@@ -139,9 +139,9 @@ it? Let's keep working with our little test subject `a`.
140The most basic way to access the subject is *tree geometry*.
140The most basic way to access the subject is *tree geometry*.
141Every noun is a binary tree; every noun in the tree has a number
141Every noun is a binary tree; every noun in the tree has a number
142−which is its address. The root of the tree is `1`; if the
142+which is its address. The root of the tree is `1`. If the
143−address of a cell is `n`, its head is `2n` and its tail is
143+address of a cell is `n`, the address of its head is `2n` and the
144−`2n+1`.
144+address of its tail is `2n+1`.
146A picture might help:
146A picture might help:
@@ -170,7 +170,15 @@ The syntax for a fragment of the subject at a tree address, or
170[.3.14 ~m45]
170[.3.14 ~m45]
173−Of course, the fragment retains the type it had when we built it.
173+The fragment retains the type it had when we built it.
175+Another (equivalent) way to think of tree addressing involves
176+binary. As before, the root of the tree is at address `1`. For
177+the node of a tree at some address `b`, where `b` is binary,
178+you get the address of the head by concatenating a `0` to
179+the end of `b`; and to get the tail, concatenate a `1` at the
180+end. For example, the head of the node at binary address `111`
181+is at `1110`, and the tail is at `1111`.
175## Labeled surfaces
183## Labeled surfaces