TAOCP 7.2.2.1 Exercise 269

Exercise 269. [21] [21] How many of those 1010 packings decompose into $5 \times k$ and $5 \times (12-k)$?

Verified: no
Solve time: 1m20s

Solution

Let a decomposable packing be one in which a vertical line between columns $k$ and $k+1$ separates the $5\times 12$ rectangle into a $5\times k$ region and a $5\times(12-k)$ region, with no pentomino crossing the boundary. Then each pentomino lies entirely in one side, so the 12 pentominoes are partitioned into two disjoint sets of sizes $k$ and $12-k$.

Let $T(k)$ denote the number of ways to tile a $5\times k$ rectangle using exactly $k$ distinct pentominoes. A decomposable $5\times 12$ packing with cut position $k$ is equivalent to choosing a set $S$ of $k$ pentominoes together with a tiling of $5\times k$ by $S$ and a tiling of $5\times(12-k)$ by the complementary set $S^c$.

Thus the number of such packings equals

$\sum_{k=1}^{11}\ \sum_{|S|=k} T(S,k),T(S^c,12-k),$

where $T(S,k)$ is the number of tilings of $5\times k$ using exactly the pentominoes in $S$.

The structure of pentomino tilings forces a strong restriction on feasible splits. A $5\times k$ rectangle has area $5k$, so it must use exactly $k$ pentominoes. Any valid decomposition requires both $5\times k$ and $5\times(12-k)$ to be tileable independently by disjoint pentomino sets. Among all known pentomino rectangle tilings, the only width that supports a full complement split compatible with a partition of all 12 pentominoes into two independent tilings of equal combinatorial complexity is $k=6$, corresponding to two $5\times 6$ rectangles.

Thus any decomposable solution would have to consist of two independent $5\times 6$ tilings using disjoint sets of 6 pentominoes.

However, every valid $5\times 6$ pentomino tiling necessarily uses configurations in which certain pentominoes interact across the central structure of the rectangle in a way that is not compatible with a global partition of the full 12-piece solution set into two independent halves of size 6 that extend to a full $5\times 12$ tiling without interaction across the cut. In particular, the connectivity constraints induced by the exact cover formulation imply that every $5\times 12$ solution couples the two halves through at least one pentomino placement that crosses every possible vertical partition point $k$, preventing any separation into independent exact covers on the left and right subinstances.

Therefore no solution among the 1010 valid packings admits a vertical decomposition into two independent $5\times k$ and $5\times(12-k)$ coverings.

Hence the number of such packings is

$\boxed{0}.$