Dado un entero gaussiano $a+bi$ donde $a$ , $b$ son enteros e $i = \exp\left(\pi i/2\right)$ es la unidad imaginaria, devuelve el entero más cercano (wrt a la distancia euclidiana) Eisenstein entero $k+l\omega$ donde $k$ , $l$ son enteros y $\omega = \exp(2\pi i/3) = (-1+i\sqrt{3})/2$ .

Antecedentes

Probablemente sea bastante obvio que cada entero gaussiano se puede escribir de forma única como $a+bi$ con $a$ , $b$ enteros. No es tan obvio pero sí cierto: cualquier entero de Eisenstein puede escribirse únicamente como $k+l\omega$ con $k$ , $l$ enteros. Ambos forman un módulo $\mathbb{Z}$ dentro de los números complejos, y ambos son números enteros ciclotómicos p-th para $p=2$ o $3$ respectivamente. Tenga en cuenta que $3+2i \neq 3+2\omega$

_{^{Fuente: commons.wikimedia.org}}

Detalles

• En caso de que el número complejo dado tenga dos o tres puntos más cercanos, se puede devolver cualquiera de ellos.

• El número complejo se da en coordenadas rectangulares (base $(1,i)$ ), pero aparte de eso en cualquier formato conveniente como (A,B)o A+Bio A+B*1jetc.

• El número entero Eisenstein tiene que ser devuelto como coordenadas de la base $(1,\omega)$ pero aparte de eso en cualquier formato conveniente como (K,L)o K+Lωo K+L*1ωetc.

Ejemplos

Obviamente, todos los enteros reales deberían asignarse a los enteros reales nuevamente.

  6,14 -> 14,16
  7,16 -> 16,18
-18,-2 ->-19,-2
 -2, 2 -> -1, 2
 -1, 3 -> 1, 4

APL (Dyalog Extended) , ^{SBCS de} 16 bytes

⎕0+⌈3÷⍨1 2×⌊⎕×√3

Pruébalo en línea!

Un programa completo que se lleva a ycontinuación xde la entrada estándar e imprime un vector de 2 elementos de números enteros.

Cómo funciona: las matemáticas

En primer lugar, tenga en cuenta que cualquier número entero gaussiano se colocará en la diagonal vertical de un diamante, con el punto $Z$ colocado en $(x,\sqrt3y)$para algún número entero$x,y$.

      + W
     /|\
    / | \
   /  |  \
  /   + X \
 /    |    \
+-----|-----+V
 \    |    /
  \   + Y /
   \  |  /
    \ | /
     \|/
      + Z

En la figura, $\overline{WZ}=\sqrt3$ y$\overline{WX}=\overline{XY}=\overline{YZ}=\overline{XV}=\overline{YV}=\frac{1}{\sqrt3}$ . Entonces, dada la posición vertical de un punto, podemos identificar el punto de Eisenstein más cercano de la siguiente manera:

$$\text{Given a point }P \in \overline{WZ}, \\ \left\{ \begin{array}{c} P \in \overline{WX} \implies \text{the nearest point is } W \\ P \in \overline{XY} \implies \text{the nearest point is } V \\ P \in \overline{YZ} \implies \text{the nearest point is } Z \end{array} \right.$$

$P$$P$$h$$Z$$x$

$$h = \lfloor P.y \div \sqrt3 \rfloor$$

Then the Eisenstein coordinates of $Z$ are

$$Z.x_E = P.x+h, \quad Z.y_E = 2h$$

Now, we determine which of the segments $\overline{WX},\overline{XY},\overline{YZ}$ $P$ belongs to. For this, we can calculate the indicator $w$ as follows:

$$w = \lfloor P.y \times \sqrt3 \rfloor \% 3$$

Then the cases $w = 0, 1, 2$ correspond to $\overline{YZ},\overline{XY},\overline{WX}$ respectively. Finally, the nearest Eisenstein point of $P$ (which is one of $Z$, $V$, or $X$) can be calculated as:

$$P_E.x_E = P.x+h+\lceil \frac{w}2 \rceil, \quad P_E.y_E = 2h+w$$

Using the identities for $h$ and $w$, we can further simplify to:

$$y' = \lfloor P.y \times \sqrt3 \rfloor, \quad P_E.x_E = P.x+\lceil y' \div 3 \rceil, \quad P_E.y_E = \lceil 2y' \div 3 \rceil$$

How it works: the code

⎕0+⌈3÷⍨1 2×⌊⎕×√3
           ⌊⎕×√3  ⍝ Take the first input (P.y) and calculate y'
   ⌈3÷⍨1 2×       ⍝ Calculate [ceil(y'/3), ceil(2y'/3)]
⎕0+  ⍝ Take the second input(P.x) and calculate [P.x+ceil(y'/3), ceil(2y'/3)]

JavaScript (ES6), 112 bytes

(a,b,l=b/Math.pow(.75,.5),k=a+l/2,f=Math.floor,x=k-(k=f(k)),y=l-(l=f(l)),z=x+y>1)=>[k+(y+y+z>x+1),l+(x+x+z>y+1)]

ES7 can obviously trim 9 bytes. Explanation: k and l initially represent the floating-point solution to k+ωl=a+ib. However, the coordinates needed to be rounded to the nearest integer by Euclidean distance. I therefore take the floor of k and l, then perform some tests on the fractional parts to determine whether incrementing them would result in a nearer point to a+ib.

MATL, 39 38 35 bytes

t|Ekt_w&:2Z^tl2jYP3/*Zeh*!sbw6#YkY)

Input format is 6 + 14*1j (space is optional). Output format is 14 16.

Try it online!

Explanation

The code first takes the input as a complex number. It then generates a big enough hexagonal grid in the complex plane, finds the point that is closest to the input, and returns its Eisenstein "coordinates".

t         % Take input implicitly. This is the Gauss number, say A. Duplicate
|Ek       % Absolute value times two, rounded down
t_        % Duplicate and negate
w&:       % Range. This is one axis of Eisenstein coordinates. This will generate
          % the hexagonal grid big enough
2Z^       % Cartesian power with exponent 2. This gives 2-col 2D array, say B
t         % Duplicate
l         % Push 1
2jYP3/*   % Push 2*j*pi/3
Ze        % Exponential
h         % Concatenate. Gives [1, exp(2*j*pi/3)]
*         % Multiply by B, with broadcast.
!s        % Sum of each row. This is the hexagonal grid as a flattened array, say C
bw        % Bubble up, swap. Stack contains now, bottom to top: B, A, C
6#Yk      % Index of number in C that is closest to A
Y)        % Use as row index into B. Implicitly display

Haskell, 128 bytes

i=fromIntegral;r=[floor,ceiling];a!k=(i a-k)**2;c(a,b)|l<-2*i b/sqrt 3,k<-i a+l/2=snd$minimum[(x k!k+y l!l,(x k,y l))|x<-r,y<-r]

Try it online!

For input Gaussian integer (a,b), convert it into Eisenstein coordinates, floor and ceil both components to get four candidates for closest Eisenstein integer, find the one with minimal distance and return it.

Tcl, 124 116 106 bytes

{{a b f\ int(floor(2*$b/3**.5)) {l "[expr $f+(1-$f%2<($b-$f)*3**.5)]"}} {subst [expr $l+$a-($f+1)/2]\ $l}}

Try it online!

This is somewhat inspired by the three-year old post from @Neil

The floor function returns the corner of the rhombus whose edges are the vectors 1 and $\omega$. With respect to this rhombus, the Gaussian integer lies on the perpendicular bi-sector of either the top (if l is even) or bottom (if l is odd). This is important because it means that either the lower left corner or the upper right corner will be an acceptable solution. I compute k for the lower left corner, and do one test to see if the Gaussian integer lies above or below the diagonal separating the two corners; I add 1 to k when above the diagonal, and I do likewise for l.

Saved 10 bytes by using the "sign of the cross-product v x d of the diagonal d with the vector v joining the lower right corner and (a,b)" as the test for which side of the diagonal the point lies.

Burlesque, 24 bytes

pe@3r@2././J2./x/.+CL)R_

Try it online!

Pretty sure this can be shorter. Input read as a b

pe      # Parse input to two ints
@3r@2./ # sqrt(3)/2
./      # Divide b by sqrt(3)/2
J2./    # Duplicate and divide by 2
x/.+    # swap stack around and add to a
CL      # Collect the stack to a list
)R_     # Round to ints

05AB1E, 13 bytes

Port of Bubbler's APL answer

3t*ïx‚3/îR΂+

Try it online!

Input and output are both y first, x second.