二次规划（QP）样条路径优化
原地址：https://daobook.github.io/apollo/docs/specs/qp_spline_path_optimizer_cn.html
二次规划（QP）+样条插值
1. 目标函数
1.1 获得路径长度
路径定义在station-lateral坐标系中。s的变化区间为从车辆当前位置点到默认路径的长度。
1.2 获得样条段
将路径划分为n段，每段路径用一个多项式来表示。
1.3 定义样条段函数
每个样条段 i 都有沿着参考线的累加距离 d i d_i di。每段的路径默认用5介多项式表示。
l = f i ( s ) = a i 0 + a i 1 ⋅ s + a i 2 ⋅ s 2 + a i 3 ⋅ s 3 + a i 4 ⋅ s 4 + a i 5 ⋅ s 5 ( 0 ≤ s ≤ d i ) l = f_i(s) = a_{i0} + a_{i1} \cdot s + a_{i2} \cdot s^2 + a_{i3} \cdot s^3 + a_{i4} \cdot s^4 + a_{i5} \cdot s^5 (0 \leq s \leq d_{i}) l=fi(s)=ai0+ai1⋅s+ai2⋅s2+ai3⋅s3+ai4⋅s4+ai5⋅s5(0≤s≤di)
1.4 定义每个样条段优化目标函数
c o s t = ∑ i = 1 n ( w 1 ⋅ ∫ 0 d i ( f i ′ ) 2 ( s ) d s + w 2 ⋅ ∫ 0 d i ( f i ′ ′ ) 2 ( s ) d s + w 3 ⋅ ∫ 0 d i ( f i ′ ′ ′ ) 2 ( s ) d s ) cost = \sum_{i=1}^{n} \Big( w_1 \cdot \int\limits_{0}^{d_i} (f_i')^2(s) ds + w_2 \cdot \int\limits_{0}^{d_i} (f_i'')^2(s) ds + w_3 \cdot \int\limits_{0}^{d_i} (f_i^{\prime\prime\prime})^2(s) ds \Big) cost=i=1∑n(w1⋅0∫di(fi′)2(s)ds+w2⋅0∫di(fi′′)2(s)ds+w3⋅0∫di(fi′′′)2(s)ds)
1.5 将开销（cost）函数转换为QP公式
QP公式:
m i n i m i z e 1 2 ⋅ x T ⋅ H ⋅ x + f T ⋅ x s . t . L B ≤ x ≤ U B A e q x = b e q A x ≥ b \begin{aligned} minimize & \frac{1}{2} \cdot x^T \cdot H \cdot x + f^T \cdot x \\ s.t. \qquad & LB \leq x \leq UB \\ & A_{eq}x = b_{eq} \\ & Ax \geq b \end{aligned} minimizes.t.21⋅xT⋅H⋅x+fT⋅xLB≤x≤UBAeqx=beqAx≥b
下面是将开销（cost）函数转换为QP公式的例子：
f i ( s ) ＝ ∣ 1 s s 2 s 3 s 4 s 5 ∣ ⋅ ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ f_i(s) ＝ \begin{vmatrix} 1 & s & s^2 & s^3 & s^4 & s^5 \end{vmatrix} \cdot \begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix} fi(s)＝
1ss2s3s4s5
⋅
ai0ai1ai2ai3ai4ai5
且
f i ′ ( s ) = ∣ 0 1 2 s 3 s 2 4 s 3 5 s 4 ∣ ⋅ ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ f_i'(s) = \begin{vmatrix} 0 & 1 & 2s & 3s^2 & 4s^3 & 5s^4 \end{vmatrix} \cdot \begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix} fi′(s)=
012s3s24s35s4
⋅
ai0ai1ai2ai3ai4ai5
且
f i ′ ( s ) 2 = ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ ⋅ ∣ 0 1 2 s 3 s 2 4 s 3 5 s 4 ∣ ⋅ ∣ 0 1 2 s 3 s 2 4 s 3 5 s 4 ∣ ⋅ ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ f_i'(s)^2 = \begin{vmatrix} a_{i0} & a_{i1} & a_{i2} & a_{i3} & a_{i4} & a_{i5} \end{vmatrix} \cdot \begin{vmatrix} 0 \\ 1 \\ 2s \\ 3s^2 \\ 4s^3 \\ 5s^4 \end{vmatrix} \cdot \begin{vmatrix} 0 & 1 & 2s & 3s^2 & 4s^3 & 5s^4 \end{vmatrix} \cdot \begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix} fi′(s)2=
ai0ai1ai2ai3ai4ai5
⋅
012s3s24s35s4
⋅
012s3s24s35s4
⋅
ai0ai1ai2ai3ai4ai5
然后得到，
∫ 0 d i f i ′ ( s ) 2 d s ＝ ∫ 0 d i ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ ⋅ ∣ 0 1 2 s 3 s 2 4 s 3 5 s 4 ∣ ⋅ ∣ 0 1 2 s 3 s 2 4 s 3 5 s 4 ∣ ⋅ ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ d s \int\limits_{0}^{d_i} f_i'(s)^2 ds ＝ \int\limits_{0}^{d_i} \begin{vmatrix} a_{i0} & a_{i1} & a_{i2} & a_{i3} & a_{i4} & a_{i5} \end{vmatrix} \cdot \begin{vmatrix} 0 \\ 1 \\ 2s \\ 3s^2 \\ 4s^3 \\ 5s^4 \end{vmatrix} \cdot \begin{vmatrix} 0 & 1 & 2s & 3s^2 & 4s^3 & 5s^4 \end{vmatrix} \cdot \begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix} ds 0∫difi′(s)2ds＝0∫di
ai0ai1ai2ai3ai4ai5
⋅
012s3s24s35s4
⋅
012s3s24s35s4
⋅
ai0ai1ai2ai3ai4ai5
ds
从聚合函数中提取出常量得到，
∫ 0 d i f ′ ( s ) 2 d s ＝ ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ ⋅ ∫ 0 d i ∣ 0 1 2 s 3 s 2 4 s 3 5 s 4 ∣ ⋅ ∣ 0 1 2 s 3 s 2 4 s 3 5 s 4 ∣ d s ⋅ ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ \int\limits_{0}^{d_i} f'(s)^2 ds ＝ \begin{vmatrix} a_{i0} & a_{i1} & a_{i2} & a_{i3} & a_{i4} & a_{i5} \end{vmatrix} \cdot \int\limits_{0}^{d_i} \begin{vmatrix} 0 \\ 1 \\ 2s \\ 3s^2 \\ 4s^3 \\ 5s^4 \end{vmatrix} \cdot \begin{vmatrix} 0 & 1 & 2s & 3s^2 & 4s^3 & 5s^4 \end{vmatrix} ds \cdot \begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix} 0∫dif′(s)2ds＝
ai0ai1ai2ai3ai4ai5
⋅0∫di
012s3s24s35s4
⋅
012s3s24s35s4
ds⋅
ai0ai1ai2ai3ai4ai5
＝ ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ ⋅ ∫ 0 d i ∣ 0 0 0 0 0 0 0 1 2 s 3 s 2 4 s 3 5 s 4 0 2 s 4 s 2 6 s 3 8 s 4 10 s 5 0 3 s 2 6 s 3 9 s 4 12 s 5 15 s 6 0 4 s 3 8 s 4 12 s 5 16 s 6 20 s 7 0 5 s 4 10 s 5 15 s 6 20 s 7 25 s 8 ∣ d s ⋅ ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ ＝\begin{vmatrix} a_{i0} & a_{i1} & a_{i2} & a_{i3} & a_{i4} & a_{i5} \end{vmatrix} \cdot \int\limits_{0}^{d_i} \begin{vmatrix} 0 & 0 &0&0&0&0\\ 0 & 1 & 2s & 3s^2 & 4s^3 & 5s^4\\ 0 & 2s & 4s^2 & 6s^3 & 8s^4 & 10s^5\\ 0 & 3s^2 & 6s^3 & 9s^4 & 12s^5&15s^6 \\ 0 & 4s^3 & 8s^4 &12s^5 &16s^6&20s^7 \\ 0 & 5s^4 & 10s^5 & 15s^6 & 20s^7 & 25s^8 \end{vmatrix} ds \cdot \begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix} ＝ ai0ai1ai2ai3ai4ai5 ⋅0∫di 000000012s3s24s35s402s4s26s38s410s503s26s39s412s515s604s38s412s516s620s705s410s515s620s725s8 ds⋅ ai0ai1ai2ai3ai4ai5
最后得到，
∫ 0 d i f i ′ ( s ) 2 d s = ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ ⋅ ∣ 0 0 0 0 0 0 0 d i d i 2 d i 3 d i 4 d i 5 0 d i 2 4 3 d i 3 6 4 d i 4 8 5 d i 5 10 6 d i 6 0 d i 3 6 4 d i 4 9 5 d i 5 12 6 d i 6 15 7 d i 7 0 d i 4 8 5 d i 5 12 6 d i 6 16 7 d i 7 20 8 d i 8 0 d i 5 10 6 d i 6 15 7 d i 7 20 8 d i 8 25 9 d i 9 ∣ ⋅ ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ \int\limits_{0}^{d_i} f'_i(s)^2 ds =\begin{vmatrix} a_{i0} & a_{i1} & a_{i2} & a_{i3} & a_{i4} & a_{i5} \end{vmatrix} \cdot \begin{vmatrix} 0 & 0 & 0 & 0 &0&0\\ 0 & d_i & d_i^2 & d_i^3 & d_i^4&d_i^5\\ 0& d_i^2 & \frac{4}{3}d_i^3& \frac{6}{4}d_i^4 & \frac{8}{5}d_i^5&\frac{10}{6}d_i^6\\ 0& d_i^3 & \frac{6}{4}d_i^4 & \frac{9}{5}d_i^5 & \frac{12}{6}d_i^6&\frac{15}{7}d_i^7\\ 0& d_i^4 & \frac{8}{5}d_i^5 & \frac{12}{6}d_i^6 & \frac{16}{7}d_i^7&\frac{20}{8}d_i^8\\ 0& d_i^5 & \frac{10}{6}d_i^6 & \frac{15}{7}d_i^7 & \frac{20}{8}d_i^8&\frac{25}{9}d_i^9 \end{vmatrix} \cdot \begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix} 0∫difi′(s)2ds= ai0ai1ai2ai3ai4ai5 ⋅ 0000000didi2di3di4di50di234di346di458di5610di60di346di459di5612di6715di70di458di5612di6716di7820di80di5610di6715di7820di8925di9 ⋅ ai0ai1ai2ai3ai4ai5
请注意我们最后得到一个6介的矩阵来表示5介样条插值的衍生开销。
应用同样的推理方法可以得到2介，3介样条插值的衍生开销。
2 约束条件
2.1 初始点约束
假设第一个点为 ( s 0 s_0 s0, l 0 l_0 l0), ( s 0 s_0 s0, l 0 ′ l'_0 l0′) and ( s 0 s_0 s0, l 0 ′ ′ l''_0 l0′′)，其中 l 0 l_0 l0 , l 0 ′ l'_0 l0′ and l 0 ′ ′ l''_0 l0′′表示横向的偏移，并且规划路径的起始点的第一，第二个点的衍生开销可以从 f i ( s ) f_i(s) fi(s), f i ′ ( s ) f'_i(s) fi′(s), f i ( s ) ′ ′ f_i(s)'' fi(s)′′计算得到。
将上述约束转换为QP约束等式，使用等式：
A e q x = b e q A_{eq}x = b_{eq} Aeqx=beq
下面是转换的具体步骤：
f i ( s 0 ) = ∣ 1 s 0 s 0 2 s 0 3 s 0 4 s 0 5 ∣ ⋅ ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ = l 0 f_i(s_0) = \begin{vmatrix} 1 & s_0 & s_0^2 & s_0^3 & s_0^4&s_0^5 \end{vmatrix} \cdot \begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5}\end{vmatrix} = l_0 fi(s0)=
1s0s02s03s04s05
⋅
ai0ai1ai2ai3ai4ai5
=l0
且
f i ′ ( s 0 ) = ∣ 0 1 2 s 0 3 s 0 2 4 s 0 3 5 s 0 4 ∣ ⋅ ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ = l 0 ′ f'_i(s_0) = \begin{vmatrix} 0& 1 & 2s_0 & 3s_0^2 & 4s_0^3 &5 s_0^4 \end{vmatrix} \cdot \begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix} = l'_0 fi′(s0)=
012s03s024s035s04
⋅
ai0ai1ai2ai3ai4ai5
=l0′
且
f i ′ ′ ( s 0 ) = ∣ 0 0 2 3 × 2 s 0 4 × 3 s 0 2 5 × 4 s 0 3 ∣ ⋅ ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ = l 0 ′ ′ f''_i(s_0) = \begin{vmatrix} 0&0& 2 & 3\times2s_0 & 4\times3s_0^2 & 5\times4s_0^3 \end{vmatrix} \cdot \begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix} = l''_0 fi′′(s0)=
0023×2s04×3s025×4s03
⋅
ai0ai1ai2ai3ai4ai5
=l0′′
其中，i是包含 s 0 s_0 s0的样条段的索引值。
2.2 终点约束
和起始点相同，终点 ( s e , l e ) (s_e, l_e) (se,le) 也应当按照起始点的计算方法生成约束条件。
将起始点和终点组合在一起，得出约束等式为：
∣ 1 s 0 s 0 2 s 0 3 s 0 4 s 0 5 0 1 2 s 0 3 s 0 2 4 s 0 3 5 s 0 4 0 0 2 3 × 2 s 0 4 × 3 s 0 2 5 × 4 s 0 3 1 s e s e 2 s e 3 s e 4 s e 5 0 1 2 s e 3 s e 2 4 s e 3 5 s e 4 0 0 2 3 × 2 s e 4 × 3 s e 2 5 × 4 s e 3 ∣ ⋅ ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ = ∣ l 0 l 0 ′ l 0 ′ ′ l e l e ′ l e ′ ′ ∣ \begin{vmatrix} 1 & s_0 & s_0^2 & s_0^3 & s_0^4&s_0^5 \\ 0&1 & 2s_0 & 3s_0^2 & 4s_0^3 & 5s_0^4 \\ 0& 0&2 & 3\times2s_0 & 4\times3s_0^2 & 5\times4s_0^3 \\ 1 & s_e & s_e^2 & s_e^3 & s_e^4&s_e^5 \\ 0&1 & 2s_e & 3s_e^2 & 4s_e^3 & 5s_e^4 \\ 0& 0&2 & 3\times2s_e & 4\times3s_e^2 & 5\times4s_e^3 \end{vmatrix} \cdot \begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix} = \begin{vmatrix} l_0\\ l'_0\\ l''_0\\ l_e\\ l'_e\\ l''_e\\ \end{vmatrix} 100100s010se10s022s02se22se2s033s023×2s0se33se23×2ses044s034×3s02se44se34×3se2s055s045×4s03se55se45×4se3 ⋅ ai0ai1ai2ai3ai4ai5 = l0l0′l0′′lele′le′′
2.3 平滑节点约束
该约束的目的是使样条的节点更加平滑。假设两个段 s e g k seg_k segk 和 s e g k + 1 seg_{k+1} segk+1互相连接，且 s e g k seg_k segk的累计值s为 s k s_k sk。计算约束的等式为：
f k ( s k ) = f k + 1 ( s 0 ) f_k(s_k) = f_{k+1} (s_0) fk(sk)=fk+1(s0)
下面是计算的具体步骤：
∣ 1 s k s k 2 s k 3 s k 4 s k 5 ∣ ⋅ ∣ a k 0 a k 1 a k 2 a k 3 a k 4 a k 5 ∣ = ∣ 1 s 0 s 0 2 s 0 3 s 0 4 s 0 5 ∣ ⋅ ∣ a k + 1 , 0 a k + 1 , 1 a k + 1 , 2 a k + 1 , 3 a k + 1 , 4 a k + 1 , 5 ∣ \begin{vmatrix} 1 & s_k & s_k^2 & s_k^3 & s_k^4&s_k^5 \\ \end{vmatrix} \cdot \begin{vmatrix} a_{k0} \\ a_{k1} \\ a_{k2} \\ a_{k3} \\ a_{k4} \\ a_{k5} \end{vmatrix} = \begin{vmatrix} 1 & s_{0} & s_{0}^2 & s_{0}^3 & s_{0}^4&s_{0}^5 \\ \end{vmatrix} \cdot \begin{vmatrix} a_{k+1,0} \\ a_{k+1,1} \\ a_{k+1,2} \\ a_{k+1,3} \\ a_{k+1,4} \\ a_{k+1,5} \end{vmatrix}
1sksk2sk3sk4sk5
⋅
ak0ak1ak2ak3ak4ak5
=
1s0s02s03s04s05
⋅
ak+1,0ak+1,1ak+1,2ak+1,3ak+1,4ak+1,5
然后
∣ 1 s k s k 2 s k 3 s k 4 s k 5 − 1 − s 0 − s 0 2 − s 0 3 − s 0 4 − s 0 5 ∣ ⋅ ∣ a k 0 a k 1 a k 2 a k 3 a k 4 a k 5 a k + 1 , 0 a k + 1 , 1 a k + 1 , 2 a k + 1 , 3 a k + 1 , 4 a k + 1 , 5 ∣ = 0 \begin{vmatrix} 1 & s_k & s_k^2 & s_k^3 & s_k^4&s_k^5 & -1 & -s_{0} & -s_{0}^2 & -s_{0}^3 & -s_{0}^4&-s_{0}^5\\ \end{vmatrix} \cdot \begin{vmatrix} a_{k0} \\ a_{k1} \\ a_{k2} \\ a_{k3} \\ a_{k4} \\ a_{k5} \\ a_{k+1,0} \\ a_{k+1,1} \\ a_{k+1,2} \\ a_{k+1,3} \\ a_{k+1,4} \\ a_{k+1,5} \end{vmatrix} = 0
1sksk2sk3sk4sk5−1−s0−s02−s03−s04−s05
⋅
ak0ak1ak2ak3ak4ak5ak+1,0ak+1,1ak+1,2ak+1,3ak+1,4ak+1,5
=0
将 s 0 s_0 s0 = 0代入等式。
同样地，可以为下述等式计算约束等式：
f k ′ ( s k ) = f k + 1 ′ ( s 0 ) f k ′ ′ ( s k ) = f k + 1 ′ ′ ( s 0 ) f k ′ ′ ′ ( s k ) = f k + 1 ′ ′ ′ ( s 0 ) f'_k(s_k) = f'_{k+1} (s_0) \\ f''_k(s_k) = f''_{k+1} (s_0) \\ f'''_k(s_k) = f'''_{k+1} (s_0) fk′(sk)=fk+1′(s0)fk′′(sk)=fk+1′′(s0)fk′′′(sk)=fk+1′′′(s0)
2.4 点采样边界约束
在路径上均匀的取样m个点，检查这些点上的障碍物边界。将这些约束转换为QP约束不等式，使用不等式：
A x ≥ b Ax \geq b Ax≥b
首先基于道路宽度和周围的障碍物找到点 ( s j , l j ) (s_j, l_j) (sj,lj)的下边界 l l b , j l_{lb,j} llb,j，且 j ∈ [ 0 , m ] j\in[0, m] j∈[0,m]。计算约束的不等式为：
∣ 1 s 0 s 0 2 s 0 3 s 0 4 s 0 5 1 s 1 s 1 2 s 1 3 s 1 4 s 1 5 . . . . . . . . . . . . . . . . . . 1 s m s m 2 s m 3 s m 4 s m 5 ∣ ⋅ ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ ≥ ∣ l l b , 0 l l b , 1 . . . l l b , m ∣ \begin{vmatrix} 1 & s_0 & s_0^2 & s_0^3 & s_0^4&s_0^5 \\ 1 & s_1 & s_1^2 & s_1^3 & s_1^4&s_1^5 \\ ...&...&...&...&...&... \\ 1 & s_m & s_m^2 & s_m^3 & s_m^4&s_m^5 \\ \end{vmatrix} \cdot \begin{vmatrix}a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix} \geq \begin{vmatrix} l_{lb,0}\\ l_{lb,1}\\ ...\\ l_{lb,m}\\ \end{vmatrix} 11...1s0s1...sms02s12...sm2s03s13...sm3s04s14...sm4s05s15...sm5 ⋅ ai0ai1ai2ai3ai4ai5 ≥ llb,0llb,1...llb,m
同样地，对上边界 l u b , j l_{ub,j} lub,j，计算约束的不等式为：
∣ − 1 − s 0 − s 0 2 − s 0 3 − s 0 4 − s 0 5 − 1 − s 1 − s 1 2 − s 1 3 − s 1 4 − s 1 5 . . . . . . − . . . . . . . . . . . . − 1 − s m − s m 2 − s m 3 − s m 4 − s m 5 ∣ ⋅ ∣ a i 0 a i 1 a i 2 a i 3 a i 4 a i 5 ∣ ≥ − 1 ⋅ ∣ l u b , 0 l u b , 1 . . . l u b , m ∣ \begin{vmatrix} -1 & -s_0 & -s_0^2 & -s_0^3 & -s_0^4&-s_0^5 \\ -1 & -s_1 & -s_1^2 & -s_1^3 & -s_1^4&-s_1^5 \\ ...&...-&...&...&...&... \\ -1 & -s_m & -s_m^2 & -s_m^3 & -s_m^4&-s_m^5 \\ \end{vmatrix} \cdot \begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix} \geq -1 \cdot \begin{vmatrix} l_{ub,0}\\ l_{ub,1}\\ ...\\ l_{ub,m}\\ \end{vmatrix}
−1−1...−1−s0−s1...−−sm−s02−s12...−sm2−s03−s13...−sm3−s04−s14...−sm4−s05−s15...−sm5
⋅
ai0ai1ai2ai3ai4ai5
≥−1⋅
lub,0lub,1...lub,m