and problem-solving techniques 2 Some mathematical principles and - - PDF document

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Problem Solving Skills (14021601-3 )

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Part 2 Some mathematical principles and problem-solving techniques

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Some mathematical principles and problem-solving techniques

Invariant Principle

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Gauss

When Karl Friedrich Gauss was still in elementary school (around the age of nine), a lazy teacher used to give some “hard” problems to his pupils, so that they would be busy solving them while he read the morning paper! One day, the teacher asked the students to calculate the total of all integer numbers from 1 to 100; he was sure he would be able to read for the next half hour! To his great surprise, young Gauss was ready in a couple of minutes.

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Gauss

By grouping all the numbers from 1 to 100 into pairs: (1 and 100), (2 and 99), (3 and 98), and so on till (50 and 51) he noticed that the total for each pair is 101. This was the invariant. Given that there were only 50 pairs, a simple multiplication of 101 × 50 = 5,050 gave him the answer!

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Tennis tournament

There is a tennis tournament with 937 players. The player who wins a game advances further, while the loser leaves the competition. How many games are required to complete the tournament?

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Tennis tournament

There are many possible approaches for finding the solution… But if you notice the invariant: number of players left in the tournament + number of games played = 937 you arrive at solution immediately…

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Puzzle

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After one exchange, what amount is greater? Amount of water in the juice or amount of juice in the water?

Puzzle

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Puzzle – A model

• Volume of a glass: V
• Volume of teaspoon x
• The ratio of juice in the water:

x / (V + x)

• Amount of juice coming back:

x × x / (V + x)

• Etc.

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Puzzle – Another model

• Volume of water w and juice j on the return

trip of the teaspoon

• Volume of teaspoon x = w + j
• Amount of water in the juice:

w

• Amount of juice in the water:

x – j = w

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Puzzle – Invariance principle

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Puzzle

Invariance principle is very helpful in solving problems where there is repetition in the statement of the problem. The invariance principle simply advises to look for things that do not change (i.e. invariants). For water & juice problem, the volumes in both glasses do not change…

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Puzzle – Invariance principle

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Puzzle – Invariance principle

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Puzzle

A rectangular chocolate bar consists of m × n small rectangles and you wish to break it into its constituent parts. At each step, you can only pick up one piece and break it along any of its vertical or horizontal lines. How should you break the chocolate bar using the minimum number of steps (breaks)?

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Puzzle

The invariance principle simply advises to look for things that do not change (i.e. invariants). For chocolate bar puzzle, the number of pieces is always greater by 1 than the number of breaks which has been made…

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Puzzle

Number of breaks Number of pieces 1 1 2 2 3 3 4 . . . . . . 62 63

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Invariance principle

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Question

A tetromino is a figure made from 4 squares of the same size. Consider five different tetrominoes, called O-, Z-, L-, T-, and I-tetrominoes (given below in this order):

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Question

(a) Is it possible to cover a rectangular board of 11 × 15 squares with the above tetrominoes? Justify your answer. (b) Is it possible to cover a rectangular board

• f

10 × 30 squares with T- tetrominoes only? Justify your answer.

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Invariance principle again

• Puzzle. A knight fights a dragon with 100 heads. This

knight can cut off 15, 17, 20, or 5 heads with one blow of his sword, however, in each of these cases a number of new heads will grow immediately: 24, 2, 14, or 17 heads, respectively. The dragon dies only if all his heads are cut off. Is it possible to kill the dragon?

Any Question ???

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Some mathematical principles and problem-solving techniques

Divide and Conquer & Backtracking

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Divide and Conquer

Split the problem P into sub-problems P1 , P2 , …, Pk For each sub-problem P1 , P2 , …, Pk : If size (Pi) is small then solve Pi ,

• therwise split the sub-problem Pi

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Puzzle

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Puzzle

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Puzzle

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Puzzle

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Backtracking

Backtracking is a refinement of the brute force approach, which systematically searches for a solution to a problem among all available options. It does so by assuming that the solutions are represented by vectors (v1, ..., vk)

• f values and by systematically checking the domains of

the vectors until the solutions are found.

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Puzzle

Place 8 queens on the chessboard in such a way that none

• f the queens attacks

any other

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Puzzle

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Puzzle

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Puzzle

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Puzzle

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Puzzle

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Puzzle

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Puzzle

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Puzzle

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Puzzle

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Puzzle

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Puzzle

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Puzzle

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Puzzle

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Puzzle

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Backtracking…

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More backtracking…

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Continuing

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Continuing

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Even more backtracking

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Forward again

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Puzzle : A solution

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Backtracking

You have used backtracking in your last homework:

• If s = 1 then g = 81/6 which is not a whole number.
• Consider the case, then, when n = 4, i.e., there were

4 events. You’ll use backtracking in solving many problems to come…

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Knapsack problem

There is a number n of items. Each item has some weight (w1, w2, …, wn). Each item has a value, often called profit (p1, p2, …, pn). The thief’s dilemma is to maximize the total value of the items while not making the total weight exceed W. How to find the best subset of items?

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Knapsack problem – example

Items: 1 2 3 4 5 6 7 Weights: 12 10 9 13 8 11 10 Values: $25 $21 $18 $30 $16 $23 $22 The total weight W which can be carried away by a thief is 32. How to find the best subset of items?

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Knapsack problem

There are many possible approaches for such knapsack problem:

• A greedy approach
• Dynamic programming
• Integer programming
• Branch and bound
• Heuristic methods
• Modern heuristic methods
• Backtracking…

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Knapsack problem

$0 32

value capacity left

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Knapsack problem

$0 32

value capacity left

$25 20

Item #1

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Knapsack problem

$0 32

value capacity left

$25 20

Item #1 Item #2

$46 10

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Knapsack problem

$0 32

value capacity left

$25 20

Item #1 Item #2

$46 10

Item #7

$68

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Knapsack problem

$0 32 $25 20

Item #1 Item #2

$46 10

Item #7

$68

Item #4

$30 17

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Puzzle

 There are 100 pebbles on the table. There are

two players, A and B, who move alternatively. Player A moves first. The rules of the game are the same for both players: at each move they can remove one, two, three, four, or five

• pebbles. The winner is the one who take the

last pebbles. What is the winning strategy for player A, if one exists?

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Games

We will discuss games towards the end of this course if time permits – but you should see already a huge applicability

• f

backtracking in a variety of strategic games (including checkers, chess, tic-tac-toe, etc).

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Any Question ???

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Some mathematical principles and problem-solving techniques

Constraints

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Processing constraints

• Constraints are helpful as they can lead you

towards the solution of the problem…

• Models are usually simple: A number of

variables correspond to some objects and each variable has a domain.

• The process of solving a problem is the

process

• f

narrowing the domain

• f

variables. Example: Sherlock Holmes stories…

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Two men meet on the street (they have not seen each other for many years): A: All three of my sons celebrate their birthday today. Can you tell me how old each one is? B: Yes, but you have to tell me something about them… A: The product of their ages is 36. B: I need more info… A: The sum of their ages is equal to the number of windows in the building next to us… B: I need more info… A: My oldest son has blue eyes. B: That is sufficient!

Puzzle

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Puzzle

Remember 3 rules?

• Do you understand the problem?
• Do you have any intuition on the solution?
• Can we build a model?

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Puzzle – A model

• Age of the first son: x
• Age of the second son: y
• Age of the third son: z

Safe assumption: x ≥ y ≥ z dom(x) = dom(y) = dom(z) = {1, 2,…, 36}

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36 1 1 18 2 1 12 3 1 9 4 1 9 2 2 6 6 1 6 3 2 4 3 3

Puzzle

• The product of their ages is 36:

x y z

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36 + 1 + 1 = 38 18 + 2 + 1 = 21 12 + 3 + 1 = 16 9 + 4 + 1 = 14 9 + 2 + 2 = 13 6 + 6 + 1 = 13 6 + 3 + 2 = 11 4 + 3 + 3 = 10

Puzzle

• The sum of their ages is equal to the number
• f windows in the building next to us…

x y z

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There are five houses, each of a different color and inhabited by men of different nationalities, with one unique pet, drink, and car. Some facts are given:

• 1. The Englishman lives in the red house.
• 2. The Spaniard owns the dog.
• 3. The man in the green house drinks cocoa.
• 4. The Ukrainian drinks eggnog.
• 5. The green house is immediately to the right (your right) of the ivory house.
• 6. The owner of the Oldsmobile also owns snails.
• 7. The owner of the Ford lives in the yellow house.
• 8. The man in the middle house drinks milk.
• 9. The Norwegian lives in the first house on the left.
• 10. The man who owns the Chevrolet lives in the house next to the house where the

man owns a fox.

• 11. The Ford owner's house is next to the house where the horse is kept.
• 12. The Mercedes-Benz owner drinks orange juice.
• 13. The Japanese drives a V
• lkswagen.
• 14. The Norwegian lives next to the blue house.

Puzzle

Who owns the zebra?

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Puzzle – a model

House 1 2 3 4 5 Color Drink Country Car Pet

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Puzzle – a model

There are:

• five colors: blue, green, ivory, red, and yellow;
• five drinks: cocoa, eggnog, milk, orange, and

water;

• five nationalities: Englishman, Japanese,

Norwegian, Spaniard, and Ukrainian;

• five cars: Chevrolet, Ford, Mercedes, Oldsmobile,

and Volkswagen;

• five pets: dog, fox, horse, snails, and zebra.

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Puzzle – arbitrary arrangement

House 1 2 3 4 5 Color

blue yellow ivory red green

Drink

milk eggnog

• range

water cocoa

Country

Englishman Ukrainian Norwegian Japanese Spaniard

Car

Mercedes Chevrolet Oldsmobile Ford V

• lkswagen

Pet

zebra dog snails horse fox

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• 1. The Englishman lives in the red house.
• 2. The Spaniard owns the dog.
• 3. The man in the green house drinks cocoa.
• 4. The Ukrainian drinks eggnog.
• 5. The green house is immediately to the right (your right) of the ivory house.
• 6. The owner of the Oldsmobile also owns snails.
• 7. The owner of the Ford lives in the yellow house.
• 8. The man in the middle house drinks milk.
• 9. The Norwegian lives in the first house on the left.
• 10. The man who owns the Chevrolet lives in the house next to the house where the

man owns a fox.

• 11. The Ford owner's house is next to the house where the horse is kept.
• 12. The Mercedes-Benz owner drinks orange juice.
• 13. The Japanese drives a V
• lkswagen.
• 14. The Norwegian lives next to the blue house.

Puzzle

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Puzzle

House 1 2 3 4 5 Color

blue14

Drink

milk8

Country

Norwegian9

Car Pet

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• 1. The Englishman lives in the red house.
• 2. The Spaniard owns the dog.
• 3. The man in the green house drinks cocoa.
• 4. The Ukrainian drinks eggnog.
• 5. The green house is immediately to the right (your right) of the ivory house.
• 6. The owner of the Oldsmobile also owns snails.
• 7. The owner of the Ford lives in the yellow house.
• 8. The man in the middle house drinks milk.
• 9. The Norwegian lives in the first house on the left.
• 10. The man who owns the Chevrolet lives in the house next to the house where the

man owns a fox.

• 11. The Ford owner's house is next to the house where the horse is kept.
• 12. The Mercedes-Benz owner drinks orange juice.
• 13. The Japanese drives a V
• lkswagen.
• 14. The Norwegian lives next to the blue house.

Puzzle

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Puzzle

House 1 2 3 4 5 Color

yellow1,5 blue14

Drink

milk8

Country

Norwegian9

Car Pet

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• 1. The Englishman lives in the red house.
• 2. The Spaniard owns the dog.
• 3. The man in the green house drinks cocoa.
• 4. The Ukrainian drinks eggnog.
• 5. The green house is immediately to the right (your right) of the ivory house.
• 6. The owner of the Oldsmobile also owns snails.
• 7. The owner of the Ford lives in the yellow house.
• 8. The man in the middle house drinks milk.
• 9. The Norwegian lives in the first house on the left.
• 10. The man who owns the Chevrolet lives in the house next to the house where the

man owns a fox.

• 11. The Ford owner's house is next to the house where the horse is kept.
• 12. The Mercedes-Benz owner drinks orange juice.
• 13. The Japanese drives a V
• lkswagen.
• 14. The Norwegian lives next to the blue house.

Puzzle

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Puzzle

House 1 2 3 4 5 Color

yellow1,5 blue14

Drink

milk8

Country

Norwegian9

Car

Ford7

Pet

horse11

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Puzzle

Now we can consider two cases for two possible sequences of colors for houses 3, 4, and 5:

• ivory, green, red;
• red, ivory, green;

Let’s consider the first possibility: ivory, green, red.

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Puzzle

House 1 2 3 4 5 Color

yellow1,5 blue14 ivory green red

Drink

milk8

Country

Norwegian9

Car

Ford7

Pet

horse11

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• 1. The Englishman lives in the red house.
• 2. The Spaniard owns the dog.
• 3. The man in the green house drinks cocoa.
• 4. The Ukrainian drinks eggnog.
• 5. The green house is immediately to the right (your right) of the ivory house.
• 6. The owner of the Oldsmobile also owns snails.
• 7. The owner of the Ford lives in the yellow house.
• 8. The man in the middle house drinks milk.
• 9. The Norwegian lives in the first house on the left.
• 10. The man who owns the Chevrolet lives in the house next to the house where the

man owns a fox.

• 11. The Ford owner's house is next to the house where the horse is kept.
• 12. The Mercedes-Benz owner drinks orange juice.
• 13. The Japanese drives a V
• lkswagen.
• 14. The Norwegian lives next to the blue house.

Puzzle

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Puzzle

House 1 2 3 4 5 Color

yellow1,5 blue14 ivory green red

Drink

milk8

Country

Norwegian9 Englishman1

Car

Ford7

Pet

horse11

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• 1. The Englishman lives in the red house.
• 2. The Spaniard owns the dog.
• 3. The man in the green house drinks cocoa.
• 4. The Ukrainian drinks eggnog.
• 5. The green house is immediately to the right (your right) of the ivory house.
• 6. The owner of the Oldsmobile also owns snails.
• 7. The owner of the Ford lives in the yellow house.
• 8. The man in the middle house drinks milk.
• 9. The Norwegian lives in the first house on the left.
• 10. The man who owns the Chevrolet lives in the house next to the house where the

man owns a fox.

• 11. The Ford owner's house is next to the house where the horse is kept.
• 12. The Mercedes-Benz owner drinks orange juice.
• 13. The Japanese drives a V
• lkswagen.
• 14. The Norwegian lives next to the blue house.

Puzzle

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Puzzle

House 1 2 3 4 5 Color

yellow1,5 blue14 ivory green red

Drink

milk8 cocoa3

Country

Norwegian9 Englishman1

Car

Ford7

Pet

horse11

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• 1. The Englishman lives in the red house.
• 2. The Spaniard owns the dog.
• 3. The man in the green house drinks cocoa.
• 4. The Ukrainian drinks eggnog.
• 5. The green house is immediately to the right (your right) of the ivory house.
• 6. The owner of the Oldsmobile also owns snails.
• 7. The owner of the Ford lives in the yellow house.
• 8. The man in the middle house drinks milk.
• 9. The Norwegian lives in the first house on the left.
• 10. The man who owns the Chevrolet lives in the house next to the house where the

man owns a fox.

• 11. The Ford owner's house is next to the house where the horse is kept.
• 12. The Mercedes-Benz owner drinks orange juice.
• 13. The Japanese drives a V
• lkswagen.
• 14. The Norwegian lives next to the blue house.

Puzzle

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Puzzle

House 1 2 3 4 5 Color

yellow1,5 blue14 ivory green red

Drink

eggnog4 milk8 cocoa3

Country

Norwegian9 Ukrainian4 Englishman1

Car

Ford7

Pet

horse11

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• 1. The Englishman lives in the red house.
• 2. The Spaniard owns the dog.
• 3. The man in the green house drinks cocoa.
• 4. The Ukrainian drinks eggnog.
• 5. The green house is immediately to the right (your right) of the ivory house.
• 6. The owner of the Oldsmobile also owns snails.
• 7. The owner of the Ford lives in the yellow house.
• 8. The man in the middle house drinks milk.
• 9. The Norwegian lives in the first house on the left.
• 10. The man who owns the Chevrolet lives in the house next to the house where the

man owns a fox.

• 11. The Ford owner's house is next to the house where the horse is kept.
• 12. The Mercedes-Benz owner drinks orange juice.
• 13. The Japanese drives a V
• lkswagen.
• 14. The Norwegian lives next to the blue house.

Puzzle

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Puzzle

House 1 2 3 4 5 Color

yellow1,5 blue14 ivory green red

Drink

eggnog4 milk8 cocoa3

• range12

Country

Norwegian9 Ukrainian4 Englishman1

Car

Ford7 Mercedes12

Pet

horse11

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• 1. The Englishman lives in the red house.
• 2. The Spaniard owns the dog.
• 3. The man in the green house drinks cocoa.
• 4. The Ukrainian drinks eggnog.
• 5. The green house is immediately to the right (your right) of the ivory house.
• 6. The owner of the Oldsmobile also owns snails.
• 7. The owner of the Ford lives in the yellow house.
• 8. The man in the middle house drinks milk.
• 9. The Norwegian lives in the first house on the left.
• 10. The man who owns the Chevrolet lives in the house next to the house where the

man owns a fox.

• 11. The Ford owner's house is next to the house where the horse is kept.
• 12. The Mercedes-Benz owner drinks orange juice.
• 13. The Japanese drives a V
• lkswagen.
• 14. The Norwegian lives next to the blue house.

Puzzle

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Contradiction! When we assumed that the sequence of colors for houses 3, 4, and 5 is: ivory, green, red it is impossible to satisfy the remaining constraints

• f the problem.

Thus: backtrack! The second sequence: red, ivory, green must be true…

Puzzle

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Puzzle

House 1 2 3 4 5 Color

yellow1,5 blue14 red ivory green

Drink

milk8

Country

Norwegian9

Car

Ford7

Pet

horse11

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The rest is easy… We can continue in a similar fashion and we can easily complete the table… Then we can answer the question: “Who owns the zebra?”

Puzzle

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Puzzle

House 1 2 3 4 5 Color

yellow1,5 blue14 red ivory green

Drink

water eggnog milk8

• range

cocoa

Country

Norwegian9 Ukrainian Englishman Spaniard Japanese

Car

Ford7 Chevrolet Oldsmobile Mercedes V

• lkswagen

Pet

fox horse11 snails dog zebra

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A standard Sudoku puzzle is a table of 3 × 3 boxes, each of which contains a 3 × 3 array

• f cells; altogether Sudoku table looks like

9 × 9 grid.

Sudoku

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Sudoku

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At the start of the puzzle some cells of the grid are filled with integer numbers from 1 to 9 and the

• thers are empty. Our task is to fill in the empty

cells in such a way that:

• Every row of the whole 9 × 9 grid must contain

all the numbers from 1 to 9.

• Every column of the whole 9 × 9 grid must

contain all the numbers from 1 to 9.

• Every 3 × 3 box must contain all the numbers

from 1 to 9.

Sudoku

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100

Puzzle

7 6 2 4 5 5 8 2 6 9 3 8 8 9 5 1 2 6 3 3 1 9 9 3 6 4 5 9 8 4 2 9 6 7

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Puzzle

7 6 2 4 5 5 8 2 6 9 3 8 8 9 5 1 2 6 3 3 1 9 9 3 6 4 5 9 8 4 4 2 9 6 7

102

• Mr. and Mrs. Smith invited 4 married couples for dinner. When

everyone arrived, they greeted each other by a handshake. These were the rules: – no one shakes his/her own hand – no handshake between married couple – no pair does it twice – there are pairs which did not shake their hands

• Mr. Smith asked all people in the room, how many times

did they shake their hands, and all answers he got were different. How many hands did Mrs. Smith shake?

Puzzle

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Puzzle

Remember 3 rules?

• Do you understand the problem?
• Do you have any intuition on the solution?
• Can we build a model?

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Puzzle – a model

• Mr. Smith

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• Mr. Smith

1 2 3 4 5 6 7 8

Puzzle – a model

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106

• Mr. Smith

1 2 3 4 5 6 7 8

Puzzle – a model

107

• Mr. Smith

1 2 3 4 5 6 7 8

Puzzle – a model

108

• Mr. Smith

1 2 3 4 5 6 7 8

Puzzle – a model

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• Mr. Smith

1 2 3 4 5 6 7 8

Puzzle – a model

110

• Mr. Smith

1 2 3 4 5 6 7 8

Puzzle – a model

111

• Mr. Smith

1 2 3 4 5 6 7 8

• Mrs. Smith

Puzzle – a model

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112

Puzzle

• Mrs. Brown, Mrs. White, and Mrs. Green went to a

hairdresser where they experimented with different colors and haircuts. When they left the hairdresser, the lady with green hair said: – “Have you noticed that although our hair colors match our names, none of us has the same hair color as our name?”

• Mrs. Brown replied:

– “Indeed, you are right! This is remarkable!”

What hair color did each lady have?

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Puzzle

Rule #1: Do we understand the problem? Rule #2: Reject your intuition “It is impossible to tell!” Rule #3: Construct a model by listing all the variables, constraints, and objectives. The letters B, W, and G mark three colors: (B) brown, (W) white, and (G) green. If we knew nothing apart that three ladies selected some colors for their hair from this list, the model would be:

• Mrs. Brown:

B or W or G

• Mrs. White:

B or W or G

• Mrs. Green:

B or W or G

114

Puzzle

Initial model:

• Mrs. Brown:

B or W or G

• Mrs. White:

B or W or G

• Mrs. Green:

B or W or G As none of the ladies’ hair color matched their name (this is the first fact, hence the first constraint), the possible arrangements are:

• Mrs. Brown:

W or G

• Mrs. White:

B or G

• Mrs. Green:

B or W

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115

Puzzle

Now we are ready for considering the other constraints. As the lady with green hair made a remark that was answered by a different lady, Mrs. Brown, so the immediate inference is that Mrs. Brown does not have green hair! And this is all we need to solve this puzzle, as:

• Mrs. Brown:

W or G

• Mrs. White:

B or G

• Mrs. Green:

B or W

116

Puzzle

Now:

• Mrs. Brown:

W

• Mrs. White:

B or G

• Mrs. Green:

B or W implies:

• Mrs. Brown:

W

• Mrs. White:

B or G

• Mrs. Green:

B which in turn implies:

• Mrs. Brown:

W

• Mrs. White:

G

• Mrs. Green:

B

117

Puzzle

Find a number that consists of six digits such that the result of multiplying that number with any number in the range from two to six, inclusive, is a number that consists of the same six digits as the original number but in a different order.

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118

Puzzle

We are after a six-digit number x that satisfies the given property, and we can mark each digit by a separate variable: a is the first digit, b is the second digit, and so on. x → a b c d e f Of course, for each variable, there are 10 available digits: from 0 to 9.

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Puzzle

x → a b c d e f 2x → * * * * * * 3x → * * * * * * 4x → * * * * * * 5x → * * * * * * 6x → * * * * * *

120

Puzzle

x → a b c d e f 2x → * * * * * * 3x → * * * * * * 4x → * * * * * * 5x → * * * * * * 6x → * * * * * *

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121

Puzzle

x → 1 b c d e f 2x → * * * * * * 3x → * * * * * * 4x → * * * * * * 5x → * * * * * * 6x → * * * * * *

122

Puzzle

What are the consequences of x → 1 b c d e f ? Since the six digits for x, 2x, 3x, 4x, 5x, and 6x are the same, and each of these numbers starts with a digit greater than the first digit of the previous number, then

• all six digits are different!
• also, the digit 0 is not included.

123

Puzzle

So we know that all digits are from a range from 1 to 9, and they are different, i.e. a ≠ b, a ≠ c, a ≠ d, a ≠ e, a ≠ f, b ≠ c, and so on. Now, what about f: x → 1 b c d e f ?

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124

Puzzle

f = 1 f = 2 f = 3 f = 4 f = 5 f = 6 f = 7 f = 8 f = 9

125

Puzzle

f = 1 – impossible! f = 2 – impossible! f = 3 – ? f = 4 – impossible! f = 5 – impossible! f = 6 – impossible! f = 7 – ? f = 8 – impossible! f = 9 – ?

126

Puzzle

So f = 3, f = 7, or f = 9…

• If f = 3:

x → 1 b c d e 3 then the last digits of 2x, 3x, 4x, 5x, and 6x are 6, 9, 2, 5, and 8. Impossible (1 is not included).

• If f = 9:

x → 1 b c d e 9 then the last digits of 2x, 3x, 4x, 5x, and 6x are 8, 7, 6, 5, and 4. Impossible (1 is not included).

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127

Puzzle

x → 1 b c d e 7 2x → * * * * * * 3x → * * * * * * 4x → * * * * * * 5x → * * * * * * 6x → * * * * * *

128

Puzzle

Actually, as f = 7, we know much more… The digits we are after are: 7, 4, 1, 8, 5, and 2 In sorted order: 1, 2, 4, 5, 7, and 8, which means that:

129

Puzzle

x → 1 b c d e 7 2x → 2 * * * * 4 3x → 4 * * * * 1 4x → 5 * * * * 8 5x → 7 * * * * 5 6x → 8 * * * * 2

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130

Puzzle

Further observation: all columns contain different digits (i.e. a permutation of 1, 2, 4, 5, 7, and 8). Why? Can we take advantage of this observation?

131

Puzzle

Can we take advantage of this observation? Yes, we can!

132

Puzzle

x → 1 b c d e 7 2x → 2 * * * * 4 3x → 4 * * * * 1 4x → 5 * * * * 8 5x → 7 * * * * 5 6x → 8 * * * * 2 _______________________ 21x = 2 7 7 7 7 7 7

45

133

Puzzle

As 21x = 2,777,777 then x = 142,857

134

Puzzle

• Mr. White, Mr. Brown, and Mr. Green took

part in a shooting competition where each of them fired 6 shots. The competition resulted in a three-way draw, as each participant got 71 points. The first two shots of Mr. White gave him 22 points and the first shot of Mr. Green gave him 3 points. Who hit the bull’s eye?

135

Puzzle

1 2 3 5 10 20 25 50

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136

Puzzle

We know the scores for all 18 shots made by the three competitors; these are (from the best to the worst): 50, 25, 25, 20, 20, 20, 10, 10, 10, 5, 5, 3, 3, 2, 2, 1, 1, 1.

137

Puzzle

Let us denote the scores of Mr. White, Mr. Brown, and Mr. Green by:

• Mr. White: w1, w2, w3, w4, w5, w6
• Mr. Brown: b1, b2, b3, b4, b5, b6
• Mr. Green: g1, g2, g3, g4, g5, g6

Q: Is w1 the score of the first shot by Mr. White or rather the score of his best shot?

138

Puzzle

It does not really matter that much (both approaches would lead to a solution) but it is important to define these precisely to model the

• constraints. So, let us assume that all shots of

all three competitors are sorted from the best to the worst: w1 ≥ w2 ≥ w3 ≥ w4 ≥ w5 ≥ w6 b1 ≥ b2 ≥ b3 ≥ b4 ≥ b5 ≥ b6 g1 ≥ g2 ≥ g3 ≥ g4 ≥ g5 ≥ g6

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139

Puzzle

What do we know? w1 + w2 + w3 + w4 + w5 + w6 = b1 + b2 + b3 + b4 + b5 + b6 = g1 + g2 + g3 + g4 + g5 + g6 = 71

140

Puzzle

We should convince ourselves (it is time- consuming exercise) that the only way to partition the set of scores into three disjoint subsets of six shots with a total of 71 for each subset is: 25, 20, 20, 3, 2, 1 25, 20, 10, 10, 5, 1 50, 10, 5, 3, 2, 1

141

Puzzle

Anything else?

• There exist 1 ≤ i, k ≤ 6 such that

wi + wk = 22, as two shots of Mr. White gave him 22 points.

• There exist 1 ≤ i ≤ 6 such that gi = 3, as
• ne shot of Mr. Green gave him 3 points.

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142

Puzzle

The first row must represent results of Mr. White, as it is the only row that contains two numbers whose sum is 22. Then the third row must be of Mr. Green, as it is the only remaining row that contains 3. Thus it was Mr. Green who hit the bull’s eye …

143

Puzzle

• 31 participants
• ne main organizer
• it is

it is necessary to test, whether all participants are logicians…

• each participant gets from the organizer a dot of some colour
• n his forehead
• the organizer assures one participant (who expressed a

concern) that everyone will be able to guess the colour of his/her dot at some stage

Annual International Conference on Logic:

144

• 1st time: 4 people left
• 2nd time: all with red dots left
• 3rd time: no one left
• 4th time: at least one left
• shortly afterwards: a participant, who expressed concern left

together with his sister; both had dots of different colour… At that time there were still some people left…

How many times did the bell ring?

This is what happened. When the bell ring:

Puzzle

49

145

Puzzle

What does it mean? What did they see? What did they know at that time? When the bell ring the 1st time, 4 people left…

146

Puzzle

Hint: You are one of the participants. You look around and you see only one participant with yellow dot. What can you claim? When the bell ring the 1st time, 4 people left…

Any Question ???

50

148

Some mathematical principles and problem-solving techniques

Optimisation What is the best arrangement?

149

Optimization

Many problems fall into category of optimization

• problems. These problems require finding the best

solution among many possible solutions. There is hardly a real-world problem without an optimization

• component. For example, how should we get to a

particular destination in the shortest possible time? How should we schedule orders on a production line to minimize the production cost? How should we cut components from a piece of metal to minimize the waste? And so on…

150

A(1), B(2), C(5), D(10) Four travellers approached a bridge… How should the travelers schedule the crossing

• f the bridge to minimize the total time?

Puzzle

51

151

A possible solution: A & B ↑ 2 A ↓ 1 A & C ↑ 5 A ↓ 1 A & D ↑ 10 A(1), B(2), C(5), D(10)

Rule #2

Puzzle

152

Puzzle

Find the shortest possible travel distance for a salesman that must visit every city in his territory (exactly once) and then return home. The diagram below represents a seven-city version of this problem:

5 1 4 2 7 6 3

153

Puzzle

1 2 3 4 5 6 7 1 19 27 17 21 31 2 16 15 18 14 22 26 3 25 14 20 18 13 4 16 20 11 5 19 12 17 12 17 6 33 24 20 23 15 7 29 14 16

52

154

Puzzle

Different tours may have different costs (total distances). For example, if we follow the tour: 2 – 3 – 7 – 6 – 1 – 4 – 5 – 2 the quality measure (i.e. the total distance) of this tour is 15 + 13 + 16 + 33 + 17 + 11 + 12 = 117. If the salesman follows a different tour: 2 – 1 – 3 – 7 – 6 – 5 – 4 – 2 then the total distance would be 127… Which tour provides the shortest total distance?

155

Puzzle

• A 10-city problem has 362,880 possible

solutions.

• A 20-city problem has approximately 1017

possible solutions.

• A 50-city problem has approximately 1064

possible solutions.

156

Puzzle 6.5

y x 48 288 99 33 x = 18 y = 45

x + 6y ≤ 288 3x + y ≤ 99 maximize $20x + $30y

This diagram explains the principles of linear programming

53

157

Puzzle

From a piece of rectangular paper 20×30 centimeters, cut corners in such a way that after bending the sides

• f the paper we get box of the largest possible volume:

20 cm 30 cm

158

Puzzle

V = (20 – 2x)(30 – 2x)x = = 4x3 – 100x2 + 600x This is example of a non-linear problem – nonlinear programming techniques must be used. Solution: x ≈ 3.92 cm, and V ≈ 1056.31 cm3

159

Making connections…

• Make an efficient connection:

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160

length = 3

A possibility

161

length = 2.83

Another possibility…

162

length = 3.14

And another possibility….

55

163

length = 2.73 n points – so called Steiner’s problem… still unsolved…

A solution

164

Find a point F such that the total distance AF + BF + CF is minimum

How did it start?

F A C B

165

Puzzle

A rectangular chocolate bar consists of m × n small rectangles and you wish to break it into its constituent parts. At each step, you can only pick up one piece and break it along any of its vertical or horizontal lines. How should you break the chocolate bar using the minimum number of steps (breaks)?

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166

Key questions

Key questions for any optimization problem: (1) What is a solution (potential/candidate solution versus final solution)? (2) What is feasible solution? Infeasible one? (3) What is representation of a solution? (4) What is the quality measure of a solution?

167

Optimization

all solutions (search space)

168

Optimization

quality measure all solutions (search space)

57

169

Optimization

feasible solution infeasible solution quality measure Optimization task: Find the highest quality feasible solution…

170

feasible solution quality measure Optimization task: Find the highest quality feasible solution…

Optimization

another quality measure infeasible solution

171

Puzzle

Suppose you wish to know which floors in a 36-story building are safe to drop eggs from and which will cause the eggs to break on landing (using a special container for the eggs). We eliminate chance and possible differences between different eggs (e.g. one egg breaks when dropped from the 7th floor and another egg survives a drop from the 20th floor) by making a few (reasonable!) assumptions (Rule #1):

58

172

Puzzle

• An egg that survives a drop can be used again (no

harm is done and the egg is not weaker).

• A broken egg can not be used again for any

experiment.

• The effect of a fall is the same for all eggs.
• If an egg breaks when dropped from some floor, it

would break also if dropped from a higher floor.

• If an egg survives a fall when dropped from some

floor, it would survive also if dropped from a lower floor.

173

Puzzle

Further, there are no pre-existing assumptions concerning when the egg will

• break. It is possible that a drop from the

first floor in the special container would break an egg. It is also possible that a drop from the 36th floor in the special container would not break an egg.

174

Puzzle

Now, if only one egg is available for experimentation, we have no choice. To

• btain the required result, we have to start

by dropping the egg from the first floor. If it breaks, we know the answer. If it survives, we drop it from the second floor and continue upward until the egg breaks. The worst-case scenario would require 36 drops to determine the egg-breaking floor.

59

175

Puzzle

Suppose we have two eggs. What is the least (minimum) number of egg drops required to determine the egg-breaking floor?

176

Puzzle

Most people when given this puzzle try to start somewhere in the middle of the building, e.g. dropping the first egg from the half-height. Most likely they use – more

• r less consciously – a technique called

binary search, where we divide a sample in half (or as close to half as possible) and – based on the outcome – proceed further.

177

Puzzle

Is the binary search a good approach?

If we drop the first egg from, say, the 18th floor, there are two possible outcomes:

• The egg breaks. In this case we have to test

every floor starting from floor 1. In the worst case, we drop the second egg 17 times to determine the egg-breaking floor....

• The egg does not break. We have to examine

remaining 18 floors; however, we have still two eggs for experimentation…

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178

Puzzle

These two cases suggest that we should start lower than the 18th floor: if the first egg breaks, we will have a shorter segment of floors to experiment with; if the first egg does not break, the segment would be longer but we would have still two eggs for our experiments…

179

Key questions, again

What is representation of a solution? a b c A drop from floor level a. The thick line indicates that the egg breaks and the thin line that it does not. In the former case, we can drop the second egg from the level b, whereas in the latter case we drop the first egg from the level c.

180

Example

What is the quality measure of a solution? 14 1 22 15 30

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181

Puzzle

The quality measure of a solution is determined by the longest branch of the strategy tree: the shorter the longest branch is, the better the solution… So the problem is to find a strategy tree where its longest branch is the shortest possible!

182

Puzzle

8 1 15 2 9 21 3 10 16 26 4 11 17 22 30 5 12 18 23 27 33 6 13 19 24 28 31 35 7 14 20 25 29 32 34 36

183

Puzzle

Now, suppose we have three eggs. What is the least (minimum) number of egg drops required to determine the egg- breaking floor? Suppose we have k eggs and the building has n floors. What is the least (minimum) number

• f

egg drops required to determine the egg-breaking floor?

62

Any Question ???

185

Some mathematical principles and problem-solving techniques

Probability: Coins, dice, boxes, and bears

186

Two bears

Let us consider the case of two bears – one white and

• ne black. We ask three similar questions and suggest

thinking about the answers before reading further. The questions are:

 What is the probability that both bears are males?  What is the probability that both bears are males if you

were told that one of them is male?

 What is the probability that both bears are males if you

were told that the white one is male?

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187

Two bears

Rule #1: Be sure you understand the problem, and all the basic terms and expressions used to define it. There are two bears, but where did they come from? Were they selected from a large population of bears? If so, we should know the distribution of sexes and colors in this population… Since this information was not given in the problem, we have to assume that among all bears, the two sexes and two colors (black or white) are equally likely.

188

Two bears

Rule #2: Do not rely on your intuition too much; solid calculations are far more reliable Rule #3: Solid calculations and reasoning are more meaningful when you build a model of the problem by defining its variables, constraints, and objectives.

189

Two bears

 What is the probability that both bears are males?

Let’s list all the possible outcomes, which are: (white, black) = (f f ), (f m), (m f ), and (m m) where m is male and f is female. Answer: 1/4

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190

Two bears

 What is the probability that both bears are males if

you were told that one of them is male? We know that all possible outcomes are: (white, black) = (f f ), (f m), (m f ), and (m m) Answer: 1/3

191

Two bears

 What is the probability that both bears are males if

you were told that the white one is male? We know that all possible outcomes are: (white, black) = (f f ), (f m), (m f ), and (m m) Answer: 1/2

192

Puzzle

There are three cards in a bag:

• The first card has the symbol X written on both sides,
• The second card has the symbol O written on both sides,
• The third card has an X on one side and an O on the other.

You draw one card at random and examine one side of this card. You see an X.

What is the probability that there is also an X

• n the other side?

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193

Puzzle

Again: Rule #1, Rule #2, and Rule #3… Clue: Altogether, there are 6 symbols on these three cards: three Xs and three Os. Model: a, b, c X d, e, f O Card #1 Card #2 Card #3 a b c d e f

194

Puzzle

Model: a, b, c X d, e, f O Card #1 Card #2 Card #3 a b c d e f Now we can do some reasoning: “You draw one card at random and examine one side of this card. You see an X.” So you saw either a, b, or c… Answer: 2/3

195

Puzzle

 A boy is often late for school. When approached by

his teacher, he explained that it is not his fault. Then he provided some details. His father takes him from home to the bus stop every morning. The bus is supposed to leave at 8:00 am, but this departure time is only approximate. The bus arrives at the stop anytime between 7:58 and 8:02 and immediately

• departs. The boy and his father aim at arriving at the

bus stop at 8:00, however, due to variable traffic conditions they arrive anytime between 7:55 and 8:01. This is why the boy misses the bus so often.

 Can you determine how often the boy is late for

school?

66

196

Puzzle – the model

As we know, the two variables of the problem are:

 x: arrival/departure time of the bus,

7:58 ≤ x ≤ 8:02.

 y: arrival time of the boy,

7:55 ≤ y ≤ 8:01.

197

Puzzle – one model

8:02 8:01 8:00 7:59 7:58 7:55 7:56 7:57 7:58 7:59 8:00 8:01 x y

6/35 ≈ 17% Possible arrivals: every minute…

198

Puzzle – another model

8:02 8:01 8:00 7:59 7:58 7:55 7:56 7:57 7:58 7:59 8:00 8:01 y x

81/925 ≈ 18.5% Possible arrivals: every 10 sec…

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199

Puzzle - solution

8:02 8:01 8:00 7:59 7:58 7:55 7:56 7:57 7:58 7:59 8:00 8:01 x y

The line x = y divides the rectangle into two areas: area x < y (dark part) when the boy is late for the bus, and area x ≥ y (light part), when the boy is on time.

Exact answer: 18.75%

201

Martin Gardner, 1960s “Everything we do, everything that happens around us, obeys the laws of probability. We can no more escape them than we can escape gravity. ‘Probability,’a philosopher

• nce said, ‘is the very guide of life.’We are

all gamblers who go through life making countless bets on the outcome of countless actions.”

Remark

202

Multiplication principle: If some choice can be made in n different ways and some subsequent choice can be made in m different ways, then there are n × m different ways these choices can be made in succession.

Multiplication principle

68

203

Multiplication principle, examples: A woman has 5 blouses and 4 skirts, then she has 5 × 4 = 20 choices of outfit. Rolling three dice: there are 6 × 6 × 6 = 216 possible outcomes. There are 6 × 5 × 4 = 120 possible outcomes in which the numbers on dice differ…

Multiplication principle

204

Julius Caesar

The assumption is that Shakespeare’s account of the circumstances of Julius Caesar death is accurate: Julius Caesar gasped: “You too, Brutus” at his last breath. Further, we assume that today – after more than 2,000 years from that event – all molecules present at that time are distributed uniformly in the atmosphere today. Then the question is: What are the chances that you have just inhaled a molecule which Julius Caesar exhaled in his dying breath?

205

Our intuition is that the probability is very small, to say the least. After all, the numbers of molecules exhaled by Caesar was a tiny fraction of all molecules available… However, get ready for a surprising answer! (Rule #2)

Julius Caesar

69

206

As usual, we will introduce some variables and build a model (Rule #3). So let us say that:

• There are n molecules available in the

atmosphere.

• Julius Caesar exhaled p molecules.
• We have just inhaled q molecules.

Now we are ready to build a model for the required probability.

Julius Caesar

207

Let us consider a single (randomly chosen) molecule in the atmosphere. The probability that this is a molecule exhaled by Julius Caesar is precisely p/n. Conversely, the probability that this is not a molecule exhaled by Julius Caesar is precisely 1 – p/n. So, if we inhale a single molecule, the probability that this molecule was not a molecule exhaled by Julius Caesar is 1 – p/n.

Julius Caesar

208

And if we inhale 2 molecules, the probability that none of them was exhaled by Julius Caesar – by multiplication principle – is (1 – p/n)2. If we inhale q molecules, again, by multiplication principle, the probability that none of them was exhaled by Julius Caesar is (1 – p/n)q. So finally, the probability that we did inhale a molecule which Julius Caesar exhaled in his dying breath is given as: P = 1 – (1 – p/n)q

Julius Caesar

70

209

We now have a model and are ready for the

• calculations. Estimating the values of our

variables: n ≈ 1044 p ≈ q ≈ 2.2 × 1022 the probability P is larger than 0.99. This is absolutely amazing: the chance that we have just inhaled a molecule which Julius Caesar exhaled in his dying breath is more than 99%!

Julius Caesar

210

Coincidences

People have tendency to underestimate the frequency of coincidences. We have already talked about during lecture #6:

211

Coincidences

• Christopher Columbus discovered New World

in 1492 and his fellow Italian Enrico Fermi discovered the new world of atom in 1942.

• President Kennedy’s secretary was named

Lincoln, while President Lincoln’s secretary was named Kennedy.

• Each word in the name Ronald Wilson

Reagan (former US President) has 6 letters (thus a connection with “666”).

71

212

Birthdays

Since the year has 366 days (including February 29), there would have to be 367 people gathered together in order for us to be absolutely certain that at least two people in the group have the same birthday (Dirichlet’s box principle).

213

Puzzle

How many people should we have in a room to make a bet (with a better than 50% chance

• f winning) that there are at least two people

in the room with the same birthday?

214

Puzzle

An initial guess might be 183, about half of 365… and the surprising answer is that there need be only 23. In other words: “More than half of the time that 23 randomly selected people are gathered together, two or more of them will share a birthday.”

72

215

Puzzle

By multiplication principle, the number of ways in which five dates can be chosen (with a possible repetition) is: 365 × 365 × 365 × 365 × 365 and without repetitions: 365 × 364 × 363 × 362 × 361

216

Puzzle

By dividing 365 × 364 × 363 × 362 × 361 by 3655 we get the probability that five people chosen at random will have no birthday in common. And if we subtract this probability from 1, we get the complementary probability that at least two of the five people have a birthday in common.

217

Puzzle

A similar calculation using 23 rather than 5 yields ½, as the probability that at least two people of 23 people will have a common birthday.

73

218

In a casino we are not sure which table we should play at:

• Table A: we win when we throw one six

when 6 dice are rolled, or

• Table B: we win when we throw two sixes

when 12 dice are rolled?

Puzzle

219

Of course, it is possible to extend the question to “or throw exactly three sixes when 18 dice are rolled,” etc. Also, we can ask a slightly different, but similar question: It is is more likely to throw at least one six when 6 dice are rolled, or to throw at least two sixes when 12 dice are rolled? But let’s start with the original question…

Puzzle

220

The intuition usually fails here. It does not seem like there is much difference between these two scenarios (Table A vs. Table B), because the expected outcomes are “one six” and “two sixes,” respectively, but the probabilities of these events are in fact very different!

Puzzle

74

221

Table A: the probability of throwing precisely 1 six out of 6 dice is:

= 0.4019

Table B: the probability of throwing precisely 2 sixes out of 12 dice is

= 0.2961

which is quite a difference!

6 5 6 5 6 5 6 5 6 5 6 1 1 6               6 5 6 5 6 5 6 5 6 5 6 5 6 5 6 5 6 5 6 5 6 1 6 1 2 12                    

Puzzle

222

We can extend this to the general case by noting that the probability of throwing precisely k sixes with n dice is: × (1/6)k × (5/6)n – k

        k n

Puzzle

223

The “at least” cases are quite easy. To calculate the probability of throwing at least one six when 6 dice are rolled it is sufficient to calculate the probability of throwing no sixes and subtract the result from 1.

Puzzle

75

224

So, the probability of throwing no sixes when 6 dice are rolled is

= 0.3349

hence, the probability of throwing at least one six when 6 dice are rolled is:

1 – 0.3349 = 0.6651

6 5 6 5 6 5 6 5 6 5 6 5     

Puzzle

225

Similarly, to calculate the probability of throwing at least two sixes when 12 dice are rolled, it is sufficient to calculate the probability of throwing no sixes and the probability of throwing precisely

• ne six and subtracting these results from 1.

Puzzle

226

So, the probability of throwing no sixes when 12 dice are rolled is

= 0.1122

and the probability of throwing precisely one six when 12 dice are rolled is

= 0.2692

6 5 6 5 6 5 6 5 6 5 6 5 6 5 6 5 6 5 6 5 6 5 6 5            6 5 6 5 6 5 6 5 6 5 6 5 6 5 6 5 6 5 6 5 6 5 6 1 1 12                    

Puzzle

76

227

…hence the probability of throwing at least two sixes when 12 dice are rolled is

1 – 0.1122 – 0.2692 = 0.6186

Puzzle

228

Lessons learned:

• Note that there is a significant difference in

probabilities for throwing precisely one six (while rolling 6 dice) versus throwing precisely two sixes (while throwing 12 dice). These probabilities are 0.4019 and 0.2961, respectively (approximately 40% versus 30%).

Puzzle

229

Lessons learned:

• On the other hand, the difference in probabilities

for throwing at least one six (while rolling 6 dice) versus throwing at least two sixes (while throwing 12 dice) are relatively small: 0.6651 versus 0.6186 (approximately 67% versus 62%).

Puzzle

77

230

Lessons learned:

• For most people these results are quite counter-
• intuitive. Somehow, our intuition suggests that

“expected” should translate into a “probability of 50% or higher,” which of course need not be the case.

Puzzle

Any Question ???