On the Complexity of Allocation Problems with Probabilistic Players
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On the Complexity of Allocation Problems with Probabilistic Players
Rishab Nithyanand
Research Proficiency ExaminationSummer 2012
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Presentation Outline
• Introduction
• The Password Allocation Problem
• The Weapon-Target Allocation Problem
• Conclusions and Future Work
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Traditional Allocation Problems
• Given:– resources (r1, r2, …, rn) and tasks (t1, t2, …, tk)– objective function F
• Goal:– Find allocation for which F is optimal
• Constraint:– at most one task per resource
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Allocation Problems with Probabilistic Players
• Given:– resources (r1, r2, …, rn) and tasks (t1, t2, …, tk)
– resource ri completes task tj with probability pij
– objective function F
• Goal:– Find allocation for which E[F] is optimal
• Constraint:– at most one task per resource
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• Users have a large set of accounts– some are very valuable
– and some are less valuable
• Passwords are hard to remember– [Vu, 2006]: Average users remember upto 6 unique passwords.
• [Perito, 2011]: Internet accounts are easily linkable by pseudonyms.– Compromise of one account ) compromise of all accounts allocated the same password.– Some accounts (eg., email) are gateway accounts.
• Problem: – What allocation results in minimum expected loss?
The Password Allocation Problem
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• I don’t care. I’m super secure, phish-proof, and use 40 char long passwords!– People do stupid things!
• July 12, 2012: Yahoo lost 45000 unhashed passwords.– All passwords are equal.
• Compromise probability is only server dependent.
• June 5, 2012: 6.5 million hashed passwords stolen.– Some passwords are uncrackable.
• Compromise probability is server and password dependent.
The Password Allocation Problem
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PA as a Parallel Job Allocation Problem
• Given a set of programs to be executed and a (smaller) set of machines.
• Each program may cause a system failure with some probability.– This may be machine independent (i.e., all machines are the same).
• Parallel Processing Constraint: Failure of one of the programs ) failure of all programs on the system.
• Problem:– How should programs be allocated to machines to maximize expected
throughput?On the Complexity of Allocation Problems with Probabilistic Players
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The Weapon-Target Allocation Problem
• Military offense allocation problem.
• Given a set of weapons and a set of enemy targets.
• Not all weapons destroy their targets– Enemy interception– Mechanical failures
• Probability of failure depends on the weapon-target pair– Placement of defenses against weapons– Distance from allocated weapon
• Problem: – What allocation maximizes expected damage to the enemy targets?
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The Weapon-Target Allocation Problem
Research Timeline:
• Formulation: Allan Manne (Stanford) [1958]
• NP-Completeness: Lloyd and Witsenhausen (Bell Labs) [1988]
• Analysis, Variants: Hosein (MIT) [1987-1992], Athans (Bell Labs) [1989-1992]
• Approximation (heuristics): 1977 – today
• Best approximations: Ahuja (UF), Orlin (MIT) [2007]
• (Existence of) Constant-factor approximations: ??
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Presentation Outline
• Introduction
• The Password Allocation Problem
• The Weapon-Target Allocation Problem
• Conclusions and Future Work
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The PA Problem: Definition
• Problem Instance:– n accounts: a1, a2, …, an
– k passwords: PW1, PW2, …, PWk
– ai has value vi and compromise probability qi = (1-pi)• i.e., compromise probability is independent of password strength
• Compromise of one account 2 PWj ) compromise of all accounts 2 PWj
• Constraint: Every account receives exactly one password.
• Goal: Minimize expected loss through password compromise– Equivalent to maximizing expected survival value (or, Expected Gain (EG)).
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• Allocation matrix: X = {xij}– xij = 1 ) account aj is allocated password PWi
– xij = 0, otherwise
• Objective Function (Expected Gain): [to be maximized]
• Constraint: – Every account is allocated exactly one password.
The PA Problem: Mathematical Formulation
kX
i=1
0@nY
j =1((pj )xi j )
nX
j =1(xi j vj )
1A
kX
i=1
0@nY
j =1((pj )xi j )
nX
j =1(xi j vj )
1A (1)xi j =
½ 1 if account j is allocated to password i0 otherwise
maximize : F =kX
i=1
0@nY
j =1((pj )x i j )
nX
j =1(xi j vj )
1A
subj ect to :kX
i=1xi j =1; 8j 2 f1;:::;ng
F is theexpected gain (EG).
F = P ki=1³ Q n
j =1((pj )xi j )P nj =1(xi j vj )
´F = P ki=1
³ Q nj =1((pj )xi j )
P nj =1(xi j vj )
´
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Complexity of PA
Theorem: PA with 2 passwords PA2 2 NP Complete.Proof: Part I: Formulating the Decision Version (PA2)
• Instance: – P = {p1,…,pn} where pi 2 (0,1)
– V = {v1,…,vn}– r
• Is there a partition of N ={1,…,n} into S1 and S2 such that:
• Clearly PA2 2 NP.
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Complexity of PA
Theorem: PA with 2 passwords PA2 2 NP Complete.Proof: Part II: Finding the known hard problem
• The Partition Problem:– Instance: Q = {q1, …, qn}, qi 2 Z+
– Is there a partition of Q into Q1 and Q2 such that:
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Complexity of PA
Theorem: PA with 2 passwords PA2 2 NP Complete.Proof: Part III: Making the Transformation
• Convert Partition instance to PA2 instance in poly-time.
• Given: Q = {q1, …, qn}
• Construct PA2 instance as follows:
• What is x?– For now, just a rational 2 (0,1)
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Complexity of PA
Theorem: PA with 2 passwords PA2 2 NP Complete.Proof: Part IV: Why it works• Solving equations:
• Gives us the following solutions:
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Complexity of PATheorem: PA with 2 passwords PA2 2 NP Complete.Proof: Part IV: Why it works• We will eliminate the solutions where VS1
VS2.
– As a result our solver will return that the constructed PA2 instance is a yes instance iff the Partition instance is a yes instance.
• Eliminating solution 1: – Recall our transformation:
– When we have:• Since x < 1
– Therefore, solution 1 can never occur.
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Complexity of PATheorem: PA with 2 passwords PA2 2 NP Complete.Proof: Part IV: Why it works• Eliminating solution 2:
– Recall our transformation:
– We need to ensure that when :
– We will find an x such that
– Therefore, when , solution 2 can never occur ) PA2 2 NP Complete.
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Efficiently Solvable Cases
The case of n = k• Optimal Strategy: Allocate exactly one account to each
password.
• Proof of optimality: – Since and , we have:
– This means an account contributes the most to the EG when it has its own password.
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The case of identical accountsWe have The problem reduces to:
where xi = number of accounts allocated to pi
• Optimal Strategy: Assign accounts (sequentially) to the password for which the EG increases the most.
• Proof of Optimality: Greedy argument – we always stay on par or ahead of any feasible solution.
Efficiently Solvable Cases
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A Special Case
The Case of Correlated Values and ProbabilitiesWe have:
and where .
• Property of Optimal Solution:
• Proof (Sketch):
a1 ai aj ak an
PW1 PWl PWm PWk
EG(PWl) = (vi + vk + …) pi pk ….. EG(PWm) = (vj + …) pi …..
If we have: pi > pj > pk , vi > vj > vk, and pi/qi > vj/vi then…
EG’(PWl) = (vi + vj + …) pi pj ….. EG’(PWm) = (vk + …) pk …..
EG(PWl) + EG(PWm) < EG’(PWl) + EG’(PWm)
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Presentation Outline
• Introduction
• The Password Allocation Problem
• The Weapon-Target Allocation Problem
• Conclusions and Future Work
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The Single Round WTA Problem: Definition
• Problem Instance:– n targets: t1, t2, …, tn
– k weapons: w1, w2, …, wk
– wi destroys tj with probability qij
• i.e., kill probability is weapon and target dependent
• Constraint: Each weapon is allocated to exactly one target.
• Goal: Minimize expected survival of enemy targets– Equivalent to maximizing expected damage to enemy targets.
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SWTA Problem Assumptions
• Given kill probabilities are independent of other pairs.– Isolates problem from geometric and geographic factors.
• Either a target is destroyed completely or survives completely.– qij (kill probability) = 1-pij (survival probability)
• Damage is surveyed after weapons are fired.– Models short battles with limited ammunition – does not consider enemy
retreats
• No fractional allocations may be made.– A weapon can only be allocated to a single target
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• Allocation matrix: X = {xij}– xij = 1 ) weapon wj is allocated to target ti
– xij = 0, otherwise
• Objective Function (Survival Value): [to be minimized]
• Constraint: – Every weapon is allocated to exactly one target.
The SWTA Problem: Mathematical Formulation
kX
i=1
0@nY
j =1((pj )xi j )
nX
j =1(xi j vj )
1A
kX
i=1
0@nY
j =1((pj )xi j )
nX
j =1(xi j vj )
1A (1)xi j =
½ 1 if account j is allocated to password i0 otherwise
maximize : F =kX
i=1
0@nY
j =1((pj )x i j )
nX
j =1(xi j vj )
1A
subj ect to :kX
i=1xi j =1; 8j 2 f1;:::;ng
F is theexpected gain (EG).
F = P ki=1³ Q n
j =1((pj )xi j )P nj =1(xi j vj )
´F = P ki=1
³ Q nj =1((pj )xi j )
P nj =1(xi j vj )
´
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Complexity of SWTA
Theorem: SWTA with 2 targets (SWTA2) 2 NP Complete.Proof: Part I: Formulating the Decision Version
• Instance: – P = {pij} where pij 2 (0,1) – r
• Is there a 0-1 matrix X such that:– The sum of the survival probabilities of the 2 targets is less than r
– and every weapon is allocated to at least one target.
• Clearly SWTA2 2 NP.
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Complexity of SWTA
Theorem: SWTA with 2 targets (SWTA2) 2 NP Complete.Proof: Part II: Finding the known hard problem
• The Rational Product Dichotomy (Fractional Subset Product):– Instance: Q = {q1, …, qn}, qi 2 (0,1)– Is there a partition of N={1, …, n} into S1 and S2 such that:
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Complexity of SWTA
Theorem: SWTA with 2 targets (SWTA2) 2 NP Complete.Proof: Part III: Making the Transformation
• Convert RPD instance to SWTA2 instance in poly-time.
• Given: Q = {q1, …, qn}
• Construct SWTA2 instance as follows:
Where pij is the survival probability of target i after a strike by weapon j.
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Complexity of SWTATheorem: SWTA with 2 targets (SWTA2) 2 NP Complete.Proof: Part IV: Why it works• Our SWTA2 solver will return yes iff
Where qi is the ith rational in the given RPD instance.
• By AGMI: – Therefore, can never occur.– SWTA2 solver returns yes iff
• By AGMI: – SWTA2 solver returns yes iff
which is a yes instance of RPD.
• SWTA2 2 NP Complete
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Efficiently Solvable Cases
The Case of Identical Weapons and TargetsWe have all weapon-target pairs with same survival probability p
i.e.,The problem reduces to:
subject towhere xi is the number of weapons allocated to target i.
• Optimal Strategy: Divide weapons as evenly as possible.
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If dividing k weapons evenly is not optimal. Then:
Target i Target j
xi weapons xj = d+xi weapons
But, switching one of the weapons target gives us:
1+xi weapons xj = d-1+xi weapons
Since p2(0,1) and xi < d+xi - 1
Therefore, switching targets strictly decreases the net survival value ) solution is not optimal
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Efficiently Solvable Cases
The Case of Equal WeaponsWe have one type of weapon – so all weapons destroy target i
with the same probability – pi.Problem reduces to:
• Optimal Strategy: Assign weapons to the target for which the objective function (i.e., pi xi) decreases the most.
• Proof of Optimality: By induction.– When allocating one weapon to n weapons trivially true.
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Assume Xk is the optimal solution for k weapons to n targets
Xk = <x1, x2, …, xn>
Let Xk+1 be the solution returned for k+1 weapons to n targets
Xk+1 = <x1, x2, …, xm+1, …, xn>
Where ±m · ±i 8 i 2 {1, …, n}
± =
p m x m
£ (p
m-1
)
Zk+1 = <z1, z2, …, zn>
Let Zk+1 be any other solution
Since Zk+1 Xk+1, there is a j where zk+1(j) > xk+1(j) ¸ xk (j)
Zk = <z1, z2, …, zj-1, …, zn>
Let Zk be the same solution with one less weapon for target j.
± =
p j (zj -1
) £ (p
j-1)
X*k+1 = <x1, x2, …, xj+1, …, xn>
Let X*k+1 be Xk with one more weapon allocated to target j.
± = pj xj £ (p
j -1)
·
· Since xj < zj
··
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Efficiently Solvable Cases
The Case of One Weapon per TargetWe have each of the n targets getting at most one weapon – i.e.,
As a result:
(1) is true since xij 2 {0,1}(2) is true since there is only one xij = 1 for each target.
Therefore:
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On the Complexity of Allocation Problems with Probabilistic Players 36
Efficiently Solvable Cases
The Case of One Weapon per Target• This can now be written as:
• Which is the transportation problem with:– costij = -qij
– k supply nodes with supply = 1– n demand nodes with demand = 1
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SWTA Approximation HeuristicsThree main techniques:• Integer constraint relaxation
– Allow fractional allocations of weapons to targets.– Solve resulting LP .– Use randomized rounding to obtain approximate solution to integer problem.
• Modeling as network flow problems– Create a graph of weapons and targets.– Each edge between a weapon and target has a cost approximately equal to the change
in objective function.• Approximate due to non-linear nature
– Set appropriate constraints (eg., supply/demand, capacity).– Solve network flow problem using MCMF, MF, TP algorithms (as is appropriate).
• Localized search– Start with a feasible solution of reasonable quality.– Perform swaps and multi-swaps yielding better solutions.
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On the Complexity of Allocation Problems with Probabilistic Players 38
Presentation Outline
• Introduction
• The Password Allocation Problem
• The Weapon-Target Allocation Problem
• Conclusions and Future Work
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On the Complexity of Allocation Problems with Probabilistic Players 39
Conclusions and Future Work• The Password Allocation Problem
– Also models parallel processing allocation problems.
– NP Complete even when all passwords are equal
– Has several efficiently solvable cases
• Analysis for cases with varying passwords.
• Approximation Techniques – Heuristics– Boundable algorithms (??)
• Online version of the problem
Varia
bilit
y of
pas
swor
dsVariability of accounts
Equal accounts and PWs
Equal #accounts and PWs
Correlated accounts
2 P
2 NP Complete2 ?
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On the Complexity of Allocation Problems with Probabilistic Players 40
Conclusions and Future Work• The Weapon-Target Allocation
Problem– NP Complete even for the single-
round version.– There are special poly-time solvable
cases.– General approaches to making
approximate solutions.• Most current work ignores
analysis – too much focus on heuristics (unboundable)!– Existence of constant-factor
bounds?– Almost no analysis for multi-round
variant.
Varia
bilit
y of
wea
pons
Variability of targets
All equal targets
Equal weapons and targets
All equal weapons
One weapon per target
2 P
2 NP Complete