SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Figure 6: Comparison of min-pay and min-budget contracts under MLRP for varying validation set size  $m$ . The delegation setting is similar to the one depicted in Fig. 5: two binomial-outcome actions  $(p_1 = 0.5, p_2 = 0.8)$  and varying  $m$ . Action 1 has zero cost  $c_1 = 0$ , and action 2 has unit cost  $c_2 = 1$ . In the left and center plots, the cost of action 2 is represented by an orange line  $(c_2 = 1)$  representing the lower bound as in Fig. 3 (Center). (Left) Expected pay  $\mathbb{E}_{j \sim f_2}[t_j]$ . (Center) Required budget  $\| t \|_{\infty}$ . (Right) Probability of getting any payment  $\mathbb{P}_{j \sim f_2}[t_j > 0]$ .
The following equivalent definition is useful:
Claim 4. Let  $P, Q \in \Delta(\Omega)$ . It holds:
$$
\left\| P - Q \right\| _ {\mathrm {T V}} = \sum_ {j \in \Omega} \left(Q _ {j} - P _ {j}\right) ^ {+}
$$
where  $x^{+} = \max \{x,0\}$
Proof. By definition:
Decompose the  $L_{1}$  norm:
$$
\frac {1}{2} \| P - Q \| _ {1} = \frac {1}{2} \sum_ {j \in \Omega} \left(\left(Q _ {j} - P _ {j}\right) ^ {+} + \left(P _ {j} - Q _ {j}\right) ^ {+}\right)
$$
As  $\sum_{j\in \Omega}P_j = \sum_{j\in \Omega}Q_j = 1$  , it holds that:
$$
\sum_ {j \in \Omega} (Q _ {j} - P _ {j}) ^ {+} = \sum_ {j \in \Omega} (P _ {j} - Q _ {j}) ^ {+}
$$
and therefore  $\| P - Q\|_{\mathrm{TV}} = \sum_{j\in \Omega}\left(Q_j - P_j\right)^+$  as required.
B.4.1 Optimal two-action contract
Theorem 5 (Two-action min-budget contract; formal statement of Theorem 1). When  $n = 2$ , the optimal min-budget contract  $t^*$  is given by:
$$
t _ {j} ^ {*} = \frac {c}{\left\| F _ {2} - F _ {1} \right\| _ {\mathrm {T V}}} \mathbb {1} \left[ F _ {2, j} \geq F _ {1, j} \right] \tag {18}
$$
Proof. We prove this claim using LP duality, by showing that the optimal primal objective corresponding to  $t^*$  is identical to a feasible solution of the dual LP.
The primal LP (Eq. (6)) for two actions is given by:
$$
\min  _ {t \in \mathbb {R} _ {\geq 0} ^ {| \Omega |}, B \in \mathbb {R} _ {\geq 0}} B \tag {19}
$$
$$
\forall j \in \Omega : t _ {j} \leq B \tag {BUDGET}
$$
$$
\sum_ {j \in \Omega} \left(F _ {2, j} - F _ {1, j}\right) t _ {j} \geq c \tag {IC}
$$
As  $t^*$  is bounded, the optimal objective  $B^*$  corresponds to the maximal possible payout of  $t^*$ :
$$
B ^ {*} = \max  _ {j \in \Omega} t _ {j} ^ {*} = \frac {c}{\| F _ {2} - F _ {1} \| _ {\mathrm {T V}}} \tag {20}
$$
and therefore the (BUDGET) constraint is satisfied. For the (IC) constraint, denote by  $\Omega_{\geq}$  the following set:
$$
\Omega_ {\geq} = \{j \in \Omega \mid F _ {2, j} \geq F _ {1, 2} \}
$$
using this notation, we note that  $t_j^* > 0$  if and only if  $j \in \Omega_{\geq}$ . Plugging into the constraint, we obtain:
$$
\begin{array}{l} \sum_ {j \in \Omega} \left(F _ {2, j} - F _ {1, j}\right) t _ {j} ^ {*} = \sum_ {j \in \Omega_ {\geq}} \left(F _ {2, j} - F _ {1, j}\right) B ^ {*} \\ = c \frac {\sum_ {j \in \Omega_ {\geq}} \left(F _ {2 , j} - F _ {1 , j}\right)}{\| F _ {2} - F _ {1} \| _ {\mathrm {T V}}} \\ = c \\ \end{array}
$$
Where  $\| F_2 - F_1\|_{\mathrm{TV}} = \sum_{j\in \Omega_\geq}\left(F_{2,j} - F_{1,j}\right)$  by definition. This shows that  $t^*$  is a feasible solution for the primal LP.
To prove optimality, we show that the dual linear program attains an identical objective. By Claim 2, the dual LP for two actions is given by:
$$
\max_{\substack{\lambda \in \mathbb{R}_{\geq 0},\mu \in \mathbb{R}_{\geq 0}^{\left|\Omega\right|}}}c\lambda
$$
$$
s. t.
$$
$$
\forall j \in \Omega : \left(F _ {2, j} - F _ {1, j}\right) \lambda \leq \mu_ {j} \tag {21}
$$
Denote the vector  $\vec{v}_j(\lambda) \in \mathbb{R}_{\geq 0}^{|\Omega|}$ :
$$
\vec {v} _ {j} (\lambda) = \lambda (F _ {2, j} - F _ {1, j}) ^ {+}
$$
We note that outcomes  $j'$  for which  $F_{2,j'} \leq F_{1,j'}$  (formally  $j' \in \Omega \setminus \Omega_{\geq}$ ) correspond to constraints which are satisfied for any  $\lambda \geq 0$ . Otherwise  $j \in \Omega_{\geq}$ , and in these cases  $\lambda$  can be increased until  $\vec{v}(\lambda)$  saturates the simplex constraint  $\sum_{j \in \Omega} \mu_j = 1$ . The simplex constraint is binding for a value  $\lambda^* > 0$  satisfying:
$$
\sum_ {j \in \Omega} \lambda^ {*} (F _ {2, j} - F _ {1, j}) ^ {+} = 1
$$
and therefore the optimal value  $\lambda^{*}$  of the dual LP (Eq. (21)) is:
$$
c \lambda^ {*} = \frac {c}{\sum_ {j \in \Omega} \left(F _ {2 , j} - F _ {1 , j}\right) ^ {+}} = \frac {c}{\| F _ {2} - F _ {1} \| _ {\mathrm {T V}}} \tag {22}
$$
Where the second equality is given by claim 4. The dual objective in Eq. (22) is identical to the primal objective attained by the contract  $t^*$  in Eq. (20), and therefore  $t^*$  is an optimal contract by strong LP duality.
B.4.2 Contracts and hypothesis tests
We recall the formal definition of the Neyman-Pearson lemma:
Lemma 3 (Neyman-Pearson, [e.g., 45, 4.3]). Let  $P, Q \in \Delta(\mathcal{X})$  be two probability measures over an arbitrary set  $\mathcal{X}$ . Then for any hypothesis test  $\psi: \mathcal{X} \to \{0,1\}$ , it holds:
$$
P (\psi (x) = 1) + Q (\psi (x) = 0) \geq 1 - \| P - Q \| _ {\mathrm {T V}}. \tag {23}
$$
Moreover, equality holds for the Likelihood Ratio test  $\psi^{*}(x) = \mathbb{1}[q(x)\geq p(x)]$
In our analysis, we will assume that the space  $\mathcal{X}$  is finite. We also note that Lemma 3 is stated for decision rules with binary output, but its domain can be extended to fractional decision functions  $\psi : \mathcal{X} \to [0,1]$  without loss of generality: As the sum of errors is linear in  $\psi$ , the optimal fractional decision rule is a solution to the linear program  $\min_{\psi \in [0,1]^{\mathcal{X}}} (P(\psi = 1) + Q(\psi = 0))$  where  $P(\psi = 1) = \sum_{x \in \mathcal{X}} p_x \psi(x)$  and  $Q(\psi = 0) = \sum_{x \in \mathcal{X}} q_x (1 - \psi(x))$ . The feasible region of this linear program is the hypercube  $[0,1]^{\mathcal{X}}$  and its vertices are the set of binary decision rules  $\{0,1\}^{\mathcal{X}}$ , which also includes the optimal binary rule  $\psi^*$  given by Lemma 3. As every feasible linear program attains its optimum on a vertex, the binary  $\psi^*$  is also the optimal among fractional rules.
Proof of Theorem 2. For a given two-action contract, denote  $P_{j} = F_{1,j}$ ,  $Q_{j} = F_{2,j}$ , and  $c = c_{2} - c_{1}$ . We recall that the statistical min-budget LP for two actions is given by Lemma 2:
s.t. (24)
$$
\sum_ {j \in \Omega} P _ {j} \phi_ {j} + \sum_ {j \in \Omega} Q _ {j} (1 - \phi_ {j}) \leq 1 - c \beta
$$
and the Neyman-Pearson lemma is given by Eq. (23).
Given a min-budget contract design problem, apply Theorem 1 to obtain its optimal solution, as given by Eq. (18):
$$
B ^ {*} = \frac {c}{\| P - Q \| _ {\mathrm {T V}}}
$$
$$
t _ {j} ^ {*} = \frac {c}{\| P - Q \| _ {\mathrm {T V}}} \mathbb {1} [ Q _ {j} \geq P _ {j} ]
$$
Applying the transformation from Definition 3 on the optimal contract, we obtain equivalently:
$$
\beta^ {*} = \left(B ^ {*}\right) ^ {- 1} = \frac {\| P - Q \| _ {\mathrm {T V}}}{c}
$$
$$
\phi_ {j} ^ {*} = t _ {j} ^ {*} / B ^ {*} = \mathbb {1} \left[ Q _ {j} \geq P _ {j} \right]
$$
Note that  $\phi_j^*$  is a maximum-likelihood decision rule, similar to the optimal critical function in the Neyman-Pearson lemma. Additionally, by optimality of the contract-design optimization variable  $\beta^{*}$ , any feasible solution  $\phi^\prime$  of Eq. (24) satisfies:
$$
\sum_ {j = 1} ^ {m} P _ {j} \phi_ {j} ^ {\prime} + \sum_ {j = 1} ^ {m} Q _ {j} \left(1 - \phi_ {j} ^ {\prime}\right) \geq 1 - c \beta^ {*} = 1 - \| P - Q \| _ {\mathrm {T V}}
$$
with equality satisfied by the maximum likelihood rule  $\phi^{*}$ . Therefore, the min-budget optimality of the contract  $t^{*}$  implies the power optimality of the hypothesis test  $\phi^{*}$ .
Conversely, let  $\phi : \Omega \to [0,1]$  be an maximal-power hypothesis test. This provides a lower bound on the constraint in Eq. (24):
$$
\sum_ {j \in \Omega} P _ {j} \phi_ {j} + \sum_ {j \in \Omega} Q _ {j} (1 - \phi_ {j}) \geq 1 - \| P - Q \| _ {\mathrm {T V}}
$$
By the Neyman-Pearson lemma, the bound is tight for the maximum likelihood rule  $\phi^{\star} = \mathbb{1}[Q\geq P]$  and therefore the optimal objective  $\beta^{\star}$  satisfies:
$$
1 - c \beta^ {\star} = 1 - \| P - Q \| _ {\mathrm {T V}}
$$
and hence  $\beta^{\star} = \frac{\|P - Q\|_{\mathrm{TV}}}{c}$ . Applying the transformation in Definition 3 yields the optimal contract  $t_j^\star = \frac{c}{\|P - Q\|_{\mathrm{TV}}}\mathbb{1}[Q_j\geq P_j]$ , showing that the corresponding contract is min-budget optimal if the corresponding hypothesis has optimal statistical power.
Remark 1. Since the proof of Theorem 1 is independent of the Neyman-Pearson lemma, the argument proving the optimality of  $\phi^{*}$  in the proof above implies the Neyman-Pearson lemma for finite  $\mathcal{X}$ .
B.5 More than two actions
Definition 5 (All-or-nothing contract). Let  $B > 0$ . An all-or-nothing contract  $t:\Omega \to \mathbb{R}_{\geq 0}$  satisfies  $t_j\in \{0,B\}$  for all  $j\in \Omega$
B.5.1 Binary outcomes
In this section, we show that every binary-outcome min-budget contract is all-or-nothing. This is a corollary of a more general lemma:
Lemma 4. For any feasible min-budget contract  $t^*$ , there exists  $j_0 \in \Omega$  such that  $t_{j_0}^* = 0$ .
Proof. By contradiction, assume that  $t_j^* > 0$  for all  $j$ , and denote  $j_0 = \operatorname{argmin}_j t_j^*$ . Denote  $a = \min_j t_j$ , and note that  $a > 0$ . Define the contract  $\tilde{t}$  as follows:
$$
\forall j: \tilde {t} _ {j} = t _ {j} - a
$$
By definition,  $\tilde{t}_j\geq 0$  for all  $j$ , and  $\tilde{t}_{j_0} = 0$ . Since  $t^*$  is a feasible min-budget contract, it satisfies the min-budget LP in Eq. (6), and in particular the (IC) constraint:
$$
\forall i \in [ n - 1 ]: \sum_ {j} F _ {i, j} t _ {j} ^ {*} - c _ {i} \leq \sum_ {j} F _ {n, j} t _ {j} ^ {*} - c _ {n}
$$
Plugging in the definition  $\tilde{t} = t^{*} - a$  and using the fact that  $\sum_{j}F_{i,j} = 1$  for all  $i\in [n]$ , we obtain:
$$
\forall i \in [ n - 1 ]: \sum_ {j} F _ {i, j} \tilde {t} _ {j} - c _ {i} \leq \sum_ {j} F _ {n, j} \tilde {t} _ {j} - c _ {n}
$$
and therefore  $\tilde{t}$  satisfies the (IC) constraint as well. This is a contradiction, since  $\tilde{t}$  is a feasible contract with a lower required budget. From this we conclude that the optimal contract must satisfy  $t_j = 0$  for some  $j$ .
Corollary 1. In any min-budget contract design setting with two outcomes, the optimal contract is all-or-nothing.
B.5.2 Hardness: Basic definitions and construction
In this section, we show that finding optimal all-or-nothing contracts is NP-hard in the general case, by reduction from 3SAT.
Definition 6 (Maximin design matrix). For a contract design setting with action set  $\mathcal{A}$ , outcome space  $\Omega$ , target action  $a^* \in \mathcal{A}$ , and  $c_a < c_{a^*}$  for all  $a \in \mathcal{A} \setminus \{a^*\}$ , the maximin design matrix  $A$  is defined as:
$$
A _ {a \omega} = \frac {F _ {a ^ {*} , \omega} - F _ {a , \omega}}{c _ {a ^ {*}} - c _ {a}} \tag {25}
$$
Claim 5. For a contract design problem, denote the maximin design matrix by  $A$ . An all-or-nothing min-budget contract  $t^*$  can be written as  $t^* = \phi^* / \beta^*$ , where:
$$
\phi^{*} = \operatorname *{argmax}_{\phi \in \{0,1\}^{\left|\Omega \right|}}\min_{a\in \mathcal{A}\setminus \{a^{*}\}}(A\phi)_{a}
$$
$$
\beta^ {*} = \min  _ {a \in A \backslash \{a ^ {*} \}} \left(A \phi^ {*}\right) _ {a} \tag {26}
$$
Proof. By Lemma 2, Eq. (11), the min-budget contract design LP is equivalent to:
$$
\max_{\substack{\phi \in [0,1]^{| \Omega |},\beta \in \mathbb{R}_{\geq 0}}}\beta
$$
$$
\forall a \in \mathcal {A} \backslash \left\{a ^ {*} \right\}: \sum_ {\omega \in \Omega} \frac {F _ {a ^ {*} , \omega} - F _ {a , \omega}}{c _ {a ^ {*}} - c _ {a}} \phi_ {\omega} \geq \beta
$$
Where  $\beta = B^{-1}$ . As  $A_{a,\omega} = \frac{F_{a^{*},\omega} - F_{a,\omega}}{c_{a^{*}} - c_{a}}$ , it holds that  $\sum_{\omega \in \Omega} \frac{F_{a^{*},\omega} - F_{a,\omega}}{c_{a^{*}} - c_{a}} \phi_{\omega} = A\phi$ . The LP above is therefore equivalent to:
$$
\forall a \in \mathcal {A} \backslash \left\{a ^ {*} \right\}: A _ {a} \phi \geq \beta
$$
as  $\beta$  is a lower bound for every constraint, its optimal value is the maximal minimum attainable through variation of  $\phi$ :
$$
\max  _ {\phi \in [ 0, 1 ] ^ {| \Omega |}} \min  _ {a \in \mathcal {A} \backslash \{a ^ {*} \}} (A \phi) _ {a}
$$
and restricting the optimization space of  $\phi$  to  $\{0,1\}^{|\Omega|}$  yields the desired result.
Definition 7 (3-CNF - Conjunctive normal form). A 3-CNF formula over  $m$  variables and  $n$  clauses is a boolean-valued function  $\psi : \{0,1\}^m \to \{0,1\}$  of the form:
$$
\psi (x _ {1}, \dots , x _ {m}) = \bigwedge_ {i = 1} ^ {n} \left(z _ {i 1} \vee z _ {i 2} \vee z _ {i 3}\right)
$$
where  $z_{ik}\in \{x_1,\ldots ,x_m,\neg x_1,\ldots ,\neg x_m\}$
We assume that variables in each clause are distinct.  $\psi$  is satisfiable if and only if there exists  $x\in \{0,1\} ^m$  such that  $\psi (x) = 1$
Definition 8 (Number of positives in clause i). Given a 3-CNF  $\psi$  and an assignment  $x\in \{0,1\} ^m$  denote by  $\sigma_{i}(\psi ,x)\in \{0,\ldots ,3\}$  the number of variables  $z_{ik}$  in clause  $i$  which evaluate to 1 under assignment  $x$ .
Claim 6. A formula  $\psi$  is satisfiable if and only if there exists an assignment  $x \in \{0,1\}^m$  such that  $\min_{i \in [n]} \sigma_i(\psi, x) \geq 1$ .
Proof. If  $\psi$  is satisfiable then there exists  $x\in \{0,1\} ^m$  such that  $\psi (x) = 1$ . Since  $\psi$  is a 3-CNF, every clause  $i$  must evaluate to 1, and therefore  $\sigma_{i}(\psi ,x)\geq 1$  for all  $i\in [n]$ . Conversely, if there exist an assignment  $x$  such that  $\min_{i\in [n]}\sigma_i(\psi ,x)\geq 1$  then by definition  $x$  satisfies every clause, and therefore the whole formula  $\psi$ .
B.5.3 Hardness: Reduction from 3SAT
Given a 3-CNF  $\psi$  with  $n$  clauses and  $m$  variables, we define a min-budget contract design problem over actions  $\mathcal{A} = [n + 1]$  and outcome space  $\Omega = [m] \cup \{\text{pos, neg, const}\}$ . The target action is  $a^* = n + 1$ , and the cost associated with action  $i$  is:
$$
c _ {i} = \left\{ \begin{array}{l l} 1 & i = n + 1 \\ 0 & \text {o t h e r w i s e} \end{array} \right. \tag {27}
$$
For simplicity of notations, let  $\varepsilon > 0$ , which will be assigned a suitable value later in the construction. The outcome distribution of the target action is denoted by  $Q$ , and defined as:
$$
\forall j \in [ m ]: Q _ {j} = \frac {\varepsilon}{m}
$$
$$
Q _ {\text {p o s}} = 1 - \varepsilon \left(1 + \frac {3}{m}\right) \tag {28}
$$
$$
Q _ {\mathrm {n e g}} = 0
$$
$$
Q _ {\mathrm {c o n s t}} = \frac {3 \varepsilon}{m}
$$
Note that  $Q$  is well-defined for  $m \geq 3$  and  $\varepsilon \leq \frac{1}{2}$ .
For each  $i \in [n]$ , denote the number of negated variables in clause  $i$  by  $k_{i} \in \{0, \dots, 3\}$ . The distribution corresponding to the  $i$ -th action is denoted by  $P^{(i)}$ , and defined as:
$$
\forall j \in [ m ]: P _ {j} ^ {(i)} = \left\{ \begin{array}{l l} 0 & x _ {j} \text {e x i s t s i n c l a u s e} i \\ \frac {2 \varepsilon}{m} & \neg x _ {j} \text {e x i s t s i n c l a u s e} i \\ \frac {\varepsilon}{m} & \text {o t h e r w i s e} \end{array} \right.
$$
$$
P _ {\text {p o s}} ^ {(i)} = 0 \tag {29}
$$
$$
P _ {\mathrm {n e g}} ^ {(i)} = 1 - \varepsilon \left(1 + \frac {k _ {i}}{m}\right)
$$
$$
P _ {\mathrm {c o n s t}} ^ {(i)} = (3 - k _ {i}) \frac {\varepsilon}{m}
$$
The distributions  $P^{(i)}$  are well-defined when  $m \geq 3$  and  $\varepsilon \leq \frac{1}{2}$ . For concreteness, set:
$$
\varepsilon = \frac {1}{1 0} \tag {30}
$$
and note that  $Q_{\mathrm{pos}}$  and  $P_{\mathrm{neg}}^{(i)}$  are both strictly positive for  $m \geq 3$  and this value of  $\varepsilon$ .
Plugging equations (27, 28, 29) into Definition 6, the maximin design matrix  $A^{\psi} \in \mathbb{R}^{n \times (m + 3)}$  corresponding to the contract design problem above is given by:
$$
\forall j \in [ m ]: A _ {i, j} ^ {\psi} = \left\{ \begin{array}{l l} \frac {\varepsilon}{m} & x _ {j} \text {e x i s t s i n c l a u s e i a n d i s n o t n e g a t e d} \\ - \frac {\varepsilon}{m} & x _ {j} \text {e x i s t s i n c l a u s e i a n d i s n e g a t e d} \\ 0 & \text {o t h e r w i s e} \end{array} \right.
$$
$$
A _ {i, \text {p o s}} ^ {\psi} = Q _ {\text {p o s}} \tag {31}
$$
$$
A _ {i, \mathrm {n e g}} ^ {\psi} = - P _ {\mathrm {n e g}} ^ {(i)}
$$
$$
A _ {i, \mathrm {c o n s t}} ^ {\psi} = k _ {i} \frac {\varepsilon}{m}
$$
Definition 9 (Assignment normalized contract). For an assignment  $x \in \{0,1\}^m$ , the corresponding normalized contract  $\phi^x \in \{0,1\}^{m+3}$  is:
$$
\forall j \in [ m ]: \phi_ {j} ^ {x} = x _ {j}
$$
$$
\phi_ {\text {p o s}} ^ {x} = 1 \tag {32}
$$
$$
\phi_ {\mathrm {n e g}} ^ {x} = 0
$$
$$
\phi_ {\mathrm {c o n s t}} ^ {x} = 1
$$
Claim 7. Let  $\psi$  be a 3-CNF, and  $x\in \{0,1\} ^m$  an assignment. For all  $i\in [n]$
$$
\left(A ^ {\psi} \phi^ {x}\right) _ {i} = \frac {\varepsilon}{m} \sigma_ {i} (\psi , x) + Q _ {\mathrm {p o s}}
$$
Proof. To prove this claim, plug  $\phi_x$  from Definition 9 into  $A^{\psi}$  defined in Eq. (31). To simplify notations, here we denote  $A = A^{\psi}$ ,  $\phi = \phi^{x}$ ,  $z_{i} = \{z_{i1}, z_{i2}, z_{i3}\}$ . We obtain:
$$
\begin{array}{l} (A\phi)_{i} = \sum_{\omega \in [m]\cup \{\mathrm{pos, neg, const}\}}A_{i,\omega}\phi_{\omega} \\ = \qquad \qquad \underbrace {\sum_ {j \in [ m ]} A _ {i , j} \phi_ {j}} _ {= Q _ {\mathrm {p o s}}} + \underbrace {A _ {i , \mathrm {p o s}} \phi_ {\mathrm {p o s}}} _ {= 1} + A _ {i, \mathrm {n e g}} \underbrace {\phi_ {\mathrm {n e g}}} _ {= 0} + \underbrace {A _ {i , \mathrm {c o n s t}} \phi_ {\mathrm {c o n s t}}} _ {= k _ {i} \frac {\varepsilon}{m}} \\ = \frac {\varepsilon}{m} \left(\sum_ {x _ {j}} \in z _ {i} \phi_ {j} - \sum_ {\neg x _ {j ^ {\prime}}} \in z _ {i} \phi_ {j ^ {\prime}}\right) \\ = \frac {\varepsilon}{m} \left(\sum_ {x _ {j} \in z _ {i}} \phi_ {j} + \sum_ {\neg x _ {j ^ {\prime}} \in z _ {i}} \left(1 - \phi_ {j ^ {\prime}}\right)\right) + Q _ {\text {p o s}} \\ = \frac {\varepsilon}{m} \sigma_ {i} (\psi , x) + Q _ {\mathrm {p o s}} \\ \end{array}
$$
Claim 8. For a given 3-CNF  $\psi$ , denote by  $\phi^{*}$  the optimal solution for Eq. (26), with  $A = A^{\psi}$ . There exists an assignment  $x^{*}$  such that  $\phi^{*} = \phi^{x^{*}}$ .
Proof. For the maximin matrix  $A^{\psi}$  defined in Eq. (31), the optimization objective in Eq. (26) is:
$$
\min  _ {i \in [ n ]} \left(A ^ {\psi} \phi\right) _ {i}
$$
By the choice of  $\varepsilon$  in Eq. (30),  $A^{\psi}$  satisfies the following for all  $i\in [n]$ :
$$
A _ {i, \mathrm {p o s}} = Q _ {\mathrm {p o s}} > 0
$$
$$
A _ {i, \mathrm {n e g}} = - P _ {\mathrm {n e g}} ^ {(i)} <   0
$$
$$
A _ {i, \mathrm {c o n s t}} = k _ {i} \frac {\varepsilon}{m} > 0
$$
Thus, any optimal solution  $\phi^{*}$  must satisfy:
$$
\phi_ {\mathrm {p o s}} ^ {*} = 1
$$
$$
\phi_ {\mathrm {n e g}} ^ {*} = 0
$$
$$
\phi_ {\mathrm {c o n s t}} ^ {*} = 1
$$
Otherwise, the value of any  $\left(A^{\psi}\phi\right)_i$  would strictly increase by changing the value of  $\psi^{*}$  at the coordinates  $\{\mathrm{pos},\mathrm{neg},\mathrm{const}\}$  to the values specified above. This would increase the optimization objective  $\min_{a\in \mathcal{A}\backslash \{a^{*}\}}\left(A^{\psi}\phi\right)_a$ , in contradiction to the optimality of  $\phi^{*}$ .
Hence, denoting  $x_{j}^{*} = \phi_{j}^{*}$  for all  $j \in [m]$ , we obtain  $\phi^{*} = \phi^{x^{*}}$  as required.
Claim 9. A 3-CNF  $\psi$  is satisfiable if and only if the optimization problem in Eq. (26) with  $A = A^{\psi}$  satisfies  $\beta^{*} \geq Q_{\mathrm{pos}} + \frac{\varepsilon}{m}$ .
Proof. If  $\psi$  is satisfiable by assignment  $x \in \{0,1\}^m$ , then  $\sigma_i(\psi, x) \geq 1$  for all  $i \in [n]$  by Claim 6. Observe that  $\beta^* = \left(A^\psi \phi^x\right)_{i^*}$  for some  $i^* \in [n]$  by construction. Let  $\phi^x$  denote the vector corresponding to the satisfying assignment, according to Definition 9. Apply Claim 7 to obtain:
$$
\begin{array}{l} \beta^ {*} = \left(A ^ {\psi} \phi^ {x}\right) _ {i ^ {*}} \\ = \frac {\varepsilon}{m} \sigma_ {i ^ {*}} (\psi , x) + Q _ {\mathrm {p o s}} \\ \geq Q _ {\mathrm {p o s}} + \frac {\varepsilon}{m} \\ \end{array}
$$
Conversely, assume the optimal solution to Eq. (26) satisfies  $\beta^{*} \geq Q_{\mathrm{pos}} + \frac{\varepsilon}{m}$ , and denote the vector attaining the optimal solution by  $\phi^{*}$ . By Claim 8, there exists an assignment  $x^{*}$  such that  $\phi^{*} = \phi^{x^{*}}$ . Combining the lower bound on the value of the optimal solution with the result of Claim 7, we obtain that the following holds for all  $i \in [n]$ :
$$
\left(A ^ {\psi} \phi^ {x ^ {*}}\right) _ {i} = \frac {\varepsilon}{m} \sigma_ {i} (\psi , x ^ {*}) + Q _ {\mathrm {p o s}} \geq \frac {\varepsilon}{m} + Q _ {\mathrm {p o s}}
$$
Therefore, it must hold that  $\sigma_{i}(\psi ,x^{*})\geq 1$  , and thus  $x^{*}$  satisfies  $\psi$  according to Claim 6.
B.5.4 Proof of hardness
Proof of Theorem 3. By reduction from 3SAT. Given a 3-CNF  $\psi$  with  $n$  clauses and  $m$  variables, construct in polynomial time the corresponding matrix  $A^{\psi}$  as defined by Eq. (31), and apply an all-or-nothing min-budget contract solver according to Eq. (26) to obtain the optimal solution. The validity of the LP is given by Eq. (26). By Claim 9, the formula  $\psi$  is satisfiable if and only if the value attained in the optimization is at least  $\frac{\varepsilon}{m} + Q_{\mathrm{pos}}$ , where  $m$  is the number of variables in  $x$ , and  $\varepsilon, Q_{\mathrm{pos}}$  are constants defined in Eq. (30), Eq. (28) respectively.
B.6 The single binding action algorithm
To prove the soundness of SBA, we first make the following definition:
Definition 10 ((i' - IC) relaxation). Consider a MIN-BUDGET LP with target action  $i \in \mathcal{A}$ , as given by Eq. (6). For any action  $i' \neq i$ , the ( $i'$ -IC) relaxation of the min-budget problem is given by eliminating all (IC) constraints except for the one corresponding to action  $i'$ .
Proof of Proposition 2. Denote the target action by  $i \in \mathcal{A}$ . On each iteration of the loop, the algorithm considers some action  $i' \neq i$ , and applies Theorem 1 on the  $(i' - \mathrm{IC})$  relaxation of the original MIN-BUDGET LP. Since the relaxed LP has only one (IC) constraint, its optimal solution, denoted by  $t^*(i', i)$ , is given by Eq. (5).
By Definition 10, the feasible region of the original LP lies within feasible region corresponding to its  $(i^{\prime}$  -IC) relaxation. Thus, if an optimal solution of the relaxed LP lies within the feasible region of the original LP, then it is also a global optimum of the original LP.  $t^{*}(i^{\prime},i)$  lies within the feasible region of the original LP is it satisfies the remaining (IC) constraints-a condition equivalent to the notation  $a(t^{*}(i^{\prime},i)) = i$  by Eq. (2). The SBA algorithm terminates successfully only in such cases, and therefore it is sound.