SlowGuess/NeurIPS_2023_OCR huggingface dataset:
1.  $|F|\geq k$  and  $t_k\leq m_k$  ; and
2.  $|\mathcal{H}_{t_k}|\geq (1 / m_k)^2\cdot |\mathcal{H}_1|$
The base case is immediate for  $t_1 = 1 \in F$ . For the induction step, assuming that Items 1 and 2 hold for some  $k \in \mathbb{N}$  such that  $m_{k+1} \leq n$ , we show that they also hold for  $k + 1$ . For Item 1, assume for contradiction that  $t \notin F$  for all  $t$  such that  $t_k < t \leq m_{k+1}$ .
For each  $t$ ,  $t_k < t \leq m_{k+1}$ , the definition of  $r_t$  and the adversary's labeling strategy imply that the label  $y_t$  agrees with at least a  $(1 - \varepsilon_t)$ -majority of the functions in the version space  $\mathcal{H}_t$ . Hence,
$$
\begin{array}{l} \left| \mathcal {H} _ {m _ {k + 1}} \right| \geq \left| \mathcal {H} _ {t _ {k}} \right| \cdot \prod_ {t = t _ {k} + 1} ^ {m _ {k + 1}} \left(1 - \varepsilon_ {t}\right) \\ = \left| \mathcal {H} _ {t _ {k}} \right| \cdot \left(1 - 1 / m _ {k + 1}\right) ^ {m _ {k + 1} - t _ {k}} \\ \geq \left| \mathcal {H} _ {t _ {k}} \right| \cdot \left(1 - 1 / m _ {k + 1}\right) ^ {m _ {k + 1}} \\ \geq \left| \mathcal {H} _ {1} \right| \cdot \left(1 / m _ {k}\right) ^ {2} \cdot \left(1 - 1 / m _ {k + 1}\right) ^ {m _ {k + 1}} \quad (\text {I n d u c t i o n h y p o t h e s i s f o r I t e m 2}) \\ \geq \left| \mathcal {H} _ {1} \right| \cdot \left(1 / m _ {k}\right) ^ {2} \cdot (1 / 4) \\ = \left| \mathcal {H} _ {1} \right| \cdot 2 ^ {- 2 ^ {2 k + 1} - 2}. \tag {6} \\ \end{array}
$$
Observe that for every  $t \in [n]$ , if  $x_{t}$  is a node with depth  $\ell$  in  $T$  (i.e., the shortest path from the root to  $x_{t}$  contains  $\ell$  edges), then there exists an 'active' node  $x_{\ell}^{*}$  with the same depth  $\ell$  in  $T$  such that the version space  $\mathcal{H}_t$  contains only functions from  $\mathcal{H}_1$  that are consistent with the labels along the path from the root of  $T$  to  $x_{\ell}^{*}$ . Namely,  $\mathcal{H}_t$  is a subset of the  $2^{d - \ell + 1}$  functions in  $\mathcal{H}_1$  that witness the shattering of the subtree  $T_{\ell}$  of  $T$  that is rooted at  $x_{\ell}^{*}$ . In particular, the depth (distance from the root) of node  $x_{m_{k + 1}}$  is  $\log \left(2^{2^{(k + 1)}}\right) = 2^{2k + 2}$ , so
$$
\left| \mathcal {H} _ {m _ {k + 1}} \right| \leq 2 ^ {d - 2 ^ {2 k + 2} + 1} = 2 ^ {d + 1} \cdot 2 ^ {- 2 ^ {2 k + 2}} = \left| \mathcal {H} _ {1} \right| \cdot 2 ^ {- 2 ^ {2 k + 2}}. \tag {7}
$$
Combining Eqs. (6) and (7) yields  $2^{-2^{2k+1}-2} \leq 2^{-2^{2k+2}}$ , which is a contradiction. This establishes Item 1. Item 2 follows by a similar calculation, which accounts for the fact that at time  $t_{k+1}$  the adversary forces a mistake, and this reduces the version space by a factor of at most  $\varepsilon_{t_{k+1}}$ :
$$
\begin{array}{l} \left| \mathcal {H} _ {t _ {k + 1}} \right| \geq \left| \mathcal {H} _ {t _ {k}} \right| \cdot \left(\prod_ {t = t _ {k} + 1} ^ {t _ {k + 1} - 1} \left(1 - \varepsilon_ {t}\right)\right) \cdot \varepsilon_ {t _ {k + 1}} \\ \geq \left| \mathcal {H} _ {t _ {k}} \right| \cdot \left(1 - 1 / m _ {k + 1}\right) ^ {m _ {k + 1}} \cdot \left(1 / m _ {k + 1}\right) \\ \geq \left| \mathcal {H} _ {t _ {k}} \right| \cdot (1 / 4) \cdot (1 / m _ {k + 1}) \\ \geq \left| \mathcal {H} _ {1} \right| \cdot \left(1 / m _ {k}\right) ^ {2} \cdot \left(1 / 4\right) \cdot \left(1 / m _ {k + 1}\right) \quad (\text {I n d u c t i o n h y p o t h e s i s f o r I t e m 2}) \\ = \left| \mathcal {H} _ {1} \right| \cdot 2 ^ {- 2 \cdot 2 ^ {2 k}} \cdot (1 / 4) \cdot 2 ^ {- 2 ^ {2 k + 2}} = \left| \mathcal {H} _ {1} \right| \cdot 2 ^ {- 6 \cdot 2 ^ {2 k} - 2} \\ \geq \left| \mathcal {H} _ {1} \right| \cdot 2 ^ {- 8 \cdot 2 ^ {2 k}} = \left| \mathcal {H} _ {1} \right| \cdot 2 ^ {- 2 \cdot 2 ^ {2 (k + 1)}} = \left| \mathcal {H} _ {1} \right| \cdot (1 / m _ {k + 1}) ^ {2}. \\ \end{array}
$$
To complete the proof, let  $k^* = \min \left\{\lfloor \log (d) / 2 \rfloor, \lfloor \log \log (n) / 2 \rfloor \right\}$ . Then  $m_{k^*} \leq 2^d < 2^{d + 1} - 1 = n_T$ , so  $T$  contains at least  $m_{k^*}$  nodes. Additionally,  $m_{k^*} \leq n$ , so Item 1 implies that  $|F| \geq k^*$ , namely, the adversary can force at least  $k^*$  mistakes, as desired.
B Multiclass Trichotomy
The following generalization of the Littlestone dimerion is due to [DSBS15].
Definition B.1 (Multiclass Littlestone Dimension). Let  $\mathcal{X}$  and  $\mathcal{Y}$  be sets and let  $d \in \mathbb{N}$ . A Littlestone tree of depth  $d$  with domain  $\mathcal{X}$  and label set  $\mathcal{Y}$  is a set
$$
T = \left\{\left(x _ {u}, y _ {u \circ 0}, y _ {u \circ 1}\right) \in \mathcal {X} \times \mathcal {Y} \times \mathcal {Y}: u \in \bigcup_ {s = 0} ^ {d} \{0, 1 \} ^ {s} \wedge y _ {u \circ 0} \neq y _ {u \circ 1} \right\}, \tag {8}
$$
where  $\circ$  denotes string concatenation. Let  $\mathcal{H} \subseteq \mathcal{Y}^{\mathcal{X}}$ . We say that  $\underline{\mathcal{H}}$  shatters a tree  $T$  as in Eq. (8) if for every  $u \in \{0,1\}^{d+1}$  there exists  $h_u \in \mathcal{H}$  such that
$$
\forall i \in [ d + 1 ]: h \left(x _ {u _ {\leq i - 1}}\right) = y _ {u _ {\leq i}}.
$$
The Littlestone dimension of  $\mathcal{H}$ , denoted  $\mathsf{LD}(\mathcal{H})$ , is the supremum over all  $d \in \mathbb{N}$  such that there exists a Littlestone tree of depth  $d$  with domain  $\mathcal{X}$  and label set  $\mathcal{Y}$  that is shattered by  $\mathcal{H}$ .
The Natarajan dimension is a popular generalization of the VC dimension to the multiclass setting.
Definition B.2 ([Nat89]). Let  $\mathcal{X}$  and  $\mathcal{Y}$  be sets, let  $\mathcal{H} \subseteq \mathcal{Y}^{\mathcal{X}}$ , let  $d \in \mathbb{N}$ , and let  $X = \{x_{1}, \ldots, x_{d}\} \subseteq \mathcal{X}$ . We say that  $\mathcal{H}$  Natarajan-shatters  $X$  if there exist  $f_{0}, f_{1}: X \to \mathcal{Y}$  such that:
1.  $\forall x\in X:f_0(x)\neq f_1(x)$  ; and
2.  $\forall A\subseteq X\exists h\in \mathcal{H}\forall x\in X:h(x) = f_{\mathbb{1}(x\in A)}(x).$
The Natarajan dimension of  $\mathcal{H}$  is  $\mathsf{ND}(\mathcal{H}) = \sup \{|X|:X\subseteq \mathcal{X}$  finite  $\wedge$  Natarajan-shatters  $X\}$ .
We show the following generalization of Theorem 4.1 for the multiclass setting.
Theorem B.3 (Formal Version of Theorem 5.1). Let  $\mathcal{X}$  and  $\mathcal{Y}$  be sets with  $k = |\mathcal{Y}| < \infty$ , let  $\mathcal{H} \subseteq \mathcal{Y}^{\mathcal{X}}$ , and let  $n \in \mathbb{N}$  such that  $n \leq |\mathcal{X}|$ .
1. If  $\mathsf{ND}(\mathcal{H}) = \infty$  then  $M(\mathcal{H}, n) = n$ .
2. Otherwise, if  $\mathsf{ND}(\mathcal{H}) = d < \infty$  and  $\mathsf{LD}(\mathcal{H}) = \infty$  then
$$
\max  \left\{\min  \{d, n \}, \lfloor \log (n) \rfloor \right\} \leq M (\mathcal {H}, n) \leq O (d \log (n k / d)). \tag {9}
$$
The  $\Omega (\cdot)$  and  $O(\cdot)$  notations hide universal constants that do not depend on  $\mathcal{X}$ ,  $\mathcal{Y}$  or  $\mathcal{H}$ .
3. Otherwise, there exists a number  $C(\mathcal{H}) \in \mathbb{N}$  (that depends on  $\mathcal{X}$ ,  $\mathcal{Y}$  and  $\mathcal{H}$  but does not depend on  $n$ ) such that  $M(\mathcal{H}, n) \leq C(\mathcal{H})$ .
The proof of Theorem B.3 uses the following generalization of the Sauer-Shelah-Perles lemma.
Theorem B.4 ([Nat89]; Corollary 5 in [HL95]). Let  $d, n, k \in \mathbb{N}$ , let  $\mathcal{X}$  and  $\mathcal{Y}$  be sets of cardinality  $n$  and  $k$  respectively, and let  $\mathcal{H} \subseteq \mathcal{Y}^{\mathcal{X}}$  such that  $\mathrm{ND}(\mathcal{H}) \leq d$ . Then
$$
| \mathcal {H} | \leq \sum_ {i = 0} ^ {d} \binom {n} {i} \binom {k + 1} {2} ^ {i} \leq \left(\frac {e n k ^ {2}}{d}\right) ^ {d}.
$$
Proof of Theorem B.3. Items 1 and 3 and the  $\min\{d, n\}$  lower bound in Item 2 follow similarly to the corresponding items in Theorem 4.1. The upper bound in Item 2 also follows similarly to the corresponding item in Theorem 4.1, except that it uses Theorem B.4 instead of the Sauer-Shelah-Perles lemma. The  $\lfloor \log(n) \rfloor$  lower bound in Item 2 follows from Theorem D.5 and Claim D.4.  $\square$
C Combinatorics of Trees
In this section we present a simple lemma from Ramsey theory about trees that is used for proving Theorem D.5. We start with a generalized definition of subtrees.
Definition C.1. Let  $X$  be a finite set and let  $(X, \preceq)$  be a partial order relation. For  $p, c \in X$ , we say that  $c$  is a child of  $p$  if  $p \preceq c$  and there does not exist  $m \in X$  such that  $p \preceq m \preceq c$ . We say that  $z \in X$  is a leaf if there exists no  $x \in X$  such that  $z \preceq x$ .  $(X, \preceq)$  is a binary tree if every non-leaf  $x \in X$  has precisely 2 children. The depth of  $z \in X$  is the largest  $d \in \mathbb{N}$  for which there exist distinct  $x_1, \ldots, x_d \in X$  such that  $x_1 \preceq x_2 \preceq \dots \preceq x_d \preceq z$ . For  $d \in \mathbb{N}$ , we say that  $(X, \preceq)$  is a complete binary tree of depth  $d$  if  $(X, \preceq)$  is a binary tree and all the leaves in  $X$  have depth  $d$ . We say that a partial order  $(X', \preceq')$  is a subtree of  $(X, \preceq)$  if  $X' \subseteq X$ , and  $\forall a, b \in X': a \preceq' b \iff a \preceq b$ .
Lemma C.2 (Lemma 16 in [ALMM19]). Let  $p, q \in \mathbb{R}$  be non-negative such that  $p + q \in \mathbb{N}$ . Let  $T = (X, \preceq)$  be a complete binary tree of depth  $d = p + q - 1$ , and let  $f: X \to \{0, 1\}$ . Then at least one of the following statements holds:
-  $T$  has a 0-monochromatic complete binary subtree of depth at least  $p$ . Namely, there exists  $T' = (X', \preceq')$  such that  $T'$  is a subtree of  $T$ ,  $T'$  is a complete binary tree of depth at least  $p$ , and  $f(x) = 0$  for all  $x \in X'$ .
-  $T$  has a 1-monochromatic complete binary tree subtree of depth at least  $q$ .
For completeness, we include a proof of this lemma.
Proof of Lemma C.2. We prove the claim by induction on the depth  $d$ . The base case of  $d = 0$  (a tree with a single node) is immediate. For the induction step, let  $a$  denote the root of  $T$ , and let  $T_{\ell}$
and  $T_{r}$  denote the subtrees of  $T$  of depth  $d - 1$  consisting of all descendants of the left and right child of  $a$  respectively. Assume that  $f(a) = 0$ . If  $T_{\ell}$  or  $T_{r}$  contain a 1-monochromatic subtree of depth at least  $q$ , then we are done. Otherwise, by the induction hypothesis, both trees contain a 0-monochromatic subtree of depth at least  $p - 1$ . Joining these two subtrees as children of the root  $a$  yields a 0-monochromatic subtree of depth at least  $p$ , as desired. The proof for the case  $f(a) = 1$  is similar.
We use the following corollary of Lemma C.2.
Lemma C.3. Let  $k, d \in \mathbb{N}$ . Let  $T = (X, \preceq)$  be a complete binary tree of depth  $d \in \mathbb{N}$ , and let  $f: X \to [k]$ . Then  $T$  has an  $f$ -monochromatic complete binary subtree  $T' = (X', \preceq')$  of depth at least
$$
d ^ {\prime} = \frac {d + 1}{2 ^ {\lceil \log (k) \rceil}}.
$$
Namely, there exists  $T'$  such that  $T'$  is a subtree of  $T$ ,  $T'$  is a complete binary tree of depth at least  $d'$ , and  $|\{f(a) : a \in X'\}| = 1$ .
Proof of Lemma C.3. We will show that for any  $b \in \mathbb{N}$ , if  $k \leq 2^b$  then there exists an  $f$ -monochromatic subtree of  $T$  of depth at least
$$
\frac {d + 1}{2 ^ {b}}.
$$
This implies the lemma, which corresponds to the special case  $b = \lceil \log (k) \rceil$ .
We proceed by induction on  $b$ . The base case  $b = 1$  follows from Lemma C.2. For the induction step, we assume that the claim holds for  $b$  and prove that it holds for  $b + 1$ . Namely, we show that if  $f: X \to [k]$  and  $k \leq 2^{b + 1}$  then there exists an  $f$ -monochromatic subtree of depth at least  $(d + 1) / 2^{b + 1}$ .
Define  $g: X \to \{1,2\}$  by  $g(x) = 1 + (f(x) \mod 2)$ . By Lemma C.2, there exists a  $g$ -monochromatic complete binary subtree  $T_0 = (X_0, \preceq)$  of  $T$  of depth at least  $(d + 1)/2$ . In particular  $|\{f(x): x \in X_0\}| \leq 2^b$ . By invoking the induction hypotheses on  $T_0$ , there exists a complete binary subtree of  $T_0$  that is  $f$ -monochromatic and has depth at least
$$
\frac {\frac {d + 1}{2} + 1}{2 ^ {b}} > \frac {d + 1}{2 ^ {b + 1}},
$$
D Multiclass Threshold Bounds
Definition D.1. Let  $\mathcal{X}$  and  $\mathcal{Y}$  be sets, let  $X = \{x_{1},\ldots ,x_{t}\} \subseteq \mathcal{X}$ , and let  $\mathcal{H}\subseteq \mathcal{V}^{\mathcal{X}}$ . We say that  $X$  is threshold-shattered by  $\mathcal{H}$  if there exist distinct  $y_0,y_1\in \mathcal{V}$  and functions  $h_1,\dots ,h_t\in \mathcal{H}$  such that  $h_i(x_j) = y_{1(j\leq i)}$ . The threshold dimension of  $\mathcal{H}$ , denoted  $\mathsf{T}\mathsf{D}(\mathcal{H})$ , is the supremum of the set of integers  $t$  for which there exists a threshold-shattered set of cardinality  $t$ .
We introduce the following generalization of the threshold dimension.
Definition D.2. Let  $\mathcal{X}$  and  $\mathcal{Y}$  be sets, let  $X = \{x_{1},\ldots ,x_{t}\} \subseteq \mathcal{X}$ , and let  $\mathcal{H}\subseteq \mathcal{V}^{\mathcal{X}}$ . We say that  $X$  is multi-class threshold-shattered by  $\mathcal{H}$  if there exist  $y_{1},y_{1}^{\prime}\dots,y_{t},y_{t}^{\prime}\in \mathcal{V}$  such that  $y_{i}\neq y_{j}^{\prime}$  for all  $i,j\in [t]$ , and there exist functions  $h_1,\ldots ,h_t\in \mathcal{H}$  such that
$$
h _ {i} (x _ {j}) = \left\{ \begin{array}{l l} y _ {i} & (j \leq i) \\ y _ {j} ^ {\prime} & (j > i). \end{array} \right.
$$
The multi-class threshold dimension of  $\mathcal{H}$ , denoted  $\mathrm{MTD}(\mathcal{H})$ , is the supremum of the set of integers  $t$  for which there exists a threshold-shattered set of cardinality  $t$ .
See Table 1 for an illustration of this definition.
Claim D.3. Let  $\mathcal{X}$  and  $\mathcal{Y}$  be sets,  $k = |\mathcal{Y}| < \infty$ , and let  $\mathcal{H} \subseteq \mathcal{Y}^{\mathcal{X}}$ . Then  $\mathrm{TD}(\mathcal{H}) \geq \lfloor \mathrm{MTD}(\mathcal{H}) / k^2 \rfloor$ . Proof of Claim D.3. The proof follows from two applications of the pigeonhole principle.
x1	x1	x3	x4	x5
h1	y1	y'2	y'3	y'4	y'5
h2	y2	y2	y'3	y'4	y'5
h3	y3	y3	y3	y'4	y'5
h4	y4	y4	y4	y4	y'5
h5	y5	y5	y5	y5	y5
Table 1: An illustration of Definition D.2. The table shows a collection of points  $\{x_{1},\ldots ,x_{5}\}$  that are multi-class threshold shattered by functions  $\{h_1,\dots,h_5\}$ .
Claim D.4. Let  $\mathcal{X}$  and  $\mathcal{Y}$  be sets, let  $\mathcal{H} \subseteq \mathcal{Y}^{\mathcal{X}}$  such that  $d = \mathrm{TD}(\mathcal{H}) < \infty$ , and let  $n \in \mathbb{N}$ . Then
The proof of Claim D.4 is similar to that of Claim 3.4.
Theorem D.5. Let  $\mathcal{X}$  and  $\mathcal{Y}$  be sets with  $k = |\mathcal{Y}| < \infty$ , let  $\mathcal{H} \subseteq \mathcal{Y}^{\mathcal{X}}$ . If  $\operatorname{LD}(\mathcal{H}) = \infty$  then  $\operatorname{MTD}(\mathcal{H}) = \infty$ .
Proof of Theorem D.5. Let  $f_{k}(d)$  be the largest number such that every class with Littlestone dimension  $d$  has multi-class threshold dimension at least  $f_{k}(d)$ . We show by induction on  $d$  that  $f_{k}$  satisfies the following recurrence relation:
$$
f _ {k} (d) \geq \left\{ \begin{array}{l l} 1 & d = 1 \\ 1 + f _ {k} (\lceil d / 2 k \rceil - 1) & d > 1 \end{array} \right..
$$
In particular, this implies that  $f_{k}(d) \xrightarrow{d \to \infty} \infty$ , which implies the theorem.
The base case  $d = \mathsf{LD}(\mathcal{H}) = 1$  is immediate. For the induction step, we assume the relation holds whenever  $\mathsf{LD}(\mathcal{H}) \in [d - 1]$ , and prove that it holds for  $\mathsf{LD}(\mathcal{H}) = d$ . Let  $T$  be a Littlestone tree of depth  $d$  that is shattered by  $\mathcal{H}$ . Fix  $h \in \mathcal{H}$ . Then  $h$  is a  $k$ -cloning of the nodes of  $T$ . By Lemma C.3, there exists an  $h$ -monochromatic subtree  $T' \subseteq T$  of depth at least  $\lceil d / 2k \rceil$ . Let  $y$  be the color assigned by  $h$  to all nodes of  $T'$ .  $T'$  is shattered by  $\mathcal{H}$ , so there exists a child  $c$  of the root  $x$  of  $T'$  such that edge from  $x$  to  $c$  is labeled by some value  $y' \neq y$ . Let  $\mathcal{H}' = \{g \in \mathcal{H} : g(x) = y'\}$ .  $\mathcal{H}'$  shatters the subtree rooted at  $c$ , so  $\mathsf{LD}(\mathcal{H}') \geq \lceil d / 2k \rceil - 1$ . By the induction hypothesis, there exist  $x_1, \ldots, x_s$  for  $s = f_k(\lceil d / 2k \rceil - 1)$  that are multi-class threshold shattered by functions  $h_1, \ldots, h_s \in \mathcal{H}'$ .
By construction, the set  $X = \{x_{1},\ldots ,x_{s},x_{s + 1} = x\}$  is multi-class threshold shattered by  $\{h_1,\dots ,h_s,h_{s + 1} = h\}$ , because  $h_{s + 1}(x_j) = y$  for all  $j\in [s + 1]$ , and  $h_i(x_{s + 1}) = y'$  for all  $i\in [s]$ . Hence,  $f_{k}(d)\geq s + 1 = 1 + f_{k}(\lceil d / 2k\rceil -1)$ , as desired.
E Proof of Agnostic Lower Bound
The lower bound in Theorem 6.1 is derived using an anti-concentration technique from Lemma 14 of [BPS09]. Specifically, this technique uses the following inequality.
Theorem E.1 (Khinchine's inequality; Lemma 8.2 in [CL06]). Let  $k \in \mathbb{N}$ , and let  $\sigma_1, \sigma_2, \ldots, \sigma_k$  be random variables sampled independently and uniformly at random from  $\{\pm 1\}$ . Then
$$
\mathbb {E} \left[ \left| \sum_ {i \in k} \sigma_ {k} \right| \right] \geq \sqrt {k / 2}.
$$
Proof of lower bound in Theorem 6.1. Let  $d = \mathsf{VC}(\mathcal{H})$ . Let  $\{x_1^*,\ldots ,x_d^*\} \subseteq \mathcal{X}$  be a set of cardinality  $d$  that is VC-shattered by  $\mathcal{H}$ . Let  $k\in \mathbb{N}$  be the largest integer such that  $kd\leq n$ .
Let  $x \in \mathcal{X}^n$  be a sequence consisting of  $k$  copies of the shattered set, namely,
$$
\left(x _ {1}, \dots , x _ {k d}\right) = \left(x _ {1} ^ {1}, x _ {1} ^ {2}, \dots , x _ {1} ^ {k}, x _ {2} ^ {1}, x _ {2} ^ {2}, \dots , x _ {2} ^ {k}, \dots , x _ {d} ^ {1}, x _ {d} ^ {2}, \dots , x _ {d} ^ {k}\right),
$$
such that  $x_{i}^{j} = x_{i}^{*}$  for all  $i \in [d]$  and  $j \in [k]$ . If  $kd < n$  then the remaining  $n - kd$  elements of  $x$  may be arbitrary.
Consider a randomized adversary that selects the sequence  $x$ , and chooses all labels to be i.i.d. uniform random bits. For each  $i \in [d]$  and  $j \in [k]$ , let  $y_{i}^{j} = y_{(i-1)k+j}$  and  $\hat{y}_{i}^{j} = \hat{y}_{(i-1)k+j}$  denote, respectively, the labels and the predictions corresponding to  $x_{i}^{j}$ . Then for any learner strategy  $A$ ,
$$
\begin{array}{l} \underset {y \sim \mathrm {U} (\{0, 1 \} ^ {n})} {\mathbb {E}} \left[ R (A, \mathcal {H}, x, y) \right] \\ = \underset {y \sim \mathrm {U} (\{0, 1 \} ^ {n})} {\mathbb {E}} \left[ \underset {\hat {y}} {\mathbb {E}} \left[ \sum_ {i \in [ n ]} \mathbb {1} \left(\hat {y} _ {t} \neq y _ {t}\right) \right] - \underset {h \in \mathcal {H}} {\min } \sum_ {i \in [ n ]} \mathbb {1} \left(h (x _ {t}) \neq y _ {t}\right) \right] \\ = \frac {n}{2} - \underset {y \sim \mathrm {U} (\{0, 1 \} ^ {n})} {\mathbb {E}} \left[ \min  _ {h \in \mathcal {H}} \sum_ {i \in [ n ]} \mathbb {1} (h (x _ {t}) \neq y _ {t}) \right] \quad (y _ {i} \perp \hat {y} _ {i}) \\ \geq \frac {k d}{2} - \underset {y \sim \mathrm {U} \left(\{0, 1 \} ^ {k d}\right)} {\mathbb {E}} \left[ \min  _ {h \in \mathcal {H}} \sum_ {i \in [ d ]} \sum_ {j \in [ k ]} \mathbb {1} \left(h \left(x _ {i} ^ {j}\right) \neq y _ {i} ^ {j}\right) \right] (10) \\ = \frac {k d}{2} - \underset {y \sim \mathrm {U} (\{0, 1 \} ^ {k d})} {\mathbb {E}} \left[ \sum_ {i \in [ d ]} \min  _ {h \in \mathcal {H}} \sum_ {j \in [ k ]} \mathbb {1} \left(h \left(x _ {i} ^ {j}\right) \neq y _ {i} ^ {j}\right) \right] \quad (\mathcal {H} \text {s h a t t e r s} \left\{x _ {1} ^ {*}, \dots , x _ {d} ^ {*} \right\}) \\ = \sum_ {i = 1} ^ {d} \frac {k}{2} - \underset {y _ {i} \sim \mathrm {U} (\{0, 1 \} ^ {k})} {\mathbb {E}} \left[ \min  _ {h \in \mathcal {H}} \sum_ {j \in [ k ]} \mathbb {1} \left(h (x _ {i} ^ {j}) \neq y _ {i} ^ {j}\right) \right] \\ = \sum_ {i = 1} ^ {d} \frac {k}{2} - \underset {y _ {i} \sim \mathrm {U} (\{0, 1 \} ^ {k})} {\mathbb {E}} [ \min  \left\{r _ {i}, k - r _ {i} \right\} ] \quad (\text {L e t} r _ {i} = \sum_ {j \in [ k ]} y _ {i} ^ {j}) \\ = \sum_ {i = 1} ^ {d} \underset {y _ {i} \sim \mathrm {U} (\{0, 1 \} ^ {k})} {\mathbb {E}} \left[ \left| \frac {k}{2} - r _ {i} \right| \right] \\ = \sum_ {i = 1} ^ {d} \underset {y _ {i} \sim \mathrm {U} (\{0, 1 \} ^ {k})} {\mathbb {E}} \left[ \left| \frac {k}{2} - \sum_ {j \in [ k ]} \left(\frac {1}{2} + \frac {\sigma_ {i} ^ {j}}{2}\right) \right| \right] \quad (\text {L e t} \sigma_ {i} ^ {j} = \left\{ \begin{array}{l l} 1 & y _ {i} ^ {j} = 1 \\ - 1 & y _ {i} ^ {j} = 0 \end{array} \right.) \\ = \frac {1}{2} \sum_ {i = 1} ^ {d} \mathbb {E} _ {y _ {i} \sim \mathrm {U} (\{0, 1 \} ^ {k})} \left[ \left| \sum_ {j \in [ k ]} \sigma_ {i} ^ {j} \right| \right] \\ \geq \frac {1}{2} \sum_ {i = 1} ^ {d} \sqrt {\frac {k}{2}} = \frac {d \sqrt {k}}{2 \sqrt {2}} = \Omega (\sqrt {n d}), (11) \\ \end{array}
$$
where the final inequality is Khinchne's inequality (Theorem E.1). To see that Inequality (10) holds, let  $h^* \in \operatorname{argmin}_{h \in \mathcal{H}} \sum_{i=1}^{kd} \mathbb{1}(h(x_t) \neq y_t)$ , and then
$$
\begin{array}{l} \underset {y \sim \mathrm {U} (\{0, 1 \} ^ {n})} {\mathbb {E}} \left[ \min  _ {h \in \mathcal {H}} \sum_ {i \in [ n ]} \mathbb {1} (h (x _ {t}) \neq y _ {t}) \right] \\ \leq \underset {y \sim \mathrm {U} (\{0, 1 \} ^ {n})} {\mathbb {E}} \left[ \sum_ {i = 1} ^ {k d} \mathbb {1} \left(h ^ {*} \left(x _ {t}\right) \neq y _ {t}\right) \right] + \underset {y \sim \mathrm {U} (\{0, 1 \} ^ {n})} {\mathbb {E}} \left[ \sum_ {i = k d + 1} ^ {n} \mathbb {1} \left(h ^ {*} \left(x _ {t}\right) \neq y _ {t}\right) \right] \\ \leq \underset {y \sim \mathrm {U} (\{0, 1 \} ^ {n})} {\mathbb {E}} \left[ \sum_ {i = 1} ^ {k d} \mathbb {1} \left(h ^ {*} \left(x _ {t}\right) \neq y _ {t}\right) \right] + \frac {n - k d}{2}. \quad \left(\left\{y _ {t} \right\} _ {t > k d} \perp h ^ {*}\right) \\ \end{array}
$$
In particular, Eq. (11) implies that for every learner strategy  $A$  there exists  $y \in \{0, 1\}^n$  such that  $R(A, \mathcal{H}, x, y) = \Omega(\sqrt{nd})$ .