Another slit rectangle IIT system

One more observation on Integrated Information Theory (IIT), in Aaronson’s simplified formulation.

Let R_M, N be a wide rectangular grid of points (x,y) with x and y integers such that 0 ≤ x < M and 0 ≤ y < N. Suppose M ≫ 4N and M is divisible by four. Let R_M, N, t be the grid R_M, N with all points with coordinates (M/4,y) where y ≥ tN removed. This is a grid with a “bottleneck” at x = M/4.

Let S_M, N, t be a system with a binary cell at each coordinate of R_M, N, t evolving according to the rule that at the next time step, each cell’s value changes to the xor of the up-to-four neighboring cells’ values (near the boundaries and the slit, the count will be less than four).

My intuitions about IIT say that the measure of integrated information Φ(S_M, N, t) will be equal to exactly 2N bits when t = 1, and will stay at 2N bits as we decrease t, until t is below or around 1/4, at which point it will jump to 2ceil(tN) bits. This shows two problems with this version of IIT. First, we as we cut a small slit in the rectangle, we should always be decreasing the amount of integrated information in the system—not just suddenly when the slit reaches around 3/4 of the width of the rectangle. Second, we would expect the amount of integrated information to vary much more continuously rather than suddenly jump from 2N to N/2 bits at around t = 1/4.

The problem here is the rather gerrymandered character in which IIT minimizes one quantity to generate an optimal decomposition of the system, and then defines the measure of integrated information using another quantity.

Specifically, we calculate Φ by finding a cut of the system into two subsystems A and B that minimizes Φ(A,B)/min (|A|,|B|), and intuitively, there are two types of candidates for an optimal cut if M ≫ 4N:

• cut the grid around x = M/2 into two equally sized portions using a cut that snips each horizontal line at exactly one cell (a vertical line is the most obvious option, but a diagonal cut will give the same Φ(A,B) value); the Φ(A,B) value is twice the vertical length of the cut, namely 2N bits

• cut the grid around the bottleneck into two portions, A and B, where A contains slightly more than a quarter of the cells in the system using a cut that follows the same rule as above: more precisely, we start the cut at the top of the bottleneck, and cut diagonally down and to the right (the result is that A is an (M/4) by N rectangle together with an isosceles triangle with equal sides of size tN; the triangle’s area is swamped by the rectangle’s area because M ≫ 4N); the Φ(A,B) value is (maybe modulo an off-by-one error) twice the vertical length of the cut, namely 2ceil(tN) bits.

Assuming these intuitions are right, when t is close to 1, the first type of cut results in a smaller Φ(A,B)/min (|A|,|B|) value, but when t is below or around 1/4, the second type of cut results in a smaller value.