Table 1.

The detailed steps to find solutions for SA−U4 with $T=20$.

${\mathit{q}}^{\prime }$	${\mathit{L}}_{{\mathit{q}}^{\prime }}$	$\mathfrak{C}\left({\mathbb{D}}_{{\mathit{v}}_{{\mathit{qq}}^{\prime }}}\right)$
$q^{\prime }=2$	$min\left[\mathfrak{C}\left({\mathbb{D}}_{v_{12}}\right)\right]=-4+L_2=7$	$\mathfrak{C}\left({\mathbb{D}}_{v_{12}}\right)=[7,23]$
	→$L_2=11$
$q^{\prime }=3$	$min\left[\mathfrak{C}\left({\mathbb{D}}_{v_{23}}\right)\right]=-6+L_3-L_2=24$	$\mathfrak{C}\left({\mathbb{D}}_{v_{23}}\right)=[24,40]$
	→$L_3=41,L_3-L_2=30$
	→$L_3-L_1=41$	$\mathfrak{C}\left({\mathbb{D}}_{v_{13}}\right)=[37,57]$
$q^{\prime }=4$	$min\left[\mathfrak{C}\left({\mathbb{D}}_{v_{43}}\right)\right]=-4+L_4-L_3=58$	$\mathfrak{C}\left({\mathbb{D}}_{v_{34}}\right)=[58,70]$
	→$L_4=103,L_4-L_3=62$
	→$L_4-L_2=92$	$\mathfrak{C}\left({\mathbb{D}}_{v_{24}}\right)=[88,104]$
Update	$L_3=41-(88-70-1)=24$	$\mathfrak{C}\left({\mathbb{D}}_{v_{23}}\right)=[8,24]$
	$L_4=103-2(88-70-1)=69$	$\mathfrak{C}\left({\mathbb{D}}_{v_{13}}\right)=[20,40]$
	→$L_4-L_3=45$	$\mathfrak{C}\left({\mathbb{D}}_{v_{34}}\right)=[41,53]$
$q^{\prime }=4$	→$L_4-L_2=58$	$\mathfrak{C}\left({\mathbb{D}}_{v_{24}}\right)=[54,70]$
	→$L_4-L_1=69$	$\mathfrak{C}\left({\mathbb{D}}_{v_{14}}\right)=[65,89]$