Class 18: Hardness of Auctions

Class 18: Hardness of Auctions
https://uvammm.github.io/class18

Markets, Mechanisms, and Machines
University of Virginia
cs4501/econ4559 Spring 2019
David Evans and Denis Nekipelov
https://uvammm.github.io/

Transcript

1. MARKETS, MECHANISMS, MACHINES University of Virginia, Spring 2019
   Class 18:
   Hardness of
   Auctions
   21 March 2019
   cs4501/econ4559 Spring 2019
   David Evans and Denis Nekipelov
   https://uvammm.github.io
   Trial Auction starts Now!
   

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2. Trial Auction Starts Now!
   1
   

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3. Recap: Multi-Unit Auction
   2
   Input: Valuations for the players: !"
   , … , !%
   .
   Output: Allocation, '"
   , … , '%
   where ∑
   )
   ')
   ≤ ' that maximizes ∑
   )
   !)
   ')
   .
   Query model:
   !)
   0 = 0, ∀. ∈ 0, … , ' − 1 : !)
   . ≤ !)
   (. + 1)
   Single-Minded Step Bids:
   !)
   . = 0 for . < .)
   ∗; !)
   . = <)
   ∗ for . ≥ .)
   ∗.
   Downward Sloping:
   ∀. ∈ 0, … , ' − 1 : !)
   . + 1 − !)
   . ≤ !)
   . − !)
   (. − 1)
   No polynomial (in log ') time solution
   Know polynomial (in log ') time solution
   ???
   

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4. Second most Interesting Complexity Class?
   3
   NP
   Non-deterministic Polynomial Time
   

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5. q
   1
   1 1 1
   0
   1 0 …
   q
   0
   1 1 1
   0
   0 0 …
   q
   6
   0 1 1
   0
   1 0 …
   Deterministic Turing Machine Nondeterministic Turing Machine
   q
   0
   1 1 1
   0
   0 0 …
   Tries all possible transitions; accepts if
   any path leads to accepting state.
   q
   1
   1 1 1
   0
   1 0 …
   q
   7
   1 1 1
   0
   0 0 …
   

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6. Computability:
   Is NDTM more powerful than DTM?
   

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7. Computability:
   Is NDTM more powerful than DTM?
   No! We can simulate a NDTM with a DTM.
   Use a tape to keep track of which paths to try
   (breadth-first search, not depth first!)
   

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8. Complexity:
   Are there problems that are “hard” to solve with a
   DTM that are “easy” to solve with a NDTM?
   

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9. Complexity Classes
   NP
   P
   Problems that a DTM can
   solve in polynomial time.
   Problems that a NDTM can
   solve in polynomial time.
   

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10. P
    NP P
    NP
    We know P ⊆ NP: no need to take advantage of non-determinism.
    

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12. P
    NP P
    NP
    Option 1: There are problems in
    Class NP that are not tractable
    Option 2: All problems in
    Class NP are tractable
    P = NP?
    P ⊊ NP P = NP
    

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13. Assuming P ⊊ NP, how can we prove a
    problem is not in P?
    12
    

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14. Class NP-Complete
    13
    

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15. NP-Complete
    A problem B is in NP-Complete iff:
    2. There is a polynomial-time reduction
    from every problem A Î NP to B.
    1. B Î NP
    B
    NP
    B
    NP
    What does NP-Complete look like?
    

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16. NP-Hard
    A problem B is in NP-Hard iff:
    There is a polynomial-time reduction
    from every problem A Î NP to B.
    B
    NP
    What does NP-Hard look like?
    

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17. NP-Hard (if P Ì NP)
    P
    NP-C
    Option 1a: P Ì NP,
    NP-C È P Ì NP
    Option 1b: P Ì NP,
    NP-C È P = NP
    NP-Hard
    P
    NP-C
    

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18. NP-Hard (if P = NP)
    P
    NP
    Option 2: P = NP
    NP-C
    ≈ NP-Complete
    

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19. What’s a Reduction?
    18
    There is a polynomial-time reduction
    from every problem A Î NP to B.
    B
    NP
    

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20. Valid Reductions
    Transform ! into "
    Transformation must not
    take too long: finish in
    polynomial time
    Transformation must not
    expand input size too
    much: polynomial in
    original input size There is a polynomial-time reduction
    from every problem A Î NP to B.
    B
    NP
    Invalid transformation: “do an exponential search to find the answer, and output that”
    

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21. Reduction Proofs
    Goal: Prove A does not have some property Y.
    We already know B does not have property Y.
    Proof by Contradiction:
    Assume ! has some property ".
    Since ! has ", there is a solution $ to ! that satisfies ".
    Show how $ could be used to solve %.
    

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22. Reduction Proofs
    Goal: Prove A does not have some property Y.
    We already know B does not have property Y.
    Proof by Contradiction:
    Assume ! has some property ".
    Since ! has ", there is a solution $ to ! that satisfies ".
    Show how $ could be used to solve %.
    Since we know B does not satisfy Y, but having S would imply B satisfies Y,
    S cannot exist. Thus, S cannot exist, and A does not have Y.
    

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23. NP-Hardness Reduction Proofs
    Goal: Prove A does not have some property Y.
    We already know B does not have property Y.
    Proof by Contradiction:
    Assume ! has some property ".
    Since ! has ", there is a solution $ to ! that satisfies ".
    Show how $ could be used to solve %.
    NP-Hardness:
    " = “is NP-Hard”, % = a known NP-Hard problem,
    $ = “a TM that solves ! in polynomial-time”
    

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24. Proving NP-Hardness
    Show there is a polynomial-time
    reduction from every problem A
    Î NP to B.
    B
    NP
    Show there is a polynomial-
    time reduction from one
    problem X Î NP-Hard to B.
    X
    NP
    B
    This assumes we already know some problem X that is in NP-Hard.
    To get the first one, we need to prove it the hard way!
    

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26. KNAPSACK Problem
    Input: !"
    , $"
    , !%
    , $%
    , … , !'
    , $'
    !(
    , $(
    ∈ ℝ+, maximum weight ,
    Output: - ⊆ {1, … , 1} that maximizes ∑
    ( ∈4
    $(
    subject to ∑
    ( ∈4
    !(
    ≤ ,
    25
    Known (but won’t prove):
    KNAPSACK problem is in NP-Hard.
    

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27. Multi-Unit Auction: Input Representation
    26
    Input: Valuations for the players: !"
    , … , !%
    .
    Output: Allocation, '"
    , … , '%
    where ∑
    )
    ')
    ≤ ' that maximizes ∑
    )
    !)
    ')
    .
    Query model:
    !)
    0 = 0, ∀. ∈ 0, … , ' − 1 : !)
    . ≤ !)
    (. + 1)
    Single-Minded Step Bids:
    !)
    . = 0 for . < .)
    ∗; !)
    . = <)
    ∗ for . ≥ .)
    ∗.
    Downward Sloping:
    ∀. ∈ 0, … , ' − 1 : !)
    . + 1 − !)
    . ≤ !)
    . − !)
    (. − 1)
    No polynomial (in log ') time solution
    Know polynomial (in log ') time solution
    ???
    

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28. Proof: Multi-Unit Auction with
    Single-Minded Bids is NP-Hard
    27
    Goal: Prove A does not have some property Y.
    We already know B does not have property Y.
    Proof by Contradiction:
    Assume ! has some property ".
    Since ! has ", there is a solution $ to ! that satisfies ".
    Show how $ could be used to solve %.
    NP-Hardness:
    " = “is NP-Hard”, % = a known NP-Hard problem,
    $ = “a TM that solves ! in polynomial-time”
    

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29. Proof: Multi-Unit Auction with
    Single-Minded Bids is NP-Hard
    28
    Goal: Prove Multi-Unit Auction with Single-Minded Bids is NP-Hard.
    We already know KNAPSACK is NP-Hard.
    Proof by Contradiction:
    Assume !"# is not NP-Hard.
    Since !"# is not NP-Hard, there is a polynomial-time solution %
    to !"#.
    Show how % could be used to solve &'#(%#)&.
    NP-Hardness:
    * = “is NP-Hard”, + = KNAPSACK,
    % = “a TM that solves !"# in polynomial-time”
    

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30. Reduction
    29
    MUA-SMB
    Input: Valuations for the players: !"
    , … , !%
    .
    !'
    ( = 0 for ( < ('
    ∗; !'
    ( = 1'
    ∗ for ( ≥ ('
    ∗.
    Output: Allocation, 3"
    , … , 3%
    where
    ∑
    '
    3'
    ≤ 3 that maximizes ∑
    '
    !'
    3'
    .
    KNAPSACK
    Input: 1"
    , !"
    , 16
    , !6
    , … , 1%
    , !%
    1'
    , !'
    ∈ ℝ9, maximum weight :
    Output: ; ⊆ {1, … , ?} that maximizes
    ∑
    ' ∈A
    !'
    subject to ∑
    ' ∈A
    1'
    ≤ :
    

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31. Multi-Unit Auction: Input Representation
    30
    Input: Valuations for the players: !"
    , … , !%
    .
    Output: Allocation, '"
    , … , '%
    where ∑
    )
    ')
    ≤ ' that maximizes ∑
    )
    !)
    ')
    .
    Query model:
    !)
    0 = 0, ∀. ∈ 0, … , ' − 1 : !)
    . ≤ !)
    (. + 1)
    Single-Minded Step Bids:
    !)
    . = 0 for . < .)
    ∗; !)
    . = <)
    ∗ for . ≥ .)
    ∗.
    Downward Sloping:
    ∀. ∈ 0, … , ' − 1 : !)
    . + 1 − !)
    . ≤ !)
    . − !)
    (. − 1)
    No polynomial (in log ') time solution
    Know polynomial (in log ') time solution
    NP-Hard: no polynomial time solution unless P=NP
    

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32. 31
    Trial Auction Results
    

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33. “If P = NP, then the world would be a
    profoundly different place than we usually
    assume it to be. There would be no special
    value in ‘creative leaps’, no fundamental gap
    between solving a problem and recognizing the
    solution once it’s found. Everyone who could
    appreciate a symphony would be Mozart;
    everyone who could follow a step-by-step
    argument would be Gauss...”
    Scott Aaronson
    Charge
    Project 4:
    Final auction
    next Thursday
    Project 4 reports
    Due next Friday
    

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