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Pick's theorem

Pick's theorem is a fundamental result in discrete geometry that expresses the area of a simple polygon with vertices at integer coordinates (a lattice polygon) in terms of the number of lattice points strictly inside the polygon and the number of lattice points on its boundary.^[1] The theorem states that the area A satisfies A = I + \frac{B}{2} - 1, where I is the number of interior lattice points and B is the number of boundary lattice points.^[1] This formula reveals that the area of such polygons is always a multiple of \frac{1}{2}, providing a discrete analog to continuous area calculations.^[2] Named after the Austrian mathematician Georg Alexander Pick (1859–1942), the theorem was first published in 1899 in his paper "Geometrisches zur Zahlentheorie," appearing in the Sitzungsberichte des Lotos in Prague.^[2] Pick, a professor at the German University of Prague, contributed to various fields including topology and number theory, but this result connects geometric properties of lattice figures to arithmetic counts.^[2] Although formulated over a century ago, the theorem remained relatively obscure until it was popularized in 1950 by Hugo Steinhaus in the English edition of his book Mathematical Snapshots, which highlighted its intuitive appeal and applications.^[3] Pick's theorem has broad implications in lattice point geometry and Ehrhart theory, where it serves as a foundational tool for counting lattice points in polytopes and understanding their volumes.^[1] It applies to convex polygons and can be extended to non-convex ones without self-intersections, as well as to regions with holes via adjustments to the formula.^[2] The theorem is particularly useful in computational geometry for verifying areas without coordinate-based integration and in educational contexts, such as geoboards, to explore the interplay between discrete points and continuous measures.^[2]

Introduction and Statement

Historical Context

Georg Alexander Pick, an Austrian mathematician born on August 10, 1859, in Vienna, discovered the theorem that bears his name in 1899 while investigating areas of lattice polygons.^[4] He published his findings in the paper "Geometrisches zur Zahlenlehre" (Geometric Number Theory), appearing in the Sitzungsberichte des Lotos in Prag, volume 47, pages 311–319.^[4] This work addressed the determination of polygonal areas using integer lattice points, a topic emerging in late 19th-century discrete geometry amid growing interest in lattice point enumeration and number-theoretic applications to geometric figures.^[5] Pick's academic career centered at the German University of Prague (now Charles University), where he served as an assistant to physicist Ernst Mach starting in 1880, became a lecturer in 1881, and advanced to full professor of mathematics in 1892, retiring in 1927.^[4] Over his tenure, he supervised around 20 doctoral students, including the notable mathematician Charles Loewner, and contributed broadly to fields like linear algebra, geometry, and complex analysis, authoring 67 papers.^[4] Despite these accomplishments, Pick's 1899 theorem received little attention during his lifetime, overshadowed in the evolving landscape of geometric number theory.^[5] The theorem's significance was largely overlooked until the mid-20th century, when Polish mathematician Hugo Steinhaus revived it in his popular book Mathematical Snapshots (English edition, 1969), bringing it to wider international recognition.^[4] Pick, who died on July 26, 1942, in the Theresienstadt concentration camp during the Nazi occupation, thus saw his key geometric contribution gain prominence only posthumously.^[4]

Mathematical Statement

Pick's theorem relates the area of a simple lattice polygon to the number of lattice points lying on its boundary and in its interior. A lattice polygon is defined as a polygon whose vertices lie on points with integer coordinates in the Euclidean plane. The theorem assumes the polygon is simple, meaning it is non-self-intersecting, and has a closed boundary that forms a Jordan curve. Let P denote such a lattice polygon. The interior lattice points of P, denoted by I, are the lattice points that lie strictly inside the polygon. The boundary lattice points of P, denoted by B, are the lattice points that lie on the edges of the polygon, including the vertices. The area A(P) of P is then given by the formula

A(P) = I + \frac{B}{2} - 1.

This relation was first established by Georg Alexander Pick in his 1899 paper.^[1]^[6] Geometrically, the term I accounts for the full unit area contributions from lattice points entirely enclosed within the polygon, each corresponding to a complete unit square in the lattice. The term \frac{B}{2} adjusts for the partial area contributions along the boundary, where each boundary lattice point effectively adds a half-unit area due to the linear interpolation between vertices on lattice edges. The subtractive term -1 provides a topological correction that ensures the formula accurately reflects the enclosed area for a closed polygonal region.^[1]

Proofs

Proof Using Euler's Polyhedral Formula

One approach to proving Pick's theorem utilizes Euler's polyhedral formula, V - E + F = 2, applied to the planar graph formed by a triangulation of the lattice polygon. Consider a simple lattice polygon P with interior lattice points I and boundary lattice points B. Triangulate the interior of P by drawing non-intersecting diagonals between existing lattice points, ensuring that all interior lattice points are vertices and that the resulting triangles are primitive—meaning each has no lattice points in its interior or on its edges other than its three vertices, and thus has area \frac{1}{2}.^[7] This triangulation yields a connected planar graph G embedded in the plane, where the vertices of G consist of all lattice points inside and on the boundary of P, so V = I + B. The edges of G include the boundary edges (divided into primitive segments between consecutive boundary lattice points, totaling B such edges) and the internal edges (diagonals and edges of the triangles). Let t denote the number of triangular faces (internal to P), and let F = t + 1 account for these triangles plus the unbounded exterior face.^[7] To relate these quantities, count the edges via the faces: each of the t triangles has three edges, each internal edge is shared by two triangles, and each of the B boundary edges is incident to one triangle. This gives the relation $3t = 2(E - B) + B = 2E - B, so t = \frac{2E - B}{3}.^[7] Apply Euler's formula to G: V - E + F = 2, substituting F = t + 1 yields I + B - E + t + 1 = 2, or t = E - I - B + 1. Equating the expressions for t gives \frac{2E - B}{3} = E - I - B + 1. Multiplying through by 3 produces $2E - B = 3E - 3I - 3B + 3, rearranging terms yields -E + 2B - 3 = -3I, or E = 3I + 2B - 3 .^[7] Substitute back to find t = (3I + 2B - 3) - I - B + 1 = 2I + B - 2. Since the area A of P is the sum of the areas of the t primitive triangles, A = t \cdot \frac{1}{2} = I + \frac{B}{2} - 1.^[7] This proof assumes P is a simple polygon (no self-intersections or holes) with vertices on lattice points, and that a triangulation incorporating all interior lattice points into primitive triangles exists; such a triangulation is always possible for simple lattice polygons. The result holds under these conditions but may require modification for polygons with holes, where the Euler characteristic adjusts to account for additional components.^[7]

Proofs via Double Counting and Induction

One approach to proving Pick's theorem employs double counting in the context of a triangulation of the lattice polygon into elementary lattice triangles, focusing on the total angle sum to determine the number of such triangles. Consider a simple lattice polygon P triangulated into N elementary triangles, where each elementary triangle has no interior lattice points and an area of \frac{1}{2}. The total interior angle sum of P is N \pi, since each triangle contributes \pi radians. At each interior lattice point, the surrounding angles sum to $2\pi, while at each boundary lattice point (except the vertices), they sum to \pi; however, the B boundary points collectively contribute B \pi, but the polygon's exterior requires adjusting for the vertices by subtracting $2\pi to account for the full turn around the boundary. Thus, N \pi = 2\pi I + \pi B - 2\pi, simplifying to N = 2I + B - 2. The area A of P is then A = N \cdot \frac{1}{2} = I + \frac{1}{2} B - 1.^[8] A key component supporting this double counting is the lemma that the area of an elementary lattice triangle (with no interior points) is \frac{1}{2}. To establish this, note that any such triangle can be shown to have exactly three boundary lattice points (its vertices), and further analysis via the shoelace formula or parallelogram decomposition confirms the area as \frac{1}{2}. More generally, for a lattice triangle with vertices at integer coordinates, the area is \frac{1}{2} |ad - bc| where the vertices are (0,0), (a,b), and (c,d), and if there are no interior points, this evaluates to \frac{1}{2}. Additionally, the number of lattice points on a boundary edge from (x_1, y_1) to (x_2, y_2) is \gcd(|x_2 - x_1|, |y_2 - y_1|) + 1, which links boundary contributions to the overall count without interior points in primitive cases.^[9]^[10] An inductive proof of Pick's theorem proceeds by first verifying the base case for lattice triangles and then extending to polygons via triangulation, updating the interior and boundary counts incrementally. For the base case of a lattice triangle, the area A satisfies A = I + \frac{1}{2} B - 1 directly from the determinant formula and the gcd lemma for edges, as I = 0 for elementary triangles yields A = \frac{1}{2} B - 1 = \frac{1}{2} when B = 3. Assume the theorem holds for all lattice polygons with fewer than n vertices. For an n-gon P (n \geq 4), select an ear—a triangle formed by three consecutive vertices—and remove it to obtain a smaller polygon Q with n-1 vertices. The area A_P = A_Q + A_T, where T is the ear triangle with I_T = 0 and B_T = 3 + k (accounting for shared edge points k \geq 0, assuming no additional lattice points on the ear's other two edges). The interior points update as I_P = I_Q + I_T + k, where k is the number of lattice points strictly interior to the shared edge (the diagonal), which are interior to P but boundary to Q and T. The boundary points update as B_P = B_Q + B_T - 2 - 2k (subtracting the shared endpoints double-counted and the k points on the shared edge, which become interior to P). Substituting the inductive hypothesis for Q and the base case for T yields A_P = (I_Q + \frac{1}{2} B_Q - 1) + (\frac{1}{2} B_T - 1) = I_P + \frac{1}{2} B_P - 1, confirming the formula. This ear-clipping process ensures the triangulation preserves the relation without introducing new interior points prematurely.^[11]^[9] In the inductive step for general polygons, the triangulation into F = 2I + B - 2 fundamental triangles (each of area \frac{1}{2}) is justified by double counting edges or angles in the graph: the total edges satisfy $3(F + 1) = 2E + B (interior faces plus exterior), and vertices V = I + B, leading via Euler's characteristic to the count of F, though here the focus remains on combinatorial adjustments rather than topology. When combining two adjacent polygons sharing an edge with X interior points on that edge, the areas add as A = A_1 + A_2, with I = I_1 + I_2 + X and B = B_1 + B_2 - 2X - 2 (subtracting the shared vertices and edge interiors), so the formula holds inductively: A = (I_1 + \frac{1}{2} B_1 - 1) + (I_2 + \frac{1}{2} B_2 - 1) = I + \frac{1}{2} B - 1.^[9]^[11] These methods offer an elementary alternative to topological proofs, relying solely on basic geometric decomposition, point counting, and simple algebraic manipulations, making them accessible with minimal prerequisites beyond lattice properties.^[8]

Examples and Applications

Illustrative Examples

Pick's theorem, which relates the area A of a simple lattice polygon to the number of interior lattice points I and boundary lattice points B via the formula A = I + \frac{B}{2} - 1, can be illustrated through straightforward computations on basic polygons.^[12] These examples demonstrate the process of enumerating I and B, where interior points are lattice points strictly inside the polygon, and boundary points include all lattice points on the edges, with careful counting to avoid overlaps at vertices.^[13] Consider the unit square with vertices at (0,0), (1,0), (1,1), and (0,1). There are no interior lattice points, so I = 0. Each edge has length 1 with no intermediate lattice points, yielding 4 boundary points at the vertices, so B = 4. Applying the formula gives A = 0 + \frac{4}{2} - 1 = 1, matching the known area of the square.^[14] To count boundary points systematically, the number of lattice points on an edge from (x_1, y_1) to (x_2, y_2) (including endpoints) is \gcd(|x_2 - x_1|, |y_2 - y_1|) + 1; here, each edge has \gcd(1,0) = 1 or similar, confirming no extras. Interior points can be enumerated by checking all lattice points within the bounding box and verifying containment, but none qualify in this case.^[12] A right triangle with vertices at (0,0), (3,0), and (0,1) provides another simple case. No lattice points lie strictly inside, so I = 0. For boundary points: the base edge has \gcd(3,0) + 1 = 4 points; the height edge has \gcd(0,1) + 1 = 2 points; the hypotenuse has \gcd(3,1) + 1 = 2 points. Summing these overcounts the three vertices (each shared by two edges), so B = (4 + 2 + 2) - 3 = 5. The formula yields A = 0 + \frac{5}{2} - 1 = 1.5, consistent with the geometric area \frac{1}{2} \times 3 \times 1 = 1.5.^[15] This counting method highlights a common pitfall: failing to subtract shared vertices when summing per-edge points, which would incorrectly inflate B to 8 here. For interior enumeration, the bounding box spans x = 0 to $3, y = 0 to $1, but points like (1,0) and (2,0) are on the boundary, leaving none inside.^[12] For a more involved example, consider a 2-by-3 rectangle with vertices at (0,0), (3,0), (3,2), and (0,2). Interior points are the lattice points at (1,1) and (2,1), so I = 2. Each horizontal edge has \gcd(3,0) + 1 = 4 points, and each vertical edge has \gcd(0,2) + 1 = 3 points; subtracting the 4 overcounted vertices gives B = (4 + 4 + 3 + 3) - 4 = 10. The formula computes A = 2 + \frac{10}{2} - 1 = 6, aligning with the rectangle's area of $3 \times 2 = 6.^[14] Enumerating interiors involves checking the 2-by-1 grid inside the bounds, confirming the two points; boundary counting avoids pitfalls like treating collinear points on longer edges as separate by relying on the gcd formula, which inherently includes all without duplication beyond vertices.^[13]

Practical Applications

Pick's theorem finds significant utility in computer graphics, particularly for computing the exact area of lattice polygons during rasterization processes on integer grids. In rendering algorithms, where polygons are approximated by pixels aligned to grid points, the theorem enables precise area determination by counting interior and boundary lattice points, avoiding floating-point arithmetic errors that could arise from traditional methods like the shoelace formula. This approach is especially valuable for irregular shapes in 2D graphics pipelines, ensuring accurate pixel filling without approximation.^[16] In robotics and path planning, Pick's theorem supports the estimation of obstacle-free regions within lattice-based maps, which are common in discrete environments for autonomous navigation. By modeling navigable areas as lattice polygons, the theorem allows robots to quantify free space efficiently, aiding in collision avoidance and trajectory optimization on grid representations of terrain or indoor spaces. For instance, in hybrid aerial-ground systems, it helps evaluate coverage areas for path feasibility.^[17]^[18] Within number theory and combinatorics, the theorem aids in counting lattice points for optimization problems, such as approximating solutions in integer programming where feasible regions are bounded by lattice polygons. It provides a direct link between point enumerations and area measures, facilitating bounds on the number of integer solutions in polyhedral sets and supporting algorithms for lattice-based packing or covering problems. This connection is fundamental in discrete optimization, where exact area computations inform the density of feasible lattice points.^[19]^[20] In geographic information systems (GIS), Pick's theorem has been applied to discrete terrain modeling by using it with Freeman-encoded polygons on raster grids for accurate area assessments in cartographic applications. This is crucial for vector-to-raster conversions in environmental modeling, where lattice points represent sampled terrain data, allowing exact quantification of land parcels or elevation zones without interpolation errors.^[21]

Generalizations and Extensions

To Higher Dimensions

The extension of Pick's theorem to three dimensions reveals that no simple formula exists for the volume of a lattice polyhedron in terms of only the number of interior lattice points I and total boundary lattice points B, unlike the two-dimensional case. This limitation is illustrated by Reeve tetrahedra, which have vertices at (0,0,0), (1,0,0), (0,1,0), and (1,1,r) for positive integers r. These tetrahedra contain no interior lattice points (I = 0) and exactly four boundary lattice points (the vertices), yet their volumes are r/6, varying with r. For instance, when r=1, the volume is $1/6; for r=5, it is $5/6. This counterexample demonstrates that boundary point counts alone cannot determine the volume, as the same lattice point configuration yields different volumes depending on the height parameter r. Instead, the appropriate framework for higher dimensions is Ehrhart theory, where the Ehrhart polynomial L_P(t) counts the lattice points in the t-fold dilation tP of a d-dimensional lattice polytope P. For d=3, this polynomial takes the form

L_P(t) = V t^3 + \frac{S}{2} t^2 + c_1 t + 1,

where V is the volume of P, S is the total surface area (sum of the areas of all faces), and the linear coefficient c_1 is more intricate, depending on the lengths of edges, dihedral angles between faces, and often expressible via Dedekind sums. Setting t=1 gives the total lattice points I + B = L_P(1), but inverting for V requires knowledge of the full polynomial or additional boundary data beyond mere point counts. For the Reeve tetrahedron, the Ehrhart polynomial is explicitly \frac{r}{6} t^3 + t^2 + \left(2 - \frac{r}{6}\right) t + 1, confirming the volume as the leading coefficient while highlighting the variability in lower terms. In general n dimensions, the Ehrhart polynomial L_P(t) is of degree n, with leading coefficient equal to the volume of P (normalized so the unit cube has volume 1), and successive coefficients corresponding to normalized volumes of the boundary faces of decreasing dimension: the coefficient of t^{n-k} relates to the (n-k-1)-dimensional surface measure. Pick's theorem emerges as the precise degree-2 instance of this theory, where the quadratic term gives the area and the linear term half the boundary length, directly yielding A = I + B/2 - 1 upon reciprocity (relating interior points to the polynomial evaluated at -t).^[22] Higher-dimensional extensions face increasing complexity due to the exponential growth in boundary structure: faces, edges, and vertices contribute to intricate coefficients, precluding a closed-form expression akin to Pick's theorem that relies solely on global interior and boundary counts. Computational methods or full Ehrhart polynomials are typically required to relate lattice points to volumes in dimensions greater than 2.^[23]

To Non-Lattice and Rational Polygons

Pick's theorem can be extended to polygons with rational vertex coordinates through a scaling transformation that reduces the problem to the lattice case. If the vertices of the polygon lie on the rational grid (1/d)\mathbb{Z}^2 for some positive integer d, scaling all coordinates by d produces a lattice polygon P' with integer vertices. The area of P' is then computed using the standard Pick's theorem: A(P') = I' + B'/2 - 1, where I' is the number of interior lattice points and B' is the number of boundary lattice points of P'. The original area is A(P) = A(P') / d^2, yielding an exact result.^[24] This scaling method is illustrated by the triangle with vertices at (0,0), (1.5,0), and (0,1), which has rational coordinates with denominator d=2. Scaling by 2 gives the lattice triangle with vertices (0,0), (3,0), and (0,2). This scaled triangle has I' = 1 interior point (1,1) and B' = 6 boundary points: (0,0), (1,0), (2,0), (3,0), (0,1), (0,2). Applying Pick's theorem yields A(P') = 1 + 6/2 - 1 = 3. Thus, the original area is 3 / 4 = 0.75, matching the direct computation (1/2) \times 1.5 \times 1 = 0.75.^[24] For polygons with real (irrational) vertex coordinates, no exact analog of Pick's theorem exists, but approximations can be obtained by lattice point counting on a fine grid. Consider a grid with spacing 1/N for large integer N; scaling the polygon by N approximates a lattice case, where the number of grid points inside and on the boundary provides an estimate of the scaled area via a Pick's-like formula, and the original area is recovered by dividing by N^2. Error bounds for this approximation arise from Ehrhart-Macdonald reciprocity, which relates lattice point counts in dilates to interior points and provides quasi-polynomial estimates with discrepancies of order O(1/N).^[25] A key tool for bounding these approximations is Blichfeldt's theorem, which states that any bounded measurable set S in the plane with area A(S) > k can be translated to contain at least k+1 lattice points. This implies lower bounds on lattice point counts relative to area, enabling error estimates for non-lattice polygons by considering translates that maximize point inclusion and relating them to boundary effects. These extensions sacrifice the exactness of the original theorem, as real-coordinate cases rely on asymptotic or probabilistic estimates rather than closed forms. While seminal results like Ehrhart-Macdonald reciprocity offer theoretical bounds, computational software such as LattE integrale implements efficient algorithms (based on Barvinok's method) for lattice point enumeration in dilated polytopes, addressing practical limitations for large N, though exact non-lattice formulas remain elusive.^[26]

Ehrhart Theory

Ehrhart theory generalizes Pick's theorem by providing a polynomial that enumerates lattice points in scaled versions of lattice polytopes across any dimension. For a d-dimensional lattice polytope P \subset \mathbb{R}^d, the Ehrhart polynomial L_P(t) is defined as the number of lattice points in the dilation tP, where t is a positive integer:

L_P(t) = \#(tP \cap \mathbb{Z}^d).

Ehrhart's theorem states that L_P(t) extends to a polynomial in t of degree d, with leading coefficient equal to the Euclidean volume of P.^[27]^[25] This framework was pioneered by the French mathematician Eugène Ehrhart in the 1960s, building on earlier lattice point counting ideas like Pick's theorem to establish a systematic algebraic structure for such enumerations.^[28] In the specific case of two-dimensional lattice polygons, the Ehrhart polynomial directly encodes Pick's theorem: L_P(1) = I + B, where I is the number of interior lattice points and B is the number on the boundary. The explicit form is L_P(t) = A t^2 + \frac{B}{2} t + 1, with A the area of P; substituting t=1 yields Pick's formula A = I + \frac{B}{2} - 1.^[25]^[27] Ehrhart polynomials exhibit several fundamental properties. By the Ehrhart–Macdonald reciprocity theorem, L_{\mathrm{int}(P)}(t) = (-1)^d L_P(-t), linking counts in the polytope to those in its relative interior; in particular, evaluating at negative integers yields signed interior point enumerations, and L_P(-1) = (-1)^d I, where I is the number of interior lattice points of P.^[29] Additionally, the coefficients of L_P(t) are non-negative integers, and in the h^*-polynomial representation, they form a unimodal sequence under conditions like those for Gorenstein* polytopes or normal lattice polytopes.^[30]^[31] For extensions beyond full-dimensional lattice polytopes, Ehrhart theory employs quasi-polynomials, where the coefficients are periodic functions of t with period determined by the denominators in the polytope's vertex coordinates. This applies to rational polytopes and non-full-dimensional cases embedded in higher-dimensional space, with recent work characterizing maximal periods and structural properties of these quasi-polynomials.^[32]^[33]

Shoelace Formula Connections

The shoelace formula computes the area A of a simple polygon with vertices (x_i, y_i) for i = 1, \dots, n, where (x_{n+1}, y_{n+1}) = (x_1, y_1), via the expression

A = \frac{1}{2} \left| \sum_{i=1}^n (x_i y_{i+1} - x_{i+1} y_i) \right|.

This method relies on the ordered list of vertex coordinates and works for any real-number coordinates.^[34] For lattice polygons, where all vertices lie on integer lattice points, the shoelace formula produces an area that is always a multiple of \frac{1}{2}, since the summation yields an even integer due to the integer inputs.^[35] Pick's theorem provides an alternative: A = I + \frac{B}{2} - 1, where I is the number of interior lattice points and B is the number of boundary lattice points. Both approaches compute the identical area for any given lattice polygon, as verified through direct application— for instance, a lattice polygon with I = 6 and B = 6 has area 8 by Pick's theorem and the same value when shoelace is applied to its vertices.^[35] Their equivalence stems from the fact that the polygon area is uniquely determined; a proof sketch involves decomposing the lattice polygon into primitive triangles (with no interior points) using the vertices, applying the shoelace formula to each to sum the total area, and showing that the contributions aggregate to I + \frac{B}{2} - 1 by accounting for how lattice points on edges and inside contribute to the summed determinants.^[14] Unlike shoelace, which requires enumerating all vertices, Pick's theorem bypasses this if I and B are known, making it advantageous in scenarios where point counting is simpler than coordinate tabulation, such as discrete optimization problems on lattices.^[36] Conversely, shoelace excels for general continuous polygons without lattice constraints.^[37] The shoelace formula originated in the 18th century, described by Albrecht Ludwig Friedrich Meister in 1769 based on earlier trapezoid methods, well before Georg Pick formulated his theorem in 1899.^[38] In modern computational geometry, the two are often combined, with shoelace providing vertex-based area estimates that Pick's theorem refines or validates using lattice point counts for efficiency in grid-based algorithms.^[36]

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Theorem 1.2. (a) The roots of Ehrhart polynomials of lattice d-polytopes are bounded in norm by 1+(d + 1)!. (b) All real roots of Ehrhart polynomials of d- ...
[28]
[1407.0255] Stanley's Major Contributions to Ehrhart Theory - ar5iv
Abstract. This expository paper features a few highlights of Richard Stanley's extensive work in Ehrhart theory, the study of integer-point enumeration in ...Missing: higher | Show results with:higher
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[PDF] Combinatorial Reciprocity Theorems - matthias beck
Theorem (Ehrhart 1962) ehrP(k) is a polynomial in k. Theorem (Macdonald 1971) (−1) dim P. ehrP(−k) enumerates the interior lattice points in kP ...
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[PDF] Unimodality Problems in Ehrhart Theory - arXiv
Nov 29, 2017 · It is immediate that the polynomial (1 + z)d is real-rooted, hence the binomial coefficients are both log-concave and unimodal. Many additional ...
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[PDF] Ehrhart Polynomials - matthias beck
(An orientation α of Γ and a k -coloring x are compatible if xj ě xi whenever there is an edge oriented from i to j. An orientation is acyclic if.
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Rational polytopes with Ehrhart coefficients of arbitrary period
Ehrhart states that the number of integer lattice points in the dilation of a rational polytope by a positive integer k is a quasi-polynomial function of k — ...
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[PDF] Ehrhart quasi-polynomials of almost integral polytopes - arXiv
Aug 30, 2023 · To complete the proof, we show that if P is not centrally symmetric, then there exists a rational vector c such that L(P,c)(t) 6= L(P,−c)(t).
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Shoelace Formula -- from Wolfram MathWorld
The shoelace formula, also known as Gauss's area formula, the shoelace algorithm, shoelace method, or surveyor's formula, is a name sometimes given to the ...<|separator|>
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[PDF] The Shoelace Formula - Theorem of the Day
Then t = Aco − Acl 2A∆ . For our example polygon, again using Pick, Acl = 5/2, Aco = 8 − 5/2 = 11/2 and A∆ = 5/2. This gives t = (11/2 − 5/2)/5 = 3/5 (a little ...Missing: equivalence | Show results with:equivalence
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Pick's Theorem - area of lattice polygons - CP-Algorithms
Jun 8, 2022 · Pick's theorem provides a way to compute the area of this polygon through the number of vertices that are lying on the boundary and the number of vertices that ...
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What's the difference between Pick's formula and the Shoelace ...
Nov 28, 2020 · Pick's theorem works only if the polygon vertices have integer coordinates. Shoelace formula works with any real coordinates.
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[PDF] Who Invented the Shoelace Formula? - Theorem of the Day
... history “The formula was described by. Albrecht Ludwig Friedrich Meister (1724–1788) in 1769 and is based on the trapezoid formula which was described by ...