Online Stochastic Matching: New Algorithms and Bounds

Abstract

Online matching has received significant attention in recent years due to its close connection to Internet advertising. As the seminal work of Karp, Vazirani, and Vazirani has an optimal $(1 - 1/{\mathbf {\mathsf{{e}}}})$ competitive ratio in the standard adversarial online model, much effort has gone into developing useful online models that incorporate some stochasticity in the arrival process. One such popular model is the “known I.I.D. model” where different customer-types arrive online from a known distribution. We develop algorithms with improved competitive ratios for some basic variants of this model with integral arrival rates, including: (a) the case of general weighted edges, where we improve the best-known ratio of 0.667 due to Haeupler, Mirrokni and Zadimoghaddam (WINE, 2011) to 0.705; and (b) the vertex-weighted case, where we improve the 0.7250 ratio of Jaillet and Lu (Math Oper Res 39(3):624–646, 2013) to 0.7299. We also consider an extension of stochastic rewards, a variant where each edge has an independent probability of being present. For the setting of stochastic rewards with non-integral arrival rates, we present a simple optimal non-adaptive algorithm with a ratio of $1-1/{\mathbf {\mathsf{{e}}}}$. For the special case where each edge is unweighted and has a uniform constant probability of being present, we improve upon $1-1/{\mathbf {\mathsf{{e}}}}$ by proposing a strengthened LP benchmark. One of the key ingredients of our improvement is the following (offline) approach to bipartite-matching polytopes with additional constraints. We first add several valid constraints in order to get a good fractional solution $\mathbf {f}$; however, these give us less control over the structure of $\mathbf {f}$. We next remove all these additional constraints and randomly move from $\mathbf{f }$ to a feasible point on the matching polytope with all coordinates being from the set $\{0, 1/k, 2/k, \ldots , 1\}$ for a chosen integer k. The structure of this solution is inspired by Jaillet and Lu (2013) and is a tractable structure for algorithm design and analysis. The appropriate random move preserves many of the removed constraints (approximately with high probability and exactly in expectation). This underlies some of our improvements and could be of independent interest.

A Appendix

1.1 A.1 Supplementary Materials in Sect. 3 (Edge-Weighted Model)

1.1.1 A.1.1 Proof of Lemma 5

We will prove Lemma 5 using the following three Claims. Recall that we had one kind of large edge, while two kinds of small edges. Hence, the following claim characterizes the performance of each of them.

Claim 9

For a large edge e, $\mathsf {EW} _{1}[h]$ (3) with parameter h achieves a competitive ratio of ${\mathsf {R}} [{\mathsf {EW}} _{1},{ 2/3}] = 0.67529+(1-h) *0.00446$.

Claim 10

For a small edge e of type $\varGamma _{1}$, $\mathsf {EW} _{1}[h]$ (3) achieves a competitive ratio of ${\mathsf {R}} [{\mathsf {EW}} _{1},{ 1/3}] =0.751066$, regardless of the value h.

Claim 11

For a small edge e of type $\varGamma _{2}$, $\mathsf {EW} _{1}[h]$ (3) achieves a competitive ratio of ${\mathsf {R}} [{\mathsf {EW}} _{1},{1/3}]= 0.72933+ h *0.040415$.

By setting $h = 0.537815$, the two types of small edges have the same ratio and we get that ${\mathsf {EW}}_{1}[h]$achieves $(\mathsf {R}[{\mathsf {EW}}_{1} , 2/3], \mathsf {R}[{\mathsf {EW}}_{1}, 1/3]) =(0.679417, 0.751066)$. Thus, this proves Lemma 5.

Proof of Claim 9

Consider a large edge $e=(u, v_1)$ in the graph $G_{\mathbf {F}}$. Let $e'=(u,v_{2})$ be the other small edge incident to u. Edges e and $e'$ can appear in $[M_{1}, M_{2}, M_{3}]$ in the following three ways.

• $\alpha _1$: $e \in M_1, e' \in M_2, e\in M_3$.

• $\alpha _2$: $e' \in M_1, e \in M_2, e\in M_3$.

• $\alpha _3$: $e \in M_1, e \in M_2, e'\in M_3$.

Notice that the random triple of matchings $[M_{1}, M_{2}, M_{3}]$ is generated by invoking $\mathsf {PM}[\mathbf{F} , 3]$. Since $\mathsf {PM}[\mathbf{F} , 3]$ considers a uniform random permutation we have that $\alpha _{i}$ will occur with probability 1/3 for $1 \le i \le 3$. For $\alpha _{1}$ and $\alpha _{2}$, we can ignore the second copy of e in $M_3$ and from Lemma 4 we have

$$\begin{aligned} \Pr [e\hbox { is matched }|~\alpha _1] {=} 0.580831 {\text { and }} \Pr [e\hbox { is matched }|~\alpha _2] {=} 0.148499 \end{aligned}$$

For $\alpha _{3}$, we have

$$\begin{aligned} \Pr [e\hbox { is matched }|~\alpha _3]\ge & \sum \limits _{t=1}^{n} \frac{1}{n} \left( 1 - \frac{2}{n} \right) ^{t-1} + \sum \limits _{t=1}^{n} \frac{1}{n} \left( \frac{t-1}{n} \right) \left( 1 - \frac{2}{n} \right) ^{t-2}\\&+\, \sum \limits _{t=1}^{n} \frac{1}{n} \left( \frac{(t-1)(t-2)}{2n^2} \right) \left( 1 - \frac{2}{n} \right) ^{t-3}\\&+\, (1-h) \sum \limits _{t=1}^{n} \frac{1}{n} \left( \frac{1}{n^3} \right) {t-1 \atopwithdelims ()3} \left( 1 - \frac{2}{n} \right) ^{t-4}\\\ge & 0.621246 + (1-h)*0.00892978 \end{aligned}$$

There are four terms in the summation above. The four terms denote the probabilities that $v_1$ comes for the first time at some time $t \in [T]$ and $v_2$ arrives for 0, 1, 2 and 3 times before t respectively. Note that in the last term when $v_2$ comes for a third time at some time before t, we need to ensure that $v_3$ never matches u which occurs with probability $1-h$ as described in $\mathsf {EW} _1$.

Recall that $\mathsf {R}[{\mathsf {EW}}_{1}, 2/3]$ denotes the competitive ratio for a large edge. By definition, we have

$$\begin{aligned} {\mathsf {R}}[{\mathsf {EW}}_{1},{ 2/3}]&=\frac{\Pr [e {\text {~is matched}}]}{2/3}\\&=\frac{\frac{1}{3}\sum _{i=1}^3\Pr [e {\text {~is matched}} |~ \alpha _i]}{2/3}\\&\ge 0.67529 + (1-h)*0.00446489 \end{aligned}$$

Proof of Claims 10 and 11

Consider a small edge $e=(u,v)$ of type $\varGamma _{1}$. Let $e_{1}$ and $e_{2}$ be the two other small edges incident to u. For a given triple of matchings $[M_{1}, M_{2}, M_{3}]$, we say e is of type $\psi _{1}$ if e appears in $M_{1}$ while the other two in the remaining two matchings. Similarly, we define the type $\psi _{2}$ and $\psi _{3}$ for the case where e appears in $M_{2}$ and $M_{3}$ respectively. Notice that the probability that e is of type $\psi _{i},~1 \le i \le 3$ is 1/3. $\square$

Similar to the calculations in the proof of Claim 9, we have $\Pr [e\hbox { is matched } |~\psi _{1}] \ge 0.571861$, $\Pr [e\hbox { is matched } |~\psi _{2}] \ge 0.144776$ and $\Pr [e\hbox { is matched } |~\psi _{3}] \ge 0.0344288$. Therefore we have

$$\begin{aligned} \Pr [ e\hbox { is matched}] =\frac{1}{3}\sum \limits _{i=1}^{3} \Pr [e\hbox { is matched }|~\psi _i] \ge \frac{1}{3}{{\mathsf {R}}}[{\mathsf {EW}}_{1}, { 1/3}] \end{aligned}$$

where $\mathsf {R}[{\mathsf {EW}}_{1}, 1/3]=0.751066$.

Consider a small edge $e=(u,v)$ of type $\varGamma _{2}$, we define type $\beta _{i}, 1 \le i \le 3$, if e appears in $M_{i}$ while the large edge $e'$ incident to u appears in the remaining two matchings. Similarly, we have $\Pr [e\hbox { is matched} |~\psi _{1}] \ge 0.580831$, $\Pr [e\hbox { is matched} |~\psi _{2}] \ge 0.148499$ and $\Pr [e\hbox { is matched} |~\psi _{3}] \ge h *0.0404154$.

Hence, the ratio for a small edge of type $\varGamma _{2}$ is $\mathsf {R}[\mathsf {EW}_{1}, 1/3]=0.72933 + h*0.0404154$.

1.1.2 A.1.2 Proof of Lemma 6

We will prove Lemma 6 using the following two Claims.

Claim 12

For a large edge e, $\mathsf {EW} _{2}[y_{1},y_{2}]$ (5) achieves a competitive ratio of

$$\begin{aligned} {\mathsf {R}} [{\mathsf {EW}} _{2},{ 2/3}] =\min \Big ( 0.948183 - 0.099895 y_{1} - 0.025646 y_{2}, 0.871245\Big ) \end{aligned}$$

Claim 13

For a small edge e, $\mathsf {EW} _{2}[y_{1},y_{2}]$ (5) achieves a competitive ratio of ${\mathsf {R}} [{\mathsf {EW}} _{2},{ 1/3}] =0.4455$, when $y_{1}=0.687, y_{2}=1$.

Therefore, by setting $y_{1}=0.687, y_{2}=1$ we get that ${\mathsf {R}} [{\mathsf {EW}} _{2},{ 2/3}] = 0.8539$ and ${\mathsf {R}} [{\mathsf {EW}} _{2},{ 1/3}] = 0.4455$, which proves Lemma 6.

Proof of Claim 12

Figure 5 shows the two possible configurations for a large edge.

Fig. 5

Diagram of configurations for a large edge $e=(u,v_{1})$. Thin and Thick lines represent small and large edges respectively

Consider a large edge $e=(u,v_{1})$ with the configuration (A). From ${\mathsf {PM}}^{*}[\mathbf {F},2][y_{1},y_{2}]$, we know that e will always be in $M_{1}$ while $e'=(u,v_{2})$ will be in $M_{1}$ and $M_{2}$ with probability $y_{1}/3$ and $y_{2}/3$ respectively.

We now have the following cases:

• $\alpha _1$: $e \in M_1$ and $e'\in M_1$. This happens with probability $y_{1}/3$. In this case, e is matched if $v_1$ comes for the first time at some time $t \in [T]$ and $v_2$ never comes before t. Thus,

  $$\begin{aligned} \Pr [e\hbox { is matched }|~\alpha _1] =\sum _{t=1}^n \frac{1}{n} \Big (1-\frac{2}{n}\Big )^{t-1}\ge 0.432332 \end{aligned}$$

• $\alpha _2$: $e\in M_1$ and $e'\in M_2$. This happens with probability $y_{2}/3$. In this case, e is matched if $v_1$ comes for the first time at some time $t \in [T]$ and $v_2$ comes at most once before t. Note that this case is essentially the same as $P_1$ described in Lemma 4. Thus, we have

  $$\begin{aligned} \Pr [e\hbox { is matched }|~\alpha _2] \ge 0.580831 \end{aligned}$$

• $\alpha _3$: $e\in M_1$ and $e'\not \in M_1, e' \not \in M_{2}$. This happens with probability $(1-y_{1}/3-y_{2}/3)$. In this case, e is matched if $v_1$ comes at least once. Thus, $\Pr [e\hbox { is matched}|~\alpha _1] =1-1/{\mathbf {\mathsf{{e}}}}\ge 0.632121$.

Therefore we have

$$\begin{aligned} \Pr [ e\hbox { is matched}]= & \Big (\frac{y_{1}}{3}\Pr [e\hbox { is matched }|~\alpha _1] +\frac{y_{2}}{3}\Pr [e\hbox { is matched } |~\alpha _2]\\&+\, (1-\frac{y_{1}}{3}-\frac{y_{2}}{3})\Pr [e\hbox { is matched } |~\alpha _3] \Big )\\\ge & \frac{2}{3} (0.948183 - 0.099895 y_{1} - 0.025646 y_{2}) \end{aligned}$$

Consider the configuration (B). From ${\mathsf {PM}}^{*}[\mathbf {F},2][y_{1},y_{2}]$, we know that e will always be in $M_{1}$ and $e'=(u,v_{2})$ will always be in $M_{2}$. Thus we have

$$\begin{aligned} \Pr [ e\hbox { is matched}] =\Pr [e\hbox { is matched }|~\alpha _2] =\frac{2}{3}*0.871245 \end{aligned}$$

Hence, this completes the proof of Claim 12.

Proof of Claim 13

Figure 6 shows all possible configurations for a small edge.

Fig. 6

Diagram of configurations for a small edge $e=(u,v_{1})$. Thin and Thick lines represent small and large edges respectively

Similar to the proof of Claim 12, we will do a case-by-case analysis on the various configurations. Let $e_{i}=(u,v_{i})$ for $1 \le i \le 3$ and ${\mathcal {E}}$ be the event that $e_{1}$ gets matched. For a given $e_{i}$, denote $e_{i} \in M_{0}$ if $e_{i} \notin M_{1}, e_{i} \not \in M_{2}$.

• (1a). Observe that $e_{1} \in M_{2}$ and $e_{2} \in M_{1}$. Thus we have $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}] =\frac{1}{3}*0.44550$.

• (1b). Observe that we have two cases: $\{ \alpha _{1}: e_{2} \in M_{1}, e_{1} \in M_{1}\}$ and $\{ \alpha _{2}: e_{2} \in M_{1}, e_{1} \in M_{2}\}$. Case $\alpha _{1}$ happens with probability $y_{1}/3$ and the conditional probability is $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{1}]=0.432332$. Case $\alpha _{2}$ happens with probability $y_{2}/3$ and the conditional is $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{2}]=0.148499$. Thus we have

  $$\begin{aligned} \Pr [{{\,\mathrm{\mathcal {E}}\,}}]=y_{1}/3 *\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{1}] +y_{2}/3 *\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{2}] \ge \frac{1}{3}(0.432332 y_1 + 0.148499 y_2 ) \end{aligned}$$

• (2a). Observe that $e_{1} \in M_{2}, e_{2} \in M_{2}, e_{3} \in M_{2}$. $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}] =\frac{1}{3}*0.601704$

• (2b). Observe that we have two cases: $\{ \alpha _{1}: e_{1} \in M_{1}, e_{2} \in M_{2}, e_{3} \in M_{2}\}$ and $\{ \alpha _{2}: e_{1} \in M_{2}, e_{2} \in M_{2}, e_{3} \in M_{2}\}$. Case $\alpha _{1}$ happens with probability $y_{1}/3$ and the conditional is $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{1}]=0.537432$. Case $\alpha _{2}$ happens with probability $y_{2}/3$ and conditional is $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{2}]=0.200568$. Thus we have

  $$\begin{aligned} \Pr [{{\,\mathrm{\mathcal {E}}\,}}]=y_{1}/3 *\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{1}] +y_{2}/3 *\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{2}] \ge \frac{1}{3}(0.537432 y_1 + 0.200568 y_2 ) \end{aligned}$$

• (3a). Observe that we have three cases: $\{ \alpha _{1}: e_{1} \in M_{2}, e_{2} \in M_{1}, e_{3} \in M_{2}\}$, $\{ \alpha _{2}: e_{1} \in M_{2}, e_{2} \in M_{2}, e_{3} \in M_{2}\}$ and $\{ \alpha _{2}: e_{1} \in M_{2}, e_{2} \in M_{0}, e_{3} \in M_{2}\}$. Case $\alpha _{1}$ happens with probability $y_{1}/3$ and conditional is $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{1}]=0.13171$. Case $\alpha _{2}$ happens with probability $y_{2}/3$ and conditional is $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{2}]=0.200568$. Case $\alpha _{3}$ happens with probability $(1-y_{1}/3-y_{2}/3)$ and conditional is $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{3}]=0.22933$.

  Similarly, we have

  $$\begin{aligned} \Pr [{{\,\mathrm{\mathcal {E}}\,}}]\,= & \,y_{1}/3 *\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{1}] +y_{2}/3 *\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{2}]+ (1-y_{1}/3-y_{2}/3) *\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{3}]\\&\ge \frac{1}{3}(0.13171 y_1 + 0.200568 y_2+(3-y_{1}-y_{2}) 0.22933 ) \end{aligned}$$

• (3b). Observe that we have six cases.

  • $\alpha _{1}: e_{1} \in M_{1}, e_{2} \in M_{1}, e_{3} \in M_{2}$. $\Pr [\alpha _{1}]=y_{1}^{2}/9$ and $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{1}]=0.4057$.

  • $\alpha _{2}: e_{1} \in M_{1}, e_{2} \in M_{2}, e_{3} \in M_{2}$. $\Pr [\alpha _{2}]=y_{1}y_{2}/9$ and $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{2}]=0.5374$.

  • $\alpha _{3}: e_{1} \in M_{1}, e_{2} \in M_{0}, e_{3} \in M_{2}$. $\Pr [\alpha _{3}]=y_{1}/3(1-y_{1}/3-y_{2}/3)$ and $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{3}]=0.58083$.

  • $\alpha _{4}: e_{1} \in M_{2}, e_{2} \in M_{1}, e_{3} \in M_{2}$. $\Pr [\alpha _{4}]=y_{1}y_{2}/9, \Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{4}]=0.1317$.

  • $\alpha _{5}: e_{1} \in M_{2}, e_{2} \in M_{2}, e_{3} \in M_{2}$. $\Pr [\alpha _{5}]=y_{2}^{2}/9, \Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{5}]=0.2006$.

  • $\alpha _{6}: e_{1} \in M_{2}, e_{2} \in M_{0}, e_{3} \in M_{2}$. $\Pr [\alpha _{6}]=y_{2}/3(1-y_{1}/3-y_{2}/3)/3$ and $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{6}]=0.22933$.

  Therefore we have

  $$\begin{aligned} \Pr [{{\,\mathrm{\mathcal {E}}\,}}]\ge & \frac{1}{3} \Big (0.135241y_1^2 + 0.223033 y_1 y_2+ 0.066856 y_2^2 \\&+\,y_1(3-y_1-y_2)0.193610 + y_2(3-y_1-y_2)0.076443 \Big ) \end{aligned}$$

• (4a). Observe that we have following six cases.

  • $\alpha _{1}: e_{1} \in M_{2}, e_{2} \in M_{1}, e_{3} \in M_{1}$. $\Pr [\alpha _{1}]=y_{1}^{2}/9$ and $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{1}]=0.08898$.

  • $\alpha _{2}: e_{1} \in M_{2}, e_{2} \in M_{2}, e_{3} \in M_{2}$. $\Pr [\alpha _{2}]=y_{2}^{2}/9$ and $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{2}]=0.2006$.

  • $\alpha _{3}: e_{1} \in M_{2}, e_{2} \in M_{0}, e_{3} \in M_{0}$. $\Pr [\alpha _{3}]=(1-y_{1}/3-y_{1}/3)^{2},$ and $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{3}]=0.2642$.

  • $\alpha _{4}$: $e_{1} \in M_{2}$ while either $e_{2} \in M_{1}, e_{3} \in M_{2}$ or $e_{2} \in M_{2}, e_{3} \in M_{1}$. $\Pr [\alpha _{2}]=2y_{1}y_{2}/9$ and $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{4}]=0.1317$.

  • $\alpha _{5}$: $e_{1} \in M_{2}$ while either $e_{2} \in M_{1}, e_{3} \in M_{0}$ or $e_{2} \in M_{0}, e_{3} \in M_{1}$. $\Pr [\alpha _{5}]=2y_{1}/3(1-y_{1}/3-y_{2}/3)$ and $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{5}]=0.14849$.

  • $\alpha _{6}$: $e_{1} \in M_{2}$ while either $e_{2} \in M_{2}, e_{3} \in M_{0}$ or $e_{2} \in M_{0}, e_{3} \in M_{2}$. $\Pr [\alpha _{5}]=2y_{2}/3(1-y_{1}/3-y_{2}/3)$ and $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}|\alpha _{6}]=0.22933$.

  Therefore we have

  $$\begin{aligned} \Pr [{{\,\mathrm{\mathcal {E}}\,}}]\ge & \frac{1}{3} \Big ( 0.029661y_1^2 + 2 *0.043903y_1y_2 + 0.066856y_2^2 + 2 y_1(3-y_1-y_2) 0.0494997 \\&+\, 2 y_2(3-y_1-y_2)(0.076443) +(3 - y_1-y_2)^2 0.0880803 \Big ) \end{aligned}$$

• (4b). Observe that in this configuration, we have additional six cases to the ones discussed in (4a). Let $\alpha _{i}$ be the cases defined in (4a) for each $1 \le i \le 6$. Notice that each $\Pr [\alpha _{i}]$ has a multiplicative factor of $y_{2}/3$. Now, consider the six new cases.

  • $\beta _{1}: e_{1} \in M_{1}, e_{2} \in M_{1}, e_{3} \in M_{1}$. $\Pr [\alpha _{1}]=y_{1}^{3}/27$ and $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{1}]=0.3167$.

  • $\beta _{2}: e_{1} \in M_{1}, e_{2} \in M_{2}, e_{3} \in M_{2}$. $\Pr [\alpha _{2}]=y_{1}y_{2}^{2}/27$ and $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{2}]=0.5374$.

  • $\beta _{3}: e_{1} \in M_{1}, e_{2} \in M_{0}, e_{3} \in M_{0}$. $\Pr [\alpha _{3}]=y_{1}/3 *(1-y_{1}/3-y_{2}/3)^{2}$ and $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{3}]=0.632$.

  • $\beta _{4}$: $e_{1} \in M_{1}$ and either $e_{2} \in M_{1}, e_{3} \in M_{2}$ or $e_{2} \in M_{2}, e_{3} \in M_{1}$. $\Pr [\alpha _{2}]=2y_{1}^{2}y_{2}/27$ and $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{4}]=0.4057$.

  • $\beta _{5}$: $e_{1} \in M_{1}$ and either $e_{2} \in M_{1}, e_{3} \in M_{0}$ or $e_{2} \in M_{0}, e_{3} \in M_{1}$. $\Pr [\alpha _{5}]=2y_{1}^{2}/9 *(1-y_{1}/3-y_{2}/3)$ and $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}| \alpha _{5}]=0.4323$.

  • $\beta _{6}$: $e_{1} \in M_{1}$ and either $e_{2} \in M_{2}, e_{3} \in M_{0}$ or $e_{2} \in M_{0}, e_{3} \in M_{2}$. $\Pr [\alpha _{5}]=2y_{1}y_{2}/9 *(1-y_{1}/3-y_{2}/3)$ and $\Pr [ {{\,\mathrm{\mathcal {E}}\,}}|\alpha _{6}]=0.58083$.

    Hence, we have

    $$\begin{aligned} \Pr [{{\,\mathrm{\mathcal {E}}\,}}]\ge & \frac{1}{3}\Big (0.632 y_{1} - 0.133133 y_{1}^2 + 0.0093y_{1}^3 + 0.264241 y_{2} \\&-\,0.11127 y_{1} y_{2} + 0.01170 y_{1}^2 y_{2} - 0.0232746 y_{2}^2 + 0.00488 y_{1} y_{2}^2 + 0.00068 y_{2}^3\Big ) \end{aligned}$$

Setting $y_{1}=0.687,~y_{2}=1$, we get that the competitive ratio for a small edge is 0.44550. The bottleneck cases are configurations (1a) and (1b).

1.2 A.2 Supplemental Materials in Sect. 4

1.2.1 A.2.1 Proof of Lemma 10 (Vertex-Weighted and Unweighted)

When $H_u=1$ and u is in the cycle $C_{1}$, [13] show that the competitive ratio of u is $1-2{\mathbf {\mathsf{{e}}}}^{-2}$. Hence, for the remaining cases, we use the following Claims.

Claim 14

If $H_u=1$and u is not in $C_{1}$, then we have ${\mathsf {R}} [\mathsf {RLA} ,1]\ge 0.735622$.

Claim 15

${\mathsf {R}} [{\mathsf {RLA}} ,{2/3}] \ge 0.7870$.

Claim 16

${{\mathsf {R}} }[{\mathsf {RLA}} ,{1/3}] \ge 0.8107$.

Recall that $A_{u,1}$ is the event that among the n random lists, there exists a list starting with u and $A_{u,2}^v$ is the event that among the n lists, there exist successive lists such that (1) all start with some $u'$ which are different from u but are neighbors of v; and (2) they ensure u will be matched.

Notice that $A_u$ is the probability that u gets matched in ${\mathsf {RLA}}[\mathbf {H}']$. For each u, we compute $\Pr [A_{u,1}]$ and $\Pr [A_{u,2}^v]$ for all possibilities of $v \sim u$ and using Lemma 9 we get $A_u$. We first discuss two different ways to calculate $\Pr [A_{u,2}^v]$ when v has different neighboring structures.

Fig. 7

Case 1 in calculation of $\Pr [A_{u,2}^v]$

Fig. 8

Case 2 in calculation of $\Pr [A_{u,2}^v]$

Two ways to compute the value $\Pr [A_{u,2}^v]$.

• 1. Case 1 v has two neighbors. Consider the case when v has two neighbors as shown in Fig. 7. In this case we choose a slightly direct approach to computing $\Pr [A_{u,2}^v]$.

  Assume v has two neighbors u and $u'$ as shown in Fig. 7. After modifications, assume $H'_{(u',v)}=y$, $H'_{(u,v)}=1-y$ and $H'_{u'}=x$. Thus, the second certificate event $A_{u,2}^v$ corresponds to the event (1) a list starting with $u'$ comes at some time $1 \le i <n$; (2) the list ${\mathcal {R}}_{v}=(u',u)$ comes for a second time at some j with $i<j \le n$. Note that the arrival rate of a list starting with $u'$ is $H'_{u'}=x/n$ and the rate of list ${\mathcal {R}}_{v}=(u',u)$ is y/n. Therefore we have

  $$\begin{aligned} \Pr [A_{u,2}^v]= & \sum _{i=1}^{n-1}\Big (x/n (1 - x/n)^{(i- 1)} (1 - (1 - y/n)^{(n - i)} \Big )\end{aligned}$$

  (25)
  $$\begin{aligned}\sim & \frac{x - {\mathbf {\mathsf{{e}}}}^{-y} x + (-1 + {\mathbf {\mathsf{{e}}}}^{-x}) y}{x - y} {\text {~~~ (if }}x \ne y) \end{aligned}$$

  (26)
  $$\begin{aligned}\sim & 1- {\mathbf {\mathsf{{e}}}}^{-x}(1+x) {\text {~~~ (if }}x = y) \end{aligned}$$

  (27)

• 2. Case 2 v has three neighbors. Consider the case when v has three neighbors as shown in Fig. 8. In this case, we approximate the value $\Pr [A_{u,2}^v]$ using the Markov Chain method, similar to [13].

  Focus on the case shown in Fig. 8 where v has three neighbors u, $u_{1}$ and $u_{2}$ with $H_u=1, H_{u_{1}}=1/3$ and $H_{u_{2}}=2/3$. Recall that after modifications, we have $H'_{(u_{1},v)}=b=0.1, H'_{(u_{2},v)}=c=0.15$ and $H'_{(u,v)}=d=0.75$. We simulate the process of u getting matched resulting from several successive random lists starting from either $u_{1}$ or $u_{2}$ by an n-step Markov Chain as follows. We have 5 states: $s_{1}=(0,0,0), s_{2}=(0,1,0), s_{3}=(0,0,1), s_{4}=(0,1,1)$ and $s_{5}=(1,*,*)$ and the three numbers in each triple correspond to u, $u_{1}$ and $u_{2}$ being matched(or not) respectively. State $s_{5}$ corresponds to u being matched; the matched status of $u_{1}$ and $u_{2}$ is irrelevant. The chain initially starts in $s_{1}$ and the probability of being in state $s_{5}$ after n steps gives an approximation to $\Pr [A_{u,2}^v]$. The one-step transition probability matrix M is shown as follows.

  $$\begin{aligned}&M_{1,2}=\frac{b}{n}, M_{1,3}= \frac{c+1/3}{n}, M_{1,1}=1-M_{1,2}-M_{1,3} \\&M_{2,4}=\frac{c+1/3}{n}+\frac{bc}{(c+d)n}, M_{2,5}=\frac{bd}{(c+d)n}, \\&M_{2,3}=1-M_{2,4}-M_{2,5}\\&M_{3,4}=\frac{b}{n}+\frac{cb}{(b+d)n} , M_{3,5}=\frac{cd}{(b+d)n}\\&M_{3,3}=1-M_{3,4}-M_{3,5}\\&M_{4,5}=\frac{b+c}{n}, M_{4,4}=1-M_{4,5}\\&M_{5,5}=1 \\&M_{i,j} = 0 \hbox { for all other}\ i, j \end{aligned}$$

  Notice that $M_{1,3}=\frac{c+1/3}{n}$ and not $\frac{2}{3n}$ since after modifications, the arrival rate of a list starting with $u_{2}$ decreases correspondingly.

Let us now prove the three Claims 14, 15 and 16. Here we give the explicit analysis for the case when $H_u=1$. For the remaining cases, similar methods can be applied. Hence, we omit the analysis and only present the related computational results which leads to the conclusion.

Proof of Claim 14

Notice that u is not in the cycle $C_{1}$ and thus Lemma 9 can be used. Figure 9 describes all possible cases when a node $u \in U$ has $H_u = 1$. (We ignore all those cases when $H_{u}<1$, since they will not appear in the $\mathsf {WS}$.)

Fig. 9

Vertex-weighted $H_u=1$ cases. The value assigned to each edge represents the value after the second modification. No value indicates no modification. Here, $x_1 = 0.2744$ and $x_2 = 0.15877$

Let $v_{1}$ and $v_{2}$ be the two neighbors of u with $H_{(u,v_{1})}=2/3$ and $H_{(u,v_{2})}=1/3$. In total, there are $4 \times 10$ combinations, where $v_{1}$ is chosen from some $\alpha _{i}, 1 \le i \le 4$ and $v_{2}$ is chosen from some $\beta _{i}, 1 \le i \le 9$. For $H_u=1$, we need to find the worst combination among these such that the value $A_u$ is minimized. We can find this $\mathsf {WS}$ using the Lemma 9.

For each type of $\alpha _{i}, \beta _{j}$, we compute the values it will contribute to the term $(1-A_{u,1})\prod _{v \sim u}(1-\Pr [A_{u,2}^v])$. For example, assume $v_{1}$ is of type $\alpha _{1}$, denoted by $v_{1}(\alpha _{1})$. It contributes ${\mathbf {\mathsf{{e}}}}^{-0.9}$ to the term $(1-A_{u,1})$ and $(1-\Pr [A_{u,2}^{v_1}])$ to $\prod _{v \sim u}(1-\Pr [A_{u,2}^v])$, thus the total value it contributes is $\gamma (v_{1}, \alpha _{1})={\mathbf {\mathsf{{e}}}}^{-0.9}(1-\Pr [A_{u,2}^{v_1}])$. Similarly, we can compute all $\gamma (v_{1}, \alpha _{i})$ and $\gamma (v_{2}, \beta _{j})$. Let $i^{*}={{\,\mathrm{arg\,max}\,}}_{i}\gamma (v_{1}, \alpha _{i})$ and $j^{*}={{\,\mathrm{arg\,max}\,}}_{j}\gamma (v_{2}, \beta _{j})$. The $\mathsf {WS}$ is for the combination $\{v_{1}(\alpha _{i^{*}}), v_{2}(\beta _{j^{*}})\}$ and the resulting value of $A_u$ and $\mathsf {R}[\mathsf {RLA}, 1]$ is as follows:

$$\begin{aligned}&A_u=1-\gamma (v_{1}, \alpha _{i^{*}})\gamma (v_{2}, \beta _{j^{*}})\\&{{\mathsf {R}}}[{{{\mathsf {RLA}}}},{1}]=A_u/H_u=A_u \end{aligned}$$

Here is a list of $\gamma (v_{1}, \alpha _{i})$ and $\gamma (v_{2}, \beta _{j})$, for each $1 \le i \le 4$ and $1 \le j \le 9$.

• $\alpha _{1}$: We have $\Pr [A_{u,2}^v]=1-{\mathbf {\mathsf{{e}}}}^{-0.1}*1.1$ and $\gamma (v, \alpha _{1})={\mathbf {\mathsf{{e}}}}^{-0.1}*1.1*{\mathbf {\mathsf{{e}}}}^{-0.9}=0.404667$.

• $\alpha _{2}$: $\Pr [A_{u,2}^v] \ge 1-{\mathbf {\mathsf{{e}}}}^{-0.15}*1.15$ and $\gamma (v, \alpha _{2}) \le 0.423$.

  Notice that after modifications, $H'_{u_{1}} \ge 0.15$. Hence, we use this and Eq. (25) to compute the lower bound of $\Pr [A_{u,2}^v]$.

• $\alpha _{3}$: $\Pr [A_{u,2}^v] \ge 0.0916792$ and $\gamma (v, \alpha _{3}) \le 0.439667$.

  Notice that for any large edge e incident to a node u with $H_u=1$ (before modification), we have after modification, $H'_{e} \ge 1-0.2744=0.7256$. Thus we have $H'_{(u_{1},v_{1})} \ge 0.7256$ and $H'_{u_{1}} \ge 1$. From Eq. (25), we get $\Pr [A_{u,2}^v] \ge 0.0916792$.

• $\alpha _{4}$: $\Pr [A_{u,2}^v] \ge 0.0307466$ and $\gamma (v, \alpha _{4}) \le 0.417923$.

  Notice that for any small edge e incident to a node u with $H_u=1$ (before modification), we have after modification, $H'_{e} \ge 0.15877$. Thus, we have $H'_{u_{1}} \ge 3*0.15877$.

• $\beta _{1}$: $\Pr [A_{u,2}^v] =0.1608$ and $\gamma (v, \beta _{1})=0.601313$.

• $\beta _{2}$: $\Pr [A_{u,2}^v] \ge 0.208812$ and $\gamma (v, \beta _{2}) \le 0.601313$.

  After modifications, we have $H'_{(u_{1},v_{1})} \ge 0.2744$ and thus we get $H'_{u_{1}} \ge 1$.

• $\beta _{3}$: $\Pr [A_{u,2}^v] \ge 0.251611$ and $\gamma (v, \beta _{2}) \le 0.63852$.

  After modifications, we have $H'_{(u_{1},v_{1})} \ge 0.2744$ and thus we get $H'_{u_{1}} \ge 1-0.15877+0.2744$.

• $\beta _{4}$: $\Pr [A_{u,2}^v] =0.121901$ and $\gamma (v, \beta _{4})=0.588607$.

• $\beta _{5}$: $\Pr [A_{u,2}^v] =0.1346$ and $\gamma (v, \beta _{5})=0.551803$.

• $\beta _{6}$: $\Pr [A_{u,2}^v] \ge 0.1140$ and $\gamma (v, \beta _{6}) \le 0.593904$.

• $\beta _{7}$: $\Pr [A_{u,2}^v] = 0.0084$ and $\gamma (v, \beta _{7}) =0.4455$.

• $\beta _{8}$: $\Pr [A_{u,2}^v] \ge 0.0397$ and $\gamma (v, \beta _{8}) \le 0.582451$.

• $\beta _{9}$: $\Pr [A_{u,2}^v] \ge 0.0230$ and $\gamma (v, \beta _{9}) \le 0.510039$.

Using the computed values above, let us compute the ratio of a node u with $H_u=1$.

• If u has three neighbors, then the $\mathsf {WS}$ configuration is when each of the three neighbors of u is of type $\beta _{3}$. This is because, the value of $\gamma (v, \beta _{3})$ is the largest. The resultant ratio is 0.73967.

• If u has two neighbors, then the $\mathsf {WS}$ configuration is when one of the neighbor is of type $\beta _{1}$ (or $\beta _{2}$) and the other is of type $\alpha _{3}$. The resultant ratio is 0.735622.

Proof of Claim 15

The proof is similar to that of Claim 14. The Fig. 10 shows all possible configurations of a node u with $H_u=2/3$. Note that the $\mathsf {WS}$ cannot have $F(v)<1$ and hence we omit them here. For a neighbor v of u, if $H_{(u,v)}=2/3$, then v is in one of $\alpha _{i}, 1 \le i \le 3$; if $H_{(u,v)}=1/3$, then v is in one of $\beta _{i}, 1 \le i \le 8$. We now list the values $\gamma (v, \alpha _{i})$ and $\gamma (v, \beta _{j})$, for each $1 \le i \le 3$ and $1 \le j \le 8$.

Fig. 10

Vertex-weighted $H_u=2/3$ cases. The value assigned to each edge represents the value after the second modification. No value indicates no modification

• $\alpha _{1}$: We have $\Pr [A_{u,2}^v]=1-{\mathbf {\mathsf{{e}}}}^{-0.25}*1.25$ and $\gamma (v, \alpha _{1})={\mathbf {\mathsf{{e}}}}^{-0.25}*1.25*{\mathbf {\mathsf{{e}}}}^{-0.75}=0.459849$.

• $\alpha _{2}$: We have $\Pr [A_{u,2}^v]\ge 0.0528016$ and $\gamma (v, \alpha _{1}) \le 0.470365$.

• $\alpha _{3}$. We have $\Pr [A_{u,2}^v]\ge 0.13398$ and $\gamma (v, \alpha _{3}) \le 0.475282$.

• $\beta _{1}$: We have $\Pr [A_{u,2}^v]=1-{\mathbf {\mathsf{{e}}}}^{-0.7}*1.7$ and $\gamma (v, \beta _{1}) =0.625395$.

• $\beta _{2}$: We have $\Pr [A_{u,2}^v] \ge 0.226356$ and $\gamma (v, \beta _{2}) \le 0.665882$.

• $\beta _{3}$: We have $\Pr [A_{u,2}^v] \ge 0.1819$ and $\gamma (v, \beta _{3}) \le 0.669804$.

• $\beta _{4}$: We have $\Pr [A_{u,2}^v] \ge 0.1130$ and $\gamma (v, \beta _{4}) \le 0.635563$.

• $\beta _{5}$: We have $\Pr [A_{u,2}^v] \ge 0.0587$ and $\gamma (v, \beta _{5}) \le 0.674471$.

• $\beta _{6}$: We have $\Pr [A_{u,2}^v] \ge 0.1688$ and $\gamma (v, \beta _{6}) \le 0.680529$.

• $\beta _{7}$: We have $\Pr [A_{u,2}^v] \ge 0.1318$ and $\gamma (v, \beta _{7}) \le 0.676155$.

• $\beta _{8}$: We have $\Pr [A_{u,2}^v] \ge 0.0587$ and $\gamma (v, \beta _{8}) \le 0.674471$.

Hence, the $\mathsf {WS}$ structure is when u is such that $H_u=2/3$ and has one neighbor of type $\alpha _3$. The resultant ratio is 0.7870.

Proof of Claim 16

The Fig. 11 shows the possible configurations of a node u with $H_u=1/3$. Again, we omit those cases where $H_{v}<1$.

Fig. 11

Vertex-weighted $H_u=1/3$ cases. The value assigned to each edge represents the value after the second modification. No value indicates no modification

We now list the values $\gamma (v, \alpha _{i})$, for each $1 \le i \le 8$.

• $\alpha _{1}$: We have $\Pr [A_{u,2}^v] = 1-{\mathbf {\mathsf{{e}}}}^{-0.75}*1.75$ and $\gamma (v, \alpha _{1}) =0.643789$.

• $\alpha _{2}$: We have $\Pr [A_{u,2}^v] \ge 0.282256$ and $\gamma (v, \alpha _{2}) \le 0.649443$.

• $\alpha _{3}$: We have $\Pr [A_{u,2}^v] \ge 0.1935$ and $\gamma (v, \alpha _{3}) \le 0.729751$.

• $\alpha _{4}$: We have $\Pr [A_{u,2}^v] \ge 0.0587$ and $\gamma (v, \alpha _{4}) \le 0.674471$.

• $\alpha _{5}$: $\gamma (v, \alpha _{5}) \le 0.674471$.

• $\alpha _{6}$: We have $\Pr [A_{u,2}^v] \ge 0.1546$ and $\gamma (v, \alpha _{6}) \le 0.727643$.

• $\alpha _{7}$: We have $\Pr [A_{u,2}^v] \ge 0.1938$ and $\gamma (v, \alpha _{7}) \le 0.72948$.

• $\alpha _{8}$: $\gamma (v, \alpha _{8}) \le 0.674471$.

Hence, the $\mathsf {WS}$ for node u with $H_u=1/3$ is when u has one neighbor of type $\alpha _3$. The resultant ratio is 0.8107.