TAOCP 7.1.4 Exercise 218

We restart from the exact cover formulation, but we now build the BDD/ZDD constructions in a way that does not rely on variable ordering to magically enforce constraints.

taocpmathematicsalgorithmsvolume-4medium

Section 7.1.4: Binary Decision Diagrams

Exercise 218. ▶ [24] [24] Apply BDD and ZDD techniques to the problem of Langford pairs, discussed at the beginning of this chapter.

7.1.4 BINARY DECISION DIAGRAMS 275

Verified: yes
Solve time: 3m12s

We restart from the exact cover formulation, but we now build the BDD/ZDD constructions in a way that does not rely on variable ordering to magically enforce constraints. Instead, all constraints are enforced either structurally (ZDD decomposition) or by explicit finite state tracking (BDD).

Reformulation as exact cover

We index potential placements by pairs $(k,i)$ with

$$ 1 \le k \le n, \qquad 1 \le i \le 2n-k-1, $$

and define the placement

$$ p_{k,i} = {(k,i),(k,i+k+1)}. $$

Introduce Boolean variables $x_{k,i}$ indicating whether placement $p_{k,i}$ is chosen.

A Langford solution is exactly a set $S \subseteq {(k,i)}$ such that

  1. For every $k$, exactly one $x_{k,i}$ is chosen.
  2. Every position $j \in {1,\dots,2n}$ is covered exactly once by the chosen placements.

We now construct a ZDD for constraint (1), a BDD for constraint (2), and then combine them.

ZDD for “exactly one placement per $k$”

For each fixed $k$, define the family

$$ \mathcal{F}k = \big{ {x{k,i}} ;:; 1 \le i \le 2n-k-1 \big}. $$

We first construct a ZDD $Z_k$ that represents “choose exactly one variable from block $k$”:

$$ Z_k = { {x_{k,i}} : i \in I_k }. $$

This is a standard ZDD construction: at the root for block $k$, we branch over all $i$, each branch selecting a singleton set containing $x_{k,i}$, with the ZDD terminals representing single-element families.

Importantly, this is not enforced by ordering. It is enforced by the structure of the ZDD, since every root-to-terminal path in $Z_k$ selects exactly one variable from the block.

We then combine all blocks using the ZDD product (disjoint variable families):

$$ Z_{\text{label}} = \bigotimes_{k=1}^n Z_k. $$

Because the variable sets for different $k$ are disjoint, the ZDD product correctly enforces that the final set contains exactly one choice per block $k$.

Thus $Z_{\text{label}}$ represents exactly the feasible selections satisfying the “one per $k$” constraint.

BDD for position constraints

We now construct a BDD over the same variables $x_{k,i}$, but we do not rely on ordering to guarantee locality. Instead, each BDD node carries an explicit finite state describing partial coverage.

State definition

Let the state be a vector

$$ \sigma = (\sigma_1,\dots,\sigma_{2n}), $$

where each $\sigma_j \in {0,1,2}$ counts how many selected placements currently cover position $j$.

A state is valid if $\sigma_j \le 1$ for all $j$. The initial state is $\sigma^{(0)} = (0,\dots,0)$. A terminal accepting state is $(1,\dots,1)$.

Transition function

When we assign $x_{k,i} = 1$, we update exactly two positions:

$$ j_1 = i, \qquad j_2 = i+k+1, $$

by

$$ \sigma_{j_1} \leftarrow \sigma_{j_1} + 1, \qquad \sigma_{j_2} \leftarrow \sigma_{j_2} + 1. $$

If any coordinate exceeds 1, the transition goes to the reject sink.

When $x_{k,i} = 0$, the state is unchanged.

BDD nodes

A BDD node is a pair

$$ (x_{k,i}^{\text{next}}, \sigma), $$

where $\sigma$ is the current coverage state after processing all previously ordered variables.

Because each update only modifies finitely many coordinates and each coordinate is ternary, the number of possible states is finite:

$$ |{0,1,2}^{2n}| = 3^{2n}. $$

Thus the BDD is well-defined as a layered graph over a finite state space.

Crucially, correctness does not depend on any relation between variable order and position indices. All positional interactions are explicitly encoded in the state transition.

The BDD accepts exactly those assignments that produce final state $(1,\dots,1)$, i.e., exact coverage.

Combination of BDD and ZDD

Both diagrams use the same variable set ${x_{k,i}}$ and the same ordering (any fixed order suffices, e.g. lexicographic in $(k,i)$). We form their conjunction by a synchronized product construction.

A node in the combined diagram is a pair

$$ (v_Z, v_B), $$

where $v_Z$ is a ZDD node and $v_B$ is a BDD node, both corresponding to the same variable.

Transitions are synchronized:

  • If $x_{k,i}=0$, follow the 0-edges in both diagrams.

  • If $x_{k,i}=1$, the move is allowed only if:

  • the ZDD allows selecting $x_{k,i}$ within its block structure, and

  • the BDD transition does not violate $\sigma_j \le 1$.

The accepting condition is that:

  • the ZDD component satisfies exactly one selection per $k$,
  • the BDD component reaches state $(1,\dots,1)$.

Correctness

Soundness follows because any accepting path chooses exactly one $x_{k,i}$ per $k$ by construction of $Z_{\text{label}}$, and the BDD ensures that every position is covered exactly once.

Completeness follows because every Langford solution induces a unique assignment of variables $x_{k,i}$, which respects both:

  • the per-block constraint (each $k$ appears exactly once), and
  • the positional constraint (each position is covered exactly once),

so the corresponding path is accepted in both structures.

Conclusion

The corrected construction removes all dependence on variable ordering for enforcing constraints. The ZDD enforces the per-$k$ cardinality constraints structurally via decomposition into disjoint choice families, and the BDD enforces positional exact coverage via an explicit finite state machine. Their synchronized product yields a correct BDD–ZDD representation of Langford pairs.