Algorithm 1: Intersection object filtering algorithm

$\Omega$: the number of objects in the database $\mathbb{Q}$ D: the array of subscripts that need to be deleted in the database $\mathbb{Q}$

1: $\epsilon =0$;

2: for $\sigma$ = 0; $\sigma$ < $\Omega$; $\sigma$++ do 3:  for $\tau$ = $\sigma$ + 1; $\tau$< $\Omega$; $\tau$++ do

4:   Calculate the intersection $V_{\cap }$ between object ${\mathsf{\phi }}_{\sigma }$ and ${\mathsf{\phi }}_{\tau }$;

5:   ${\eta }_{\sigma }=V_{\cap }/V_{\sigma }$;

6:   ${\eta }_{\tau }=V_{\cap }/V_{\tau }$;

7:  end for 8:  $\xi =\underset{\tau }{\arg \max }({\eta }_{\sigma +1},\cdots ,{\eta }_{\tau },\cdots ,{\eta }_{\Omega })$;

9:  if $({\eta }_{\sigma }>{\eta }_{\xi })\cap ({\eta }_{\sigma }>{\mathsf{\Gamma }}_{V_{\cap }})$ then

10:   $\mathrm{D}(\epsilon )=\sigma$; 11:   $\epsilon ++$; 12:  elseif $({\eta }_{\sigma }\le {\eta }_{\xi })\cap ({\eta }_{\xi }>{\mathsf{\Gamma }}_{V_{\cap }})$ then 13:   $\mathrm{D}(\epsilon )=\xi$; 14:   $\epsilon ++$;

15:  end if

16: end for 17: $\epsilon =0$;

18: for $\sigma$ = 0; $\sigma$ < $\Omega$; $\sigma$++ do 19:  if $\mathrm{D}(\epsilon )=\sigma$ then 20:   Delete the object ${\mathsf{\phi }}_{\sigma }$ in the database $\mathbb{Q}$; 21:  end if

22: end for