The same physical order in both rows
Abstract
High X first and an interior ear for every a>s
Put q=2^s,h=2^(a-s),e=2^(a-s-1),r=2^(s-1), so 2e=h and 2r=q. Split X’s positive local indices into [0,a-s-1) and [a-s-1,a-1). The low interval has mass e-1 and the high interval has mass e(q-1). In particular the low interval is empty at a=s+1; no positive label is invented there. The terminal order is X,U,V,W. Use high X before X-, then native U, then native V, then native W followed by low X and W-. The original natural word is range’(a+2s-1,s+1) ++ range(3s-1) ++ range’(3s,a-s-1) ++ [3s-1]. The first argument of range’ is a start and its second is a length. Its complete physical inventory follows by rearranging consecutive intervals. There are exactly 3s entries before V-, so inserting k immediately before that terminal is the numeric cut 3s. The actual fresh value is A. The uniform strict endpoint inequalities bound both actual rows throughout these same positive bags.