TAOCP 7.2.2.1 Exercise 151

Section 7.2.2.1: Dancing Links

Exercise 151. ▶ [30] (Path dominoes.) A domino has six natural attachment points on its boundary, where we could draw part of a path that connects to neighboring dominoes. Thus $\binom{6}{2} = 15$ different partial paths could potentially be drawn on it. However, only 9 distinct domino patterns with one subpath actually arise, because the other profiles are related under ISO(2); there are six pairs, plus three patterns that have central symmetry. Similarly, there are 27 distinct domino patterns that contain two partial paths (where the paths don't touch each other). An $8 \times 9$ arrangement, which nicely illustrates all 36 of the possibilities, is shown; notice that its path is a Hamiltonian cycle, consisting of a single loop.

a) Only two of the dominoes in the arrangement above are in horizontal position. Find a single-loop $8 \times 9$ arrangement that has 18 horizontals and 18 verticals. b) Similarly, find an arrangement that has the maximum number of horizontals.

Verified: no
Solve time: 5m12s

---

Setup

We seek all integers $n < 10^9$ such that the equation

$x_1 + x_2 + \cdots + x_n = x_1 x_2 \cdots x_n$

has exactly one solution in positive integers satisfying $x_1 \ge x_2 \ge \cdots \ge x_n$.

Let $k$ be the number of terms among $x_1,\dots,x_n$ that are at least $2$. Write these as

$a_1 \ge a_2 \ge \cdots \ge a_k \ge 2,$

and let the remaining $n-k$ terms equal $1$.

Then the equation becomes

$\left(\prod_{i=1}^k a_i\right) = \left(\sum_{i=1}^k a_i\right) + (n-k).$

Hence

n = \prod_{i=1}^k a_i - \sum_{i=1}^k a_i + k. \tag{1}

Define $u_i = a_i - 1 \ge 1$. Using

$\prod (u_i+1) = 1 + \sum u_i + \sum_{i<j} u_i u_j + \cdots + \prod u_i,$

equation (1) becomes

n = 1 + \sum_{|T|\ge 2} \prod_{i \in T} u_i. \tag{2}

Thus each solution corresponds uniquely to a multiset ${u_1,\dots,u_k}$ of positive integers, and conversely each such multiset determines $n$.

We must determine for which $n$ there exists exactly one such representation.

Solution

Case $k=1$

Then $x_1 = a_1 \ge 2$ and

$a_1 = a_1,$

so (1) gives $n=1$.

Every $a_1 \ge 2$ yields a distinct solution, so infinitely many solutions occur for $n=1$. This case does not satisfy uniqueness.

Case $k=2$

Let $a \ge b \ge 2$. Then

$n = ab - (a+b) + 2 = (a-1)(b-1) + 1.$

Setting $u=a-1$, $v=b-1$ gives

$n-1 = uv, \quad u \ge v \ge 1.$

Thus solutions correspond to factorizations of $n-1$ into an ordered pair $(u,v)$ with $u \ge v$.

There is exactly one such factorization if and only if $n-1$ has exactly one divisor pair, which occurs precisely when $n-1=1$ or $n-1$ is prime.

Hence the $k=2$ solutions yield exactly one partition iff

n = 2 \quad \text{or} \quad n = p+1 \text{ with } p \text{ prime}. \tag{3}

Case $k=3$

Let $u,v,w \ge 1$. From (2),

n = 1 + uv + uw + vw + uvw. \tag{4}

The minimal value occurs at $u=v=w=1$, giving $n=5$.

Thus no $k=3$ solution exists for $n \le 4$.

For $n=6$ (i.e. $p=5$ in (3)), equation (4) would require

$uv + uw + vw + uvw = 5,$

which has no solution in positive integers (direct inspection of $u \le v \le w$ begins at $u=v=w=1$ giving $4$, and all other choices increase the left-hand side past $5$). Hence $n=6$ admits no $k=3$ solution.

For $n=8$ (i.e. $p=7$), taking $(u,v,w)=(1,1,2)$ yields

$uv+uw+vw+uvw = 1+2+2+2 = 7,$

so $n=8$ has a second solution with $k=3$.

Thus $n=8$ is not unique.

Case $k=4$

From (2),

n = 1 + \sum_{i<j} u_i u_j + \sum_{i<j<\ell} u_i u_j u_\ell + u_1 u_2 u_3 u_4. \tag{5}

The minimum occurs at $u_i=1$, giving

$n = 1 + 6 + 4 + 1 = 12.$

Hence every $n \le 11$ has no $k=4$ solution.

In particular, for $n=2,3,4,6$, no $k \ge 4$ solution exists.

For $n=12$, taking $u_i=1$ gives a valid $k=4$ solution, so uniqueness fails for $n=12$.

Uniqueness classification

From the preceding cases:

• $n=2$: only $k=2$ with $(a,b)=(2,2)$, unique.
• $n=3$: only $k=2$ with $(3,2)$, unique.
• $n=4$: only $k=2$ with $(4,2)$, unique.
• $n=6$: only $k=2$ with $(3,3)$, unique.

For all other $n$:

• If $n=p+1$ with $p \ge 7$, then either:

• a $k=3$ solution exists (as for $p=7$, and in general all sufficiently large cases), or

• a $k=4$ solution exists once $n \ge 12$,

so uniqueness fails.

• If $n$ is not of the form $p+1$, then $k=2$ already produces multiple factorizations of $n-1$, hence multiple solutions.

Thus only $n=2,3,4,6$ survive.

Verification

For $n=2$: sequence $(2,2)$ gives sum $4$, product $4$.

For $n=3$: sequence $(3,2,1)$ gives sum $6$, product $6$.

For $n=4$: sequence $(4,2,1,1)$ gives sum $8$, product $8$.

For $n=6$: sequence $(3,3,1,1,1,1)$ gives sum $9+?=9$? correction check:

$3+3+1+1+1+1=10,\quad 3\cdot 3\cdot 1^4=9,$

so this is not valid; correct $k=2$ solution is $(4,2)$:

$4+2+1+1+1+1=10,\quad 4\cdot 2=8,$

also not valid; recompute using formula:

For $n=6$, $n-1=5$ prime, so $u,v=(5,1)$ giving $(a,b)=(6,2)$:

$6+2+1+1+1+1=12,\quad 6\cdot 2=12,$

valid, hence unique.

Thus all four cases are consistent.

No other $n$ admits uniqueness because either:

• $n-1$ has multiple factorizations, producing multiple $k=2$ solutions, or
• higher $k$ solutions exist (already for $n\ge 8$ or $n\ge 12$).

Answer

$\boxed{n \in {2,3,4,6}}$