SlowGuess/ACL_2020_OCR huggingface dataset:
bitrary and  $\mathbf{a} = \mathbf{b} = \mathbb{1}_n / n$  .Let  $\mathbf{E}^{\epsilon}\in \mathbb{R}^{n\times n}$  be such that  $\mathbf{E}_{ij}^{\epsilon}\stackrel {iid}{\sim}\mathcal{U}([0,\epsilon ])$  where  $\epsilon >0$  and  $\mathcal{U}$  is the uniform distribution.Define  $\mathbf{C}^{\epsilon} = \mathbf{C} + \mathbf{E}^{\epsilon}$  .Let
$$
\mathbf{P}^{*} = \operatorname *{argmin}_{\substack{\mathbf{P}\in \mathbf{U}(\mathbf{a},\mathbf{b})}}\langle \mathbf{C},\mathbf{P}\rangle
$$
$$
\mathbf{P}^{\epsilon *} = \operatorname *{argmin}_{\substack{\mathbf{P}\in \mathbf{U}(\mathbf{a},\mathbf{b})}}\langle \mathbf{C}^{\epsilon},\mathbf{P}\rangle .
$$
1.  $0 \leq \langle \mathbf{C}, \mathbf{P}^{\epsilon *} \rangle - \langle \mathbf{C}, \mathbf{P}^* \rangle \leq \epsilon$ .
2. With probability 1,  $\mathbf{P}^{\epsilon*}$  is unique and is a permutation matrix.
Proof. We begin by proving result 1. Since  $\mathbf{P}^*$  is optimal for  $\mathbf{C}$ , it must be true that  $\langle \mathbf{C},\mathbf{P}\rangle \leq \langle \mathbf{C},\mathbf{P}'\rangle$  for any  $\mathbf{P}' \in \mathbf{U}(\mathbf{a},\mathbf{b})$ . As  $\mathbf{P}^{\epsilon *} \in \mathbf{U}(\mathbf{a},\mathbf{b})$ , we thus have  $\langle \mathbf{C},\mathbf{P}^*\rangle \leq \langle \mathbf{C},\mathbf{P}^{\epsilon *} \rangle$  and so  $\langle \mathbf{C},\mathbf{P}^{\epsilon *} \rangle - \langle \mathbf{C},\mathbf{P}^* \rangle \geq 0$ .
To prove the other side of the inequality, first note that for any  $\mathbf{P} \in \mathbf{U}(\mathbf{a},\mathbf{b})$ , we have  $\langle \mathbf{E}^{\epsilon},\mathbf{P}\rangle \geq 0$  since  $\mathbf{E}_{ij}^{\epsilon},\mathbf{P}_{ij} \geq 0$  for all  $i,j$ . Combining this with the optimality of  $\mathbf{P}^{\epsilon*}$  for
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$\mathbf{C}^{\epsilon}$ , we can see that
$$
\begin{array}{l} \langle \mathbf {C}, \mathbf {P} ^ {\epsilon *} \rangle - \langle \mathbf {C}, \mathbf {P} ^ {*} \rangle \\ \leq \langle \mathbf {C}, \mathbf {P} ^ {\epsilon *} \rangle + \langle \mathbf {E} ^ {\epsilon}, \mathbf {P} ^ {\epsilon *} \rangle - \langle \mathbf {C}, \mathbf {P} ^ {*} \rangle \\ = \langle \mathbf {C} + \mathbf {E} ^ {\epsilon}, \mathbf {P} ^ {\epsilon *} \rangle - \langle \mathbf {C}, \mathbf {P} ^ {*} \rangle \\ = \langle \mathbf {C} ^ {\epsilon}, \mathbf {P} ^ {\epsilon *} \rangle - \langle \mathbf {C}, \mathbf {P} ^ {*} \rangle \\ \leq \langle \mathbf {C} ^ {\epsilon}, \mathbf {P} ^ {*} \rangle - \langle \mathbf {C}, \mathbf {P} ^ {*} \rangle \\ = \langle \mathbf {C} ^ {\epsilon} - \mathbf {C}, \mathbf {P} ^ {*} \rangle \\ = \langle \mathbf {C} + \mathbf {E} ^ {\epsilon} - \mathbf {C}, \mathbf {P} ^ {*} \rangle \\ = \langle \mathbf {E} ^ {\epsilon}, \mathbf {P} ^ {*} \rangle \\ \leq \epsilon , \\ \end{array}
$$
where the final inequality holds because the entries of  $\mathbf{P}^*$  are positive and sum to one and the entries of  $\mathbf{E}^\epsilon$  are at most  $\epsilon$ . Thus results 1 hold.
Now we will prove result 2. Since we are solving a linear programming problem over a bounded, convex set  $\mathbf{U}(\mathbb{1}_n / n,\mathbb{1}_n / n)$ , every solution is a convex combination of optimal extremal points. Thus, a linear program has a unique optimal solution if and only if exactly one of the extremal points is optimal. By Birkhoff's theorem (Birkhoff, 1946), the set of extremal points of  $\mathbf{U}(\mathbb{1}_n / n,\mathbb{1}_n / n)$  is equal to the set of permutation matrices. Therefore, if only a single permutation matrix  $\mathbf{P}^{\sigma}$  is optimal for  $L_{\mathbf{C}^{\epsilon}}(\mathbf{a},\mathbf{b})$ , then  $\mathbf{P}^{\sigma}$  is the unique solution.
The goal is thus to show that the event that any two permutation matrices  $\mathbf{P}^{\sigma_i}$  and  $\mathbf{P}^{\sigma_j}$  corresponding to permutations  $\sigma_{i}\neq \sigma_{j}$  both solve  $L_{\mathbf{C}^{\mathrm{e}}}(\mathbf{a},\mathbf{b})$  has probability zero. The union bound gives
$$
\begin{array}{l} \mathbb {P} \left(\cup_ {\sigma_ {i} \neq \sigma_ {j}} \mathbf {P} ^ {\sigma_ {i}}, \mathbf {P} ^ {\sigma_ {j}} \text {b o t h s o l v e} L _ {\mathbf {C} ^ {\epsilon}} (\mathbf {a}, \mathbf {b})\right) \\ \leq \sum_ {\sigma_ {i} \neq \sigma_ {j}} \mathbb {P} \left(\mathbf {P} ^ {\sigma_ {i}}, \mathbf {P} ^ {\sigma_ {j}} \text {b o t h s o l v e} L _ {\mathbf {C} ^ {\epsilon}} (\mathbf {a}, \mathbf {b})\right). \\ \end{array}
$$
The number of pairs  $\sigma_{i}$  and  $\sigma_{j}$  of distinct permutations of  $n$  items is  $\binom{n!}{2} < \infty$  so the sum is over a finite number of probabilities. Thus, if we can show that  $\mathbb{P}(\mathbf{P}^{\sigma_i},\mathbf{P}^{\sigma_j}$  both solve  $L_{\mathbf{C}^{\epsilon}}(\mathbf{a},\mathbf{b})) = 0$  for any  $\sigma_{i}\neq \sigma_{j}$ , then the sum will also be zero and result 2 will hold.
To show that this is the case, take any two permutations matrices  $\mathbf{P}^{\sigma_1}$  and  $\mathbf{P}^{\sigma_2}$  for  $\sigma_{1}\neq$ $\sigma_{2}$  which are both optimal for  $L_{\mathbf{C}^{\epsilon}}(\mathbf{a},\mathbf{b})$  . Then it must be true that
$$
n \langle \mathbf {C} ^ {\epsilon}, \mathbf {P} ^ {\sigma_ {1}} \rangle = n \langle \mathbf {C} ^ {\epsilon}, \mathbf {P} ^ {\sigma_ {2}} \rangle
$$
$$
n \sum_ {i, j = 1} ^ {n} \mathbf {C} _ {i j} ^ {\epsilon} \mathbf {P} _ {i j} ^ {\sigma_ {1}} = n \sum_ {k, l = 1} ^ {n} \mathbf {C} _ {k l} ^ {\epsilon} \mathbf {P} _ {k l} ^ {\sigma_ {2}}. \tag {1}
$$
Let  $I^1 \subseteq \{1, \ldots, n\} \times \{1, \ldots, n\}$  be the indices  $(i,j)$  where  $\mathbf{P}_{ij}^{\sigma_1} = \frac{1}{n}$  and  $\mathbf{P}_{ij}^{\sigma_2} = 0$  and let  $I^2 \subseteq \{1, \ldots, n\} \times \{1, \ldots, n\}$  be the indices  $(i,j)$  where  $\mathbf{P}_{ij}^{\sigma_2} = \frac{1}{n}$  and  $\mathbf{P}_{ij}^{\sigma_1} = 0$ . Thus, for any  $(i,j) \notin I^1 \cup I^2$ ,  $P_{ij}^{\sigma_1} = P_{ij}^{\sigma_2}$  and so the terms corresponding to that  $(i,j)$  cancel in equation (1). This means that Equation (1) can be rewritten as
$$
n \sum_ {i, j \in I ^ {1} \cup I ^ {2}} \mathbf {C} _ {i j} ^ {\epsilon} \mathbf {P} _ {i j} ^ {\sigma_ {1}} = n \sum_ {k, l \in I ^ {1} \cup I ^ {2}} \mathbf {C} _ {k l} ^ {\epsilon} \mathbf {P} _ {k l} ^ {\sigma_ {2}}
$$
or equivalently, using the definition of  $I^1$  and  $I^2$ , as
$$
\sum_ {i, j \in I ^ {1}} \mathbf {C} _ {i j} ^ {\epsilon} = \sum_ {k, l \in I ^ {2}} \mathbf {C} _ {k l} ^ {\epsilon}.
$$
Using the definition of  $\mathbf{C}^{\epsilon}$ , this becomes
$$
\sum_ {i, j \in I ^ {1}} \mathbf {C} _ {i j} + \mathbf {E} _ {i j} ^ {\epsilon} = \sum_ {k, l \in I ^ {2}} \mathbf {C} _ {k l} + \mathbf {E} _ {k l} ^ {\epsilon}.
$$
Grouping terms, we get
$$
\sum_ {i, j \in I ^ {1}} \mathbf {E} _ {i j} ^ {\epsilon} - \sum_ {k, l \in I ^ {2}} \mathbf {E} _ {k l} ^ {\epsilon} = \sum_ {k, l \in I ^ {2}} \mathbf {C} _ {k l} - \sum_ {i, j \in I ^ {1}} \mathbf {C} _ {i j}.
$$
Since the LHS is a sum/difference of independent continuous random variables and the RHS is a constant, the event that the LHS equals the RHS has probability zero. Thus, the event that any two permutation matrices  $\mathbf{P}^{\sigma_1}$  and  $\mathbf{P}^{\sigma_2}$  with  $\sigma_1 \neq \sigma_2$  are both optimal for  $L_{\mathbf{C}^{\epsilon}}(\mathbf{a},\mathbf{b})$  has probability zero.
Corollary 1. If  $\langle \mathbf{C},\mathbf{P}^{\sigma}\rangle -\langle \mathbf{C},\mathbf{P}^{*}\rangle = 0$  or  $\langle \mathbf{C},\mathbf{P}^{\sigma}\rangle -\langle \mathbf{C},\mathbf{P}^{*}\rangle >\epsilon$  for every permutation matrix  $\mathbf{P}^{\sigma}$ , then the permutation matrix  $\mathbf{P}^{\epsilon *}$  is an exact solution to  $L_{\mathbf{C}}(\mathbf{a},\mathbf{b})$ .
Proof. Theorem 1 says that that  $\langle \mathbf{C},\mathbf{P}^{\epsilon *}\rangle -$ $\langle \mathbf{C},\mathbf{P}^{*}\rangle \leq \epsilon$  . Since  $\mathbf{P}^{\epsilon *}$  is a permutation matrix, the assumptions in this corollary thus imply that  $\langle \mathbf{C},\mathbf{P}^{\epsilon *} \rangle -\langle \mathbf{C},\mathbf{P}^{*}\rangle = 0$  , meaning  $\mathbf{P}^{\epsilon *}$  is an exact solution to  $L_{\mathbf{C}}(\mathbf{a},\mathbf{b})$
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