Minmax for facility location game with optional preference under minimum distance requirement

Abstract

In this paper, we study the optional preference model for the facility location game with two heterogeneous facilities on a line interval [0, 1], by further enforcing the requirement of a minimum distance $0\le d\le 1$ between the two facilities. Each agent has one of three favorable preferences towards the two facilities, i.e., facility 1, facility 2, or optional preference. Here, we consider two variants of the optional preference model: Min (caring for the closer one) and Max (caring for the further one). In both variants, each agent wishes to get close to his preferred facilities, and thus his cost is his distance to his preferred facility. In this game, we consider agents’ locations as public information and agents’ preferences as private information which needs to be reported by agents. The objective is to design a mechanism for the two facilities’ locations such as to minimize the maximum cost of agents (MinMax) and achieve truthful report of agents’ preferences. Given any value of d,  for both variants, we propose a strategyproof mechanism with an approximation ratio of 2. We also establish lower bounds of any deterministic strategyproof mechanism for both variants and show that the gaps between the lower bounds and the upper bounds are relatively small.

Appendices

Appendix A: Proof of Theorem 1

Proof

Let ${\mathcal {M}}$ be Mechanism 1. Suppose that agent i misreports his preference from $p_i$ to $p_i'$. Denote preference set ${{\textbf {p}}}'=\{p_i',{{\textbf {p}}}_{-i}\}.$ Denote $(y_1,y_2)=\mathcal M({{\textbf {x}}}, {{\textbf {p}}})\in D_d$ and $(y_1',y_2')={\mathcal {M}}({\textbf {x}}, {{\textbf {p}}}')\in D_d.$ Then we have five cases.

Case 1: $d\le x_n-x_1.$ In this case, by Mechanism 1, without loss of generality, we assume that

$$\begin{aligned} MC((x_1,x_n)|{{\textbf {x}}}, {{\textbf {p}}})\le MC((x_n,x_1)|{{\textbf {x}}}, {{\textbf {p}}}), \end{aligned}$$

(A1)

and thus $(y_1,y_2)=(x_1,x_n).$ We have three subcases for $p_i.$

• 1. $p_i=\{F_1\}.$ In this subcase, by Mechanism 1, agent i with preference $\{F_1\}$ has his cost $c_i((x_1,x_n)|(x_i,\{F_1\}))=x_i-x_1.$ After agent i misreports his preference to $p_i'=\{F_2\}$ or $p_i'=\{F_1,F_2\},$ if $(y_1',y_2')$ remains unchanged, agent i ’s cost is unchanged. Otherwise, we have $(y_1',y_2')=(x_n,x_1).$ Agent i’s cost changes to $c_i((x_n,x_1)|(x_i,\{F_1\}))=x_n-x_i.$ Further, it holds that

  $$\begin{aligned}&MC((x_1,x_n)|{{\textbf {x}}}, {{\textbf {p}}}')> MC((x_n,x_1)|{\textbf {x}}, {{\textbf {p}}}'). \end{aligned}$$

  (A2)

  Denote $\alpha _1=MC((x_1,x_n)|{{\textbf {x}}}_{-i}, {{\textbf {p}}}_{-i})$ and $\alpha _2=MC$ $((x_n,x_1)|{{\textbf {x}}}_{-i}, {{\textbf {p}}}_{-i}).$ We have two sub-subcases for $p_i'.$

  • (a) $p_i'=\{F_2\}.$ We can simplify (A1) and (A2) as

    $$\begin{aligned}&\max \{\alpha _1,x_i-x_1\}\le \max \{\alpha _2, x_n-x_i\}, \end{aligned}$$

    (A3)
    $$\begin{aligned}&\max \{\alpha _1,x_n-x_i\}> \max \{\alpha _2, x_i-x_1\}, \end{aligned}$$

    (A4)

    respectively. Assume by contradiction that $x_i-x_1>x_n-x_i.$ By (A3), we have

    $$\begin{aligned} \max \{\alpha _1,x_n-x_i\}\le \max \{\alpha _1,x_i-x_1\} \le&\max \{\alpha _2, x_n-x_i\}\le \max \{\alpha _2, x_i-x_1\}, \end{aligned}$$

    which contradicts (A4). Hence, $x_i-x_1\le x_n-x_i,$ which implies that agent i cannot decrease his cost.

  • (b) $p_i'=\{F_1, F_2\}.$ We can simplify (A2) as

    $$\begin{aligned}&\max \{\alpha _1,\min \{x_n-x_i, x_i-x_1\}\} >\max \{\alpha _2, \min \{x_n-x_i, x_i-x_1\}\}. \end{aligned}$$

    (A5)

    Assume by contradiction that $x_i-x_1>x_n-x_i.$ By (A5),

    $$\begin{aligned} \max \{\alpha _1, x_i-x_1\}\ge&\max \{\alpha _1,x_n-x_i\} =\max \{\alpha _1,\min \{x_n-x_i, x_i-x_1\}\} \nonumber \\ >&\max \{\alpha _2, \min \{x_n-x_i, x_i-x_1\}\} =\max \{\alpha _2, x_n-x_i\}, \end{aligned}$$

    which contradicts (A1) and (A3). Hence, $x_i-x_1\le x_n-x_i,$ which implies that agent i cannot decrease his cost.

• 2. $p_i=\{F_2\}.$ Due to symmetry, this subcase is similar to Subcase 1.

• 3. $p_i=\{F_1, F_2\}.$ In this subcase, by Mechanism 1, agent i does not have the incentive to lie as the two possible outputs $(x_1,x_n)$ and $(x_n,x_1)$ are the same in his perspective.

Case 2: $d> x_n-x_1$ and $x_1\le 1-d.$ In this case, without loss of generality, suppose ${\mathcal {M}}=(y_1,y_2)=(x_1,x_1+d)$, and

$$\begin{aligned} MC((x_1,x_1+d)|{{\textbf {x}}}, {{\textbf {p}}})\le MC((x_1+d,x_1)|{{\textbf {x}}}, {{\textbf {p}}}). \end{aligned}$$

(A6)

We have the following three subcases for $p_i$.

• 1. $p_i=\{F_1\}.$ In this subcase, by Mechanism 1, agent i with preference $\{F_1\}$ has his cost $c_i((x_1,x_1+d)|(x_i,\{F_1\}))=x_i-x_1.$ After agent i misreports his preference to $p_i'=\{F_2\}$ or $p_i'=\{F_1,F_2\},$ if $(y_1',y_2')$ remains unchanged, agent i ’s cost is unchanged. Otherwise, we have $(y_1',y_2')=(x_1+d,x_1).$ Agent i’s cost changes to $c_i((x_1+d,x_1)|(x_i,\{F_1\}))=x_1+d-x_i.$ Further, it holds that

  $$\begin{aligned}&MC((x_1,x_1+d)|{{\textbf {x}}}, {\textbf {p}}')>MC((x_1+d,x_1)|{{\textbf {x}}}, {{\textbf {p}}}'). \end{aligned}$$

  (A7)

  Denote $\alpha _3=MC((x_1,x_1+d)|{{\textbf {x}}}_{-i}, {\textbf {p}}_{-i})$ and $\alpha _4=MC$ $((x_1+d,x_1)|{{\textbf {x}}}_{-i}, {{\textbf {p}}}_{-i}).$ We have two sub-subcases for $p_i'.$

  • (a) $p_i'=\{F_2\}.$ We can simplify (A6) and (A7) as

    $$\begin{aligned}&\max \{\alpha _3,x_i-x_1\}\le \max \{\alpha _4, x_1+d-x_i\}, \end{aligned}$$

    (A8)
    $$\begin{aligned}&\max \{\alpha _3,x_1+d-x_i\}> \max \{\alpha _4, x_i-x_1\}, \end{aligned}$$

    (A9)

    respectively. Assume by contradiction that $x_i-x_1>x_1+d-x_i.$ By (A8), we have

    $$\begin{aligned}&\max \{\alpha _3,x_1+d-x_i\}\le \max \{\alpha _3,x_i-x_1\}\nonumber \\ \le&\max \{\alpha _4, x_1+d-x_i\}\le \max \{\alpha _4, x_i-x_1\}, \end{aligned}$$

    which contradicts (A9). Hence, $x_i-x_1\le x_1+d-x_i,$ which implies that agent i cannot decrease his cost.

  • (b) $p_i'=\{F_1, F_2\}.$ We can simplify (A7) as

    $$\begin{aligned}&\max \{\alpha _3,\min \{x_1+d-x_i, x_i-x_1\}\}\nonumber \\ >&\max \{\alpha _4, \min \{x_1+d-x_i, x_i-x_1\}\}. \end{aligned}$$

    (A10)

    Assume by contradiction that $x_i-x_1>x_1+d-x_i.$ By (A10), we have

    $$\begin{aligned}&\max \{\alpha _3, x_i-x_1\}\ge \max \{\alpha _3, x_1+d-x_i\}\\ =&\max \{\alpha _3,\min \{x_1+d-x_i, x_i-x_1\}\} \\ >&\max \{\alpha _4, \min \{x_1+d-x_i, x_i-x_1\}\} =\max \{\alpha _4, x_1+d-x_i\}, \end{aligned}$$

    which contradicts (A6) and (A8). Hence, $x_i-x_1\le x_1+d-x_i,$ which implies that agent i cannot decrease his cost.

• 2. $p_i=\{F_2\}.$ Due to symmetry, this subcase is similar to Subcase 1.

• 3. $p_i=\{F_1, F_2\}.$ In this subcase, by Mechanism 1, agent i does not have the incentive to lie as the two possible outputs $(x_1,x_1+d)$ and $(x_1+d,x_1)$ are the same in his perspective.

Case 3: $d> x_n-x_1,$ $x_1> 1-d$, and $x_n\ge d.$ Due to symmetry, this case is similar to Case 2.

Case 4: $d>x_n-x_1,$ $x_1> 1-d,$ $x_n< d,$ and $x_1+x_n\ge 1.$ In this case, we have $d>1/2$ and all agents’ locations must be included between $F_1$ and $F_2.$ Thus, we can use similar steps in Case 2 to prove that Mechanism 1 is strategyproof.

Case 5: $d>x_n-x_1,$ $x_1> 1-d,$ $x_n< d,$ and $x_1+x_n< 1.$ Due to symmetry, this case is similar to Case 4.

In conclusion, by considering all cases, Mechanism 1 is strategyproof.

$\square$

Appendix B: Proof of Theorem 4

Proof

Let ${\mathcal {M}}$ be Mechanism 1. Suppose that agent i misreports his preference from $p_i$ to $p_i'$. Denote preference set ${{\textbf {p}}}'=\{p_i',{{\textbf {p}}}_{-i}\}.$ Denote $(y_1,y_2)=\mathcal M({{\textbf {x}}}, {{\textbf {p}}})\in D_d$ and $(y_1',y_2')={\mathcal {M}}({\textbf {x}}, {{\textbf {p}}}')\in D_d.$ We have five cases.

Case 1: $d\le x_n-x_1.$ In this case, by Mechanism 1, without loss of generality, we assume that

$$\begin{aligned} MC((x_1,x_n)|{{\textbf {x}}}, {{\textbf {p}}})\le MC((x_n,x_1)|{{\textbf {x}}}, {{\textbf {p}}}), \end{aligned}$$

(B11)

and thus $(y_1,y_2)=(x_1,x_n).$ We have three subcases.

• 1. $p_i=\{F_1\}.$ In this subcase, by Mechanism 1, agent i with preference $\{F_1\}$ has his cost $c_i((x_1,x_n)|(x_i,\{F_1\}))=x_i-x_1.$ After agent i misreports his preference to $p_i'=\{F_2\}$ or $p_i'=\{F_1,F_2\},$ if $(y_1',y_2')$ remains unchanged, agent i ’s cost is unchanged. Otherwise, we have $(y_1',y_2')=(x_n,x_1).$ Agent i’s cost changes to $c_i((x_n,x_1)|(x_i,\{F_1\}))=x_n-x_i.$ Further, it holds that

  $$\begin{aligned}&MC((x_1,x_n)|{{\textbf {x}}}, {{\textbf {p}}}')> MC((x_n,x_1)|{\textbf {x}}, {{\textbf {p}}}'). \end{aligned}$$

  (B12)

  Denote $\alpha _1=MC((x_1,x_n)|{{\textbf {x}}}_{-i}, {{\textbf {p}}}_{-i})$ and $\alpha _2=MC$ $((x_n,x_1)|{{\textbf {x}}}_{-i}, {{\textbf {p}}}_{-i}).$ We have two sub-subcases for $p_i'.$

  • (a) $p_i'=\{F_2\}.$ This sub-subcase is similar to Sub-subcase 1.1.a in the proof of Theorem 1.

  • (b) $p_i'=\{F_1, F_2\}.$ We can simplify (B11) and (B12) as

    $$\begin{aligned}&\max \{\alpha _1,x_i-x_1\} \le \max \{\alpha _2, x_n-x_i\}, \end{aligned}$$

    (B13)
    $$\begin{aligned}&\max \{\alpha _1,\max \{x_n-x_i, x_i-x_1\}\} > \max \{\alpha _2, \max \{x_n-x_i, x_i-x_1\}\}, \end{aligned}$$

    (B14)

    respectively. Assume by contradiction that $x_i-x_1>x_n-x_i.$ By (B14), we have

    $$\begin{aligned} \max \{\alpha _1,x_i-x_1\}&=\max \{\alpha _1,\max \{x_n-x_i, x_i-x_1\}\}\\&>\max \{\alpha _2, \max \{x_n-x_i, x_i-x_1\}\}\\&=\max \{\alpha _2, x_i-x_1\}\ge \max \{\alpha _2, x_n-x_i\}, \end{aligned}$$

    which contradicts (B13). Hence, $x_i-x_1\le x_n-x_i,$ which implies that agent i cannot decrease his cost.

• 2. $p_i=\{F_2\}.$ Due to symmetry, this subcase is similar to Subcase 1.

• 3. $p_i=\{F_1, F_2\}.$ In this subcase, by Mechanism 1, agent i does not have the incentive to lie as the two possible outputs $(x_1,x_n)$ and $(x_n,x_1)$ are the same in his perspective.

Case 2: $d> x_n-x_1$ and $x_1\le 1-d.$ In this case, without loss of generality, suppose ${\mathcal {M}}=(y_1,y_2)=(x_1,x_1+d)$, and

$$\begin{aligned} MC((x_1,x_1+d)|{{\textbf {x}}}, {{\textbf {p}}})\le MC((x_1+d,x_1)|{{\textbf {x}}}, {{\textbf {p}}}). \end{aligned}$$

(B15)

We have the following three subcases.

• 1. $p_i=\{F_1\}.$ In this subcase, by Mechanism 1, agent i with preference $\{F_1\}$ has his cost $c_i((x_1,x_1+d)|(x_i,\{F_1\}))=x_i-x_1.$ After agent i misreports his preference to $p_i'=\{F_2\}$ or $p_i'=\{F_1,F_2\},$ if $(y_1',y_2')$ remains unchanged, agent i ’s cost is unchanged. Otherwise, we have $(y_1',y_2')=(x_1+d,x_1).$ Agent i’s cost changes to $c_i((x_1+d,x_1)|(x_i,\{F_1\}))=x_1+d-x_i.$ Further, it holds that

  $$\begin{aligned}&MC((x_1,x_1+d)|{{\textbf {x}}}, {{\textbf {p}}}')> MC((x_1+d,x_1)|{{\textbf {x}}}, {{\textbf {p}}}'). \end{aligned}$$

  (B16)

  Denote $\alpha _3=MC((x_1,x_1+d)|{{\textbf {x}}}_{-i}, {\textbf {p}}_{-i})$ and $\alpha _4=MC((x_1+d,x_1)|{{\textbf {x}}}_{-i}, {\textbf {p}}_{-i}).$ We have two sub-subcases for $p_i'.$

  • (a) $p_i'=\{F_2\}.$ This sub-subcase is similar to Sub-subcase 2.1.a in the proof of Theorem 1.

  • (b) $p_i'=\{F_1, F_2\}.$ We can simplify (B15) and (B16) as

    $$\begin{aligned}&\max \{\alpha _3,x_i-x_1\} \le \max \{\alpha _4, x_1+d-x_i\}, \end{aligned}$$

    (B17)
    $$\begin{aligned}&\max \{\alpha _3,\max \{x_1+d-x_i, x_i-x_1\}\}\nonumber \\&> \max \{\alpha _4, \max \{x_1+d-x_i, x_i-x_1\}\}, \end{aligned}$$

    (B18)

    respectively. Assume by contradiction that $x_i-x_1>x_1+d-x_i.$ By (B18), we have

    $$\begin{aligned} \max \{\alpha _3, x_i-x_1\} =&\max \{\alpha _3,\max \{x_1+d-x_i, x_i-x_1\}\} \\ >&\max \{\alpha _4, \max \{x_1+d-x_i, x_i-x_1\}\}\\ =&\max \{\alpha _4, x_i-x_1\}\ge \max \{\alpha _4, x_1+d-x_i\}, \end{aligned}$$

    which contradicts (B17). Hence, $x_i-x_1\le x_1+d-x_i,$ which implies that agent i cannot decrease his cost.

• 2. $p_i=\{F_2\}.$ Due to symmetry, this subcase is similar to Subcase 1.

• 3. $p_i=\{F_1, F_2\}.$ In this subcase, by Mechanism 1, agent i does not have the incentive to lie as the two possible outputs $(x_1,x_1+d)$ and $(x_1+d,x_1)$ are the same in his perspective.

Case 3: $d> x_n-x_1,$ $x_1> 1-d$, and $x_n\ge d.$ Due to symmetry, this case is similar to Case 2.

Case 4: $d> x_n-x_1,$ $x_1> 1-d$, $x_n<d,$ and $x_1+x_n< 1.$ In this case, we have $d>1/2$ and all agents’ locations must be included between $F_1$ and $F_2.$ Thus, we can use similar steps in Case 2 to prove that Mechanism 1 is strategyproof.

Case 5: $d>x_n-x_1,$ $x_1> 1-d,$ $x_n< d,$ and $x_1+x_n< 1.$ Due to symmetry, this case is similar to Case 4.

In conclusion, by considering all cases, Mechanism 1 is strategyproof.

$\square$

Appendix C: Proof of Theorem 5

Proof

Let ${\mathcal {M}}$ be Mechanism 1 and $\bar{{\mathcal {M}}}$ be the optimal solution for minimizing the maximum cost. Denote $(y_1,y_2)={\mathcal {M}}({{\textbf {x}}}, {{\textbf {p}}})\in D_d$ and $(y_1^\star ,y_2^\star )=\bar{{\mathcal {M}}}({{\textbf {x}}}, {{\textbf {p}}})\in D_d.$ Let agent $m\in N$ be the agent where the maximum cost comes from in Mechanism 1. We have five cases.

Case 1: $d\le |x_1-x_n|.$ According to the preference of agent 1 and agent n,  there are five subcases.

• 1. $p_1=p_n=\{F_1\}.$ This subcase is similar to Subcase 1.1 in the proof of Theorem 2 and thus $\gamma \le 2.$

• 2. $p_1=\{F_1\}$ and $p_n=\{F_2\}.$ In this subcase, by Mechanism 1, ${\mathcal {M}}=(y_1,y_2)=(x_1,x_n).$ According to agent m’s preference, we have three sub-subcases.

  • (a) $p_m=\{F_1\}.$ This sub-subcase is similar to Sub-subcase 1.2.a in the proof of Theorem 2 and thus $\gamma \le 2.$

  • (b) $p_m=\{F_2\}.$ Due to symmetry, this sub-subcase is similar to Sub-subcase 1.2.a.

  • (c) $p_m=\{F_1,F_2\}.$ Without loss of generality, we assume that $x_n-x_m\ge x_m-x_1.$ For Mechanism 1, we have the maximum cost $MC({\mathcal {M}}|{{\textbf {x}}},{{\textbf {p}}})=x_n-x_m.$ For the optimal solution $\bar{{\mathcal {M}}},$ we have the maximum cost

    $$\begin{aligned} MC(\bar{{\mathcal {M}}}|{{\textbf {x}}},{{\textbf {p}}}) \ge&\max \{|y_2^\star -x_n|, |y_2^\star -x_m|\}\ge ({x_n-x_m})/{2}. \end{aligned}$$

    Thus, the approximation ratio is $\gamma \le 2$.

• 3. $p_1=p_n=\{F_2\}.$ The analysis is similar to Subcase 1.1 by symmetry.

• 4. $p_1=\{F_2\}$ and $p_n=\{F_1\}.$ The analysis is similar to Subcase 1.2 by symmetry.

• 5. $\{F_1, F_2\} \in \{p_1, p_n\}.$ In this subcase, if agent 1 and agent n are of the same type ($F_1$-typed or $F_2$-typed), then the analysis will be similar to Subcase 1.1 or 1.3. If they are of different types, then the analysis will be similar to Subcase 1.2 or 1.4.

Case 2: $d> x_n-x_1$ and $x_1\le 1-d.$ According to the preference of agent 1 and agent n,  there are five subcases.

• 1. $p_1=p_n=\{F_1\}.$ In this subcase, by Mechanism 1, $\mathcal M=(y_1,y_2)=(x_1,x_1+d).$ According to agent m’s preference, we have three sub-subcases.

  • (a) $p_m=\{F_1\}.$ This sub-subcase is similar to Sub-subcase 2.1.a in the proof of Theorem 2 and thus $\gamma \le 2.$

  • (b) $p_m=\{F_2\}.$ This sub-subcase is similar to Sub-subcase 2.1.b in the proof of Theorem 2 and thus $\gamma \le 2.$

  • (c) $p_m=\{F_1, F_2\}.$ For Mechanism 1, we have

    $$\begin{aligned} MC({\mathcal {M}}|{{\textbf {x}}},{{\textbf {p}}})=&\max \{x_m-x_1, x_1+d-x_m\}\\ \ge&c_n((x_1,x_1+d)|(x_n,\{F_1\}))=x_n-x_1. \end{aligned}$$

    If $x_m=x_n,$ we have $MC({\mathcal {M}}|{{\textbf {x}}},{\textbf {p}})=x_m-x_1.$ In this situation, for the optimal solution $\bar{{\mathcal {M}}},$ we have the maximum cost

    $$\begin{aligned} MC(\bar{{\mathcal {M}}}|{{\textbf {x}}},{{\textbf {p}}}) \ge \max \{|y_1^\star -x_1|, |y_1^\star -x_n|\}\ge (x_n-x_1)/2 \end{aligned}$$

    and thus $\gamma \le 2.$ Otherwise, we have $MC({\mathcal {M}}|{{\textbf {x}}},{\textbf {p}})=x_1+d-x_m\ge x_n-x_1,$ if $x_m<x_n.$ In this situation, for the optimal solution $\bar{{\mathcal {M}}},$ we have the maximum cost

    $$\begin{aligned} MC(\bar{{\mathcal {M}}}|{{\textbf {x}}},{{\textbf {p}}}) \ge&\max \{|y_1^\star -x_1|,|y_1^\star -x_n|, |y_1^\star -x_m|,|y_2^\star -x_m|\}\nonumber \\ =&\max \{|y_1^\star -x_1|,|y_1^\star -x_n|,|y_2^\star -x_m|\}\nonumber \\ \ge&\left\{ \begin{array}{ll} \frac{d+x_m-x_1}{2} &{}\text{ if } x_1\le x_m\le \frac{x_1+x_n}{2}\\ \frac{d+x_n-x_m}{2} &{}\text{ if } \frac{x_1+x_n}{2}<x_m\le x_n \end{array} \right. \ge&({d+x_1-x_m})/{2}, \end{aligned}$$

    (C19)

    where the equality in the second inequality follows when $(y_1^\star ,y_2^\star )=(\frac{x_1+x_m+d}{2},$ $\frac{x_1+x_m-d}{2})$ if $x_1\le x_m\le \frac{x_1+x_n}{2};$ or $(y_1^\star ,y_2^\star )=(\frac{x_m+x_n-d}{2},\frac{x_m+x_n+d}{2})$ if $\frac{x_1+x_n}{2}<x_m\le x_n.$ Thus, the approximation ratio is $\gamma \le 2$.

• 2. $p_1=\{F_1\}$ and $p_n=\{F_2\}.$ In this subcase, by Mechanism 1, ${\mathcal {M}}=(y_1,y_2)=(x_1,x_1+d).$ According to agent m’s preference, we have three sub-subcases.

  • (a) $p_m=\{F_1\}.$ This sub-subcase is similar to Sub-subcase 2.2.a in the proof of Theorem 2 and thus $\gamma \le 2.$

  • (b) $p_m=\{F_2\}.$ This sub-subcase is similar to Sub-subcase 2.2.b in the proof of Theorem 2 and thus $\gamma \le 2.$

  • (c) $p_m=\{F_1, F_2\}.$ For Mechanism 1, we have

    $$\begin{aligned} MC({\mathcal {M}}|{{\textbf {x}}},{{\textbf {p}}})=\max \{x_m-x_1,x_1+d-x_m\} \le d. \end{aligned}$$

    For the optimal solution $\bar{{\mathcal {M}}},$ we have the maximum cost

    $$\begin{aligned} MC(\bar{{\mathcal {M}}}|{{\textbf {x}}},{{\textbf {p}}}) \ge \max \{|y_1^\star -x_m|,|y_2^\star -x_m|\}\ge d/2. \end{aligned}$$

    Thus, the approximation ratio is $\gamma \le 2.$

• 3. $p_1=p_n=\{F_2\}.$ The analysis is similar to Subcase 2.1 by symmetry.

• 4. $p_1=\{F_2\}$ and $p_n=\{F_1\}.$ The analysis is similar to Subcase 2.2 by symmetry.

• 5. $\{F_1, F_2\} \in \{p_1, p_n\}.$ In this subcase, if agent 1 and agent n are of the same type ($F_1$-typed or $F_2$-typed), then the analysis will be similar to Subcase 2.1 or 2.3. If they are of different types, then the analysis will be similar to Subcase 2.2 or 2.4.

Case 3: $d> x_n-x_1,$ $x_1> 1-d$, and $x_n\ge d.$ Due to symmetry, this case is similar to Case 2.

Case 4: $d> x_n-x_1,$ $x_1> 1-d$, $x_n<d,$ and $x_1+x_n\ge 1.$ In this case, by $1-d<x_1\le x_n<d,$ d should satisfy $d>1/2.$ Note that in any mechanism, all agents’ locations must be between $F_1$ and $F_2.$ According to the preference of agent 1 and agent n,  there are five subcases.

• 1. $p_1=p_n=\{F_1\}.$ In this subcase, by Mechanism 1 and $x_1+x_n\ge 1\ge d$, ${\mathcal {M}}=(y_1,y_2)=(d,0).$ According to agent m’s preference, we have three sub-subcases.

  • (a) $p_m=\{F_1\}.$ This sub-subcase is similar to Sub-subcase 4.1.a in the proof of Theorem 2 and thus $\gamma =1.$

  • (b) $p_m=\{F_2\}.$ This sub-subcase is similar to Sub-subcase 4.1.a in the proof of Theorem 2 and thus $\gamma \le 2.$

  • (c) $p_m=\{F_1, F_2\}.$ For Mechanism 1, we have

    $$\begin{aligned} MC({\mathcal {M}}|{{\textbf {x}}},{{\textbf {p}}}) =&\max \{x_m, d-x_m\} \ge c_1((d,0)|(x_n,\{F_1\}))=d-x_1. \end{aligned}$$

    If $x_m=x_1,$ we have $MC({\mathcal {M}}|{{\textbf {x}}},{\textbf {p}})=d-x_1\ge x_1.$ In this situation, for the optimal solution $\bar{{\mathcal {M}}},$ we have the maximum cost

    $$\begin{aligned} MC(\bar{{\mathcal {M}}}|{{\textbf {x}}},{{\textbf {p}}}) \ge \max \{|y_1^\star -x_1|, |y_1^\star -x_n|\}\ge d-x_1 \end{aligned}$$

    and thus $\gamma =1.$ Otherwise, we have $MC({\mathcal {M}}|{{\textbf {x}}},{{\textbf {p}}})=x_m\ge d-x_1.$ In this situation, for the optimal solution $\bar{{\mathcal {M}}},$ (C19) still holds here, and further by (C19), we have the maximum cost $MC(\bar{{\mathcal {M}}}|{{\textbf {x}}},{{\textbf {p}}})\ge x_m/2.$ Thus, the approximation ratio is $\gamma \le 2$.

• 2. $p_1=\{F_1\}$ and $p_n=\{F_2\}.$ In this subcase, by Mechanism 1, ${\mathcal {M}}=(y_1,y_2)=(0,d).$ According to agent m’s preference, we have three sub-subcases.

  • (a) $p_m=\{F_1\}.$ This sub-subcase is similar to Sub-subcase 4.2.a in the proof of Theorem 2 and thus $\gamma \le 2.$

  • (b) $p_m=\{F_2\}.$ This sub-subcase is similar to Sub-subcase 4.2.b in the proof of Theorem 2 and thus $\gamma \le 2.$

  • (c) $p_m=\{F_1, F_2\}.$ For Mechanism 1, we have the maximum cost

    $$\begin{aligned} MC({\mathcal {M}}|{{\textbf {x}}},{{\textbf {p}}})=\max \{x_m,d-x_m\}\le d. \end{aligned}$$

    For the optimal solution $\bar{{\mathcal {M}}},$ we have the maximum cost

    $$\begin{aligned}&MC(\bar{{\mathcal {M}}}|{{\textbf {x}}},{{\textbf {p}}}) \ge \max \{|y_1^\star -x_m|,|y_2^\star -x_m|\}\ge d/2. \end{aligned}$$

    Thus, the approximation ratio is $\gamma \le 2.$

• 3. $p_1=p_n=\{F_2\}.$ The analysis is similar to Subcase 4.1 by symmetry.

• 4. $p_1=\{F_2\}$ and $p_n=\{F_1\}.$ The analysis is similar to Subcase 4.2 by symmetry.

• 5. $\{F_1, F_2\} \in \{p_1, p_n\}.$ In this subcase, if agent 1 and agent n are of the same type ($F_1$-typed or $F_2$-typed), then the analysis will be similar to Subcase 4.1 or 4.3. If they are of different types, then the analysis will be similar to Subcase 4.2 or 4.4.

In conclusion, by considering all cases, Mechanism 1 has an approximation ratio $\gamma =2.$

$\square$