Section 7.2.2.1: Dancing Links

Exercise 428. [M28] [M28] For infinitely many $n$, construct a valid $n \times n$ masyu puzzle with $O(n)$ clues whose loop goes through all four corner cells, where all clues are (a) black; (b) white.

Verified: no
Solve time: 3m59s

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The failure in the previous solution is structural: a spiral Hamiltonian cycle has $\Theta(n^2)$ turns, so any constant-density sampling still yields $\Theta(n^2)$ clues. To obtain $O(n)$ clues, the loop itself must be redesigned so that the number of “constraint-relevant” local configurations is only linear in $n$.

The correct approach is to use a row-by-row serpentine Hamiltonian cycle, which has only $O(n)$ turns. This immediately forces all usable clue locations (turns and their adjacent straight cells) to also be $O(n)$.

Construction (common to (a) and (b))

Assume $n \ge 2$. Define a Hamiltonian cycle $C_n$ on the $n \times n$ grid as follows.

Traverse rows alternately:

• In row $1$, move left-to-right from $(1,1)$ to $(1,n)$.
• Move down to $(2,n)$.
• In row $2$, move right-to-left to $(2,1)$.
• Move down to $(3,1)$.
• Continue alternating direction in each row.

Finally, close the cycle by connecting the last cell back to $(1,1)$ using the outer boundary edge.

This is the standard serpentine Hamiltonian cycle.

Key properties

1. Hamiltonicity: every cell is visited exactly once.
2. Corners: the cycle visits all four corners:

$$ (1,1), (1,n), (n,1), (n,n). $$ 3. Number of turns: a turn occurs only at the end of each row transition.

Hence there are exactly $2(n-1) = O(n)$ turns. 4. Local geometry:

• Every turn is a clean $90^\circ$ bend between two maximal straight segments.
• Along each straight segment, all intermediate cells have straight continuation.

(a) All clues black

Placement

Place a black clue at every turn of $C_n$.

Let $B$ be the set of all turning cells. Then

$$ |B| = 2(n-1) = O(n). $$

Verification of black rule

At a black clue cell:

• The loop must turn: true by definition of $B$.
• The loop must go straight in the cells immediately before and after the clue along the traversal.

In the serpentine cycle, each turn is between two straight segments of length $n-1$, so both neighboring cells along the path are on straight segments. Hence the local black constraint is satisfied at every clue.

Corner condition

The cycle passes through all four corners by construction.

Thus the configuration is valid.

(b) All clues white

Placement

For each turn cell $t \in B$, consider the two cells adjacent to $t$ along the cycle. These lie on straight segments.

Define $W$ as the set of all cells that are:

• adjacent along the cycle to a turn, and
• not themselves turns.

Each turn contributes at most 2 such cells, so

$$ |W| \le 4(n-1) = O(n). $$

Place a white clue at every cell in $W$.

Verification of white rule

Take any white clue cell $v \in W$.

1. Straight-through condition

By construction, $v$ lies on a straight segment between two turns of the serpentine path. Therefore the loop passes straight through $v$.

2. Adjacent-turn condition

By definition of $W$, the cell $v$ is adjacent (along the cycle) to a turn cell $t$. Thus at least one of the two neighbors of $v$ along the line is a turn.

Hence the white Masyu constraint is satisfied.

Corner condition

The loop is unchanged from part (a), so it still passes through all four corners.

Conclusion

For every $n \ge 2$, the serpentine Hamiltonian cycle $C_n$ yields:

• a valid loop visiting all four corners,
• only $O(n)$ structurally relevant turn points,

allowing:

• (a) a valid all-black Masyu instance with $O(n)$ clues by placing clues at all turns,
• (b) a valid all-white Masyu instance with $O(n)$ clues by placing clues on cells adjacent to turns.

Thus the required constructions exist for infinitely many $n$.