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Writing loops: a first example

Writing loops: a first example. Programming Fundamentals 17 Feliks Kluźniak. The Fundamental Invariance Theorem for the Iterative Construct (a.k.a. The MainTheorem of Computing Science): Let P be an invariant of the statement S . If P is true just before the loop while B do

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Writing loops: a first example

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  1. Writing loops: a first example Programming Fundamentals 17 Feliks Kluźniak

  2. The Fundamental Invariance Theorem for the Iterative Construct (a.k.a. The MainTheorem of Computing Science): Let P be an invariant of the statement S. If P is true just before the loop while B do S od then not B and P will be true just after the loop, provided the loop terminates. Writing loops: a first example

  3. Consider the problem of multiplying integers, on a machine like the A1: we have addition, subtraction, shifts, but no multiplication instruction. Writing loops: a first example

  4. Consider the problem of multiplying integers, on a machine like the A1: we have addition, subtraction, shifts, but no multiplication instruction. So we must write a program... For simplicity, let us limit our attention only to positive integers. Writing loops: a first example

  5. Consider the problem of multiplying integers, on a machine like the A1: we have addition, subtraction, shifts, but no multiplication instruction. So we must write a program... For simplicity, let us limit our attention only to positive integers. We can also assume none of them is zero: the computation is trivial in that case. Writing loops: a first example

  6. Given integers A > 0 and B > 0 find an integer r such that r = A * B . Writing loops: a first example

  7. Given integers A > 0 and B > 0 find an integer r such that r = A * B . Clearly, we will need a loop. So we must find an invariant P such that P and not C implies r = A * B and C is suitable for the condition of the loop. % P while C do ..... od % P and not C, therefore r = A * B Writing loops: a first example

  8. Given integers A > 0 and B > 0 find an integer r such that r = A * B . Clearly, we will need a loop. So we must find an invariant P such that P and not C implies r = A * B and C is suitable for the condition of the loop. Moreover, P must be relatively easy to establish. % P while C do ..... od % P and not C, therefore r = A * B Writing loops: a first example

  9. Given integers A > 0 and B > 0 find an integer r such that r = A * B . Clearly, we will need a loop. So we must find an invariant P such that P and not C implies r = A * B and C is suitable for the condition of the loop. Moreover, P must be relatively easy to establish. We could try to manipulate some variables a and b in such a way that P : A * B = r + a * b and b >= 0 Clearly, P and b = 0 would imply r = A * B , Writing loops: a first example

  10. Given integers A > 0 and B > 0 find an integer r such that r = A * B . Clearly, we will need a loop. So we must find an invariant P such that P and not C implies r = A * B and C is suitable for the condition of the loop. Moreover, P must be relatively easy to establish. We could try to manipulate some variables a and b in such a way that P : A * B = r + a * b and b >= 0 Clearly, P and b = 0 would imply r = A * B , and b != 0 is an excellent boolean expression (easy to evaluate), Writing loops: a first example

  11. Given integers A > 0 and B > 0 find an integer r such that r = A * B . Clearly, we will need a loop. So we must find an invariant P such that P and not C implies r = A * B and C is suitable for the condition of the loop. Moreover, P must be relatively easy to establish. We could try to manipulate some variables a and b in such a way that P : A * B = r + a * b and b >= 0 Clearly, P and b = 0 would imply r = A * B , and b != 0 is an excellent boolean expression (easy to evaluate), and P is very easily established. How? Writing loops: a first example

  12. Given integers A > 0 and B > 0 find an integer r such that r = A * B . Clearly, we will need a loop. So we must find an invariant P such that P and not C implies r = A * B and C is suitable for the condition of the loop. Moreover, P must be relatively easy to establish. We could try to manipulate some variables a and b in such a way that P : A * B = r + a * b and b >= 0 Clearly, P and b = 0 would imply r = A * B , and b != 0 is an excellent boolean expression (easy to evaluate), and P is very easily established by a, b, r := A, B, 0 . Writing loops: a first example

  13. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 ... od % P and b = 0, therefore r = A * B Writing loops: a first example

  14. Given integers A > 0 and B > 0 find an integer r such that r = A * B . • P : A * B = r + a * b and b >= 0 . • a, b, r := A, B, 0 ; % P • while b != 0 do • % P and b != 0 • ... • od • % P and b = 0, therefore r = A * B • So now we must write the body of the loop in such a way that: • the loop will terminate; • P will be an invariant. Writing loops: a first example

  15. Given integers A > 0 and B > 0 find an integer r such that r = A * B . • P : A * B = r + a * b and b >= 0 . • a, b, r := A, B, 0 ; % P • while b != 0 do • % P and b != 0 • ... • od • % P and b = 0, therefore r = A * B • So now we must write the body of the loop in such a way that: • the loop will terminate; • P will be an invariant. • We can easily ensure termination by always strictly decreasing the value of b . Writing loops: a first example

  16. Given integers A > 0 and B > 0 find an integer r such that r = A * B . • P : A * B = r + a * b and b >= 0 . • a, b, r := A, B, 0 ; % P • while b != 0 do • % P and b != 0 • ... • od • % P and b = 0, therefore r = A * B • So now we must write the body of the loop in such a way that: • the loop will terminate; • P will be an invariant. • We can easily ensure termination by always strictly decreasing the value of b . Why will this ensure termination? Writing loops: a first example

  17. Given integers A > 0 and B > 0 find an integer r such that r = A * B . • P : A * B = r + a * b and b >= 0 . • a, b, r := A, B, 0 ; % P • while b != 0 do • % P and b != 0 • ... • od • % P and b = 0, therefore r = A * B • So now we must write the body of the loop in such a way that: • the loop will terminate; • P will be an invariant. • We can easily ensure termination by always strictly decreasing the value of b . Because, if we preserve P, b is bounded from below! Writing loops: a first example

  18. Given integers A > 0 and B > 0 find an integer r such that r = A * B . • P : A * B = r + a * b and b >= 0 . • a, b, r := A, B, 0 ; % P • while b != 0 do • % P and b != 0 • ... • od • % P and b = 0, therefore r = A * B • So now we must write the body of the loop in such a way that: • the loop will terminate; • P will be an invariant. • We can easily ensure termination by always strictly decreasing the value of b . For example, by subtracting 1... Writing loops: a first example

  19. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination ... .... od % P and b = 0, therefore r = A * B Writing loops: a first example

  20. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) .... od % P and b = 0, therefore r = A * B Writing loops: a first example

  21. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) .... od % P and b = 0, therefore r = A * B Now, how do we restore P ? Writing loops: a first example

  22. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) .... od % P and b = 0, therefore r = A * B Now, how do we restore P ? We have almost no choice, but to use: r + a * b = (r + a) + a * (b – 1) Writing loops: a first example

  23. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) r := r + a % restores P od % P and b = 0, therefore r = A * B Writing loops: a first example

  24. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) r := r + a % restores P od % P and b = 0, therefore r = A * B a b r 2 5 0 Writing loops: a first example

  25. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) r := r + a % restores P od % P and b = 0, therefore r = A * B a b r 2 5 0 2 4 2 Writing loops: a first example

  26. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) r := r + a % restores P od % P and b = 0, therefore r = A * B a b r 2 5 0 2 4 2 2 3 4 Writing loops: a first example

  27. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) r := r + a % restores P od % P and b = 0, therefore r = A * B a b r 2 5 0 2 4 2 2 3 4 2 2 6 Writing loops: a first example

  28. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) r := r + a % restores P od % P and b = 0, therefore r = A * B a b r 2 5 0 2 4 2 2 3 4 2 2 6 2 1 8 Writing loops: a first example

  29. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) r := r + a % restores P od % P and b = 0, therefore r = A * B a b r 2 5 0 2 4 2 2 3 4 2 2 6 2 1 8 2 0 10 Writing loops: a first example

  30. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) r := r + a % restores P od % P and b = 0, therefore r = A * B This algorithm is not very surprising: a * b = a + a + .... + a b times Writing loops: a first example

  31. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) r := r + a % restores P od % P and b = 0, therefore r = A * B This algorithm is not very surprising. It is also extremely inefficient! Consider, e.g., A = 1 and B = 32 000. Writing loops: a first example

  32. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) r := r + a % restores P od % P and b = 0, therefore r = A * B This algorithm is not very surprising. It is also extremely inefficient! Consider, e.g., A = 1 and B = 32 000. We must do better! When did we last make a decision? Writing loops: a first example

  33. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) .... od % P and b = 0, therefore r = A * B Can we decrease b in a more clever way? (The difficulty is in restoring P.) Writing loops: a first example

  34. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) .... od % P and b = 0, therefore r = A * B If b is an even number, then we can divide it by 2 simply by shifting! Writing loops: a first example

  35. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) .... od % P and b = 0, therefore r = A * B If b is an even number, then we can divide it by 2 simply by shifting! How would we then restore P ? b:= b div 2 Writing loops: a first example

  36. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) .... od % P and b = 0, therefore r = A * B If b is an even number, then we can divide it by 2 simply by shifting! How would we then restore P ? b:= b div 2 ; a := a * 2 Writing loops: a first example

  37. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 b := b – 1 ; % ensures termination (and violates P !) .... od % P and b = 0, therefore r = A * B If b is an even number, then we can divide it by 2 simply by shifting! How would we then restore P ? b:= b div 2 ; a := a * 2 a * b = (a * 2) * (b / 2) and b / 2 = b div 2 if b is even! Writing loops: a first example

  38. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 if b is odd then % just check the lowest bit b := b – 1; r := r + a % P else b := b div 2 ; % just a shift a := a * 2 % just a shift, restores P fi % P od % P and b = 0, therefore r = A * B

  39. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 if b is odd then % just check the lowest bit b := b – 1; r := r + a % P else b := b div 2 ; % just a shift a := a * 2 % just a shift, restores P fi % P od % P and b = 0, therefore r = A * B • a b r • 5 0

  40. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 if b is odd then % just check the lowest bit b := b – 1; r := r + a % P else b := b div 2 ; % just a shift a := a * 2 % just a shift, restores P fi % P od % P and b = 0, therefore r = A * B • a b r • 5 0 • 2 4 2

  41. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 if b is odd then % just check the lowest bit b := b – 1; r := r + a % P else b := b div 2 ; % just a shift a := a * 2 % just a shift, restores P fi % P od % P and b = 0, therefore r = A * B • a b r • 5 0 • 2 4 2 • 2 2

  42. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 if b is odd then % just check the lowest bit b := b – 1; r := r + a % P else b := b div 2 ; % just a shift a := a * 2 % just a shift, restores P fi % P od % P and b = 0, therefore r = A * B • a b r • 5 0 • 2 4 2 • 2 2 • 1 2

  43. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 if b is odd then % just check the lowest bit b := b – 1; r := r + a % P else b := b div 2 ; % just a shift a := a * 2 % just a shift, restores P fi % P od % P and b = 0, therefore r = A * B • a b r • 5 0 • 2 4 2 • 2 2 • 1 2 • 8 0 10

  44. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 if b is odd then % just check the lowest bit b := b – 1; r := r + a % P else b := b div 2 ; % just a shift a := a * 2 % just a shift, restores P fi % P od % P and b = 0, therefore r = A * B In general this is much faster! What is the maximum number of iterations?

  45. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 if b is odd then % just check the lowest bit b := b – 1; r := r + a % P else b := b div 2 ; % just a shift a := a * 2 % just a shift, restores P fi % P od % P and b = 0, therefore r = A * B In general his is much faster! 39 (for a 20 bit word)

  46. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 if b is odd then % just check the lowest bit b := b – 1; r := r + a % P else b := b div 2 ; % just a shift a := a * 2 % just a shift, restores P fi % P od % P and b = 0, therefore r = A * B 39 (for a 20 bit word) Can we make it even faster?

  47. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 if b is odd then % just check the lowest bit b := b – 1; r := r + a % P fi; b := b div 2 ; % just a shift a := a * 2 % just a shift,P od % P and b = 0, therefore r = A * B b must be even ! Writing loops: a first example

  48. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 if b is odd then % just check the lowest bit b := b – 1; r := r + a % P fi; b := b div 2 ; % just a shift a := a * 2 % just a shift,P od % P and b = 0, therefore r = A * B What if b = 0 ? Writing loops: a first example

  49. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 if b is odd then % just check the lowest bit b := b – 1; r := r + a % P fi; b := b div 2 ; % just a shift a := a * 2 % just a shift,P od % P and b = 0, therefore r = A * B What if b = 0 ? We make use of the fact that: 0 div 2 = 0 a * 0 = a * 2 * 0 Writing loops: a first example

  50. Given integers A > 0 and B > 0 find an integer r such that r = A * B . P : A * B = r + a * b and b >= 0 . a, b, r := A, B, 0 ; % P while b != 0 do % P and b != 0 if b is odd then % just check the lowest bit b := b – 1; r := r + a % P fi; b := b div 2 ; % just a shift a := a * 2 % just a shift,P od % P and b = 0, therefore r = A * B This is exactly the algorithm we use when multiplying by hand! Writing loops: a first example

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