1 / 20

Loops II

Loops II. For Loop. For <index variable> = <vector of values> execute statements for each value of index end for x = [1 4 7 9] disp(num2str(x^2)); end 1 16 49 81. Drawing a Line. for cX = [10 11 12 13 14 15 16 17 18 19 20] putPixel(cX, 10, g) end a better way?.

cyndi
Download Presentation

Loops II

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Loops II

  2. For Loop For <index variable> = <vector of values> execute statements for each value of index end for x = [1 4 7 9] disp(num2str(x^2)); end 1 16 49 81

  3. Drawing a Line for cX = [10 11 12 13 14 15 16 17 18 19 20] putPixel(cX, 10, g) end a better way? for cX = 10:1:20 putPixel(cX, 10, g) end for cX = 10:20 putPixel(cX, 10, g) end

  4. Adding numbers 1-5 i = 1; sum = 0; while ( i <= 5) sum = sum + i i = i + 1; end sum = 0; for i = 1:5 sum = sum + i end using for loop ?

  5. Adding Elements of a Vector • Given vector v, sum up its contents sum = 0; i = 1; while ( i <= length(v) ) sum = sum + v(i) i = i + 1; end sum = 0; for i = 1:length(v) sum = sum + v(i) end

  6. Minimum Element of a Vector min = inf; index = 0; for i = 1:length(v) if (v(i) < min) min = v(i) index = i end end min = inf; for i = 1:length(v) if (v(i) < min) min = v(i) end end What if we need the location as well?

  7. Drawing a Rectangle putPixel(10, 10, g); putPixel(10, 11, g); putPixel(10, 12, g); putPixel(11, 10, g); putPixel(11, 11, g); putPixel(11, 12, g); putPixel(12, 10, g); putPixel(12, 11, g); putPixel(12, 12, g); putPixel(10, 10, g); putPixel(11, 10, g); putPixel(12, 10, g); putPixel(10, 11, g); putPixel(11, 11, g); putPixel(12, 11, g); putPixel(10, 12, g); putPixel(11, 12, g); putPixel(12, 12, g);

  8. In general, for (x1,y1)-(x2, y2) for cY = y1:y2 for cX = x1:x2 putPixel(cX, cY, g); end end for cX = x1:x2 for cY = y1:y2 putPixel(cX, cY, g); end end

  9. Draw a Right Triangle putPixel(10, 10, g); putPixel(10, 11, g); putPixel(10, 12, g); putPixel(11, 11, g); putPixel(11, 12, g); putPixel(12, 12, g); (x1, y1) (x2, y2)

  10. In General, scanning through x first … (x1, y1) for cX = x1:x2 for cY = ?:y2 putPixel(cX, xY, g); end end (cX, ?) dy (cX, y2) (x2, y2) cX dx

  11. Triangle Solution dx = x2 - x1; dy = y2 - y1; for cX = x1:x2 yStart = y1 + round( cX * dy / dx); for cY = yStart:y2 putPixel(cX, cY, g); end end

  12. All Subset of Size 2 • Given a vector v, list all the possible subsets of elements of v of size 2. One idea: For a given element, list all the other elements together with it, effectively listing all 2-element subsets that contain that element Repeat this for all elements of the vector (set)

  13. First Try For all elements x of v do list all the other elements together with x For all elements x of v do For all elements y of v except x do list (x, y)

  14. First Try … for i=1:length(v) for j=1:length(v) if (i ~= j) disp([num2str(v(i)) ' ' num2str(v(j))]) end end end

  15. First Try? • For v = 1:4, our program produces: 1 2 1 3 1 4 2 1 2 3 2 4 3 1 3 2 3 4 4 1 4 2 4 3

  16. An idea … • For two valid index values k,l, we list two entries : v(k)-v(l) and v(l)-v(k) • One of the listings is extra and not needed • We can put a new rule to apply before making a listing that will eliminate one of the duplicates

  17. Second Try … for i=1:length(v) for j=1:length(v) if (i ~= j && i < j) disp([num2str(v(i)) ' ' num2str(v(j))]) end end end

  18. Eliminate useless iterations • We don’t need to go over all the i,j combinations • Only cases where i < j are used • For a given i value, iterating the j from 1 till i is useless

  19. Solution for i=1:length(v) for j=i+1:length(v) disp([num2str(v(i)) ' ' num2str(v(j))]) end end

  20. Closest to Target • Given a vector v, and a target number n, Find the the separate elements of vector v such that their sum is closest to number n. • Example: v = [10, 5, 20, 7, 13, 3 ] , n=22 The pair whose sum comes closest to 22 is 20 and 3.

More Related