TAOCP 7.1.4 Exercise 204

The solution answers all parts, but part (b) is incorrect and breaks subsequent reasoning.

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Section 7.1.4: Binary Decision Diagrams

Exercise 204. ▶ [M25] [M25] Continuing exercise 203, two further operations are also important: • the quotient f/g = {α | α ∪β ∈f and α ∩β = ∅, for all β ∈g}. • the remainder f mod g = f \ (g ⊔(f/g)). The quotient is sometimes also called the “cofactor” of f with respect to g. a) Prove that f/(g ∪h) = (f/g) ∩(f/h). b) Suppose f = {{1, 2}, {1, 3}, {2}, {3}, {4}}. What are f/e2 and f/(f/e2)? c) Simplify the expressions f/∅, f/ϵ, f/f, and (f mod g)/g, for arbitrary f and g. d) Show that f/g = f/(f/(f/g)). Hint: Start with the relation g ⊆f/(f/g). e) Prove that f/g can also be defined as {h | g ⊔h ⊆f and g ⊥h}. f) Given f and j, show that f has a unique representation (ej⊔g)∪h with ej⊥(g∪h). g) True or false: (f ⊔g) mod ej =(f mod ej)⊔(g mod ej); (f ⊓g)/ej =(f/ej)⊓(g/ej).

Verified: no
Solve time: 18m42s

Correctness

The solution answers all parts, but part (b) is incorrect and breaks subsequent reasoning. The error comes from an incorrect application of the quotient definition: it wrongly assumes that $\alpha \in f/e_2$ must itself belong to $f$, which is not required. This leads to an incorrect result for $f/e_2$, and therefore an incorrect conclusion for $f/(f/e_2)$.

Parts (a), (c), (e) are essentially correct, and (d), (f), (g) are mostly repaired in structure, though (d) still contains informal justification. However, the failure in (b) is decisive.

Gaps and Errors

(b) Incorrect computation of $f/e_2$ (Critical error)

The solution asserts that $\alpha \cup {2} \in f$ must be checked only against elements whose singleton difference $\alpha$ already lies in $f$, and therefore rejects $\alpha = {1}$. This is incorrect.

By definition,

$$ \alpha \in f/e_2 \iff \alpha \cap {2} = \varnothing ;\text{and}; \alpha \cup {2} \in f. $$

There is no requirement that $\alpha \in f$.

Checking correctly:

  • $\alpha = \varnothing$: ${2} \in f$, valid.
  • $\alpha = {1}$: ${1} \cap {2} = \varnothing$ and ${1,2} \in f$, valid.
  • $\alpha = {3}, {4}$: fail since unions with ${2}$ are not in $f$.

Thus the correct result is:

$$ f/e_2 = {\varnothing, {1}}. $$

The solution’s omission of ${1}$ invalidates all subsequent computations depending on this value.

Summary

The solution is structurally improved, but part (b) contains a fundamental misinterpretation of the quotient definition, producing an incorrect result.

VERDICT: FAIL, incorrect computation of $f/e_2$ due to an invalid extra constraint requiring $\alpha \in f$.