SLIDE 1
randomized computation Sometimes randomness helps in computation. - - PowerPoint PPT Presentation
randomized computation Sometimes randomness helps in computation. - - PowerPoint PPT Presentation
randomized computation Sometimes randomness helps in computation. randomized computation Augment our usual Turing machines with a read-only tape of random bits . start 0 2 accept 1 # # 0 1 1 0
SLIDE 2
SLIDE 3
randomized computation
The notion of acceptance changes. One-sided error [two variants]: Two-sided error: ๐ฆ โ ๐ โ Pr ๐ accepts ๐ฆ = 1 ๐ฆ โ ๐ โ Pr ๐ accepts ๐ฆ <
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What does it mean for a TM ๐ with access to random bits to decide a language ๐? ๐ฆ โ ๐ โ Pr ๐ accepts ๐ฆ >
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๐ฆ โ ๐ โ Pr ๐ accepts ๐ฆ <
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๐ฆ โ ๐ โ Pr ๐ accepts ๐ฆ >
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๐ฆ โ ๐ โ Pr ๐ accepts ๐ฆ = 0 In both cases, we can make the error much smaller by repeating a few times.
SLIDE 4
randomized computation
For instance, the two-sided error variant of ๐ is called ๐๐๐. ๐๐๐ stands for โbounded-error probabilistic polynomial timeโ Derandomization problem: Can every efficient randomized algorithm be replaced by a deterministic one? We donโt know the answer, though it is expected to be โyes.โ This is the question of ๐ vs. ๐๐๐.
SLIDE 5
randomized computation
Recall the complexity class ๐ = SPACE log ๐ The randomized one-sided error variant of ๐ is called ๐๐. ๐ฆ โ ๐ต โ Pr ๐ accepts ๐ฆ โฅ
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๐ฆ โ ๐ต โ Pr ๐ accepts ๐ฆ = 0 A language ๐ต is in ๐๐ if there is a randomized ๐ log ๐ space Turing machine such that [one way access to random tape]: The question of whether ๐ = ๐๐ is also open. We know that ๐๐ โ ๐๐ โ SPACE log ๐ 2
Nisan showed how to do this using a pseudo-random number generator for space-bounded machines.
SLIDE 6
undirected connectivity
Recall the language: DIRPATH = ๐ป, ๐ก, ๐ข โถ ๐ป is a directed graph with a directed ๐ก โ ๐ข path We saw that DIRPATH is ๐๐-complete. Thus ๐ = ๐๐ if and only if DIRPATH โ ๐ด. What about the language? PATH = ๐ป, ๐ก, ๐ข โถ ๐ป is an ๐ฏ๐จ๐๐ฃ๐ฌ๐๐๐ฎ๐๐ graph with an ๐ก โ ๐ข path
SLIDE 7
undirected connectivity
What about the language? PATH = ๐ป, ๐ก, ๐ข โถ ๐ป is an ๐ฏ๐จ๐๐ฃ๐ฌ๐๐๐ฎ๐๐ graph with an ๐ก โ ๐ข path Is PATH โ ๐?
SLIDE 8
random walks
Letโs show that PATH โ ๐๐. One step of random walk: Move to a uniformly random neighbor.
SLIDE 9
random walks
What we will show: If ๐ป = (๐, ๐น) is a graph with ๐ vertices and ๐ edges, then a random walker started at ๐ก โ ๐ who takes 4๐๐ steps will visit every vertex in the connected component of ๐ก with probability at least ยฝ. In particular, if there is an ๐ก โ ๐ข path in ๐ป, she will visit ๐ข with probability at least ยฝ.
SLIDE 10
random walks
What we will show: If ๐ป = (๐, ๐น) is a graph with ๐ vertices and ๐ edges, then a random walker started at ๐ก โ ๐ who takes 4๐๐ steps will visit every vertex in the connected component of ๐ก with probability at least ยฝ. In particular, if there is an ๐ก โ ๐ข path in ๐ป, she will visit ๐ข with probability at least ยฝ. ๐๐ Algorithm: Start a random walker at ๐ก and use the random bits to simulate a random walk for 4๐๐ steps (can count this high in ๐ log ๐ space). If we visit ๐ข at any point, accept. Otherwise, reject. If there is an ๐ก โ ๐ข path in ๐ป: Pr accept ๐ป, ๐ก, ๐ข โฅ
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If no ๐ก โ ๐ข path in ๐ป: Pr accept ๐ป, ๐ก, ๐ข = 0
SLIDE 11
cover time of a graph
Cover time of a graph = expected # of steps before all vertices are visited.
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cover time of a graph
We will show that if ๐ป has ๐ vertices and ๐ then the expected time before the random walk covers ๐ป (from any starting point) is 2๐๐. Then it must be that with probability at least ยฝ, we cover ๐ป after 4๐๐ steps. (This fact is called โMarkovโs inequality.โ)
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heat dispersion on a graph
SLIDE 14
spectral embedding
๐ค2
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hitting times and commute times
For vertices ๐ฃ, ๐ค โ ๐, the hitting time is ๐ผ ๐ฃ, ๐ค = expected time for random walk to hit ๐ค starting at ๐ฃ ๐ฃ ๐ค The commute time between ๐ฃ and ๐ค is ๐ท ๐ฃ, ๐ค = ๐ผ ๐ฃ, ๐ค + ๐ผ(๐ค, ๐ฃ)
SLIDE 16
hitting times and commute times
For vertices ๐ฃ, ๐ค โ ๐, the hitting time is ๐ผ ๐ฃ, ๐ค = expected time for random walk to hit ๐ค starting at ๐ฃ ๐ฃ ๐ค The commute time between ๐ฃ and ๐ค is ๐ท ๐ฃ, ๐ค = ๐ผ ๐ฃ, ๐ค + ๐ผ(๐ค, ๐ฃ) Claim: If {๐ฃ, ๐ค} is an edge of ๐ป, then ๐ท ๐ฃ, ๐ค โค 2๐.
SLIDE 17
hitting times and commute times
๐ก Claim: If {๐ฃ, ๐ค} is an edge of ๐ป, then ๐ท ๐ฃ, ๐ค โค 2๐. Letโs use the claim to show that the cover time of ๐ป is at most 2๐(๐ โ 1).
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hitting times and commute times
๐ก Claim: If {๐ฃ, ๐ค} is an edge of ๐ป, then ๐ท ๐ฃ, ๐ค โค 2๐. Letโs use the claim to show that the cover time of ๐ป is at most 2๐(๐ โ 1). Choose an arbitrary spanning tree of ๐ป.
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hitting times and commute times
๐ก Claim: If {๐ฃ, ๐ค} is an edge of ๐ป, then ๐ท ๐ฃ, ๐ค โค 2๐. Letโs use the claim to show that the cover time of ๐ป is at most 2๐(๐ โ 1). Choose an arbitrary spanning tree of ๐ป. Consider the โtour around the spanning treeโ
This is a path ๐ก = ๐ฃ0, ๐ฃ1, ๐ฃ2, โฆ , ๐ฃ2๐ = ๐ก Where ๐ฃ๐, ๐ฃ๐+1 is an edge for each ๐, and each edge appears exactly twice.
SLIDE 20
hitting times and commute times
๐ก Claim: If {๐ฃ, ๐ค} is an edge of ๐ป, then ๐ท ๐ฃ, ๐ค โค 2๐. Letโs use the claim to show that the cover time of ๐ป is at most 2๐(๐ โ 1). Choose an arbitrary spanning tree of ๐ป. Consider the โtour around the spanning treeโ
Claim: Cover time of ๐ป is at most ๐ท ๐ฃ0, ๐ฃ1 + ๐ท ๐ฃ1, ๐ฃ2 + ๐ท ๐ฃ2, ๐ฃ3 + โฏ + ๐ท ๐ฃ2๐โ1, ๐ฃ2๐
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hitting times and commute times
๐ก Claim: If {๐ฃ, ๐ค} is an edge of ๐ป, then ๐ท ๐ฃ, ๐ค โค 2๐. Letโs use the claim to show that the cover time of ๐ป is at most 2๐(๐ โ 1). Choose an arbitrary spanning tree of ๐ป. Consider the โtour around the spanning treeโ
Claim: Cover time of ๐ป is at most
๐ฃ,๐ค โ๐
๐ท(๐ฃ, ๐ค) โค 2๐(๐ โ 1)
SLIDE 22
high school physics
Claim: If {๐ฃ, ๐ค} is an edge of ๐ป, then ๐ท ๐ฃ, ๐ค โค 2๐.
SLIDE 23
high school physics
Claim: If {๐ฃ, ๐ค} is an edge of ๐ป, then ๐ท ๐ฃ, ๐ค โค 2๐. View the graph ๐ป as an electrical network with unit resistances on the edges.
When a potential difference is applied between the nodes from an external source, current flows in the network in accordance with Kirchoffโs and Ohmโs laws. K1: The total current flowing into a vertex = total current flowing out K2: The sum of potential differences around any cycle is zero. Ohm: The current flowing along any edge {๐ฃ, ๐ค} is equal to (potential ๐ฃ โ potential ๐ค )/resistance(๐ฃ, ๐ค) The effective resistance between ๐ฃ and ๐ค is defined as the potential difference required to send one unit of current from ๐ฃ to ๐ค.
๐ฃ ๐ค
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