2. Content Lecture 2 Algorithm
Part 2
The Towers of Hanoi
Permutation
Backtracking and Branch-And-Bound
The n-queens problem
Maze
Travelling Salesman
NP-complete class
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3. The Towers of Hanoi
Overview
The towers of Hanoi is a mathematical game or puzzle
Developed in 1883 by a French mathematician Édouard
Lucas
The Game
There are 3 rods with n disks in different sizes
At first all disks are placed at the first rod in order so that a
bigger size disk is below a smaller size disk
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4. The Towers of Hanoi
The objective of the puzzle is to move the entire stack to
another rod, obeying the following rules:
Only one disk may be moved at a time
Each move consists of taking the upper disk from one of
the rods and sliding it onto another rod
You can put the disk on top of other disks which may
already be present on that rod
But no disk may be placed on top of a smaller disk
Therefore it is only allowed to place a smaller disk on the
top of a bigger disk
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5. The Towers of Hanoi
Example
3 rods
3 disks
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6. The Towers of Hanoi
To solve this game that means moving n disks from rod 1 to
3 the following process is considered:
the first n-1 disks are moved from rod 1 to rod 2
than the last disk remaining on 1 are moved to 3
than n-1 disks are moved from 2 to 3
That means the roles of the three rods are always different
In total you need 2n – 1 moves
This is minimal
Example: for 3 disk you need 23 – 1 moves = 7
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7. The Towers of Hanoi
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Move disks a and
b to rod 2
Move disk c from
rod 1 to 3
Move disks a and
b from rod 2 to 3
Move disk a from
1 to 3
Move disk b from
rod 1 to 2
Move disk a from
3 to 2
1 2 3 1 2 3
How to move disk c from 1 to 3: How to move disks a and b from 1 to 2:
Note: the biggest disk is c, the middle disk is b and the smallest disk is a!
8. Permutation
A permutation is a sequence of n different object in a row
The order of the objects is altered by swapping the
elements so that each possibility is reached
There are n! different permutations for n objects
Example
A permutation of the three objects a, b, c would be:
abc bac cab
acb bca cba
n = 3
3! = 3*2 *1 = 6 permutations
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10. Permutation
Solution
Two solution methods exist for this problem
Method 1
For each permutation a1 a2 … an-1 you generate n new ones by
putting the number n at all possible places:
n a1 a2 … an-1 a1 n a2 …. an-1 …. a1 a2 … an-1 n
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11. Permutation
Example
From the permutation 1 we got 2 new permutation using Method 1 by
placing the new number 2 at all possible places:
21 12
From the permutation 21 we got three new permutation using Method 1
by placing the new number 3 at all possible places:
321 231 213
From the permutation 12 we got three new permutation using Method 1
by placing the new number 3 at all possible places:
312 132 123
These are all 6 possible permutation of the numbers 1, 2 and 3
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12. Permutation
From the permutation 321 we got four new permutation using
Method 1 by placing the new number 4 at all possible places:
4321 3421 3241 3214
From the permutation 231 we got four new permutation using
Method 1 by placing the new number 4 at all possible places:
4231 2431 2341 2314
If you do this for all 6 given permutation of the numbers 1,2
and 3 you will receive 24 permutation of the numbers 1,2,3 and
4
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13. Permutation
Method 2
For each permutation a1 a2 … an-1 an integer k with 1 ≤ k ≤ n is
added and each ai is increased by 1 if ai ≥ k.
Example
From the permutation 231 we got 4 new permutation using Method
2 by adding 1, 2, 3, and 4:
3421 3412 2413 2314
Note
The second method is more efficient if the permutation are placed
in an array because disarrange of parts of the array is not necessary
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15. The n-queens problem
This is a part of the chess game (pawn,
king, queen, bishop, knight, rook)
It considers only the queen token
A queen on a chessboard threaten all
fields
in the same row
in the same column
on both diagonals
Challenge: Find a position for n
queens on a n x n chessboard such that
they not threaten each other
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16. The n-queens problem
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Example
n = 4
4 queens are placed on a 4 x 4 chessboard
not considered the threating
You can place any queen in any row or
column
For the first queen you have 16 possibilities,
for the second queen 15, for the third 14 and
for the fourth 13
16*15*14*13 = 43680
Therefore if you try any possibility you
would need a lot of time
How can you solve this problem?
17. The n-queens problem
Idea
Build up a solution step by step by putting each queen one
by one on the chessboard so that it is not threaten by any
other queen
If there is no possible place left for the queen k the k-1
queen is removed from the chessboard and placed on
another not already tried position
This is done as long as a solution is found or all possibilities
are tried
This is an example of a method called Backtracking!
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18. The n-queens problem
Definition
A method in which every possibility is tested and in the case
of a death end the step before is withdrawn and a new variant
is tested is called a backtracking method.
Used to solve constraint satisfaction problems like
crosswords and puzzles
Often the most efficient solution for parsing and
optimization problems
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19. The n-queens problem
Backtracking is a systematical search in the whole set to
find acceptable states
In the most cases in backtracking it is very important to see
as early as possible death ends to avoid a performance
increase
A death end is a state in the given problem where you can
not do the next step without violating any of the criteria of
the given problem
Therefore you have to go back to the previous state
(backtracking)
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20. The n-queens problem
A death end in the n-queen
problem for n = 8 would be the
following illustration:
It is not possible to set another
queen
To find a solution you have to
even remove all queens beside
the first one
Only if the second queen is
placed from c7 to e7 (or f7/g7)
a solution can be found
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21. The n-queens problem
If you try a step by step method for backtracking you have to
consider that
The numbers of possible moves are limited in a clever way
To check immediately if the partial solution still fulfill the
necessary criteria which are needed for the whole solution
In the case of the n-queen problem you have (n2!)/(n2–n)!
possibilities to place n queens on a chessboard (not
considered the threating)
For n = 8 this are 64!/56! ~ 1014 possibilities
Extreme performance killing case if you try every possibility!
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22. The n-queens problem
If you look at the n-queen problem it is therefore necessary
to:
Consider for the queen k only the kth row on the
chessboard for the next position
To check for each new queen immediately if she is
threaten by the other queens on the chessboard
If you consider only the kth row for the queen k than the
number of possibilities is reduced to nn
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23. The n-queens problem
Example n = 4
• In this case: 44 = 256.
• If you threw away all
position where even less
then n queens threaten
each other than only 17
possibilities are left
• Under these possibilities
there are only 2 solutions
and 4 death ends
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24. The n-queens problem
Chose the data structure
The efficiency of the backtracking algorithm depends strongly
on the chosen data structure
How efficient is
The check if a solution is reached
The identification of possible steps in a given situation
The check of the usefulness of a partial solution
The execution of a step
If possible each step should have an effort of O(1)
You can improve for example the implementation of the n-queen
problem if you save the amount of threaten row, columns and
diagonals
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25. Backtracking
Other examples for back tracking
knapsack problem
A knapsack/rucksack can carry a weight W. Given n objects with a certain
value and weight. Problem: Objects should be chosen in such a way that a
maximal value will be acquired but the total weight of the knapsack is not
exceeded
Dye Problem
Given is a topographic map with N countries which are going to be dyed
with c different colours. Problem: To find a colour scheme so that all
countries with the same border having a different colour.
Solitaire
Sudoku
The figures 1 to 9 are placed in a 9x9 matrix divided into 9 3x3 fields after
particular rules
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28. Maze
A maze is an intricate and complex
network of interconnecting paths
Sometimes placed in gardens
Other name: labyrinth
Starting point S somewhere in the
maze
Problem
To find the way out of a maze
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S
29. Maze
Solution
Mark each point used as starting for a
new path (like crumb of bread)
You move in one direction until you
reach a wall
Than you turn either to the right or to
the left
If you reach a death end you unmark
the point and you go a step back and
try the other direction (left/right)
To find the shortest way out you can
use the branch–and–bound method
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. .. . . .
.. .. .. . .
. . .. . .
. . .. . .
. . S ..
.. .. .. ..
30. Travelling Salesman
Description
Be n a number of place in a n x n distance
matrix M where Mi,j is the distance between
the places i and j
Seek for the route with the minimal length
such that all places are reached exactly one
time and then return to the starting point
If you seek for an optimization or improvement
than a lot of branches can be cut off which are
only produce inefficient solutions
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31. Travelling Salesman
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Definition
A Branch-and-bound method is a general algorithm for
finding an optimal solution of various optimization
problems.
It consists of a systematic enumeration of all possible
solutions where inefficient solutions are eliminated by using
upper and lower estimated bounds
Be L the solution space and c: L R a function and lo a
solution to find such that:
c(lo) ≤ k for a given bound k or
c(lo) ≤ c(l) for all l ϵ L (global minimum)
32. Travelling Salesman
It is reasonable to find another function c’: L’ R such
that c’ is an estimate efficient lower bound for all part
solution l’ ϵ L’. l’ is derived from l
The function c’ can be used to decide if it is efficient to
pursue this part solution l’ or if it would be better to cut
this brunch
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33. Travelling Salesman
In the travelling salesman problem L is the amount of all
permutation over all the places 1…n and the function c the
length of a route
The function for a part solution can consider the already
travelled distance and an approximation for the length of
the route to the not reached places
If you just try every permutation the effort is O(n!) (or
O((n-1)!)
The branch-and-bound method helps to reduce the effort
but this is not a grantee
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34. Travelling Salesman
There is still no algorithm to really reduce the effort
The problem is therefore computationally difficult but a
large number of heuristics and exact methods are known,
so that some instances with tens of thousands of cities can
be solved
This problem belongs to a set of so called NP-complete
problems
The theory is that if you solve one of the NP-complete
problems all of them are solvable
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35. Travelling Salesman
Other branch-and-bound problems
Canadian traveller problem
Vehicle routing problem
Route inspection problem
Set TSP problem
Seven Bridges of Königsberg
Traveling repairman problem (minimum latency
problem)
Traveling tourist problem
Tube Challenge
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36. NP-complete class
Algorithm can provide an answer in polynomial time are
called class P or just P
NP problems are Nondeterministic Polynomial Time
(NP) problems
NP-hard problems can be described by: “at least as hard as
the hardest problem in NP”
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37. NP-complete class
In computational complexity theory, the complexity class
NP-complete (NP-C or NPC) is a class of decision
problems. A problem L is NP-complete if it has two
properties:
It is in the set of Nondeterministic Polynomial Time
(NP) problems: Any given solution to L can be verified
quickly (in polynomial time)
It is also in the set of NP-hard problems: Any NP-
problem can be converted into L by a transformation of
the inputs in polynomial time
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38. NP-complete class
Euler diagram for P,
NP, NP-complete,
and NP-hard set of
problems
It is still unknown if
P = NP or P ≠NP
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39. NP-complete class
A given solution to such a problem can be verified quickly
There is no known way so far to locate a solution in the first
place that means no fast solution is known to the NP-
complete problems
This implies that the time required solving the problem
using any known algorithm increase very quickly
Even for moderately large version of many of these
problems the required time reaches into billions of years
Therefore one of the unsolved problems in computer
science today is the so called P versus NP problem
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In computer science and linguistics, parsing, or, more formally, syntactic analysis, is the process Parsing: of analyzing a text, made of a sequence of tokens (for example, words), to determine its grammatical structure with respect to a given (more or less) formal grammar.
