The document discusses three recurrent problems: the Tower of Hanoi problem, the lines in the plane problem, and the Josephus problem. It explains that recurrent problems can be solved using recursion, where the solution to a problem relies on previously solved smaller instances of the same problem. It provides details on the recursive solutions and recurrence relations for calculating the number of moves in the Tower of Hanoi problem and the maximum number of regions created by lines in a plane. It also outlines the general steps to solve a recurrent problem by finding a closed-form expression through a recurrence relation and mathematical induction.

1. Recurrent Problem

2. Motivation
We will explore three sample problems that give a feel for what’s to come.
They have two traits in common:
 Investigated repeatedly
 Solutions all use the idea of recurrence
• Solution to each problem depends on the solutions to smaller instances of the same problem.

3. Problems with Recursive Solution
Easy to solve small problem.
Refers to previous steps and so on becomes hard to solve problems that are
relatively large.

4. Closed Form
An expression that can be expressed analytically in terms of a finite
number of certain "well-known" functions.
 Elementary Functions
o Constants
o One variable
o Finite number of exponentials or logarithms or roots of polynomial
o Four elementary operations (+ – × ÷).

5. Steps to Solve A Problem
Look at small cases.
Find and prove a mathematical expression for the quantity of interest.
Recurrence Relation.
Find and prove a closed form for our mathematical expression. Prove
the closed form using mathematical induction

6. Recurrent Problems
The Tower of Hanoi
The Lines in The Plane
The Josephus Problem

7. Tower of Hanoi Problem
Invented by the French mathematician Edouard Lucas in 1883.
Given a tower of n disks, initially stacked in decreasing size on one of three pegs.
The objective is
 Transfer the entire tower to one of the other pegs.
 Move only one disk at a time.
 Never move a larger one onto a smaller.

8. Tower of Hanoi : 3 Disks START
BEG AUX END

9. Tower of Hanoi : 3 Disks FINISH
BEG AUX END

10. Tower of Hanoi : Recursive Solution
Let’s call the three peg BEG(Source), AUX(Auxiliary) and END(Destination).
1) Move the top N – 1 disks from the BEG to AUX tower using END
2) Move the Nth disk from BEG to END tower
3) Move the N – 1 disks from AUX tower to END tower using BEG

11. Tower of Hanoi : 3 Disks
BEG AUX END
Move n-1 disks from BEG to AUX using END

12. Tower of Hanoi : 3 Disks
BEG AUX END

13. Tower of Hanoi : 3 Disks
BEG AUX END

14. Tower of Hanoi : 3 Disks
BEG AUX END
Move largest one from BEG to END

15. Tower of Hanoi : 3 Disks
BEG AUX END
Move n-1 disks from AUX to END using BEG

16. Tower of Hanoi : 3 Disks
BEG AUX END

17. Tower of Hanoi : 3 Disks
BEG AUX END

18. Tower of Hanoi : 3 Disks
BEG AUX END

19. Tower of Hanoi : No. of Moves, Tn
• for 1 disk it takes 1 move to transfer 1 disk from BEG to END;
• for 2 disks, it will take 3 moves: 2 Tn-1 + 1 = 2(1) + 1 = 3
• for 3 disks, it will take 7 moves: 2 Tn-1 + 1 = 2(3) + 1 = 7
• for 4 disks, it will take 15 moves: 2 Tn-1 + 1 = 2(7) + 1 = 15
• for 5 disks, it will take 31 moves: 2 Tn-1 + 1 = 2(15) + 1 = 31
• for 6 disks... ?

20. Tower of Hanoi : No. of Moves, Tn
• Explicit Pattern
Number of Disks Number of Moves
1 1
2 3
3 7
4 15
5 31
T0 = 0
Tn = 2 Tn-1 + 1,n>0

21. Tower of Hanoi : Recursive Solution Difficulties
T7 = ?
T7 = 2 T6 + 1
= 2(2 T5 + 1) + 1
= 4(2 T4 + 1) + 3
= 8(2 T3 + 1) + 7
= 16(2 T2 + 1) + 15
= 32(2 T1 + 1) + 31
= 64(2 T0 + 1) + 63
= 64 × 0 + 64+ 63
= 127
T50 = ?
T50 = 2 T49 + 1
= 2(2 T48 + 1) + 1
= 4(2 T47 + 1) + 3
= 8(2 T46 + 1) + 7
……………………….
=?

22. Tower of Hanoi : Finding Closed Form
• Explicit Pattern
Number of Disks Number of Moves
1 1
2 3
3 7
4 15
5 31
• Powers of two help reveal the pattern
Number of Disks Number of Moves
1 21 - 1 = 2 - 1 = 1
2 22 - 1 = 4 - 1 = 3
3 23 - 1 = 8 - 1 = 7
4 24 - 1 = 16 - 1 = 15
5 25 - 1 = 32 - 1 = 31
Tn = 2 n - 1
A Closed Form

23. Tower of Hanoi : Proving Closed Form
Basis : For n= 0, T0 = 2n – 1 = 20 – 1 = 0
Induction : Let for n = n-1 the expression is true, then Tn-1 = 2n-1 – 1
Hypothesis : For n = n ,
Tn = 2 Tn-1 + 1
= 2 (2n-1 – 1 ) + 1
= 2n-1+1 – 2 + 1
= 2n – 1

24. 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 Ln of regions defined by n
lines in the plane?

25. Lines in The Plane : Recursive Solution
Draw lines so that they intersect each others
Calculate maximum number of regions created by lines

26. Lines in The Plane : N = 0
L0 = 1

27. Lines in The Plane : N = 1
L1 = 2

28. Lines in The Plane : N = 2
L2 = 4

29. Lines in The Plane : N = 3
L3 = 7

30. Lines in The Plane : N = 4
L4 = 11

31. Lines in The Plane : Maximum No. of Regions, Ln
• for 0 line it creates 1 region;
• for 1 lines, it creates 2 region; : Ln-1 + n = 1 + 1 = 2
• for 2 lines, it creates 4 region; : Ln-1 + n = 2 + 2 = 4
• for 3 lines, it will take 7 region: Ln-1 + n = 4 + 3 = 7
• for 4 lines, it will take 10 region: Ln-1 + n = 7 + 4 = 11
• for 5 lines... ?
L0 = 1
Ln = Ln-1 + n, n>0

32. Lines in The Plane : Recursive Solution Difficulties
L7 = ?
L7 = L6 + 7
= L5 + 6 + 7
= L4 + 5+ 13
= L3 + 4 + 18
= L2 + 3 + 22
= L1 + 2 + 25
= L0 + 1 + 27
= 1+ 28
= 29
L50 = ?
L50 = L49 + 50
= L48 + 49+ 50
= L47 + 48+ 99
= L46 + 47 +14 7
……………………….
=?

33. Lines in The Plane: Finding Closed Form
Ln = Ln-1 + n
= Ln-2 + (n-1) + n
= Ln-3 + (n-2) + (n-1) + n
= Ln-2 + (n-3) + (n-2) + (n-1) + n
……………………..
= L0 + 1 + 2 + …… + (n-3) + (n-2) + (n-1) + n
= 1 +
𝒏(𝒏+𝟏)
𝟐

34. Lines in The Plane : Proving Closed Form
Basis : For n= 0, L0 = 1 +
𝟎(𝟎+𝟏)
𝟐
= 1
Induction : Let for n = n-1 the expression is true, then Ln-1 = 1 +
𝒏(𝒏 − 𝟏)
𝟐
Hypothesis : For n = n ,
Ln = Ln-1 + n
= 1 +
𝒏(𝒏 − 𝟏)
𝟐
+ n
= 1 +
𝒏𝟐
− 𝒏 + 𝟐𝒏
𝟐
= 1 +
𝒏𝟐
+ 𝒏
𝟐
= 1 +
𝒏(𝒏 + 𝟏)
𝟐

35. Josephus Problem
Do it Yourself
Solve Exercises

36. The End