Tree-like resolution complexity of two planar problems
Dmitry Itsykson, Anna Malova, Vsevolod Oparin, Dmitry Sokolov
aa r X i v : . [ c s . CC ] D ec Tree-like resolution complexity of two planar problems
Dmitry Itsykson ∗ , Anna Malova † , Vsevolod Oparin ‡ , and DmitrySokolov § St. Petersburg Department of V.A. Steklov Institute of Mathematics ofthe Russian Academy of Sciences St. Petersburg Academic University of the Russian Academy of SciencesJune 9, 2018
Abstract
We consider two CSP problems: the first CSP encodes 2D Sperner’s lemma forthe standard triangulation of the right triangle on n small triangles; the secondCSP encodes the fact that it is impossible to match cells of n × n square to arrows(two horizontal, two vertical and four diagonal) such that arrows in two cells witha common edge differ by at most 45 ◦ , and all arrows on the boundary of the squaredo not look outside (this fact is a corollary of the Brower’s fixed point theorem).We prove that the tree-like resolution complexities of these CSPs are 2 Θ( n ) . ForSperner’s lemma our result implies Ω( n ) lower bound on the number of request tocolors of vertices that is enough to make in order to find a trichromatic triangle;this lower bound was originally proved by Crescenzi and Silvestri.CSP based on Sperner’s lemma is related with the PPAD-complete problem.We show that CSP corresponding to arrows is also related with a PPAD-completeproblem. The resolution proof system is one of the simplest and well-studied proof systems. Thereare well known methods of proving lower and upper bounds on the complexity of severaltypes of formulas. However there are no known universal methods that may be used todetermine the asymptotic resolution complexity of a given family of formulas. ∗ [email protected], partially supported by the RFBR grant 14-01-00545, by the President’s grantMK-2813.2014.1, by the Government of the Russia (grant 14.Z50.31.0030) and by the RAS program offundamental research † [email protected] ‡ [email protected], partially supported by ANR NAFIT 008-01 § [email protected], partially supported by the RFBR grant 12-01-31239-mol-a, by the Presi-dent’s grant MK-2813.2014.1, by the Government of the Russia (grant 14.Z50.31.0030) k -elementalphabet creates k -ary tree. Every node of this tree corresponds to a variable, and edges,going to children, correspond to k different substitutions of the variable. Every leaf ofthe tree contains a constraint that is falsified by substitution made on the path from theroot to that leaf. We call such tree as contradiction search tree. It is well known that theminimal size of a contradiction search tree for an unsatisfiable CSP is equal to the sizeof the minimal tree-like resolution proof of CSP.Every unsatisfiable constraint satisfaction problem F under k -element alphabet hasthree parameters: S ( F ) is the minimal size of resolution proof of F , S T ( F ) is the minimalsize of tree-like resolution proof, and d ( F ) is the minimal depth of contradiction searchtree. These parameters are connected by trivial inequalities: S ( F ) ≤ S T ( F ) ≤ k d ( F ) .The paper [2] presents the family of formulas that exponentially separate S and S T , andpaper [8] gives an example of family F n such that S T ( F n ) = O ( n ) and d ( F n ) = Ω( n/ log n ).The number d ( F ) is the worst-case lower bound on the number of requests to the variablesof CSP that is necessary to do in order to find a falsified constraint.We consider the constraint satisfaction problem that codes 2D Sperner lemma: namelyis impossible to color vertices of the standard triangulation of the right triangle in threecolors such that vertices of the right triangle are colored in different colors and all verticeson the side are colored in two colors and there are no small triangles cored with threedifferent colors. We prove that that size of tree-like refutation of this CSP is at least2 Ω( n ) , where n is the number of points on the side of the triangle. Previous result byCrescenzi and R. Silvestri [4] gives a linear lower bound on the depth of resolution prooffor the same CSP.We also consider the constraint satisfaction problem that codes that it is impossible tomatch cells of the n × n square with arrows from the set {← , տ , ↑ , ր , → , ց , ↓ , ւ} suchthat arrows in every two adjacent cells differ by at most 45 ◦ and boundary arrows are notdirected outside the square. The impossibility is a corollary of the Brower’s fixed pointtheorem. We prove that tree-like resolution complexity of this CSP is equal to 2 Θ( n ) .Our research is motivated by the investigation of complexity classesPPA , PPAD , PPADS , PPP that are subclasses of TFNP function problems thatare guaranteed to have a solution because of unsatisfiability of certain CSP [6]. It isknown that Sperner’s Lemma corresponds to PPAD-complete problem. We show thatCSP with arrows also corresponds to PPAD-complete problem.
Let X = { x , x , . . . , x n } be a finite number of variables that can take values from a finitealphabet W = { w , w , . . . , w k } . Let S be the set of constraints for X : every constraintdepends on some subset X ′ ⊆ X and defines a set of values that variables from X ′ cantake simultaneously. A constraint satisfaction problem (CSP) is a triplet h X, W, S i . Aconstraint satisfaction problem is satisfiable if there is a set of values for X that satisfiesall constraints from S , otherwise we call CSP unsatisfiable.2 partial substitution is a map from X to W ∪ {∗} . We say that substitution ρ setsvalue for variable x if ρ ( x ) = ∗ . A substitution is full if values of all variables are set.Substition ρ falsifies constraint C ∈ S if values of all variables, that constraint C depends on, are defined by ρ and constraint C forbids such a setting of values by ρ . A contradiction search tree for an unsatisfiable CSP is a rooted k -arity tree such thatvertices are marked with variables and edges, that connect a vertex marked with x to itschildren, are marked with substitutions x := w for every w ∈ W . Every leaf of the treeis marked with a constraint that is refuted by the substitution made along the path fromthe root to the leaf.A nogood is a constraint of the following form: ¬ ( x = a ∧ · · · ∧ x ℓ = a ℓ ), where x , . . . , x ℓ ∈ X, a , . . . , a ℓ ∈ W . In the case of binary alphabet W = { , } nogoods areequivalent to clauses. If constraint depends on ℓ variables, then it can be represented asa conjunction of at most | W | ℓ nogoods.The resolution proof system can be generalized to CSP if all its constraints are rep-resented as conjunctions of nogoods. A resolution proof for some CSP formula φ is asequence of nogoods C , C , . . . , C t , that ends with empty nogood (cid:3) . Every nogood iseither contained in φ or can be derived from k previous nogoods by the resolution rule.Let { N a } a ∈ W be the set of nogoods in the following form: N a = ¬ ( x = a ∧ α a ) for all a ∈ W . Then nogood ¬ ( V a ∈ W α a ) is a resolvent (a result of resolution rule) of nogoods { N a } a ∈ W . A resolution proof is called tree-like if there exists a k -ary tree such that itleaves are marked with nogoods of initial formula φ and nogood in any other vertex v isa resolvent of nogoods that are written in children of v ; the root of the tree is markedwith the empty nogood (cid:3) . Proposition 2.1 ([1][5]) . The size of the minimal tree-like resolution proof for everyunsatisfiable CSP formula φ is equal to the size of minimal contradiction search tree for φ . Consider CSP φ under a finite alphabet W = { w , w , . . . , w k } that depends on variables X = { x , x , . . . , x n } and consists of constraints C , C , . . . , C m .We consider the following game that will be used for proving lower bounds on the sizeof contradiction search trees. A game is defined by an unsatisfiable CSP formula φ ; twoplayers Alice and Bob play as follows. Alice secretly from Bob chooses a contradictionsearch tree for φ and put a token in the root of the tree. Alice asks Bob about the valueof a variable that corresponds a vertex that contains the token. Bob either returns avalue of asked variable or suggest to Alice choose the value by herself from some subsetof W of cardinality at least 2. In the second case we say that Bob moves ChooseAny.Alice moves the token according the Bob’s answer, if Bob moves ChooseAny, then Alicechooses a value herself from the set proposed by Bob. The game is over if the token isin a leaf, that is a contradiction is found. The goal of Bob is to maximize the number ofChooseAny answers along the path from the root to a leaf. Lemma 3.1.
Let for CSP formula φ there exist such a strategy of Bob that for all strate-gies of Alice, Bob moves ChooseAny at least t times. Then the size of any contradictionsearch tree for φ is at least 2 t . 3 roof. Consider some contradiction search tree T for φ . We construct probabilistic distri-bution on the leaves of T that corresponds to Bob’s strategy and the following randomizedstrategy of Alice. Alice chooses the tree T and if Bob moves ChooseAny, Alice chooses avalue at random with equal probabilities. By the statement of the Lemma the probabilitythat the game will finish in every particular leaf is at most 2 − t . Since with probability 1the game will finish in a leaf, the number of leafs is at least 2 t .In applications of Lemma 3.1 it is convenient to describe the strategy of Bob inthe following terms: Bob has a partial substitution that assigns values to variables thatAlice asked for and probably to some other variables. The current substitution should notfalsify constraints. If Alice asks for a variable that is determined by the substitution, Bobreturns the assigned value, otherwise Bob moves ChooseAny and extends the substitutionby the value that Alice chooses and probably substitutes values to some other variables. Let us consider triangle T , each side of whom is divided on n − p , . . . , p n , p n = q . . . , q n , q n = r , . . . , r n = p as it is shown on Figure 1.The standard triangulation is obtained by joining the following pairs of vertices:1. p i and q n − ( i − , ∀ i (2 ≤ i ≤ n − p i and r n − ( i − , ∀ i (2 ≤ i ≤ n − q i and r n − ( i − , ∀ i (2 ≤ i ≤ n − S . A vertex coloring c : S → { Red, Green, Blue } is said to be Sperner’s coloring if vertices p , q ,and r are labeled by three different colors and each vertex on each edge of triangle T iscolored by only one of two colors of the ends of this edge.In the triangulation we will discuss only small triangles, i.e. triangles without crossingsinside. A triangle in the triangulation is said to be trichromatic if all its vertices arecolored with three different colors. Lemma 3.2 (Sperner [7]) . If c is a Sperner coloring, the triangulation contains a trichro-matic triangle.Let denote the unsatisfable CSP corresponding to Lemma 3.2 by φ n . Variablescorrespond to the vertices of the triangulation and can take values from alpabet { Red, Green, Blue } . Three constraints fix particular colors for vertices of triangle T and we set a constraint for each vertex on each side of T , which fixes a set of two possiblecolors. Also for each small triangle in the triangulation we set a constraint which forbidsits trichromatic coloring. The unsatisfability of φ n immediately follows from Sperner’slemma. Lemma 3.3.
For CSP φ n there exists such a strategy of Bob that gives Ω( n ) ChooseAnyanswers for any strategy of Alice. 4efore the proof we associate variables of the CSP with vertices of the triangulationand a partial substitution with a coloring. Some vertices can be non-colored, so valuesof the corresponding vertices has not been set. Proof.
We provide the strategy of Bob and then prove the lower bound.As it is described in the previous section, Bob maintains a coloring. If Alice asks Bobfor a colored vertex, Bob returns its color.Bob maintains a path P which starts from r ⌈ n ⌉ and stops at a crossing vertex f . Path P goes along edges of the triangulation and can use only right or right-down directions.Bob keeps in mind two segments which start from vertex f and go in right and right-downdirections to the border of the triangle. We denote them by RP and DP , respectively.We denote the parallelogram, surrounded by RP , DP and the border of triangle T , by H Buffer is an area that is strictly in parallelogram H and contains all points whosedistance to RP or DP are at most two. Note that the buffer does not include points on RP and DP .Informally Bob hides a trichromatic triangle in parallelogram H . Formally he main-tains the following invariants (see Figure 1):1. The border vertices are colored according the following(a) c ( p ) = c ( r i ) = Blue , for any i , with ⌈ n ⌉ − ≤ i ≤ n − c ( q n ) = c ( r i ) = Green , for any i , with 2 ≤ i ≤ ⌈ n ⌉ (c) c ( p i ) = c ( q j ) = Red , for any i , j , with 2 ≤ i ≤ n , 2 ≤ j ≤ n −
12. Vertices in the buffer are empty. It means Alice has not asked for them.3. Any vertex on path P is green, and any vertex just below the path is blue.4. Any vertex above P ∪ RP is colored, and each blue vertex is surrounded by redvertices.5. Any vertex left to P ∪ DP is colored, and each green vertex is surrounded by redvertices.6. Any vertex on RP is green, any vertex on DP is blue.7. Any green or blue vertex in H , excluding RP and DP , is surrounded by red vertices.8. All uncolored vertices are in H .It is easy to see that if all invariants are hold, then there are no trichromatic trianglein the current coloring.Absence of such a triangle above P ∪ RP (including border P ∪ RP ) follows fromInvariant 4. There is not adjacent vertices colored on blue and green. Similarly absenceof a trichromatic triangle left to P ∪ DP follows from Invariant 5. Invariant 3 guaranteesthat vertices on P can not lie on a trichromatic triangle, because all these vertices don’thave adjacent red vertex below. Finally Invariants 2 and 7 implies that there are notrichromatic triangles in H . 5 RPRP r ⌈ n ⌉ Buffer f B u ff e r p = r n q = p n r = q n Figure 1: Invariants6 2 2 2 2 2 2 2 2 2 2 2 23 33 33 33 3332 2 2 2 244 44 44 44 f RP D P Figure 2: CasesWe will prove that if after a query of Alice, Bob cannot support these invariants, hehas alredy given Ω( n ) ChooseAny answers.Initially f = r ⌈ n ⌉ , P contains only f . Segments RP and DP are defined by thecurrent position of vertex f . In the current coloring all vertices in area above RP arecolored by green and ones left to DP by blue.Let Alice ask for a non-colored vertex. By Invariant 8 such a vertex lies in parallelo-gram H . Bob answers accordingly the following cases, which are shown on Figure 2.In Case 4 Bob returns ChooseAny(Green, Blue) and surrounds selected vertex by redones to hold Invariant 7.In cases 1 – 3 Bob has to move the buffer to hold Invariant 2. In order to explainhow to move teh buffer, we introduce two new notions: a slot and an expanded slot. Theparallelogram H is a union of horizontal and diagonal segments. The crossings of thesesegments are vertices of the triangulation. For the horizontal segments we skip first threeones on the top and divide all others onto the groups with four segments in each. We dothe same thing with the vertical segments skipping three leftmost ones. We ignore thelast segments of each type if their number is at most 3. An expanded slot is a union of two groups of horizontal and diagonal groups of seg-ments.For an expanded slot we consider the parallelogram which is formed by the bordersof triangle T , the next to the leftmost segment in the vertical group and the next to topsegment in horizontal one. A slot is a subset of the considered expanded slot which alsolies in the described parallelogram.The left and top borders of a slot can be used for new RP and DP , respectively. Therest of the slot can be used for the new buffer.We say that a group of segments is clear if Alice has not requested any vertex in thisgroup.If Alice’s request fits Cases 1 – 3, Bob looks for the first clear horizontal and diagonalgroups of segments starting from vertex f , forms an expanded slot and extends path P asit will be described below. The new end of path P defines new position of parallelogram H and segments RP and DP .Bob adds all vertices, which lie outside of new parallelogram H or belong segment RP and DP , in the partial substitution as it is described in Invariants 47 xpanded slot Figure 3: The First Extension of path P and 5.Now we are ready to describe how Bob should answer on Alice’s requests.1. Bob answers ChooseAny(Green, Blue). If Alice chooses green, Bob extends P accordinally Figure 4, and Figure 3 otherwise.2. Bob answers ChooseAny(Green, Red) and extends P as showed on Figure 4 regard-less Alice’s choice.3. Bob answers ChooseAny(Blue, Red) and extends P as showed on Figure 3 regardlessAlice’s choice.4. Bob answers ChooseAny(Green, Blue) and colors all uncolored vertices adjacent torequested vertex by red.Note that all of these assignments do not violate the invariants. Path P is extendedby empty vertices or colored by green and all adjacent vertices lies below P are emptyor colored by blue. The fourth case guarantees satisfying Invariant 7, and 4 and 5 in thefuture.At the start of the game the number of clear groups of segments in both directionsis Ω( n ). In a turn Alice can spoil at most two groups in each direction. Every timeAlice spoils a group, Bob either gives ChooseAny answer or finishes the game. The lattermeans that Bob can not hold all invariants.Every time Bob moves the buffer he skips only spoiled groups. So at the end of thegame all slots in at least one direction will be spoiled. The latter guarantees that Bobgives Ω( n ) ChooseAny answers.Remind that Bob’s strategy guarantees that a trichromatic triangle can not be founduntil all invariants are hold. So for any strategy of Alice Bob gives Ω( n ) ChooseAnyanswers. 8 lot Figure 4: The Second Extension of path P We consider one more CSP, denote it as ψ n , that corresponds to the game on a n × n square.The variables correspond to the cells of the square and can take values from set W = {← , տ , ↑ , ր , → , ց , ↓ , ւ} .There are two types of constraints. 1) Two arrows in two cells with a common edgediffer by at most 45 ◦ . 2) All arrows on the boundary of the n × n square should not bedirected outside of the square. We set a constraint for each pair of adjacent cells andeach cell on the border.The unsatisfiability of ψ n follows from Brouwer’s theorem. Theorem 3.1.
CSP ψ n is unsatisfiable. Corollary 3.1.
For the CSP ψ n there exists a contradiction search tree of depth O ( n )and therefore of size 2 O ( n ) .For proofs of Theorem 3.1 and Corollary 3.1 see Appendix A. Theorem 3.2.
The size of any contradiction search tree for CSP ψ n is 2 Ω( n ) , and thereforethe depth is Ω( n ).The proof of Theorem 3.2 follows from Lemma 3.1 and following one. Lemma 3.4.
For CSP ψ n there exists such a strategy of Bob that Bob gives Ω( n )ChooseAny answers for any strategy of Alice. Proof.
Starting from the left side Bob mentally split the square on vertical strips n × slots . Since n may not be divisible by 8, there may be a strip of size atmost n ×
7, which remains unsplited.Let us divide a slot into two strips of size n ×
4. We say that left one is a bufferposition. At the beginning of the game Bob divides the rightmost slot into to strips n × ← . 9uring the game Bob may move the buffer, but the new position for the buffer isdetermined in the same way. Bob divides a slot in two parts and chooses left one.Bob maintains the following invariants: 1) The buffer is empty. 2) The area, to theright of the buffer, is filled completely. The leftmost strip n × ← .3) All neighbours of empty cells are either empty or filled by → . 4) Let k be the number of slots to the right of the buffer. Then Bob gave at least k answersChooseAny for cells to the right of the buffer. 5) No constraints are falsified.Note that all invariants are hold at the begining.The Bob’s strategy has the following type: 1) If Alice requests already filled cell, thenBob returns its value. 2) If Alice requests empty cell, then Bob moves ChooseAny andprobably fill some cells in order to maintain invariant. 3) If it is impossible to maintaininvariants, Bob moves in arbitrary way.We specify the second step and prove that if Bob fails to maintain invariants, then hehas already made Ω( n ) ChooseAny answers.Let Alice request an empty cell c . We consider three cases: 1) the cell c is to the leftof the buffer and does not share edges with cells from the buffer; 2) the cell c is in thebuffer; 3) the cell c is in the n × → , ց ) if c is on the upper boundary andChooseAny( → , ր ), otherwise.Bob assigns → to empty neighbors of c . It is easy to see that such assignments donot violate invariants.In the second case c lies in the buffer. We consider two cases a) there are no emptyslots left of the buffer; b) there is an empty slot left of the buffer.a) Assume that there is no empty slot left of the buffer. Every of t slots left of thebuffer has a filled cell. Since for any request to an empty cell left of the buffer Bob returnsChooseAny and assigns values for no more than two slots, Bob has already made at least t ChooseAny answers. Also Bob has made k ChooseAny answers for cells to the rightof the buffer, where k is the number of slots to the right of the buffer. Finally, Bob hasmade t + k = Ω( n ) ChooseAny answers.b) If there is an empty slot left of the buffer, Bob chooses the rightmost one anddenote it by S . He divides the slot into two strips n × B . We denote the old buffer by B ′ . We denote the area between S and B ′ by M .Bob is looking for an empty horizontal strip H of height 8 in M , that is not adjacent tolower and upper boundary. Note that if there is no such a strip, then Bob has made Ω( n )answers ChooseAny for cells in M .Bob assigns → to all empty cells from M \ H and fills the parts S \ B and M ∩ H asit is shown in Figure 5.We split the old buffer B ′ into three parts: above, on, and below strip H . Bob assingscells from B ′ in two ways: with Pattern 1 or Pattern 2, shown on Figure 5. Note thatthese two patterns has the following properties: there is no cell in strip H , which is filledwith the same arrow in both patterns. So Bob can answer ChooseAny suggesting twovariants according Patterns 1 or 2.It is easy to see that all invariants are hold and constraints are not violated after suchassignments. Let k be the number of slots that are to the right of buffer B ′ and s is thenumber of slots between old and new buffers. There are k + s + 1 slots to the right ofbuffer B . Bob made at least k ChooseAny answers in the cells that are to the right of10 · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · · · · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · · ................................................ →→→→→→→→ց↓ւ←ր↑տ← →→→→→→→→ց↓ւ←ր↑տ← →→→→→→→→ց↓ւ←ր↑տ← →→→→→→→→ց↓ւ←ր↑տ←← ւ ↓ ց← ւ ↓ ց← ւ ↓ ց← ւ ↓ ց← ւ ↓ ց← ւ ↓ ↓← ւ ւ ւ← ← ← ←← ← ← ←← տ տ տ← տ ↑ ↑← տ ↑ ր← տ ↑ ր← տ ↑ ր← տ ↑ ր← տ ↑ ր ←←←←←←←←←←←←←←←←
Old BufferNew Buffer P a tt e r n Pattern 1 → ց ↓ ւ→ ց ↓ ւ· · ·· · ·· · ·· · ·→ ց ↓ ւ→ ց ↓ ւց ↓ ւ ւ↓ ւ ւ ււ ւ ւ ււ ւ ւ ւ← ← ← ←տ ← ← ←↑ տ ← ←ր ↑ տ ←→ ր ↑ տ→ ր ↑ տ· · ·· · ·· · ·· · ·→ ր ↑ տ→ ր ↑ տ
Pattern 2 ց ↓ ւ ←ց ↓ ւ ←· · ·· · ·· · ·· · ·ց ↓ ւ ←ց ↓ ւ ←↓ ւ ← ←ւ ← ← ←← ← ← ←← ← ← ←տ տ տ տ↑ տ տ տր ↑ տ տ→ ր ↑ տր ↑ տ ←ր ↑ տ ←· · ·· · ·· · ·· · ·ր ↑ տ ←ր ↑ տ ←
Figure 5: Assignments between buffers, patterns for old bufferbuffer B ′ , all s slots between B and B ′ are nonempty. The first request to every of thoseslots got the answer ChooseAny and after recovering of the invariants at most two slotsmay be affected. Thus we have at least s answers ChooseAny for these s slots. Finally,the number of ChooseAny answers to the right of buffer B ′ is at least s + k + 1 > s + k +12 .In the third case c is a left neighbour of the buffer. In this case Bob gives answersChooseAny( → , ց ) as it was in the first case. Bob has to fill the neighbours of cell c by → , but some of them lie in the buffer. Thus we move the buffer as in the second case, butfill it with Pattern 1 (see Figure 5). So we do not violate the constraints. The numberof ChooseAny answers are estimated as follows: k answers to the right of B ′ and at least s +12 others (in contrast to the second case, there is no ChooseAny in B ′ but Bob gavethe ChooseAny answer in at least half of s + 1 remaining slots). Thus there are at least s + k +12 ChooseAny answers to the right of the new buffer B . In this section we revisit the problems from the previous one, but with another compu-tational model. In the first step we will give basic definitions of computational classes
FNP , TFNP and
PPAD . It is already known that
TFNP -problem, based on Sperner’slemma, is
PPAD -complete [3]. This section is devoted to
TFNP -problem, based on ar-rows’ game. We denote it as
ARROWS and give the strict definition below. The main result,presented in this section, that
ARROWS is PPAD -complete.
Definition 4.1 ( FNP and
TFNP ) . Let R ⊂ Σ ∗ × Σ ∗ be a polynomial-time computablerelation such that there exists a polynomial p that for every ( x, y ) ∈ R we have y ≤ p ( | x | ).The NP search problem Q R specified by R is to given input x ∈ Σ ∗ find y ∈ Σ ∗ suchthat ( x, y ) ∈ R , if such y exists, or return ‘no’ otherwise. We use FNP to denote theclass of NP search problems. An NP search problem is said to be total if for every x ,there exists an y such that ( x, y ) ∈ R . We use TFNP to denote the class of total NP search problems. Definition 4.2 (Polynomial Reduction) . A search problem Q R ∈ TFNP is polynomial-time reducible to a search problem Q R ∈ TFNP if there exists a pair of polynomial-time11omputable functions ( f, g ) such that for every input x in Q R , if y satisfies ( f ( x ) , y ) ∈ R ,then ( x, g ( y )) ∈ R . Definition 4.3 ( LEAFD ) . The input of the problem is a pair ( M, k ) where M is thedescription of a polynomial-time Turing machine which satisfies1. for every v ∈ { , } k , M ( v ) is an ordered pair ( u , u ) where u , u ∈ { , } k ∪ { no } ;2. M (0 k ) = ( no, k ) and the first component of M (1 k ) is 0 k . M generates a directed graph G = h V, E i where V = { , } k in the following way. Anedge uv appears in E iff v is the second component of M ( u ) and u is the first componentof M ( v ). The output is a directed leaf (in-degree + out-degree = 1) of graph G which isdifferent from 0 k . PPAD [6] is the set of total NP search problems that are polynomial-time reducibleto LEAFD . By definition,
LEAFD is complete for
PPAD .For every n ≥
1, let B n = { p = ( p , p ) ∈ Z | ≤ p < n and 0 ≤ p < n } .The boundary of B n is then the set of points p ∈ B n with p i ∈ { , n − } for some i ∈ { , } . For every p ∈ Z , we define K p = { q = ( q , q ) ∈ Z | q ∈ [ p , p + 1] and q ∈ [ p , p + 1] } , the 2 × p .An arrow assignment of B n is a function f from B n to A = {↑ , տ , ← , ւ , ↓ , ց , → , ր} .It is said to be valid if for every p ∈ B n on the boundary B n
1. if p = 0, then f ( p ) ∈ {ր , ↑ , տ} ;2. if p = n −
1, then f ( p ) ∈ {ց , ↓ , ւ} ;3. if p = 0, then f ( p ) ∈ {ց , → , ր} ;4. if p = n −
1, then f ( p ) ∈ {ւ , ← , տ} .Notice that for every corner there is only one valid arrow. Definition 4.4 ( ARROWS ) . The input of the problem
ARROWS is a pair ( F, k ) where F is the description of a polynomial-time Turing machine which generates a valid arrowassignment f on B k . Here f ( p ) = F ( p ) ∈ A for every p ∈ B k . The output is a pair ofpoints (p, q) ∈ B k such that p and q are adjacent by edge and the angle between theirarrows more than 45 degree.In order to prove that ARROWS ∈ PPAD we use − BROUWER problem which is verysimilar to
ARROWS . It is known that − BROUWER is PPAD -complete [3].A 3-coloring of B n is a function g from B n to { , , } . It is said to be valid if forevery p on the boundary of B n ,1. if p = 0, then g ( p ) = 2;2. if p = 0 and p = 0, then g ( p ) = 0;3. otherwise, g ( p ) = 1.The search problem − BROUWER is then defined as follows.12 efinition 4.5 ( − BROUWER , [3]) . The input of the problem − BROUWER is a pair( F, k ) where F is the description of a polynomial-time Turing machine which generatesa valid 3-coloring g on B k . Here g ( p ) = F ( p ) ∈ { , , } for every p ∈ B k . The outputis a point p ∈ B k such that K p is trichromatic, that is, K p has all the three colors. Lemma 4.1.
ARROWS ∈ PPAD . Proof.
We divide all arrows in three groups {↓ , ւ , ←} , {→ , ց} , {ր , ↑ , տ} and assigneach group value 0, 1, and 2, respectively. րց տւ→ ←↑↓→ ←↑↓→ ←↑↓→ ←↑↓→ ←↑↓→ ←↑↓→ ←↑↓ Figure 6: Square boundaryNow we surround the current square by arrows as it is shown on Figure 6, and convertnew instance to − BROUWER problem replacing arrows by values 0, 1, and 2 as they aredivided on groups above.Notice that built coloring is valid, and there should be a trichromatic square K p ,which will be an output of − BROUWER problem. By construction square K p can nottouch the border of the big square.On the other hand if square 2 × PPAD -hardness we use planar version of
LEAFD problem.For every n ≥ V n = { u = ( u , u ) | ≤ u < n and 0 ≤ u < n } . Definition 4.6 ( RLEAFD ) . The input instance is a pair ( K, k ) where K is the descriptionof a polynomial-time Turing machine which satisfies:1. For every u ∈ V k K ( u ) is an ordered pair ( u , u ) where u , u ∈ V k ∪ { no } ;2. K ((0 , no, (1 , K ((1 , , K generates a directed graph G = ( V k , E ) on the grid in the following way. ( u, v ) isan edge iff v is the second component of K ( u ), u is the first component of K ( v ) and | u − v | + | u − v | = 1.The output is a directed leaf (with in-degree + out-degree = 1) of graph G which isdifferent from the origin (0 , RLEAFD problem is
PPAD -complete [3].
Lemma 4.2.
ARROWS is PPAD -hard. 13 roof.
We build a polynomial reduction from
RLEAFD to ARROWS using the following gad-gets. Let G = h V k , E i be a graph from the defition of RLEAFD . We build the followinggrid as it is shown on Figure 7. Empty rectangle on the figure contains 2 k × k blocks,where every block is a 9 × i, j ) according the information aboutvertex ( i, j ) in graph G and edges that are incident to ( i, j ) (see Figure 8). ց ↓ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւց ↓ ւ ←ց ↓ ւ ←ց ↓ ւ ←ց ↓ ւ ←ց ↓ ւ ←ց ↓ ւ ←ց ↓ ւ ← ←←←←←←←←←←←←←←←տ տ տ տ տ տ տ տ տ տ տ տտ· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · · · · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·· · ·ր ↑ տր ↑ տ ←ց ↓ ւ ւց ↓ ↓ ↓ց ց ց ց→ → → →ր ր ր րր ↑ ↑ ↑ր ↑ տ տ տ B l o c k , Figure 7: Border gadgetIf the degree of the node is zero, then each cell in the block contains ← . Otherwisewe build a path of → that goes through the center of the block in direction of thecorresponding path in graph G (see Figure 9).It is easy to see that there are no any conflicts on the borders of the blocks andwith the border gadget. So the only places with conflicts are ends of the paths in graph G . A value in any cell can be computed in polynomial time using a description of apolynomial-time Turing Machine from the instance of RLEAFD ; it is also easy to computein polynomial time a leaf in
RLEAFD from the coordinates of the conflict in
ARROWS .The proof of the next theorem follows from Lemma 4.1 and 4.2.
Theorem 4.1.
ARROWS is PPAD -complete.
Acknowledgements
The authors are grateful to Alexander Shen for fruitfull discussions and the statement ofthe problem and also thank Mikhail Slabodkin for helpfull comments.
References [1] Andrew B. Baker. Intelligent backtracking on constraint satisfaction problems: Ex-perimental and theoretical results, 1995.[2] Maria Luisa Bonet, Juan Luis Esteban, Nicola Galesi, and Jan Johannsen. On therelative complexity of resolution refinements and cutting planes proof systems.
SIAMJ. Comput. , 30(5):1462–1484, May 2000.143] Xi Chen and Xiaotie Deng. On the complexity of 2d discrete fixed point problem.
Electronic Colloquium on Computational Complexity (ECCC) , 13(037), 2006.[4] P. Crescenzi and R. Silvestri. Sperner’s lemma and robust machines.
ComputationalComplexity , 7(2):163–173, May 1998.[5] Cho Yee Joey Hwang. A Theoretical Comparison of Resolution Proof Systems forCSP Algorithms. Master’s thesis, Simon Fraser University, 2004.[6] Christos H. Papadimitriou. On the complexity of the parity argument and otherinefficient proofs of existence.
J. Comput. Syst. Sci. , 48(3):498–532, June 1994.[7] E. Sperner. Fifty years of further development of a combinatorial lemma.
W Forster(Ed.), Numerical Solution of Highly Nonlinear Problems, North-Holland, Amsterdam,New York , 8:183–217, 1980.[8] Alasdair Urquhart. The depth of resolution proofs.
Studia Logica , 99(1-3):249–364,2011. 15
Proofs from Section 3.2
The unsatisfiability of ψ n follows from Brower’s theorem, but we give an alternative proofthat will be used in proving upper bound on the size of a contradiction search tree.Let us consider arrows in two adjacent by edge cells and suppose the first arrow shouldbe rotate on x degree in the clockwise way to get the second one. We also suppose that x is minimal in absolute value angle. The rotation between two arrows is signed x .Now let us consider a closed path on the board, that goes through cells with commonedges. We go along the path and calculate the sum of rotations between neughbouringarrows. Since the path returns back to the initial cell, the total rotation is divisible by360 ◦ . Lemma A.1.
Consider a rectangle, with arrows in its cells, such that the total rotationon the closed path going along the boundary of the rectangle is nonzero. Then there aretwo cells with the common edge such that the angle between arrows in them is more than45 ◦ . Proof.
We give the proof by induction on the size of rectangle. The base of the inductionis 2 × R with total rotation along the boundary S . Wedivide rectangle R on two parts as it is shown on Figure 10. Let S and S be totalrotation along the boundary of smaller rectangles R and R . Note that S + S = S .Hence for one of the smaller rectangle the total rotation is nonzero. Theorem A.1.
CSP formula ψ n is unsatisfiable. Proof.
We prove that the total rotation along the boundary of n × n sqaure is 360 ◦ .Consider upper left cell A and lower right cell B . Let the total rotation along the upperpath from A to B is l + 360 ◦ k , where 0 ≤ l < ◦ , k ∈ Z . An arrow in the cell A belongs to set {→ , ց , ↓} and one in the cell B belongs to set {← , տ , ↑} , therefore l = 0.Note that if k = 0, then on the path from A to B there exists ր that contradicts to theboundary constraints. So we have that the total rotation from A to B is positive and lessthan 360 ◦ . Similarly the total rotation along lower path from B to A is positive and lessthan 360 ◦ . Thus the total rotation along the closed path is positive and less than 720 ◦ .But the total rotation should be divisible by 360 ◦ , therefore it equals 360 ◦ .The Theorem follows from Lemma A.1. Corollary A.1.
For the CSP formula ψ n there exists a contradiction search tree of depth O ( n ) and therefore of size 2 O ( n ) . Proof.
Make a request to all boundary cells. After it we split a square into two approxi-mate equal parts by a column. Since the total rotation on the boundary of two parts isnonzero, then by Lemma A.1 one of these parts has nonzero rotation and therefore con-tains a contradiction. We split the contradictory part by a row and reduce the problem ofa contradiction search to a rectangle of size at most ( n + 1) × ( n + 1) with known arrowson the boundary. So we make approximately 1 . n requests and reduce the problem totwo times smaller problem. The depth of the resulting tree is O ( n ).16 ← ւ ↓ ց → ր ↑ տ ←← ւ ↓ ց → ր ↑ տ տ← ւ ↓ ց → ր ↑ ↑ ↑← ւ ↓ ց → ր ր ր ր← ւ ↓ ց → → → → →← ւ ↓ ց ց ց ց ց ց← ւ ↓ ↓ ↓ ↓ ↓ ↓ ↓← ւ ւ ւ ւ ւ ւ ւ ւ← ← ← ← ← ← ← ← ← ⇒ ← ← ← ← ← ← ← ← ←տ տ տ տ տ տ տ տ տ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ր ր ր ր ր ր ր ր ր→ → → → → → → → →ց ց ց ց ց ց ց ց ց↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ւ ւ ւ ւ ւ ւ ւ ւ ւ← ← ← ← ← ← ← ← ← ⇒ ← տ ↑ ր → ց ↓ ւ ←տ տ ↑ ր → ց ↓ ւ ←↑ ↑ ↑ ր → ց ↓ ւ ←ր ր ր ր → ց ↓ ւ ←→ → → → → ց ↓ ւ ←ց ց ց ց ց ց ↓ ւ ←↓ ↓ ↓ ↓ ↓ ↓ ↓ ւ ←ւ ւ ւ ւ ւ ւ ւ ւ ←← ← ← ← ← ← ← ← ←⇒ ← ւ ↓ ց → ր ↑ տ ←← ւ ↓ ց → ր ↑ տ ←← ւ ↓ ց → ր ↑ տ ←← ւ ↓ ց → ր ↑ տ ←← ւ ↓ ց → ր ↑ տ ←← ւ ↓ ց → ր ↑ տ ←← ւ ↓ ց → ր ↑ տ ←← ւ ↓ ց → ր ↑ տ ←← ւ ↓ ց → ր ↑ տ ← ⇒ ← ← ← ← ← ← ← ← ←← ← ← ← ← ← ← ← ←← ← ← ← ← ← ← ← ←← ← ← ← ← ← ← ← ←← ← ← ← ← ← ← ← ←← ← ← ← ← ← ← ← ←← ← ← ← ← ← ← ← ←← ← ← ← ← ← ← ← ←← ← ← ← ← ← ← ← ← ⇒ ← տ ↑ ր → ց ↓ ւ ←← տ ↑ ր → ց ↓ ւ ←← տ ↑ ր → ց ↓ ւ ←← տ ↑ ր → ց ↓ ւ ←← տ ↑ ր → ց ↓ ւ ←← տ ↑ ր → ց ↓ ւ ←← տ ↑ ր → ց ↓ ւ ←← տ ↑ ր → ց ↓ ւ ←← տ ↑ ր → ց ↓ ւ ←⇒ ← ← ← ← ← ← ← ← ←← ւ ւ ւ ւ ւ ւ ւ ւ← ւ ↓ ↓ ↓ ↓ ↓ ↓ ↓← ւ ↓ ց ց ց ց ց ց← ւ ↓ ց → → → → →← ւ ↓ ց → ր ր ր ր← ւ ↓ ց → ր ↑ ↑ ↑← ւ ↓ ց → ր ↑ տ տ← ւ ↓ ց → ր ↑ տ ← ⇒ ← ← ← ← ← ← ← ← ←ւ ւ ւ ւ ւ ւ ւ ւ ւ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ց ց ց ց ց ց ց ց ց→ → → → → → → → →ր ր ր ր ր ր ր ր ր↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑տ տ տ տ տ տ տ տ տ← ← ← ← ← ← ← ← ← ⇒ ← ← ← ← ← ← ← ← ←ւ ւ ւ ւ ւ ւ ւ ւ ←↓ ↓ ↓ ↓ ↓ ↓ ↓ ւ ←ց ց ց ց ց ց ↓ ւ ←→ → → → → ց ↓ ւ ←ր ր ր ր → ց ↓ ւ ←↑ ↑ ↑ ր → ց ↓ ւ ←տ տ ↑ ր → ց ↓ ւ ←← տ ↑ ր → ց ↓ ւ ←⇒ ← տ ↑ ր → ց ↓ ւ ←← տ ↑ ր → ց ↓ ւ ւ← տ ↑ ր → ց ↓ ↓ ↓← տ ↑ ր → ց ց ց ց← տ ↑ ր → → → → →← տ ↑ ր ր ր ր ր ր← տ ↑ ↑ ↑ ↑ ↑ ↑ ↑← տ տ տ տ տ տ տ տ← ← ← ← ← ← ← ← ← ⇒ ← ւ ↓ ց → ր ↑ տ ←ւ ւ ↓ ց → ր ↑ տ ←↓ ↓ ↓ ց → ր ↑ տ ←ց ց ց ց → ր ↑ տ ←→ → → → → ր ↑ տ ←ր ր ր ր ր ր ↑ տ ←↑ ↑ ↑ ↑ ↑ ↑ ↑ տ ←տ տ տ տ տ տ տ տ ←← ← ← ← ← ← ← ← ← ⇒ ← տ ↑ ր → ց ↓ ւ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր ր ր ↑ տ ←← տ ↑ ↑ ↑ ↑ ↑ տ ←← տ տ տ տ տ տ տ ←← ← ← ← ← ← ← ← ←⇒ ← ← ← ← ← ← ← ← ←← տ տ տ տ տ տ տ տ← տ ↑ ↑ ↑ ↑ ↑ ↑ ↑← տ ↑ ր ր ր ր ր ր← տ ↑ ր → → → → →← տ ↑ ր → ց ց ց ց← տ ↑ ր → ց ↓ ↓ ↓← տ ↑ ր → ց ↓ ւ ւ← տ ↑ ր → ց ↓ ւ ← ⇒ ← ← ← ← ← ← ← ← ←տ տ տ տ տ տ տ տ ←↑ ↑ ↑ ↑ ↑ ↑ ↑ տ ←ր ր ր ր ր ր ↑ տ ←→ → → → → ր ↑ տ ←ց ց ց ց → ր ↑ տ ←↓ ↓ ↓ ց → ր ↑ տ ←ւ ւ ↓ ց → ր ↑ տ ←← ւ ↓ ց → ր ↑ տ ← ⇒ ← ← ← ← ← ← ← ← ←← տ տ տ տ տ տ տ ←← տ ↑ ↑ ↑ ↑ ↑ տ ←← տ ↑ ր ր ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր → ց ↓ ւ ←⇒ ← ← ← ← ← ← ← ← ←← տ տ տ տ տ տ տ տ← տ ↑ ↑ ↑ ↑ ↑ ↑ ↑← տ ↑ ր ր ր ր ր ր← տ ↑ ր → → → → →← տ ↑ ր ր ր ր ր ց← տ ↑ ↑ ↑ ↑ ↑ ↑ ↓← տ տ տ տ տ տ տ ւ← ← ← ← ← ← ← ← ← ⇒ ← ← ← ← ← ← ← ← ←տ տ տ տ տ տ տ տ ←↑ ↑ ↑ ↑ ↑ ↑ ↑ տ ←ր ր ր ր ր ր ↑ տ ←→ → → → → ր ↑ տ ←ց ր ր ր ր ր ↑ տ ←↓ ↑ ↑ ↑ ↑ ↑ ↑ տ ←ւ տ տ տ տ տ տ տ ←← ← ← ← ← ← ← ← ← ⇒ ← ւ ↓ ց → ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր ր ր ↑ տ ←← տ ↑ ↑ ↑ ↑ ↑ տ ←← տ տ տ տ տ տ տ ←← ← ← ← ← ← ← ← ←⇒ ← ← ← ← ← ← ← ← ←← տ տ տ տ տ տ տ ւ← տ ↑ ↑ ↑ ↑ ↑ ↑ ↓← տ ↑ ր ր ր ր ր ց← տ ↑ ր → → → → →← տ ↑ ր ր ր ր ր ր← տ ↑ ↑ ↑ ↑ ↑ ↑ ↑← տ տ տ տ տ տ տ տ← ← ← ← ← ← ← ← ← ⇒ ← ← ← ← ← ← ← ← ←ւ տ տ տ տ տ տ տ ←↓ ↑ ↑ ↑ ↑ ↑ ↑ տ ←ց ր ր ր ր ր ↑ տ ←→ → → → → ր ↑ տ ←ր ր ր ր ր ր ↑ տ ←↑ ↑ ↑ ↑ ↑ ↑ ↑ տ ←տ տ տ տ տ տ տ տ ←← ← ← ← ← ← ← ← ← ⇒ ← ← ← ← ← ← ← ← ←← տ տ տ տ տ տ տ ←← տ ↑ ↑ ↑ ↑ ↑ տ ←← տ ↑ ր ր ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← ւ ↓ ց → ր ↑ տ ← Figure 8: Block gadget17 ← տ ↑ ր → ց ↓ ւ ←← տ ↑ ր → ց ↓ ւ ւ← տ ↑ ր → ց ↓ ↓ ↓← տ ↑ ր → ց ց ց ց← տ ↑ ր → → → → →← տ ↑ ր ր ր ր ր ր← տ ↑ ↑ ↑ ↑ ↑ ↑ ↑← տ տ տ տ տ տ տ տ← ← ← ← ← ← ← ← ← ← ւ ↓ ց → ր ↑ տ ←ւ ւ ↓ ց → ր ↑ տ ←↓ ↓ ↓ ց → ր ↑ տ ←ց ց ց ց → ր ↑ տ ←→ → → → → ր ↑ տ ←ր ր ր ր ր ր ↑ տ ←↑ ↑ ↑ ↑ ↑ ↑ ↑ տ ←տ տ տ տ տ տ տ տ ←← ← ← ← ← ← ← ← ←← ← ← ← ← ← ← ← ←ւ ւ ւ ւ ւ ւ ւ ւ ←↓ ↓ ↓ ↓ ↓ ↓ ↓ ւ ←ց ց ց ց ց ց ↓ ւ ←→ → → → → ց ↓ ւ ←ր ր ր ր → ց ↓ ւ ←↑ ↑ ↑ ր → ց ↓ ւ ←տ տ ↑ ր → ց ↓ ւ ←← տ ↑ ր → ց ↓ ւ ← ← ← ← ← ← ← ← ← ←← տ տ տ տ տ տ տ ←← տ ↑ ↑ ↑ ↑ ↑ տ ←← տ ↑ ր ր ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← տ ↑ ր → ր ↑ տ ←← ւ ↓ ց → ր ↑ տ ←ր ↑ տ տ տ տ տ տ տ տ տ տ տ տ տ տ տ տ տ տ տ տ տր ↑ տ տ ←ր ↑ տ տ ←ր ↑ ↑ ↑ ←ր ր ր ր ←→ → → → ←ց ց ց ց ←ց ↓ ւ ↓ ←ց ↓ ւ ւ ←ց ↓ ւ ← ←ց ↓ ւ ← ←ց ↓ ւ ← ←ց ↓ ւ ← ←ց ↓ ւ ← ←ց ↓ ւ ← ←ց ↓ ւ ← ←ց ↓ ւ ← ←ց ↓ ւ ← ←ց ↓ ւ ← ←ց ↓ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւ ւ
Figure 9: Example (1 , , , , ,
5) (2 , , , , ,
5) (3 , , , , ,
5) (4 , , , , ,
5) (5 , , , , ,
5) (6 , , , , ,
5) (7 , , , , , R R Figure 10: Division of R into R and R2