SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \mathcal {M} ^ {\intercal} (u) \mathcal {T} (H ^ {A}, v) \mathcal {M} (u) = \left(\sum_ {i, j} m _ {i j} (u) \mathbf {e} _ {j} \mathbf {e} _ {i} ^ {\intercal}\right) \left(\sum_ {i, j} t _ {i j} \mathbf {e} _ {i} \mathbf {e} _ {j} ^ {\intercal}\right) \left(\sum_ {i, j} m _ {i j} (u) \mathbf {e} _ {i} \mathbf {e} _ {j} ^ {\intercal}\right) \\ = \sum_ {i, j} t _ {i j} \left(\sum_ {k} m _ {i k} (u) \mathbf {e} _ {k}\right) \left(\sum_ {l} m _ {j l} (u) \mathbf {e} _ {l}\right) ^ {\intercal} \\ := \sum_ {i, j} t _ {i j} \mathbf {f} _ {i} (u) \mathbf {f} _ {j} (u) ^ {\intercal}. \\ \end{array}
$$
Thus it is enough to show that  $\sum_{i,j} s_{ij} \mathbf{e}_i \mathbf{e}_j^{\top}$  and  $\sum_{i,j} t_{ij} \mathbf{f}_i(u) \mathbf{f}_j(u)^{\top}$  are the same under the basis transformation  $\mathbf{f}_i(u) = \sum_k m_{ik}(u) \mathbf{e}_k$ . From here, we briefly write  $\mathbf{f}_i$  instead of  $\mathbf{f}_i(u)$ , and  $\mathbf{f}_{-1} = \mathbf{0}$ . Note that  $u_i(\mathbf{e}_i - \mathbf{e}_{i+1}) = (\mathbf{f}_{N-i} - \mathbf{f}_{N-i-1})$ ,  $0 \leq i \leq N$  by definition of  $\mathcal{M}(u)$ . Therefore, we have
$$
\begin{array}{l} \frac {1}{L} \mathcal {H} ^ {\intercal} \sum_ {i = 0} ^ {N} u _ {i} (\mathbf {e} _ {0} + \dots + \mathbf {e} _ {i}) (\mathbf {e} _ {i} - \mathbf {e} _ {i + 1}) ^ {\intercal} = \frac {1}{L} \mathcal {H} ^ {\intercal} \sum_ {i = 0} ^ {N} (\mathbf {e} _ {0} + \dots + \mathbf {e} _ {i}) (\mathbf {f} _ {N - i} - \mathbf {f} _ {N - i - 1}) ^ {\intercal} \\ = \frac {1}{L} \mathcal {H} ^ {\intercal} \left[ \sum_ {i = 0} ^ {N} e _ {i} \mathbf {f} _ {N - i} ^ {\intercal} \right]. \\ \end{array}
$$
Therefore, we can rewrite (19) as follows:
$$
\begin{array}{l} \mathcal {S} (H, u) = - \frac {1}{2 L} \mathbf {f} _ {N} \mathbf {f} _ {N} ^ {\intercal} + \frac {1}{2 L} \mathcal {H} ^ {\intercal} \left[ \sum_ {i = 0} ^ {N} \mathbf {e} _ {i} \mathbf {f} _ {N - i} ^ {\intercal} \right] + \frac {1}{2 L} \sum_ {i = 0} ^ {N} \mathbf {f} _ {N - i} \mathbf {e} _ {i} ^ {\intercal} \mathcal {H} \\ + \frac {1}{2 L} \sum_ {i = 0} ^ {N} \left[ (\mathbf {f} _ {N - i} - \mathbf {f} _ {N - i - 1}) \mathbf {e} _ {i} ^ {\intercal} + \mathbf {e} _ {i} (\mathbf {f} _ {N - i} \mathbf {f} _ {N - i - 1}) ^ {\intercal} \right] - \frac {u _ {N}}{2 L} \mathbf {e} _ {N} \mathbf {e} _ {N} ^ {\intercal} \\ = \underbrace {- \frac {1}{2 L} \mathbf {f} _ {N} \mathbf {f} _ {N} ^ {\intercal}} _ {\mathbf {A} _ {1}} + \underbrace {\frac {1}{2 L} \left[ \sum_ {i , j} h _ {i , j} \mathbf {e} _ {j} \mathbf {f} _ {N - i} ^ {\intercal} \right] + \frac {1}{2 L} \left[ \sum_ {i , j} h _ {i , j} \mathbf {f} _ {N - i} \mathbf {e} _ {j} ^ {\intercal} \right]} _ {\mathbf {B} _ {1}} \tag {27} \\ + \underbrace {\frac {1}{2 L} \sum_ {i = 0} ^ {N} \left[ (\mathbf {f} _ {N - i} - \mathbf {f} _ {N - i - 1}) \mathbf {e} _ {i} ^ {\intercal} + \mathbf {e} _ {i} (\mathbf {f} _ {N - i} - \mathbf {f} _ {N - i - 1}) ^ {\intercal} \right]} _ {\mathbf {C} _ {1}} - \underbrace {\frac {u _ {N}}{2 L} \mathbf {e} _ {N} \mathbf {e} _ {N} ^ {\intercal}} _ {\mathbf {D} _ {1}}. \\ \end{array}
$$
Similarly, by using
$$
\mathbf {e} _ {N - i} - \mathbf {e} _ {N - i + 1} = \frac {1}{u _ {N - i}} \left(\mathbf {f} _ {i} - \mathbf {f} _ {i - 1}\right) = v _ {i} \left(\mathbf {f} _ {i} - \mathbf {f} _ {i - 1}\right),
$$
we can rewrite (22) as follows:
$$
\begin{array}{l} \mathcal {M} ^ {\intercal} (u) \mathcal {T} \left(H ^ {A}, v\right) \mathcal {M} (u) \\ = \frac {1}{2 L} \sum_ {i = 0} ^ {N} \left(\mathbf {e} _ {N - i + 1} - \mathbf {e} _ {N - i}\right) \left(\mathbf {f} _ {i - 1} - \mathbf {f} _ {N}\right) ^ {\intercal} + \left(\mathbf {f} _ {i - 1} - \mathbf {f} _ {N}\right) \left(\mathbf {e} _ {N - i + 1} - \mathbf {e} _ {N - i}\right) ^ {\intercal} \\ - \frac {1}{2 u _ {N} L} \mathbf {f} _ {0} \mathbf {f} _ {0} ^ {\intercal} - \frac {1}{2 L} \mathbf {f} _ {N} \mathbf {f} _ {N} ^ {\intercal} + \frac {1}{2 L} \left(\mathcal {H} ^ {A}\right) ^ {\intercal} \left[ \sum_ {i = 0} ^ {N} \mathbf {f} _ {i} \mathbf {e} _ {N - i} ^ {\intercal} \right] + \frac {1}{2 L} \left[ \sum_ {i = 0} ^ {N} \mathbf {e} _ {N - i} \mathbf {f} _ {i} ^ {\intercal} \right] \mathcal {H} ^ {A} \\ = - \underbrace {\frac {1}{2 u _ {N} L} \mathbf {f} _ {0} \mathbf {f} _ {0} ^ {\intercal}} _ {\mathbf {D} _ {2}} + \underbrace {\frac {1}{2 L} \left[ \sum_ {i , j} h _ {N - i , N - j} \mathbf {f} _ {j} \mathbf {e} _ {N - i} ^ {\intercal} \right] + \frac {1}{2} \left[ \sum_ {i , j} h _ {N - j , N - i} \mathbf {e} _ {N - i} \mathbf {f} _ {j} ^ {\intercal} \right]} _ {\mathbf {B} _ {2}} \\ + \underbrace {\left[ \frac {1}{2 L} \sum_ {i = 0} ^ {N} \left(\mathbf {e} _ {N - i + 1} - \mathbf {e} _ {N - i}\right) \mathbf {f} _ {i - 1} ^ {\intercal} + \mathbf {f} _ {i - 1} \left(\mathbf {e} _ {N - i + 1} - \mathbf {e} _ {N - i}\right) ^ {\intercal} - \frac {1}{2 L} \left(\mathbf {e} _ {0} \mathbf {f} _ {N} ^ {\intercal} + \mathbf {f} _ {N} \mathbf {e} _ {0} ^ {\intercal}\right) \right]} _ {\mathbf {C} _ {2}} - \underbrace {\frac {1}{2 L} \mathbf {f} _ {N} \mathbf {f} _ {N} ^ {\intercal}} _ {\mathbf {A} _ {2}}. \tag {28} \\ \end{array}
$$
For the final step, we compare (28) and (27) term-by-term, by showing  $\mathbf{X}_1 = \mathbf{X}_2$  for  $X = A, B, C, D$ .
-  $A_{1} = A_{2}$  comes directly.
-  $B_{1} = B_{2}$  comes from changing the summation index  $i \to N - i$  and  $j \to N - j$ .
-  $C_1 = C_2$  comes from the expansion of summation.
-  $D_{1} = D_{2}$  comes from  $\mathbf{f}_0 = u_N\mathbf{e}_N$
$$
\mathcal {S} (H, u) = \mathcal {M} ^ {\intercal} (u) \mathcal {T} \left(H ^ {A}, v\right) \mathcal {M} (u),
$$
Remark 1. We can interpret  $\mathcal{M}$  as a basis transformation, where
$$
u _ {N} \mathbf {e} _ {N} = \mathbf {f} _ {0}, \quad u _ {i} \left(\mathbf {e} _ {i} - \mathbf {e} _ {i + 1}\right) = \mathbf {f} _ {N - i} - \mathbf {f} _ {N - i - 1} \quad i = 0, 1, \dots , N - 1. \tag {29}
$$
Remark 2. To clarify, the quantifier  $[\forall \nabla f(x_0),\dots ,\nabla f(x_N)]$  in (C1) means  $\nabla f(x_0),\ldots ,\nabla f(x_N)$  can be any arbitrary vectors in  $\mathbb{R}^d$ . This is different from  $[\nabla f(x_0),\dots ,\nabla f(x_N)$  be gradient of some  $f\colon \mathbb{R}^d\to \mathbb{R}$ . The same is true for (C2).
B Omitted calculation of Section 2
B.1 Calculation of  $H$  matrices
OGM and OGM-G. [26, Proposition 3] provides a recursive formula for (OGM) as follows:
$$
h _ {k + 1, i} = \left\{ \begin{array}{l l} \frac {\theta_ {k} - 1}{\theta_ {k + 1}} h _ {k, i} & i = 0, \dots , k - 2 \\ \frac {\theta_ {k} - 1}{\theta_ {k + 1}} (h _ {k, k - 1} - 1) & i = k - 1 \\ 1 + \frac {2 \theta_ {k} - 1}{\theta_ {k + 1}} & i = k. \end{array} \right. \tag {30}
$$
If  $k > i$
$$
h _ {k + 1, k - 1} = \frac {\theta_ {k} - 1}{\theta_ {k + 1}} \frac {2 \theta_ {k - 1} - 1}{\theta_ {k}},
$$
$$
h _ {k + 1, i} = \frac {\theta_ {k} - 1}{\theta_ {k + 1}} h _ {k, i} = \dots = \left(\prod_ {l = i + 2} ^ {k} \frac {\theta_ {l} - 1}{\theta_ {l + 1}}\right) h _ {i + 1, i - 1} = \left(\prod_ {l = i + 1} ^ {k} \frac {\theta_ {l} - 1}{\theta_ {l + 1}}\right) \frac {2 \theta_ {i} - 1}{\theta_ {i + 1}}.
$$
Thus  $H_{\mathrm{OGM}}$  can be calculated as
$$
H _ {\mathrm {O G M}} (k + 1, i + 1) = \left\{ \begin{array}{l l} 0 & i > k \\ 1 + \frac {2 \theta_ {k} - 1}{\theta_ {k + 1}} & i = k \\ \left(\prod_ {l = i + 1} ^ {k} \frac {\theta_ {l} - 1}{\theta_ {l + 1}}\right) \frac {2 \theta_ {i} - 1}{2 \theta_ {i + 1}} & i <   k. \end{array} \right.
$$
Recursive formula [29] of (OGM-G) is as following:
$$
h _ {k + 1, i} = \left\{ \begin{array}{l l} \frac {\theta_ {N - i - 1} - 1}{\theta_ {N - i}} h _ {k + 1, i + 1} & i = 0, \dots , k - 2 \\ \frac {\theta_ {N - k} - 1}{\theta_ {N - k + 1}} (h _ {k + 1, i} - 1) & i = k - 1 \\ 1 + \frac {2 \theta_ {N - k - 1} - 1}{\theta_ {N - k}} & i = k. \end{array} \right. \tag {31}
$$
$$
h _ {k + 1, k - 1} = \frac {\theta_ {N - k} - 1}{\theta_ {N - k + 1}} \frac {2 \theta_ {N - k - 1} - 1}{\theta_ {N - k}},
$$
$$
h _ {k + 1, i} = \frac {\theta_ {N - i - 1} - 1}{\theta_ {N - i}} h _ {k + 1, i + 1} = \dots = \left(\prod_ {l = N - k + 1} ^ {N - i - 1} \frac {\theta_ {l} - 1}{\theta_ {l + 1}}\right) h _ {k + 1, k - 1} = \left(\prod_ {l = N - k} ^ {N - i - 1} \frac {\theta_ {l} - 1}{\theta_ {l + 1}}\right) \frac {2 \theta_ {N - k - 1} - 1}{\theta_ {N - k}}.
$$
Thus  $H_{\mathrm{OGM - G}}$  can be calculated as
$$
H _ {\mathrm {O G M - G}} (k + 1, i + 1) = \left\{ \begin{array}{l l} 0 & i > k \\ 1 + \frac {2 \theta_ {N - k - 1} - 1}{\theta_ {N - k}} & i = k \\ \left(\prod_ {l = N - k} ^ {N - i - 1} \frac {\theta_ {l} - 1}{\theta_ {l + 1}}\right) \frac {2 \theta_ {N - k - 1} - 1}{\theta_ {N - k}} & i <   k \end{array} \right.,
$$
which gives  $H_{\mathrm{OGM - G}} = H_{\mathrm{OGM}}^A$ .
Gradient Descent. For (GD),  $H(i + 1, k + 1) = h_{i + 1,k} = h\delta_{i + 1,k + 1}$ , where  $\delta_{i,j}$  is the Kronecker delta. Therefore,  $H_{\mathrm{GD}} = H_{\mathrm{GD}}^A$ .
OBL- $\mathbf{F}_{\flat}$  and OBL- $\mathbf{G}_{\flat}$ . Recall  $\gamma = \sqrt{\frac{N(N + 1)}{2}}$ . We obtain the recursive formula of the  $H$  matrix of (OBL- $\mathbf{F}_{\flat}$ ).
$$
h _ {k + 1, i} = \left\{ \begin{array}{l l} \frac {k}{k + 3} h _ {k, i} & k = 0, \dots , N - 2, i = 0, \dots , k - 2 \\ 1 + \frac {2 k}{k + 3} & k = 0, \dots , N - 2, i = k \\ \frac {k}{k + 3} (h _ {k, k - 1} - 1) & k = 1, \dots , N - 2, i = k - 1 \\ 1 + \frac {N - 1}{\gamma + 1} & k = N - 1, i = N - 1 \\ \frac {N - 1}{2 (\gamma + 1)} (h _ {N - 1, N - 2} - 1) & k = N - 1, i = N - 2 \\ \frac {N - 1}{2 (\gamma + 1)} h _ {N - 1, i} & k = N - 1, i = 0, \dots , N - 3 \end{array} . \right. \tag {32}
$$
By using the above formula, we obtain
$$
H _ {\mathrm {O B L - F} _ {b}} (k + 1, i + 1) = \left\{ \begin{array}{l l} 1 + \frac {2 k}{k + 3} & \quad k = 0, \ldots , N - 2,   i = k \\ \frac {2 i (i + 1) (i + 2)}{(k + 1) (k + 2) (k + 3)} & \quad k = 0, \ldots , N - 2,   i = 0, \ldots , k - 1 \\ 1 + \frac {N - 1}{\gamma + 1} & \quad k = N - 1,   i = N - 1 \\ \frac {i (i + 1) (i + 2)}{(\gamma + 1) N (N + 1)} & \quad k = N - 1,   i = 0, \ldots , N - 2 \end{array} \right..
$$
Similarly, we achieve the following recursive formula of the  $H$  matrix of  $(\mathrm{OBL - G_b})$
$$
h _ {k + 1, i} = \left\{ \begin{array}{l l} \frac {N - k - 1}{N - k + 2} h _ {k, i} & k = 1, \dots , N - 1, i = 0, \dots , k - 2 \\ 1 + \frac {2 (N - k - 1)}{N - k + 2} & k = 1, \dots , N - 1, i = k \\ \frac {N - k - 1}{N - k + 2} \left(h _ {k, k - 1} - 1\right) & k = 1, \dots , N - 1, i = k - 1 \\ 1 + \frac {N - 1}{\gamma + 1} & k = 0, i = 0 \end{array} . \right. \tag {33}
$$
By using the above recursive formula, we obtain
$$
H _ {\mathrm {O B L - G} _ {b}} (k + 1, i + 1) = \left\{ \begin{array}{l l} 1 + \frac {N - 1}{\gamma + 1} & k = 0,   i = 0 \\ \frac {(N - k - 1) (N - k) (N - k + 1)}{(\gamma + 1) N (N + 1)} & k = 1, \ldots , N - 1,   i = 0 \\ 1 + \frac {2 (N - k - 1)}{N - k + 2} & k = 1, \ldots , N - 1,   i = k \\ \frac {2 (N - k - 1) (N - k) (N - k + 1)}{(N - i) (N - i + 1) (N - i + 2)} & k = 1, \ldots , N - 1,   i = 1, \ldots , k - 1 \end{array} \right..
$$
Thus  $H_{\mathrm{OBL - F_b}} = H_{\mathrm{OBL - G_b}}^A$
B.2 Calculation of energy functions
Calculation of  $\mathbf{U}$  and  $\mathbf{V}$  with  $H$  matrix In this paragraph, we calculate  $\mathbf{U}$  and  $\mathbf{V}$ . Recall (13) and (14).
First, we put  $x_{k + 1} - x_k = -\frac{1}{L}\sum_{i = 0}^k h_{k + 1,i}\nabla f(x_i)$  to (11). We have
$$
\begin{array}{l} \mathbf {U} = - \frac {1}{2 L} \left\| \sum_ {i = 0} ^ {N} (u _ {i} - u _ {i - 1}) \nabla f (x _ {i}) \right\| ^ {2} + \sum_ {i = 0} ^ {N} (u _ {i} - u _ {i - 1}) \left(\langle \nabla f (x _ {i}), x _ {0} - x _ {i} \rangle + \frac {1}{2 L} \| \nabla f (x _ {i}) \| ^ {2}\right) \\ + \sum_ {i = 0} ^ {N - 1} u _ {i} \left(\left\langle \nabla f \left(x _ {i + 1}\right), x _ {i} - x _ {i + 1} \right\rangle + \frac {1}{2 L} \| \nabla f (x _ {i}) - \nabla f (x _ {i + 1}) \| ^ {2}\right) \\ = - \frac {1}{2 L} \left\| \sum_ {i = 0} ^ {N} (u _ {i} - u _ {i - 1}) \nabla f (x _ {i}) \right\| ^ {2} + \sum_ {i = 0} ^ {N} \frac {u _ {i} - u _ {i - 1}}{L} \left(\left\langle \nabla f (x _ {i}), \sum_ {l = 0} ^ {i - 1} \sum_ {j = 0} ^ {l} h _ {l + 1, j} \nabla f (x _ {j}) \right\rangle + \frac {1}{2} \| \nabla f (x _ {i}) \| ^ {2}\right) \\ + \sum_ {i = 0} ^ {N - 1} \frac {u _ {i}}{L} \left(\left\langle \nabla f (x _ {i + 1}), \sum_ {j = 0} ^ {i} h _ {i + 1, j} \nabla f (x _ {j}) \right\rangle + \frac {1}{2} \| \nabla f (x _ {i}) - \nabla f (x _ {i + 1}) \| ^ {2}\right). \\ \end{array}
$$
Byarranging,weobtain
$$
\mathbf {U} = \sum_ {0 \leq j \leq i \leq N} \frac {s _ {i , j}}{L} \left\langle \nabla f (x _ {i}), \nabla f (x _ {j}) \right\rangle
$$
$$
s _ {i, j} = \left\{ \begin{array}{l l} - \frac {1}{2} \left(u _ {N} - u _ {N - 1}\right) ^ {2} + \frac {1}{2} u _ {N} & j = i, i = N \\ - \frac {1}{2} \left(u _ {i} - u _ {i - 1}\right) ^ {2} + u _ {i} & j = i, i = 0, \dots , N - 1 \\ u _ {i} h _ {i, i - 1} - u _ {i - 1} - \left(u _ {i} - u _ {i - 1}\right) \left(u _ {i - 1} - u _ {i - 2}\right) & j = i - 1 \\ \left(u _ {i} - u _ {i - 1}\right) \sum_ {l = j} ^ {i - 1} h _ {l + 1, j} + u _ {i - 1} h _ {i, j} - \left(u _ {i} - u _ {i - 1}\right) \left(u _ {j} - u _ {j - 1}\right) & j = 0, \dots , i - 2. \end{array} \right. \tag {34}
$$
Recall that we defined  $u_{-1} = 0$ . Next, put  $y_{k + 1} - y_k = -\frac{1}{L}\sum_{i = 0}^{k}h_{k + 1,i}\nabla f(y_i)$  to (12). We have
$$
\begin{array}{l} \mathbf {V} = \mathcal {V} _ {N} - \frac {1}{2 L} \| \nabla f (y _ {N}) \| ^ {2} \\ = \frac {v _ {0}}{2 L} \| \nabla f (y _ {N}) \| ^ {2} + \sum_ {i = 0} ^ {N - 1} v _ {i + 1} \left(\langle \nabla f (y _ {i + 1}), y _ {i} - y _ {i + 1} \rangle + \frac {1}{2 L} \| \nabla f (y _ {i}) - \nabla f (y _ {i + 1}) \| ^ {2}\right) \\ + \sum_ {i = 0} ^ {N - 1} \left(v _ {i + 1} - v _ {i}\right) \left(\left\langle \nabla f \left(y _ {i}\right), y _ {N} - y _ {i} \right\rangle + \frac {1}{2 L} \| \nabla f \left(y _ {i}\right) - \nabla f \left(y _ {N}\right) \| ^ {2}\right) - \frac {1}{2 L} \| \nabla f \left(y _ {N}\right) \| ^ {2} \\ = \frac {v _ {0} - 1}{2 L} \| \nabla f (y _ {N}) \| ^ {2} + \sum_ {i = 0} ^ {N - 1} v _ {i + 1} \left(\left\langle \nabla f (y _ {i + 1}), \frac {1}{L} \sum_ {j = 0} ^ {i} h _ {i + 1, j} \nabla f (y _ {j}) \right\rangle + \frac {1}{2 L} \| \nabla f (y _ {i}) - \nabla f (y _ {i + 1}) \| ^ {2}\right) \\ + \sum_ {i = 0} ^ {N - 1} (v _ {i + 1} - v _ {i}) \left(\left\langle \nabla f (y _ {i}), - \frac {1}{L} \sum_ {l = i} ^ {N - 1} \sum_ {j = 0} ^ {l} h _ {l + 1, j} \nabla f (y _ {j}) \right\rangle + \frac {1}{2 L} \| \nabla f (y _ {i}) - \nabla f (y _ {N}) \| ^ {2}\right). \\ \end{array}
$$
$$
\mathbf {V} = \sum_ {0 \leq j \leq i \leq N} \frac {t _ {i , j}}{L} \left\langle \nabla f (y _ {i}), \nabla f (y _ {j}) \right\rangle
$$
$$
t _ {i, j} = \left\{ \begin{array}{l l} \frac {v _ {1}}{2} + \frac {v _ {1} - v _ {0}}{2} - \left(v _ {1} - v _ {0}\right) \sum_ {l = 0} ^ {N - 1} h _ {l + 1, 0} & i = 0, j = i \\ \frac {v _ {i + 1} + v _ {i}}{2} + \frac {v _ {i + 1} - v _ {i}}{2} - \left(v _ {i + 1} - v _ {i}\right) \sum_ {l = i} ^ {N - 1} h _ {l + 1, i} & i = 1, \dots , N - 1, j = i \\ \frac {v _ {0} - 1}{2} + \frac {v _ {N}}{2} + \sum_ {i = 0} ^ {N - 1} \frac {v _ {i + 1} - v _ {i}}{2} & i = N, j = i \\ v _ {i} h _ {i, i - 1} - v _ {i} - \left(v _ {i + 1} - v _ {i}\right) \sum_ {l = i} ^ {N - 1} h _ {l + 1, i - 1} - \left(v _ {i} - v _ {i - 1}\right) \sum_ {l = i} ^ {N - 1} h _ {l + 1, i} & i = 1, \dots , N - 1, j = i - 1 \\ v _ {N} h _ {N, N - 1} - v _ {N} - \left(v _ {N} - v _ {N - 1}\right) & i = N, j = i - 1 \\ v _ {i} h _ {i, j} - \left(v _ {i + 1} - v _ {i}\right) \sum_ {l = i} ^ {N - 1} h _ {l + 1, j} - \left(v _ {j + 1} - v _ {j}\right) \sum_ {l = i} ^ {N - 1} h _ {l + 1, i} & i = 2, \dots , N - 1, j = 0, \dots , i - 2 \\ v _ {N} h _ {N, j} - \left(v _ {j + 1} - v _ {j}\right) & i = N, j = 0, \dots , N - 2 \end{array} . \right. \tag {35}
$$
Now we calculate  $\{s_{ij}\}$  and  $\{t_{ij}\}$  for  $[(OGM), (OGM-G)]$ $[(OBL-F_b), (OBL-G_b)]$  and  $[(GD), (GD)]$ .
B.2.1 Calculation of energy function of (OGM) and (OGM-G)
We will show  $s_{ij} = 0$  and  $t_{ij} = 0$  for all  $i,j$ . By the definition of  $\{u_i\}_{i = -1}^N$  and  $\{\theta_i\}_{i = -1}^N$ ,
$$
u _ {i} - u _ {i - 1} = 2 \theta_ {i} ^ {2} - 2 \theta_ {i - 1} ^ {2} = 2 \theta_ {i} \quad 0 \leq i \leq N - 1,
$$
$$
u _ {N} - u _ {N - 1} = \theta_ {N} ^ {2} - 2 \theta_ {N - 1} ^ {2} = \theta_ {N}.
$$
$$
s _ {i, i} = \left\{ \begin{array}{l l} - \frac {1}{2} (u _ {N} - u _ {N - 1}) ^ {2} + \frac {1}{2} u _ {N} = - \theta_ {N} + \theta_ {N} = 0 & j = i,   i = N \\ - \frac {1}{2} (u _ {i} - u _ {i - 1}) ^ {2} + u _ {i} = - \theta_ {i} + \theta_ {i} = 0 & j = i,   i = 0, \ldots , N - 1. \end{array} \right.
$$
Now we claim that  $s_{ij} = 0$  when  $j \neq i$ . In the case that  $i = j + 1$ , we have
$$
\begin{array}{l} s _ {i, i - 1} = u _ {i} h _ {i, i - 1} - u _ {i - 1} - \left(u _ {i} - u _ {i - 1}\right) \left(u _ {i - 1} - u _ {i - 2}\right) \\ = u _ {i} \left(\frac {2 \theta_ {i - 1} - 1}{\theta_ {i}} + 1\right) - u _ {i - 1} - (u _ {i} - u _ {i - 1}) (u _ {i - 1} - u _ {i - 2}) \\ = \left\{ \begin{array}{l} 2 \theta_ {i} ^ {2} \left(\frac {2 \theta_ {i - 1} - 1}{\theta_ {i}} + 1\right) - 2 \theta_ {i - 1} ^ {2} - 4 \theta_ {i} \theta_ {i - 1} \quad 0 \leq i \leq N - 1 \\ \theta_ {N} ^ {2} \left(\frac {2 \theta_ {N - 1} - 1}{\theta_ {N}} + 1\right) - 2 \theta_ {N - 1} ^ {2} - 2 \theta_ {N - 1} \theta_ {N} \quad i = N \end{array} \right. \\ = 0. \\ \end{array}
$$
We show  $s_{ij} = 0$  for  $j \neq i$  with induction on  $i$ , i.e., proving
$$
\left[ s _ {i, j} = \left(u _ {i} - u _ {i - 1}\right) \sum_ {l = j} ^ {i - 1} h _ {l + 1, j} + u _ {i - 1} h _ {i, j} - \left(u _ {i} - u _ {i - 1}\right) \left(u _ {j} - u _ {j - 1}\right) = 0, \quad j = 0, \dots , i - 2 \right]. \tag {36}
$$
First we prove (36) for  $i = j + 2$
$$
\begin{array}{l} (u _ {j + 2} - u _ {j + 1}) \left(h _ {j + 1, j} + h _ {j + 2, j}\right) + u _ {j + 1} h _ {j + 2, j} - \left(u _ {j + 2} - u _ {j + 1}\right) \left(u _ {j} - u _ {j - 1}\right) \\ = (u _ {j + 2} - u _ {j + 1}) h _ {j + 1, j} + u _ {j + 2} h _ {j + 2, j} - (u _ {j + 2} - u _ {j + 1}) (u _ {j} - u _ {j - 1}) \\ = \left\{ \begin{array}{l} 2 \theta_ {j + 2} h _ {j + 1, j} + 2 \theta_ {j + 2} ^ {2} h _ {j + 2, j} - 4 \theta_ {j + 2} \theta_ {j} \quad 0 \leq j \leq N - 3 \\ \theta_ {N} h _ {N - 1, N - 2} + \theta_ {N} ^ {2} h _ {N, N - 2} - 2 \theta_ {N} \theta_ {N - 2} \quad j = N - 2 \end{array} \right. \\ = 0. \\ \end{array}
$$
Next, assume (36) for  $i = i_0$ . When  $i = i_0 + 1$ ,
$$
\begin{array}{l} (u _ {i _ {0} + 1} - u _ {i _ {0}}) \sum_ {l = j} ^ {i _ {0}} h _ {l + 1, j} + u _ {i _ {0}} h _ {i _ {0} + 1, j} - (u _ {i _ {0} + 1} - u _ {i _ {0}}) (u _ {j} - u _ {j - 1}) \\ = \left(u _ {i _ {0} + 1} - u _ {i _ {0}}\right) \left(\sum_ {l = j} ^ {i _ {0} - 1} h _ {l + 1, j} + h _ {i _ {0} + 1, j}\right) + u _ {i _ {0}} h _ {i _ {0} + 1, j} - \left(u _ {i _ {0} + 1} - u _ {i _ {0}}\right) \left(u _ {j} - u _ {j - 1}\right) \\ = \left(u _ {i _ {0} + 1} - u _ {i _ {0}}\right) \left(\frac {\left(u _ {i _ {0}} - u _ {i _ {0} - 1}\right) \left(u _ {j} - u _ {j - 1}\right) - u _ {i _ {0} - 1} h _ {i _ {0}, j}}{u _ {i _ {0}} - u _ {i _ {0} - 1}} + h _ {i _ {0} + 1, j}\right) + u _ {i _ {0}} h _ {i _ {0} + 1, j} \\ - \left(u _ {i _ {0} + 1} - u _ {i _ {0}}\right) \left(u _ {j} - u _ {j - 1}\right) \\ = u _ {i _ {0} + 1} h _ {i _ {0} + 1, j} - \frac {u _ {i _ {0} - 1} \left(u _ {i _ {0} + 1} - u _ {i _ {0}}\right)}{u _ {i _ {0}} - u _ {i _ {0} - 1}} h _ {i _ {0}, j} \\ = \left\{ \begin{array}{l l} 2 \theta_ {i _ {0} + 1} ^ {2} h _ {i _ {0} + 1, j} - \frac {4 \theta_ {i _ {0} - 1} ^ {2} \theta_ {i _ {0} + 1}}{2 \theta_ {i _ {0}}} h _ {i _ {0}, j} & 0 \leq i _ {0} \leq N - 2 \\ \theta_ {N} ^ {2} h _ {i _ {0} + 1, j} - \frac {2 \theta_ {N - 2} ^ {2} \theta_ {N}}{2 \theta_ {N - 1}} h _ {i _ {0}, j} & i _ {0} = N - 1 \end{array} \right. \\ = 0 \\ \end{array}
$$
where the second equality comes from the induction hypothesis, and the third equality comes from (30). In sum, we proved  $s_{ij} = 0$  for every  $i$  and  $j$ , which implies  $\mathbf{U} = 0$ .
Next, we will claim that  $t_{ij} = 0$  for all  $i,j$ . Firstly, explicit formula of  $H_{\mathrm{OGM - G}}(k + 1,i + 1)$  first. When  $k > i$ ,
$$
\begin{array}{l} H _ {\mathrm {O G M - G}} (k + 1, i + 1) = \frac {\theta_ {N - k} - 1}{\theta_ {N - k + 1}} \frac {\theta_ {N - k + 1} - 1}{\theta_ {N - k + 2}} \dots \frac {\theta_ {N - i - 1} - 1}{\theta_ {N - i}} \frac {2 \theta_ {N - k - 1} - 1}{\theta_ {N - k}} \\ = \frac {\theta_ {N - k} ^ {2} - \theta_ {N - k}}{\theta_ {N - k} \theta_ {N - k + 1}} \frac {\theta_ {N - k + 1} ^ {2} - \theta_ {N - k + 1}}{\theta_ {N - k + 1} \theta_ {N - k + 2}} \dots \frac {\theta_ {N - i - 1} ^ {2} - \theta_ {N - i - 1}}{\theta_ {N - i - 1} \theta_ {N - i}} \frac {2 \theta_ {N - k - 1} - 1}{\theta_ {N - k}} \\ = \frac {\theta_ {N - k - 1} ^ {2}}{\theta_ {N - k} \theta_ {N - k + 1}} \frac {\theta_ {N - k} ^ {2}}{\theta_ {N - k + 1} \theta_ {N - k + 2}} \dots \frac {\theta_ {N - i - 2} ^ {2}}{\theta_ {N - i - 1} \theta_ {N - i}} \frac {2 \theta_ {N - k - 1} - 1}{\theta_ {N - k}} \\ = \frac {\theta_ {N - k - 1} ^ {2} (2 \theta_ {N - k - 1} - 1)}{\theta_ {N - i - 1} ^ {2} \theta_ {N - i}}. \\ \end{array}
$$
To calculate  $\{t_{i,j}\}$ , it is enough to deal with the sum  $\sum_{l = i}^{N - 1}h_{l + 1,j}$ , which can be expressed as
$$
\sum_ {l = i} ^ {N - 1} h _ {l + 1, j} = \left\{ \begin{array}{l l} \frac {\theta_ {N} + 1}{2} & i = 0, j = i \\ \theta_ {N - i} & i = 1, \dots , N - 1, j = i \\ \frac {\theta_ {N - i - 1} ^ {4}}{\theta_ {N - j} \theta_ {N - j - 1} ^ {2}} & i = 1, \dots , N - 1, j = 0, \dots , i - 1 \end{array} . \right. \tag {37}
$$
By inserting (37) in (35),  $[t_{ij} = 0, \forall i,j]$  is obtained, which implies  $\mathbf{V} = 0$ . (37) and (35) are also stated in [29, Lemma 6.1].
B.2.2 Calculation of energy function of  $(\mathrm{OBL - F_b})$  and  $(\mathrm{OBL - G_b})$
First we calculate  $\{s_{ij}\}$  for  $(\mathrm{OBL - F_b})$ . Recall  $u_{i} = \frac{(i + 1)(i + 2)}{2}$  for  $0\leq i\leq N - 1$  and  $u_{N} = \gamma^{2} + \gamma$  where  $\gamma = \sqrt{N(N + 1) / 2}$ . When  $j = i$ ,
$$
\begin{array}{l} s _ {i, i} = \left\{ \begin{array}{l l} - \frac {1}{2} \left(u _ {N} - u _ {N - 1}\right) ^ {2} + \frac {1}{2} u _ {N} & i = N \\ - \frac {1}{2} \left(u _ {i} - u _ {i - 1}\right) ^ {2} + u _ {i} & 0 \leq i \leq N - 1 \end{array} \right. \\ = \left\{ \begin{array}{l} \frac {\gamma}{2} = \frac {u _ {N} - u _ {N - 1}}{2} \quad i = N \\ - \frac {1}{2} (i + 1) ^ {2} + \frac {(i + 1) (i + 2)}{2} = \frac {u _ {i} - u _ {i - 1}}{2} \quad 0 \leq i \leq N - 1 \end{array} \right. \\ \end{array}
$$
Now we claim that  $s_{ij} = 0$  when  $j \neq i$ . In the case  $j = i - 1$ , we have
$$
\begin{array}{l} s _ {i, i - 1} = u _ {i} h _ {i, i - 1} - u _ {i - 1} - \left(u _ {i} - u _ {i - 1}\right) \left(u _ {i - 1} - u _ {i - 2}\right) \\ = \left\{ \begin{array}{l l} \frac {(i + 1) (i + 2)}{2} h _ {i, i - 1} - \frac {i (i + 1)}{2} - (i + 1) i & 0 \leq i \leq N - 1 \\ (\gamma^ {2} + \gamma) h _ {N, N - 1} - \frac {N (N + 1)}{2} - \gamma N & i = N \end{array} \right. \\ = 0. \\ \end{array}
$$
$$
\left[ s _ {i, j} = \left(u _ {i} - u _ {i - 1}\right) \sum_ {l = j} ^ {i - 1} h _ {l + 1, j} + u _ {i - 1} h _ {i, j} - \left(u _ {i} - u _ {i - 1}\right) \left(u _ {j} - u _ {j - 1}\right) = 0 \quad j = 0, \dots , i - 2 \right]. \tag {38}
$$
(38) holds when  $i = j + 2$  since
$$
\begin{array}{l} \left(u _ {j + 2} - u _ {j + 1}\right) \left(h _ {j + 1, j} + h _ {j + 2, j}\right) + u _ {j + 1} h _ {j + 2, j} - \left(u _ {j + 2} - u _ {j + 1}\right) \left(u _ {j} - u _ {j - 1}\right) \\ = (u _ {j + 2} - u _ {j + 1}) h _ {j + 1, j} + u _ {j + 2} h _ {j + 2, j} - (u _ {j + 2} - u _ {j + 1}) (u _ {j} - u _ {j - 1}) \\ = \left\{ \begin{array}{l} (j + 3) h _ {j + 1, j} + \frac {(j + 3) (j + 4)}{2} h _ {j + 2, j} - (j + 3) (j + 1) \quad 0 \leq j \leq N - 3 \\ \gamma h _ {N - 1, N - 2} + (\gamma^ {2} + \gamma) h _ {N, N - 2} - \gamma \quad j = N - 2 \end{array} \right. \\ = 0. \\ \end{array}
$$
Assume (38) for  $i = i_0$ . For  $i = i_0 + 1$ ,