- 90 Views
- Uploaded on
- Presentation posted in: General

On the Complexity of Search Problems George Pierrakos

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

On the Complexity of Search ProblemsGeorge Pierrakos

Mostly based on:

On the Complexity of the Parity Argument and Other Insufficient Proofs of Existence [Pap94]

On total functions, existence theorems and computational complexity [MP91]

How easy is local search? [JPY88]

Computational Complexity [Pap92]

The complexity of computing a Nash equilibrium [DGP06]

The complexity of pure Nash equilibria [FPT04]

slides and scribe notes from many people…

- Generally on Search Problems
- The Class TFNP
- Subclasses of TFNP part I: PPA, PPAD
- Problems in PPA, PPAD
- Completeness in PPAD

- Subclasses of TFNP part II: PPP, PLS
- PPAD-completeness of NASH & the complexity of computing equilibria in congestion games

- Generally on Search Problems
- The Class TFNP
- Subclasses of TFNP part I: PPA, PPAD
- Problems in PPA, PPAD
- Completeness in PPAD

- Subclasses of TFNP part II: PPP, PLS
- PPAD-completeness of NASH & the complexity of computing equilibria in congestion games

- SAT
- Input: boolean CNF-formula φ
- Output: “yes” or “no”

- FSAT
- Input: boolean CNF-formula φ
- Output: satisfying assignment or “no” if none exist

They are definitely not easier:

a poly-time algorithm for FSAT can be easily tweaked to give a poly-time algorithm for SAT

…and vice versa, FSAT “reduces” to SAT:

we can figure out a satisfying assignment by running poly-time algorithm for SAT n-times

- SAT is self-reducible but TSP is not…
- Here is how we solve FTSP using TSP:
- Use binary search to figure out the cost C of the optimum tour, using TSP algorithm
- Run TSP algorithm n2 times, always with cost parameter C, each time setting the weight of some edge to C+1

- L €NP iff there exists poly-time computable RL(x,y) s.t.
X € L y { |y| ≤ p(|x|) & RL(x,y) }

- Note how RL defines the problem-language L

- The corresponding search problem ΠR(L) €FNP is:
given an x find any y s.t. RL(x,y) and reply “no” if none exist

- FSAT € FNP… what about FTSP?
- Are all FNP problems self-reducible like FSAT? [open?]

- FP is the subclass of FNP where we only consider problems for which a poly-time algorithm is known

- A function problem ΠR reduces to a function problem ΠS if there exist log-space computable string functions f and g, s.t.
R(x,g(y)) S(f(x),y)

- intuitively f reduces problem ΠR to ΠS
- and g transforms a solution of ΠS to one of ΠR

- Standard notion of completeness works fine…

- A proof a-la-Cook shows that FSAT is FNP-complete
- Hence, if FSAT € FP then FNP = FP
- But we showed self-reducibility for SAT, so the theorem follows:
- Theorem: FP = FNP iff P=NP
- So, why care for function problems anyway??

- Generally on Search Problems
- The Class TFNP
- Subclasses of TFNP part I: PPA, PPAD
- Problems in PPA, PPAD
- Completeness in PPAD

- Subclasses of TFNP part II: PPP, PLS
- PPAD-completeness of NASH & the complexity of computing equilibria in congestion games

- What happens if the relation R is total?
i.e., for each x there is at least one y s.t. R(x,y)

- Define TFNP to be the subclass of FNP where the relation R is total
- TFNP contains problems that always have a solution, e.g. factoring, fix-point theorems, graph-theoretic problems, …
- How do we know a solution exists?
By an “inefficient proof of existence”, i.e. non-(efficiently)-constructive proof

- The idea is to categorize the problems in TFNP based on the type of inefficient argument that guarantees their solution

- FP TFNP FNP
- TFNP = F(NP coNP)
- NP = problems with “yes” certificate y s.t. R1(x,y)
- coNP = problems with “no” certificate z s.t. R2(x,y)
- for TFNP F(NP coNP) take R = R1 U R2
- for F(NP coNP) TFNP take R1 = R and R2 = ø

- There is an FNP-complete problem in TFNP iff NP = coNP
- : If NP = coNP then trivially FNP = TFNP
- : If the FNP-complete problem ΠR is in TFNP then:FSAT reduces to ΠR via f and g, hence any unsatisfiable formula φ has a “no” certificate y, s.t. R(f(φ),y) (y exists since ΠR is in TFNP) and g(y)=“no”

- TFNP is a semantic complexity class no complete problems!
- note how telling whether a relation is total is undecidable (and not even RE!!)

- Generally on Search Problems
- The Class TFNP
- Subclasses of TFNP part I: PPA, PPAD
- Problems in PPA, PPAD
- Completeness in PPAD

- Subclasses of TFNP part II: PPP, PLS
- PPAD-completeness of NASH & the complexity of computing equilibria in congestion games

- PPA stands for polynomial parity argument:
- Every graph has an even number of odd-degree nodes
- Every graph of degree two or less has an even number of leaves
- Both arguments turn out to be equally generic

- PPAD stands for polynomial parity argument directed
- Parity argument in directed graphs, where we now seek for a sink or a source

- Thm: any graph with odd degrees has an even number of HC through edge xy
- Proof Idea:
- take a HC
- remove edge (1,2) & take a HP
- fix endpoint 1 and start “rotating” from the other end
- each HP has two “valid” neighbors (d=3) except for those paths with endpoints 1,2
- Evoke the parity argument: at least two (different) such HPs must exist: add edge (1,2) et voila!

- time: worst case exponential…

- Given a triangulation of a triangle (castle) s.t.
- each original vertex obtains its own color
- no vertex on edge ij contains color 3-i-j
there exists a trichromatic triangle

- Proof Idea:
- add external 0-1 edges
- traverse triangles (“rooms”) using 0-1 edges (“doors”)
- If a “room” has no “door” out then it is trichromatic

- Note that
- the process cannot exit the castle and cannot fold upon itself
- our tour leaves 0 nodes to the right, hence direction, hence PPAD…

- Let A be a problem and M the (associated) poly-time TM
- Let x be an input of A
- Let Cx= Σp(|x|) be the configuration space of x, i.e. the set of all strings of length at most p(|x|)
- On input c € Cx machine M outputs in time O(p(n)) a set of at most two configurations M(x,c)
- M(x,c) may well be empty if c is “rubbish”

- c and c’ are neighbors ([c,c’] € G(x)) iff c € M(x,c) and c’ € M(x,c’)
- G(x) is symmetric with degree at most 2
- It is the implicit search graph of the problem

- Let M(x, 0…0) = {1…1} and 0…0 € M(x, 1…1), hence 0…0 is the standard leaf
- PPA is the class of problems defined as follows:
“Given x, find a leaf of G(x) other than the standard leaf 0…0”

Take A to be ANOTHER HAMILTON CYCLE. We show that A € PPA

- x is the input of A:
- a graph and a Hamilton Cycle

- Cxis the configuration space of x:
- Cxcontains encodings of “valid” Hamilton Paths (poly-space) and other irrelevant stuff

- On input c € Cx machine M outputs in time O(p(n)) a set M(x,c) of:
- the at most two neighbors of a valid Hamilton Path, or possibly nothing

- c and c’ are neighbors ([c,c’] € G(x)) iff c € M(x,c) and c’ € M(x,c’)
- remember the graph of neighboring “valid” Hamilton Paths

- the standard leaf is the initial Hamilton Path (from removal of (1,2))

- Same as PPA but…
- M(x,c) = In(x,c) U Out (x,c)is an ordered pair of configurations with|In(x,c)| ≤ 1 & |Out (x,c)| ≤ 1
- G(x) is now directed:[c, c’] € G(x) iff c € In(x,c’) and c’ € Out(x,c)
- We search for a node (other than the standard leaf) with |In(x,c)| + |Out (x,c)| = 1,i.e. any source or sink other than the standard one

Take A to be 2D-SPERNER. We show that A € PPAD

- x is the input of A:
- a triangle, a triangulation and a valid color assignment
- CAUTION: restricted input “promise” problems: we usually need some sort of syntactic guarantee of the validity of the input

- Cxis the configuration space of x:
- Cxcontains encodings of all sub-triangles in the triangulation

- M(x,c) = In(x,c) U Out (x,c) where:
- In(x,c) is the castle room where we came from
- Out(x,c) is the castle room where we are heading to

- c and c’ are neighbors ([c,c’] € G(x)) iff c € M(x,c) and c’ € M(x,c’)
- remember our tour through the triangular castle

- the standard leaf is the node lying outside the triangle

- PPA as defined above makes use of the “even leaves” argument
- Why not modify the definition as follows (PPA’):
- …on input c € Cx machine M outputs in time O(p(n)) a set M(x,c) s.t. |M(x,c)| = poly(n)…
- …“Given x, find an odd-degree node of G(x) other than the standard leaf 0…0”…
and make use of the (more general?) parity argument?

- Quite surprisingly: PPA’ = PPA

- Obviously PPA PPA’
- For the inverse:
- For each node c of degree k ≥ 1, we create new nodes (c,i), where i = 0,…,k div 2
- each node (c,i) is connected to the two nodes (c’,j) s.t. the lexicographically 2i, 2i+1 edges out of c are the 2j, 2j+1 edges into c’
- nonstandard leaves of the new graph correspond to nonstandard odd-degree nodes of the former

- Proposition 1: FP PPAD PPA FNP
- Proposition 2:
- Consider the more natural definition of PPA, where the leaf we search for is the one connected to the standard leaf (PPA’’)
- Then PPA’’ = PPAD’’ = FPSPACE
- just show FPSPACE PPA’’
- any problem in FPSPACE can be solved by a reversible TM, i.e. one where each configuration has at most one predecessor and one successor [Be84]
- define M(x,c) to return the (at most) two configurations that are the predecessor and successor of c on input x
- hence, the nonstandard leaf on the same component as the standard leaf is the halting configuration with the desired output

- Brouwer’s fix-point theorem: let S be any n-dimensional simplex and let φ: SSbe a continuous function. Then φ has a fix point, i.e. an x* s.t. φ(x*) = x*
- Proof for the special case of a triangle, using Sperner lemma
- Let T1, …, Ti,… be a sequence of finer triangulations
- Define labels L1, …, Li,… for the triangulations as depicted: Li is a valid labeling for Ti
- Consider the sequence of centers c1, …, ci,… of the resulting trichromatic triangles t1, …, ti,…(guaranteed by Sperner’s Lemma)
- This is an infinite sequence in a closed convex space, hence it has a subsequence converging at x*: a point which is arbitrarily closed to the centroid of a trichromatic triangle

- x* is a fix point of φ: if not, then we can find a small triangle containing x* but not φ(x*). Because of the continuity of φ, we can show that this cannot be labeled with 3 colors
- Also need to take care of continuity for representation in TM

- Theorem[Nash51]: every game has a mixed Nash equilibrium
- Original proof of Nash uses Kakutani’s fix point theorem; KAKUTANI is in PPAD by a reduction from BROWER
- Nash’s theorem can be proved using Brower’s fix-point theorem too
- In [Pap94] they show that NASH is in PPAD by a reduction from LINEAR COMPLENTARITY

- LINEAR COMPLEMENTARITY
- We are given an nxn matrix A and an n-vector b
- We are seeking vectors x,y ≥ 0 s.t.
- Ax + b = y
- x·y = 0, i.e. at least one of xi, yi is 0

- Solved by Lemke’s algorithm:
- some times it does not even converge
- if it converges it might well be (worst case) exponential

- In general, the problem is NP-complete (?)
- For the case of P-matrices A (matrices satisfying det(AS) > 0 for all non-empty subsets S of {1,…,n}) the problem has a unique solution and Lemke’s algorithm converges to it.
- Is checking for the P-property for general matrices in P? Open…

- PPAD formulation of L-COMPL: Given A and b, find a solution x, y to the linear complementarity problem or a subset S s.t. det(AS) ≤ 0

- Example: from bimatrix games to a linear complementarity problem
- Let
- are the probability and payoff of player 1 for row i
- are the probability and payoff of player 2 for column j
- is the payoff of player 1
- is the payoff of player 2

- Approach 1:
- implies that the player has equal expected payoff from both his choices (else he would pick the one maximizing his profit with probability 1 and drop the other)
- Hence we can write down the following linear system:
- To compute all equilibria we need to solve a linear system (in poly-time) for all different supports of players 1 and 2 in time 2n2mP(n,m)(support = subset of strategies played with p>0)

- Approach 2: reduce to L-COMPL

Use Lemke’s algorithm: “double pivoting”

- Given a graph G and a HP in it, find either another HP in G or in its complement Gc
- Construct a bipartite graph, whose odd-degree nodes are HP of G and Gc
- “left” nodes: all HP of the complete graph on the nodes of G
- “right” nodes: all subgraphs of Gc -including the empty subgraph
- Edge from h (hpath) on left side to s (subgraph) on the right side iff s is a proper subgraph of h
- Claim: the only odd-degree nodes are the HP of G and Gc on the left side
- If node on the left is not a HP of G or Gc then it contains a non-empty subset of edges of Gc and will be adjacent to all proper subsets of this set: these are 2k
- A HP of G on the left will only be adjacent to the empty subgraph
- A HP of Gc on the left will be adjacent to its 2n-1-1 proper subgraphs (excluding itself) and will also have odd degree
- Any subgraph (node on the right) will be connected to all its possible connections to a HP (usually 0): for graphs with more than 5 nodes this number is even (can be proved via an easy calculation…)

- PPA:
- HAMILTON DECOMPOSITION
- TSP NONADJACENCY
- CHEVALLEY MOD 2
- TOTALLY EVEN SUBGRAPH
- CUBIC SUBGRAPH

- PPAD:
- k-D SPERNER
- KAKUTANI
- COMPETITIVE & EXCHANGE EQUILIBRIUM
- TUCKER
- BORSUK-ULAM, DISCRETE HAM SANDWICH, NECKLAGE SPLITTING…

- Generally on Search Problems
- The Class TFNP
- Subclasses of TFNP part I: PPA, PPAD
- Problems in PPA, PPAD
- Completeness in PPAD

- Subclasses of TFNP part II: PPP, PLS
- PPAD-completeness of NASH & the complexity of computing equilibria in congestion games

- PPP stands for Polynomial Pigeonhole Principle
- Definition:
- A problem A is in PPP if it is defined in terms of a poly-time TM M
- Given input x define the set of “solutions” S(x) = Σp(|x|)and
- For any y € S(x), M(x,y) € S(x)
- Desired output:
- y € S(x), s.t. M(x,y) = 0…0 or
- y, y’ € S(x) s.t. M(x,y) = M(x,y’)

- One of the two is guaranteed to exist by pigeonhole principle

- Input: Boolean circuit C with n inputs and n output
- Output: either find an input y s.t. C(y) = 0n or find two inputs y, y’ s.t. C(y) = C(y’)
- Obviously PIGEONHOLE CIRCUIT € PPP:
- M is the poly-time TM that simulates circuit C on input y
- Input x is the description of the circuit C
- S(x) = {0,1}n and for any y € S(x), M(x,y) € S(x)

- And any problem in PPP can be evaluated by such a circuit: just tweak the proof of CIRCUIT-VALUE P-completeness

- Monotone circuits are the circuits that do not contain “no” gates
- Property of monotone circuits:swapping xi from false to true cannot change any output bit from true to false
- Proof: trivial by induction on the depth of the tree

- MONOTONE PIGEONHOLE CIRCUIT € FP
- Proof:
- We need to consider only the input sequences of the form 0i1n-i, i=0,…,n, i.e. only n input sequences
- Start with 1n and start swapping bits to 0. The only way to have n distinct outputs is if: output starts at 0n and each time one output bit swaps to 1
- Hence, either output = 0n for some input, or two inputs have same output

- Proof:

- Input: n positive integers with sum less than 2n-1
- Output: two subsets with the same sum
- EQUAL SUMS in PPP:
- Two such subsets trivially exist since there are more subsets than possible sums
- The formal definitions utilizes a TM M that given the input x (set of all sets) and the description of a set y, outputs in polynomial time the sum of the set’s elements. Since the sum of all elements is less than 2n-1 we cannot have M(x,y) = 0…0
- Is EQUAL SUMS PPP-complete?

- One-way permutation π is a poly-time algorithm mapping Σn to itself, such that for all x≠y, π(x) ≠ π(y), s.t. it is hard to recover x from π(x).
- Prop1: If PPP=FP then one-way permutations do not exist
- Proof:
- suppose PIGEONHOLE CIRCUIT is in FP
- We wish to invert an alleged one-way permutation π, i.e. given y, find x s.t. y=π(x)
- Turn the algorithm for π into a circuit (CIRCUIT VAL is P-complete) and add bitwise XOR gates between output y and x; call this circuit C
- Apply the poly-time algorithm for PIGEONHOLE CIRCUIT to C. Since π is a permutation, the algorithm cannot return two inputs giving the same output, hence it returns y, s.t. C(y) = 0, i.e. π(x) = y

- Proof:

- Prop2: PPAD PPP
- Proof:
- Reformulate any problem in PPAD with graph G as a problem in PPP by use of π(x) defined as
- π(x) = x if x is a sink
- π(x) = G2(x) if x is a nonstandard source
- π(x) = G(x) otherwise

- Trivially, from any confluence of values of π we can recover a nonstandard source or a sink of G

- Reformulate any problem in PPAD with graph G as a problem in PPP by use of π(x) defined as

- Proof:

- PLS originally defined in [Johnson, et. al. ‘88]: “find some local optimum in a reasonable search space”.
- Basic ingredient: a problem with a search space, i.e. a set of feasible solutions which has a neighborhood structure
- Every problem L (e.g. TSP) in PLS has a set of instances DL (e.g. encodings of adjacency matrix)
- Every instance x in DL has a finite set of solutions FL(x) (e.g. tours) considered wlog polynomial in size
- For each solution s in FL(x) we have
- A poly-time cost function c(x,s) that given an instance x and a solution s, it computes its cost (e.g. length of a tour)
- A poly-time neighbor function N(x,s) FL(x) that given an instance x and a solution s, returns its neighboring solutions (e.g. a tour that differs from the original one in at most two edges)

- Remaining constraints are expressed in terms of 3 polynomial time algorithms AL, BL, CL
- AL, given an instance x in DL produces a particular standard solution of L, AL(x) € FL(x) (e.g. a standard tour for TSP)
- BL, given an instance x in DL and a string s determines whether s is a solution (s € FL(x)?) and if so, it computes its cost c(x,s)
- CL, given an instance x € DL and a string s € FL(x) has two possible outcomes, depending on s:
- If there is any solution s’ € N(x,s) s.t. c(x,s’) < c(x,s) it outputs s’
- Otherwise it outputs “no”, which implies that s is locally optimal

- The sense of local searching is obvious: starting at some arbitrary solution start searching locally for a better solution and move towards it.
- Standard local search algorithm:
Given x, use AL to produce a starting solution s= AL(x)

Repeat until locally optimal solution is found:

Apply algorithm CL to x and s

If CL yields a better cost neighbor s’ of then set s=s’

- Since the solution space is finite the above algorithm always halts
- If FL(x) is polynomially bounded then the above algorithm is polynomial
- However, things can turn out pretty bad:
- Finding a local optimum reachable from a specific state can be proven PSPACE-complete
- There are instances with states exponentially far from any local optimum

- Correct question to ask: “Given x, find some locally optimum solution s”

- PLS-reduction: we must map the local optima of one problem to those of another, thus pairtaining the neighboring structure. Hence problem L reduces to K if there exist poly-time functions f and g s.t.
- f maps instances x of L to instances f(x) of K
- g maps (solutions of f(x), x) pairs to solutions of x
- for all instances x of L, if s is a local optimum for instance f(x) of K, then g(s,x) is a local optimum for x

- Basic PLS-complete problem:
- weighted CIRCUIT – SAT under input bitflips
- some relatives of TSP, MAXCUT and:

- POS-NAE-3SAT: [Shaeffer&Yannakakis-91]
Given an instance of NAE-3SAT with weights on its clauses and with positive literals only, find a truth assignment, whose total weight (i.e. the weight of the unsatisfied and totally satisfied clauses) cannot be improved by flipping a variable

- Generally on Search Problems
- The Class TFNP
- Subclasses of TFNP part I: PPA, PPAD
- Problems in PPA, PPAD
- Completeness in PPAD

- Subclasses of TFNP part II: PPP, PLS
- PPAD-completeness of NASH & the complexity of computing equilibria in congestion games

- Idea: generalization of Selfish Task Allocation
- Intuition:
- Route traffic through a network
- Resource allocation

- We have
- a set of resources E (network edges…?)
- a set of players N
- the resource delays de

- For each player i
- an action set Ai 2E
- a weight (traffic demand) wi

- let A=(a1,…, an), ai€ Ai, be a strategy profile
- denote by
- Pe(A) the set of players picking resource e in profile A
- ne(A) the number of players picking resource e in profile A, ne(A)=|Pe(A)|

- define the load of resource e to be
L(e) = Σiin P(e)wi

- for each strategy profile A, each player incurs a total delay/cost
ci(A) = Σein aiL(e)

- A Pure Nash Equilibrium (PNE) is a strategy profile A=(a1,…, an), s.t. for all i € N and for all ai’ € Ai
c(a1,…, ai,…, an) ≤ c(a1,…, ai’,…, an)

- Notice the local optimality flavor of a PNE

- Exact Potential for a c.g. is a function Φ:Ε→R s.t. for all i € N, ai’ € Ai and A
ci(A) - ci(A-i, ai’) = Φ(A) - Φ(A-i, ai’)

- It is easy to verify that every game that admits an exact potential possesses a PNE

- Symmetric vs AsymmetricPlayers have the same action set and the same weights, i.e. they are indistinguishable
- Network vs General Congestion Games
- the resources are graph edges
- each player has a source and a sink vertex (si,ti)
- the action sets Ai are si-ti paths

- Single vs multi-commodity (network) games
- the (si,ti) pairs are common for all players

- Theorem [Rosenthal ’73]:Every unweighted congestion game is an exact potential game
- The inverse holds [Mondere&Shapley]:
Every finite exact potential game is isomorphic to an unweighted c.g.

- So it is only for c.g. that we can find an exact potential
- Does this mean that the potential technique (for guaranteeing existence of a local minimum) can be used only for c.g.?

- We need to be less strict about the potential function.
- Consider the party affiliation game (HAPPYNET):
- n players (nodes in graph)
- Symmetric preference relations between players-nodes
- 2 actions per player: {-1,1}
- State S is a mapping V{-1,1} (players pick sides)
- payoff of i: sgn[S(i) ΣjS(j)wij] (if the payoff is positive then we say “node i is happy”)
- The problem: “Find a state in which all nodes are happy”
- intuition: everyone wants to be with his friends
- application: convergence of Hopfield NN

- HAPPYNET always admits a solution
- Proof: by use of a (general) potential function
- φ[S] = Σi,jS(i)S(j)wij
- Suppose that node i is unhappy, i.e. S(i) ΣjS(j)wij = - δ < 0
- Consider the state S’ where S is flipped, i.e. S’(i) = S(i)Obviously φ[S’] = φ[S] +2δ, i.e. the potential function is increased by at least two when we flip any unhappy node
- Since φ takes values in [-W, W] where W is the sum of all weights, and it is increased by two in each iteration, this process has to end at a state with no unhappy nodes, i.e. at a solution

- How many steps? Polynomial in W, exponential in |W|

- Proof: by use of a (general) potential function
- Connection with congestion games:
- The party affiliation game is NO congestion game therefore it admits NO exact potential; it admits a general potential s.t. sgn( ci(A) - ci(A’) ) = sgn( Φ(A) – Φ(A’) )
- Using thiswe proved that the problem always has a solution, i.e. the underlying game always admits a PNE, which can be derived from the local optima of siΣjsj wij.

- Main results here from [Fabrikant/ Papadimitriou/ Talwar STOC 2004]
- They proved that the problem of finding a PNE of unweighted games:
- is in P for symmetric network c.g. [irrelevant]
- is PLS-complete for
- General asymmetric c.g.
- General symmetric c.g.
- Asymmetric network c.g. [too damn hard]

- POS-NAE-3SAT ≤PLS General Asymmetric c.g.
- variable x → player x
- clause c → resources ec, ec’
- Ax→ {{ec|x in ec}, {ec’|x in ec’}}
- x plays ec if x=0 in POS-NAE-3SAT
- x plays ec’ if x=1 in POS-NAE-3SAT

- de(1)=de(2)=0, de(3)=w(c)
- Every PNE of the c.g. corresponds to a local minimum of POS-NAE-3SAT, since bitflip corresponds to single-player deviation

- General Asymmetric c.g. ≤PLS General Symmetric c.g.
- Augment the network by adding n additional resources e1,…, en
- dei(k)=0 if k=1 and
- dei(k)=M otherwise (M suff. large number)

- A = Ux { sU{ex} | s in Ax}
- same number of players so any solution that uses a ex twice has an unused ex, so it is no PNE
- at PNE players arbitrarily take on the “roles” of the players in the general asymmetric game

- Augment the network by adding n additional resources e1,…, en

FNP

TFNP

PLSPPPPPA

PPAD

FP