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Announcements. Next lecture will be given by Dr. Leigh Johnston. 35. 30. 25. 20. T(n). 15. 10. 5. 0. 0. 5. 10. 15. 20. n. Master Theorem Case 3: When the Regularity Condition Fails. Alternative Iterative Algorithms for Binary Search. function mid=bsearch1(L,p,q,key)

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    1. Announcements • Next lecture will be given by Dr. Leigh Johnston.

    2. 35 30 25 20 T(n) 15 10 5 0 0 5 10 15 20 n Master Theorem Case 3: When the Regularity Condition Fails

    3. Alternative Iterative Algorithms for Binary Search function mid=bsearch1(L,p,q,key) while q>p mid=floor((p+q)/2); if key<=L(mid) q=mid; else p=mid+1; end end if key==L(p) mid=p; else mid=0; end function mid=bsearch2(L,p,q,key) while q>p mid=floor((p+q)/2); if key==L(mid) return elseif key<L(mid) q=mid-1; else p=mid+1; end end %NB: q may be invalid here if key==L(p) mid=p; else mid=0; end

    4. Central Algorithmic Techniques Recursion

    5. Different Representationsof Recursive Algorithms Views Pros

    6. Code Representation of an Algorithm MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Pros and Cons?

    7. Runs on computers Precise and succinct Perception that being able to code is the only thing needed to get a job. (false) I am not a computer I need a higher level of intuition. Prone to bugs Language dependent Code Representation of an Algorithm Pros: Cons:

    8. Different Representationsof Recursive Algorithms Views Pros

    9. X = 2133 Y = 2312 ac = bd = (a+b)(c+d) = XY = MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames Stack Frame: A particular execution of one routine on one particular input instance.

    10. X = 2133 Y = 2312 ac = ? bd = (a+b)(c+d) = XY = MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    11. X = 2133 Y = 2312 ac = ? bd = (a+b)(c+d) = XY = X = 21 Y = 23 ac = bd = (a+b)(c+d) = XY = MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    12. X = 2133 Y = 2312 ac = ? bd = (a+b)(c+d) = XY = X = 21 Y = 23 ac = ? bd = (a+b)(c+d) = XY = MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    13. X = 2133 Y = 2312 ac = ? bd = (a+b)(c+d) = XY = X = 21 Y = 23 ac = ? bd = (a+b)(c+d) = XY = X = 2 Y = 2 XY= MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    14. X = 2133 Y = 2312 ac = ? bd = (a+b)(c+d) = XY = X = 21 Y = 23 ac = ? bd = (a+b)(c+d) = XY = X = 2 Y = 2 XY = 4 MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    15. X = 2133 Y = 2312 ac = ? bd = (a+b)(c+d) = XY = X = 21 Y = 23 ac = 4 bd = (a+b)(c+d) = XY = MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    16. X = 2133 Y = 2312 ac = ? bd = (a+b)(c+d) = XY = X = 21 Y = 23 ac = 4 bd = ? (a+b)(c+d) = XY = MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    17. X = 2133 Y = 2312 ac = ? bd = (a+b)(c+d) = XY = X = 21 Y = 23 ac = 4 bd = ? (a+b)(c+d) = XY = X = 1 Y = 3 XY = 3 MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    18. X = 2133 Y = 2312 ac = ? bd = (a+b)(c+d) = XY = X = 21 Y = 23 ac = 4 bd = 3 (a+b)(c+d) = XY = MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    19. X = 2133 Y = 2312 ac = ? bd = (a+b)(c+d) = XY = X = 21 Y = 23 ac = 4 bd = 3 (a+b)(c+d) = ? XY = MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    20. X = 2133 Y = 2312 ac = ? bd = (a+b)(c+d) = XY = X = 21 Y = 23 ac = 4 bd = 3 (a+b)(c+d) = ? XY = X = 3 Y = 5 XY = 15 MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    21. X = 2133 Y = 2312 ac = ? bd = (a+b)(c+d) = XY = X = 21 Y = 23 ac = 4 bd = 3 (a+b)(c+d) = 15 XY = ? MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    22. X = 2133 Y = 2312 ac = ? bd = (a+b)(c+d) = XY = X = 21 Y = 23 ac = 4 bd = 3 (a+b)(c+d) = 15 XY = 483 MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    23. X = 2133 Y = 2312 ac = 483 bd = (a+b)(c+d) = XY = MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    24. X = 2133 Y = 2312 ac = 483 bd = ? (a+b)(c+d) = XY = MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    25. X = 2133 Y = 2312 ac = 483 bd = ? (a+b)(c+d) = XY = X = 33 Y = 12 ac = ? bd = (a+b)(c+d) = XY = 15 MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    26. X = 2133 Y = 2312 ac = 483 bd = ? (a+b)(c+d) = XY = X = 33 Y = 12 ac = ? bd = (a+b)(c+d) = XY = 15 X = 3 Y = 1 XY = 3 MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    27. X = 2133 Y = 2312 ac = 483 bd = ? (a+b)(c+d) = XY = X = 33 Y = 12 ac = 3 bd = ? (a+b)(c+d) = XY = 15 MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    28. X = 2133 Y = 2312 ac = 483 bd = ? (a+b)(c+d) = XY = X = 33 Y = 12 ac = 3 bd = ? (a+b)(c+d) = XY = 15 X = 3 Y = 2 XY = 6 MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    29. X = 2133 Y = 2312 ac = 483 bd = ? (a+b)(c+d) = XY = X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = ? XY = 15 MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames

    30. X = 2133 Y = 2312 ac = 483 bd = ? (a+b)(c+d) = XY = X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = ? XY = 396 MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Stack of Stack Frames An so on ….

    31. Trace what actually occurs in the computer Concrete. It is what students attempt to describe when told not to give code. Described in words it is impossible to follow who is doing what. Does not explain why it works. Demonstrates for only one of many inputs. Stack of Stack Frames Representation of an Algorithm Pros: Cons:

    32. Different Representationsof Recursive Algorithms Views Pros

    33. X = 2133 Y = 2312 ac = 483 bd = 396 (a+b)(c+d) = 1890 XY = 4931496 X = 21 Y = 23 ac = 4 bd = 3 (a+b)(c+d) = 15 XY = 483 X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = 18 XY = 396 X = 54 Y = 35 ac = 15 bd = 20 (a+b)(c+d) = 72 XY = 1890 X = 2 Y = 2 XY=4 X = 1 Y = 3 XY=3 X = 3 Y = 5 XY=15 X = 3 Y = 1 XY=3 X = 3 Y = 2 XY=6 X = 6 Y = 3 XY=18 X = 5 Y = 3 XY=15 X = 4 Y = 5 XY=20 X = 9 Y = 8 XY=72 MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Tree of Stack Frames

    34. View the entire computation. Good for computing the running time. Must describe entire tree. For each stack frame input instance computation solution returned. who calls who Structure of tree may be complex. Stack of Stack Frames Representation of an Algorithm Pros: Cons:

    35. Different Representationsof Recursive Algorithms Views Pros

    36. X = 2133 Y = 2312 ac = 483 bd = 396 (a+b)(c+d) = 1890 XY = 4931496 X = 21 Y = 23 ac = 4 bd = 3 (a+b)(c+d) = 15 XY = 483 X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = 18 XY = 396 X = 54 Y = 35 ac = 15 bd = 20 (a+b)(c+d) = 72 XY = 1890 X = 2 Y = 2 XY=4 X = 1 Y = 3 XY=3 X = 3 Y = 5 XY=15 X = 3 Y = 1 XY=3 X = 3 Y = 2 XY=6 X = 6 Y = 3 XY=18 X = 5 Y = 3 XY=15 X = 4 Y = 5 XY=20 X = 9 Y = 8 XY=72 MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Friends & Strong Induction One Friend for each stack frame. Each worries only about his job.

    37. X = 2133 Y = 2312 ac = 483 bd = 396 (a+b)(c+d) = 1890 XY = 4931496 X = 21 Y = 23 ac = 4 bd = 3 (a+b)(c+d) = 15 XY = 483 X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = 18 XY = 396 X = 54 Y = 35 ac = 15 bd = 20 (a+b)(c+d) = 72 XY = 1890 X = 2 Y = 2 XY=4 X = 1 Y = 3 XY=3 X = 3 Y = 5 XY=15 X = 3 Y = 1 XY=3 X = 3 Y = 2 XY=6 X = 6 Y = 3 XY=18 X = 5 Y = 3 XY=15 X = 4 Y = 5 XY=20 X = 9 Y = 8 XY=72 MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Friends & Strong Induction Worry about one step at a time. Imagine that you are one specific friend.

    38. X = 33 Y = 12 ac = bd = (a+b)(c+d) = XY = MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Friends & Strong Induction • Consider your input instance

    39. X = 33 Y = 12 ac = ? bd = ? (a+b)(c+d) = XY = MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Friends & Strong Induction • Consider your input instance • Allocate work • Construct one or more subinstances X = 3 Y = 2 XY=? X = 3 Y = 1 XY=? X = 6 Y = 3 XY=?

    40. X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = 18 XY = MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Friends & Strong Induction • Consider your input instance • Allocate work • Construct one or more subinstances • Assume by magic your friends give you the answer for these. X = 3 Y = 2 XY=6 X = 3 Y = 1 XY=3 X = 6 Y = 3 XY=18

    41. X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = 18 XY = 396 MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Friends & Strong Induction • Consider your input instance • Allocate work • Construct one or more subinstances • Assume by magic your friends give you the answer for these. • Use this help to solve your own instance. X = 3 Y = 2 XY=6 X = 3 Y = 1 XY=3 X = 6 Y = 3 XY=18

    42. X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = 18 XY = 396 MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Friends & Strong Induction • Consider your input instance • Allocate work • Construct one or more subinstances • Assume by magic your friends give you the answer for these. • Use this help to solve your own instance. • Do not worry about anything else. • Who your boss is. • How your friends solve their instance. X = 3 Y = 2 XY=6 X = 3 Y = 1 XY=3 X = 6 Y = 3 XY=18

    43. X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = 18 XY = 396 MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Friends & Strong Induction • This technique is often • referred to as • Divide and Conquer X = 3 Y = 2 XY=6 X = 3 Y = 1 XY=3 X = 6 Y = 3 XY=18

    44. MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Friends & Strong Induction • Consider generic instances. MULT(X,Y): Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f MULT(X,Y): If |X| = |Y| = 1 then RETURN XY • Remember! • Do not worry about • Who your boss is. • How your friends solve their instance. bd (a+b)(c+d) ac

    45. X = 2133 Y = 2312 ac = 483 bd = 396 (a+b)(c+d) = 1890 XY = 4931496 X = 21 Y = 23 ac = 4 bd = 3 (a+b)(c+d) = 15 XY = 483 X = 33 Y = 12 ac = 3 bd = 6 (a+b)(c+d) = 18 XY = 396 X = 54 Y = 35 ac = 15 bd = 20 (a+b)(c+d) = 72 XY = 1890 X = 2 Y = 2 XY=4 X = 1 Y = 3 XY=3 X = 3 Y = 5 XY=15 X = 3 Y = 1 XY=3 X = 3 Y = 2 XY=6 X = 6 Y = 3 XY=18 X = 5 Y = 3 XY=15 X = 4 Y = 5 XY=20 X = 9 Y = 8 XY=72 MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Friends & Strong Induction This solves the problem for every possible instance.

    46. Friends & Strong Induction • Recursive Algorithm: • Assume you have an algorithm that works. • Use it to write an algorithm that works.

    47. Circular Argument! Friends & Strong Induction • Recursive Algorithm: • Assume you have an algorithm that works. • Use it to write an algorithm that works. If I could get in, I could get the key. Then I could unlock the door so that I can get in.

    48. Friends & Strong Induction • Recursive Algorithm: • Assume you have an algorithm that works. • Use it to write an algorithm that works. To get into my house I must get the key from a smaller house

    49. X = 33 Y = 12 ac = ? bd = ? (a+b)(c+d) = XY = MULT(X,Y): If |X| = |Y| = 1 then RETURN XY Break X into a;b and Y into c;d e = MULT(a,c) and f =MULT(b,d) RETURN e 10n + (MULT(a+b, c+d) – e - f) 10n/2 + f Friends & Strong Induction • Allocate work • Construct one or more subinstances • Each subinstance must be • a smaller instance • to the same problem. X = 3 Y = 2 XY=? X = 3 Y = 1 XY=? X = 6 Y = 3 XY=?

    50. Use brute force to get into the smallest house. Friends & Strong Induction • Recursive Algorithm: • Assume you have an algorithm that works. • Use it to write an algorithm that works.