Recurrent Problems. ITT9131 Konkreetne Matemaatika. Chapter One The Tower of Hanoi Lines in the Plane The Josephus Problem

Transcription

1 Recurrent Problems ITT93 Konkreetne Matemaatika Chapter One The Tower of Hanoi Lines in the Plane The Josephus Problem

2 Contents The Tower of Hanoi Lines in the Plane 3 The Josephus Problem 4 Intermezzo: Structural induction

3 Next section The Tower of Hanoi Lines in the Plane 3 The Josephus Problem 4 Intermezzo: Structural induction

4 The Tower of Hanoi The Tower of Hanoi puzzle was invented by the French mathematician Edouard Lucas in 883. Using mathematical induction one can prove that For the Tower of Hanoi puzzle with n 0 (and 3 pegs), the minimum number of moves needed T n = n. Let's look at the example borrowed from Martin Hofmann and Berteun Damman.

5 The Tower of Hanoi The Tower of Hanoi puzzle was invented by the French mathematician Edouard Lucas in 883. Using mathematical induction one can prove that For the Tower of Hanoi puzzle with n 0 (and 3 pegs), the minimum number of moves needed T n = n. Let's look at the example borrowed from Martin Hofmann and Berteun Damman.

6 The Tower of Hanoi The Tower of Hanoi puzzle was invented by the French mathematician Edouard Lucas in 883. Using mathematical induction one can prove that For the Tower of Hanoi puzzle with n 0 (and 3 pegs), the minimum number of moves needed T n = n. Let's look at the example borrowed from Martin Hofmann and Berteun Damman.

7 Tower of Hanoi 5 Discs 3 4 5

8 Tower of Hanoi 5 Discs Moved disc from pole to pole 3.

9 Tower of Hanoi 5 Discs Moved disc from pole to pole.

10 Tower of Hanoi 5 Discs Moved disc from pole 3 to pole.

11 Tower of Hanoi 5 Discs Moved disc from pole to pole 3.

12 Tower of Hanoi 5 Discs Moved disc from pole to pole.

13 Tower of Hanoi 5 Discs Moved disc from pole to pole 3.

14 Tower of Hanoi 5 Discs Moved disc from pole to pole 3.

15 Tower of Hanoi 5 Discs Moved disc from pole to pole.

16 Tower of Hanoi 5 Discs Moved disc from pole 3 to pole.

17 Tower of Hanoi 5 Discs Moved disc from pole 3 to pole.

18 Tower of Hanoi 5 Discs Moved disc from pole to pole.

19 Tower of Hanoi 5 Discs Moved disc from pole 3 to pole.

20 Tower of Hanoi 5 Discs Moved disc from pole to pole 3.

21 Tower of Hanoi 5 Discs Moved disc from pole to pole.

22 Tower of Hanoi 5 Discs Moved disc from pole 3 to pole.

23 Tower of Hanoi 5 Discs Moved disc from pole to pole 3.

24 Tower of Hanoi 5 Discs Moved disc from pole to pole.

25 Tower of Hanoi 5 Discs Moved disc from pole to pole 3.

26 Tower of Hanoi 5 Discs Moved disc from pole to pole 3.

27 Tower of Hanoi 5 Discs Moved disc from pole to pole.

28 Tower of Hanoi 5 Discs Moved disc from pole 3 to pole.

29 Tower of Hanoi 5 Discs Moved disc from pole 3 to pole.

30 Tower of Hanoi 5 Discs Moved disc from pole to pole.

31 Tower of Hanoi 5 Discs Moved disc from pole to pole 3.

32 Tower of Hanoi 5 Discs Moved disc from pole to pole 3.

33 Tower of Hanoi 5 Discs Moved disc from pole to pole.

34 Tower of Hanoi 5 Discs Moved disc from pole 3 to pole.

35 Tower of Hanoi 5 Discs Moved disc from pole to pole 3.

36 Tower of Hanoi 5 Discs Moved disc from pole to pole.

37 Tower of Hanoi 5 Discs Moved disc from pole to pole 3.

38 Tower of Hanoi 5 Discs Moved disc from pole to pole 3.

39 Tower of Hanoi 5 Discs OK 3 4 5

40 A recursive solution in Python #!/ usr /bin /env python3 import os def hanoi (n, start = ' ', step = ' ', stop = '3 ' ): ''' Solve the Hanoi tower with n disks, from start peg to stop peg, using step peg as a spool ''' if n > 0: hanoi (n -, start, stop, step ) move (n, start, stop ) hanoi (n -, step, start, stop ) def move (n, start, stop ): ''' Display move of disk n from start to stop ''' print ( " Disk % d : % s -> % s " % (n, start, stop )) if name == ' main ': n = int ( input ( 'How many disks? ' )) hanoi ( n ) Question: why does this program show that T n = n?

41 A recursive solution in Python #!/ usr /bin /env python3 import os def hanoi (n, start = ' ', step = ' ', stop = '3 ' ): ''' Solve the Hanoi tower with n disks, from start peg to stop peg, using step peg as a spool ''' if n > 0: hanoi (n -, start, stop, step ) move (n, start, stop ) hanoi (n -, step, start, stop ) def move (n, start, stop ): ''' Display move of disk n from start to stop ''' print ( " Disk % d : % s -> % s " % (n, start, stop )) if name == ' main ': n = int ( input ( 'How many disks? ' )) hanoi ( n ) Question: why does this program show that T n = n?

42 Warmup: What is wrong with this proof? Theorem All horses are the same color. Proof The thesis is clearly true for n =, so let n >. Put the n horses on a line. By inductive hypothesis, the n leftmost horses are the same color, and so do the n rightmost horses. Then the n horses in the middle are the same color. The rst and last horse must then be that color. Solution: For n = there are no n horses in the middle.

43 Warmup: What is wrong with this proof? Theorem All horses are the same color. Proof The thesis is clearly true for n =, so let n >. Put the n horses on a line. By inductive hypothesis, the n leftmost horses are the same color, and so do the n rightmost horses. Then the n horses in the middle are the same color. The rst and last horse must then be that color. Solution: For n = there are no n horses in the middle.

44 Next section The Tower of Hanoi Lines in the Plane 3 The Josephus Problem 4 Intermezzo: Structural induction

45 Lines in the Plane Problem Popularly: How many slices of pizza can a person obtain by making n straight cuts with a pizza knife? Academically: What is the maximum number L n of regions dened by n lines in the plane? Solved rst in 86, by the Swiss mathematician Jacob Steiner.

46 Lines in the Plane small cases V V V L 0 = L = V V V V V 3 V V V 3 V 4 V 4 V 4 L = 4 L 3 = L + 3 = 7

47 Lines in the Plane small cases V V V L 0 = L = V V V V V 3 V V V 3 V 4 V 4 V 4 L = 4 L 3 = L + 3 = 7

48 Lines in the Plane generalization Observation: The n-th line (for n > 0) increases the number of regions by k i it splits k of the "old regions" i it hits the previous lines in k dierent places.

49 Lines in the Plane generalization Observation: The n-th line (for n > 0) increases the number of regions by k i it splits k of the "old regions" i it hits the previous lines in k dierent places.

50 Lines in the Plane generalization Observation: The n-th line (for n > 0) increases the number of regions by k i it splits k of the "old regions" i it hits the previous lines in k dierent places.

51 Lines in the Plane generalization Observation: The n-th line (for n > 0) increases the number of regions by k i it splits k of the "old regions" i it hits the previous lines in k dierent places. k must be less or equal to n. Why?

52 Lines in the Plane generalization Observation: The n-th line (for n > 0) increases the number of regions by k i it splits k of the "old regions" i it hits the previous lines in k dierent places. k = 3; places k = ; places

53 Lines in the Plane generalization () Therefore the new line can intersect the n "old" lines in at most "n- " dierent points, we have established the upper bound L n L n + n for n > 0. If n-th line is not parallel to any of the others (hence it intersects them all), and doesn't go through any of the existing intersection points (hence it intersects them all in dierent places) then we get the recurrent equation: L 0 = ; L n = L n + n for n > 0. n L n

54 Lines in the Plane generalization () Therefore the new line can intersect the n "old" lines in at most "n- " dierent points, we have established the upper bound L n L n + n for n > 0. If n-th line is not parallel to any of the others (hence it intersects them all), and doesn't go through any of the existing intersection points (hence it intersects them all in dierent places) then we get the recurrent equation: L 0 = ; L n = L n + n for n > 0. n L n

55 Lines in the Plane solving recurrence Observation: L n = L n + n = L n + (n ) + n = L n 3 + (n ) + (n ) + n = L (n ) + (n ) + n = + S n, where S n = (n ) + n

56 Lines in the Plane solving recurrence () Evaluation of S n = (n ) + n. Recurrent equation: S 0 = 0; S n = S n + n for n > 0. Solution (Gauss, 786): S n = (n ) + n +S n = n + (n ) S n = (n + ) + (n + ) + + (n + ) + (n + ) S n = n(n + ) n(n + ) S n =

57 Lines in the Plane solving recurrence () Evaluation of S n = (n ) + n. Recurrent equation: S 0 = 0; S n = S n + n for n > 0. Solution (Gauss, 786): S n = (n ) + n +S n = n + (n ) S n = (n + ) + (n + ) + + (n + ) + (n + ) S n = n(n + ) n(n + ) S n =

58 Lines in the Plane solving recurrence () Evaluation of S n = (n ) + n. Recurrent equation: S 0 = 0; S n = S n + n for n > 0. Solution (Gauss, 786): S n = (n ) + n +S n = n + (n ) S n = (n + ) + (n + ) + + (n + ) + (n + ) S n = n(n + ) n(n + ) S n =

59 Lines in the Plane solving recurrence (3) Theorem: Closed formula for L n L n = n(n + ) +, for n > 0. Proof (by induction ). Basis: L 0 = 0(0+) + =. Step: Let assume L n = n(n+) + and evaluate L n+ = L n + n + n(n + ) = + + n + n(n + ) + + n = + n(n + ) + (n + ) = + (n + )(n + ) = +. Q.E.D.

60 Next section The Tower of Hanoi Lines in the Plane 3 The Josephus Problem 4 Intermezzo: Structural induction

61 The Josephus Problem Legend: During the Jewish-Roman war, Flavius Josephus, a famous historian of the rst century, was among a band of 4 Jewish rebels trapped in a cave by the Romans. Preferring suicide to capture, the rebels decided to form a circle and, proceeding around it, to kill every third remaining person until no one was left. But Josephus, together his friend, wanted to avoid being killed. So he quickly calculated where he and his friend should stand in the vicious circle

62 The Josephus Problem Our variation of the problem: we start with n people numbered to n around a circle we eliminate every second remaining person until only one survives Task is to compute the survivor's number, J(n) Example, n = The elimination order is, 4, 6, 8, 0, 3, 7,, 9. So, we have J(0) =

63 The Josephus Problem Our variation of the problem: we start with n people numbered to n around a circle we eliminate every second remaining person until only one survives Task is to compute the survivor's number, J(n) Example, n = The elimination order is, 4, 6, 8, 0, 3, 7,, 9. So, we have J(0) =

64 The Josephus Problem Our variation of the problem: we start with n people numbered to n around a circle we eliminate every second remaining person until only one survives Task is to compute the survivor's number, J(n) Example, n = The elimination order is, 4, 6, 8, 0, 3, 7,, 9. So, we have J(0) =

65 The Josephus Problem Our variation of the problem: we start with n people numbered to n around a circle we eliminate every second remaining person until only one survives Task is to compute the survivor's number, J(n) Example, n = The elimination order is, 4, 6, 8, 0, 3, 7,, 9. So, we have J(0) =

66 The Josephus Problem Our variation of the problem: we start with n people numbered to n around a circle we eliminate every second remaining person until only one survives Task is to compute the survivor's number, J(n) Example, n = The elimination order is, 4, 6, 8, 0, 3, 7,, 9. So, we have J(0) =

67 The Josephus Problem Our variation of the problem: we start with n people numbered to n around a circle we eliminate every second remaining person until only one survives Task is to compute the survivor's number, J(n) Example, n = The elimination order is, 4, 6, 8, 0, 3, 7,, 9. So, we have J(0) =

68 The Josephus Problem Our variation of the problem: we start with n people numbered to n around a circle we eliminate every second remaining person until only one survives Task is to compute the survivor's number, J(n) Example, n = The elimination order is, 4, 6, 8, 0, 3, 7,, 9. So, we have J(0) =

69 The Josephus Problem Our variation of the problem: we start with n people numbered to n around a circle we eliminate every second remaining person until only one survives Task is to compute the survivor's number, J(n) Example, n = The elimination order is, 4, 6, 8, 0, 3, 7,, 9. So, we have J(0) =

70 The Josephus Problem Our variation of the problem: we start with n people numbered to n around a circle we eliminate every second remaining person until only one survives Task is to compute the survivor's number, J(n) Example, n = The elimination order is, 4, 6, 8, 0, 3, 7,, 9. So, we have J(0) =

71 The Josephus Problem Our variation of the problem: we start with n people numbered to n around a circle we eliminate every second remaining person until only one survives Task is to compute the survivor's number, J(n) Example, n = The elimination order is, 4, 6, 8, 0, 3, 7,, 9. So, we have J(0) =

72 The Josephus Problem Our variation of the problem: we start with n people numbered to n around a circle we eliminate every second remaining person until only one survives Task is to compute the survivor's number, J(n) Example, n = The elimination order is, 4, 6, 8, 0, 3, 7,, 9. So, we have J(0) =

73 The Josephus Problem small numbers Evaluate J(n) for small n: n J(n) Properties J(n) is always odd; Recurrent equation J() = ; J(n) = J(n), for n ; J(n + ) = J(n) +, for n. 3 Closed formula J( m + l) = l +, for m 0 and 0 l < m.

74 The Josephus Problem small numbers Evaluate J(n) for small n: n J(n) Properties J(n) is always odd; Recurrent equation J() = ; J(n) = J(n), for n ; J(n + ) = J(n) +, for n. 3 Closed formula J( m + l) = l +, for m 0 and 0 l < m.

75 The Josephus Problem small numbers Evaluate J(n) for small n: n J(n) Properties J(n) is always odd; Recurrent equation J() = ; J(n) = J(n), for n ; J(n + ) = J(n) +, for n. 3 Closed formula J( m + l) = l +, for m 0 and 0 l < m.

76 The Josephus Problem recurrent equation () Case n = m. 0 First trip eliminates all even numbers. Then we change numbers and repeat: Old number k New number k' or k = k That correspondance between "old" and "new numbers" gives us that: J(n) = J(n)

77 The Josephus Problem recurrent equation () Case n = m. 0 First trip eliminates all even numbers. Then we change numbers and repeat: Old number k New number k' or k = k That correspondance between "old" and "new numbers" gives us that: J(n) = J(n)

78 The Josephus Problem recurrent equation () Case n = m +. First trip eliminates all even numbers. Then we change numbers and repeat: Old number k New number k' or k = k + That correspondence between "old" and "new numbers" givs us that: J(n + ) = J(n) +.

79 The Josephus Problem recurrent equation () Case n = m First trip eliminates all even numbers. Then we change numbers and repeat: Old number k New number k' or k = k + That correspondence between "old" and "new numbers" givs us that: J(n + ) = J(n) +.

80 The Josephus Problem application of recurrence The equation J() = ; J(n) = J(n), for n ; J(n + ) = J(n) +, for n. can be used for computing function for large arguments. For example J(86) = J(43) = 53 J(43) = J() + = 7 J() = J() = 3 J() = 7

81 The Josephus Problem closed formula Teoreem J( m + l) = l +, for m 0 and 0 l < m. Proof by induction over m. Basis If m = 0 then also l = 0, and J() = ; Step If m > 0 and m + l = n, then l is even and J( m +l) = J( m +l/) = (l/+) = l+ If m + l = n +, then considering that J(n + ) = + J(n) = + (l ) + = l + Q.E.D.

82 The Josephus Problem closed formula () Closed formula can be used for computing function J(n): Example We have 030 = 0 + 6, so J(030) = 6 + = 3.

83 Next section The Tower of Hanoi Lines in the Plane 3 The Josephus Problem 4 Intermezzo: Structural induction

84 Structural induction Premises Let S be a set having the following features: A set S B of basic cases is contained in S. Finitely many operations u i : S m i S, i =,...,n, exist such that, if x,...,x mi S, then u i (x,...,x mi ) S. 3 Nothing else belongs to S. Technique Let P be a property such that: Each base case x S B has property P. For every i =,...,n and every x,...,x mi S, if each value x,...,x mi has property P, then u i (x,...,x mi ) has property P. Then every element of S has property P.

85 Structural induction Premises Let S be a set having the following features: A set S B of basic cases is contained in S. Finitely many operations u i : S m i S, i =,...,n, exist such that, if x,...,x mi S, then u i (x,...,x mi ) S. 3 Nothing else belongs to S. Technique Let P be a property such that: Each base case x S B has property P. For every i =,...,n and every x,...,x mi S, if each value x,...,x mi has property P, then u i (x,...,x mi ) has property P. Then every element of S has property P.

86 Mathematical induction as a special case of structural induction Premises The set S = N of natural numbers is constructed as follows: A set S B = {0} of basic cases is contained in N. A single operation, the successor, s : N N, exists such that, if n N, then s(n) N. 3 Nothing else belongs to N. Technique Let P be a property such that 0 has property P. For every n N, if n has property P, then s(n) has property P. Then every n N has property P.

87 Mathematical induction as a special case of structural induction Premises The set S = N of natural numbers is constructed as follows: A set S B = {0} of basic cases is contained in N. A single operation, the successor, s : N N, exists such that, if n N, then s(n) N. 3 Nothing else belongs to N. Technique Let P be a property such that 0 has property P. For every n N, if n has property P, then s(n) has property P. Then every n N has property P.