kolmogorov complexity and topological arguments
DESCRIPTION
Talk in 2013TRANSCRIPT
. . . . . .
JAC 2012
Topological arguments and Kolmogorovcomplexity
Andrei Romashenko (joint work with Alexander Shen)LIRMM, CNRS & UM2, Montpellier; on leave from
ИППИ РАН, Москва
Supported by ANR NAFIT grant
. . . . . .
Apologies
I off-topicmessage: not only general topology can be useful incomputer science
I no blackboard
. . . . . .
Apologies
I off-topic
message: not only general topology can be useful incomputer science
I no blackboard
. . . . . .
Apologies
I off-topicmessage: not only general topology can be useful incomputer science
I no blackboard
. . . . . .
Apologies
I off-topicmessage: not only general topology can be useful incomputer science
I no blackboard
. . . . . .
Conditional complexity as distance
I C(x|y), conditional complexity of x given y, minimallength of a program that maps y to x
I depends on the programming language, is minimal upto O(1) for some “optimal” languages; one of them isfixed
I C(x|y) measures “how far is x from y” in a sense, but notsymmetric
I task: given string x and number n, find y such thatC(x|y) = n+ O(1) and C(y|x) = n+ O(1)
I not always possible: C(x) should be at least n
. . . . . .
Conditional complexity as distance
I C(x|y), conditional complexity of x given y, minimallength of a program that maps y to x
I depends on the programming language, is minimal upto O(1) for some “optimal” languages; one of them isfixed
I C(x|y) measures “how far is x from y” in a sense, but notsymmetric
I task: given string x and number n, find y such thatC(x|y) = n+ O(1) and C(y|x) = n+ O(1)
I not always possible: C(x) should be at least n
. . . . . .
Conditional complexity as distance
I C(x|y), conditional complexity of x given y, minimallength of a program that maps y to x
I depends on the programming language, is minimal upto O(1) for some “optimal” languages; one of them isfixed
I C(x|y) measures “how far is x from y” in a sense, but notsymmetric
I task: given string x and number n, find y such thatC(x|y) = n+ O(1) and C(y|x) = n+ O(1)
I not always possible: C(x) should be at least n
. . . . . .
Conditional complexity as distance
I C(x|y), conditional complexity of x given y, minimallength of a program that maps y to x
I depends on the programming language, is minimal upto O(1) for some “optimal” languages; one of them isfixed
I C(x|y) measures “how far is x from y” in a sense, but notsymmetric
I task: given string x and number n, find y such thatC(x|y) = n+ O(1) and C(y|x) = n+ O(1)
I not always possible: C(x) should be at least n
. . . . . .
Conditional complexity as distance
I C(x|y), conditional complexity of x given y, minimallength of a program that maps y to x
I depends on the programming language, is minimal upto O(1) for some “optimal” languages; one of them isfixed
I C(x|y) measures “how far is x from y” in a sense, but notsymmetric
I task: given string x and number n, find y such thatC(x|y) = n+ O(1) and C(y|x) = n+ O(1)
I not always possible: C(x) should be at least n
. . . . . .
Conditional complexity as distance
I C(x|y), conditional complexity of x given y, minimallength of a program that maps y to x
I depends on the programming language, is minimal upto O(1) for some “optimal” languages; one of them isfixed
I C(x|y) measures “how far is x from y” in a sense, but notsymmetric
I task: given string x and number n, find y such thatC(x|y) = n+ O(1) and C(y|x) = n+ O(1)
I not always possible: C(x) should be at least n
. . . . . .
M. Vyugin theorem and its extension
I Theorem: if C(x) > 2n, there exists y such thatC(x|y) = n+ O(1) and C(y|x) = n+ O(1).
I proof uses a game argumentI in fact C(x) > n+ O(log n) is enoughI but for completely different reasonsI simple topological fact: if a continuous mapping of acircle S1 to R2 turns around some point O, then any itscontinuous extension to a mapping of a disk D2 covers O
I strangely, for C(x) ≫ n this argument does not work(only for C(x) ≤ poly(n))
I so C(x) ≥ n+ O(log n) is enough, but two essentiallydifferent arguments are needed at both ends
. . . . . .
M. Vyugin theorem and its extension
I Theorem: if C(x) > 2n, there exists y such thatC(x|y) = n+ O(1) and C(y|x) = n+ O(1).
I proof uses a game argumentI in fact C(x) > n+ O(log n) is enoughI but for completely different reasonsI simple topological fact: if a continuous mapping of acircle S1 to R2 turns around some point O, then any itscontinuous extension to a mapping of a disk D2 covers O
I strangely, for C(x) ≫ n this argument does not work(only for C(x) ≤ poly(n))
I so C(x) ≥ n+ O(log n) is enough, but two essentiallydifferent arguments are needed at both ends
. . . . . .
M. Vyugin theorem and its extension
I Theorem: if C(x) > 2n, there exists y such thatC(x|y) = n+ O(1) and C(y|x) = n+ O(1).
I proof uses a game argument
I in fact C(x) > n+ O(log n) is enoughI but for completely different reasonsI simple topological fact: if a continuous mapping of acircle S1 to R2 turns around some point O, then any itscontinuous extension to a mapping of a disk D2 covers O
I strangely, for C(x) ≫ n this argument does not work(only for C(x) ≤ poly(n))
I so C(x) ≥ n+ O(log n) is enough, but two essentiallydifferent arguments are needed at both ends
. . . . . .
M. Vyugin theorem and its extension
I Theorem: if C(x) > 2n, there exists y such thatC(x|y) = n+ O(1) and C(y|x) = n+ O(1).
I proof uses a game argumentI in fact C(x) > n+ O(log n) is enough
I but for completely different reasonsI simple topological fact: if a continuous mapping of acircle S1 to R2 turns around some point O, then any itscontinuous extension to a mapping of a disk D2 covers O
I strangely, for C(x) ≫ n this argument does not work(only for C(x) ≤ poly(n))
I so C(x) ≥ n+ O(log n) is enough, but two essentiallydifferent arguments are needed at both ends
. . . . . .
M. Vyugin theorem and its extension
I Theorem: if C(x) > 2n, there exists y such thatC(x|y) = n+ O(1) and C(y|x) = n+ O(1).
I proof uses a game argumentI in fact C(x) > n+ O(log n) is enoughI but for completely different reasons
I simple topological fact: if a continuous mapping of acircle S1 to R2 turns around some point O, then any itscontinuous extension to a mapping of a disk D2 covers O
I strangely, for C(x) ≫ n this argument does not work(only for C(x) ≤ poly(n))
I so C(x) ≥ n+ O(log n) is enough, but two essentiallydifferent arguments are needed at both ends
. . . . . .
M. Vyugin theorem and its extension
I Theorem: if C(x) > 2n, there exists y such thatC(x|y) = n+ O(1) and C(y|x) = n+ O(1).
I proof uses a game argumentI in fact C(x) > n+ O(log n) is enoughI but for completely different reasonsI simple topological fact: if a continuous mapping of acircle S1 to R2 turns around some point O, then any itscontinuous extension to a mapping of a disk D2 covers O
I strangely, for C(x) ≫ n this argument does not work(only for C(x) ≤ poly(n))
I so C(x) ≥ n+ O(log n) is enough, but two essentiallydifferent arguments are needed at both ends
. . . . . .
M. Vyugin theorem and its extension
I Theorem: if C(x) > 2n, there exists y such thatC(x|y) = n+ O(1) and C(y|x) = n+ O(1).
I proof uses a game argumentI in fact C(x) > n+ O(log n) is enoughI but for completely different reasonsI simple topological fact: if a continuous mapping of acircle S1 to R2 turns around some point O, then any itscontinuous extension to a mapping of a disk D2 covers O
I strangely, for C(x) ≫ n this argument does not work(only for C(x) ≤ poly(n))
I so C(x) ≥ n+ O(log n) is enough, but two essentiallydifferent arguments are needed at both ends
. . . . . .
M. Vyugin theorem and its extension
I Theorem: if C(x) > 2n, there exists y such thatC(x|y) = n+ O(1) and C(y|x) = n+ O(1).
I proof uses a game argumentI in fact C(x) > n+ O(log n) is enoughI but for completely different reasonsI simple topological fact: if a continuous mapping of acircle S1 to R2 turns around some point O, then any itscontinuous extension to a mapping of a disk D2 covers O
I strangely, for C(x) ≫ n this argument does not work(only for C(x) ≤ poly(n))
I so C(x) ≥ n+ O(log n) is enough, but two essentiallydifferent arguments are needed at both ends
. . . . . .
Why topology can be useful
I simple example: imagine we want C(x|y) = n and knowthat C(x) ≥ n.
I let y be x, then C(x|y) = O(1)I let us remove bits in y one by one (e.g., from right toleft)
I C(x|y) then changes but gradually: C(x|y0) and C(x|y1)are C(x|y) + O(1)
I at the end y is empty, and C(x|y) = C(x) ≥ nI discrete intermediate value theorem guarantees thatC(x|y) = n+ O(1) for some y on the way
. . . . . .
Why topology can be useful
I simple example: imagine we want C(x|y) = n and knowthat C(x) ≥ n.
I let y be x, then C(x|y) = O(1)I let us remove bits in y one by one (e.g., from right toleft)
I C(x|y) then changes but gradually: C(x|y0) and C(x|y1)are C(x|y) + O(1)
I at the end y is empty, and C(x|y) = C(x) ≥ nI discrete intermediate value theorem guarantees thatC(x|y) = n+ O(1) for some y on the way
. . . . . .
Why topology can be useful
I simple example: imagine we want C(x|y) = n and knowthat C(x) ≥ n.
I let y be x, then C(x|y) = O(1)
I let us remove bits in y one by one (e.g., from right toleft)
I C(x|y) then changes but gradually: C(x|y0) and C(x|y1)are C(x|y) + O(1)
I at the end y is empty, and C(x|y) = C(x) ≥ nI discrete intermediate value theorem guarantees thatC(x|y) = n+ O(1) for some y on the way
. . . . . .
Why topology can be useful
I simple example: imagine we want C(x|y) = n and knowthat C(x) ≥ n.
I let y be x, then C(x|y) = O(1)I let us remove bits in y one by one (e.g., from right toleft)
I C(x|y) then changes but gradually: C(x|y0) and C(x|y1)are C(x|y) + O(1)
I at the end y is empty, and C(x|y) = C(x) ≥ nI discrete intermediate value theorem guarantees thatC(x|y) = n+ O(1) for some y on the way
. . . . . .
Why topology can be useful
I simple example: imagine we want C(x|y) = n and knowthat C(x) ≥ n.
I let y be x, then C(x|y) = O(1)I let us remove bits in y one by one (e.g., from right toleft)
I C(x|y) then changes but gradually: C(x|y0) and C(x|y1)are C(x|y) + O(1)
I at the end y is empty, and C(x|y) = C(x) ≥ nI discrete intermediate value theorem guarantees thatC(x|y) = n+ O(1) for some y on the way
. . . . . .
Why topology can be useful
I simple example: imagine we want C(x|y) = n and knowthat C(x) ≥ n.
I let y be x, then C(x|y) = O(1)I let us remove bits in y one by one (e.g., from right toleft)
I C(x|y) then changes but gradually: C(x|y0) and C(x|y1)are C(x|y) + O(1)
I at the end y is empty, and C(x|y) = C(x) ≥ n
I discrete intermediate value theorem guarantees thatC(x|y) = n+ O(1) for some y on the way
. . . . . .
Why topology can be useful
I simple example: imagine we want C(x|y) = n and knowthat C(x) ≥ n.
I let y be x, then C(x|y) = O(1)I let us remove bits in y one by one (e.g., from right toleft)
I C(x|y) then changes but gradually: C(x|y0) and C(x|y1)are C(x|y) + O(1)
I at the end y is empty, and C(x|y) = C(x) ≥ nI discrete intermediate value theorem guarantees thatC(x|y) = n+ O(1) for some y on the way
. . . . . .
O(log n) precision is easy
I to get C(y|x) = n we need to put some n bits of newinformation (that is not in x) into y
I to get C(x|y) = n we need to put in y all the informationabout x except for n bits
I let p be the shortest program for x, so |p| = C(x) ≥ nI p is incompressibleI let y be p without n bitsI plus some random n bits (independent from p)I then both C(x|y) and C(y|x) are n+ O(log n)I O(1) cannot be obtained in this way (since all thearguments about random and independent bits workwith O(log n) precision only)
. . . . . .
O(log n) precision is easy
I to get C(y|x) = n we need to put some n bits of newinformation (that is not in x) into y
I to get C(x|y) = n we need to put in y all the informationabout x except for n bits
I let p be the shortest program for x, so |p| = C(x) ≥ nI p is incompressibleI let y be p without n bitsI plus some random n bits (independent from p)I then both C(x|y) and C(y|x) are n+ O(log n)I O(1) cannot be obtained in this way (since all thearguments about random and independent bits workwith O(log n) precision only)
. . . . . .
O(log n) precision is easy
I to get C(y|x) = n we need to put some n bits of newinformation (that is not in x) into y
I to get C(x|y) = n we need to put in y all the informationabout x except for n bits
I let p be the shortest program for x, so |p| = C(x) ≥ nI p is incompressibleI let y be p without n bitsI plus some random n bits (independent from p)I then both C(x|y) and C(y|x) are n+ O(log n)I O(1) cannot be obtained in this way (since all thearguments about random and independent bits workwith O(log n) precision only)
. . . . . .
O(log n) precision is easy
I to get C(y|x) = n we need to put some n bits of newinformation (that is not in x) into y
I to get C(x|y) = n we need to put in y all the informationabout x except for n bits
I let p be the shortest program for x, so |p| = C(x) ≥ n
I p is incompressibleI let y be p without n bitsI plus some random n bits (independent from p)I then both C(x|y) and C(y|x) are n+ O(log n)I O(1) cannot be obtained in this way (since all thearguments about random and independent bits workwith O(log n) precision only)
. . . . . .
O(log n) precision is easy
I to get C(y|x) = n we need to put some n bits of newinformation (that is not in x) into y
I to get C(x|y) = n we need to put in y all the informationabout x except for n bits
I let p be the shortest program for x, so |p| = C(x) ≥ nI p is incompressible
I let y be p without n bitsI plus some random n bits (independent from p)I then both C(x|y) and C(y|x) are n+ O(log n)I O(1) cannot be obtained in this way (since all thearguments about random and independent bits workwith O(log n) precision only)
. . . . . .
O(log n) precision is easy
I to get C(y|x) = n we need to put some n bits of newinformation (that is not in x) into y
I to get C(x|y) = n we need to put in y all the informationabout x except for n bits
I let p be the shortest program for x, so |p| = C(x) ≥ nI p is incompressibleI let y be p without n bits
I plus some random n bits (independent from p)I then both C(x|y) and C(y|x) are n+ O(log n)I O(1) cannot be obtained in this way (since all thearguments about random and independent bits workwith O(log n) precision only)
. . . . . .
O(log n) precision is easy
I to get C(y|x) = n we need to put some n bits of newinformation (that is not in x) into y
I to get C(x|y) = n we need to put in y all the informationabout x except for n bits
I let p be the shortest program for x, so |p| = C(x) ≥ nI p is incompressibleI let y be p without n bitsI plus some random n bits (independent from p)
I then both C(x|y) and C(y|x) are n+ O(log n)I O(1) cannot be obtained in this way (since all thearguments about random and independent bits workwith O(log n) precision only)
. . . . . .
O(log n) precision is easy
I to get C(y|x) = n we need to put some n bits of newinformation (that is not in x) into y
I to get C(x|y) = n we need to put in y all the informationabout x except for n bits
I let p be the shortest program for x, so |p| = C(x) ≥ nI p is incompressibleI let y be p without n bitsI plus some random n bits (independent from p)I then both C(x|y) and C(y|x) are n+ O(log n)
I O(1) cannot be obtained in this way (since all thearguments about random and independent bits workwith O(log n) precision only)
. . . . . .
O(log n) precision is easy
I to get C(y|x) = n we need to put some n bits of newinformation (that is not in x) into y
I to get C(x|y) = n we need to put in y all the informationabout x except for n bits
I let p be the shortest program for x, so |p| = C(x) ≥ nI p is incompressibleI let y be p without n bitsI plus some random n bits (independent from p)I then both C(x|y) and C(y|x) are n+ O(log n)I O(1) cannot be obtained in this way (since all thearguments about random and independent bits workwith O(log n) precision only)
. . . . . .
Putting pieces together
I let p be the shortest program for x, so |p| = C(x) ≥ nI let q be a random (incompressible) string of length 2nwhen p is known (independent from p)
I for every k ∈ [0, C(x)] and every l ∈ [0, 2n] consider
yk,l = (k-bit prefix of p, l-bit prefix of q)
I mapping (k, l) 7→ (C(x|yk,l), C(yk,l|x))
|p|
|q|
C(x)A B
CD2m
A′B′
C′ D′
C(x|y)
C(y|x)
C(x)
2n
(n,n)
. . . . . .
Putting pieces togetherI let p be the shortest program for x, so |p| = C(x) ≥ n
I let q be a random (incompressible) string of length 2nwhen p is known (independent from p)
I for every k ∈ [0, C(x)] and every l ∈ [0, 2n] consider
yk,l = (k-bit prefix of p, l-bit prefix of q)
I mapping (k, l) 7→ (C(x|yk,l), C(yk,l|x))
|p|
|q|
C(x)A B
CD2m
A′B′
C′ D′
C(x|y)
C(y|x)
C(x)
2n
(n,n)
. . . . . .
Putting pieces togetherI let p be the shortest program for x, so |p| = C(x) ≥ nI let q be a random (incompressible) string of length 2nwhen p is known (independent from p)
I for every k ∈ [0, C(x)] and every l ∈ [0, 2n] consider
yk,l = (k-bit prefix of p, l-bit prefix of q)
I mapping (k, l) 7→ (C(x|yk,l), C(yk,l|x))
|p|
|q|
C(x)A B
CD2m
A′B′
C′ D′
C(x|y)
C(y|x)
C(x)
2n
(n,n)
. . . . . .
Putting pieces togetherI let p be the shortest program for x, so |p| = C(x) ≥ nI let q be a random (incompressible) string of length 2nwhen p is known (independent from p)
I for every k ∈ [0, C(x)] and every l ∈ [0, 2n] consider
yk,l = (k-bit prefix of p, l-bit prefix of q)
I mapping (k, l) 7→ (C(x|yk,l), C(yk,l|x))
|p|
|q|
C(x)A B
CD2m
A′B′
C′ D′
C(x|y)
C(y|x)
C(x)
2n
(n,n)
. . . . . .
Putting pieces togetherI let p be the shortest program for x, so |p| = C(x) ≥ nI let q be a random (incompressible) string of length 2nwhen p is known (independent from p)
I for every k ∈ [0, C(x)] and every l ∈ [0, 2n] consider
yk,l = (k-bit prefix of p, l-bit prefix of q)
I mapping (k, l) 7→ (C(x|yk,l), C(yk,l|x))
|p|
|q|
C(x)A B
CD2m
A′B′
C′ D′
C(x|y)
C(y|x)
C(x)
2n
(n,n)
. . . . . .
Putting pieces togetherI let p be the shortest program for x, so |p| = C(x) ≥ nI let q be a random (incompressible) string of length 2nwhen p is known (independent from p)
I for every k ∈ [0, C(x)] and every l ∈ [0, 2n] consider
yk,l = (k-bit prefix of p, l-bit prefix of q)
I mapping (k, l) 7→ (C(x|yk,l), C(yk,l|x))
|p|
|q|
C(x)A B
CD2m
A′B′
C′ D′
C(x|y)
C(y|x)
C(x)
2n
(n,n)
. . . . . .
Topological details
I mapping is defined on a grid (rectangle)I and maps neighbor points to a points at O(1) distanceI “Lipschitz continuity”I covers (n, n) with O(1) precisionI reduction to continuous version: interpolation ontriangles (linear)
I preimage may be not in the grid, but neighbor gridpoint gives O(1)-precision
I Alternative: repeat the proof for discrete case
. . . . . .
Topological details
I mapping is defined on a grid (rectangle)
I and maps neighbor points to a points at O(1) distanceI “Lipschitz continuity”I covers (n, n) with O(1) precisionI reduction to continuous version: interpolation ontriangles (linear)
I preimage may be not in the grid, but neighbor gridpoint gives O(1)-precision
I Alternative: repeat the proof for discrete case
. . . . . .
Topological details
I mapping is defined on a grid (rectangle)I and maps neighbor points to a points at O(1) distance
I “Lipschitz continuity”I covers (n, n) with O(1) precisionI reduction to continuous version: interpolation ontriangles (linear)
I preimage may be not in the grid, but neighbor gridpoint gives O(1)-precision
I Alternative: repeat the proof for discrete case
. . . . . .
Topological details
I mapping is defined on a grid (rectangle)I and maps neighbor points to a points at O(1) distanceI “Lipschitz continuity”
I covers (n, n) with O(1) precisionI reduction to continuous version: interpolation ontriangles (linear)
I preimage may be not in the grid, but neighbor gridpoint gives O(1)-precision
I Alternative: repeat the proof for discrete case
. . . . . .
Topological details
I mapping is defined on a grid (rectangle)I and maps neighbor points to a points at O(1) distanceI “Lipschitz continuity”I covers (n, n) with O(1) precision
I reduction to continuous version: interpolation ontriangles (linear)
I preimage may be not in the grid, but neighbor gridpoint gives O(1)-precision
I Alternative: repeat the proof for discrete case
. . . . . .
Topological details
I mapping is defined on a grid (rectangle)I and maps neighbor points to a points at O(1) distanceI “Lipschitz continuity”I covers (n, n) with O(1) precisionI reduction to continuous version: interpolation ontriangles (linear)
I preimage may be not in the grid, but neighbor gridpoint gives O(1)-precision
I Alternative: repeat the proof for discrete case
. . . . . .
Topological details
I mapping is defined on a grid (rectangle)I and maps neighbor points to a points at O(1) distanceI “Lipschitz continuity”I covers (n, n) with O(1) precisionI reduction to continuous version: interpolation ontriangles (linear)
I preimage may be not in the grid, but neighbor gridpoint gives O(1)-precision
I Alternative: repeat the proof for discrete case
. . . . . .
Topological details
I mapping is defined on a grid (rectangle)I and maps neighbor points to a points at O(1) distanceI “Lipschitz continuity”I covers (n, n) with O(1) precisionI reduction to continuous version: interpolation ontriangles (linear)
I preimage may be not in the grid, but neighbor gridpoint gives O(1)-precision
I Alternative: repeat the proof for discrete case
. . . . . .
Comments
I 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
I it is not necessary (unlike for original Vyugin argument)to have the same targets for C(x|y) and C(y|x)
I other applications of the same type of argument: forevery x, y that are almost independent (I(x : y) is smallcompared to C(x) and C(y)) one can find z such thatC(x|z) = C(x)/2 + O(1) and C(y|z) = C(y)/2 + O(1)
I similar statement for halving complexity of three ormore strings by adding a condition:
I under the assumption of independence (can beweakened but not eliminated).
I an open problem in the general case (problematic case:x and y are very close to each other, but not completelyidentical)
. . . . . .
Comments
I 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
I it is not necessary (unlike for original Vyugin argument)to have the same targets for C(x|y) and C(y|x)
I other applications of the same type of argument: forevery x, y that are almost independent (I(x : y) is smallcompared to C(x) and C(y)) one can find z such thatC(x|z) = C(x)/2 + O(1) and C(y|z) = C(y)/2 + O(1)
I similar statement for halving complexity of three ormore strings by adding a condition:
I under the assumption of independence (can beweakened but not eliminated).
I an open problem in the general case (problematic case:x and y are very close to each other, but not completelyidentical)
. . . . . .
Comments
I 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
I it is not necessary (unlike for original Vyugin argument)to have the same targets for C(x|y) and C(y|x)
I other applications of the same type of argument: forevery x, y that are almost independent (I(x : y) is smallcompared to C(x) and C(y)) one can find z such thatC(x|z) = C(x)/2 + O(1) and C(y|z) = C(y)/2 + O(1)
I similar statement for halving complexity of three ormore strings by adding a condition:
I under the assumption of independence (can beweakened but not eliminated).
I an open problem in the general case (problematic case:x and y are very close to each other, but not completelyidentical)
. . . . . .
Comments
I 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
I it is not necessary (unlike for original Vyugin argument)to have the same targets for C(x|y) and C(y|x)
I other applications of the same type of argument: forevery x, y that are almost independent (I(x : y) is smallcompared to C(x) and C(y)) one can find z such thatC(x|z) = C(x)/2 + O(1) and C(y|z) = C(y)/2 + O(1)
I similar statement for halving complexity of three ormore strings by adding a condition:
I under the assumption of independence (can beweakened but not eliminated).
I an open problem in the general case (problematic case:x and y are very close to each other, but not completelyidentical)
. . . . . .
Comments
I 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
I it is not necessary (unlike for original Vyugin argument)to have the same targets for C(x|y) and C(y|x)
I other applications of the same type of argument: forevery x, y that are almost independent (I(x : y) is smallcompared to C(x) and C(y)) one can find z such thatC(x|z) = C(x)/2 + O(1) and C(y|z) = C(y)/2 + O(1)
I similar statement for halving complexity of three ormore strings by adding a condition:
I under the assumption of independence (can beweakened but not eliminated).
I an open problem in the general case (problematic case:x and y are very close to each other, but not completelyidentical)
. . . . . .
Comments
I 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
I it is not necessary (unlike for original Vyugin argument)to have the same targets for C(x|y) and C(y|x)
I other applications of the same type of argument: forevery x, y that are almost independent (I(x : y) is smallcompared to C(x) and C(y)) one can find z such thatC(x|z) = C(x)/2 + O(1) and C(y|z) = C(y)/2 + O(1)
I similar statement for halving complexity of three ormore strings by adding a condition:
I under the assumption of independence (can beweakened but not eliminated).
I an open problem in the general case (problematic case:x and y are very close to each other, but not completelyidentical)
. . . . . .
Comments
I 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
I it is not necessary (unlike for original Vyugin argument)to have the same targets for C(x|y) and C(y|x)
I other applications of the same type of argument: forevery x, y that are almost independent (I(x : y) is smallcompared to C(x) and C(y)) one can find z such thatC(x|z) = C(x)/2 + O(1) and C(y|z) = C(y)/2 + O(1)
I similar statement for halving complexity of three ormore strings by adding a condition:
I under the assumption of independence (can beweakened but not eliminated).
I an open problem in the general case (problematic case:x and y are very close to each other, but not completelyidentical)
. . . . . .
Thanks!
. . . . . .
Original game argument
I If C(x) > 3n, there exists y such that C(x|y) and C(y|x)are n+ O(1)
I we replaced 2n by 3n to simplify explanations (and inany case this is already covered)
I we present some gameI then show why winning this game is enoughI and finally show how to win the game
. . . . . .
Original game argument
I If C(x) > 3n, there exists y such that C(x|y) and C(y|x)are n+ O(1)
I we replaced 2n by 3n to simplify explanations (and inany case this is already covered)
I we present some gameI then show why winning this game is enoughI and finally show how to win the game
. . . . . .
Original game argument
I If C(x) > 3n, there exists y such that C(x|y) and C(y|x)are n+ O(1)
I we replaced 2n by 3n to simplify explanations (and inany case this is already covered)
I we present some gameI then show why winning this game is enoughI and finally show how to win the game
. . . . . .
Original game argument
I If C(x) > 3n, there exists y such that C(x|y) and C(y|x)are n+ O(1)
I we replaced 2n by 3n to simplify explanations (and inany case this is already covered)
I we present some game
I then show why winning this game is enoughI and finally show how to win the game
. . . . . .
Original game argument
I If C(x) > 3n, there exists y such that C(x|y) and C(y|x)are n+ O(1)
I we replaced 2n by 3n to simplify explanations (and inany case this is already covered)
I we present some gameI then show why winning this game is enough
I and finally show how to win the game
. . . . . .
Original game argument
I If C(x) > 3n, there exists y such that C(x|y) and C(y|x)are n+ O(1)
I we replaced 2n by 3n to simplify explanations (and inany case this is already covered)
I we present some gameI then show why winning this game is enoughI and finally show how to win the game
. . . . . .
Dating agency and its task
I two countable sets X and YI game starts with a perfect matching, i.e., one to onecorrespondence between X and Y.
I An element of X or Y can refuse the current partner,then the current relationship (x, y) is dissolved
I y then becomes free; the agency may eitherI find a new pair for x from the dissolved pair (among free
elements of Y not tried with x previously) orI declare x hopeless and do not try to find a pair for x anymore
(#free in Y incremented)I the refusals appear (and are processed by the agency)one at a time
I each element can produce < N refusals (parameter ofthe game), but no restrictions for #(being refused)
I agency obligations:I ≤ 2N attempts for each elementI ≤ 2N3 hopeless elements; all others in X are ultimately
connected to some y ∈ Y and this connection lasts forever
. . . . . .
Dating agency and its taskI two countable sets X and Y
I game starts with a perfect matching, i.e., one to onecorrespondence between X and Y.
I An element of X or Y can refuse the current partner,then the current relationship (x, y) is dissolved
I y then becomes free; the agency may eitherI find a new pair for x from the dissolved pair (among free
elements of Y not tried with x previously) orI declare x hopeless and do not try to find a pair for x anymore
(#free in Y incremented)I the refusals appear (and are processed by the agency)one at a time
I each element can produce < N refusals (parameter ofthe game), but no restrictions for #(being refused)
I agency obligations:I ≤ 2N attempts for each elementI ≤ 2N3 hopeless elements; all others in X are ultimately
connected to some y ∈ Y and this connection lasts forever
. . . . . .
Dating agency and its taskI two countable sets X and YI game starts with a perfect matching, i.e., one to onecorrespondence between X and Y.
I An element of X or Y can refuse the current partner,then the current relationship (x, y) is dissolved
I y then becomes free; the agency may eitherI find a new pair for x from the dissolved pair (among free
elements of Y not tried with x previously) orI declare x hopeless and do not try to find a pair for x anymore
(#free in Y incremented)I the refusals appear (and are processed by the agency)one at a time
I each element can produce < N refusals (parameter ofthe game), but no restrictions for #(being refused)
I agency obligations:I ≤ 2N attempts for each elementI ≤ 2N3 hopeless elements; all others in X are ultimately
connected to some y ∈ Y and this connection lasts forever
. . . . . .
Dating agency and its taskI two countable sets X and YI game starts with a perfect matching, i.e., one to onecorrespondence between X and Y.
I An element of X or Y can refuse the current partner,then the current relationship (x, y) is dissolved
I y then becomes free; the agency may eitherI find a new pair for x from the dissolved pair (among free
elements of Y not tried with x previously) orI declare x hopeless and do not try to find a pair for x anymore
(#free in Y incremented)I the refusals appear (and are processed by the agency)one at a time
I each element can produce < N refusals (parameter ofthe game), but no restrictions for #(being refused)
I agency obligations:I ≤ 2N attempts for each elementI ≤ 2N3 hopeless elements; all others in X are ultimately
connected to some y ∈ Y and this connection lasts forever
. . . . . .
Dating agency and its taskI two countable sets X and YI game starts with a perfect matching, i.e., one to onecorrespondence between X and Y.
I An element of X or Y can refuse the current partner,then the current relationship (x, y) is dissolved
I y then becomes free; the agency may either
I find a new pair for x from the dissolved pair (among freeelements of Y not tried with x previously) or
I declare x hopeless and do not try to find a pair for x anymore(#free in Y incremented)
I the refusals appear (and are processed by the agency)one at a time
I each element can produce < N refusals (parameter ofthe game), but no restrictions for #(being refused)
I agency obligations:I ≤ 2N attempts for each elementI ≤ 2N3 hopeless elements; all others in X are ultimately
connected to some y ∈ Y and this connection lasts forever
. . . . . .
Dating agency and its taskI two countable sets X and YI game starts with a perfect matching, i.e., one to onecorrespondence between X and Y.
I An element of X or Y can refuse the current partner,then the current relationship (x, y) is dissolved
I y then becomes free; the agency may eitherI find a new pair for x from the dissolved pair (among free
elements of Y not tried with x previously) or
I declare x hopeless and do not try to find a pair for x anymore(#free in Y incremented)
I the refusals appear (and are processed by the agency)one at a time
I each element can produce < N refusals (parameter ofthe game), but no restrictions for #(being refused)
I agency obligations:I ≤ 2N attempts for each elementI ≤ 2N3 hopeless elements; all others in X are ultimately
connected to some y ∈ Y and this connection lasts forever
. . . . . .
Dating agency and its taskI two countable sets X and YI game starts with a perfect matching, i.e., one to onecorrespondence between X and Y.
I An element of X or Y can refuse the current partner,then the current relationship (x, y) is dissolved
I y then becomes free; the agency may eitherI find a new pair for x from the dissolved pair (among free
elements of Y not tried with x previously) orI declare x hopeless and do not try to find a pair for x anymore
(#free in Y incremented)
I the refusals appear (and are processed by the agency)one at a time
I each element can produce < N refusals (parameter ofthe game), but no restrictions for #(being refused)
I agency obligations:I ≤ 2N attempts for each elementI ≤ 2N3 hopeless elements; all others in X are ultimately
connected to some y ∈ Y and this connection lasts forever
. . . . . .
Dating agency and its taskI two countable sets X and YI game starts with a perfect matching, i.e., one to onecorrespondence between X and Y.
I An element of X or Y can refuse the current partner,then the current relationship (x, y) is dissolved
I y then becomes free; the agency may eitherI find a new pair for x from the dissolved pair (among free
elements of Y not tried with x previously) orI declare x hopeless and do not try to find a pair for x anymore
(#free in Y incremented)I the refusals appear (and are processed by the agency)one at a time
I each element can produce < N refusals (parameter ofthe game), but no restrictions for #(being refused)
I agency obligations:I ≤ 2N attempts for each elementI ≤ 2N3 hopeless elements; all others in X are ultimately
connected to some y ∈ Y and this connection lasts forever
. . . . . .
Dating agency and its taskI two countable sets X and YI game starts with a perfect matching, i.e., one to onecorrespondence between X and Y.
I An element of X or Y can refuse the current partner,then the current relationship (x, y) is dissolved
I y then becomes free; the agency may eitherI find a new pair for x from the dissolved pair (among free
elements of Y not tried with x previously) orI declare x hopeless and do not try to find a pair for x anymore
(#free in Y incremented)I the refusals appear (and are processed by the agency)one at a time
I each element can produce < N refusals (parameter ofthe game), but no restrictions for #(being refused)
I agency obligations:I ≤ 2N attempts for each elementI ≤ 2N3 hopeless elements; all others in X are ultimately
connected to some y ∈ Y and this connection lasts forever
. . . . . .
Dating agency and its taskI two countable sets X and YI game starts with a perfect matching, i.e., one to onecorrespondence between X and Y.
I An element of X or Y can refuse the current partner,then the current relationship (x, y) is dissolved
I y then becomes free; the agency may eitherI find a new pair for x from the dissolved pair (among free
elements of Y not tried with x previously) orI declare x hopeless and do not try to find a pair for x anymore
(#free in Y incremented)I the refusals appear (and are processed by the agency)one at a time
I each element can produce < N refusals (parameter ofthe game), but no restrictions for #(being refused)
I agency obligations:
I ≤ 2N attempts for each elementI ≤ 2N3 hopeless elements; all others in X are ultimately
connected to some y ∈ Y and this connection lasts forever
. . . . . .
Dating agency and its taskI two countable sets X and YI game starts with a perfect matching, i.e., one to onecorrespondence between X and Y.
I An element of X or Y can refuse the current partner,then the current relationship (x, y) is dissolved
I y then becomes free; the agency may eitherI find a new pair for x from the dissolved pair (among free
elements of Y not tried with x previously) orI declare x hopeless and do not try to find a pair for x anymore
(#free in Y incremented)I the refusals appear (and are processed by the agency)one at a time
I each element can produce < N refusals (parameter ofthe game), but no restrictions for #(being refused)
I agency obligations:I ≤ 2N attempts for each element
I ≤ 2N3 hopeless elements; all others in X are ultimatelyconnected to some y ∈ Y and this connection lasts forever
. . . . . .
Dating agency and its taskI two countable sets X and YI game starts with a perfect matching, i.e., one to onecorrespondence between X and Y.
I An element of X or Y can refuse the current partner,then the current relationship (x, y) is dissolved
I y then becomes free; the agency may eitherI find a new pair for x from the dissolved pair (among free
elements of Y not tried with x previously) orI declare x hopeless and do not try to find a pair for x anymore
(#free in Y incremented)I the refusals appear (and are processed by the agency)one at a time
I each element can produce < N refusals (parameter ofthe game), but no restrictions for #(being refused)
I agency obligations:I ≤ 2N attempts for each elementI ≤ 2N3 hopeless elements; all others in X are ultimately
connected to some y ∈ Y and this connection lasts forever
. . . . . .
Why computable winning strategy is enough
I X = Y = B∗
I initial matching: identity (x, x)I u refuses v if C(v|u) < n (here u may be in X or in Y)I less than N = 2n refusals for each uI computable behaviorI agency produces O(N3) = O(23n) hopeless elements ofcomplexity 3n+O(1) (identified by 3n+O(1) bit ordinalnumber)
I for every x that is not hopeless its final partner y hasC(y|x) and C(x|y) at most n+ O(1): determined by aordinal number that is O(N) = 2n+O(1)
I but both complexities are at least n, otherwise refused
. . . . . .
Why computable winning strategy is enough
I X = Y = B∗
I initial matching: identity (x, x)I u refuses v if C(v|u) < n (here u may be in X or in Y)I less than N = 2n refusals for each uI computable behaviorI agency produces O(N3) = O(23n) hopeless elements ofcomplexity 3n+O(1) (identified by 3n+O(1) bit ordinalnumber)
I for every x that is not hopeless its final partner y hasC(y|x) and C(x|y) at most n+ O(1): determined by aordinal number that is O(N) = 2n+O(1)
I but both complexities are at least n, otherwise refused
. . . . . .
Why computable winning strategy is enough
I X = Y = B∗
I initial matching: identity (x, x)
I u refuses v if C(v|u) < n (here u may be in X or in Y)I less than N = 2n refusals for each uI computable behaviorI agency produces O(N3) = O(23n) hopeless elements ofcomplexity 3n+O(1) (identified by 3n+O(1) bit ordinalnumber)
I for every x that is not hopeless its final partner y hasC(y|x) and C(x|y) at most n+ O(1): determined by aordinal number that is O(N) = 2n+O(1)
I but both complexities are at least n, otherwise refused
. . . . . .
Why computable winning strategy is enough
I X = Y = B∗
I initial matching: identity (x, x)I u refuses v if C(v|u) < n (here u may be in X or in Y)
I less than N = 2n refusals for each uI computable behaviorI agency produces O(N3) = O(23n) hopeless elements ofcomplexity 3n+O(1) (identified by 3n+O(1) bit ordinalnumber)
I for every x that is not hopeless its final partner y hasC(y|x) and C(x|y) at most n+ O(1): determined by aordinal number that is O(N) = 2n+O(1)
I but both complexities are at least n, otherwise refused
. . . . . .
Why computable winning strategy is enough
I X = Y = B∗
I initial matching: identity (x, x)I u refuses v if C(v|u) < n (here u may be in X or in Y)I less than N = 2n refusals for each u
I computable behaviorI agency produces O(N3) = O(23n) hopeless elements ofcomplexity 3n+O(1) (identified by 3n+O(1) bit ordinalnumber)
I for every x that is not hopeless its final partner y hasC(y|x) and C(x|y) at most n+ O(1): determined by aordinal number that is O(N) = 2n+O(1)
I but both complexities are at least n, otherwise refused
. . . . . .
Why computable winning strategy is enough
I X = Y = B∗
I initial matching: identity (x, x)I u refuses v if C(v|u) < n (here u may be in X or in Y)I less than N = 2n refusals for each uI computable behavior
I agency produces O(N3) = O(23n) hopeless elements ofcomplexity 3n+O(1) (identified by 3n+O(1) bit ordinalnumber)
I for every x that is not hopeless its final partner y hasC(y|x) and C(x|y) at most n+ O(1): determined by aordinal number that is O(N) = 2n+O(1)
I but both complexities are at least n, otherwise refused
. . . . . .
Why computable winning strategy is enough
I X = Y = B∗
I initial matching: identity (x, x)I u refuses v if C(v|u) < n (here u may be in X or in Y)I less than N = 2n refusals for each uI computable behaviorI agency produces O(N3) = O(23n) hopeless elements ofcomplexity 3n+O(1) (identified by 3n+O(1) bit ordinalnumber)
I for every x that is not hopeless its final partner y hasC(y|x) and C(x|y) at most n+ O(1): determined by aordinal number that is O(N) = 2n+O(1)
I but both complexities are at least n, otherwise refused
. . . . . .
Why computable winning strategy is enough
I X = Y = B∗
I initial matching: identity (x, x)I u refuses v if C(v|u) < n (here u may be in X or in Y)I less than N = 2n refusals for each uI computable behaviorI agency produces O(N3) = O(23n) hopeless elements ofcomplexity 3n+O(1) (identified by 3n+O(1) bit ordinalnumber)
I for every x that is not hopeless its final partner y hasC(y|x) and C(x|y) at most n+ O(1): determined by aordinal number that is O(N) = 2n+O(1)
I but both complexities are at least n, otherwise refused
. . . . . .
Why computable winning strategy is enough
I X = Y = B∗
I initial matching: identity (x, x)I u refuses v if C(v|u) < n (here u may be in X or in Y)I less than N = 2n refusals for each uI computable behaviorI agency produces O(N3) = O(23n) hopeless elements ofcomplexity 3n+O(1) (identified by 3n+O(1) bit ordinalnumber)
I for every x that is not hopeless its final partner y hasC(y|x) and C(x|y) at most n+ O(1): determined by aordinal number that is O(N) = 2n+O(1)
I but both complexities are at least n, otherwise refused
. . . . . .
How to win the game
I each element not currently matched keeps “experience”=(#refusalssent, #refusals received)
I the first is < N; the second a priori is unbounded, but also will bekept < N due to agency strategy
I when (x, y) is terminated, numbers updatedI invariant: in all pairs people have matching experiences (#sent =
#received for the other)I corollary: #refusals received < NI new partner for x is found if possible (=there is y ∈ Y with
matching experience not tried earlier with x)I otherwise x is declared hopelessI invariant: for matching experiences the number of non-matched
people in X and Y are the sameI ≤ 2N attempts for each (experience increases each time)I there are N2 experience classes; if class reaches 2N, it stops
growing since y can be always found in the class (< 2N are triedearlier with given x), so O(N3) hopeless
. . . . . .
How to win the gameI each element not currently matched keeps “experience”=(#refusals
sent, #refusals received)
I the first is < N; the second a priori is unbounded, but also will bekept < N due to agency strategy
I when (x, y) is terminated, numbers updatedI invariant: in all pairs people have matching experiences (#sent =
#received for the other)I corollary: #refusals received < NI new partner for x is found if possible (=there is y ∈ Y with
matching experience not tried earlier with x)I otherwise x is declared hopelessI invariant: for matching experiences the number of non-matched
people in X and Y are the sameI ≤ 2N attempts for each (experience increases each time)I there are N2 experience classes; if class reaches 2N, it stops
growing since y can be always found in the class (< 2N are triedearlier with given x), so O(N3) hopeless
. . . . . .
How to win the gameI each element not currently matched keeps “experience”=(#refusals
sent, #refusals received)I the first is < N; the second a priori is unbounded, but also will be
kept < N due to agency strategy
I when (x, y) is terminated, numbers updatedI invariant: in all pairs people have matching experiences (#sent =
#received for the other)I corollary: #refusals received < NI new partner for x is found if possible (=there is y ∈ Y with
matching experience not tried earlier with x)I otherwise x is declared hopelessI invariant: for matching experiences the number of non-matched
people in X and Y are the sameI ≤ 2N attempts for each (experience increases each time)I there are N2 experience classes; if class reaches 2N, it stops
growing since y can be always found in the class (< 2N are triedearlier with given x), so O(N3) hopeless
. . . . . .
How to win the gameI each element not currently matched keeps “experience”=(#refusals
sent, #refusals received)I the first is < N; the second a priori is unbounded, but also will be
kept < N due to agency strategyI when (x, y) is terminated, numbers updated
I invariant: in all pairs people have matching experiences (#sent =#received for the other)
I corollary: #refusals received < NI new partner for x is found if possible (=there is y ∈ Y with
matching experience not tried earlier with x)I otherwise x is declared hopelessI invariant: for matching experiences the number of non-matched
people in X and Y are the sameI ≤ 2N attempts for each (experience increases each time)I there are N2 experience classes; if class reaches 2N, it stops
growing since y can be always found in the class (< 2N are triedearlier with given x), so O(N3) hopeless
. . . . . .
How to win the gameI each element not currently matched keeps “experience”=(#refusals
sent, #refusals received)I the first is < N; the second a priori is unbounded, but also will be
kept < N due to agency strategyI when (x, y) is terminated, numbers updatedI invariant: in all pairs people have matching experiences (#sent =
#received for the other)
I corollary: #refusals received < NI new partner for x is found if possible (=there is y ∈ Y with
matching experience not tried earlier with x)I otherwise x is declared hopelessI invariant: for matching experiences the number of non-matched
people in X and Y are the sameI ≤ 2N attempts for each (experience increases each time)I there are N2 experience classes; if class reaches 2N, it stops
growing since y can be always found in the class (< 2N are triedearlier with given x), so O(N3) hopeless
. . . . . .
How to win the gameI each element not currently matched keeps “experience”=(#refusals
sent, #refusals received)I the first is < N; the second a priori is unbounded, but also will be
kept < N due to agency strategyI when (x, y) is terminated, numbers updatedI invariant: in all pairs people have matching experiences (#sent =
#received for the other)I corollary: #refusals received < N
I new partner for x is found if possible (=there is y ∈ Y withmatching experience not tried earlier with x)
I otherwise x is declared hopelessI invariant: for matching experiences the number of non-matched
people in X and Y are the sameI ≤ 2N attempts for each (experience increases each time)I there are N2 experience classes; if class reaches 2N, it stops
growing since y can be always found in the class (< 2N are triedearlier with given x), so O(N3) hopeless
. . . . . .
How to win the gameI each element not currently matched keeps “experience”=(#refusals
sent, #refusals received)I the first is < N; the second a priori is unbounded, but also will be
kept < N due to agency strategyI when (x, y) is terminated, numbers updatedI invariant: in all pairs people have matching experiences (#sent =
#received for the other)I corollary: #refusals received < NI new partner for x is found if possible (=there is y ∈ Y with
matching experience not tried earlier with x)
I otherwise x is declared hopelessI invariant: for matching experiences the number of non-matched
people in X and Y are the sameI ≤ 2N attempts for each (experience increases each time)I there are N2 experience classes; if class reaches 2N, it stops
growing since y can be always found in the class (< 2N are triedearlier with given x), so O(N3) hopeless
. . . . . .
How to win the gameI each element not currently matched keeps “experience”=(#refusals
sent, #refusals received)I the first is < N; the second a priori is unbounded, but also will be
kept < N due to agency strategyI when (x, y) is terminated, numbers updatedI invariant: in all pairs people have matching experiences (#sent =
#received for the other)I corollary: #refusals received < NI new partner for x is found if possible (=there is y ∈ Y with
matching experience not tried earlier with x)I otherwise x is declared hopeless
I invariant: for matching experiences the number of non-matchedpeople in X and Y are the same
I ≤ 2N attempts for each (experience increases each time)I there are N2 experience classes; if class reaches 2N, it stops
growing since y can be always found in the class (< 2N are triedearlier with given x), so O(N3) hopeless
. . . . . .
How to win the gameI each element not currently matched keeps “experience”=(#refusals
sent, #refusals received)I the first is < N; the second a priori is unbounded, but also will be
kept < N due to agency strategyI when (x, y) is terminated, numbers updatedI invariant: in all pairs people have matching experiences (#sent =
#received for the other)I corollary: #refusals received < NI new partner for x is found if possible (=there is y ∈ Y with
matching experience not tried earlier with x)I otherwise x is declared hopelessI invariant: for matching experiences the number of non-matched
people in X and Y are the same
I ≤ 2N attempts for each (experience increases each time)I there are N2 experience classes; if class reaches 2N, it stops
growing since y can be always found in the class (< 2N are triedearlier with given x), so O(N3) hopeless
. . . . . .
How to win the gameI each element not currently matched keeps “experience”=(#refusals
sent, #refusals received)I the first is < N; the second a priori is unbounded, but also will be
kept < N due to agency strategyI when (x, y) is terminated, numbers updatedI invariant: in all pairs people have matching experiences (#sent =
#received for the other)I corollary: #refusals received < NI new partner for x is found if possible (=there is y ∈ Y with
matching experience not tried earlier with x)I otherwise x is declared hopelessI invariant: for matching experiences the number of non-matched
people in X and Y are the sameI ≤ 2N attempts for each (experience increases each time)
I there are N2 experience classes; if class reaches 2N, it stopsgrowing since y can be always found in the class (< 2N are triedearlier with given x), so O(N3) hopeless
. . . . . .
How to win the gameI each element not currently matched keeps “experience”=(#refusals
sent, #refusals received)I the first is < N; the second a priori is unbounded, but also will be
kept < N due to agency strategyI when (x, y) is terminated, numbers updatedI invariant: in all pairs people have matching experiences (#sent =
#received for the other)I corollary: #refusals received < NI new partner for x is found if possible (=there is y ∈ Y with
matching experience not tried earlier with x)I otherwise x is declared hopelessI invariant: for matching experiences the number of non-matched
people in X and Y are the sameI ≤ 2N attempts for each (experience increases each time)I there are N2 experience classes; if class reaches 2N, it stops
growing since y can be always found in the class (< 2N are triedearlier with given x), so O(N3) hopeless
. . . . . .
Thanks
I to the organizers who accepted to consider thesearguments
I to Misha Vyugin and Andrej Muchnik who invented thegame argument and its generalization for severalstrings yi
I to Laurent Bienvenu who convinced us to write thissimple argument down
I to all colleagues (ESCAPE team in Marseille andMontpellier, participants of Kolmogorov seminar inMoscow)
I to the audience for following the talk to that point :-)
. . . . . .
Thanks
I to the organizers who accepted to consider thesearguments
I to Misha Vyugin and Andrej Muchnik who invented thegame argument and its generalization for severalstrings yi
I to Laurent Bienvenu who convinced us to write thissimple argument down
I to all colleagues (ESCAPE team in Marseille andMontpellier, participants of Kolmogorov seminar inMoscow)
I to the audience for following the talk to that point :-)
. . . . . .
Thanks
I to the organizers who accepted to consider thesearguments
I to Misha Vyugin and Andrej Muchnik who invented thegame argument and its generalization for severalstrings yi
I to Laurent Bienvenu who convinced us to write thissimple argument down
I to all colleagues (ESCAPE team in Marseille andMontpellier, participants of Kolmogorov seminar inMoscow)
I to the audience for following the talk to that point :-)
. . . . . .
Thanks
I to the organizers who accepted to consider thesearguments
I to Misha Vyugin and Andrej Muchnik who invented thegame argument and its generalization for severalstrings yi
I to Laurent Bienvenu who convinced us to write thissimple argument down
I to all colleagues (ESCAPE team in Marseille andMontpellier, participants of Kolmogorov seminar inMoscow)
I to the audience for following the talk to that point :-)
. . . . . .
Thanks
I to the organizers who accepted to consider thesearguments
I to Misha Vyugin and Andrej Muchnik who invented thegame argument and its generalization for severalstrings yi
I to Laurent Bienvenu who convinced us to write thissimple argument down
I to all colleagues (ESCAPE team in Marseille andMontpellier, participants of Kolmogorov seminar inMoscow)
I to the audience for following the talk to that point :-)
. . . . . .
Thanks
I to the organizers who accepted to consider thesearguments
I to Misha Vyugin and Andrej Muchnik who invented thegame argument and its generalization for severalstrings yi
I to Laurent Bienvenu who convinced us to write thissimple argument down
I to all colleagues (ESCAPE team in Marseille andMontpellier, participants of Kolmogorov seminar inMoscow)
I to the audience for following the talk to that point :-)