Smart search: a practical facet of theoretical mathematics Mikhail - - PowerPoint PPT Presentation

Smart search: a practical facet of theoretical mathematics

Mikhail Volkov

Ural Federal University, Ekaterinburg, Russia

Where am I from?

Part I: Smart Search

A typical search problem (Data Mining):

◮ A huge data base ◮ Strict time limit ◮ Time depends on the numbers of queries

Part I: Smart Search

A typical search problem (Data Mining):

◮ A huge data base ◮ Strict time limit ◮ Time depends on the numbers of queries

How can one quickly retrieve data? Theoretical mathematics offers a solution!! [G´ al, Miltersen. ICALP 2003]

100 prisoners problem

In order to get freedom, 100 prisoners 0, . . . , 99 have to solve the following problem.

100 prisoners problem

In order to get freedom, 100 prisoners 0, . . . , 99 have to solve the following problem. In a room there is a cupboard with 100 drawers. Each drawer contains the number of exactly one prisoner in random order.

100 prisoners problem

In order to get freedom, 100 prisoners 0, . . . , 99 have to solve the following problem. In a room there is a cupboard with 100 drawers. Each drawer contains the number of exactly one prisoner in random order. The prisoners enter the room one after another. Each prisoner may open and look into 50 drawers in any order and the drawers are closed again afterwards.

100 prisoners problem

In order to get freedom, 100 prisoners 0, . . . , 99 have to solve the following problem. In a room there is a cupboard with 100 drawers. Each drawer contains the number of exactly one prisoner in random order. The prisoners enter the room one after another. Each prisoner may open and look into 50 drawers in any order and the drawers are closed again afterwards.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

100 prisoners problem

In order to get freedom, 100 prisoners 0, . . . , 99 have to solve the following problem. In a room there is a cupboard with 100 drawers. Each drawer contains the number of exactly one prisoner in random order. The prisoners enter the room one after another. Each prisoner may open and look into 50 drawers in any order and the drawers are closed again afterwards.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 8

100 prisoners problem

In order to get freedom, 100 prisoners 0, . . . , 99 have to solve the following problem. In a room there is a cupboard with 100 drawers. Each drawer contains the number of exactly one prisoner in random order. The prisoners enter the room one after another. Each prisoner may open and look into 50 drawers in any order and the drawers are closed again afterwards.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 8 76

100 prisoners problem

In order to get freedom, 100 prisoners 0, . . . , 99 have to solve the following problem. In a room there is a cupboard with 100 drawers. Each drawer contains the number of exactly one prisoner in random order. The prisoners enter the room one after another. Each prisoner may open and look into 50 drawers in any order and the drawers are closed again afterwards.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 8 76 13

100 prisoners problem

In order to get freedom, 100 prisoners 0, . . . , 99 have to solve the following problem. In a room there is a cupboard with 100 drawers. Each drawer contains the number of exactly one prisoner in random order. The prisoners enter the room one after another. Each prisoner may open and look into 50 drawers in any order and the drawers are closed again afterwards.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 8 76 13 42

100 prisoners problem

In order to get freedom, 100 prisoners 0, . . . , 99 have to solve the following problem. In a room there is a cupboard with 100 drawers. Each drawer contains the number of exactly one prisoner in random order. The prisoners enter the room one after another. Each prisoner may open and look into 50 drawers in any order and the drawers are closed again afterwards.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 8 76 13 42

If during this search every (!) prisoner finds his number in one of the drawers, all prisoners are pardoned.

100 prisoners problem

In order to get freedom, 100 prisoners 0, . . . , 99 have to solve the following problem. In a room there is a cupboard with 100 drawers. Each drawer contains the number of exactly one prisoner in random order. The prisoners enter the room one after another. Each prisoner may open and look into 50 drawers in any order and the drawers are closed again afterwards.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 8 76 13 42

If during this search every (!) prisoner finds his number in one of the drawers, all prisoners are pardoned. Before the first prisoner enters the room, the prisoners may discuss their strategy, afterwards no communication is possible.

Naive approach

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How big is the chance to get pardoned?

Naive approach

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 8 76 13 42

How big is the chance to get pardoned? If a prisoner opens 50 drawers at random, he finds his number with the probability

50 100 = 1 2

Naive approach

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 8 76 13 42

How big is the chance to get pardoned? If a prisoner opens 50 drawers at random, he finds his number with the probability

50 100 = 1 2

Then the probability that everyone will find his number is

1 2100 ≈ 0.000000000000000000000000000000

• 30 zeroes

8

Naive approach

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 8 76 13 42

How big is the chance to get pardoned? If a prisoner opens 50 drawers at random, he finds his number with the probability

50 100 = 1 2

Then the probability that everyone will find his number is

1 2100 ≈ 0.000000000000000000000000000000

• 30 zeroes

8 Is there a better strategy?

Naive approach

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 8 76 13 42

How big is the chance to get pardoned? If a prisoner opens 50 drawers at random, he finds his number with the probability

50 100 = 1 2

Then the probability that everyone will find his number is

1 2100 ≈ 0.000000000000000000000000000000

• 30 zeroes

8 Is there a better strategy? Yes!

Naive approach

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 8 76 13 42

How big is the chance to get pardoned? If a prisoner opens 50 drawers at random, he finds his number with the probability

50 100 = 1 2

Then the probability that everyone will find his number is

1 2100 ≈ 0.000000000000000000000000000000

• 30 zeroes

8 Is there a better strategy? Yes! And the success probability is more than 30%

Cycle-following strategy

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

Cycle-following strategy:

Cycle-following strategy

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Cycle-following strategy: First open the drawer that bears your own number

Cycle-following strategy

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 17 89

Cycle-following strategy: First open the drawer that bears your own number Then open the drawer with the number found in the previous drawer

Cycle-following strategy

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 17 89 66

Cycle-following strategy: First open the drawer that bears your own number Then open the drawer with the number found in the previous drawer Repeat the last step (until success or until reaching the limit of 50 drawers)

Cycle-following strategy

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 17 89 66 3

Cycle-following strategy: First open the drawer that bears your own number Then open the drawer with the number found in the previous drawer Repeat the last step (until success or until reaching the limit of 50 drawers)

Cycle-following strategy

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 17 89 66 3 42

Cycle-following strategy: First open the drawer that bears your own number Then open the drawer with the number found in the previous drawer Repeat the last step (until success or until reaching the limit of 50 drawers)

Cycle-following strategy

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 17 89 66 3 42

Cycle-following strategy: First open the drawer that bears your own number Then open the drawer with the number found in the previous drawer Repeat the last step (until success or until reaching the limit of 50 drawers) Why are success chances more than 30%?

Cycle-following strategy

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 17 89 66 3 42

Cycle-following strategy: First open the drawer that bears your own number Then open the drawer with the number found in the previous drawer Repeat the last step (until success or until reaching the limit of 50 drawers) Why are success chances more than 30%? Drawers and numbers in them represent a random permutation

• f the numbers 0, 1, . . . , 99

Cycle-following strategy

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 17 89 66 3 42

Cycle-following strategy: First open the drawer that bears your own number Then open the drawer with the number found in the previous drawer Repeat the last step (until success or until reaching the limit of 50 drawers) Why are success chances more than 30%? Drawers and numbers in them represent a random permutation

• f the numbers 0, 1, . . . , 99

Prisoners win if this permutation has no cycle of length > 50

Justification

How many permutations with this property do exist?

Justification

How many permutations with this property do exist? How many permutations of 2n numbers without cycle of length k > n do exist?

Justification

How many permutations with this property do exist? How many permutations of 2n numbers without cycle of length k > n do exist? Let us calculate the number of permutations that have a cycle C of a fixed length k > n:

Justification

How many permutations with this property do exist? How many permutations of 2n numbers without cycle of length k > n do exist? Let us calculate the number of permutations that have a cycle C of a fixed length k > n:

◮ there are

2n

k

• =

(2n)! k!(2n−k)! ways to choose k numbers out of 2n

(these are k numbers that will in C)

Justification

How many permutations with this property do exist? How many permutations of 2n numbers without cycle of length k > n do exist? Let us calculate the number of permutations that have a cycle C of a fixed length k > n:

◮ there are

2n

k

• =

(2n)! k!(2n−k)! ways to choose k numbers out of 2n

(these are k numbers that will in C)

◮ there are (k − 1)! ways to put k numbers in the cycle C

Justification

How many permutations with this property do exist? How many permutations of 2n numbers without cycle of length k > n do exist? Let us calculate the number of permutations that have a cycle C of a fixed length k > n:

◮ there are

2n

k

• =

(2n)! k!(2n−k)! ways to choose k numbers out of 2n

(these are k numbers that will in C)

◮ there are (k − 1)! ways to put k numbers in the cycle C ◮ there are (2n − k)! ways to put the remaining 2n − k numbers

Justification

How many permutations with this property do exist? How many permutations of 2n numbers without cycle of length k > n do exist? Let us calculate the number of permutations that have a cycle C of a fixed length k > n:

◮ there are

2n

k

• =

(2n)! k!(2n−k)! ways to choose k numbers out of 2n

(these are k numbers that will in C)

◮ there are (k − 1)! ways to put k numbers in the cycle C ◮ there are (2n − k)! ways to put the remaining 2n − k numbers

Altogether: 2n k

• ·(k −1)!·(2n−k)! =

(2n)! k!(2n − k)! ·(k −1)!·(2n−k)! = (2n)! k

Justification

How many permutations with this property do exist? How many permutations of 2n numbers without cycle of length k > n do exist? Let us calculate the number of permutations that have a cycle C of a fixed length k > n:

◮ there are

2n

k

• =

(2n)! k!(2n−k)! ways to choose k numbers out of 2n

(these are k numbers that will in C)

◮ there are (k − 1)! ways to put k numbers in the cycle C ◮ there are (2n − k)! ways to put the remaining 2n − k numbers

Altogether: 2n k

• ·(k −1)!·(2n−k)! =

(2n)! k!(2n − k)! ·(k −1)!·(2n−k)! = (2n)! k A permutation of 2n numbers has at most one cycle of length k > n

Justification

How many permutations with this property do exist? How many permutations of 2n numbers without cycle of length k > n do exist? Let us calculate the number of permutations that have a cycle C of a fixed length k > n:

◮ there are

2n

k

• =

(2n)! k!(2n−k)! ways to choose k numbers out of 2n

(these are k numbers that will in C)

◮ there are (k − 1)! ways to put k numbers in the cycle C ◮ there are (2n − k)! ways to put the remaining 2n − k numbers

Altogether: 2n k

• ·(k −1)!·(2n−k)! =

(2n)! k!(2n − k)! ·(k −1)!·(2n−k)! = (2n)! k A permutation of 2n numbers has at most one cycle of length k > n The probability that a permutation chosen at random from all (2n)! permutations has a cycle of length k is 1

k

Justification (continued)

A permutation of 2n numbers has at most one cycle of length k > n

Justification (continued)

A permutation of 2n numbers has at most one cycle of length k > n The probability p that a permutation has no cycles of length > n is 1 − 1 n + 1 − 1 n + 2 − · · · − 1 2n

Justification (continued)

A permutation of 2n numbers has at most one cycle of length k > n The probability p that a permutation has no cycles of length > n is 1 − 1 n + 1 − 1 n + 2 − · · · − 1 2n It is known that 1 + 1

2 + 1 3 + · · · + 1 i = Hi > ln i

Justification (continued)

A permutation of 2n numbers has at most one cycle of length k > n The probability p that a permutation has no cycles of length > n is 1 − 1 n + 1 − 1 n + 2 − · · · − 1 2n It is known that 1 + 1

2 + 1 3 + · · · + 1 i = Hi > ln i

and Hi − Hj < ln i − ln j (for i > j)

Justification (continued)

A permutation of 2n numbers has at most one cycle of length k > n The probability p that a permutation has no cycles of length > n is 1 − 1 n + 1 − 1 n + 2 − · · · − 1 2n It is known that 1 + 1

2 + 1 3 + · · · + 1 i = Hi > ln i

and Hi − Hj < ln i − ln j (for i > j)

• Slight deviation towards a classic open problem: Hi − ln i

tends to a constant called the Euler–Mascheroni constant, 0.57721566490153286060651209008240243104215933593992. . . Its arithmetical nature is yet unknown!

Justification (continued)

A permutation of 2n numbers has at most one cycle of length k > n The probability p that a permutation has no cycles of length > n is 1 − 1 n + 1 − 1 n + 2 − · · · − 1 2n It is known that 1 + 1

2 + 1 3 + · · · + 1 i = Hi > ln i

and Hi − Hj < ln i − ln j (for i > j)

• Slight deviation towards a classic open problem: Hi − ln i

tends to a constant called the Euler–Mascheroni constant, 0.57721566490153286060651209008240243104215933593992. . . Its arithmetical nature is yet unknown!

• Hence

p = 1 − (H2n − Hn)

Justification (continued)

A permutation of 2n numbers has at most one cycle of length k > n The probability p that a permutation has no cycles of length > n is 1 − 1 n + 1 − 1 n + 2 − · · · − 1 2n It is known that 1 + 1

2 + 1 3 + · · · + 1 i = Hi > ln i

and Hi − Hj < ln i − ln j (for i > j)

• Slight deviation towards a classic open problem: Hi − ln i

tends to a constant called the Euler–Mascheroni constant, 0.57721566490153286060651209008240243104215933593992. . . Its arithmetical nature is yet unknown!

• Hence

p = 1 − (H2n − Hn) > 1 − (ln 2n − ln n)

Justification (continued)

A permutation of 2n numbers has at most one cycle of length k > n The probability p that a permutation has no cycles of length > n is 1 − 1 n + 1 − 1 n + 2 − · · · − 1 2n It is known that 1 + 1

2 + 1 3 + · · · + 1 i = Hi > ln i

and Hi − Hj < ln i − ln j (for i > j)

• Slight deviation towards a classic open problem: Hi − ln i

tends to a constant called the Euler–Mascheroni constant, 0.57721566490153286060651209008240243104215933593992. . . Its arithmetical nature is yet unknown!

• Hence

p = 1 − (H2n − Hn) > 1 − (ln 2n − ln n) = 1 − ln 2 ≈ 0.3118

Conclusion (of Part I)

The advantages of the cycle-following strategy:

Conclusion (of Part I)

The advantages of the cycle-following strategy: It guarantees fast search in approx. 30% of cases

Conclusion (of Part I)

The advantages of the cycle-following strategy: It guarantees fast search in approx. 30% of cases It can be proven to be optimal (Curtin and Warshauer, 2006)

Conclusion (of Part I)

The advantages of the cycle-following strategy: It guarantees fast search in approx. 30% of cases It can be proven to be optimal (Curtin and Warshauer, 2006) Intellect opens the gateway to freedom

Conclusion (of Part I)

The advantages of the cycle-following strategy: It guarantees fast search in approx. 30% of cases It can be proven to be optimal (Curtin and Warshauer, 2006) Intellect opens the gateway to freedom “The best ideas come from universities, from places full of youth, knowledge and freedom”, Eric Schmidt, the executive chairman of Google in 2011 to 2015)

Question (to Part I)

Suppose that the prison’s director knows the above cycle-following strategy and intentionally distributes numbers into drawers such that the resulting permutation has a very long cycle.

Question (to Part I)

Suppose that the prison’s director knows the above cycle-following strategy and intentionally distributes numbers into drawers such that the resulting permutation has a very long cycle. Then, if prisoners will use the cycle-following strategy, they will definitely lose!

Question (to Part I)

Suppose that the prison’s director knows the above cycle-following strategy and intentionally distributes numbers into drawers such that the resulting permutation has a very long cycle. Then, if prisoners will use the cycle-following strategy, they will definitely lose! Do they have any chance to win under such pre-condition?

Question (to Part I)

Suppose that the prison’s director knows the above cycle-following strategy and intentionally distributes numbers into drawers such that the resulting permutation has a very long cycle. Then, if prisoners will use the cycle-following strategy, they will definitely lose! Do they have any chance to win under such pre-condition? Solution: Yes!

Question (to Part I)

Suppose that the prison’s director knows the above cycle-following strategy and intentionally distributes numbers into drawers such that the resulting permutation has a very long cycle. Then, if prisoners will use the cycle-following strategy, they will definitely lose! Do they have any chance to win under such pre-condition? Solution: Yes! Trick: RANDOMIZE!

Question (to Part I)

Suppose that the prison’s director knows the above cycle-following strategy and intentionally distributes numbers into drawers such that the resulting permutation has a very long cycle. Then, if prisoners will use the cycle-following strategy, they will definitely lose! Do they have any chance to win under such pre-condition? Solution: Yes! Trick: RANDOMIZE! Prisoners should create a random permutation σ and memorize it.

Question (to Part I)

Suppose that the prison’s director knows the above cycle-following strategy and intentionally distributes numbers into drawers such that the resulting permutation has a very long cycle. Then, if prisoners will use the cycle-following strategy, they will definitely lose! Do they have any chance to win under such pre-condition? Solution: Yes! Trick: RANDOMIZE! Prisoners should create a random permutation σ and memorize it. The prisoner no. x should open the drawer σ(x); if the drawer contains the number y, the next drawer to open should be σ(y), etc.

Question (to Part I)

Suppose that the prison’s director knows the above cycle-following strategy and intentionally distributes numbers into drawers such that the resulting permutation has a very long cycle. Then, if prisoners will use the cycle-following strategy, they will definitely lose! Do they have any chance to win under such pre-condition? Solution: Yes! Trick: RANDOMIZE! Prisoners should create a random permutation σ and memorize it. The prisoner no. x should open the drawer σ(x); if the drawer contains the number y, the next drawer to open should be σ(y),

• etc. Thus, the cycle-following strategy is applied not to the

director’s “bad” permutation but to its composition with σ.

Question (to Part I)

Suppose that the prison’s director knows the above cycle-following strategy and intentionally distributes numbers into drawers such that the resulting permutation has a very long cycle. Then, if prisoners will use the cycle-following strategy, they will definitely lose! Do they have any chance to win under such pre-condition? Solution: Yes! Trick: RANDOMIZE! Prisoners should create a random permutation σ and memorize it. The prisoner no. x should open the drawer σ(x); if the drawer contains the number y, the next drawer to open should be σ(y),

• etc. Thus, the cycle-following strategy is applied not to the

director’s “bad” permutation but to its composition with σ. The composition of any permutation with a random one can be shown to be a random permutation.

Part II: Smart Retrieval

The Problem

Given a set E of size m (huge!) and f : E → {0, 1}. Want: Data structure R that, given y ∈ E reproduces f (y).

Part II: Smart Retrieval

The Problem

Given a set E of size m (huge!) and f : E → {0, 1}. Want: Data structure R that, given y ∈ E reproduces f (y).

Memory Requirement of R

Want: (1 + ε)m bits for some small constant ε.

Part II: Smart Retrieval

The Problem

Given a set E of size m (huge!) and f : E → {0, 1}. Want: Data structure R that, given y ∈ E reproduces f (y).

Memory Requirement of R

Want: (1 + ε)m bits for some small constant ε. Observe that the entries of E may be relatively big so storing the array of all pairs (y, f (y)) is not a solution (requires m(max{|y|} + 1) bits and does not admit a speedy retrieval.)

Part II: Smart Retrieval

The Problem

Given a set E of size m (huge!) and f : E → {0, 1}. Want: Data structure R that, given y ∈ E reproduces f (y).

Memory Requirement of R

Want: (1 + ε)m bits for some small constant ε. Observe that the entries of E may be relatively big so storing the array of all pairs (y, f (y)) is not a solution (requires m(max{|y|} + 1) bits and does not admit a speedy retrieval.)

Example Data Set: Names annotated with gender

E = { Ana, Bea, Cal, Dan, Eli, Fen } → → → → → → f : 1 1 1

Retrieval is Tricky: Watson’s Mistake

Watson is the famous supercomputer by IBM.

Retrieval is Tricky: Watson’s Mistake

Watson is the famous supercomputer by IBM.

Retrieval is Tricky: Watson’s Mistake

Watson is the famous supercomputer by IBM. In 2011, it won in the intellectual TV-show “Jeopardy!” against two human champions.

Retrieval is Tricky: Watson’s Mistake

Watson is the famous supercomputer by IBM. In 2011, it won in the intellectual TV-show “Jeopardy!” against two human champions. However, Watson wasn’t perfect and did a few mistakes.

Retrieval is Tricky: Watson’s Mistake

Watson is the famous supercomputer by IBM. In 2011, it won in the intellectual TV-show “Jeopardy!” against two human champions. However, Watson wasn’t perfect and did a few mistakes. For instance, it failed to guess a correct answer from the following clue in the “U.S. Cities” category: “Its largest airport was named for a World War II hero; its second-largest, for a World War II battle”.

Retrieval is Tricky: Watson’s Mistake

Watson is the famous supercomputer by IBM. In 2011, it won in the intellectual TV-show “Jeopardy!” against two human champions. However, Watson wasn’t perfect and did a few mistakes. For instance, it failed to guess a correct answer from the following clue in the “U.S. Cities” category: “Its largest airport was named for a World War II hero; its second-largest, for a World War II battle”. The answer is Chicago – for the city’s O’Hare and Midway airports.

Retrieval is Tricky: Watson’s Mistake

Watson is the famous supercomputer by IBM. In 2011, it won in the intellectual TV-show “Jeopardy!” against two human champions. However, Watson wasn’t perfect and did a few mistakes. For instance, it failed to guess a correct answer from the following clue in the “U.S. Cities” category: “Its largest airport was named for a World War II hero; its second-largest, for a World War II battle”. The answer is Chicago – for the city’s O’Hare and Midway airports. It is important to realize that Watson had all the necessary data in its 15-terabyte databank of human knowledge!

Retrieval is Tricky: Watson’s Mistake

Watson is the famous supercomputer by IBM. In 2011, it won in the intellectual TV-show “Jeopardy!” against two human champions. However, Watson wasn’t perfect and did a few mistakes. For instance, it failed to guess a correct answer from the following clue in the “U.S. Cities” category: “Its largest airport was named for a World War II hero; its second-largest, for a World War II battle”. The answer is Chicago – for the city’s O’Hare and Midway airports. It is important to realize that Watson had all the necessary data in its 15-terabyte databank of human knowledge! The failure was due to its inability to retrieve the relevant data quickly enough!

Retrieval using Linear Systems

Pick n = (1 + ε)m and h: E → {1, . . . , n}3, a hash function.

Retrieval using Linear Systems

Pick n = (1 + ε)m and h: E → {1, . . . , n}3, a hash function. h should be easy to compute and satisfy y1 = y2 → h(y1) = h(y2).

Retrieval using Linear Systems

Pick n = (1 + ε)m and h: E → {1, . . . , n}3, a hash function. h should be easy to compute and satisfy y1 = y2 → h(y1) = h(y2). This yields A ∈ Fm×n: Input Hash Values (Ana, 1) h(Ana) = (1, 3, 9) (Bea, 1) h(Bea) = (2, 3, 4) (Cal, 0) h(Cal) = (3, 6, 8) (Dan, 0) h(Dan) = (5, 8, 9) (Eli, 0) h(Eve) = (2, 8, 9) (Fen, 1) h(Fen) = (1, 5, 6)        

1

1

2 3

1

4 5 6 7 8 9

1 0 1 1 1 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 1 0 0 1 1 0 1 0 0 0 0 0 1 1 1 0 0 0 1 1 0 0 0        

Retrieval using Linear Systems

Pick n = (1 + ε)m and h: E → {1, . . . , n}3, a hash function. h should be easy to compute and satisfy y1 = y2 → h(y1) = h(y2). This yields A ∈ Fm×n: Input Hash Values (Ana, 1) h(Ana) = (1, 3, 9) (Bea, 1) h(Bea) = (2, 3, 4) (Cal, 0) h(Cal) = (3, 6, 8) (Dan, 0) h(Dan) = (5, 8, 9) (Eli, 0) h(Eve) = (2, 8, 9) (Fen, 1) h(Fen) = (1, 5, 6)        

1

1

2 3

1

4 5 6 7 8 9

1 0 1 1 1 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 1 0 0 1 1 0 1 0 0 0 0 0 1 1 1 0 0 0 1 1 0 0 0         · x =         1 1 1        

Retrieval using Linear Systems

Pick n = (1 + ε)m and h: E → {1, . . . , n}3, a hash function. h should be easy to compute and satisfy y1 = y2 → h(y1) = h(y2). This yields A ∈ Fm×n: Input Hash Values (Ana, 1) h(Ana) = (1, 3, 9) (Bea, 1) h(Bea) = (2, 3, 4) (Cal, 0) h(Cal) = (3, 6, 8) (Dan, 0) h(Dan) = (5, 8, 9) (Eli, 0) h(Eve) = (2, 8, 9) (Fen, 1) h(Fen) = (1, 5, 6)        

1

1

2 3

1

4 5 6 7 8 9

1 0 1 1 1 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 1 0 0 1 1 0 1 0 0 0 0 0 1 1 1 0 0 0 1 1 0 0 0         · x =         1 1 1        

Theorem

For ε > 0.09 such systems are solvable with high probability For ε < 0.09 such systems are not solvable with high probability

Retrieval using Linear Systems

Pick n = (1 + ε)m and h: E → {1, . . . , n}3, a hash function. h should be easy to compute and satisfy y1 = y2 → h(y1) = h(y2). This yields A ∈ Fm×n: Input Hash Values (Ana, 1) h(Ana) = (1, 3, 9) (Bea, 1) h(Bea) = (2, 3, 4) (Cal, 0) h(Cal) = (3, 6, 8) (Dan, 0) h(Dan) = (5, 8, 9) (Eli, 0) h(Eve) = (2, 8, 9) (Fen, 1) h(Fen) = (1, 5, 6)        

1

1

2 3

1

4 5 6 7 8 9

1 0 1 1 1 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 1 0 0 1 1 0 1 0 0 0 0 0 1 1 1 0 0 0 1 1 0 0 0         ·

              1 1              

=         1 1 1        

Theorem

For ε > 0.09 such systems are solvable with high probability For ε < 0.09 such systems are not solvable with high probability

Theorem

For ε > 0.23 the system is solvable in O(m) time with high probability.

Retrieval using Linear Systems

Pick n = (1 + ε)m and h: E → {1, . . . , n}3, a hash function. h should be easy to compute and satisfy y1 = y2 → h(y1) = h(y2). This yields A ∈ Fm×n: Input Hash Values (Ana, 1) h(Ana) = (1, 3, 9) (Bea, 1) h(Bea) = (2, 3, 4) (Cal, 0) h(Cal) = (3, 6, 8) (Dan, 0) h(Dan) = (5, 8, 9) (Eli, 0) h(Eve) = (2, 8, 9) (Fen, 1) h(Fen) = (1, 5, 6)        

1

1

2 3

1

4 5 6 7 8 9

1 0 1 1 1 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 1 0 0 1 1 0 1 0 0 0 0 0 1 1 1 0 0 0 1 1 0 0 0         ·

              1 1              

=         1 1 1        

The retrieval data structure R = (h, x)

query(y) := 3

i=1

x[hi(y)] – Constant time!! (sum of just 3 bits) Memory Requirement: ∼ 1.09m bits

(or ∼ 1.23m bits with O(m) construction).

Conclusion

We are doing pure math because it makes fun, not because it may have usage.

Conclusion

We are doing pure math because it makes fun, not because it may have usage. But it does have usage . . .

Conclusion

We are doing pure math because it makes fun, not because it may have usage. But it does have usage . . . . . . and this makes fun!