On Rossian physical indeterminacy

Further to understanding what Ross(*) means by 'the indeterminacy of the physical'.  Here is the operation table of a three-bit wide adder.  I have arbitrarily labelled the inputs and outputs.

⊕		a	b	c	d	e	f	g	h	i
a		i	d	b	d	f	c	d	a	g
b		d	d	d	d	d	d	d	b	d
c		b	d	d	d	g	d	d	c	d
d		d	d	d	d	d	d	d	d	d
e		f	d	g	d	a	i	b	e	c
f		c	d	d	d	i	g	d	f	b
g		d	d	d	d	b	d	d	g	d
h		a	b	c	d	e	f	g	h	i
i		g	d	a	d	c	b	d	i	d

This does not, prima facie, look like addition.  However, symmetry about the main diagonal shows that ⊕ is commutative.  Much laborious checking reveals that ⊕ is also associative.  More obviously, the element h under ⊕ behaves as an identity element:  h⊕x=x and x⊕h=x, for every x.  Further, the element d under ⊕ acts as an absorbing element: d⊕x=d and x⊕d=d, for every x.   Reordering the table with h first and d last gives us this table.

⊕		h	a	b	c	e	f	g	i	d
h		h	a	b	c	e	f	g	i	d
a		a	i	d	b	f	c	d	g	d
b		b	d	d	d	d	d	d	d	d
c		c	b	d	d	g	d	d	d	d
e		e	f	d	g	a	i	b	c	d
f		f	c	d	d	i	g	d	b	d
g		g	d	d	d	b	d	d	d	d
i		i	g	d	d	c	b	d	d	d
d		d	d	d	d	d	d	d	d	d

Perhaps the next thing to note is that the element d crops up a good deal.  Why this is will emerge shortly.  But note that, in combination under ⊕ with the other elements, h produces one d, e produces two ds, a three ds, f four ds, i five ds, c six ds, g seven ds, and b eight ds.  Re-arranging the table in order of increasing 'd-productivity' gives us the following.

⊕		h	e	a	f	i	c	g	b	d
h		h	e	a	f	i	c	g	b	d
e		e	a	f	i	c	g	b	d	d
a		a	f	i	c	g	b	d	d	d
f		f	i	c	g	b	d	d	d	d
i		i	c	g	b	d	d	d	d	d
c		c	g	b	d	d	d	d	d	d
g		g	b	d	d	d	d	d	d	d
b		b	d	d	d	d	d	d	d	d
d		d	d	d	d	d	d	d	d	d

Some structure is starting to emerge.  Note now that

a=e⊕e,
a⊕e=e⊕e⊕e=f,
f⊕e=e⊕e⊕e⊕e=i,
i⊕e=e⊕e⊕e⊕e⊕e=c.
c⊕e=e⊕e⊕e⊕e⊕e⊕e=g,
g⊕e=e⊕e⊕e⊕e⊕e⊕e⊕e=b,
b⊕e=e⊕e⊕e⊕e⊕e⊕e⊕e⊕e,

so we might relabel a as 2e, f as 3e, i as 4e, and so on. We can also relabel h as Z, e as 1U, and d as ⟂.  This gives us a final version of the combination table for ⊕ as this:

⊕		Z	1U	2U	3U	4U	5U	6U	7U	⟂
Z		Z	1U	2U	3U	4U	5U	6U	7U	⟂
1U		1U	2U	3U	4U	5U	6U	7U	⟂	⟂
2U		2U	3U	4U	5U	6U	7U	⟂	⟂	⟂
3U		3U	4U	5U	6U	7U	⟂	⟂	⟂	⟂
4U		4U	5U	6U	7U	⟂	⟂	⟂	⟂	⟂
5U		5U	6U	7U	⟂	⟂	⟂	⟂	⟂	⟂
6U		6U	7U	⟂	⟂	⟂	⟂	⟂	⟂	⟂
7U		7U	⟂	⟂	⟂	⟂	⟂	⟂	⟂	⟂
⟂		⟂	⟂	⟂	⟂	⟂	⟂	⟂	⟂	⟂

We can see now that the table is generated by the elements Z and 1U.  Any non-Z element can be reached by combining a number of 1Us with ⊕, and in general, nU⊕mU=(n+m)U, unless n+m >7, in which case the result is ⟂.    Compare now the table for ⊕ with that of addition, +, shown below.

+		0	1	2	3	4	5	6	7	...
0		0	1	2	3	4	5	6	7	...
1		1	2	3	4	5	6	7	8	...
2		2	3	4	5	6	7	8	9	...
3		3	4	5	6	7	8	9	10	...
4		4	5	6	7	8	9	10	11	...
5		5	6	7	8	9	10	11	12	...
6		6	7	8	9	10	11	12	13	...
7		7	8	9	10	11	12	13	14	...
:		:	:	:	:	:	:	:	:	:

But for a relabelling of the elements the tables are identical in the top left corner.  There are, however, two significant differences between them,

• the ⊕ table is finite whereas the + table is infinite, as indicated by the ellipses,
• the ⊕ table contains mysterious ⟂ entries in certain places.

It should be clear that ⟂ is the adder's signal that it can't give an answer because the correct result requires more than the adder's three bits for its encoding.  This is the equivalent of the human calculator running out of paper while doing a sum.  That the + table is infinite may be the source of Ross's notion of indeterminacy.  For it would seem that the ⊕ table could be extended in infinitely many ways that are not compatible with +.  There appears to be an unanswerable question, Which of these extensions is the ⊕ table 'really' a small part of?      




(*)  Immaterial Aspects of Thought, The Journal of Philosophy, Vol. 89, No. 3, (Mar., 1992), pp. 136-150 (available at http://www.jstor.org/stable/2026790)