![. . . . . .
Putting pieces together
let p be the shortest program for x, so |p| = C(x) ≥ n
let q be a random (incompressible) string of length 2n
when p is known (independent from p)
for every k ∈ [0, C(x)] and every l ∈ [0, 2n] consider
yk,l = (k-bit prefix of p, l-bit prefix of q)](https://image.slidesharecdn.com/talk-130922050737-phpapp02/85/Kolmogorov-complexity-and-topological-arguments-39-320.jpg)
![. . . . . .
Putting pieces together
let p be the shortest program for x, so |p| = C(x) ≥ n
let q be a random (incompressible) string of length 2n
when p is known (independent from p)
for every k ∈ [0, C(x)] and every l ∈ [0, 2n] consider
yk,l = (k-bit prefix of p, l-bit prefix of q)
mapping (k, l) → (C(x|yk,l), C(yk,l|x))](https://image.slidesharecdn.com/talk-130922050737-phpapp02/85/Kolmogorov-complexity-and-topological-arguments-40-320.jpg)
![. . . . . .
Putting pieces together
let p be the shortest program for x, so |p| = C(x) ≥ n
let q be a random (incompressible) string of length 2n
when p is known (independent from p)
for every k ∈ [0, C(x)] and every l ∈ [0, 2n] consider
yk,l = (k-bit prefix of p, l-bit prefix of q)
mapping (k, l) → (C(x|yk,l), C(yk,l|x))
|p|
|q|
C(x)
A B
CD
2m
AB
C D
C(x|y)
C(y|x)
C(x)
2n
(n,n)](https://image.slidesharecdn.com/talk-130922050737-phpapp02/85/Kolmogorov-complexity-and-topological-arguments-41-320.jpg)
Kolmogorov complexity and topological arguments
- 1. . . . . . . JAC 2012 Topological arguments and Kolmogorov complexity Andrei Romashenko (joint work with Alexander Shen) LIRMM, CNRS & UM2, Montpellier; on leave from ИППИ РАН, Москва Supported by ANR NAFIT grant
- 2. . . . . . . Apologies
- 3. . . . . . . Apologies off-topic
- 4. . . . . . . Apologies off-topic message: not only general topology can be useful in computer science
- 5. . . . . . . Apologies off-topic message: not only general topology can be useful in computer science no blackboard
- 6. . . . . . . Conditional complexity as distance
- 7. . . . . . . Conditional complexity as distance C(x|y), conditional complexity of x given y, minimal length of a program that maps y to x
- 8. . . . . . . Conditional complexity as distance C(x|y), conditional complexity of x given y, minimal length of a program that maps y to x depends on the programming language, is minimal up to O(1) for some “optimal” languages; one of them is fixed
- 9. . . . . . . Conditional complexity as distance C(x|y), conditional complexity of x given y, minimal length of a program that maps y to x depends on the programming language, is minimal up to O(1) for some “optimal” languages; one of them is fixed C(x|y) measures “how far is x from y” in a sense, but not symmetric
- 10. . . . . . . Conditional complexity as distance C(x|y), conditional complexity of x given y, minimal length of a program that maps y to x depends on the programming language, is minimal up to O(1) for some “optimal” languages; one of them is fixed C(x|y) measures “how far is x from y” in a sense, but not symmetric task: given string x and number n, find y such that C(x|y) = n + O(1) and C(y|x) = n + O(1)
- 11. . . . . . . Conditional complexity as distance C(x|y), conditional complexity of x given y, minimal length of a program that maps y to x depends on the programming language, is minimal up to O(1) for some “optimal” languages; one of them is fixed C(x|y) measures “how far is x from y” in a sense, but not symmetric task: given string x and number n, find y such that C(x|y) = n + O(1) and C(y|x) = n + O(1) not always possible: C(x) should be at least n
- 12. . . . . . . M. Vyugin theorem and its extension
- 13. . . . . . . M. Vyugin theorem and its extension Theorem: if C(x) > 2n, there exists y such that C(x|y) = n + O(1) and C(y|x) = n + O(1).
- 14. . . . . . . M. Vyugin theorem and its extension Theorem: if C(x) > 2n, there exists y such that C(x|y) = n + O(1) and C(y|x) = n + O(1). proof uses a game argument
- 15. . . . . . . M. Vyugin theorem and its extension Theorem: if C(x) > 2n, there exists y such that C(x|y) = n + O(1) and C(y|x) = n + O(1). proof uses a game argument in fact C(x) > n + O(log n) is enough
- 16. . . . . . . M. Vyugin theorem and its extension Theorem: if C(x) > 2n, there exists y such that C(x|y) = n + O(1) and C(y|x) = n + O(1). proof uses a game argument in fact C(x) > n + O(log n) is enough but for completely different reasons
- 17. . . . . . . M. Vyugin theorem and its extension Theorem: if C(x) > 2n, there exists y such that C(x|y) = n + O(1) and C(y|x) = n + O(1). proof uses a game argument in fact C(x) > n + O(log n) is enough but for completely different reasons simple topological fact: if a continuous mapping of a circle S1 to R2 turns around some point O, then any its continuous extension to a mapping of a disk D2 covers O
- 18. . . . . . . M. Vyugin theorem and its extension Theorem: if C(x) > 2n, there exists y such that C(x|y) = n + O(1) and C(y|x) = n + O(1). proof uses a game argument in fact C(x) > n + O(log n) is enough but for completely different reasons simple topological fact: if a continuous mapping of a circle S1 to R2 turns around some point O, then any its continuous extension to a mapping of a disk D2 covers O strangely, for C(x) ≫ n this argument does not work (only for C(x) ≤ poly(n))
- 19. . . . . . . M. Vyugin theorem and its extension Theorem: if C(x) > 2n, there exists y such that C(x|y) = n + O(1) and C(y|x) = n + O(1). proof uses a game argument in fact C(x) > n + O(log n) is enough but for completely different reasons simple topological fact: if a continuous mapping of a circle S1 to R2 turns around some point O, then any its continuous extension to a mapping of a disk D2 covers O strangely, for C(x) ≫ n this argument does not work (only for C(x) ≤ poly(n)) so C(x) ≥ n + O(log n) is enough, but two essentially different arguments are needed at both ends
- 20. . . . . . . Why topology can be useful
- 21. . . . . . . Why topology can be useful simple example: imagine we want C(x|y) = n and know that C(x) ≥ n.
- 22. . . . . . . Why topology can be useful simple example: imagine we want C(x|y) = n and know that C(x) ≥ n. let y be x, then C(x|y) = O(1)
- 23. . . . . . . Why topology can be useful simple example: imagine we want C(x|y) = n and know that C(x) ≥ n. let y be x, then C(x|y) = O(1) let us remove bits in y one by one (e.g., from right to left)
- 24. . . . . . . Why topology can be useful simple example: imagine we want C(x|y) = n and know that C(x) ≥ n. let y be x, then C(x|y) = O(1) let us remove bits in y one by one (e.g., from right to left) C(x|y) then changes but gradually: C(x|y0) and C(x|y1) are C(x|y) + O(1)
- 25. . . . . . . Why topology can be useful simple example: imagine we want C(x|y) = n and know that C(x) ≥ n. let y be x, then C(x|y) = O(1) let us remove bits in y one by one (e.g., from right to left) C(x|y) then changes but gradually: C(x|y0) and C(x|y1) are C(x|y) + O(1) at the end y is empty, and C(x|y) = C(x) ≥ n
- 26. . . . . . . Why topology can be useful simple example: imagine we want C(x|y) = n and know that C(x) ≥ n. let y be x, then C(x|y) = O(1) let us remove bits in y one by one (e.g., from right to left) C(x|y) then changes but gradually: C(x|y0) and C(x|y1) are C(x|y) + O(1) at the end y is empty, and C(x|y) = C(x) ≥ n discrete intermediate value theorem guarantees that C(x|y) = n + O(1) for some y on the way
- 27. . . . . . . O(log n) precision is easy
- 28. . . . . . . O(log n) precision is easy to get C(y|x) = n we need to put some n bits of new information (that is not in x) into y
- 29. . . . . . . O(log n) precision is easy to get C(y|x) = n we need to put some n bits of new information (that is not in x) into y to get C(x|y) = n we need to put in y all the information about x except for n bits
- 30. . . . . . . O(log n) precision is easy to get C(y|x) = n we need to put some n bits of new information (that is not in x) into y to get C(x|y) = n we need to put in y all the information about x except for n bits let p be the shortest program for x, so |p| = C(x) ≥ n
- 31. . . . . . . O(log n) precision is easy to get C(y|x) = n we need to put some n bits of new information (that is not in x) into y to get C(x|y) = n we need to put in y all the information about x except for n bits let p be the shortest program for x, so |p| = C(x) ≥ n p is incompressible
- 32. . . . . . . O(log n) precision is easy to get C(y|x) = n we need to put some n bits of new information (that is not in x) into y to get C(x|y) = n we need to put in y all the information about x except for n bits let p be the shortest program for x, so |p| = C(x) ≥ n p is incompressible let y be p without n bits
- 33. . . . . . . O(log n) precision is easy to get C(y|x) = n we need to put some n bits of new information (that is not in x) into y to get C(x|y) = n we need to put in y all the information about x except for n bits let p be the shortest program for x, so |p| = C(x) ≥ n p is incompressible let y be p without n bits plus some random n bits (independent from p)
- 34. . . . . . . O(log n) precision is easy to get C(y|x) = n we need to put some n bits of new information (that is not in x) into y to get C(x|y) = n we need to put in y all the information about x except for n bits let p be the shortest program for x, so |p| = C(x) ≥ n p is incompressible let y be p without n bits plus some random n bits (independent from p) then both C(x|y) and C(y|x) are n + O(log n)
- 35. . . . . . . O(log n) precision is easy to get C(y|x) = n we need to put some n bits of new information (that is not in x) into y to get C(x|y) = n we need to put in y all the information about x except for n bits let p be the shortest program for x, so |p| = C(x) ≥ n p is incompressible let y be p without n bits plus some random n bits (independent from p) then both C(x|y) and C(y|x) are n + O(log n) O(1) cannot be obtained in this way (since all the arguments about random and independent bits work with O(log n) precision only)
- 36. . . . . . . Putting pieces together
- 37. . . . . . . Putting pieces together let p be the shortest program for x, so |p| = C(x) ≥ n
- 38. . . . . . . Putting pieces together let p be the shortest program for x, so |p| = C(x) ≥ n let q be a random (incompressible) string of length 2n when p is known (independent from p)
- 39. . . . . . . Putting pieces together let p be the shortest program for x, so |p| = C(x) ≥ n let q be a random (incompressible) string of length 2n when p is known (independent from p) for every k ∈ [0, C(x)] and every l ∈ [0, 2n] consider yk,l = (k-bit prefix of p, l-bit prefix of q)
- 40. . . . . . . Putting pieces together let p be the shortest program for x, so |p| = C(x) ≥ n let q be a random (incompressible) string of length 2n when p is known (independent from p) for every k ∈ [0, C(x)] and every l ∈ [0, 2n] consider yk,l = (k-bit prefix of p, l-bit prefix of q) mapping (k, l) → (C(x|yk,l), C(yk,l|x))
- 41. . . . . . . Putting pieces together let p be the shortest program for x, so |p| = C(x) ≥ n let q be a random (incompressible) string of length 2n when p is known (independent from p) for every k ∈ [0, C(x)] and every l ∈ [0, 2n] consider yk,l = (k-bit prefix of p, l-bit prefix of q) mapping (k, l) → (C(x|yk,l), C(yk,l|x)) |p| |q| C(x) A B CD 2m AB C D C(x|y) C(y|x) C(x) 2n (n,n)
- 42. . . . . . . Topological details
- 43. . . . . . . Topological details mapping is defined on a grid (rectangle)
- 44. . . . . . . Topological details mapping is defined on a grid (rectangle) and maps neighbor points to a points at O(1) distance
- 45. . . . . . . Topological details mapping is defined on a grid (rectangle) and maps neighbor points to a points at O(1) distance “Lipschitz continuity”
- 46. . . . . . . Topological details mapping is defined on a grid (rectangle) and maps neighbor points to a points at O(1) distance “Lipschitz continuity” covers (n, n) with O(1) precision
- 47. . . . . . . Topological details mapping is defined on a grid (rectangle) and maps neighbor points to a points at O(1) distance “Lipschitz continuity” covers (n, n) with O(1) precision reduction to continuous version: interpolation on triangles (linear)
- 48. . . . . . . Topological details mapping is defined on a grid (rectangle) and maps neighbor points to a points at O(1) distance “Lipschitz continuity” covers (n, n) with O(1) precision reduction to continuous version: interpolation on triangles (linear) preimage may be not in the grid, but neighbor grid point gives O(1)-precision
- 49. . . . . . . Topological details mapping is defined on a grid (rectangle) and maps neighbor points to a points at O(1) distance “Lipschitz continuity” covers (n, n) with O(1) precision reduction to continuous version: interpolation on triangles (linear) preimage may be not in the grid, but neighbor grid point gives O(1)-precision Alternative: repeat the proof for discrete case
- 50. . . . . . . Comments
- 51. . . . . . . Comments why we need C(x) be polynomial? if C(x) is very large, the value of k may contain a lot of information about z
- 52. . . . . . . Comments why we need C(x) be polynomial? if C(x) is very large, the value of k may contain a lot of information about z it is not necessary (unlike for original Vyugin argument) to have the same targets for C(x|y) and C(y|x)
- 53. . . . . . . Comments why we need C(x) be polynomial? if C(x) is very large, the value of k may contain a lot of information about z it is not necessary (unlike for original Vyugin argument) to have the same targets for C(x|y) and C(y|x) other applications of the same type of argument: for every x, y that are almost independent (I(x : y) is small compared to C(x) and C(y)) one can find z such that C(x|z) = C(x)/2 + O(1) and C(y|z) = C(y)/2 + O(1)
- 54. . . . . . . Comments why we need C(x) be polynomial? if C(x) is very large, the value of k may contain a lot of information about z it is not necessary (unlike for original Vyugin argument) to have the same targets for C(x|y) and C(y|x) other applications of the same type of argument: for every x, y that are almost independent (I(x : y) is small compared to C(x) and C(y)) one can find z such that C(x|z) = C(x)/2 + O(1) and C(y|z) = C(y)/2 + O(1) similar statement for halving complexity of three or more strings by adding a condition:
- 55. . . . . . . Comments why we need C(x) be polynomial? if C(x) is very large, the value of k may contain a lot of information about z it is not necessary (unlike for original Vyugin argument) to have the same targets for C(x|y) and C(y|x) other applications of the same type of argument: for every x, y that are almost independent (I(x : y) is small compared to C(x) and C(y)) one can find z such that C(x|z) = C(x)/2 + O(1) and C(y|z) = C(y)/2 + O(1) similar statement for halving complexity of three or more strings by adding a condition: under the assumption of independence (can be weakened but not eliminated).
- 56. . . . . . . Comments why we need C(x) be polynomial? if C(x) is very large, the value of k may contain a lot of information about z it is not necessary (unlike for original Vyugin argument) to have the same targets for C(x|y) and C(y|x) other applications of the same type of argument: for every x, y that are almost independent (I(x : y) is small compared to C(x) and C(y)) one can find z such that C(x|z) = C(x)/2 + O(1) and C(y|z) = C(y)/2 + O(1) similar statement for halving complexity of three or more strings by adding a condition: under the assumption of independence (can be weakened but not eliminated). an open problem in the general case (problematic case: x and y are very close to each other, but not completely identical)
- 57. . . . . . . Thanks!
- 58. . . . . . . Original game argument
- 59. . . . . . . Original game argument If C(x) > 3n, there exists y such that C(x|y) and C(y|x) are n + O(1)
- 60. . . . . . . Original game argument If C(x) > 3n, there exists y such that C(x|y) and C(y|x) are n + O(1) we replaced 2n by 3n to simplify explanations (and in any case this is already covered)
- 61. . . . . . . Original game argument If C(x) > 3n, there exists y such that C(x|y) and C(y|x) are n + O(1) we replaced 2n by 3n to simplify explanations (and in any case this is already covered) we present some game
- 62. . . . . . . Original game argument If C(x) > 3n, there exists y such that C(x|y) and C(y|x) are n + O(1) we replaced 2n by 3n to simplify explanations (and in any case this is already covered) we present some game then show why winning this game is enough
- 63. . . . . . . Original game argument If C(x) > 3n, there exists y such that C(x|y) and C(y|x) are n + O(1) we replaced 2n by 3n to simplify explanations (and in any case this is already covered) we present some game then show why winning this game is enough and finally show how to win the game
- 64. . . . . . . Dating agency and its task
- 65. . . . . . . Dating agency and its task two countable sets X and Y
- 66. . . . . . . Dating agency and its task two countable sets X and Y game starts with a perfect matching, i.e., one to one correspondence between X and Y.
- 67. . . . . . . Dating agency and its task two countable sets X and Y game starts with a perfect matching, i.e., one to one correspondence between X and Y. An element of X or Y can refuse the current partner, then the current relationship (x, y) is dissolved
- 68. . . . . . . Dating agency and its task two countable sets X and Y game starts with a perfect matching, i.e., one to one correspondence between X and Y. An element of X or Y can refuse the current partner, then the current relationship (x, y) is dissolved y then becomes free; the agency may either
- 69. . . . . . . Dating agency and its task two countable sets X and Y game starts with a perfect matching, i.e., one to one correspondence between X and Y. An element of X or Y can refuse the current partner, then the current relationship (x, y) is dissolved y then becomes free; the agency may either find a new pair for x from the dissolved pair (among free elements of Y not tried with x previously) or
- 70. . . . . . . Dating agency and its task two countable sets X and Y game starts with a perfect matching, i.e., one to one correspondence between X and Y. An element of X or Y can refuse the current partner, then the current relationship (x, y) is dissolved y then becomes free; the agency may either find a new pair for x from the dissolved pair (among free elements of Y not tried with x previously) or declare x hopeless and do not try to find a pair for x anymore (#free in Y incremented)
- 71. . . . . . . Dating agency and its task two countable sets X and Y game starts with a perfect matching, i.e., one to one correspondence between X and Y. An element of X or Y can refuse the current partner, then the current relationship (x, y) is dissolved y then becomes free; the agency may either find a new pair for x from the dissolved pair (among free elements of Y not tried with x previously) or declare x hopeless and do not try to find a pair for x anymore (#free in Y incremented) the refusals appear (and are processed by the agency) one at a time
- 72. . . . . . . Dating agency and its task two countable sets X and Y game starts with a perfect matching, i.e., one to one correspondence between X and Y. An element of X or Y can refuse the current partner, then the current relationship (x, y) is dissolved y then becomes free; the agency may either find a new pair for x from the dissolved pair (among free elements of Y not tried with x previously) or declare x hopeless and do not try to find a pair for x anymore (#free in Y incremented) the refusals appear (and are processed by the agency) one at a time each element can produce < N refusals (parameter of the game), but no restrictions for #(being refused)
- 73. . . . . . . Dating agency and its task two countable sets X and Y game starts with a perfect matching, i.e., one to one correspondence between X and Y. An element of X or Y can refuse the current partner, then the current relationship (x, y) is dissolved y then becomes free; the agency may either find a new pair for x from the dissolved pair (among free elements of Y not tried with x previously) or declare x hopeless and do not try to find a pair for x anymore (#free in Y incremented) the refusals appear (and are processed by the agency) one at a time each element can produce < N refusals (parameter of the game), but no restrictions for #(being refused) agency obligations:
- 74. . . . . . . Dating agency and its task two countable sets X and Y game starts with a perfect matching, i.e., one to one correspondence between X and Y. An element of X or Y can refuse the current partner, then the current relationship (x, y) is dissolved y then becomes free; the agency may either find a new pair for x from the dissolved pair (among free elements of Y not tried with x previously) or declare x hopeless and do not try to find a pair for x anymore (#free in Y incremented) the refusals appear (and are processed by the agency) one at a time each element can produce < N refusals (parameter of the game), but no restrictions for #(being refused) agency obligations: ≤ 2N attempts for each element
- 75. . . . . . . Dating agency and its task two countable sets X and Y game starts with a perfect matching, i.e., one to one correspondence between X and Y. An element of X or Y can refuse the current partner, then the current relationship (x, y) is dissolved y then becomes free; the agency may either find a new pair for x from the dissolved pair (among free elements of Y not tried with x previously) or declare x hopeless and do not try to find a pair for x anymore (#free in Y incremented) the refusals appear (and are processed by the agency) one at a time each element can produce < N refusals (parameter of the game), but no restrictions for #(being refused) agency obligations: ≤ 2N attempts for each element ≤ 2N3 hopeless elements; all others in X are ultimately connected to some y ∈ Y and this connection lasts forever
- 76. . . . . . . Why computable winning strategy is enough
- 77. . . . . . . Why computable winning strategy is enough X = Y = B∗
- 78. . . . . . . Why computable winning strategy is enough X = Y = B∗ initial matching: identity (x, x)
- 79. . . . . . . Why computable winning strategy is enough X = Y = B∗ initial matching: identity (x, x) u refuses v if C(v|u) < n (here u may be in X or in Y)
- 80. . . . . . . Why computable winning strategy is enough X = Y = B∗ initial matching: identity (x, x) u refuses v if C(v|u) < n (here u may be in X or in Y) less than N = 2n refusals for each u
- 81. . . . . . . Why computable winning strategy is enough X = Y = B∗ initial matching: identity (x, x) u refuses v if C(v|u) < n (here u may be in X or in Y) less than N = 2n refusals for each u computable behavior
- 82. . . . . . . Why computable winning strategy is enough X = Y = B∗ initial matching: identity (x, x) u refuses v if C(v|u) < n (here u may be in X or in Y) less than N = 2n refusals for each u computable behavior agency produces O(N3 ) = O(23n ) hopeless elements of complexity 3n + O(1) (identified by 3n + O(1) bit ordinal number)
- 83. . . . . . . Why computable winning strategy is enough X = Y = B∗ initial matching: identity (x, x) u refuses v if C(v|u) < n (here u may be in X or in Y) less than N = 2n refusals for each u computable behavior agency produces O(N3 ) = O(23n ) hopeless elements of complexity 3n + O(1) (identified by 3n + O(1) bit ordinal number) for every x that is not hopeless its final partner y has C(y|x) and C(x|y) at most n + O(1): determined by a ordinal number that is O(N) = 2n+O(1)
- 84. . . . . . . Why computable winning strategy is enough X = Y = B∗ initial matching: identity (x, x) u refuses v if C(v|u) < n (here u may be in X or in Y) less than N = 2n refusals for each u computable behavior agency produces O(N3 ) = O(23n ) hopeless elements of complexity 3n + O(1) (identified by 3n + O(1) bit ordinal number) for every x that is not hopeless its final partner y has C(y|x) and C(x|y) at most n + O(1): determined by a ordinal number that is O(N) = 2n+O(1) but both complexities are at least n, otherwise refused
- 85. . . . . . . How to win the game
- 86. . . . . . . How to win the game each element not currently matched keeps “experience”=(#refusals sent, #refusals received)
- 87. . . . . . . How to win the game each element not currently matched keeps “experience”=(#refusals sent, #refusals received) the first is < N; the second a priori is unbounded, but also will be kept < N due to agency strategy
- 88. . . . . . . How to win the game each element not currently matched keeps “experience”=(#refusals sent, #refusals received) the first is < N; the second a priori is unbounded, but also will be kept < N due to agency strategy when (x, y) is terminated, numbers updated
- 89. . . . . . . How to win the game each element not currently matched keeps “experience”=(#refusals sent, #refusals received) the first is < N; the second a priori is unbounded, but also will be kept < N due to agency strategy when (x, y) is terminated, numbers updated invariant: in all pairs people have matching experiences (#sent = #received for the other)
- 90. . . . . . . How to win the game each element not currently matched keeps “experience”=(#refusals sent, #refusals received) the first is < N; the second a priori is unbounded, but also will be kept < N due to agency strategy when (x, y) is terminated, numbers updated invariant: in all pairs people have matching experiences (#sent = #received for the other) corollary: #refusals received < N
- 91. . . . . . . How to win the game each element not currently matched keeps “experience”=(#refusals sent, #refusals received) the first is < N; the second a priori is unbounded, but also will be kept < N due to agency strategy when (x, y) is terminated, numbers updated invariant: in all pairs people have matching experiences (#sent = #received for the other) corollary: #refusals received < N new partner for x is found if possible (=there is y ∈ Y with matching experience not tried earlier with x)
- 92. . . . . . . How to win the game each element not currently matched keeps “experience”=(#refusals sent, #refusals received) the first is < N; the second a priori is unbounded, but also will be kept < N due to agency strategy when (x, y) is terminated, numbers updated invariant: in all pairs people have matching experiences (#sent = #received for the other) corollary: #refusals received < N new partner for x is found if possible (=there is y ∈ Y with matching experience not tried earlier with x) otherwise x is declared hopeless
- 93. . . . . . . How to win the game each element not currently matched keeps “experience”=(#refusals sent, #refusals received) the first is < N; the second a priori is unbounded, but also will be kept < N due to agency strategy when (x, y) is terminated, numbers updated invariant: in all pairs people have matching experiences (#sent = #received for the other) corollary: #refusals received < N new partner for x is found if possible (=there is y ∈ Y with matching experience not tried earlier with x) otherwise x is declared hopeless invariant: for matching experiences the number of non-matched people in X and Y are the same
- 94. . . . . . . How to win the game each element not currently matched keeps “experience”=(#refusals sent, #refusals received) the first is < N; the second a priori is unbounded, but also will be kept < N due to agency strategy when (x, y) is terminated, numbers updated invariant: in all pairs people have matching experiences (#sent = #received for the other) corollary: #refusals received < N new partner for x is found if possible (=there is y ∈ Y with matching experience not tried earlier with x) otherwise x is declared hopeless invariant: for matching experiences the number of non-matched people in X and Y are the same ≤ 2N attempts for each (experience increases each time)
- 95. . . . . . . How to win the game each element not currently matched keeps “experience”=(#refusals sent, #refusals received) the first is < N; the second a priori is unbounded, but also will be kept < N due to agency strategy when (x, y) is terminated, numbers updated invariant: in all pairs people have matching experiences (#sent = #received for the other) corollary: #refusals received < N new partner for x is found if possible (=there is y ∈ Y with matching experience not tried earlier with x) otherwise x is declared hopeless invariant: for matching experiences the number of non-matched people in X and Y are the same ≤ 2N attempts for each (experience increases each time) there are N2 experience classes; if class reaches 2N, it stops growing since y can be always found in the class (< 2N are tried earlier with given x), so O(N3 ) hopeless
- 96. . . . . . . Thanks
- 97. . . . . . . Thanks to the organizers who accepted to consider these arguments
- 98. . . . . . . Thanks to the organizers who accepted to consider these arguments to Misha Vyugin and Andrej Muchnik who invented the game argument and its generalization for several strings yi
- 99. . . . . . . Thanks to the organizers who accepted to consider these arguments to Misha Vyugin and Andrej Muchnik who invented the game argument and its generalization for several strings yi to Laurent Bienvenu who convinced us to write this simple argument down
- 100. . . . . . . Thanks to the organizers who accepted to consider these arguments to Misha Vyugin and Andrej Muchnik who invented the game argument and its generalization for several strings yi to Laurent Bienvenu who convinced us to write this simple argument down to all colleagues (ESCAPE team in Marseille and Montpellier, participants of Kolmogorov seminar in Moscow)
- 101. . . . . . . Thanks to the organizers who accepted to consider these arguments to Misha Vyugin and Andrej Muchnik who invented the game argument and its generalization for several strings yi to Laurent Bienvenu who convinced us to write this simple argument down to all colleagues (ESCAPE team in Marseille and Montpellier, participants of Kolmogorov seminar in Moscow) to the audience for following the talk to that point :-)