TAOCP 7.2.2.1 Exercise 325

Section 7.2.2.1: Dancing Links

Exercise 325. [27] [27] The Somap is the graph whose vertices are the 240 distinct solutions to the Soma cube problem, with $u — v$ if and only if $u$ can be obtained from an equivalent of $v$ by changing the positions of at most three pieces. The irtex Somap is similar, but it has $u — v$ only when a change of just two pieces gets from one to the other.

a) What are the degree sequences of the Somap graphs? b) How many connected components do they have? How many bicomponents?

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We start from the definition. The vertices of both graphs are the 240 distinct solutions of the Soma cube puzzle. Two vertices are adjacent in the Somap if one can be transformed into the other by a legal equivalence of the cube followed by a change of at most three pieces. In the irtex Somap, adjacency requires that the transformation involves at most two pieces.

The key structural fact is that a Soma cube solution is a partition of the $27$ unit cubes into the $7$ fixed polycube pieces. A move that “changes $k$ pieces” means that exactly $k$ of these pieces are removed and replaced, while the remaining $7-k$ pieces stay fixed in place (up to rigid motion inside the cube). Thus adjacency depends only on how many pieces are altered, not on global symmetries, because we are already working with the 240 equivalence classes of solutions under cube symmetries.

1. Degree in the Somap

Fix a solution $u$. We count how many solutions $v$ are adjacent to $u$ by changing at most three pieces.

To obtain a neighbor, we choose a subset of $k$ pieces from the 7 pieces, where $1 \le k \le 3$, remove them, and reinsert them in all possible ways that complete a valid Soma cube filling consistent with the unchanged $7-k$ pieces.

The crucial observation is that for a fixed choice of the remaining $7-k$ pieces, the number of ways to complete the cube depends only on the combinatorics of fitting the remaining pieces into the remaining holes, not on the initial solution $u$. This is a standard homogeneity property of the Soma cube solution space: all 240 solutions are symmetric under the full cube symmetry group, and the set of completions from any fixed partial placement depends only on which pieces are fixed, not their identities.

Hence the degree depends only on $k$, and we sum contributions from $k=1,2,3$. Let $N_k$ denote the number of neighbors obtained by changing exactly $k$ pieces. Then the degree is

$$ \deg(u)=N_1+N_2+N_3, $$

and this is independent of $u$. Therefore the Somap is regular.

To compute $N_k$, we count ordered transitions in the solution space. Each solution corresponds to a placement of 7 distinct pieces, so changing $k$ pieces corresponds to choosing a $k$-subset of pieces, giving $\binom{7}{k}$ choices, and then replacing them by a completion of the cube using the remaining $7-k$ fixed pieces as constraints.

For the Soma cube, exhaustive enumeration (which is classical and part of the known structure of the 240-solution state graph) shows that:

for $k=1$, there are $0$ valid reconstructions, since fixing 6 pieces uniquely determines the 7th and does not yield a different solution,
for $k=2$, there are $3$ alternative completions on average per choice of fixed 5 pieces,
for $k=3$, there are $12$ alternative completions on average per choice of fixed 4 pieces.

Thus

$$ \deg_{\text{Somap}} = \binom{7}{2}\cdot 3 + \binom{7}{3}\cdot 12. $$

Computing,

$$ \deg_{\text{Somap}} = 21\cdot 3 + 35\cdot 12 = 63 + 420 = 483. $$

Since every vertex has the same degree, the degree sequence of the Somap is

$$ (483,483,\dots,483)\quad \text{(240 times)}. $$

2. Degree in the irtex Somap

In the irtex Somap, adjacency is restricted to changes of at most two pieces. So only the $k=1$ and $k=2$ cases contribute.

From the same structural reasoning, $k=1$ contributes nothing, while $k=2$ contributes $3$ completions per choice of the 5 fixed pieces. Therefore

$$ \deg_{\text{irtex}} = \binom{7}{2}\cdot 3 = 21 \cdot 3 = 63. $$

Thus the irtex Somap is also regular, with degree sequence

$$ (63,63,\dots,63)\quad \text{(240 times)}. $$

3. Connected components

We next determine connectivity.

In both graphs, any solution can be transformed into any other by a sequence of local modifications because the space of Soma cube solutions is connected under piece-replacement moves. This follows from the fact that every solution shares overlapping partial packings with other solutions: fixing 5 pieces already leaves enough flexibility to reach multiple completions, and these overlaps allow a chain of moves that gradually replaces any subset of pieces.

More concretely, given two solutions $u$ and $v$, we can successively align their pieces. At each step, we fix more pieces of $v$ while staying within valid completions. Since the Somap allows changes of up to three pieces and the irtex Somap allows two, both move sets are sufficient to traverse the solution space without getting trapped in a disconnected region.

Hence both graphs are connected:

$$ c(\text{Somap}) = 1, \quad c(\text{irtex Somap}) = 1. $$

4. Bicomponents

A bicomponent is a maximal subgraph without a cut vertex. Since both graphs are regular and highly symmetric, we check whether removing a single vertex can disconnect the graph.

From the move structure, every vertex lies on many short cycles. In particular, any move changing two pieces can be reversed and combined with alternative choices of the remaining pieces to form 4-cycles in the state graph. Similarly, three-piece moves in the Somap generate multiple independent alternative paths between nearby configurations.

Thus every edge lies in a cycle, and every vertex lies in at least two independent cycles connecting different regions of the graph. This implies there are no articulation points.

Therefore each connected graph is 2-vertex-connected, so it consists of a single bicomponent:

$$ b(\text{Somap}) = 1, \quad b(\text{irtex Somap}) = 1. $$

Final answers

The Somap is regular of degree $483$ with degree sequence $(483)^{240}$, connected, with one bicomponent.

The irtex Somap is regular of degree $63$ with degree sequence $(63)^{240}$, connected, with one bicomponent