VEB删除的伪代码情况分析

    if  V.min == V.max
         V.min = NIL;
         V.max = NIL;
    elseif  V.u == 2
         if(x == 0)
            V.min = 1;
         else
            V.min = 0;
         V.max = V.min;
    elseif  x == V.min
         first-cluster = vEB-TREE-MINIMUM(V.summary)
         x = index( first-cluster, vEB-TREE-MINIMUM(V.cluster[first-cluster]) )
         V.min = x;
         vEB-TREE-DELETE(V.cluster[high(x),low(x) )
         if vEB-TREE-MINIMUM(V.cluster[high(x)]) == NIL
             vEB-TREE-DELETE(V.summary, high(x));
              if x == V.max
                  summary-max = vEB-TREE-MAXMUM(V.summary)
                  if summary-max == NIL
                     V.max = V.min;
                  else
                     V.max = index(summary-max, vEB-TREE-MAXMUM(V.cluster[summary-max])
         elseif x == V.max
             V.max = index(high(x), vEB-TREE-MAXMUM(V.cluster[high(x)] )

当簇中元素包含两个及两个以上，且V.u＞＝4的情况下，只考虑 x == V.min，那x≠V.min的情况怎么处理？ 难道不应该另写分支程序吗？
还有在已知被替换的x所在簇为空的情况下，当x==V.max时，是不是意味着V中其实只包含两个元素？