TAOCP 7.1.3 Exercise 165

Let the $3\times 3$ configuration at time $t$ be represented by a bit matrix $X(t) = (x_{ij}(t))_{1 \le i,j \le 3}$, where each $x_{ij}(t) \in {0,1}$.

taocpmathematicsalgorithmsvolume-4medium

Section 7.1.3: Bitwise Tricks and Techniques

Exercise 165. [21] [21] (R. A. Kirsch.) Discuss the computation of the 3×3 cellular automaton with X(t+1) = custer(X(t)) = ∼X(t) & (X(t) N | X(t) W | X(t) E | X(t) S ).

Verified: yes
Solve time: 1m07s

Solution

Let the $3\times 3$ configuration at time $t$ be represented by a bit matrix $X(t) = (x_{ij}(t)){1 \le i,j \le 3}$, where each $x{ij}(t) \in {0,1}$. The neighborhood operators $N,W,E,S$ act by shifting indices within this fixed $3\times 3$ domain, with boundary values taken as $0$ outside the grid.

Thus for each position $(i,j)$,

$X(t)N(i,j) = x{i-1,j}(t), \quad X(t)W(i,j) = x{i,j-1}(t), \quad X(t)E(i,j) = x{i,j+1}(t), \quad X(t)S(i,j) = x{i+1,j}(t),$

with undefined indices contributing $0$.

The update rule is

$X(t+1) = \sim X(t) ,&, \bigl(X(t)_N \mid X(t)_W \mid X(t)_E \mid X(t)_S\bigr).$

Fix a cell $(i,j)$. The value $X(t+1){ij}$ equals $1$ exactly when $x{ij}(t)=0$ and at least one of its four orthogonal neighbors is $1$. This follows directly from the bitwise semantics of $\sim$ and $\mid$ applied componentwise by equation (1) and (2), since complement and disjunction act independently on each coordinate.

To compute $X(t+1)$ using bitwise operations on packed data, encode the $3\times 3$ array into a word

$X(t) = \sum_{i=1}^3 \sum_{j=1}^3 x_{ij}(t),2^{3(i-1)+(j-1)}.$

Define shift operators corresponding to the four directions:

$X_N(t) = X(t) \gg 3, \quad X_S(t) = X(t) \ll 3,$

$X_W(t) = \bigl(X(t) \gg 1\bigr) ,&, M_W, \quad X_E(t) = \bigl(X(t) \ll 1\bigr) ,&, M_E,$

where $M_W$ masks away wraparound bits entering from the right boundary and $M_E$ masks away wraparound bits entering from the left boundary. These masks depend only on the fixed $3\times 3$ geometry and ensure that shifts do not create spurious adjacency across row boundaries.

With these definitions,

$X(t)_N \mid X(t)_W \mid X(t)_E \mid X(t)_S$

is computed by four shifts, three bitwise disjunctions, and two mask applications, and then the final state is obtained by one bitwise complementation and one bitwise conjunction.

For each bit position $(i,j)$, the value of the right-hand side equals $1$ precisely when at least one neighbor bit is $1$ and the central bit is $0$. This matches the stated rule for $custer(X(t))$ since

$custer(X(t)) = \sim X(t) ,&, (X(t)_N \mid X(t)_W \mid X(t)_E \mid X(t)_S).$

Thus the entire $3\times 3$ cellular automaton update is computed by a constant number of word operations: four shifts, four bitwise OR/AND/NOT operations, and fixed masking to enforce the grid boundary. Each operation acts simultaneously on all nine cells, so the update is performed in parallel bitwise fashion within a single word.

This completes the analysis of the computation of the cellular automaton rule. ∎