(s+1)\Delta +t}\ge \frac{1}{k}, \forall r+1< t\le \Delta -1,s\in [k-2].\end{aligned}$$

Consequently

$$\begin{aligned}x_{s\Delta +r+1}\le x_{(s+1)\Delta +r+1}, \forall s\in [k-2],\\\sum _{z=r+1}^{t-1}x_{s\Delta +z}+\frac{1}{2}x_{s\Delta +t}\le \sum _{z=r+1}^{t-1}x_{(s+1)\Delta +z}+\frac{1}{2}x_{(s+1)\Delta +t}, \forall r+1< t\le \Delta -1,s\in [k-2],\end{aligned}$$

implying condition (2). The discussion for Player 2 is similar.

On the other hand, suppose that Player 1’s strategy X satisfies conditions (1) and (2) . Denote $(XM)_i$ as the i-th component of XM. Then, due to condition (1) , $(XM)_i=1/k$ for all $i=s\Delta +t$ with $s\in [k-1], t\in [r]$.

As for the other components, combining with the inequalities ( 9 ) and condition (2) , we know that $(XM)_i\ge 1/k$ for all $i=s\Delta +t$ with $s\in [k-2]$ and $r+1\le t\le \Delta -1.$ In summary, the inequality system $XM\ge \frac{1}{k}e$ is valid. Therefore X is Player 1’s equilibrium strategy. The analysis for Player 2 can be similarly done. $\square$

Lemma 8

If $\Delta =2$ or $r=0$, then the payoff matrix M can be simplified as

$$\begin{aligned} M^* =\begin{bmatrix} M_1&{}&{}&{}&{}&{}\\ 0.5&{}0.5&{}&{}&{}&{}\\ &{}M_2&{}&{}&{}&{}\\ &{}0.5&{}0.5&{}&{}&{}\\ &{}&{}\ddots &{}\ddots &{}\ddots &{}\\ &{}&{}&{}&{}&{}M_k\\ \end{bmatrix}, \end{aligned}$$

where each $M_i$ is a $\Delta -1$ dimensional column vector with all-one elements for all $1\le i\le k$.

Proof

Using the same method that is adopted in Lemma 7, we know that

$$\begin{aligned}x_0=x_A=0,\quad y_{s\Delta +t}=0,\forall s\in [k-2], t\in [\Delta -1]\setminus \{0\},\end{aligned}$$

which proves the lemma. $\square$

Proof of Theorem 5(ii)

Suppose $r=0$. For all $s\in [k-1]$, denote $T_s^2=\{s\Delta \}$. Due to Lemma 8, these sets are all the equivalence classes for Player 2, which are also equitable. Therefore Player 2’s equilibrium strategy is unique.

It follows that the strategies in set $T_s^2$ for Player 2 are not redundant. As a consequence, condition (1) must be satisfied. Furthermore, to prevent Player 2’s redundant strategies being adopted, condition (2) should also be satisfied for Player 1.

As for sufficiency, suppose now ${\mathcal {X}}$ and ${\mathcal {Y}}$ satisfy conditions (1) and (2). Using the same method as in the proofs of Theorem 4 and Theorem 5 (i), we have $XM\ge \frac{1}{k}e$ and $MY'\le \frac{1}{k}e'$. As a consequence, ${\mathcal {X}}$ is optimal for Player 1 and ${\mathcal {Y}}$ is optimal for Player 2. Therefore, the theorem is valid. $\square$

Lemma 9

If $r=\Delta -2$, then the payoff matrix M can be simplified as

$$\begin{aligned} M^* =\begin{bmatrix} M_1&{}0.5&{}&{}&{}&{}\\ &{}0.5 &{}M_2&{}0.5&{}&{}\\ &{}&{}\ddots &{}\ddots &{}\ddots &{}\\ &{}&{}&{}&{}0.5&{}M_k\\ \end{bmatrix}, \end{aligned}$$

where $M_i$ is a $\Delta -1$ dimensional row vector with all-one elements.

Proof

Using the same method that is adopted in Lemma 7, we know that

$$\begin{aligned}x_{s\Delta +t}=0, \forall s\in [k], t\in [\Delta -2].\end{aligned}$$

Note that Player 2 does not have any redundant strategies in this case. This completes the proof. $\square$

Proof of Theorem 5(iii)

The proof is similar to that of Theorem 5(ii) and therefore is omitted. $\square$

1.3 Proof of Theorem 6

As in the proofs of Theorems 4 and  5, we first obtain a simplified game as stated in Lemma 10 below. Furthermore, we prove that both players place positive probabilities over the pure strategies in this simplified game. Finally, the uniqueness of the solution can be obtained by applying the elementary theory of linear algebra.

Lemma 10

If $\Delta \ge 2$ and $r=\Delta -1$, then the payoff matrix can be simplified as

$$\begin{aligned} M^* =\begin{bmatrix} 1&{}0.5&{}&{}&{}&{}&{}&{}\\ 0.5&{}1&{}0.5&{}&{}&{}&{}&{}\\ &{}0.5&{}1&{}0.5&{}&{}&{}&{}\\ &{}&{}&{}\ddots &{}\ddots &{}&{}&{}\\ &{}&{}&{}&{}&{}&{}0.5&{}1\\ \end{bmatrix}, \end{aligned}$$

which is a tridiagonal matrix that is positive definite and hence invertible.

Proof

Recall from Theorem 3 that the following profile is a Nash equilibrium:

$$\begin{aligned}\begin{array}{l}{\mathcal {X}}^*\sim \frac{1}{k(k+1)}\left[ \sum _{i=1}^{k} \Big ((k+1-i)I_{i\Delta -1}+iI_{i\Delta }\Big )\right] ,\\ {\mathcal {Y}}^*\sim \frac{1}{k(k+1)}\left[ \sum _{j=1}^{k} \Big ((k+1-j)I_{ (j-1)\Delta }+jI_{j\Delta -1}\Big )\right] .\end{array}\end{aligned}$$

Denote

$$\begin{aligned}\begin{array}{l}{R_1}=\{i\Delta -1, i\Delta : 1\le i\le k\},\\ R_2=\{(j-1)\Delta , j\Delta -1: 1\le j\le k\}.\end{array}\end{aligned}$$

It suffices to show that the supports of all equilibrium strategies of Player 1 and Player 2 are subsets of $R_1$ and $R_2$, respectively. Suppose on the contrary that there is an equilibrium strategy ${\mathcal {Y}}$ for Player 2 whose support contains $i\notin R_2$. Then $({\mathcal {X}}^*,{\mathcal {Y}})$ is also a Nash equilibrium. However, at shown in the proof of Theorem 2, $H({\mathcal {X}}^*,i)$ for $i\notin R_2$ is strictly bigger than the equilibrium value, a contradiction. It can be proved in a similar way that there is no equilibrium strategy ${\mathcal {X}}$ for Player 1 whose support contains $i\notin R_1$.

Since $M^*$ is a tridiagonal matrix, we can get the following recursive formula through the Laplace expansion.

$$\begin{aligned}det(M_{n})=det(M_{n-1})-\frac{1}{2}\times \frac{1}{2}det(M_{n-2}),\end{aligned}$$

where the subscript ($n\ge 2$) indicates the dimension of the matrix. Using the facts that $det(M_1)=1$ and $det(M_2)=3/4$, we have $det(M_n)=\frac{n+1}{2^n}>0.$ Since the cardinalities of $R_1$ and $R_2$ are always even, $M^*$ is invertible. This finishes the proof. $\square$

Proof of Theorem 6

The notation in the proof of Lemma 10 is still valid. Due to Lemma 10, we assume that any equilibrium strategy of Player 1 (resp. Player 2) is a distribution on $R_1$ (resp. $R_2$), and hence is of 2k dimension. We further show that $R_2$ is precisely the support of any equilibrium strategy of Player 2, i.e., all pure strategies in $R_2$ are used in an equilibrium strategy of Player 2.

Suppose on the contrary that there exists an equilibrium strategy Y of Player 2 such that pure strategy $j\in R_2$ is not used. Denote $Y_0$ as the vector obtained by removing the j-th component of Y, and $M^*_0$ the matrix obtained by removing the j-th column of $M^*$. So $Y_0$ is a $(2k-1)$-dimensional row vector and $M_0^*$ is of size $2k\times (2k-1)$.

Since $Y_0$ is an equilibrium strategy of Player 2, we must have

$$\begin{aligned}M_0^*Y_0' \le \alpha e' \equiv \beta ,\end{aligned}$$

where $\alpha$ is the equilibrium value and e is the all-one row vector. Since $X^*$ is an equilibrium strategy of Player 1, the strategy profile $(X^*, Y_0)$ must be an NE. Since it can be checked that $X^*\beta =\alpha$, no entry of $M_0^*Y_0'$ could be strictly smaller than $\alpha$, because otherwise we would have $X^*M_0^*Y_0'<\alpha$, contradicting the fact that $(X^*, Y_0)$ is an NE. This implies that $M_0^*Y_0'=\beta$ and hence the linear equation system $M_0^*Y'=\beta$ has solutions. Denote by $M_1^*$ the matrix obtained by replacing the j-th column of $M^*$ with $\beta$.

Following the fact $M^*Y^*=\beta$ we know that $\beta$ can be linearly represented by column vectors $\{m_1,m_2,\ldots ,m_j,\ldots ,m_{2k}\}$ of $M^*$. By The Substitution Theorem, we know that $rank(M_1^*)=rank(M^*)=2k$. However, $rank(M_0^* )=2k-1<rank(M_1^*)$, implying that $M_0^*Y_0=\beta$ is impossible. A contradiction.

The above contradiction implies that $R_2$ is precisely the support of all equilibrium strategies of Player 2. In a similar way, we can show that $R_1$ is precisely the support of all equilibrium strategies of Player 1. Let X and Y be arbitrary equilibrium strategies of Player 1 and Player 2, respectively. Then

$$\begin{aligned}XM^*=M^*Y'=\beta .\end{aligned}$$

Since $M^*$ is invertible, $(X^*,Y^*)$ is the unique solution. This finishes the proof. $\square$

Extreme point representation

In Sect. 5, the equilibrium strategy sets in cases (i) and (ii) are characterized by some linear inequalities, which form convex polyhedrons (Case (iii) is special, in which both players’ equilibrium strategies are unique). As suggested by an anonymous reviewer, a more desirable characterization might be by specifying their extreme points. In this appendix, we provide some incomplete results.

In this appendix, we denote by ${{\,\mathrm{\textrm{Conv}}\,}}(\textbf{E})$ the convex hull of a finite set $\textbf{E} = \{\textbf{x}_1,\ldots , \textbf{x}_n\}$. That is, ${{\,\mathrm{\textrm{Conv}}\,}}(\textbf{E}) = \left\{ \sum _{i=1}^n\lambda _i\textbf{x}_i \Big | \sum _{i=1}^{n}\lambda _i=1;\lambda _i\ge 0, 1\le i\le n\right\} .$

1.1 Case (i): $\Delta =1$

Recall that $h = \lceil A/2\rceil$. Denote

$$\begin{aligned} {\mathcal {F}}^1_x=\Big \{\textbf{x}= \big (x_{2i}, x_{2i+1}\big )'_{i\in [h-1]} \big | x_{2i}+x_{2i+1}=\frac{1}{h}, x_{2i}\le x_{2i+2}, i\in [h-1] \Big \},\\ {\mathcal {F}}^1_y=\Big \{\textbf{y}= \big (y_{2j}, y_{2j+1}\big )'_{j\in [h-1]} \big | y_{2j}+y_{2j+1}=\frac{1}{h}, y_{2j}\ge y_{2j+2}, j\in [h-1] \Big \}. \end{aligned}$$

Proposition 1

C.1 The extreme points of ${\mathcal {F}}^1_x$ and ${\mathcal {F}}^1_y$ are, respectively,

$$\begin{aligned} \textbf{E}^1_x = \left\{ \textbf{q}_x^w =\frac{1}{h}\times \left( \underbrace{\textbf{e}_2, \ldots , \textbf{e}_2}_{w}, \underbrace{\textbf{e}_1, \ldots , \textbf{e}_1}_{h-w}\right) ' \big | w\in [h]\right\} , \\ \textbf{E}^1_y = \left\{ \textbf{q}_y^w =\frac{1}{h}\times \left( \underbrace{\textbf{e}_1, \ldots , \textbf{e}_1}_{h-w},\underbrace{\textbf{e}_2, \ldots , \textbf{e}_2}_{w}\right) ' \big | w\in [h]\right\} , \end{aligned}$$

where $\textbf{e}_1=(1,0)$ and $\textbf{e}_2=(0,1)$.

Following Proposition 1 and Theorem 4:

• if A is odd, then Player 1’s equilibrium strategies can be rewritten as

  $$\begin{aligned} {\mathcal {X}}^*\sim \sum _{i=0}^{A}x_iI_i: \textbf{x}= \big (x_{2i}, x_{2i+1}\big )'_{i\in [h-1]}\in {{\,\mathrm{\textrm{Conv}}\,}}\left( \textbf{E}^1_x\right) . \end{aligned}$$

• if A is even, then Player 2’s equilibrium strategies can be rewritten as

  $$\begin{aligned} {\mathcal {Y}}^*\sim \sum _{j=0}^{B}y_jI_j: \textbf{y}= \left( y_{2j}, y_{2j+1}\right) '_{j\in [h-1]}\in {{\,\mathrm{\textrm{Conv}}\,}}\left( \textbf{E}^1_y\right) . \end{aligned}$$

Note that if A is odd (resp. even), Player 2’s (resp. Player 1’s) equilibrium strategy is not discussed in the above description, because it is unique as showed in Theorem 4.

1.2 Part 1 of Case (ii): $\Delta \ge 3$ and $1\le r\le \Delta -3$

Denote

$$\begin{aligned}{} & {} {\mathcal {F}}^2_x = \left\{ \textbf{x}= \Big (\{x_{s\Delta +r+t}\}_{t=1}^{\Delta -r -1}, s\in [k-1] \Big )'\right\} , \\{} & {} {\mathcal {F}}^2_y = \left\{ \textbf{y}= \left( \{y_{s\Delta +t}\}_{t=1}^{r}, s\in [k-1] \right) '\right\} , \end{aligned}$$

where all other entries that are not described in $\textbf{x}$ and $\textbf{y}$ are zero.

Suppose that the points in ${\mathcal {F}}^2_x$ satisfy

• (1) $\sum _{t=1}^{\Delta -r-1}x_{s\Delta +r+t}=\frac{1}{k}$ for all $s\in [k-1]$,

• (2) $\left( x_{s\Delta +r+1}\right) _{s=0}^{k-1}$ and $\left( \sum _{z=r+1}^{t-1}x_{s\Delta +z}+0.5x_{s\Delta +t}\right) _{s=0}^{k-1}$ are non-decreasing for all integer $t=r+2,r+3,\ldots , \Delta -1$,

and points in ${\mathcal {F}}^2_y$ satisfy

• (1) $\sum _{t=0}^{r}y_{s\Delta +t}=\frac{1}{k}$ for all $s\in [k-1]$,

• (2) $\left( y_{s\Delta }\right) _{s=0}^{k-1}$ and $\left( \sum _{z=0}^{t-1}y_{s\Delta +z}+0.5y_{s\Delta +t}\right) _{s=0}^{k-1}$ are non-increasing for all integer $t\in [r]\setminus \{0\}$.

Proposition 2

C.2 The extreme points of ${\mathcal {F}}^2_x$ and ${\mathcal {F}}^2_y$ are, respectively,

$$\begin{aligned} \textbf{E}^2_x = \left\{ \frac{1}{k}\left( \textbf{e}_{i_1}, \textbf{e}_{i_2}, \ldots , \textbf{e}_{i_k}\right) ', \quad 1\le i_k \le \cdots \le i_2 \le i_1 \le \Delta -r-1\right\} , \\ \textbf{E}^2_y = \left\{ \frac{1}{k}\left( \textbf{e}_{i_1}, \textbf{e}_{i_2}, \ldots , \textbf{e}_{i_k}\right) ', \quad \Delta -r-1\ge i_k \ge \cdots \ge i_2 \ge i_1 \ge 1\right\} , \end{aligned}$$

where, for all $j = 1,\ldots , k$, the $(i_j)$-th component of the row vector $e_{i_j}\in {\mathbb {R}}^{\Delta -r-1}$ is 1 and all other components are zero.

Following Proposition 2 and Theorem 5, when $\Delta \ge 3$ and $1\le r\le \Delta -3$, Player 1’s equilibrium strategies can be rewritten as

$$\begin{aligned} {\mathcal {X}}^*\sim \sum _{i=0}^{A}x_iI_i: \left( \{x_{s\Delta +r+t}\}_{t=1}^{\Delta -r -1}, s\in [k-1] \right) '\in {{\,\mathrm{\textrm{Conv}}\,}}\left( \textbf{E}_x^2\right) , \end{aligned}$$

and Player 2’s equilibrium strategies can be rewritten as

$$\begin{aligned} {\mathcal {Y}}^*\sim \sum _{j=0}^{B}y_iI_i: \left( \{y_{s\Delta +t}\}_{t=1}^{r}, s\in [k-1] \right) '\in {{\,\mathrm{\textrm{Conv}}\,}}\left( \textbf{E}_y^2\right) . \end{aligned}$$

Note that we did not completely describe the extreme points of the convex polyhedrons in Parts (2) and (3) of Case (ii). They seem to possess very complicated structures, and complete descriptions of them are unlikely to help us better understand the Blotto game.

1.3 Proof of propositions in Appendix C

Proof of Proposition 1

We only discuss $\textbf{E}^1_x$, other cases are similar. Clearly, for any $w\in [h]$, $\textbf{q}_x^w$ belongs to ${\mathcal {F}}^1_x$ and cannot be represented by any other two points of ${\mathcal {F}}^1_x$, because non-zero components of $\textbf{q}_x^w$ all attain the maximum $\frac{1}{h}$.

For any $\textbf{x} = \left( x_{2i}, x_{2i+1}\right) '_{i\in [h-1]} \in {\mathcal {F}}^1_x$,

$$\begin{aligned} \begin{aligned} \textbf{x}&= hx_0\textbf{q}_x^0 +\sum _{l = 1}^{h-1}h(x_{2l}-x_{2l-2})\textbf{q}_x^l + hx_{2h-1}{} \textbf{q}_x^h \\&= \left( \left( x_{2i}, x_{2h-2}+x_{2h-1}-x_{2i}\right) _{i\in [h-2]}, x_{2h-2},x_{2h-1}\right) ' \\&= \left( x_{2i}, x_{2i+1}\right) '_{i\in [h-1]}, \end{aligned} \end{aligned}$$

where the last equality applies $x_{2\,h-2}+x_{2\,h-1}-x_{2i} = \frac{1}{h} - x_{2i} = x_{2i+1}$. Moreover, the coefficients, which are non-negtive, satisfy that $hx_0 + \sum _{l = 1}^{h-1}h(x_{2\,l}-x_{2\,l-2}) + hx_{2\,h-1} = 1.$ That is, every point in ${\mathcal {F}}^1_x$ can be represented by a convex combination of certain $\textbf{q}_x^{w}, \forall w \in [h]$. Therefore, $\textbf{E}^1_x$ contains all the extreme points of ${\mathcal {F}}^1_x$. $\square$

Proof of Proposition 2

We only discuss $\textbf{E}_x^2$. Clearly, every point in $\textbf{E}_x^2$ belongs to ${\mathcal {F}}^2_x$ and cannot be represented by any other two points of ${\mathcal {F}}^2_x$, because non-zero components of every point in $\textbf{E}_x^2$ all attain the maximum $\frac{1}{k}$.

For any $\textbf{x} = \Big (\{x_{s\Delta +r+t}\}_{t=1}^{\Delta -r -1}, s\in [k-1] \Big )' \in {\mathcal {F}}_x$, we demonstrate constructively that $\textbf{x}$ can be represented by a convex combination of certain points in $\textbf{E}_x^2$.

Step 1.

$$\begin{aligned} \textbf{z}_1 = k\sum _{w = 1}^{\Delta - r -2}\left[ x_{r+w}\left( \textbf{e}_w, \underbrace{\textbf{e}_1,\ldots ,\textbf{e}_1}_{k-1}\right) \right] = \left( \left( \{x_{r+t}\}_{t=1}^{\Delta -r -2},0\right) ; \underbrace{\phi _{11}{} \textbf{e}_1;\ldots ; \phi _{11}{} \textbf{e}_1}_{k-1}\right) , \end{aligned}$$

where $\phi _{11} = \sum _{w = 1}^{\Delta - r -2}x_{r+w}$.

Step 2.

$$\begin{aligned} \textbf{z}_2=&\textbf{z}_1 + k(x_{\Delta +r+1}-\phi _{11}) \left( \textbf{e}_{\Delta -r-1},\textbf{e}_1,\underbrace{\textbf{e}_1,\ldots ,\textbf{e}_1}_{k-2}\right) \\&+ k\sum _{w = 2}^{\Delta - r -2}\left[ x_{\Delta +r+w}\left( \textbf{e}_{\Delta -r-1},\textbf{e}_w, \underbrace{\textbf{e}_1,\ldots ,\textbf{e}_1}_{k-2}\right) \right] \\ =&\left( \left( \{x_{r+t}\}_{t=1}^{\Delta -r -2},\phi _{21}\right) ; \left( \{x_{\Delta +r+t}\}_{t=1}^{\Delta -r -2},0\right) ; \underbrace{\phi _{22}{} \textbf{e}_1;\ldots ; \phi _{22}\textbf{e}_1}_{k-2}\right) , \end{aligned}$$

where $\phi _{21} = \phi _{22}-\phi _{11},$ and $\phi _{22} = \sum _{w = 1}^{\Delta - r -2}x_{\Delta +r+w}$.

Step 3.

$$\begin{aligned} \textbf{z}_3 =&\textbf{z}_2+ k\left( x_{2\Delta +r+1}-\phi _{22}\right) \left( \textbf{e}_{\Delta -r-1},\textbf{e}_{\Delta -r-1},\textbf{e}_1,\underbrace{\textbf{e}_1,\ldots ,\textbf{e}_1}_{k-3}\right) \\&+ k\sum _{w = 2}^{\Delta - r -2}\left[ x_{2\Delta +r+w}\left( \textbf{e}_{\Delta -r-1},\textbf{e}_{\Delta -r-1},\textbf{e}_w, \underbrace{\textbf{e}_1,\ldots ,\textbf{e}_1}_{k-3}\right) \right] \\ =&\left( \left( \{x_{r+t}\}_{t=1}^{\Delta -r -2},\phi _{31}\right) ; \left( \{x_{\Delta + r+t}\}_{t=1}^{\Delta -r -2},\phi _{32}\right) ; \left( \{x_{2\Delta +r+t}\}_{t=1}^{\Delta -r -2},0\right) ;\underbrace{\phi _{33}{} \textbf{e}_1;\ldots ; \phi _{33}\textbf{e}_1}_{k-3}\right) , \end{aligned}$$

where $\phi _{31} = \phi _{33}-\phi _{11},$ $\phi _{32} =\phi _{33}-\phi _{22},$ and $\phi _{33} = \sum _{w = 1}^{\Delta - r -2}x_{2\Delta +r+w}$.

Step $k-1$.

$$\begin{aligned} \textbf{z}_{k-1} =&\textbf{z}_{k-2} + k(x_{(k-2)\Delta +r+1}-\phi _{k-2,k-2}) \left( \underbrace{\textbf{e}_{\Delta -r-1}, \ldots ,\textbf{e}_{\Delta -r-1}}_{k-2},\textbf{e}_1,\textbf{e}_1\right) \\&+ k\sum _{w = 2}^{\Delta - r -2}\left[ x_{(k-2)\Delta +r+w}\left( \underbrace{\textbf{e}_{\Delta -r-1}, \ldots ,\textbf{e}_{\Delta -r-1}}_{k-2},\textbf{e}_w,\textbf{e}_1\right) \right] \\ =&\left( \left( \{x_{r+t}\}_{t=1}^{\Delta -r -2},\phi _{k-1,1}\right) ;\ldots ; \left( \{x_{\Delta + r+t}\}_{t=1}^{\Delta -r -2},\phi _{k-1,k-2}\right) ; \right. \\&\left. \left( \{x_{(k-2)\Delta +r+t}\}_{t=1}^{\Delta -r -2},0\right) ;\phi _{k-1,k-1}{} \textbf{e}_1\right) , \end{aligned}$$

where $\phi _{k-1, j} = \phi _{k-1,k-1}-\phi _{j,j}, \forall j= 1,\ldots , k-2$, and $\phi _{k-1, k-1} = \sum _{w = 1}^{\Delta - r -2}x_{(k-2)\Delta +r+w}$.

Step k.

$$\begin{aligned} \textbf{z}_{k} =&\textbf{z}_{k-1} + k(x_{(k-1)\Delta +r+1}-\phi _{k-1,k-1}) \left( \underbrace{\textbf{e}_{\Delta -r-1}, \ldots ,\textbf{e}_{\Delta -r-1}}_{k-1},\textbf{e}_1\right) \\&+ k\sum _{w = 2}^{\Delta - r -1}\left[ x_{(k-1)\Delta +r+w}\left( \underbrace{\textbf{e}_{\Delta -r-1}, \ldots ,\textbf{e}_{\Delta -r-1}}_{k-1},\textbf{e}_w\right) \right] \\ =&\left( \left( \{x_{r+t}\}_{t=1}^{\Delta -r -2},\phi _{k,1}\right) ;\ldots ; \left( \{x_{\Delta + r+t}\}_{t=1}^{\Delta -r -2},\phi _{k,k-2}\right) ; \right. \\&\left. \left( \{x_{(k-2)\Delta +r+t}\}_{t=1}^{\Delta -r -2},\phi _{k,k-1} \right) ; \left( \{x_{(k-1)\Delta +r+t}\}_{t=1}^{\Delta -r -1}\right) \right) , \end{aligned}$$

where $\phi _{k, j} = \sum _{w = 1}^{\Delta - r -1}x_{(k-1)\Delta +r+w}-\phi _{j,j}, \forall j= 1,\ldots , k-1$.

Moreover, all the coefficients satisfy

$$\begin{aligned}&\quad k\sum _{w = 1}^{\Delta - r -2}x_{r+w} + k\sum _{s=1}^{k-1}\left[ (x_{s\Delta + r + 1} - \phi _{s,s})+\sum _{w = 2}^{\Delta - r -2}x_{s\Delta +r+w}\right] + kx_{k\Delta -1} \\ =&\quad k\sum _{w = 1}^{\Delta - r -2}x_{r+w} +k \sum _{s=1}^{k-1}\left[ \sum _{w = 1}^{\Delta - r -2}x_{s\Delta +r+w}-\phi _{s,s}\right] + kx_{k\Delta -1} \\ =&\quad k\phi _{11} + k \sum _{s=1}^{k-2}\left( \phi _{s+1,s+1}- \phi _{s,s}\right) -\phi _{k-1,k-1}+ k\sum _{w = 1}^{\Delta - r -2}x_{(k-1)\Delta +r+w} + kx_{k\Delta -1} \\ =&\quad k\sum _{w = 1}^{\Delta - r -2}x_{(k-1)\Delta +r+w}+ kx_{k\Delta -1} \\ =&\quad k\sum _{w = 1}^{\Delta - r -1}x_{(k-1)\Delta +r+w}\\ =&\quad 1. \end{aligned}$$

Since $\textbf{x}\in {\mathcal {F}}_x$ and $\sum _{t=1}^{\Delta -r-1}x_{s\Delta +r+t}=\frac{1}{k}, \forall s\in [k-1]$, we know that $\phi _{j,j}=\frac{1}{k}- x_{j\Delta -1}$ and $\phi _{k,j}= x_{j\Delta -1}$ for all $j = 1,\ldots , k-1$. Thus, the above k steps construct a convex combination representation of each $\textbf{x}=\textbf{z}_k$ by certain points in $\textbf{E}^2_x$.

In summary, every point in ${\mathcal {F}}^2_x$ can be represented by a convex combination of certain points in $\textbf{E}_x^2$, implying that $\textbf{E}^2_x$ contains all the extreme points of ${\mathcal {F}}^2_x$. $\square$

Note that the above representation relies on the condition $\phi _{s,s} \le x_{s\Delta +r+1}, \forall s =1,\ldots , k-1$. When this condition does not hold, some coefficients might be negative. However, this condition is not restrictive, because by rearranging the order of the k steps, a similar method is applicable. We illustrate via the following example how this works.

Suppose that $A = 11$ and $B = 6$. The points in ${\mathcal {F}}_x^2$ are characterized by

$$\begin{aligned}&x_2+x_3+x_4 = \frac{1}{2}; \quad x_7+x_8+x_9 = \frac{1}{2}; \\&x_2- x_7 \le 0; \quad x_2+\frac{1}{2}x_3- x_7-\frac{1}{2}x_8 \le 0; \\&x_2+x_3+\frac{1}{2}x_4- x_7-x_8-\frac{1}{2}x_9\le 0. \end{aligned}$$

According to Proposition 2, the extreme points of ${\mathcal {F}}_x^2$ are as follows:

$$\begin{aligned} \textbf{q}_0&= \left( \frac{1}{2}, 0, 0, \frac{1}{2}, 0, 0\right) ' = \frac{1}{2}(\textbf{e}_1,\textbf{e}_1)', \quad \textbf{q}_1 = \left( 0,\frac{1}{2}, 0, \frac{1}{2}, 0, 0\right) '= \frac{1}{2}\left( \textbf{e}_2,\textbf{e}_1\right) ',\\ \textbf{q}_2&= \left( 0, 0,\frac{1}{2}, \frac{1}{2}, 0, 0\right) '= \frac{1}{2}\left( \textbf{e}_3,\textbf{e}_1\right) ', \quad \textbf{q}_3 = \left( 0,\frac{1}{2}, 0, 0,\frac{1}{2}, 0\right) '= \frac{1}{2}\left( \textbf{e}_2,\textbf{e}_2\right) ',\\ \textbf{q}_4&= \left( 0, 0,\frac{1}{2}, 0,\frac{1}{2}, 0\right) '= \frac{1}{2}\left( \textbf{e}_3,\textbf{e}_2\right) ', \quad \textbf{q}_5 = \left( 0, 0,\frac{1}{2}, 0, 0, \frac{1}{2}\right) '= \frac{1}{2}\left( \textbf{e}_3,\textbf{e}_3\right) '. \end{aligned}$$

For any $\textbf{x} = (x_2,x_3,x_4,x_7,x_8,x_9)' \in {\mathcal {F}}_x$, due to $x_2+x_3+x_4 = x_7+x_8+x_9$, exactly one of the following two assertions is true:

• (1) $x_2+x_3 \le x_7$ (i.e., $\phi _{11} = \frac{1}{2} - x_4 \le x_7$, which implies $x_4 > x_8+x_9$);

• (2) $x_2+x_3 > x_7$ (i.e., $\phi _{11} = \frac{1}{2} - x_4 > x_7$, which implies $x_4 \le x_8+x_9$).

When Assertion (1) is true, we have

$$\begin{aligned} \textbf{x}&= \underbrace{2x_2\textbf{q}_0 + 2x_3\textbf{q}_1}_{\text {Step 1}}\underbrace{+ 2(x_7-x_2-x_3)\textbf{q}_3 + 2x_8\textbf{q}_4+2x_9\textbf{q}_5}_{\text {Step 2}} \\&= (x_2, x_3,x_7+x_8+x_9-x_2-x_3,x_7,x_8,x_9)' \\&= (x_2,x_3,x_4,x_7,x_8,x_9)'. \end{aligned}$$

When Assertion (2) is true, we have

$$\begin{aligned} \textbf{x}&= \underbrace{2x_2\textbf{q}_0 + 2x_3\textbf{q}_1+ 2(x_4-x_8-x_9)\textbf{q}_2}_{\text {Step 2}} \underbrace{+ 2x_8\textbf{q}_4+2x_9\textbf{q}_5}_{\text {Step 1}} \\&= (x_2, x_3,x_4,x_2+x_3+x_4-x_8-x_9,x_8,x_9)' \\&= (x_2,x_3,x_4,x_7,x_8,x_9)'. \end{aligned}$$

That is, every point in ${\mathcal {F}}^2_x$ can be represented by a convex combination of these extreme points. Consequently, $\{\textbf{q}_i: i=0,\ldots ,5\}$ is the set of all extreme points.