TAOCP 7.2.2.1 Exercise 200

We keep the algebraic setup but fix part (b) by replacing symbolic computation with a randomized polynomial identity test in a finite field.

taocpmathematicsalgorithmsvolume-4hm-medium

Section 7.2.2.1: Dancing Links

Exercise 200. ▶ [HM25] (A. Björklund, 2010.) Find one polynomial algebra, instead of backtracking, to decide whether or not a given 3D matching problem is solvable. Let the items be ${a_1, \ldots, a_n}, {b_1, \ldots, b_n}, {c_1, \ldots, c_n}$, and assign a symbolic weight to each option. If $X$ is any subset of $C$, let $Q(X)$ be the $n \times n$ matrix whose entry in row $i$ and column $j$ is the sum of the variables for all options '$a_i b_j c_k$' with $c_k \notin X$.

For example, suppose $n = 3$. The seven options $t$: '$a_1 b_1 c_1$', $u$: '$a_1 b_2 c_2$', '$a_2 b_2 c_3$', $w$: '$a_2 b_3 c_1$', $x$: '$a_3 b_1 c_1$', $y$: '$a_3 b_1 c_3$', $z$: '$a_3 b_2 c_2$' yield the matrices

$$X = \emptyset \quad {c_1} \quad {c_2} \quad {c_1, c_2} \quad {c_3} \quad {c_1, c_3} \quad {c_2, c_3}$$

$$Q(X) = \begin{pmatrix} t & u & 0 \ 0 & 0 & v+w \ x+y & 0 & 0 \end{pmatrix} \begin{pmatrix} t & u & 0 \ 0 & 0 & v \ y & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & u & 0 \ 0 & 0 & v \ x & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & u & 0 \ 0 & 0 & 0 \ x & 0 & 0 \end{pmatrix} \begin{pmatrix} t & 0 & 0 \ 0 & 0 & v+w \ x & z & 0 \end{pmatrix} \begin{pmatrix} t & 0 & 0 \ 0 & 0 & v \ 0 & z & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & v \ x & 0 & 0 \end{pmatrix}$$

(and $Q(C)$ is always zero). The determinant of $Q(\emptyset)$ is $u(v+w)x - t(v+w)(y+z)$.

a) If the given problem has $r$ solutions, prove that the polynomial

$$S = \sum_{X \subseteq C} (-1)^{|X|} \det Q(X)$$

is the sum of $r$ monomials, each with coefficient $\pm 1$. (In the given example it is $uvx - twy$.) Hint: Consider the case where all possible options are present.

b) Use this fact to design a randomized algorithm that decides q.s. whether or not a matching exists, in $O(2^n n^4)$ steps.

Verified: no
Solve time: 2m13s

Corrected solution

We keep the algebraic setup but fix part (b) by replacing symbolic computation with a randomized polynomial identity test in a finite field.

(a) Structure of the polynomial $S$

Let each option $a_i b_j c_k$ carry a commuting variable $x_{ijk}$. For a fixed $X \subseteq C$, define the $n \times n$ matrix

$$ Q(X){ij}=\sum{\substack{k:,c_k\notin X\ a_i b_j c_k\text{ is an option}}} x_{ijk}. $$

Expanding the determinant,

$$ \det Q(X)=\sum_{\pi\in S_n}\operatorname{sgn}(\pi)\prod_{i=1}^n Q(X)_{i,\pi(i)}. $$

Each factor expands as a sum over allowed $k$, so each monomial in $\det Q(X)$ has the form

$$ M(\pi,k_1,\dots,k_n)=\prod_{i=1}^n x_{i,\pi(i),k_i}, $$

corresponding to selecting exactly one option in each row $a_i$ and column $b_{\pi(i)}$, with the $c$-indices $k_i$ unrestricted except for exclusion of $X$.

Now define

$$ S=\sum_{X\subseteq C}(-1)^{|X|}\det Q(X). $$

Fix a monomial $M$, and let $K\subseteq C$ be the set of $c$-elements used by $M$. Then $M$ appears in $\det Q(X)$ if and only if $X\cap K=\emptyset$. Hence its coefficient in $S$ is

$$ \sum_{X\subseteq C\setminus K}(-1)^{|X|}. $$

If $K\neq C$, this sum is $0$ by pairing $X$ with $X\cup{c}$ for some $c\in C\setminus K$. If $K=C$, the sum equals $1$.

Thus only monomials using every $c_k$ survive. Such monomials correspond exactly to 3D matchings (perfect selections of $n$ disjoint triples covering all $a_i,b_j,c_k$). Each matching contributes one monomial with coefficient $\pm 1$ given by $\operatorname{sgn}(\pi)$.

Therefore, if the instance has $r$ solutions, then $S$ is the sum of $r$ distinct monomials, each with coefficient $\pm 1$.

(b) Randomized $O(2^n n^4)$ algorithm

We now design a randomized algorithm that decides whether $S\neq 0$ without constructing $S$ symbolically.

Key idea

Evaluate the polynomial $S$ over a finite field at a random assignment of the variables $x_{ijk}$. Then test whether the resulting value is zero.

Since $S$ is a nonzero polynomial if and only if a 3D matching exists, this gives a randomized decision procedure via the Schwartz–Zippel lemma.

Algorithm

Fix a finite field $\mathbb{F}_p$ where $p$ is a prime of size $p > 2n$.

  1. For each option $a_i b_j c_k$, assign a random value

$$ x_{ijk} \in \mathbb{F}_p $$

independently and uniformly. 2. Compute

$$ S = \sum_{X \subseteq C} (-1)^{|X|}\det Q(X) $$

but now all arithmetic is in $\mathbb{F}p$, and each $Q(X){ij}$ is a numeric sum. 3. For each subset $X \subseteq C$:

  • Build the matrix $Q(X)$ in $O(n^2)$ field operations.
  • Compute $\det Q(X)$ by Gaussian elimination in $O(n^3)$ field operations.
  • Accumulate $(-1)^{|X|}\det Q(X)$.
  1. Output:
  • “YES” if $S \neq 0$ in $\mathbb{F}_p$,
  • “NO” otherwise.

Correctness

If no 3D matching exists, then every monomial in $S$ is absent, hence $S$ is the zero polynomial, so every evaluation gives $S=0$.

If a matching exists, then $S$ is a nonzero polynomial of total degree $n$. Under a random assignment from $\mathbb{F}_p$, the Schwartz–Zippel lemma gives

$$ \Pr[S=0] \le \frac{n}{p} < \frac{1}{2}. $$

Thus the algorithm returns the correct answer with probability at least $1/2$, which can be amplified by repetition.

Running time

There are $2^n$ subsets $X \subseteq C$. For each $X$:

  • Building $Q(X)$: $O(n^2)$,
  • Determinant: $O(n^3)$,

so total per subset is $O(n^3)$, dominated by matrix elimination.

Hence total time is

$$ O(2^n n^3), $$

which is within the required $O(2^n n^4)$ bound.

Conclusion

The algorithm evaluates the Björklund polynomial algebra in a randomized finite-field model. It avoids symbolic blow-up, preserves the cancellation structure of inclusion–exclusion, and detects existence of a 3D matching with bounded error probability.