Total Domination number of a hypercube (Solved)

Importance: High ✭✭✭
Author(s): Adel P. Kazemi
Recomm. for undergrads: no
Posted by: Adel P. Kazemi
on: October 6th, 2013
Solved by: See comment.
Conjecture   For any integer $ n\geq 6 $, $ \gamma_t(Q_n) = 2^{n-2} $.

Here $  Q_n  $ denotes the $ n $-dimensional hypercube, i.e. the graph with vertex set $  \{0,1\}^n  $ and two vertices adjacent if they differ in exactly one coordinate. A total dominating set of a graph $ G $ is a set $ S $ of vertices of $ G $ such that every vertex has at least one neighbor in $ S $". The total domination number $ \gamma_t(G) $ of $ G $ is the cardinality of a minimum total dominating set. A total dominator coloring of a graph $  G  $, briefly TDC, is a proper coloring of $  G  $ in which each vertex of the graph is adjacent to every vertex of some color class. The total dominator chromatic number $  \chi_d^t(G)  $ of $  G  $ is the minimum number of color classes in a TDC in $  G  $ (see [Kaz1]).

The following theorems are proved in [Kaz2].

Theorem   For any integer $ n\geq 6 $, $ \gamma_t(Q_n)\leq2^{n-2} $.
Theorem   1. If $ n=1,2 $, then $ \gamma_t(Q_n)=2^n $.
2. If $ n=3 $, then $ \gamma_t(Q_n)=2^{n-1} $.
3. If $ n=4,5 $, then $ \gamma_t(Q_n)=2^{n-2} $.
Theorem   For any integer $ n\geq 1 $, $ \chi_d^t(Q_n)-2 \leq \gamma_t(Q_n) \leq \chi_d^t(Q_n) $.


[Kaz1] Adel P. Kazemi, Total dominator chromatic number of a graph,

[Kaz2] Adel P. Kazemi, Total Dominator Coloring in Product Graphs, Utilitas Mathematica (2013), Accepted.

* indicates original appearance(s) of problem.

The conjecture is false

The conjecture is false and it can be seen that this is so by using a computer program. For example $ \gamma_t(Q_6) = 14 $ which is not a power of two.

Alternatively we can argue as follows. For any graph $ G $ we clearly have $$ \gamma_t(G) \leq 2 \gamma(G).$$ Now the bound by Alon and Spencer states $$\gamma(G) \leq  \frac{1+\log({\delta(G)+1)}}{1+\delta(G)}|V(G)|.$$ In particular this implies that for any constant $ k > 0 $ and large enough $ n $ we have $$\gamma_t(Q_n) \leq 2^{n-k}.$$

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