## Bijective Rotations on Hexagonal Lattice

by K. Pluta, T. Roussillon, D. Cœurjolly,

P. Romon, Y. Kenmochi, V. Ostromoukhov

21/06/2018, JIG 2018, Lyon

## Motivations

Digitized rotations on the square grid are burdened with an incompatibility between rotations and the geometry of the grid.

## Agenda

• Introduction to the Bees' Point of View

• Quick Introduction to Rigid Motions

• Bijective Digitized Rotations

• Contributions

• Conclusions & Perspectives
The beehive figure's source and author unknown (if you recognize it, please let me know). The image of the bee comes from http://karenswhimsy.com/public-domain-images (public domain)

## Contributions in Short

• Characterization of bijective digitized rotations on the regular hexagonal lattice

• Comparison of frequencies of such rotations on the square and the regular hexagonal lattices

Pure, extracted honey

## Related Studies

• Pluta, K., Romon, P., Kenmochi, Y., Passat, N.: Bijective Digitized Rigid Motions on Subsets of the Plane. Journal of Mathematical Imaging and Vision (2017)

• Pluta, K., Romon, P., Kenmochi, Y., Passat, N.: Bijectivity Certification of 3D Digitized Rotations. CTIC (2016)

• Roussillon, T., Cœurjolly, D.: Characterization of Bijective Discretized Rotations by Gaussian Integers. Research report, LIRIS UMR CNRS 5205 (2016). URL: https://hal.archives-ouvertes.fr/hal-01259826

## Introduction to the Bees' Point of View

The figure comes from http://thegraphicsfairy.com/vintage-clip-art-bees-with-honeycomb

## The Lattices

\mathbb{Z}[\omega] = \mathbb{Z} \oplus \mathbb{Z}\omega, \omega = -\frac{1}{2} + \frac{\sqrt{3}}{2} i
$\mathbb{Z}[\omega] = \mathbb{Z} \oplus \mathbb{Z}\omega, \omega = -\frac{1}{2} + \frac{\sqrt{3}}{2} i$
\mathbb{Z}[i] = \mathbb{Z} \oplus \mathbb{Z}i
$\mathbb{Z}[i] = \mathbb{Z} \oplus \mathbb{Z}i$

Eisenstein:

and Gaussian:

integers

## Properties

Property Gaussian integers Eisenstein integers
conjugate
squared modulus
units
\bar{\mathbf{x}} = a - b i
$\bar{\mathbf{x}} = a - b i$
\bar{\mathbf{x}} = (a - b) - b \omega
$\bar{\mathbf{x}} = (a - b) - b \omega$
| \mathbf{x} | = \mathbf{x} \cdot \bar{\mathbf{x}} = a^2 + b^2
$| \mathbf{x} | = \mathbf{x} \cdot \bar{\mathbf{x}} = a^2 + b^2$
| \mathbf{x} | = \mathbf{x} \cdot \bar{\mathbf{x}} = a^2 - a b + b^2
$| \mathbf{x} | = \mathbf{x} \cdot \bar{\mathbf{x}} = a^2 - a b + b^2$
\Upsilon = \{ \pm 1, \pm i \}
$\Upsilon = \{ \pm 1, \pm i \}$
\Upsilon = \{ \pm 1, \pm \omega, \pm \bar{\omega} \}
$\Upsilon = \{ \pm 1, \pm \omega, \pm \bar{\omega} \}$
 divisibility
\mathbf{y} | \mathbf{x} \text{ if } \exists \mathbf{z} \in \mathbb{Z}[\varkappa] \text{ such that } \mathbf{x} = \mathbf{y} \cdot \mathbf{z} \text{ and } \mathbf{y} \in \mathbb{Z}[\varkappa]
$\mathbf{y} | \mathbf{x} \text{ if } \exists \mathbf{z} \in \mathbb{Z}[\varkappa] \text{ such that } \mathbf{x} = \mathbf{y} \cdot \mathbf{z} \text{ and } \mathbf{y} \in \mathbb{Z}[\varkappa]$
 greatest common divisor
\gcd(\mathbf{x}, \mathbf{y}) = \mathbf{z} \in \mathbb{Z}[\varkappa]\text{ is defined as a largest }
$\gcd(\mathbf{x}, \mathbf{y}) = \mathbf{z} \in \mathbb{Z}[\varkappa]\text{ is defined as a largest }$
\mathbf{z} \in \mathbb{Z}[\varkappa]\text{ (up to multiplications by units) which }
$\mathbf{z} \in \mathbb{Z}[\varkappa]\text{ (up to multiplications by units) which }$
\text{divides both } \mathbf{x}\text{ and }\mathbf{y}
$\text{divides both } \mathbf{x}\text{ and }\mathbf{y}$
\text{Giving } a, b \in \mathbb{Z}\text{ and }\mathbf{x},\mathbf{y}, \in \mathbb{Z}[\varkappa], \varkappa \in \{i, \omega\}
$\text{Giving } a, b \in \mathbb{Z}\text{ and }\mathbf{x},\mathbf{y}, \in \mathbb{Z}[\varkappa], \varkappa \in \{i, \omega\}$

## Digitization Model

A hexagonal cell of

\kappa
$\kappa$

denoted by

\mathcal{C}_{\zeta}(\kappa),
$\mathcal{C}_{\zeta}(\kappa),$
\kappa\in \mathbb{Z}[\omega], \zeta \in \mathbb{C}.
$\kappa\in \mathbb{Z}[\omega], \zeta \in \mathbb{C}.$
\kappa
$\kappa$

## Digitization Model

The digitization operator is defined as

\mathcal{D}: \mathbb{C} \rightarrow \mathbb{Z}[\omega]
$\mathcal{D}: \mathbb{C} \rightarrow \mathbb{Z}[\omega]$

such that

\forall \mathbf{x} \in \mathbb{C}, \exists! \mathcal{D}(\mathbf{x}) \in \mathbb{Z}[\omega]
$\forall \mathbf{x} \in \mathbb{C}, \exists! \mathcal{D}(\mathbf{x}) \in \mathbb{Z}[\omega]$
\mathbf{x} \in \mathcal{C}_1(\mathcal{D}(\mathbf{x})).
$\mathbf{x} \in \mathcal{C}_1(\mathcal{D}(\mathbf{x})).$

and

\mathcal{D}(\mathbf{x})
$\mathcal{D}(\mathbf{x})$

## Quick Lesson on Rigid Motions

Or how to become a beekeeper. Part I - Equipment

The figure comes from Wikimedia. Original source  The New Student's Reference Work (public domain)

\left| \begin{array}{lllll} \mathcal{U} & : & \mathbb{C} & \to & \mathbb{C} \\ & & \mathbf{x} & \mapsto & \mathbf{a} \cdot \mathbf{x} \end{array} \right.
$\left| \begin{array}{lllll} \mathcal{U} & : & \mathbb{C} & \to & \mathbb{C} \\ & & \mathbf{x} & \mapsto & \mathbf{a} \cdot \mathbf{x} \end{array} \right.$

## Rotations on

\mathbb{C}
$\mathbb{C}$

## Properties

• Do preserve distances
• Bijective
\mathbf{a}
$\mathbf{a}$

– a unit modulus complex number

\theta = \arg(\mathbf{a})
$\theta = \arg(\mathbf{a})$

– a rotation angle

U = \mathcal{D}\ \circ \mathcal{U}_{|\mathbb{Z}[\omega]}
$U = \mathcal{D}\ \circ \mathcal{U}_{|\mathbb{Z}[\omega]}$
\mathbb{Z}[\omega]
$\mathbb{Z}[\omega]$

## Properties

- Non-injective

- Non-surjective

- Do not preserve distances

\mathcal{U}(\mathbb{Z}[\omega])
$\mathcal{U}(\mathbb{Z}[\omega])$

## Rigid Motions on

A digitized rotation is bijective if and only if

\forall \mathbf{p} \in \mathbb{Z}[\omega] \; \exists ! \mathbf{q} \in \mathbb{Z}[\omega]
$\forall \mathbf{p} \in \mathbb{Z}[\omega] \; \exists ! \mathbf{q} \in \mathbb{Z}[\omega]$

such that

\mathcal{U}(\mathbf{q}) \in \mathcal{C}_1(\mathbf{p}).
$\mathcal{U}(\mathbf{q}) \in \mathcal{C}_1(\mathbf{p}).$
\mathcal{U}
$\mathcal{U}$

- a continuous rotation

\mathcal{C}_{\mathbf{a}}(\mathbf{p})
$\mathcal{C}_{\mathbf{a}}(\mathbf{p})$

- a digitization cell centered at

\mathbf{p}
$\mathbf{p}$
\mathbb{Z}[\omega]
$\mathbb{Z}[\omega]$

- the hexagonal lattice

## Conditions for Bijectivity

and multiplied by

\mathbf{a}
$\mathbf{a}$

## Bijective Digitized Rotations on Hexagonal Lattice

The figure comes from Wikimedia. The original source The honey bee: a manual of instruction in apiculture (public domain)

The figure of bumble bee comes from http://www.ase.org.uk (public domain), The bee figure by Pearson Scott Foresman, Wikimedia.

## Set of Remainders

\forall \mathbf{p} \in \mathbb{Z}[\omega]\ \exists! \mathbf{q} \in \mathbb{Z}[\omega], S_{\mathbf{a}}(\mathbf{p},\mathbf{q}) \in \mathcal{C}_1(0)
$\forall \mathbf{p} \in \mathbb{Z}[\omega]\ \exists! \mathbf{q} \in \mathbb{Z}[\omega], S_{\mathbf{a}}(\mathbf{p},\mathbf{q}) \in \mathcal{C}_1(0)$

A 2D digitized rotation is then bijective when

## Double Surjectivity

The double surjectivity condition is then

provided that

S_{\mathbf{a}}(\mathbb{Z}[\omega], \mathbb{Z}[\omega]) \cap \mathcal{C}_1(0) = S_{\mathbf{a}}(\mathbb{Z}[\omega], \mathbb{Z}[\omega]) \cap \mathcal{C}_{\frac{\mathbf{a}}{|\mathbf{a}|}}(0).
$S_{\mathbf{a}}(\mathbb{Z}[\omega], \mathbb{Z}[\omega]) \cap \mathcal{C}_1(0) = S_{\mathbf{a}}(\mathbb{Z}[\omega], \mathbb{Z}[\omega]) \cap \mathcal{C}_{\frac{\mathbf{a}}{|\mathbf{a}|}}(0).$

## Primitive Eisenstein Integers

s, t \in \mathbb{Z}
$s, t \in \mathbb{Z}$
0 < s < t
$0 < s < t$

and

t - s
$t - s$

not being divisible by 3.

a = s^2 + 2 s t, b = t^2 + 2 s t
$a = s^2 + 2 s t, b = t^2 + 2 s t$

and

c = s^2 + t^2 + s t,
$c = s^2 + t^2 + s t,$

such that

## Set of Remainders

The double surjectivity condition is then

provided that

\check{S}_{\mathbf{a}}(\mathbb{Z}[\omega], \mathbb{Z}[\omega]) \cap \mathcal{C}_{\mathbf{a}}(0) = \check{S}_{\mathbf{a}}(\mathbb{Z}[\omega], \mathbb{Z}[\omega]) \cap \mathcal{C}_{|\mathbf{a}|}(0).
$\check{S}_{\mathbf{a}}(\mathbb{Z}[\omega], \mathbb{Z}[\omega]) \cap \mathcal{C}_{\mathbf{a}}(0) = \check{S}_{\mathbf{a}}(\mathbb{Z}[\omega], \mathbb{Z}[\omega]) \cap \mathcal{C}_{|\mathbf{a}|}(0).$

## Set of Remainders

The double surjectivity condition is then

provided that

## Proving Bijectivity

For which s and t,

or

are in the green but not in the

black hexagonal cell?

## Proving Bijectivity

From the equation of the line passing through the vertices of the green line-segment  we obtain:

## Proving Bijectivity

Then, we substitute t with s + e to arrive at

which are violated when s = 1 or when s > 1 and e = 1.

From the equation of the line passing through the vertices of the green line-segment  we obtain:

## Proving Bijectivity

At the end we have to check if

\bar{\gamma}+ \omega
$\bar{\gamma}+ \omega$

or

\bar{\gamma}-1
$\bar{\gamma}-1$

are members

of the lattice that spans the values of

S'_{\mathbf{a}}.
$S'_{\mathbf{a}}.$

## Proving Bijectivity

Step 2: Check if for s = 1 or s > 1 and e = 1, the uncommon space of the hexagonal cells does not contain Eisenstein integers.

s = 1
$s = 1$
s > 1, t = s + 1
$s > 1, t = s + 1$

## Proving Bijectivity

The Square Grid

The Hexagonal Grid

## Conclusion & Perspectives

Or extracting the pure, organic honey

The figure comes from Wikimedia. The original comes from A practical treatise on the hive and honey-bee (public domain)

• Characterization of bijective digitized rotations on the regular hexagonal lattice
• Comparison of frequencies of such rotations on the square and the regular hexagonal lattices

Conclusion

Perspectives

Investigate bijectivity of digitized rigid motions on finite subsets of Eisenstein integers.

By Kacper Pluta

# Bijective Rotations on Hexagonal Lattice

Presentation created for JIG 18 in Lyon.

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