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# Part 4.

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Part 4. The Gap and-Vanquish System. A Basic Sample. Finding the most extreme of a set S of n numbers . Time Intricacy. Time intricacy: Estimation of T(n): Accept n = 2 k , T(n) = 2T(n/2)+1 = 2(2T(n/4)+1)+1 = 4T(n/4)+2+1 : =2 k-1 T(2)+2 k-2 + … +4+2+1
Transcripts
Slide 1

Section 4 The Divide-and-Conquer Strategy

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A Simple Example Finding the greatest of a set S of n numbers

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Time Complexity Time intricacy: Calculation of T(n): Assume n = 2 k , T(n) = 2T(n/2)+1 = 2(2T(n/4)+1)+1 = 4T(n/4)+2+1 : =2 k-1 T(2)+2 k-2 + â¦ +4+2+1 =2 k-1 +2 k-2 + â¦ +4+2+1 =2 k - 1 = n-1

Slide 4

A General Divide-and-Conquer Algorithm Step 1 : If the issue size is little, tackle this issue straightforwardly; something else, split the first issue into 2 sub-issues with equivalent sizes. Step 2 : Recursively take care of these 2 sub-issues by applying this calculation. Step 3 : Merge the 2\'s arrangements sub-issues into a unique\'s answer issue.

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Time Complexity of the General Algorithm Time many-sided quality: where S(n) : time for part M(n) : time for combining b : a steady c : a consistent

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2-D Maxima Finding Problem Def : A point (x 1 , y 1 ) rules (x 2 , y 2 ) if x 1 > x 2 and y 1 > y 2 . A point is known as a maxima if no other point overwhelms it. Direct strategy : Compare each pair of focuses. Time multifaceted nature: O(n 2 )

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Divide-and-Conquer for Maxima Finding The maximal purposes of S L and S R

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The calculation: Input: A set S of n planar focuses. Yield: The maximal purposes of S. Step 1: If S contains one and only point, return it as the maxima. Something else, discover a line L opposite to the X-pivot which isolates S into S L and S R , with equivalent sizes. Step 2: Recursively locate the maximal purposes of S L and S R . Step 3: Find the biggest y-estimation of S R , meant as y R . Dispose of each of the maximal purposes of S L if its y-quality is not as much as y R .

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Time intricacy: T(n) Step 1: O(n) Step 2: 2T(n/2) Step 3: O(n) Assume n = 2 k T(n) = O(n log n)

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The Closest Pair Problem Given a set S of n focuses, discover a couple of focuses which are nearest together. 1-D rendition : Solved by sorting Time unpredictability : O(n log n) ï¼ 2-D form

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At most 6 focuses in territory A:

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The calculation: Input: A set S of n planar focuses. Yield: The separation between two nearest focuses. Step 1: Sort indicates in S concurring their y-values. Step 2: If S contains stand out point, return vastness as its separation. Step 3: Find a middle line L opposite to the X-hub to gap S into S L and S R , with equivalent sizes. Step 4: Recursively apply Steps 2 and 3 to tackle the nearest match issues of S L and S R . Let d L (d R ) indicate the separation between the nearest combine in S L (S R ). Let d = min(d L , d R ).

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Step 5: For a point P in the half-section limited by L-d and L, let its y-worth be meant as y P . For each such P, discover all focuses in the half-piece limited by L and L+d whose y-quality fall inside of y P +d and y P - d. In the event that the separation d ï¢ in the middle of P and a point in the other half-chunk is not as much as d, let d=d ï¢ . The last estimation of d is the answer. Time multifaceted nature: O(n log n) Step 1: O(n log n) Steps 2~5: ï T(n) = O(n log n)

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Concave polygon: Convex polygon: The Convex Hull Problem The raised body of an arrangement of planar focuses is the littlest arched polygon containing the focuses\' majority.

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The gap and-overcome method to take care of the issue:

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The combining technique: Select an inside point p. There are 3 successions of focuses which have expanding polar edges concerning p. (1) g, h, i, j, k (2) a, b, c, d (3) f, e Merge these 3 groupings into 1 succession: g, h, a, b, f, c, e, d, i, j, k. Apply Graham output to inspect the focuses one by one and take out the focuses which cause reflexive edges . (See the case on the following page.)

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e.g. focuses b and f should be erased. Last result:

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Divide-and-Conquer for Convex Hull Input : A set S of planar focuses Output : A curved body for S Step 1: If S contains close to five focuses, use thorough seeking to locate the arched frame and return. Step 2: Find a middle line opposite to the X-hub which partitions S into S L and S R , with equivalent sizes. Step 3: Recursively build curved frames for S L and S R , indicated as Hull(S L ) and Hull(S R ), separately.

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Step 4: Apply the combining system to consolidate Hull(S L ) and Hull(S R ) together to shape an arched structure. Time unpredictability: T(n) = 2T(n/2) + O(n) = O(n log n)

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The Voronoi Diagram Problem e.g. The Voronoi outline for three focuses Each L ij is the opposite bisector of the line.

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Definition of Voronoi Diagrams Def : Given two focuses P i , P j ï S, let H(P i ,P j ) indicate the half plane containing P i . The Voronoi polygon connected with P i is characterized as

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Given an arrangement of n focuses, the Voronoi outline comprises of all the Voronoi polygons of these focuses. The vertices of the Voronoi outline are called Voronoi focuses and its sections are called Voronoi edges .

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Delaunay Triangulation

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Example for Constructing Voronoi Diagrams Divide the focuses into two sections.

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Merging Two Voronoi Diagrams Merging along the piecewise direct hyperplane

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The Final Voronoi Diagram After combining

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Divide-and-Conquer for Voronoi Diagram Input: A set S of n planar focuses. Yield: The Voronoi graph of S. Step 1: If S contains stand out point, return. Step 2: Find a middle line L opposite to the X-pivot which separates S into S L and S R , with equivalent sizes.

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Step 3: Construct Voronoi graphs of S L and S R recursively. Signify these Voronoi graphs by VD(S L ) and VD(S R ). Step 4: Construct a partitioning piece-wise straight hyperplane HP which is the locus of focuses all the while nearest to a point in S L and a point in S R . Toss all portions of VD(S L ) which mislead the privilege of HP and all fragments of VD(S R ) that deceive the left of HP. The subsequent chart is the Voronoi graph of S. (See points of interest on the following page.)

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Merging Two Voronoi Diagrams into One Voronoi Diagram Input: (a) S L and S R where S L and S R are divided by an opposite line L. (b) VD(S L ) and VD(S R ). Yield: VD(S) where S = S L â©S R Step 1: Find the raised frames of S L and S R , meant as Hull(S L ) and Hull(S R ), separately. (A unique calculation for discovering a raised body for this situation will by given later.)

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Step 2: Find portions and which join HULL(S L ) and HULL(S R ) into a curved body (P an and P c fit in with S L and P b and P d have a place with S R ) Assume that lies above . Let x = a, y = b, SG= and HP = ï . Step 3: Find the opposite bisector of SG. Mean it by BS. Let HP = HPâª{BS}. On the off chance that SG = , go to Step 5; generally, go to Step 4.

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Step 4: The beam from VD(S L ) and VD(S R ) which BS first crosses with must be an opposite bisector of either or for some z. On the off chance that this beam is the opposite bisector of , then let SG = ; something else, let SG = . Go to Step 3. Step 5: Discard the edges of VD(S L ) which reach out to one side of HP and dispose of the edges of VD(S R ) which stretch out to one side of HP. The subsequent chart is the Voronoi outline of S = S L âªS R .

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Properties of Voronoi Diagrams Def : Given a point P and a set S of focuses, the separation in the middle of P and S is the separation in the middle of P and P i which is the closest neighbor of P in S. The HP acquired from the above calculation is the locus of focuses which keep equivalent separations to S L and S R . The HP is monotonic in y.

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# of Voronoi Edges # of edges of a Voronoi graph ï£ 3n - 6, where n is # of focuses. Thinking: # of edges of a planar chart with n vertices ï£ 3n - 6. A Delaunay triangulation is a planar diagram. Edges in Delaunay triangulation edges in Voronoi outline.

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# of Voronoi Vertices # of Voronoi vertices ï£ 2n - 4. Thinking : Let F, E and V signify # of face, edges and vertices in a planar chart. Euler â s connection: F = E - V + 2. In a Delaunay triangulation, V = n, E ï£ 3n â 6 ï F = E - V + 2 ï£ 3n - 6 - n + 2 = 2n - 4.

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Construct a Convex Hull from a Voronoi Diagram After a Voronoi outline is built, a curved structure can by found in O(n) time.

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Construct a Convex Hull from a Voronoi Diagram Step 1 : Find an interminable beam by analyzing all Voronoi edges. Step 2 : Let P i be the point to one side of the interminable beam. P i is a raised structure vertex. Inspect the Voronoi polygon of P i to locate the following endless beam. Step 3 : Repeat Step 2 until we come back to the beginning beam.

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Time Complexity Time many-sided quality for consolidating 2 Voronoi charts: Total: O(n) Step 1: O(n) Step 2: O(n) Step 3 ~ Step 5: O(n) (at most 3n - 6 edges in VD(S L ) and VD(S R ) and at most n portions in HP) Time intricacy for developing a Voronoi outline: O(n log n) b ecause T(n) = 2T(n/2) + O(n) = O(n log n)

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The Lower Bound of the Voronoi Diagram Problem The lower bound of the Voronoi graph issue is ï (n log n). sorting ïµ Voronoi chart issue The Voronoi graph for an arrangement of focuses on a straight line

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Applications of Voronoi Diagrams The Euclidean closest neighbor looking issue. The Euclidean all closest neighbor issue.

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Fast Fourier Transform (FFT) Fourier change Inverse Fourier change Discrete Fourier transform(DFT) Given a 0 , a 1 , â¦ , a n-1 , process

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FFT Algorithm Inverse DFT : DFT can be re

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