24. Divide and Conquer Algorithms. Binary Search Merge Sort Mesh Generation Recursion
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1 24. Divide and Conquer Algorithms Binary Search Merge Sort Mesh Generation Recursion
2 The Manhatten phone book has 1,000,000+ entries. How is it possible to locate a name by examining just a tiny, tiny fraction of those entries? Question
3 Answer: Repeated Halving To find the page containing Derek Jeter s number while (Phone book is longer than 1 page) Open to the middle page. if Jeter comes before the first name, Rip and throw away the 2 nd half. else end end Rip and throw away the 1 st half.
4 What Happens to Phone Book Length Original: 3000 pages After 1 rip: 1500 pages After 2 rips: 750 pages After 3 rips: 375 pages After 4 rips: 188 pages After 5 rips: 94 pages After 12 rips: 1 page
5 Binary Search The idea of repeatedly halving the size of the search space is the main idea behind the method of binary search. An item in a sorted array of length n can be located with just log 2 n comparisons.
6 Problem Search a Sorted Array For a Given Value (We assume that the array has no repeated elements.)
7 Binary Search: a = x: L: Mid: R: x(mid) <= a So throw away the left half
8 Binary Search: a = x: L: Mid: R: a < x(mid) So throw away the right half
9 Binary Search: a = x: L: Mid: R: x(mid) <= a So throw away the left half
10 Binary Search: a = x: L: Mid: R: x(mid) <= a So throw away the left half
11 Binary Search: a = x: L: Mid: R: Done because R-L = 1
12 function L = BinarySearch(a,x) % x is a row n-vector with x(1) <... < x(n) % x(1) <= a <= x(n) % x(l) <= a <= x(l+1) L = 1; R = length(x); % x(l) <= a <= x(r) while R-L > 1 mid = floor((l+r)/2); % Note that mid does not equal L or R. if a < x(mid) % x(l) <= a <= x(mid) R = mid; else % x(mid) <= a <== x(r) L = mid; end end
13 Next Problem Sort the values in an array so that they are arranged from smallest t o biggest. Chosen Method: Merge Sort This technique is an excellent example of a divide and conquer procedure.
14 Motivation A plan if you have two helpers : Split the array and have each helper sort his/her half. Merge the two sorted subarrays. And what if those two helpers each had two subhelpers? Etc
15 Subdivide the Sorting Task H E M G B K A Q F L P D R C J N H E M G B K A Q F L P D R C J N
16 Subdivide Again H E M G B K A Q F L P D R C J N H E M G B K A Q F L P D R C J N
17 And Again H E M G B K A Q F L P D R C J N H E M G B K A Q F L P D R C J N
18 And One Last Time H E M G B K A Q F L P D R C J N
19 Now Merge E H G M B K A Q F L D P C R J N H E M G B K A Q F L P D R C J N
20 And Merge Again E G H M A B K Q D F L P C J N R E H G M B K A Q F L D P C R J N
21 And Again A B E G H K M Q C D F J L N P R E G H M A B K Q D F L P C J N R
22 And One Last Time A B C D E F G H J K L M N P Q R A B E G H K M Q C D F J L N P R
23 Done! A B C D E F G H J K L M N P Q R
24 function y = MergeSort(x) % x is a column n-vector. % y is a column n-vector consisting % of the values in x sorted % from smallest to largest. n = length(x); if n==1 y = x; else m = floor(n/2); y1 = MergeSort(x(1:m)); y2 = MergeSort(x(m+1:n)); y = Merge(y1,y2); end
25 Important Sub-Problem Merging Two Sorted Arrays Into a Single Sorted Array
26 Example
27 Merge x: ix: 1 y: iy: 1 z: iz: 1 x(ix) <= y(iy)???
28 Merge x: ix: 1 y: iy: 1 z: 12 iz: 1 x(ix) <= y(iy) yes
29 Merge x: ix: 2 y: iy: 1 z: 12 iz: 2 x(ix) <= y(iy)???
30 Merge x: ix: 2 y: iy: 1 z: iz: 2 x(ix) <= y(iy) no
31 Merge x: ix: 2 y: iy: 2 z: iz: 3 x(ix) <= y(iy)???
32 Merge x: ix: 2 y: iy: 2 z: iz: 3 x(ix) <= y(iy) yes
33 Merge x: ix: 3 y: iy: 2 z: iz: 4 x(ix) <= y(iy)???
34 Merge x: ix: 3 y: iy: 2 z: iz: 4 x(ix) <= y(iy) yes
35 Merge x: ix: 4 y: iy: 2 z: iz: 5 x(ix) <= y(iy)???
36 Merge x: ix: 4 y: iy: 2 z: iz: 5 x(ix) <= y(iy) no
37 Merge x: ix: 4 y: iy: 3 z: iz: 5 x(ix) <= y(iy)???
38 Merge x: ix: 4 y: iy: 3 z: iz: 5 x(ix) <= y(iy) yes
39 Merge x: ix: 5 y: iy: 3 z: iz: 6 ix > 4
40 Merge x: ix: 5 y: iy: 3 z: iz: 6 ix > 4
41 Merge x: ix: 5 y: iy: 4 z: iz: 8 ix > 4
42 Merge x: ix: 5 y: iy: 4 z: iz: 8 ix > 4
43 Merge x: ix: 5 y: iy: 5 z: iz: 9 ix > 4
44 Merge x: ix: 5 y: iy: 5 z: iz: 9 ix > 4
45 function z = Merge(x,y) n = length(x); m = length(y); z = zeros(1,n+m); ix = 1; iy = 1; for iz=1:(n+m) if ix > n z(iz) = y(iy); iy = iy+1; elseif iy>m z(iz) = x(ix); ix = ix + 1; elseif x(ix) <= y(iy) z(iz) = x(ix); ix = ix + 1; else z(iz) = y(iy);iy = iy + 1; end end
46 New Problem Recursive Mesh Generation
47 Divide and Conquer Methods Also Show Up in Geometric Situations Chop a Region up into triangles with smnaller triangles in areas of interest.
48 Mesh Generation An Area Of Interest Step One in simulating flow around an airfoil is to generate a mesh and (say) estimate velocity at each mesh point.
49 Mesh Generation in 3D
50 Why is Mesh Generation a Divide & Conquer Process? Let Us Draw this Graphic:
51 The Basic Operation if the triangle is big enough Connect the midpoints. Color the interior triangle mauve. else Color the whole triangle yellow. end
52 At the Start
53 Apply same idea to lower left triangle Recur On this Idea
54 Recur Again
55 Recursion And Again
56
57 Now, Climb Your Way Out.
58 Etc
59 function DrawTriangle(x,y,level) if level ==5 fill(x,y,'y') else % Draw the triangle... plot([x x(1)],[y y(1)],'k') % Determine the midpoints... a = [(x(1)+x(2))/2 (x(2)+x(3))/2 (x(3)+x(1))/2]; b = [(y(1)+y(2))/2 (y(2)+y(3))/2 (y(3)+y(1))/2]; % Color the interior triangle... pause fill(a,b,'m') pause % Apply the process to the three "corner" triangles... DrawTriangle([x(1) a(1) a(3)],[y(1) b(1) b(3)],level+1) DrawTriangle([x(2) a(2) a(1)],[y(2) b(2) b(1)],level+1) DrawTriangle([x(3) a(3) a(2)],[y(3) b(3) b(2)],level+1) end
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