論文閱讀-AKS_CoRR_2011

作者 年份 近似比
Hoogeveen 1991 53\frac{5}{3}
An, Kleinberg, Shmoys 2012 1+52\frac{1+\sqrt{5}}{2}
Sebo 2013 85\frac{8}{5}
Rico Zenklusen 2019 1.5

Title: Improving Christofides’s Algorithm for the s-t path TSP

Alpha: 1+52\frac{1+\sqrt{5}}{2}

Theorem1: Hoogeveen算法的解不超過53OPTLP\frac{5}{3}OPT_{LP}

定義1: Path-TSP的HK鬆弛

mineEcexes.t.x(δ(v))={1,v=s,t2,vs,tx(δ(S)){1,S{s,t}=1,2,S{s,t}1,0xe1,eE min \sum_{e\in E} c_ex_e\\ \begin{aligned} & s.t.\\ & x(\delta(v))=\begin{cases} 1, & v=s,t\\ 2, & v\neq s,t \end{cases}\\ & x(\delta(S)) \geq \begin{cases} 1, & |S \cap \{s,t\}|=1,\\ 2, & |S \cap \{s,t\}|\neq 1,\\ \end{cases}\\ & 0 \leq x_e \leq 1, \forall e \in E \end{aligned}

其中δ(S)\delta(S)是僅有一個端點落在S中的邊的邊集, 同時X(E)=eExeX(E')=\sum_{e \in E'} x_e, 所謂的鬆弛就是最後一個0-1向量變成了實數.

定義2: 生成樹凸集

生成樹凸集由下面的不等式定義:
x(E)=V1,x(E(S))S1,SV,S2,xe0,eE \begin{aligned} & x(E)=|V|-1,\\ & x(E(S)) \leq |S|-1, \quad \forall |S| \subseteq V, |S| \geq 2,\\ & x_e \geq 0, \quad \forall e \in E \end{aligned}

其中E(S)是所有兩個端點都在S中的邊的邊集.
在這裏插入圖片描述

Lemma1: LP-relaxation的任意可行解x都在生成樹凸集中.

proof: LP-relaxation的約束滿足生成樹凸集的定義
X(E)eExe=12vVx(δ(v))=12(v2)2+2)=v1 X(E) \equiv \sum_{e\in E} x_e = \frac{1}{2}\sum_{v\in V}x(\delta(v))\\ = \frac{1}{2}(|v|-2)\cdot 2 + 2)=|v|-1
同時,
X(E(S))=12(vSx(δ(v))x(δ(S))) X(E(S))=\frac{1}{2}(\sum_{v \in S}x(\delta(v))-x(\delta(S)))
如果S{s,t}=1|S\cap \{s,t\}=1, 有X(E(S))12(1+2(S1)1)=S1X(E(S))\leq \frac{1}{2}(1+2(|S|-1)-1)=|S|-1,

如果S{s,t}=|S\cap \{s,t\}=\empty, S-1,

如果S{s,t}=2|S\cap \{s,t\}=2,S-2

定義3: 奇數集S, 如果ST|S\cap T|含有奇數個, 則S是個奇數集

Lemma2: S是一個奇數集, 如果S{s,t}=1|S\cap \{s,t\}|=1, 則Fδ(S)|F\cap \delta(S)|爲偶數, 如果S{s,t}1|S\cap \{s,t\}|\neq1, 則Fδ(S)|F\cap \delta(S)|爲奇數.

例如,

在這裏插入圖片描述

Proof: s,t 如果在S中, 它們有偶數度, 其他點有奇數度.

定義vSdegF(v)=2E(S)F+δ(S)F\sum_{v\in S}deg_F(v)=2|E(S)\cap F|+|\delta(S)\cap F|

證明如下:

1.如果S{s,t}=1|S\cap \{s,t\}=1, 假設sSs\in S, sTs\in T當且僅當degF(s)deg_F(s)even.

SoddevenS_{odd}\rightarrow even # 個奇數度的節點在S中(STodd|S\cap T| odd)

vSdegF(v)2E(s)F=δ(s)F \sum_{v\in S}deg_F(v)-2|E(s)\cap F|=|\delta(s)\cap F|
第一個子式爲偶數度, 第二個子式肯定是偶數, 則右邊也是偶數.



2. 如果S{s,t}1|S\cap \{s,t\}\neq 1,

SoddoddS_{odd}\rightarrow odd # 個奇數度的節點在S中

vSdegF(v)2E(s)F=δ(s)F \sum_{v\in S}deg_F(v)-2|E(s)\cap F|=|\delta(s)\cap F|
第一個子式爲奇數度, 第二個子式肯定是偶數, 則右邊也是奇數.

定義4: T-join LP

以下線性規劃的解是一個最小成本的T-join, 對於cost c0c\geq 0:
MineEcexes.t.x(δ(S))1,SV,SToddxe0,eE Min \sum_{e\in E}c_ex_e\\ \begin{aligned} & s.t.\\ & x(\delta(S)) \geq 1, & \forall S \subseteq V, |S\cap T| odd\\ & x_e \geq 0, & \forall e \in E \end{aligned}
對於ST|S\cap T|爲奇,
vSdegJ(v)=2E(S)J+δ(S)J \sum_{v\in S}deg_J(v)=2|E(S)\cap J|+|\delta(S)\cap J|
因爲奇數個奇數度的節點,因此等式左邊爲奇,因爲右邊第一個子式爲偶,所以第二個子式爲奇數.說明S向外連接的節點一定大於等於1.

proof of Theorem1

Step1

xx^*爲LP鬆弛的最優解OPT. cost of MSTeEcexe>OPTLP\text{cost of MST}\leq \sum_{e\in E}c_ex_e^* \equiv > OPT_{LP}, 因爲xx^*總是生成樹凸集的可行解.

XF{0,1}EX_F\in \{0,1\}^{|E|},並且
XF(e)={1,ifeF0,o.w. X_F(e)= \begin{cases} 1, \quad if e\in F \\ 0, \quad o.w. \end{cases}

claim: y=13XF+13xy=\frac{1}{3}X_F+\frac{1}{3}x^*是T-join LP的一個可行解.

則有c(FT)=c(F)+c(T)OPTLP+13c(F)+13OPTLP>53OPTLPc(F\cup T)=c(F)+c(T) \leq OPT_{LP}+\frac{1}{3}c(F)+\frac{1}{3}OPT_{LP} > \leq \frac{5}{3}OPT_{LP}

Step2

若要claim成立,需要證明如果sT|s\cap T|爲奇,則y(δ(S))1y(\delta(S)) \geq 1

如果 s{s,t}1|s\cap \{s,t\}|\neq 1y(δ(S))=13Fδ(S)+13x(δ(S))13+23=1y(\delta(S)) =\frac{1}{3}|F\cup \delta(S)|+\frac{1}{3}x^*(\delta(S)) \geq \frac{1}{3}+\frac{2}{3} = 1

【第二個部分,因爲HK relaxation成立】

如果s{s,t}=1|s\cap \{s,t\}|= 1, 則y(δ(S))=13Fδ(S)+13x(δ(S))23+13=1y(\delta(S)) =\frac{1}{3}|F\cup \delta(S)|+\frac{1}{3}x^*(\delta(S)) \geq \frac{2}{3}+\frac{1}{3} = 1

【第一個式子 lemma2】

claim 證畢.

定義5: 凸組合

xx^*爲LP的最優解, 令xFx^F表示爲一個F的邊集, 即
xF(e)={1eF0eF x_F(e)=\begin{cases} 1 \quad e \in F \\ 0 \quad e \notin F \end{cases}

因爲xx^*在生成樹凸集中,因此xx^*可以寫成生成樹F1,,FkF_1,\cdots,F_k的凸組合:
x=i=1kλixFi x^*=\sum_{i=1}^k \lambda_i x_{F_i}
其中i=1kλi=1,λi0\sum_{i=1}^k\lambda_i = 1, \lambda_i \geq 0.

對於FiF_i, 設TiT_i是其T集,JiJ_i是其最小成本T-join. 它們的和能夠構成一個解稱爲best-of-many Christofide算法的解.

Theorem2: best-of-many Christofide算法的解, 同樣滿足上限爲53OPTLP\frac{5}{3}OPT_{LP}.

下一步: 是否能夠更優?

考慮yi=αXF+βxy_i=\alpha X_F+\beta x^*, 如果是TiT_i-Join LP的可行解, 則best s-t 哈密頓路徑的長度最多不超過(1+α+β)OPTLP(1+\alpha+\beta)OPT_{LP}.

yiy_iTiT_i-Join LP的可行解分兩種情況考慮,設S 奇數集(STiodd|S\cup T_i| odd)

如果s{s,t}1|s\cap \{s,t\}|\neq 1,

y(δ(S))=αFδ(S)+βx(δ(S))α+2β y(\delta(S)) =\alpha|F\cup \delta(S)|+\beta x^*(\delta(S)) \geq \alpha+2\beta

[右邊第一個式大於等於1, 第二個式大於等於2, 同上]

我們希望α+2β1\alpha+2\beta \geq 1, 則TiT_i-join LP約束就能夠滿足.

如果
s{s,t}=1|s\cap \{s,t\}|= 1, 則
yi(δ(S))=αFδ(S)+βx(δ(S))2α+βx(δ(S)) y_i(\delta(S)) =\alpha |F\cup \delta(S)|+\beta x^*(\delta(S)) \geq 2\alpha+\beta x^*(\delta(S))

注意到我們已經假設了α+2β1\alpha+2\beta \geq 1成立, 只有2α+βx(δ(S))<12\alpha+\beta x^*(\delta(S)) < 1時, 存在問題.

注意到當α=0,β=12\alpha=0, \beta=\frac{1}{2}時, 如果x(δ(S))2x^*(\delta(S)) \geq 2, 上式成立, 並且能夠控制上限爲32OPTLP\frac{3}{2}OPT_{LP}.

因此接下來只需要關注x(δ(S))<2x^*(\delta(S)) < 2的cuts, 並且對yiy_i增加一個額外的修正來處理這些誒cuts.

定義6 τ\tau-Narrow cut

x(δ(S))<1+τ,for fixed τ1x^*(\delta(S)) < 1+\tau, \text{for fixed }\tau \leq 1, S則是τ\tau-Narrow.

只有S{s,t}=1|S\cup \{s, t\} =1能夠是τ\tau-Narrow.

定義7 τ\tau-Narrow cuts

CτC_{\tau}sSs \in S的所有τ\tau-Narrow cuts S的全集.

CτC_{\tau}的性質:

Theorem3: 如果S1,S2Cτ,S1S2S_1, S_2 \in C_{\tau}, S_1\neq S_2, 要麼S1S2S_1 \subset S_2,或S2S1S_2 \subset S1.

爲證明上述Theorem, 首先有
x(δ(S1))+x(δ(S2))x(δ(S1S2))+x(δ(S2S1)) x^*(\delta(S_1)) + x^*(\delta(S_2)) \geq x^*(\delta(S_1 - S_2)) + x^*(\delta(S_2-S_1))

Theorem proof:

假設,相反的, S1S2,S2S1S_1-S_2\neq \empty, S_2-S_1 \neq \empty.
(1+τ)+(1+τ)>x(δ(S1))+x(δ(S2))x(δ(S1S2))+x(δ(S2S1))2+2 \begin{aligned} (1 + \tau)+(1+\tau) & > x^*(\delta(S_1))+x^*(\delta(S_2)) \\ & \geq x^*(\delta(S_1-S_2)) + x^*(\delta(S_2-S_1)) \\ & \geq 2 + 2 \end{aligned}
與定義矛盾.

根據Theorem, τ\tau-Narrow cuts的結構如下:

在這裏插入圖片描述

新的修正因子

eQe_Q表示δ(Q)\delta(Q)的最小cost的邊,考慮下式
yi(δ(S))=αxFi+βx+QCτ,QTi(12αβx(δ(Q)))xeQ y_i(\delta(S)) =\alpha x_{F_i}+\beta x^* + \sum_{Q \in C_{\tau},|Q\cap T_i|}(1-2\alpha - \beta x^*(\delta(Q)))x_{e_Q}
對於α,β,τ0\alpha,\beta, \tau \geq 0,有α+2β=1\alpha + 2\beta=1並且τ=12αβ1\tau = \frac{1-2\alpha}{\beta} - 1

Theorem: yiy_i是一個TiT_i-Join LP的可行解.

proof:

對於S odd (STiodd|S\cap T_i| odd)

如果S{s,t}1|S\cap \{s,t\}|\neq 1
yi(δ(S))α+2β=1 y_i(\delta(S)) \geq \alpha + 2 \beta =1
如果S{s,t}=1|S\cap \{s,t\}|= 1

如果 S不是τ\tau-narrow
yi(δ(S))2α+β(1+τ)=1 y_i(\delta(S)) \geq 2\alpha + \beta(1+\tau) =1

如果 S是τ\tau-narrow
yi(δ(S))αFiδ(S)+βx(δ(δ(S)))+(12αβx(δ(S)))=1 y_i(\delta(S)) \geq \alpha |F_i\cap \delta(S)| + \beta x^*(\delta(\delta(S))) + (1-2\alpha - \beta x^*(\delta(S))) =1

注意到x=i=1kλixFix^*=\sum_{i=1}^k \lambda_i x_{F_i},i=1kλi=1,λi0\sum_{i=1}^k\lambda_i = 1, \lambda_i \geq 0,λi\lambda_i可以看成FiF_i的概率分佈, 是其概率. 緊接着, 有以下兩個lemma.

Lemma:

令F爲隨機採樣的生成樹FiF_i, T爲對應的點集TiT_i, QCτQ\in C_{\tau}是一個τ\tau-narrow cut.
Pr[δ(Q)F=1]2x(δ(Q))Pr[QTodd]x(δ(Q))1 Pr[|\delta(Q)\cap F|=1] \geq 2 - x^*(\delta(Q)) \\ Pr[|Q\cap T| odd] \leq x^*(\delta(Q)) - 1

proof:

x(δ(Q))=E[Fδ(Q)]Pr[Fδ(Q)=1]+2Pr[Fδ(Q)2] x^*(\delta(Q)) = E[|F \cap \delta(Q)|] \geq Pr[|F \cap \delta(Q)|=1] + 2Pr[|F \cap \delta(Q)| \geq 2]
並且Pr[Fδ(Q)=1]+Pr[Fδ(Q)2]=1Pr[|F \cap \delta(Q)|=1] + Pr[|F \cap \delta(Q)| \geq 2]=1, 因此,
Pr[Fδ(Q)=1]2x(δ(Q))Pr[Fδ(Q)2]x(δ(Q))1 Pr[|F\cap \delta(Q)|=1] \geq 2 - x^*(\delta(Q))\\ Pr[|F\cap \delta(Q)| \geq 2] \leq x^*(\delta(Q)) -1
因爲, QTiodd|Q\cap T_i| oddFiδ(Q)2|F_i \cap \delta(Q)| \geq 2,

所以Pr[QTiodd]Pr[Fδ(Q)2]x(δ(Q))1Pr[|Q\cap T_i| odd]\leq Pr[|F \cap \delta(Q)| \geq 2] \leq x^*(\delta(Q))-1

Lemma:

QCτCeQeEcexe \sum_{Q \in C_{\tau}}C_{e_Q} \leq \sum_{e \in E} c_ex_e^*

Proof:
QCτCeQcost MSTeEcexe \sum_{Q \in C_{\tau}}C_{e_Q} \leq \text{cost MST} \leq \sum_{e \in E} c_ex_e^*
構造過程, 對於QCτQ\in C_{\tau}, 將一條MST的邊e映射到Q, 每次移除一個e, 然後構造s和v.

在這裏插入圖片描述

Theorem: Best-of-Many Christofides算法是1+52\frac{1+\sqrt{5}}{2}算法.

Proof:

Best s-t Pathiλic(FiJi)=iλi[c(Fi)+αc(Fi)+βeEcexe+QCτ,QTi(12αβx(δ(Q)))ceQ](1+α+β)eEcexe+QCτ(x(δ(Q))1)(12αβx(δ(Q)))ceQ(1+α+β)eEcexe+max0z<τE(12α+β(1+z))QCτceQ(1+α+β+max0z<τE(12α+β(1+z)))eEcexe=(1+α+β+max0z<τE(βτβz))eEcexe[Maximizedatz=τ/2](1+α+β+β(τ2)2)OPTLP(2β+(3β1)24β)OPTLP \begin{aligned} & \text{Best s-t Path} \leq \sum_i \lambda_i c(F_i \cup J_i)\\ & = \sum_i \lambda_i[c(F_i) + \alpha c(F_i) + \beta\sum_{e\in E} c_e x_e^* + \sum_{Q \in C_{\tau},|Q\cap T_i|}(1-2\alpha - \beta x^*(\delta(Q)))c_{e_Q}]\\ & \leq (1+\alpha + \beta) \sum_{e\in E} c_e x_e^* + \sum_{Q\in C_{\tau}}(x^*(\delta(Q))-1)(1-2\alpha -\beta x^*(\delta(Q)))c_{e_Q}\\ & \leq (1+\alpha + \beta) \sum_{e\in E} c_e x_e^* + max_{0 \leq z < \tau}E(1-2\alpha + \beta(1+z))\sum_{Q\in C_{\tau}}c_{e_Q}\\ & \leq (1+\alpha + \beta + max_{0 \leq z < \tau}E(1-2\alpha + \beta(1+z)))\sum_{e\in E} c_e x_e^*\\ & = (1+\alpha + \beta + max_{0 \leq z < \tau}E(\beta \tau - \beta z))\sum_{e\in E} c_e x_e^* [Maximized at z=\tau/2]\\ & \leq (1+\alpha + \beta + \beta (\frac{\tau}{2})^2)OPT_{LP}\\ & \leq (2 - \beta + \frac{(3\beta -1)^2}{4\beta})OPT_{LP} \end{aligned}
在這裏插入圖片描述

證畢.

前後四篇工作的算法分析總體思路是一致的,都和wolsey分析的過程是相似, 以最後1.5的爲例,

找到一個生成樹F和一個滿足HK relaxation的點z, 證明:

  1. l(F)OPTl(F) \leq OPT,
  2. l(z)OPTl(z) \leq OPT,
  3. z/2PQTjoinz/2 \in P_{Q_{T-join}}, 其中Q_T := odd(T)\bigtriangleup{s,t}

T和一個T-join構成解,並且l(F)+l(J)l(T)+l(z)/23/2OPTl(F)+l(J) \leq l(T) + l(z)/2 \leq 3/2 OPT

而之前的工作主要是弱化了第二條的要求,使得l(z)(1+c)OPTl(z) \leq (1+c) OPT.

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