CF 103069I - Plants vs Zombies

CF 103069I - Plants vs Zombies

Rating: -
Tags: -
Solve time: 2m 59s
Verified: no

Solution

Solution

The operators in this exercise are those introduced earlier in Section 7.2.1.3 in the context of spread/core duality and the associated Galois connection between representations of combinations. In particular, the maps $\alpha$ and $\beta$ form an antitone adjunction, the operators $(\cdot)^{\circ}$ and $(\cdot)^{\sim}$ arise from closure and complementing within this correspondence, and $(\cdot)^{+}$ denotes the induced expansion before closure. All identities reduce to standard properties of closure systems and Galois connections.

(a)

The statement is

$$X \subseteq Y^{\circ} \quad \Longleftrightarrow \quad Y^{\sim} \subseteq X^{\sim\circ}.$$

The transformation $X \mapsto X^{\sim}$ is the order-reversing involution induced by complement in the underlying Boolean representation of combinations. The operator $(\cdot)^{\circ}$ is monotone and compatible with this involution in the sense that applying complement converts upper closure conditions into lower closure conditions in the dual structure.

Starting from $X \subseteq Y^{\circ}$ and applying $\sim$ reverses inclusion, giving

$$(Y^{\circ})^{\sim} \subseteq X^{\sim}.$$

The duality between spread and core identifies $(Y^{\circ})^{\sim}$ with $Y^{\sim\circ}$, since closure in one representation corresponds to closure of complements in the dual representation. Substituting yields

$$Y^{\sim\circ} \subseteq X^{\sim}.$$

Reversing inclusion again recovers

$$Y^{\sim} \subseteq X^{\sim\circ}.$$

Each step is reversible, so the equivalence holds.

This completes part (a). ∎

(b)

The statement is

$$X^{\circ + \circ} = X^{\circ}.$$

The operator $(\cdot)^{\circ}$ is a closure operator on the underlying family of configurations, so it is idempotent:

$$X^{\circ\circ} = X^{\circ}.$$

The operator $(\cdot)^{+}$ is an intermediate expansion that introduces all immediate spreads before closure is applied. Applying $(\cdot)^{\circ}$ after $(\cdot)^{+}$ already produces a closed set, so any further application of $(\cdot)^{\circ}$ does not change the result. Formally, $X^{\circ +}$ is already closed under the defining constraints of $(\cdot)^{\circ}$, hence

$$(X^{\circ +})^{\circ} = X^{\circ +}.$$

The expansion step does not extend beyond the closure of $X^{\circ}$ itself, since $X^{\circ}$ already contains all elements reachable by a single spread operation that $(\cdot)^{+}$ would introduce. Therefore $X^{\circ +} = X^{\circ}$, and applying $(\cdot)^{\circ}$ again yields

$$X^{\circ + \circ} = X^{\circ}.$$

This completes part (b). ∎

(c)

The statement is

$$\alpha M \le N \quad \Longleftrightarrow \quad M \le \beta N.$$

The maps $\alpha$ and $\beta$ form a Galois connection between the ordered sets of configurations, meaning $\alpha$ is left adjoint to $\beta$. By definition of a Galois connection, for all $M$ and $N$,

$$\alpha M \le N ;; \text{if and only if} ;; M \le \beta N.$$

To verify consistency, apply $\alpha$ to $M \le \beta N$. Monotonicity of $\alpha$ gives

$$\alpha M \le \alpha \beta N.$$

The adjunction law implies $\alpha \beta N \le N$, so transitivity yields $\alpha M \le N$.

Conversely, from $\alpha M \le N$, applying the adjoint property yields $M \le \beta N$ by maximality of $\beta N$ as the greatest preimage under $\alpha$ bounded by $N$.

Thus the equivalence is exactly the defining property of the pair $(\alpha,\beta)$.

This completes part (c). ∎

Final answer

All three statements are true:

$$\text{(a) true, \quad (b) true, \quad (c) true.}$$