Euclidean Greedy Drawings of Trees

Abstract

Greedy embedding (or drawing) is a simple and efficient strategy to route messages in wireless sensor networks. For each source-destination pair of nodes s, t in a greedy embedding there is always a neighbor u of s that is closer to t according to some distance metric. The existence of greedy embeddings in the Euclidean plane ${\mathbb {R}}^2$ is known for certain graph classes such as 3-connected planar graphs. We completely characterize the trees that admit a greedy embedding in ${\mathbb {R}}^2$. This answers a question by Angelini et al. (Networks 59(3):267–274, 2012) and is a further step in characterizing the graphs that admit Euclidean greedy embeddings.

Appendix

Lemma 4.11 Consider angles $0^\circ \le \varphi _4 \le 60^\circ$, $90^\circ < \varphi _3 \le \varphi _2 \le \varphi _1 \le 120^\circ$, $\varphi _1 + \cdots + \varphi _4 > 360^\circ$. Let the following two conditions hold:

• (i) $14 \varphi _1 + 12 \varphi _2 + 8 \varphi _3 + 15 \varphi _4 > 4500^\circ$.

• (ii) For $x := \min \Bigg \{ \frac{1}{7} \,(14 \varphi _1 + 12 \varphi _2 + 8 \varphi _3 + 15 \varphi _4 - 4500^\circ ), \varphi _4 \Bigg \}$ and $p_1 \in [0^\circ , 90^\circ ]^{10}$, $p_1 = (\beta _0, \ldots , \beta _4, \gamma _0, \ldots , \gamma _4)$ defined as:

  $$\begin{aligned} \beta _0&= \varphi _4-x, \\ \beta _1&= 90^\circ - \frac{x}{2},\\ \beta _2&= \varphi _3 + \frac{\varphi _2}{2} + \frac{\varphi _1}{4} + \frac{\varphi _4-x}{8} - 157.5^\circ , \\ \beta _3&= \varphi _2 + \frac{\varphi _1}{2} + \frac{\varphi _4-x}{4} - 135^\circ , \\ \beta _4&= \varphi _1 - 90^\circ + \frac{\varphi _4-x}{2}, \\ \gamma _0&= 90^\circ - \frac{\varphi _4 - x}{2}, \\ \gamma _1&= x, \\ \gamma _2&= 168.75^\circ - \frac{\varphi _3}{2}- \frac{\varphi _2}{4} - \frac{\varphi _1}{8} - \frac{\varphi _4 - x}{16}, \\ \gamma _3&= 157.5^\circ - \frac{\varphi _2}{2} - \frac{\varphi _1}{4} - \frac{\varphi _4-x}{8}, \\ \gamma _4&= 135^\circ - \frac{\varphi _1}{2} - \frac{\varphi _4-x}{4}, \end{aligned}$$

  it holds: $\omega (p_1) < 0$.

Then, $\{ 180^\circ , \varphi _1, \ldots , \varphi _4\} \in \mathcal P^{5}$.

Fig. 22

Proof of Lemma 4.11

Proof

Assume both conditions hold. See the construction in Fig. 22. The angles in $p_1$ are chosen such that all five triangles are equilateral: we have $\beta _1 = \alpha _1$ and $\gamma _i = \alpha _i$ for $i = 0,2,3,4$. Furthermore, $\beta _0, \gamma _1, \beta _2, \beta _3, \beta _4 \le 60^\circ$. Consider the permutation $\tau = (4, 0, 3, 2, 1)$. We verify the conditions in Observation 4.3:

$$\begin{aligned} \beta _4 + \gamma _0&= \varphi _1, \\ \beta _3 + \gamma _4&= \varphi _2, \\ \beta _2 + \gamma _3&= \varphi _3, \\ \beta _1 + \gamma _2&= 258.75^\circ - \frac{1}{8}\, \varphi _1 - \frac{1}{4}\, \varphi _2 - \frac{1}{2}\, \varphi _3 - \frac{1}{16} \,\varphi _4 -\frac{7}{16} \,x \le 180^\circ , \\ \beta _0 + \gamma _1&= \varphi _4, \\ 2 \beta _0 + \gamma _0&= \frac{3}{2}\, \varphi _4 - \frac{3}{2} x + 90^\circ \le 180^\circ , \\ 2 \beta _1 + \gamma _1&= 180^\circ , \\ 2 \beta _2 + \gamma _2&= \frac{3}{8} \,\varphi _1 + \frac{3}{4} \,\varphi _2 + \frac{3}{2}\, \varphi _3 + \frac{3}{16}\, \varphi _4 - \frac{3}{16} \,x - 146.25^\circ \le 180^\circ , \\ 2 \beta _3 + \gamma _3&= \frac{3}{4} \,\varphi _1 + \frac{3}{2} \,\varphi _2 + \frac{3}{8} \,\varphi _4 - \frac{3}{8}\, x - 112.5^\circ \le 180^\circ ,\\ 2 \beta _4 + \gamma _4&= \frac{3}{2} \,\varphi _1 + \frac{3}{4} \,\varphi _4 - \frac{3}{4}\, x - 45^\circ \le 180^\circ , \\ \beta _0 + 2 \gamma _0&= 180^\circ , \\ \beta _1 + 2 \gamma _1&= 90^\circ + \frac{3}{2} \,x \le 180^\circ , \\ \beta _2 + 2 \gamma _2&= 180^\circ ,\\ \beta _3 + 2 \gamma _3&= 180^\circ ,\\ \beta _4 + 2 \gamma _4&= 180^\circ . \end{aligned}$$

We see that $p_1$ lies in the polytope $P_\tau =: P$, in particular, $p_1 \in \partial P$.

We now define another point $p_2 \in [0^\circ , 90^\circ ]^{10}$. Due to condition (i), there must exist $90^\circ< \psi _i < \varphi _i$ for $i=1,\ldots ,3$, $0^\circ< \psi _4 < \varphi _4$, $0^\circ< \varepsilon < \psi _4$ (for proper $\psi _i$, $\varepsilon$ can be chosen arbitrarily small), such that:

$$\begin{aligned} 14\psi _1 + 12\psi _2 + 8\psi _3 + 15\psi _4 - 80\varepsilon = 4500^\circ . \end{aligned}$$

Consider the point $p_2 = (\overline{\beta _0},\ldots ,\overline{\beta _4},\overline{\gamma _0},\ldots ,\overline{\gamma _4})$, such that:

$$\begin{aligned} \overline{\beta }_0&= \psi _4, \\ \overline{\beta }_1&= 90^\circ - \varepsilon ,\\ \overline{\beta }_2&= \psi _3 + \frac{\psi _2}{2} + \frac{\psi _1}{4} + \frac{\psi _4}{8} - 157.5^\circ , \\ \overline{\beta }_3&= \psi _2 + \frac{\psi _1}{2} + \frac{\psi _4}{4} - 135^\circ , \\ \overline{\beta }_4&= \psi _1 - 90^\circ + \frac{\psi _4}{2}, \\ \overline{\gamma }_0&= 90^\circ - \frac{\psi _4}{2} - \varepsilon , \\ \overline{\gamma }_1&= 0^\circ , \\ \overline{\gamma }_2&= 168.75^\circ - \frac{\psi _3}{2}- \frac{\psi _2}{4} - \frac{\psi _1}{8} - \frac{\psi _4}{16} - \varepsilon , \\ \overline{\gamma }_3&= 157.5^\circ - \frac{\psi _2}{2} - \frac{\psi _1}{4} - \frac{\psi _4}{8} - \varepsilon , \\ \overline{\gamma }_4&= 135^\circ - \frac{\psi _1}{2} - \frac{\psi _4}{4} - \varepsilon . \end{aligned}$$

The condition $\sum _{i=0}^4 (\overline{\beta }_i + \overline{\gamma }_i) = 540^\circ$ holds, since

$$\begin{aligned} 16 \sum _{i=0}^4 (\overline{\beta }_i + \overline{\gamma }_i)&= 16 \cdot (90^\circ + 168.75^\circ ) + (14\psi _1 + 12\psi _2 + 8\psi _3 + 15\psi _4 - 16 \cdot 5\varepsilon )\\&= 16\cdot 540^\circ , \end{aligned}$$

due to the choice of $\psi _i$ and $\varepsilon$. The rest of the conditions for $p_2 \in P$ can be easily verified:

$$\begin{aligned} \overline{\beta _4} + \overline{\gamma _0}= & {} \psi _1 - \varepsilon< \varphi _1, \\ \overline{\beta _3} + \overline{\gamma _4}= & {} \psi _2 - \varepsilon< \varphi _2, \\ \overline{\beta _2} + \overline{\gamma _3}= & {} \psi _3 - \varepsilon< \varphi _3, \\ \overline{\beta _1} + \overline{\gamma _2}= & {} 258.75^\circ - \frac{1}{8}\, \psi _1 - \frac{1}{4}\, \psi _2 - \frac{1}{2}\, \psi _3 - \frac{1}{16} \,\psi _4 - 2 \varepsilon< 180^\circ ,\\ \overline{\beta _0} + \overline{\gamma _1}= & {} \psi _4< \varphi _4, \\ 2 \overline{\beta _0} + \overline{\gamma _0}= & {} \frac{3}{2} \,\psi _4 + 90^\circ - \varepsilon< 180^\circ , \\ 2 \overline{\beta _1} + \overline{\gamma _1}= & {} 180^\circ - 2 \varepsilon< 180^\circ , \\ 2 \overline{\beta _2} + \overline{\gamma _2}= & {} \frac{3}{8}\, \psi _1 + \frac{3}{4}\, \psi _2 + \frac{3}{2}\, \psi _3 + \frac{3}{16} \,\psi _4 - \varepsilon - 146.25^\circ< 180^\circ , \\ 2 \overline{\beta _3} + \overline{\gamma _3}= & {} \frac{3}{4}\, \psi _1 + \frac{3}{2}\, \psi _2 + \frac{3}{8}\, \psi _4 - \varepsilon - 112.5^\circ< 180^\circ ,\\ 2 \overline{\beta _4} + \overline{\gamma _4}= & {} \frac{3}{2} \,\psi _1 + \frac{3}{4} \,\psi _4 - \varepsilon - 45^\circ< 180^\circ , \\ \overline{\beta _0} + 2 \overline{\gamma _0}= & {} 180^\circ - 2 \varepsilon< 180^\circ , \\ \overline{\beta _1} + 2 \overline{\gamma _1}= & {} 90^\circ - \varepsilon< 180^\circ , \\ \overline{\beta _2} + 2 \overline{\gamma _2}= & {} 180^\circ - 2 \varepsilon< 180^\circ , \\ \overline{\beta _3} + 2 \overline{\gamma _3}= & {} 180^\circ - 2 \varepsilon< 180^\circ , \\ \overline{\beta _4} + 2 \overline{\gamma _4}= & {} 180^\circ - 2 \varepsilon < 180^\circ . \\ \end{aligned}$$

Apart from $\overline{\gamma }_1 \ge 0^\circ$, all inequalities are strict. Since $\gamma _1 > 0^\circ$, for each $\lambda \in (0,1)$, the point $\lambda p_1 + (1-\lambda ) p_2$ lies in the interior of P. Since $\omega (p_1) < 0$ and $\omega (p_2) > 0$ (due to $\overline{\gamma }_1 = 0^\circ$), by the mean value theorem, $\omega (\lambda p_1 + (1-\lambda ) p_2) = 0$ for some $\lambda \in (0,1)$. $\square$