SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Given  $\mathcal{T}_{\mathrm{L}}$  and a trade-off parameter  $\lambda > 1$ , consider the following combined learning objective:
$$
\widehat {\mathcal {L}} (f; \mathcal {T} _ {\mathrm {P}}, \mathcal {T} _ {\mathrm {L}}) := \frac {1}{m _ {\mathrm {P}}} \sum_ {(\boldsymbol {x}, s) \in \mathcal {T} _ {\mathrm {P}}} \ell_ {\sigma} ^ {0 1} (f (\boldsymbol {x}), s) + \frac {\lambda}{m _ {\mathrm {L}}} \sum_ {(x, y) \in \mathcal {T} _ {\mathrm {L}}} \ell^ {0 1} ([ f ] (x), y)
$$
Suppose that there is a classifier  $f_0 \in \mathcal{F}$  that can achieve a small empirical classification risk  $r_0$  given by  $\frac{1}{m_{\mathrm{L}}} \sum_{(x,y) \in \mathcal{T}_{\mathrm{L}}} \ell^{01}([f_0](x), y)^9$ . Then, any classifier  $f_1$  with an empirical classification risk greater than  $r_0 + 1 / \lambda$  will be rejected by this learning objective. This is because
$$
\widehat {\mathcal {L}} \left(f _ {1}; \mathcal {T} _ {\mathrm {P}}, \mathcal {T} _ {\mathrm {L}}\right) \geq \lambda \left(r _ {0} + \frac {1}{\lambda}\right) = \lambda r _ {0} + 1 \geq \widehat {\mathcal {L}} \left(f _ {0}; \mathcal {T} _ {\mathrm {P}}, \mathcal {T} _ {\mathrm {L}}\right)
$$
Therefore, all the possibly learned classifiers must have an empirical risk less than  $r_0 + 1 / \lambda$ . Furthermore, under standard learning theory assumptions (e.g., finite VC dimension or vanishing Rademacher complexity), for any  $\delta_0 \in (0,1)$  we can typically bound the expected classification risk with probability at least  $1 - \delta_0$  by
$$
r _ {0} + 1 / \lambda + B _ {1} \sqrt {\frac {B _ {2} + \log \left(1 / \delta_ {0}\right)}{m _ {\mathrm {L}}}} \tag {55}
$$
where  $B_{1}, B_{2}$  is a constant depending on the learning problem. As long as we can have  $m_{\mathrm{L}}$  that is large enough to make (55) less than 1 (i.e., non-vacuous), we can guarantee that our assumption holds with probability at least  $1 - \delta_0$ . The above justified the bounded risk assumption.
9Under the realizable assumption, we have that  $r_0 = 0$
C.2 Proofs
Below, we introduce the analog of Lemma 3 for the multi-classifier case. Let
$$
\mathcal {H} _ {\mathcal {F} _ {1}, \dots , \mathcal {F} _ {n}} := \left\{\left(\boldsymbol {x}, s\right) \mapsto \mathbb {1} \left\{\sigma \left([ \boldsymbol {f} ] (\boldsymbol {x})\right) \notin s \right\} \mid \boldsymbol {f} \in \mathcal {F} _ {1} \times \dots \times \mathcal {F} _ {n} \right\}
$$
Given a training sample  $(\pmb{x}, s)$ , a binary classifier from  $\mathcal{H}_{\mathcal{F}_1, \dots, \mathcal{F}_n}$  predicts whether the labels predictions for  $\pmb{x}$  are consistent with the partial label  $s$ . Suppose  $(\pmb{x}, s)$  are drawn from  $\mathcal{D}_{\mathsf{P}}$ . Then, 0 is always the gold label of this binary classification problem. The zero-one binary classification error of  $\{(x, s) \mapsto \mathbb{1} \{\sigma([f](x)) \notin s\}$  equals the zero-one partial loss defined as  $\ell_{\sigma}^{01}(f(x), s)$ .
Lemma 8. Let  $\mathrm{Nat}([\mathcal{F}_i]) = d_{[\mathcal{F}_i]} < \infty$  be the Natarajan dimension of each  $[\mathcal{F}_i]$ . Then, we can bound the VC dimension of the function class  $\mathcal{H}_{\mathcal{F}_1,\dots,\mathcal{F}_n}$  as
$$
d \leq 4 \sum_ {i = 1} ^ {n} \left(d _ {[ \mathcal {F} _ {i} ]} \log \left(M _ {i} \cdot n \cdot d _ {[ \mathcal {F} _ {i} ]}\right) + 2 d _ {[ \mathcal {F} _ {i} ]} \log c _ {i}\right) \tag {56}
$$
Proof. By Natarajan's lemma of multiclass classification, we have:
$$
2 ^ {d} \leq \prod_ {i = 1} ^ {n} \left(M _ {i} d\right) ^ {d _ {\left[ \mathcal {F} _ {i} \right]}} c _ {i} ^ {2 d _ {\left[ \mathcal {F} _ {i} \right]}} \tag {57}
$$
where  $c_{i} = |\mathcal{V}_{i}|$ . Taking the logarithm on both sides, we have
$$
d \log 2 \leq \sum_ {i = 1} ^ {n} d _ {[ \mathcal {F} _ {i} ]} \log (M _ {i} d) + 2 d _ {[ \mathcal {F} _ {i} ]} \log c _ {i}
$$
$$
\begin{array}{l} d \log 2 \leq \sum_ {i = 1} ^ {n} \left(d _ {[ \mathcal {F} _ {i} ]} \left(\log (n M _ {i} d _ {[ \mathcal {F} _ {i} ]}) + \frac {d}{n \operatorname {e d} _ {[ \mathcal {F} _ {i} ]}}\right) + 2 d _ {[ \mathcal {F} _ {i} ]} \log c _ {i}\right) \\ = \sum_ {i = 1} ^ {n} \left(d _ {[ \mathcal {F} _ {i} ]} \log (n M _ {i} d _ {[ \mathcal {F} _ {i} ]}) + 2 d _ {[ \mathcal {F} _ {i} ]} \log c _ {i}\right) + \frac {d}{\mathrm {e}} \\ \end{array}
$$
where the second step follows from the first-order Taylor series expansion of the logarithm function  $\log (M_i t)$  at the point  $t = \operatorname{ned}_{[\mathcal{F}_i]}$ . Using the fact that  $(\log 2 - 1 / \mathrm{e})^{-1} < 4$ , yields (56), concluding the proof of Lemma 8.
$$
\mathcal {R} ^ {0 1} (\boldsymbol {f}) \leq \mathcal {O} \left(\left(\mathcal {R} _ {\mathsf {P}} ^ {0 1} (\boldsymbol {f}; \sigma)\right) ^ {1 / M ^ {*}}\right) \quad a s \quad \mathcal {R} _ {\mathsf {P}} ^ {0 1} (\boldsymbol {f}; \sigma) \rightarrow 0 \tag {8}
$$
Proof. Recall that  $c_{i} = |\mathcal{Y}_{i}|$ . By the multi-unambiguity assumption, we can lower bound the partial label risk as
$$
\mathcal {R} _ {\mathsf {P}} ^ {0 1} (\boldsymbol {f}; \sigma) \geq \sum_ {i = 1} ^ {n} \quad \underbrace {\frac {(\mathcal {R} ^ {0 1} (f _ {i})) ^ {M _ {i}}}{(c _ {i} (c _ {i} - 1)) ^ {M _ {i} - 1}} \prod_ {j \neq i} (1 - \mathcal {R} ^ {0 1} (f _ {j})) ^ {M _ {j}}}
$$
Probability that each  $x \in \mathbf{x}_i$  is misclassified but each other prediction is correct.
$$
\begin{array}{l} \geq \sum_ {i = 1} ^ {n} \frac {\left(\mathcal {R} ^ {0 1} (f _ {i})\right) ^ {M _ {i}}}{(c _ {i} (c _ {i} - 1)) ^ {M ^ {*} - 1}} \prod_ {j \neq i} (1 - R) ^ {M _ {j}} \\ = \sum_ {i = 1} ^ {n} \frac {\left(\mathcal {R} ^ {0 1} \left(f _ {i}\right)\right) ^ {M _ {i}}}{\left(c _ {i} \left(c _ {i} - 1\right)\right) ^ {M ^ {*} - 1}} (1 - R) ^ {M - M _ {i}} \tag {58} \\ \geq \sum_ {i = 1} ^ {n} \frac {\left(\mathcal {R} ^ {0 1} (f _ {i})\right) ^ {M ^ {*}}}{(c _ {0} (c _ {0} - 1)) ^ {M ^ {*} - 1}} (1 - R) ^ {M - M _ {*}} \\ \geq \frac {(1 - R) ^ {M - M _ {*}}}{(n c _ {0} (c _ {0} - 1)) ^ {M ^ {*} - 1}} \left(\sum_ {i = 1} ^ {n} \mathcal {R} ^ {0 1} \left(f _ {i}\right)\right) ^ {M ^ {*}} \\ \end{array}
$$
$$
\begin{array}{l} \mathcal {R} ^ {0 1} (\boldsymbol {f}) \leq \left(\frac {\left(n c _ {0} \left(c _ {0} - 1\right)\right) ^ {M ^ {*} - 1}}{\left(1 - R\right) ^ {M - M _ {*}}} \mathcal {R} _ {\mathsf {P}} ^ {0 1} (\boldsymbol {f}; \sigma)\right) ^ {1 / M ^ {*}} \\ \leq \left(\frac {\left(n c ^ {2}\right) ^ {M ^ {*} - 1}}{\left(1 - R\right) ^ {M - M _ {*}}} \mathcal {R} _ {\mathrm {P}} ^ {0 1} (\boldsymbol {f}; \sigma)\right) ^ {1 / M ^ {*}} \tag {59} \\ = \mathcal {O} \left(\left(\mathcal {R} _ {\mathrm {P}} ^ {0 1} (\boldsymbol {f}; \sigma)\right) ^ {1 / M ^ {*}}\right) \\ \end{array}
$$
The above concludes the first part of Theorem 3.
To show the second part, from (59), we know that if the following holds:
$$
\mathcal {R} _ {\mathrm {P}} ^ {0 1} (\boldsymbol {f}; \sigma) \leq \frac {\epsilon^ {M ^ {*}} (1 - R) ^ {M - M _ {*}}}{\left(n c _ {0} ^ {2}\right) ^ {M ^ {*} - 1}} \tag {60}
$$
then,  $\mathcal{R}^{01}(\pmb {f})\leq \epsilon$  holds. By the standard bound for sample complexity based on VC-dimension, we know that for any confidence parameter  $\delta \in (0,1)$ , inequality (60) holds with probability no less than  $1 - \delta$ , if
$$
m _ {\mathrm {P}} \geq m \left(\mathcal {H} _ {\mathcal {F} _ {1}, \dots , \mathcal {F} _ {n}}, \delta , \frac {\epsilon^ {M ^ {*}} (1 - R) ^ {M - M _ {*}}}{\left(n c _ {0} ^ {2}\right) ^ {M ^ {*} - 1}}\right) \tag {61}
$$
$$
m \left(\mathcal {H} _ {\mathcal {F} _ {1}, \dots , \mathcal {F} _ {n}}, \delta , t\right) := \frac {C}{t} \left(\mathrm {V C} \left(\mathcal {H} _ {\mathcal {F} _ {1}, \dots , \mathcal {F} _ {n}}\right) \log \left(\frac {1}{t}\right) + \log \left(\frac {1}{\delta}\right)\right) \tag {62}
$$
and  $C$  is a universal constant. Combining (56) with (61) and (62), concludes the second part of Theorem 3.
Theorem 4 (Error bound under multi-unambiguity with multiple classifiers). Suppose  $\sigma$  is multi-unambiguous and each  $f_{i}$  is zero-one risk  $R$ -bounded, for  $R \in (0,1)$ . Then, for any integer  $k \geq 1$  and any  $\delta \in (0,1)$ , with probability at least  $1 - \delta$ , we have:
$$
\mathcal {R} ^ {0 1} (\boldsymbol {f}) \leq \left(\frac {n c _ {0} ^ {2 M ^ {*} - 2} (k + 1)}{(1 - R) ^ {M}} \left(\widehat {\mathcal {R}} _ {\mathsf {P}} ^ {k} (\boldsymbol {f}; \sigma ; \mathcal {T} _ {\mathsf {P}}) + \sqrt {k M} \sum_ {i = 1} ^ {n} M _ {i} \mathfrak {R} _ {m _ {\mathsf {P}} M _ {i}} (\mathcal {F} _ {i}) + \sqrt {\frac {\log (1 / \delta)}{2 m _ {\mathsf {P}}}}\right)\right) ^ {1 / M ^ {*}}
$$
$$
\boldsymbol {f} (\boldsymbol {x}) = \left[ f _ {i} ^ {y} \left(x _ {i j}\right) \right] _ {i \in [ n ], j \in [ M _ {i} ], y \in \mathcal {Y} _ {i}}
$$
be the vector that encodes all the probabilities that  $\pmb{f}$  assign to a pair  $(x, y)$  where  $x$  is from  $\pmb{x}$  and  $y$  is a possible label for  $x$ . Similarly to the single-variable case, we define
$$
\tilde {\ell} _ {\mathsf {P}} ^ {k} (\boldsymbol {f} (\boldsymbol {x}), s) := 1 - \operatorname {W M C} \left(\bigvee_ {i = 1} ^ {k} \boldsymbol {y} ^ {(i)}, \boldsymbol {f} (\boldsymbol {x})\right) \tag {63}
$$
Bound with the partial risk. From the proof of Theorem 3, we know that
$$
\mathcal {R} ^ {0 1} (\boldsymbol {f}) \leq \left(\frac {(n c ^ {2}) ^ {M ^ {*} - 1}}{(1 - R) ^ {M - M _ {*}}} \mathcal {R} _ {\mathsf {P}} ^ {0 1} (\boldsymbol {f}; \sigma)\right) ^ {1 / M ^ {*}}
$$
Bound with the top-  $k$  loss. Similarly to the single-classifier case, it can be shown that
$$
\mathcal {R} _ {\mathbb {P}} ^ {0 1} (\boldsymbol {f}; \sigma) \leq (k + 1) \mathcal {R} _ {\mathbb {P}} (\boldsymbol {f}; \tilde {\ell} _ {\mathbb {P}} ^ {k})
$$
Bound with the empirical risk. By standard Rademacher complexity bounds, given a partially labeled dataset  $\mathcal{T}_{\mathbb{P}}$  of size  $m_{\mathbb{P}}$ , for each  $\delta \in (0,1)$  and each  $f \in \mathcal{F}$ , the following holds with probability at least  $1 - \delta$ :
$$
\mathcal {R} _ {\mathsf {P}} ^ {k} (\boldsymbol {f}; \sigma) \leq \widehat {\mathcal {R}} _ {\mathsf {P}} (\boldsymbol {f}; \tilde {\ell} _ {\mathsf {P}} ^ {k}) + 2 \Re_ {m _ {\mathsf {P}}} (\mathcal {H} _ {\mathcal {F}}) + \sqrt {\frac {\log (1 / \delta)}{2 m _ {\mathsf {P}}}} \tag {64}
$$
where  $\mathcal{H}_{\mathcal{F},k}$  is a class of functions that take as input elements of the form  $(x,s)$ , with  $x \in \mathcal{X}^M$  and  $s \in S$ , and is defined as follows:
$$
\mathcal {H} _ {\mathcal {F}, k} := \left\{(\boldsymbol {x}, s) \mapsto \tilde {\ell} _ {\mathsf {P}} ^ {k} (\boldsymbol {f} (\boldsymbol {x}), s): f \in \mathcal {F} \right\} \tag {65}
$$
Denote  $\mathcal{F} = \prod_{i=1}^{M} \mathcal{F}_{i}$ . Using exactly the same argument in Lemma 7, we can show the mapping  $\pmb{f}(\pmb{x}) \mapsto \tilde{\ell}_{\mathbb{P}}^{k}(\pmb{f}(\pmb{x}), s)$  is Lipschitz with Lipschitz constant  $\sqrt{kM}$ , By the contraction lemma (Lemma 6), we have that
$$
\begin{array}{l} \mathfrak {R} _ {m} (\mathcal {H} _ {\mathcal {F}, k}) = \frac {1}{m _ {\mathrm {P}}} \mathbb {E} _ {\mathcal {T} _ {\mathrm {P}}} \mathbb {E} _ {\sigma} \left[ \sup  _ {\boldsymbol {f} \in \mathcal {F}} \sum_ {i = 1} ^ {m _ {\mathrm {P}}} \sigma_ {i} \tilde {\ell} _ {\mathrm {P}} ^ {k} (\boldsymbol {f} (\boldsymbol {x}), s) \right] \\ = \sqrt {k M} \frac {1}{m _ {\mathsf {P}}} \mathbb {E} _ {\mathcal {T}} \mathbb {E} _ {\epsilon} \left[ \sup  _ {\boldsymbol {f} \in \mathcal {F}} \sum_ {i = 1} ^ {m _ {\mathsf {P}}} \sum_ {l = 1} ^ {n} \sum_ {j = 1} ^ {M _ {l}} \sum_ {y \in \mathcal {Y} _ {i}} \epsilon_ {i l j y} f _ {l} ^ {y} (x _ {l j}) \right] \\ \leq \sqrt {k M} \sum_ {l = 1} ^ {n} M _ {l} \frac {1}{m _ {\mathsf {P}} M _ {l}} \mathbb {E} _ {\mathcal {T}} \mathbb {E} _ {\epsilon} \left[ \sup  _ {\boldsymbol {f} \in \mathcal {F}} \sum_ {i = 1} ^ {m _ {\mathsf {P}}} \sum_ {j = 1} ^ {M _ {l}} \sum_ {y \in \mathcal {Y} _ {i}} \epsilon_ {i l j y} f _ {l} ^ {y} \left(x _ {l j}\right) \right] \tag {66} \\ = \sqrt {k M} \sum_ {l = 1} ^ {n} M _ {l} \Re_ {m _ {\mathsf {P}} M _ {l}} (\mathcal {F} _ {l}) \\ = \sqrt {k M} \sum_ {i = 1} ^ {n} M _ {i} \Re_ {m _ {\mathsf {P}} M _ {i}} (\mathcal {F} _ {i}) \\ \end{array}
$$
The above implies
$$
\mathcal {R} _ {\mathrm {P}} (\boldsymbol {f}; \ell_ {\mathrm {P}, 1} ^ {k}) \leq \widehat {\mathcal {R}} _ {\mathrm {P}} (\boldsymbol {f}; \tilde {\ell} _ {\mathrm {P}} ^ {k}; \mathcal {T} _ {\mathrm {P}}) + 2 \sqrt {k M} \sum_ {i = 1} ^ {n} M _ {i} \Re_ {m _ {\mathrm {P}} M _ {i}} (\mathcal {F} _ {i}) + \sqrt {\frac {\log (1 / \delta)}{2 m _ {\mathrm {P}}}} \tag {67}
$$
Bound with cross-entropy. Finally, from the inequality  $-\log (1 - t)\geq t\forall t > 0$  , we know that
$$
\widehat {\mathcal {R}} _ {\mathrm {P}} \left(\boldsymbol {f}; \tilde {\ell} _ {\mathrm {P}} ^ {k}; \mathcal {T} _ {\mathrm {P}}\right) \leq \widehat {\mathcal {R}} _ {\mathrm {P}} \left(\boldsymbol {f}; \ell_ {\mathrm {P}} ^ {k}; \mathcal {T} _ {\mathrm {P}}\right) \tag {68}
$$
Putting the above inequalities together, the proof of Theorem 4 follows.
Theorem 5. If  $\mathcal{G}$  is unambiguous and any  $f\in \mathcal{F}$  is  $r$ -bounded, then we have:
$$
\mathcal {R} ^ {0 1} (f) \leq O \left(\mathcal {R} _ {\mathrm {P}} ^ {0 1} (f; \mathcal {G}) ^ {1 / M}\right) \quad a s \quad \mathcal {R} _ {\mathrm {P}} ^ {0 1} (f; \mathcal {G}) \rightarrow 0 \tag {10}
$$
Furthermore, suppose  $[\mathcal{F}]$  has a finite Natarajan dimension  $d_{[\mathcal{F}]}$  and the function class  $\{(y,s)\mapsto \mathbb{1}\{\sigma^{\prime}(\pmb {y})\neq s\} |\sigma^{\prime}\in \mathcal{G}\}$  has a finite VC-dimension  $d_{\mathcal{G}}$  . Then, for any  $\epsilon ,\delta \in (0,1)$  , there is a universal constant  $C_4$  such that with probability at least  $1 - \delta$  , the empirical partial risk minimizer with  $\widehat{\mathcal{R}}_{\mathsf{P}}^{01}(f;\sigma) = 0$  has a classification risk  $\mathcal{R}^{01}(f) <   \epsilon$  , if
Proof. Fix a classifier  $f \in \mathcal{F}$ . Since  $\mathcal{G}$  is unambiguous, it follows that for each pair of labels  $(l_i, l_j)$  such that  $E_{l_i, l_j}(f) > 0$  and each  $\sigma' \in \mathcal{G}$ , there exists a vector  $\pmb{y} \in \{l_i, l_j\}^M$ , such that  $\sigma'(\pmb{y}) \neq \sigma(l_i, l_i, \dots, l_i)$  holds. Therefore, we have:
$$
\begin{array}{l} \min  _ {\sigma^ {\prime} \in \mathcal {G}} \mathcal {R} _ {\mathbb {P}} ^ {0 1} (f; \sigma^ {\prime}) \geq \sum_ {i \neq j} \min  _ {0 \leq k \leq M} E _ {l _ {i}, l _ {i}} (f) ^ {k} E _ {l _ {i}, l _ {j}} (f) ^ {M - k} \\ \geq \sum_ {l _ {i} \neq l _ {j}} r ^ {M} E _ {l _ {i}, l _ {j}} ^ {M} (f) \\ \geq \frac {r ^ {M}}{c ^ {2 M - 2}} \left(\sum_ {l _ {i} \neq l _ {j}} E _ {l _ {i}, l _ {j}} (f)\right) ^ {M} \tag {69} \\ = \frac {r ^ {M}}{c ^ {2 M - 2}} \mathcal {R} ^ {0 1} (f) ^ {M} \\ \end{array}
$$
The above implies that  $\mathcal{R}^{01}(f) \leq O(\mathcal{R}_{\mathsf{P}}^{01}(f; \sigma)^{1 / M})$  holds.
To prove the second part of Theorem 5, from the above discussion we know that  $\mathcal{R}^{01}(f) \leq (c^{2M - 2}\mathcal{R}_{\mathsf{P}}^{01}(f;\sigma) / r^M)^{1 / M}$ . Therefore, if we could control  $\mathcal{R}_{\mathsf{P}}^{01}(f;\sigma) \leq \epsilon^M r^M / c^{2M - 2}$ , we would have that  $\mathcal{R}^{01}(f) \leq \epsilon$ . By the standard bound for sample complexity based on VC-dimension, then for any  $\delta \in (0,1)$ ,  $\mathcal{R}_{\mathsf{P}}^{01}(f;\sigma) \leq \epsilon^M r^M / c^{2M - 2}$  can happen with probability at least  $1 - \delta$  if
$$
m _ {\mathsf {P}} \geq m \left(\mathcal {H} _ {\mathcal {F}, \mathcal {G}}, \delta , \epsilon^ {M} r ^ {M} / c ^ {2 M - 2}\right) \tag {70}
$$
$$
m \left(\mathcal {H} _ {\mathcal {F}, \mathcal {G}}, \delta , t\right) = \frac {C}{t} \left(\operatorname {V C} \left(\mathcal {H} _ {\mathcal {F}, \mathcal {G}}\right) \log \left(\frac {1}{t}\right) + \log \left(\frac {1}{\delta}\right)\right) \tag {71}
$$
and  $C$  is a universal constant. Now, recall (21) from Lemma 3. Combining (21) with (70) and (71) yields the proof of the second part of Theorem 5.
E Results for standard PLL
E.1 Counterexamples with transition matrices
Definition 12 (Transition matrix [8, 46]). A transition matrix  $\mathbf{T}$  for a learning problem with hidden label  $Y\in \mathcal{V}$  and observation  $S\in S$  is a stochastic matrix of dimension  $|\mathcal{V}|\times |\mathcal{S}|$ , where the element in its  $i^{th}$  column and  $j^{th}$  row is the conditional probability  $\mathbb{P}(S = j|Y = i)$ . Transition matrix  $\mathbf{T}$  is invertible if it is left invertible, i.e., there is a matrix  $\mathbf{P}$  of dimension  $|\mathcal{V}|\times |\mathcal{S}|$  such that  $\mathbf{PT} = \mathbf{I}_{|\mathcal{V}|}$ .
As shown in [8, 46], if the transition is invertible, then it is possible to construct an unbiased estimator for the classification loss using the samples of the partial label  $s$ . Therefore, the invertibility of the transition is desirable since it implies one can minimize the classification risk with a partially labeled dataset alone.
A transition matrix formulation for multi-instance PLL. Consider the multi-instance PLL with a single classifier and a deterministic transition  $\sigma$ . Consider a naive formulation of the transition matrix where we view  $\mathcal{Y}^M$  as the label space. Then, if there are two different label assignments  $\pmb{y} \neq \pmb{y}'$  such that  $\sigma(\pmb{y}) \neq \sigma(\pmb{y}')$ , their corresponding column vectors in  $\mathbf{T}$  will be the same, resulting in a non-invertible transition. For example, in the MNIST with SUM2 problem, the column vectors corresponding to the label pairs  $(0,1)$  and  $(1,0)$  will be:
$$
\begin{array}{r} \boldsymbol {y} = (1, 2) \quad \boldsymbol {y} = (2, 1) \\ s = 0 \\ s = 1 \\ s = 2 \\ \vdots \\ s = 1 8 \end{array} \left[ \begin{array}{c c} 0 & 0 \\ 1 & 1 \\ 0 & 0 \\ \vdots & \vdots \\ 0 & 0 \end{array} \right] \tag {72}
$$
This fact implies that the transition  $\sigma$  must be injective to ensure invertibility, which is too strong for the partial labels in practice.
To overcome this issue, we consider an alternative formulation, where we view  $s$  as a randomized partial label for each of the  $M$  instances, where the randomness comes from the distribution of the other instances. Suppose the marginal distribution of  $Y$  is known (or can be estimated in some ways). Then, we construct a transition matrix  $\mathbf{T}_k$  for each  $k\in [M]$  by defining its  $i^{\mathrm{th}}$  column and  $j^{\mathrm{th}}$  row to be  $\mathbb{P}(s = j|Y_i = i) = \mathbb{P}(\pmb {y}|\sigma (\pmb {y}) = j,y_k = i)$ . For example, the transition matrices for the MNIST with 2sum problem are
$$
\mathbf {T} _ {1} = \mathbf {T} _ {2} = \begin{array}{c c c c c} & 0 & 1 & \dots & 9 \\ s = 0 & \left[ \begin{array}{c c c c c} \mathbb {P} (Y = 0) & 0 & \dots & 0 \\ \mathbb {P} (Y = 1) & \mathbb {P} (Y = 0) & \dots & 0 \\ \mathbb {P} (Y = 2) & \mathbb {P} (Y = 1) & \dots & 0 \end{array} \right] \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ s = 9 & \mathbb {P} (Y = 9) & \mathbb {P} (Y = 8) & \dots & \mathbb {P} (Y = 0) \\ s = 1 0 & 0 & \mathbb {P} (Y = 9) & \dots & \mathbb {P} (Y = 1) \\ s = 1 1 & 0 & 0 & \dots & \mathbb {P} (Y = 2) \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ s = 1 8 & 0 & 0 & \dots & \mathbb {P} (Y = 9) \end{array} \tag {73}
$$
We say the standard PLL problem is invertible if there exists an  $k \in [M]$  such that  $\mathbf{T}_k$  is left invertible.
$M$ -unambiguity  $\neq$  invertibility for multi-instance PLL. Consider the following example, where  $M$ -unambiguity holds but the transition matrix is not invertible for no input position. Consider  $\mathcal{V} = [3]$  and  $M = 3$ . Assume  $\sigma(1,1,1) \neq \sigma(2,2,2) \neq \sigma(3,3,3)$  so that  $\sigma$  is  $M$ -unambiguous. Now, suppose
$$
\begin{array}{l} \sigma (1, 1, 1) = \sigma (1, 2, 3) = \sigma (1, 3, 2) \\ = \sigma (2, 1, 2) = \sigma (2, 2, 1) = \sigma (2, 3, 3) \tag {74} \\ = \sigma (3, 1, 3) = \sigma (3, 2, 2) = \sigma (3, 3, 1) \\ \end{array}
$$
$$
\begin{array}{l} \sigma (1, 1, 2) = \sigma (1, 2, 1) = \sigma (1, 3, 3) \\ = \sigma (2, 1, 3) = \sigma (2, 2, 2) = \sigma (2, 3, 1) \tag {75} \\ = \sigma (3, 1, 1) = \sigma (3, 2, 3) = \sigma (3, 3, 2) \\ \end{array}
$$
$$
\begin{array}{l} \sigma (1, 1, 3) = \sigma (1, 2, 2) = \sigma (1, 3, 1) \\ = \sigma (2, 1, 1) = \sigma (2, 2, 3) = \sigma (2, 3, 2) \tag {76} \\ = \sigma (3, 1, 2) = \sigma (3, 2, 1) = \sigma (3, 3, 3) \\ \end{array}
$$
so that  $\mathcal{S} = \{\sigma(1,1,1), \sigma(2,2,2), \sigma(3,3,3)\}$ . Suppose that  $\mathbb{P}(Y = 1) = \mathbb{P}(Y = 2) = \mathbb{P}(Y = 3) = 1/3$ . Then, it can be verified that
$$
\mathbf {T} _ {1} = \mathbf {T} _ {2} = \mathbf {T} _ {3} = \begin{array}{l l l l} & 1 & 2 & 3 \\ \sigma (1, 1, 1) & \left[ \begin{array}{l l l} 1 / 3 & 1 / 3 & 1 / 3 \\ 1 / 3 & 1 / 3 & 1 / 3 \\ 1 / 3 & 1 / 3 & 1 / 3 \end{array} \right] \\ \sigma (2, 2, 2) & & & \\ \sigma (3, 3, 3) & \end{array} \tag {77}
$$
This means that all the transition matrices are non-invertible.
Invertibility  $\neq M$ -unambiguity for multi-instance PLL. Suppose  $\mathcal{Y} = [2]$  and  $\sigma(y_1, y_2) = \mathbb{1}\{y_1 = y_2\}$ . Then  $M$ -unambiguity does not hold since  $\sigma(1, 1) = \sigma(2, 2)$ . But if  $\mathbb{P}(Y = 1) = 0.1 = 1 - \mathbb{P}(Y = 2)$ , then
$$
\mathbf {T} _ {1} = \begin{array}{l l l} 0 & 1 & 2 \\ 1 & \left[ \begin{array}{l l} 0. 9 & 0. 1 \\ 0. 1 & 0. 9 \end{array} \right] \end{array} \tag {78}
$$
is an invertible transition.
E.2 Counterexamples for standard PLL
We show that for the standard PLL, the small ambiguity degree condition [28] and the invertibility of the transition matrix, see Definition 12, do not imply each other. Below, we recapitulate the small ambiguity degree condition.
Definition 13 (Ambiguity degree [28]). The ambiguity degree for standard PLL is defined as
$$
\gamma := \sup  _ {\mathcal {D} (x, y) > 0, y ^ {\prime} \neq y} \mathbb {P} _ {(x, y) \sim \mathcal {D}} \left(y ^ {\prime} \in \sigma (y)\right) \tag {79}
$$
where  $\mathcal{D}(x,y)$  is the density function at  $(x,y)$ . We say that a PLL instance satisfies the small ambiguity degree condition, if  $\gamma < 1$ .
Small ambiguity  $\neq$  invertibility. Consider the following transition matrix (where all blank elements are zero) with  $\mathcal{V} = [5]$  and  $S = 2^{\mathcal{V}}$ :
$$
\mathbf {T} = \begin{array}{l} \left\{1, 2, 3 \right\} \\ \left\{1, 4, 5 \right\} \\ \left\{2, 5 \right\} \\ \left\{3, 4 \right\} \end{array} \left[ \begin{array}{c c c c c} 1 / 2 & 2 & 3 & 4 & 5 \\ 1 / 2 & 1 / 2 & 1 / 2 & & \\ 1 / 2 & & & 1 / 2 & 1 / 2 \\ & 1 / 2 & & & 1 / 2 \\ & & 1 / 2 & 1 / 2 \end{array} \right] \tag {80}
$$
It can be verified that the small ambiguity degree is  $1/2$  but this matrix is of rank  $4 < 5$  and the invertibility condition does not hold.
Invertibility  $\nRightarrow$  small ambiguity. Consider the following transition matrix with  $\mathcal{V} = [2]$  and  $S = 2^{\mathcal{V}}$ :
$$
\mathbf {T} = \left\{ \begin{array}{l l} 1 \\ \{1, 2 \} \end{array} \right. \left[ \begin{array}{l l} 1 & 2 \\ 0 & 0 \\ 0 & 1 \end{array} \right] \tag {81}
$$
We can see that the small ambiguity degree is 1 but the above matrix is of full rank.
E.3  $M$ -ambiguity and small ambiguity
We show that our proposed  $M$ -unambiguity, see Definition 1, is an extension of the small ambiguity degree, see Definition 13. To do this, we consider a generalized setting, where  $s$  is a randomized function of  $\pmb{y}$ . In particular, similarly to standard PLL (aka superset learning problem) [28, 8], we define  $\sigma$  to be a random function  $\mathcal{Y}^M \to 2^{\mathcal{Y}^M}$  so that  $s \in 2^{\mathcal{Y}^M}$  is a random set of vectors in  $\mathcal{Y}^M$ . Below, let  $\mathcal{D}^M$  be a distribution followed by  $M$  i.i.d. random variables  $(X_i, Y_i)$ , where  $(X_i, Y_i) \sim D$  for each  $i \in [M]$ . We use  $\mathcal{D}^M(\pmb{x}, \pmb{y})$  to denote the density at  $(\pmb{x}, \pmb{y})$ , for  $\pmb{x} \in \mathcal{X}^M$  and  $\pmb{y} \in \mathcal{Y}^M$ .
Definition 14 (M-Ambiguity degree). The ambiguity degree for standard PLL subject to  $\mathcal{D}^M$  is given by