CS133 Computational Geometry Simplification Algorithms 1 Line - - PowerPoint PPT Presentation

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CS133 Computational Geometry Simplification Algorithms 1 Line - - PowerPoint PPT Presentation

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CS133 Computational Geometry Simplification Algorithms 1 Line Simplification ? 2 Line Simplification Given a line string and a distance threshold , return a simplified line string such that any point in the input line string is not

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CS133

Computational Geometry

Simplification Algorithms

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Line Simplification

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?

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Line Simplification

Given a line string and a distance threshold 𝜁, return a simplified line string such that any point in the input line string is not displaced more than 𝜁

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Douglas–Peucker Algorithm

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≀ 𝜁

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Douglas–Peucker Algorithm

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≀ 𝜁

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Douglas–Peucker Algorithm

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Douglas–Peucker Algorithm

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≀ 𝜁

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Douglas–Peucker Algorithm

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≀ 𝜁

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Douglas–Peucker Algorithm

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≀ 𝜁

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Douglas–Peucker Algorithm

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≀ 𝜁

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Douglas–Peucker Algorithm

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Douglas–Peucker Algorithm

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Douglas–Peucker Algorithm

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Running Time

function DouglasPeucker(P[], 𝜁) // Find the point with the maximum distance (=maximum cross product) cmax = 0 index = 0 for i = 2 to ( end - 1) c = 𝑄 1 𝑄[π‘œ] Γ— 𝑄 1 𝑄[𝑗] if ( c > cmax ) index = i cmax = c if (cmax / 𝑄 1 [𝑄 π‘œ > 𝜁) { R1 = DouglasPeucker(P[1..index], 𝜁) R2 = DouglasPeucker(P[index..n], 𝜁) R1.removeLast return R1 || R2 // Concatenate the two lists else return [P[1], P[n]] // Only return the first and last points

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π‘ˆ π‘œ = π‘ˆ π‘œ1 + π‘ˆ π‘œ βˆ’ π‘œ1 + 𝑃 π‘œ

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Polygon Triangulation

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Polygon Triangulation

Given a simple polygon P, break it down into a set of triangles π‘ˆ such that the union of the triangles is equal to the polygon and no two triangles intersect. That is: Ϊ‚π‘’π‘—βˆˆπ‘ˆ 𝑒𝑗 = 𝑄 and 𝑒𝑗 ∩ π‘’π‘˜ = 𝜚 for any 𝑒𝑗, π‘’π‘˜ ∈ π‘ˆ and 𝑗 β‰  π‘˜

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Polygon Triangulation

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Convex Polygons

Choose any vertex

  • n the polygon

Connect it to all

  • ther vertices to

create all diagonals Number of possible triangulations is Catalan Number π·π‘œβˆ’2

π·π‘œ =

1 π‘œ+1

2π‘œ π‘œ =

2π‘œ! π‘œ+1 !π‘œ!

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By Eric DΓ©trez (as published at www.texample.net in 2008-10-09) - The image is the result of a latex script example published at [1]. All the site is under CC-2.5-by license, as seen on every page of the wite., CC BY 2.5, https://commons.wikimedia.org/w/index.php?curid=56872963

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Number of Triangles

Number of triangles in a triangulation of a polygon of π‘œ points is π‘œ βˆ’ 2 triangles Can be proven by induction Trivial case: π‘œ = 3 General case: If it applies for all π‘œ < 𝑛, we need to prove that it applies for π‘œ = 𝑛 That is, we want to prove that if a triangulation exists, it will have 𝑛 βˆ’ 2 triangles

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Induction

π‘œ1 + π‘œ2 = 𝑛 + 1 # of triangles in 𝑄

1 = π‘œ1 βˆ’ 2, in 𝑄2 = π‘œ2 βˆ’ 2

# of triangles in 𝑄 = π‘œ1 βˆ’ 2 + π‘œ2 βˆ’ 2 + 1 = π‘œ1 + π‘œ2 βˆ’ 4 + 1 = 𝑛 + 1 βˆ’ 4 + 1 = 𝑛 βˆ’ 2

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π‘œ1 π‘œ2

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Existence of a Triangulation

Any simple polygon has at least one triangulation Proof by induction Trivial case: π‘œ = 3 General case: If there are triangulations for all polygons π‘œ < 𝑛, we need to prove that there is one for π‘œ = 𝑛

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Induction

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Dual Graph

Dual graph 𝒣 Each triangle is represented by a vertex in 𝒣 Two triangles that share an edge are connected by an edge in 𝒣

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