SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \operatorname {R e g} _ {i} \left(\pi_ {i}, H _ {- i} ^ {T}, \varepsilon\right) \leq \frac {1}{\eta} \max  _ {\mathfrak {p} \in [ N ]} \log \left(\frac {1}{\sigma_ {\mathfrak {p}}}\right) + \frac {T L ^ {2} \eta}{8} \leq \frac {K \log \left(\lceil v _ {i , 1} / \varepsilon \rceil\right)}{\eta} + \frac {T L ^ {2} \eta}{8} \\ = \frac {K \log \left(\left\lceil \sqrt {\frac {T}{K}} \right\rceil\right)}{\eta} + \frac {T (K v _ {i , 1}) ^ {2} \eta}{8} \quad (\text {S i n c e} \epsilon = v _ {i, 1} \sqrt {\frac {K}{T}} \text {a n d} L = K v _ {i, 1}.) \\ \leq \frac {K \log T}{\eta} + \frac {T \left(K v _ {i , 1}\right) ^ {2} \eta}{8}. \tag {29} \\ \end{array}
$$
For  $\eta = \sqrt{\log T} / (v_{i,1}\sqrt{KT})$ , inequality (29) gives
$$
\begin{array}{l} \operatorname {R e g} _ {i} \left(\pi_ {i}, H _ {- i} ^ {T}, \varepsilon\right) \leq \frac {K \cdot \log T}{\sqrt {\log T} / \left(v _ {i , 1} \sqrt {K T}\right)} + \frac {T \sqrt {\log T}}{8 \left(v _ {i , 1} \sqrt {K T}\right)} \cdot \left(K v _ {i, 1}\right) ^ {2} \\ = \frac {9 v _ {i , 1}}{8} \sqrt {T K ^ {3} \log T}. \tag {30} \\ \end{array}
$$
Combining inequalities (26) and (30), we get that the regret of player  $i$  when running Algorithm 2 is
$$
\begin{array}{l} \operatorname {R e g} _ {i} \left(\pi_ {i}, H _ {- i} ^ {T}\right) \leq \operatorname {R e g} _ {i} \left(\pi_ {i}, H _ {- i} ^ {T}, \varepsilon\right) + v _ {i, 1} \sqrt {T K ^ {3}} \\ \leq \frac {9 v _ {i , 1}}{8} \sqrt {T K ^ {3} \log T} + v _ {i, 1} \sqrt {T K ^ {3}} \in O \left(v _ {i, 1} \sqrt {T K ^ {3} \log T}\right). \tag {31} \\ \end{array}
$$
This completes the proof of the regret upper bound.
Polynomial time implementation. Finally we analyze the running time of the above algorithm. At every step, the computation of path kernels traverses all edges (note that each edge only appears in the sum once) and could be done in  $O(|E|) = O(Kv_{i,1}^2 / \varepsilon^2)$  time. Similarly, the update of edge probabilities  $\phi^t$  for all edges also takes  $O(|E|) = O(Kv_{i,1}^2 / \varepsilon^2)$  time.
Therefore, the overall computational complexity is  $O(TKv_{i,1}^2 /\varepsilon^2) = O(T^2)$ , where we used the choice of  $\varepsilon = v_{i,1}\sqrt{K / T}$ . This completes the proof of the theorem.
B.2 Bandit Feedback
In this section we include the main theorem and proof for the bandit setting. Recall that our algorithm for the bandit setting is the same as Algorithm 2, only with  $w^{t}(e)$  replaced by  $\widehat{w}^t (e)$ :
$$
\widehat {w} ^ {t} (e) = \overline {{w}} (e) - \frac {\overline {{w}} (e) - w ^ {t} (e)}{p ^ {t} (e)} \mathbb {1} _ {\{e \in \mathfrak {p} ^ {t} \}}, \qquad \text {w i t h} p ^ {t} (e) = \sum_ {\mathfrak {p}: e \in \mathfrak {p}} P ^ {t} (\mathfrak {p}),
$$
with  $P^t$  given in (22), and
$$
\overline {{w}} (e) = \left\{ \begin{array}{l l} v _ {i, 1} - r + j (r - s) & \text {i f} e = (z _ {r, j}, z _ {s, j + 1}), \\ v _ {i, 1} - r + K r & \text {i f} e = (z _ {r, K}, z _ {+}), \\ 0 & \text {i f} e = (z _ {-}, z _ {r, 1}). \end{array} \right.
$$
The resolution parameter and learning rate are chosen to be
$$
\varepsilon = v _ {i, 1} \min  \left\{\left(K ^ {3} \log T / T\right) ^ {1 / 4}, 1 \right\}, \quad \eta = \min  \left\{\varepsilon \sqrt {\log \left(v _ {i , 1} / \varepsilon\right) / \left(T K ^ {3} v _ {i , 1} ^ {4}\right)}, 1 / \left(K v _ {i, 1}\right) \right\}. \tag {32}
$$
Theorem 3 (restated). For each player  $i$  and time horizon  $T$ , under the bandit feedback, there is an algorithm for bidding that runs in time  $O(TK + K^{-5/4}T^{7/4})$  and guarantees the player's regret is at most  $O(\min \{v_{i,1}(T^3 K^7 \log T)^{1/4}, v_{i,1}KT\})$ .
Proof. By Lemma 3 and the choice of  $\varepsilon$  in (32), it suffices to show that the above algorithm  $\pi_i$  runs in time  $O(TKv_{i,1}^3 /\varepsilon^3)$  and achieves
$$
\operatorname {R e g} _ {i} \left(\pi_ {i}, H _ {- i} ^ {T}, \varepsilon\right) = O \left(v _ {i, 1} ^ {2} \sqrt {T K ^ {5} \log \left(v _ {i , 1} / \varepsilon\right)} / \varepsilon + v _ {i, 1} K ^ {2} \log \left(v _ {i, 1} / \varepsilon\right)\right).
$$
The claimed result then follows from the fact that the regret is always upper bounded by  $O(v_{i,1}KT)$ .
Before we proceed to the proof, we first comment on the choice of estimator  $\widehat{w}^t(e)$ . First of all, this is an unbiased estimator of  $w^t(e)$ , i.e.  $\mathbb{E}_{\mathfrak{p}^t \sim P^t}[\widehat{w}^t(e)] = w^t(e)$  for every edge  $e$  in  $G^t$ . Second, instead of using the natural importance-weighted estimator  $\widehat{w}^t(e) = w^t(e) \mathbb{1}_{\{e \in \mathfrak{p}^t\}} / p^t(e)$ , the current form in (6) is the loss-based importance-weighted estimator used for technical reasons, similar to [LS20, Eqn. (11.6)]. Third, by exploiting our DAG structure and the definition of  $w^t(e)$  in (4), we construct an edge-specific quantity  $\overline{w}(e)$  which always upper bounds  $w^t(e)$ .
We now analyze the regret of the algorithm with  $\widehat{w}^t(e)$  given by (6). The standard EXP3 analysis (see, e.g. [LS20, Chapter 11]) gives that
$$
\operatorname {R e g} _ {i} \left(\pi_ {i}, H _ {- i} ^ {T}, \varepsilon\right) \leq \frac {1}{\eta} \max  _ {\mathfrak {p}} \log \frac {1}{P ^ {1} (\mathfrak {p})} + \sum_ {t = 1} ^ {T} \mathbb {E} \left[ \frac {1}{\eta} \log \left(\sum_ {\mathfrak {p}} P ^ {t} (\mathfrak {p}) e ^ {\eta \widehat {w} ^ {t} (\mathfrak {p})}\right) - \sum_ {\mathfrak {p}} P ^ {t} (\mathfrak {p}) \widehat {w} ^ {t} (\mathfrak {p}) \right], \tag {33}
$$
where  $\widehat{w}^t (\mathfrak{p}) = \sum_{e\in \mathfrak{p}}\widehat{w}^t (e)$  is the estimated total weight of path  $\mathfrak{p}$ , and the expectation is with respect to the randomness in the estimator  $\widehat{w}^t (e)$ . For every path  $\mathfrak{p} = (z_{-},z_{r_1,1},\dots ,z_{r_K,K},z_{+})$  from the source to the sink, we have (by convention  $r_{K + 1} = 0$ ):
$$
\widehat {w} ^ {t} (\mathfrak {p}) = \sum_ {e \in \mathfrak {p}} w ^ {t} (e) \leq \sum_ {e \in \mathfrak {p}} \overline {{w}} (e) = \sum_ {j = 1} ^ {K} \left(v _ {i, 1} - (j - 1) r _ {j} + j r _ {j + 1}\right) = K v _ {i, 1}.
$$
Since  $e^x \leq 1 + x + x^2$  whenever  $x \leq 1$  and  $\log(1 + y) \leq y$  whenever  $y > -1$ , if  $\eta \leq 1 / (Kv_{i,1})$ , inequality (33) gives
$$
\begin{array}{l} \operatorname {R e g} _ {i} \left(\pi_ {i}, H _ {- i} ^ {T}, \varepsilon\right) \leq \frac {1}{\eta} \max  _ {\mathfrak {p}} \log \frac {1}{P ^ {1} (\mathfrak {p})} + \eta \sum_ {t = 1} ^ {T} \sum_ {\mathfrak {p}} P ^ {t} (\mathfrak {p}) \mathbb {E} [ \widehat {w} ^ {t} (\mathfrak {p}) ^ {2} ] \\ \leq \frac {K}{\eta} \log \left\lceil \frac {v _ {i , 1}}{\varepsilon} \right\rceil + \eta \sum_ {t = 1} ^ {T} \sum_ {\mathfrak {p}} P ^ {t} (\mathfrak {p}) \cdot (K + 1) \sum_ {e \in \mathfrak {p}} \mathbb {E} [ \widehat {w} ^ {t} (e) ^ {2} ], \\ \end{array}
$$
where the last equality uses  $\left(\sum_{i=1}^{n} x_{i}\right)^{2} \leq n \sum_{i=1}^{n} x_{i}^{2}$ , and that each path  $\mathfrak{p}$  from the source to the sink has length  $K + 1$ . To proceed, note that
$$
\begin{array}{l} \mathbb {E} [ \widehat {w} ^ {t} (e) ^ {2} ] = \overline {{w}} (e) ^ {2} \cdot (1 - p ^ {t} (e)) + \left(\overline {{w}} (e) - \frac {\overline {{w}} (e) - w ^ {t} (e)}{p ^ {t} (e)}\right) ^ {2} \cdot p ^ {t} (e) \\ = w ^ {t} (e) ^ {2} + (\overline {{w}} (e) - w ^ {t} (e)) ^ {2} \cdot \left(\frac {1}{p ^ {t} (e)} - 1\right), \\ \end{array}
$$
$$
\begin{array}{l} \sum_ {\mathfrak {p}} P ^ {t} (\mathfrak {p}) \sum_ {e \in \mathfrak {p}} \mathbb {E} [ \widehat {w} ^ {t} (e) ^ {2} ] = \sum_ {\mathfrak {p}} P ^ {t} (\mathfrak {p}) \sum_ {e \in \mathfrak {p}} \left[ w ^ {t} (e) ^ {2} + (\bar {w} (e) - w ^ {t} (e)) ^ {2} \cdot \left(\frac {1}{p ^ {t} (e)} - 1\right) \right] \\ = \sum_ {e} \left[ w ^ {t} (e) ^ {2} + (\bar {w} (e) - w ^ {t} (e)) ^ {2} \cdot \left(\frac {1}{p ^ {t} (e)} - 1\right) \right] \sum_ {\mathfrak {p}: e \in \mathfrak {p}} P ^ {t} (\mathfrak {p}) \\ = \sum_ {e} \left[ w ^ {t} (e) ^ {2} p ^ {t} (e) + (\bar {w} (e) - w ^ {t} (e)) ^ {2} (1 - p ^ {t} (e)) \right] \leq \sum_ {e} \bar {w} (e) ^ {2}, \\ \end{array}
$$
where in the middle we have used the definition  $\sum_{\mathfrak{p}:e\in \mathfrak{p}}P^{t}(\mathfrak{p}) = p^{t}(e)$  in (6). To further upper bound the above quantity, note that
$$
\begin{array}{l} \sum_ {e} \bar {w} (e) ^ {2} \leq \sum_ {j = 1} ^ {K} \sum_ {1 \leq s \leq r \leq \lceil v _ {i, 1} / \varepsilon \rceil} \left(v _ {i, 1} + (j - 1) r \varepsilon\right) ^ {2} \\ \leq \left\lceil \frac {v _ {i , 1}}{\varepsilon} \right\rceil \sum_ {j = 1} ^ {K} \sum_ {r = 1} ^ {\lceil v _ {i, 1} / \varepsilon \rceil} (V + (j - 1) r \varepsilon) ^ {2} \\ \leq 2 \left\lceil \frac {v _ {i , 1}}{\varepsilon} \right\rceil \sum_ {j = 1} ^ {K} \sum_ {r = 1} ^ {\lceil v _ {i, 1} / \varepsilon \rceil} (v _ {i, 1} ^ {2} + (j - 1) ^ {2} r ^ {2} \varepsilon^ {2}) = O \left(\frac {K ^ {3} v _ {i , 1} ^ {4}}{\varepsilon^ {2}}\right). \\ \end{array}
$$
A combination of the above inequalities shows that as long as  $\eta \leq 1 / (Kv_{i,1})$
$$
\mathrm {R e g} _ {i} (\pi_ {i}, H _ {- i} ^ {T}, \varepsilon) = O \left(\frac {K}{\eta} \log \frac {v _ {i , 1}}{\varepsilon} + \frac {\eta T K ^ {4} v _ {i , 1} ^ {4}}{\varepsilon^ {2}}\right).
$$
Next, by the choice of  $\eta$  in (32), we have
$$
\operatorname {R e g} _ {i} \left(\pi_ {i}, H _ {- i} ^ {T}, \varepsilon\right) = O \left(\frac {v _ {i , 1} ^ {2} \sqrt {T K ^ {5} \log \left(v _ {i , 1} / \varepsilon\right)}}{\varepsilon} + K ^ {2} v _ {i, 1} \log \frac {v _ {i , 1}}{\varepsilon}\right).
$$
As for the computational complexity, the only difference from the Algorithm 2 is the additional computation of the estimated weights  $\widehat{w}^t (e)$  in (6), or equivalently, the marginal probability  $p^t (e)$ . As  $P^t$  has a product structure in (22), the celebrated message passing algorithm in graphical models [WJ08, Section 2.5.1] takes  $O(|E|v_{i,1} / \varepsilon) = O(Kv_{i,1}^3 /\varepsilon^3)$  time to compute all edge marginals  $\{p^{t}(e)\}_{e\in E}$ . Therefore the overall computational complexity is  $O(TKv_{i,1}^{3} / \varepsilon^{3})$ .
B.3 Regret Lower Bound
In this section we include the theorem and proof for the regret lower bound. The construction uses the  $(K + 1)$ -st highest price, but is similar for the  $K$ -th highest price.
Theorem 4 (restated, formal). Let  $K \geq 2$ . For any policy  $\pi_i$  used by player  $i$ , there exists a bid sequence  $\{\mathbf{b}_{-i}^t\}_{t=1}^T$  for the other players such that the expected regret in (3) satisfies  $\mathbb{E}[\mathrm{Reg}_i(\pi_i, \{\mathbf{b}_{-i}^t\}_{t=1}^T)] \geq cv_{i,1}K\sqrt{T}$ , where  $c > 0$  is an absolute constant.
Proof. Without loss of generality assume that  $K = 2k$  is an even integer, and by scaling we may assume that  $v_{i,1} = 1$ . Consider the following two scenarios:
- the utility of the bidder  $i$  is  $v_{i,j} \equiv 1$ , for all  $j \in [K]$ ;
- at scenario 1, for every  $t \in [T]$ , the other bidders' bids are
$$
\mathbf {b} _ {- i} ^ {t} = \left\{ \begin{array}{l l} \left(\frac {2}{3}, \frac {2}{3}, \dots , \frac {2}{3}, 0, \dots , 0\right) & \text {w i t h p r o b a b i l i t y} 0. 5 + \delta , \\ \left(\frac {2}{3}, \frac {2}{3}, \dots , \frac {2}{3}, \frac {2}{3}, \dots , \frac {2}{3}\right) & \text {w i t h p r o b a b i l i t y} 0. 5 - \delta , \end{array} \right.
$$
where the number of non-zero entries is  $k$  in the first line and  $2k$  in the second line, and  $\delta \in (0,1/4)$  is a parameter to be determined later. The randomness used at different times is independent.
- at scenario 2, for every  $t \in [T]$ , the other bidders' bids are
$$
\mathbf {b} _ {- i} ^ {t} = \left\{ \begin{array}{l l} \left(\frac {2}{3}, \frac {2}{3}, \dots , \frac {2}{3}, 0, \dots , 0\right) & \text {w i t h p r o b a b i l i t y} 0. 5 - \delta , \\ \left(\frac {2}{3}, \frac {2}{3}, \dots , \frac {2}{3}, \frac {2}{3}, \dots , \frac {2}{3}\right) & \text {w i t h p r o b a b i l i t y} 0. 5 + \delta , \end{array} \right.
$$
We denote by  $P$  and  $Q$  the distributions of  $\{\mathbf{b}_{-i}^t\}_{t = 1}^T$  under scenarios 1 and 2, respectively, then
$$
\begin{array}{l} D _ {\mathrm {K L}} (P \| Q) = T \cdot D _ {\mathrm {K L}} (\operatorname {B e r n} (0. 5 + \delta) \| \operatorname {B e r n} (0. 5 - \delta)) \\ \leq T \cdot \chi^ {2} (\operatorname {B e r n} (0. 5 + \delta) \| \operatorname {B e r n} (0. 5 - \delta)) \\ = T \cdot \frac {(2 \delta) ^ {2}}{(0 . 5 + \delta) (0 . 5 - \delta)} \leq \frac {6 4}{3} T \delta^ {2}. \\ \end{array}
$$
Consequently, by [Tsy09, Lemma 2.6],
$$
1 - \mathrm {T V} (P, Q) \geq \frac {1}{2} \exp \left(- D _ {\mathrm {K L}} (P \| Q)\right) \geq \frac {1}{2} \exp \left(- \frac {6 4 T \delta^ {2}}{3}\right).
$$
Next we investigate the separation between these two scenarios. It is clear that
$$
\begin{array}{l} \max  _ {\mathbf {b} _ {i}} \mathbb {E} _ {P} \left[ u _ {i} \left(\mathbf {b} _ {i}; \mathbf {b} _ {- i} ^ {t}\right) \right] \geq \mathbb {E} _ {P} \left[ u _ {i} \left(\left(1, 1, \dots , 1, 0, \dots , 0\right); \mathbf {b} _ {- i} ^ {t}\right) \right] \\ = \left(\frac {1}{2} + \delta\right) \cdot k + \left(\frac {1}{2} - \delta\right) \cdot \frac {k}{3} = \frac {2 + 2 \delta}{3} \cdot k; \\ \end{array}
$$
$$
\begin{array}{l} \max  _ {\mathbf {b} _ {i}} \mathbb {E} _ {Q} \left[ u _ {i} (\mathbf {b} _ {i}; \mathbf {b} _ {- i} ^ {t}) \right] \geq \mathbb {E} _ {Q} \left[ u _ {i} ((1, 1, \dots , 1, 1, \dots , 1); \mathbf {b} _ {- i} ^ {t}) \right] \\ = \left(\frac {1}{2} - \delta\right) \cdot \frac {2 k}{3} + \left(\frac {1}{2} + \delta\right) \cdot \frac {2 k}{3} = \frac {2 k}{3}. \\ \end{array}
$$
Moreover, under  $(P + Q) / 2$  (i.e.  $\mathbf{b}_{-i}^{t}$  follows a Bern(1/2) distribution), suppose that the vector  $\mathbf{b}_i$  has  $k^{\prime}$  components smaller than 2/3. Distinguish into two scenarios:
- if  $k' < k$ , then
$$
\mathbb {E} _ {(P + Q) / 2} \left[ u _ {i} (\mathbf {b} _ {i}; \mathbf {b} _ {- i} ^ {t}) \right] \leq \frac {1}{2} \cdot \frac {2 k - k ^ {\prime}}{3} + \frac {1}{2} \cdot \frac {2 k - k ^ {\prime}}{3} \leq \frac {2 k}{3};
$$
- if  $k' \geq k$ , then
$$
\mathbb {E} _ {(P + Q) / 2} \left[ u _ {i} \left(\mathbf {b} _ {i}; \mathbf {b} _ {- i} ^ {t}\right) \right] \leq \frac {1}{2} \cdot (2 k - k ^ {\prime}) + \frac {1}{2} \cdot \frac {2 k - k ^ {\prime}}{3} \leq \frac {2 k}{3}.
$$
Therefore, it always holds that
$$
\max  _ {\mathbf {b} _ {i}} \mathbb {E} _ {(P + Q) / 2} \left[ u _ {i} \left(\mathbf {b} _ {i}; \mathbf {b} _ {- i} ^ {t}\right) \right] \leq \frac {2 k}{3}.
$$
Consequently, for each  $\mathbf{b}_i$ ,
$$
\begin{array}{l} \max  _ {\mathbf {b} _ {i} ^ {\star}} \mathbb {E} _ {P} \left[ u _ {i} \left(\mathbf {b} _ {i} ^ {\star}; \mathbf {b} _ {- i} ^ {t}\right) - u _ {i} \left(\mathbf {b} _ {i}; \mathbf {b} _ {- i} ^ {t}\right) \right] + \max  _ {\mathbf {b} _ {i} ^ {\star}} \mathbb {E} _ {Q} \left[ u _ {i} \left(\mathbf {b} _ {i} ^ {\star}; \mathbf {b} _ {- i} ^ {t}\right) - u _ {i} \left(\mathbf {b} _ {i}; \mathbf {b} _ {- i} ^ {t}\right) \right] \\ \geq \max  _ {\mathbf {b} _ {i} ^ {\star}} \mathbb {E} _ {P} \left[ u _ {i} \left(\mathbf {b} _ {i} ^ {\star}; \mathbf {b} _ {- i} ^ {t}\right) \right] + \max  _ {\mathbf {b} _ {i} ^ {\star}} \mathbb {E} _ {Q} \left[ u _ {i} \left(\mathbf {b} _ {i} ^ {\star}; \mathbf {b} _ {- i} ^ {t}\right) \right] - 2 \max  _ {\mathbf {b} _ {i}} \mathbb {E} _ {(P + Q) / 2} \left[ u _ {i} \left(\mathbf {b} _ {i}; \mathbf {b} _ {- i} ^ {t}\right) \right] \\ \geq \frac {2 + 2 \delta}{3} k + \frac {2 k}{3} - 2 \cdot \frac {2 k}{3} = \frac {2 \delta k}{3}. \\ \end{array}
$$
In other words, any bid vector  $\mathbf{b}_i$  either incurs a total regret  $(\delta kT) / 3$  under  $P$ , or incurs a total regret  $(\delta kT) / 3$  under  $Q$ .
Now the classical two-point method (see, e.g. [Tsy09, Theorem 2.2]) gives
$$
\mathbb {E} _ {(P + Q) / 2} \left[ \operatorname {R e g} _ {i} \left(\pi_ {i}, \left\{\mathbf {b} _ {- i} ^ {t} \right\} _ {t = 1} ^ {T}\right) \right] \geq \frac {\delta k T}{3} \cdot (1 - \operatorname {T V} (P, Q)) \geq \frac {\delta k T}{6} \exp \left(- \frac {6 4 T \delta^ {2}}{3}\right)
$$
for every  $\delta \in (0,1 / 4)$ . Choosing  $\delta = 1 / (8\sqrt{T})$  gives that
$$
\mathbb {E} _ {(P + Q) / 2} \left[ \operatorname {R e g} _ {i} \left(\pi_ {i}, \left\{\mathbf {b} _ {- i} ^ {t} \right\} _ {t = 1} ^ {T}\right) \right] \geq \frac {K \sqrt {T}}{9 6 e ^ {1 / 3}},
$$
i.e. the claimed lower bound holds with  $c = 1 / (96e^{1 / 3})$ .
C Appendix: Equilibrium Analysis
In this section we show that the pure Nash equilibria with price zero of the  $(K + 1)$ -st auction are the only ones that are robust to deviations by groups of players, captured through the notion of the core in the game among the bidders. The core of a game was formulated by Edgeworth [Edg81] and brought into game theory by Gillies [Gil59]. There is an extensive body of literature on the core of various games, including for auctions; see, e.g., analysis of collusion (cartel behavior) in first price auctions in [Pes00].
Our focus here is the core of the game among the bidders, where the auctioneer first sets the auction format and then the bidders can strategize and collude among themselves, without the auctioneer. Roughly speaking, a strategy profile is core-stable if no group  $C$  of players can deviate simultaneously (i.e. each player  $i \in C$  deviates to some alternative strategy profile), such that each player in  $C$  weakly improves and the improvement is strict for at least one player. For a deviation to take place, the players in  $C$  coordinate and switch their strategies simultaneously, while the players outside  $C$  keep their previous strategies. Every core-stable strategy profile is also a Nash equilibrium, since a strategy profile that is stable against deviations by groups of players is also stable against deviations by individuals.
We consider two variants of the core, with and without monetary transfers [SS69, Bon63, Sha67]. In the case with transfers, the players can make monetary payments to each other, and so a strategy profile consists of a tuple of bids and transfers. In the case without transfers, the players cannot make such transfers and their strategy is the bid vector; thus the only agreement they can make in this case is to coordinate their bids.
C.1 Core with transfers
A strategy profile in this setting is described by a tuple  $(\mathbf{b},\mathbf{t})$ , where  $\mathbf{b}$  is a bid profile and  $\mathbf{t}$  is a profile of payments (aka monetary transfers), such that  $t_{i,j} \geq 0$  is the monetary payment of player  $i$  to player  $j$ . At this strategy profile, the auctioneer runs the auction with bids  $\mathbf{b}$  and returns the outcome (price and allocation), while the players make the monetary transfers  $\mathbf{t}$  to each other.
For each profile of monetary transfers  $\mathbf{t}$ , let  $m_{i}(\mathbf{t})$  be the net amount of money that player  $i$  gets after all the transfers are made:
$$
m _ {i} (\mathbf {t}) = \sum_ {j = 1} ^ {n} t _ {j, i} - \sum_ {j = 1} ^ {n} t _ {i, j}.
$$
The utility of player  $i$  at profile  $(\mathbf{b},\mathbf{t})$  is
$$
u _ {i} (\mathbf {b}, \mathbf {t}) = m _ {i} (\mathbf {t}) + \left(\sum_ {j = 1} ^ {x _ {i} (\mathbf {b})} v _ {i, j}\right) - p \cdot x _ {i} (\mathbf {b}).
$$
**Deviations.** Since in the case of auctions the actions (e.g. bids) of a group of players can affect the utility of the players outside of the group, it is necessary to model how the players outside  $S$  react to the deviation. Such reactions have been studied in the literature on the core with externalities (see, e.g., [Koc07, Koc09]).
We consider neutral reactions, where non-deviators (i.e. players outside  $S$ ) have a mild reaction to the deviation: they maintain the same bids as before the deviation and the monetary transfer of each player  $i \in [n] \backslash S$  to each player  $j \in S$  is non-negative. The core where the deviators assume the non-derivators will have neutral reactions to the deviation is known as the neutral core  $\left[\mathrm{SMR}^{+}13\right]$ . Several other variants of the core exist, such as pessimistic core, where each deviator assumes that they will be punished in the worst possible way by the non-derivators. Such variants are also interesting to study, but next we focus on the basic case of neutral reactions.
A group of players will alternatively be called a coalition. The set of all players,  $[n]$ , will also be called sometimes the grand coalition.
A group of players that agree on a deviation are known as a blocking coalition, formally defined next.
Definition 3 (Blocking coalition, with transfers). Let  $(\mathbf{b},\mathbf{t})$  be a tuple of bids and monetary transfers. A group  $S\subseteq [n]$  of players is a blocking coalition if there exists a profile  $(\widetilde{\mathbf{b}},\widetilde{\mathbf{t}})$ , at which each player  $i\in S$  weakly improves their utility, the improvement is strict for at least one player in  $S$ , and
-  $\widetilde{b}_{i,j} = b_{i,j}$  if  $i\in [n]\setminus S,j\in [K]$ .
-  $\widetilde{t}_{i,j} = 0$  if  $i\in [n]\setminus S$  and  $j\in S$
In other words, the blocking coalition  $S$  needs to agree on their bids and transfers to each other, such that they improve their utility when the players outside  $S$  maintain their existing bids but stop payments to players in  $S$ . In fact our characterization holds even if the players outside  $S$  make any non-negative transfers to the players in  $S$ ; the case where the transfers are zero is the extreme case. If a coalition  $S$  deviates with zero transfers from players outside  $S$ , it also deviates for any transfers that are non-negative.
Definition 4 (The core with transfers). The core with transfers consists of profiles  $(\mathbf{b},\mathbf{t})$  at which there are no blocking coalitions. Such profiles are core-stable.
Theorem 6 (Core with transfers; restated). Consider  $K$  units and  $n > K$  hungry players. The core with transfers of the  $(K + 1)$ -st auction can be characterized as follows:
Proof. The proof has three steps as follows.
Step I: core transfers are zero. Assume towards a contradiction that  $(\mathbf{b},\mathbf{t})$  is core stable and  $\mathbf{t}\neq \mathbf{0}$ . Consider the directed weighted graph  $G = ([n],E,\mathbf{t})$ , where  $E$  consists of all the directed edges  $(i,j)$  and the weight of each edge is  $t_{i,j}$ . The net amount of money that each player  $i$  gets from transfers is  $m_i(\mathbf{t}) = \sum_{j = 1}^n t_{j,i} - \sum_{j = 1}^n t_{i,j}$ .
If there is a cycle  $C = (i_1, \ldots, i_k)$  such that the payments along the cycle are strictly positive:  $t_{i_1,i_2} > 0, \ldots, t_{i_{k-1},i_k} > 0$ , and  $t_{i_k,i_1} > 0$ , then some cancellations take place. That is, by subtracting  $\min\{t_{i_1,i_2}, \ldots, t_{i_{k-1},i_k}, t_{i_k,i_1}\}$  from the weight of each edge  $(i,j) \in C$ , we obtain a weighted directed graph without cycle  $C$  and where each player has the same net amount of money as in the original graph. Iterating the operation of removing cycles, we obtain a directed acyclic graph where the players have the same net amount of money as in the original graph.
Thus we can in fact assume the transfers  $\mathbf{t}$  are such that  $G$  is acyclic.
Consider a topological ordering  $(i_1, \ldots, i_n)$  of the vertices of  $G$ . Let  $j$  be the minimum index for which vertex  $i_j$  has no incoming edges and at least one outgoing edge. Then  $m_{i_j}(\mathbf{t}) < 0$ . The utility of player  $i_j$  can be upper bounded as follows:
$$
u _ {i _ {j}} (\mathbf {b}, \mathbf {t}) = m _ {i _ {j}} (\mathbf {t}) + \left(\sum_ {k = 1} ^ {x _ {i _ {j}} (\mathbf {b})} v _ {i _ {j}, k}\right) - p (\mathbf {b}) \cdot x _ {i _ {j}} (\mathbf {b}) <   \left(\sum_ {k = 1} ^ {x _ {i _ {j}} (\mathbf {b})} v _ {i _ {j}, k}\right) - p (\mathbf {b}) \cdot x _ {i _ {j}} (\mathbf {b}).
$$