New Inapproximability Bounds for TSP
arXiv:1303.6437v2 [cs.CC] 10 Jun 2013
Marek Karpinski∗
Michael Lampis†
Richard Schmied‡
Abstract
In this paper, we study the approximability of the metric Traveling Salesman Problem
(TSP) and prove new explicit inapproximability bounds for that problem. The best
up to now known hardness of approximation bounds were 185/184 for the symmetric
case (due to Lampis) and 117/116 for the asymmetric case (due to Papadimitriou and
Vempala). We construct here two new bounded occurrence CSP reductions which
improve these bounds to 123/122 and 75/74, respectively. The latter bound is the first
improvement in more than a decade for the case of the asymmetric TSP. One of our
main tools, which may be of independent interest, is a new construction of a bounded
degree wheel amplifier used in the proof of our results.
1
Introduction
The Traveling Salesman Problem (TSP) is one of the best known and most fundamental
problems in combinatorial optimization. Determining how well it can be approximated in
polynomial time is therefore a major open problem, albeit one for which the solution still
seems elusive. On the algorithmic side, the best known efficient approximation algorithm
for the symmetric case is still a 35-year old algorithm due to Christofides [C76] which
achieves an approximation ratio of 3/2. However, recently there has been a string of
improved results for the interesting special case of Graphic TSP, improving the ratio to
7/5 [OSS11, MS11, M12, SV12]. For the asymmetric case (ATSP), it is not yet known if a
constant-factor approximation is even possible, with the best known algorithm achieving
a ratio of O(log n/ log log n) [AGM+ 10].
Unfortunately, there is still a huge gap between the algorithmic results mentioned
above and the best currently known hardness of approximation results for TSP and ATSP.
For both problems, the known inapproximability thresholds are small constants (185/184
∗
Dept. of Computer Science and the Hausdorff Center for Mathematics, University of Bonn. Supported in
part by DFG grants and the Hausdorff Grant EXC59-1/2. Email: marek@cs.uni-bonn.de
†
KTH Royal Institute of Technology. Research supported by ERC Grant 226203 Email: mlampis@kth.se
‡
Dept. of Computer Science, University of Bonn. Work supported by Hausdorff Doctoral Fellowship.
Email: schmied@cs.uni-bonn.de
1
and 117/116 (cf. [L12, PV00]), respectively). In this paper, we try to improve this situation somehow by giving modular hardness reductions that slightly improve the hardness
bounds for both problems to 123/122 and 75/74, respectively. The latter bound is the first,
for more than a decade now, improvement of Papadimitriou and Vempala bound [PV00]
for the ATSP. The method of our solution differs essentially from that of [PV00] and uses
some new paradigms of the bounded occurrence optimization which could be also of independent interest in other applications. Similarly to [L12], the hope is that the modularity
of our construction, which goes through an intermediate stage of a bounded-occurrence
Constraint Satisfaction Problem (CSP), will allow an easier analysis and simplify future
improvements. Indeed, one of the main new ideas we rely on is a certain new variation of
the wheel amplifiers first defined by Berman and Karpinski [BK01] to establish inapproximability for 3-regular CSPs. This construction, which may be of independent interest,
allows us to establish inapproximability for a 3-regular CSP with a special structure. This
special structure then makes it possible to simulate many of the constraints in the produced
graph essentially “for free”, without using gadgets to represent them. Thus, even though
for the remaining constraints we mostly reuse gadgets which have already appeared in the
literature, we are still able to obtain improved bounds.
Let us now recall some of the previous work on the hardness of approximation of
TSP and ATSP. Papadimitriou and Yannakakis [PY93] were the first to construct a reduction that, combined with the PCP Theorem [ALM+ 98], gave a constant inapproximability
threshold, though the constant was not more than 1 + 10−6 for the TSP with distances
either one or two. Engebretsen [E03] gave the first explicit approximation lower bound
of 5381/5380 for the problem. The inapproximability factor was improved to 3813/3812
by Böckenhauer and Seibert [BS00], who studied the restricted version of the TSP with
distances one, two and three. Papadimitriou and Vempala [PV00] proved that it is NP hard to approximate the TSP with a factor better than 220/219. Presently, the best known
approximation lower bound is 185/184 due to Lampis [L12].
The important restriction of the TSP, in which we consider instances with distances
between cities being values in {1, . . . , B}, is often referred to as the (1, B)-TSP. The best
known efficient approximation algorithm for the (1, 2)-TSP has an approximation ratio
8/7 and is due to Berman and Karpinski [BK06]. As for lower bounds, Engebretsen and
Karpinski [EK06] gave inapproximability thresholds for the (1, B)-TSP problem of 741/740
for B = 2 and 389/388 for B = 8. More recently, Karpinski and Schmied [KS12, KS13] obtained improved inapproximability factors for the (1, 2)-TSP and the (1, 4)-TSP of 535/534
and 337/336, respectively.
For ATSP the currently best known approximation lower bound was 117/116 due to
Papadimitriou and Vempala [PV00]. When we restrict the problem to distances with values
in {1, . . . , B}, there is a simple approximation algorithm with approximation ratio B that
constructs an arbitrary tour as solution. Bläser [B04] gave an efficient approximation
algorithm for the (1, 2)-ATSP with approximation ratio 5/4. Karpinski and Schmied [KS12,
KS13] proved that it is NP -hard to approximate the (1, 2)-ATSP and the (1, 4)-ATSP within
any factor less than 207/206 and 141/140, respectively. For the case B = 8, Engebretsen
2
and Karpinski [EK06] gave an inapproximability threshold of 135/134.
Overview: In this paper we give a hardness proof which proceeds in two steps. First,
we start from the MAX-E3-LIN2 problem, in which we are given a system of linear equations mod 2 with exactly three variables in each equation and we want to find an assignment such as to maximize the number of satisfied equations. Optimal inapproximability
results for this problem were shown by Håstad [H01]. We reduce this problem to a special case where variables appear exactly 3 times and the linear equations have a particular
structure. The main tool here is a new variant of the wheel amplifier graphs of Berman
and Karpinski [BK01].
In the second step, we reduce this 3-regular CSP to TSP and ATSP. The general construction is similar in both cases, though of course we use different gadgets for the two
problems. The gadgets we use are mostly variations of gadgets which have already appeared in previous reductions. Nevertheless, we manage to obtain an improvement by
exploiting the special properties of the 3-regular CSP. In particular, we show that it is only
necessary to construct gadgets for roughly one third of the constraints of the CSP instance,
while the remaining constraints are simulated without additional cost using the consistency properties of our gadgets. This idea may be useful in improving the efficiency of
approximation-hardness reductions for other problems.
Thus, overall we follow an approach unlike that of [PV00], where the reduction is
performed in one step, and closer to [L12]. The improvement over [L12] comes mainly
from the idea mentioned above, which is made possible using the new wheel amplifiers,
as well as several other tweaks. The end result is a more economical reduction which
improves the bounds for both TSP and ATSP.
An interesting question may be whether our techniques can also be used to derive
improved inapproximability results for variants of the ATSP and TSP (cf. [EK06],[KS13]
and [KS12]) or other graph problems, such as the Steiner Tree problem.
2
Preliminaries
In the following, we give some definitions concerning directed (multi-)graphs and omit the
corresponding definitions for undirected (multi-)graphs if they follow from the directed
case. Given a directed graph G = (V
(G), E(G)) and E ′ ⊆ E(G), for e = (x, y) ∈ E(G), we
S
define V (e) = {x, y} and V (E ′ ) = e∈E ′ V (e). For convenience, we abbreviate a sequence
of edges (x1 , x2 ), (x2 , x3 ), . . . , (xn−1 , xn ) by x1 → x2 → x3 → . . . → xn−1 → xn . In the
undirected case, we use sometimes x1 − x2 − x3 − . . . − xn−1 − xn instead of {x1 , x2 },
{x2 , x3 }, . . . , {xn−1 , xn }. Given a directed (multi-)graph G, an Eulerian cycle in G is a
directed cycle that traverses all edges of G exactly once. We refer to G as Eulerian, if there
exists an Eulerian cycle in G. For a multiset ET of directed edges and v ∈ V (ET ), we
define the outdegree (indegree) of v with respect to ET , denoted by outdT (v) (indT (v)),
to be the number of edges in ET that are outgoing of (incoming to) v. The balance of
a vertex v with respect to ET is defined as balT (v) = indT (v) − outdT (v). In the case of
a multiset ET of undirected edges, we define the balance balT (v) of a vertex v ∈ V (ET )
3
to be one if the number of incident edges in ET is odd and zero otherwise. We refer to
vertices v ∈ V (ET ) with balT (v) = 0 as balanced with respect to ET . It is well known that
a (directed) (multi-)graph G = (V (G), E(G)) is Eulerian if and only if all edges are in the
same (weakly) connected component and all vertices v ∈ V (G) are balanced with respect
to E(G).
Given a multiset of edges ET , we denote by conT the number of (weakly) connected
components in the graph induced by ET . A quasi-tour ET in a (directed) graph G is
a multiset of edges from E(G) such that all vertices are balanced with respect to ET and
V (ET ) = V (G). We refer to a quasi-tour ET in G as a tour if conP
T = 1. Given a cost function
w : E(G) → R+ , the cost of a quasi-tour ET in G is defined by e∈ET w(e) + 2(conT − 1).
In the Asymmetric Traveling Salesman problem (ATSP), we are given a directed graph
G = (V (G), E(G)) with positive weights on edges and P
we want to find an ordering
v1 , . . . , vn of the vertices such as to minimize dG (vn , v1 ) + i∈[n−1] dG (vi , vi+1 ), where dG
denotes the shortest path distance in G.
In this paper, we will use the following equivalent reformulation of the ATSP: Given a
directed graph G with weights on edges, we want to find a tour ET in G, that is, a spanning
connected multi-set of edges that balances all vertices, with minimum cost.
The metric Traveling Salesman problem (TSP) is the special case of the ATSP, in which
instances are undirected graphs with positive weights on edges.
3
Bi-Wheel Amplifiers
In this section, we define the bi-wheel amplifier graphs which will be our main tool for
proving hardness of approximation for a bounded occurrence CSP with some special properties. Bi-wheel amplifiers are a simple variation of the wheel amplifier graphs given in
[BK01]. Let us first recall some definitions (see also [BK03]).
If G is an undirected graph and X ⊂ V (G) a set of vertices, we say that G is a ∆-regular
amplifier for X if the following conditions hold:
• All vertices of X have degree ∆ − 1 and all vertices of V (G)\X have degree ∆.
• For every non-empty subset U ⊂ V (G), we have the condition that |E(U, V (G)\U)| ≥
min{ |U ∩ X|, |(V (G)\U) ∩ X| }, where E(U, V (G)\U) is the set of edges with exactly
one endpoint in U.
We refer to the set X as the set of contact vertices and to V (G)\X as the set of checker
vertices. Amplifier graphs are useful in proving inapproximability for CSPs, in which every
variable appears a bounded number of times. Here, we will rely on 3-regular amplifiers.
A probabilistic argument for the existence of such graphs was given in [BK01], with the
definition of wheel amplifiers.
A wheel amplifier with 2n contact vertices is constructed as follows: first construct
a cycle on 14n vertices. Number the vertices 1, . . . , 14n and select uniformly at random
a perfect matching of the vertices whose number is not a multiple of 7. The matched
4
vertices will be our checker vertices, and the rest our contacts. It is easy to see that the
degree requirements are satisfied.
Berman and Karpinski [BK01] gave a probabilistic argument to prove that with high
probability the above construction indeed produces an amplifier graph, that is, all partitions of the sets of vertices give large cuts. Here, we will use a slight variation of this
construction, called a bi-wheel.
A bi-wheel amplifier with 2n contact vertices is constructed as follows: first construct
two disjoint cycles, each on 7n vertices and number the vertices of each 1, . . . , 7n. The
contacts will again be the vertices whose number is a multiple of 7, while the remaining
vertices will be checkers. To complete the construction, select uniformly at random a
perfect matching from the checkers of one cycle to the checkers of the other.
Intuitively, the reason that amplifiers are a suitable tool here is that, given a CSP instance, we can use a wheel amplifier to replace a variable that appears 2n times with 14n
new variables (one for each wheel vertex) each of which appears 3 times. Each appearance
of the original variable is represented by a contact vertex and for each edge of the wheel
we add an equality constraint between the corresponding variables. We can then use the
property that all partitions give large cuts to argue that in an optimal assignment all the
new vertices take the same value.
We use the bi-wheel amplifier in our construction in a similar way. The main difference is that while cycle edges will correspond to equality constraints, matching edges will
correspond to inequality constraints. The contacts of one cycle will represent the positive appearances of the original variable, and the contacts of the other the negative ones.
The reason we do this is that we can encode inequality constraints more efficiently than
equality with a TSP gadget, while the equality constraints that arise from the cycles will
be encoded in our construction “for free” using the consistency of the inequality gadgets.
Before we apply the construction however, we have to prove that the bi-wheel amplifiers still have the desired amplification properties.
Theorem 1. With high probability, bi-wheels are 3-regular amplifiers.
Proof. Exploiting the similarity between bi-wheels and the standard wheel amplifiers of
[BK01], we will essentially reuse the proof given there. First, some definitions: We say
that U is a bad set if the size of its cut is too small, violating the second property of
amplifiers. We say that it is a minimal bad set if U is bad but removing any vertex from U
gives a set that is not bad.
Recall the strategy of the proof from [BK01]: for each partition of the vertices into U
and V (G)\U, they calculate the probability (over the random matchings) that this partition
gives a minimal bad set. Then, they take the sum of these probabilities over all potentially
minimal bad sets and prove that the sum is at most γ −n for some constant γ < 1. It follows
by union bound that with high probability, no set is a minimal bad set and therefore, the
graph is a proper amplifier.
Our first observation is the following: consider a wheel amplifier on 14n vertices
where, rather than selecting uniformly at random a perfect matching among the checkers, we select uniformly at random a perfect matching from checkers with labels in the set
5
{1, . . . , 7n − 1} to checkers with labels in the set {7n + 1, . . . , 14n − 1}. This graph is almost
isomorphic to a bi-wheel. More specifically, for each bi-wheel, we can obtain a graph of
this form by rewiring two edges, and vice-versa. It easily follows that properties that hold
for this graph, asymptotically with high probability also hold for the bi-wheel.
Thus, we just need to prove that a wheel amplifier still has the amplification property
if, rather than selecting a random perfect matching, we select a random matching from
one half of the checker vertices to the other. We will show this by proving that, for each set
of vertices S, the probability that S is a minimal bad set is roughly the same in both cases.
After establishing this fact, we can simply rely on the proof of [BK01].
Recall that the wheel has 12n checker vertices. Given a set S with |S| = u, what is
the probability that exactly c edges have exactly one endpoint in S? In a standard wheel
amplifier the probability is
u 12n − u c!(u − c)!!(12n − u − c)!!
P (u, c) =
,
c
c
(12n)!!
where we denote by n!! the product of all odd natural numbers less than or equal to n, and
we assume without loss of generality that u − c is even. Let us explain this: the probability
that exactly c edges cross the cut in this graph is equal to the number of ways we can
choose their endpoints in S and in its complement, times the number of ways we can
match the endpoints, times the number of matchings of the remaining vertices, divided by
the number of matchings overall.
How much does this probability change if we only allow matchings from one half of
the checkers to the other? Intuitively, we need to consider two possibilities: one is that S
is a balanced set, containing an equal number of checkers from each side, while the other
is that S is unbalanced. It is not hard to see that if S is unbalanced, then, we can easily
establish that the cut must be large. Thus, the main interesting case is the balanced one
(and we will establish this fact more formally).
Suppose that |S| = u and S contains exactly u/2 checkers from each side. Then the
probability that there are exactly c edges crossing the cut is
u 2 12n−u 2 u−c 12n−u−c
!
c 2 2 !
′
2
2
!
P (u, c) = 2c
c
2
(6n)!
2
2
Let us explain this. If S is balanced and there are c matching edges with exactly one endpoint in S, then, exactly c/2 of them must be incident on a vertex of S on each side, since
the remaining vertices of S must have a perfect matching. Again, we pick the endpoints
on each side, and on the complement of S, select a way to match them, select matchings
on the remaining vertices and divide by the number of possible
perfect matchings.
√
n 2
−n
Using
Stirling
formulas,
it
is
not
hard
to
see
that
!
=
Θ(n!2
n). Also n!! =
2
n/2
n
′
Θ( 2 !2 ). It follows that P is roughly the same as P in this case, modulo some polynomial factors which are not significant since the probabilities we are calculating are exponentially small.
6
Let us now also show that if S is unbalanced, the probability that it is a minimal bad
set is even smaller. First, observe that if S is a minimal bad set whose cut has c edges, we
have c ≤ u/6. The reason for this is that since S is bad, then, c is smaller than the number
of contacts in S minus the number of cycle edges cut. It is not hard to see that, in each
fragment, that is, each subset of S made up of a contiguous part of the cycle, two cycle
edges are cut. Thus, the extra edges we need for the contacts the fragment contains are at
most 1/6 of its checkers.
Suppose now that S contains u/2 + k checkers on one side and u/2 − k checkers on the
other. The probability that c matching edges have one endpoint in S is
u
u
12n−u
12n−u
c
u−c ! 12n−u−c !
c
+
k
−
k
−
k
+
k
2
2
2
2
+k !
−k ! 2
P ′′ (u, c, k) = 2c
c
c
c
2
2
(6n)!
+
k
−
k
−
k
+
k
2
2
2
2
The reasoning is the same as before, except we observe that we need to select more endpoints on the side where S is larger, since after we remove checkers matched to outside
vertices S must have a perfect matching. Observe that for k = 0 this gives P ′ . We will
show that for the range of values we care about P ′′ achieves a maximum for k = 0, and
can thus be upper-bounded by (essentially) P , which is the probability that a set is bad in
the standard amplifier. The rest of the proof follows from the argument given in [BK01].
In particular, we can assume that k ≤ c/2, since 2k edges are cut with probability 1. To
show that the maximum is achieved for k = 0, we look at P ′′ (u, c, k + 1)/P ′′(u, c, k). We
will show that this is less than 1. Using the identity n+1
, we get
/ nk = n+1
k+1
k+1
P ′′ (u, c, k + 1)
=
P ′′ (u, c, k)
2k + 1
2k + 1
2k + 1
1+ u
1 + 12n−u
1− c
+k+1
−k
−k
2
2
2
Using the fact that 1 + x < ex , we end up needing to prove the following.
c
2
2k + 1
2k + 1
2k + 1
> u
+ 12n−u
+k+1
−k
−k
2
2
(1)
Combining that without loss of generality u ≤ 6n holds with the bounds of c and k we
have already mentioned, the inequality (1) is straightforward to establish.
4
Hybrid Problem
By using the bi-wheel amplifier from the previous section, we are going to prove hardness
of approximation for a bounded occurrence CSP with very special properties. This particular CSP will be well-suited for constructing a reduction to the TSP given in the next
section.
As the starting point of our reduction, we make use of the inapproximability result due
to Håstad [H01] for the MAX-E3LIN2 problem, which is defined as follows: Given a system
I1 of linear equations mod 2, in which each equation is of the form xi ⊕ xj ⊕ xk = bijk with
7
bijk ∈ {0, 1}, we want to find an assignment to the variables of I1 such as to maximize the
number of satisfied equations.
Let I1 be an instance of the MAX-E3LIN2 problem and {xi }νi=1 the set of variables, that
appear in I1 . We denote by d(i) the number of appearances of xi in I1 .
Theorem 2 (Håstad [H01]). For every ǫ > 0, there exists a constant Bǫ such that given an
instance I1 of the MAX-E3LIN2 problem with m equations and maxi∈[ν] d(i) ≤ Bǫ , it is NP hard to decide whether there is an assignment that leaves at most ǫ · m equations unsatisfied,
or all assignment leave at least (0.5 − ǫ)m equations unsatisfied.
Similarly to the work by Berman and Karpinski [BK99] (see also [BK01] and [BK03]),
we will reduce the number of occurrences of each variable to 3. For this, we will use our
amplifier construction to create special instances of the Hybrid problem, which is defined
as follows: Given a system I2 of linear equations mod 2 with either three or two variables in
each equation, we want to find an assignment such as to maximize the number of satisfied
equations.
In particular, we are going to prove the following theorem.
Theorem 3. For every constant ǫ > 0 and b ∈ {0, 1}, there exist instances of the Hybrid
problem with 31m equations such that: (i) Each variable occurs exactly three times. (ii) 21m
equations are of the form x ⊕ y = 0, 9m equations are of the form x ⊕ y = 1 and m equations
are of the form x ⊕ y ⊕ z = b . (iii) It is NP -hard to decide whether there is an assignment
to the variables that leaves at most ǫ · m equations unsatisfied, or every assignment to the
variables leaves at least (0.5 − ǫ)m equations unsatisfied.
Proof. Let ǫ > 0 be a constant and I1 an instance of the MAX-E3LIN2 problem with
maxi∈[ν] d(i) ≤ Bǫ . For a fixed b ∈ {0, 1}, we can flip some of the literals such that all
equations in the instance I1 are of the form x ⊕ y ⊕ z = b, where x, y, z are variables or
negations. By constructing three more copies of each equation, in which all possible pairs
of literals appear negated, we may assume that each variable occurs the same number of
times negated as unnegated.
Let us fix a variable xi in I1 . Then, we create 7 · d(i) = 2 · α new variables V ar(i) =
ui
α
{xj , xni
j }j=1 . In addition, we construct a bi-wheel amplifier Wi on 2 · α vertices (that is,
a bi-wheel with d(i) contact vertices) with the properties described in Theorem 1. Since
d(i) ≤ Bǫ is a constant, this can be accomplished in constant time. In the remainder,
we refer to contact and checker variables as the elements in V ar(i), whose corresponding
index is a contact and checker vertex in Wi , respectively. We denote by M(Wi ) ⊆ E(Wi )
the associated perfect matching on the set of checker vertices of Wi . In addition, we
denote by Cn (Wi ) and Cu (Wi ) the set of edges contained in the first and second cycle of
Wi , respectively.
Let us now define the equations of the corresponding instance of the Hybrid problem.
ni
For each edge {j, k} ∈ M(Wi ), we create the equation xui
j ⊕ xk = 1 and refer to equations
of this form as matching equations. On the other hand, for each edge {l, t} in the cycle
qi
Cq (Wi ) with q ∈ {u, n}, we introduce the equation xqi
l ⊕ xt = 0. Equations of this form will
8
be called cycle equations. Finally, we replace the j-th unnegated appearance of xi in I1 by
the contact variable xui
λ with λ = 7 · j, whereas the j-th negated appearance is replaced by
ni
xλ . The former construction yields m equations with three variables in the instance of the
Hybrid problem, which we will denote by I2 . Notice that each variable appears in exactly
3 equations in I2 . Clearly, we have |I2 | = 31m equations, thereof 9m matching equations,
21m cycle equations and m equations of the form x ⊕ y ⊕ z = b.
ni
A consistent assignment to V ar(i) is an assignment with xui
j = b and xj = (1 − b) for
all j ∈ [α], where b ∈ {0, 1}. A consistent assignment to the variables of I2 is an assignment that is consistent to V ar(i) for all i ∈ [ν]. By standard arguments using the amplifier
constructed in Theorem 1, it is possible to convert an assignment to a consistent assignment without decreasing the number of satisfied equations and the proof of Theorem 3
follows.
5
TSP
This section is devoted to the proof of the following theorem.
Theorem 4. It is NP -hard to approximate the TSP to within any constant approximation
ratio less than 123/122.
Let us first sketch the high-level idea of the construction. Starting with an instance of
the Hybrid problem, we will construct a graph, where gadgets represent the equations.
We will design gadgets for equations of size three (Figure 1) and for equations of size
two corresponding to matching edges of the bi-wheel (Figure 2). We will not construct
gadgets for the cycle edges of the bi-wheel; instead, the connections between the matching
edge gadgets will be sufficient to encode these extra constraints. This may seem counterintuitive at first, but the idea here is that if the gadgets for the matching edges are used in
a consistent way (that is, the tour enters and exits in the intended way) then it follows that
the tour is using all edges corresponding to one wheel and none from the other. Thus, if
we prove consistency for the matching edge gadgets, we implicitly get the cycle edges “for
free”. This observation, along with an improved gadget for size-three equations and the
elimination of the variable part of the graph, are the main sources of improvement over
the construction of [L12].
5.1 Construction
We are going to describe the construction that encodes an instance I2 of the Hybrid problem into an instance of the TSP problem. Due to Theorem 3, we may assume that the
equations with three variables in I2 are all of the form x ⊕ y ⊕ z = 0.
In order to ensure that some edges are to be used at least once in any valid tour, we
apply the following simple trick that was already used in the work by Lampis [L12]: Let e
be an edge with weight w that we want to be traversed by every tour. We remove e and
9
replace it with a path of L edges and L − 1 newly created vertices each of degree two,
where we think of L as a large constant. Each of the L edges has weight w/L and any tour
that fails to traverse at least two newly created edges is not connected. On the other hand,
a tour that traverses all but one of those edges can be extended by adding two copies of the
unused edge increasing the cost of the underlying tour by a negligible value. In summary,
we may assume that our construction contains forced edges that need to be traversed at
least once by any tour. If x and y are vertices, which are connected by a forced edge e, we
write {x, y}F or simply x−F y. In the following, we refer to unforced edges e with w(e) = 1
as simple. All unforced edges in our construction will be simple.
Let us start with the description of the corresponding graph GS : For each bi-wheel Wp ,
we will construct the subgraph Gp of GS . For each vertex of the bi-wheel, we create a
vertex in the graph and for each cycle equation x ⊕ y = 0, we create a simple edge {x, y}.
Given a matching equation between two checkers xui ⊕ xnj = 1, we connect the vertices xui
and xnj with two forced edges {xui , xnj }1F and {xui , xnj }2F . We have w({xui , xnj }iF ) = 2 for each
i ∈ {1, 2}.
Additionally, we create a central vertex s that is connected to gadgets simulating equations with three variables. Let x ⊕ y ⊕ z = 0 be the j-th equation with three variables in
I2 . We now create the graph G3S
j displayed in Figure 1 (a), where the (contact) vertices
for x, y, z have already been constructed in the cycles. The edges {γ α , γ}F with α ∈ {r, l}
and γ ∈ {x, z, y} are all forced edges with w({γ α , γ}F ) = 1.5. Furthermore, we have
w({eαj , s}F ) = 0.5 for all α ∈ {r, l}. {erj , s}F and {elj , s}F are both forced edges, whereas all
remaining edges of G3S
j are simple. This is the whole description of GS .
tj
x
elj
xl
xr
e1j
erj
y
yr
yl
xl z r
s
xr l
y
z
yr
e2j
zl
e3j
zr
zl
s
sj
(a)
(b)
Figure 1: Gadgets simulating equations with three variables in the symmetric case (a) and
in the asymmetric case (b). Dotted and straight lines represent forced and simple edges,
respectively.
10
5.2 Tour from Assignment
Given an instance I2 of the Hybrid problem and an assignment φ to the variables in I2 , we
are going to construct a tour in GS according to φ and give the proof of one direction of
the reduction. In particular, we are going to prove the following lemma.
Lemma 1. If there is an assignment to the variables of a given instance I2 of the Hybrid
problem with 31m equations and ν bi-wheels, that leaves k equations unsatisfied, then, there
exists a tour in GS with cost at most 61m + 2ν + k + 2.
Before we proceed, let us give a useful definition. Let G be an edge-weighted graph
and ET a multi-set of edges of E(G) that defines a quasi-tour. Consider a set V ′ ⊆ V (G).
The local edge cost of the set V ′ is then defined as
cT (V ′ ) =
X
X
u∈V ′ e∈ET , e={u,v}
w(e)
2
In words, for each vertex in V ′ , we count half the total weight of its incident edges used
in the quasi-tour (including multiplicities). Observe that this sum contains half the weight
of edges with one endpoint in V ′ but the full weight for edges with both endpoints in V ′
(since we count both endpoints in the sum). Also note that for two sets V1 , P
V2 , we have
cT (V1 ∪ V2 ) ≤ cT (V1 ) + cT (V2 ) (with equality for disjoint sets) and that cT (V ) = e∈ET w(e).
Proof. First, note that it is sufficient to prove that we can construct a quasi-tour of the
promised cost which uses all forced edges exactly once. Since all unforced edges have
cost 1, if we are given a quasi-tour we can connect two disconnected components by using
an unforced edge that connects them twice (this is always possible since the underlying
graph we constructed is connected). This does not increase the cost, since we added two
unit-weight edges and decreased the number of components. Repeating this results in a
connected tour.
Let {Wa }νa=1 be the associated set of bi-wheels of I2 . For a fixed bi-wheel Wp , let
{xui , xni }zi=1 be its associated set of variables. Due to the construction of instances of the
Hybrid problem in Section 4, we may assume that all equations with two variables are
satisfied by the given assignment. Thus, we have xui 6= xnj , xui = xuj and xni = xnj for all
i, j ∈ [z].
Assuming xα1 = 1 for some α ∈ {u, n}, we use once all simple edges {xαi , xαi+1 } with
i ∈ [z − 1] and the edge {xαz , xα1 }. We also use all forced edges corresponding to matching
equations once. In other words, for each biwheel we select the cycle that corresponds to
the assignment 1 and use all the simple edges from that cycle. This creates a component
that contains all checker vertices from both cycles and all contacts from one cycle.
As for the next step, we are going to describe the tour traversing G3S
j with j ∈ [m]
simulates
x ⊕ y ⊕ z = 0.
given an assignment to contact variables. Let us assume that G3S
j
3S
According to the assignment to x, y and z, we will traverse Gj as follows: In all cases, we
will use all forced edges once.
Case (x + y + z = 2): Then, we use {γ l , γ r } for all α ∈ {r, l} and γ ∈ {x, y, z} with γ = 1.
11
xui−1
xui
xui+1
xnj−1
xnj
xnj+1
xnu(i−1)
xun(j−1)
(a)
xui
xnj
xnj+1
xui+1
(b)
Figure 2: Gadget simulating equation with two variables in symmetric case (a) and in
the asymmetric case (b). Dotted and straight lines represent forced and simple edges,
respectively.
For δ ∈ {x, z, y} with δ = 0, we use {eαj , δ α } for all α ∈ {r, l}.
Case (x + y + z = b with b ∈ {0, 1}): In both cases, we traverse {γ α , eαj } for all γ ∈ {x, y, z}
and α ∈ {r, l}.
Case (x + y + z = 3): We use {γ r , γ l } with γ ∈ {y, z}. Furthermore, we include {xα , eαj }
for both α ∈ {r, l}.
Let us now analyze the cost of the edges of our quasi-tour given an assignment. For
each matching edge {xui , xnj } consider the set of vertices made up of its endpoints. Its local
cost is 5: we pay 4 for the forced edges and there are two used simple edges with one
endpoint in the set. Let us also consider the local cost for a size-three equation gadget,
where we consider the set to contain the contact vertices {x, y, z} as well the other 8
vertices of the gadget. The local cost here is 9.5 for the forced edges. We also pay 6 more
(for a total of 15.5) when the assignment satisfies the equation or 7 more when it does
not.
Thus, we have given a covering of the vertices of the graph by 9m sets of size two, m sets
of size 11 and {s}. The total edge cost is thus at most 5·9m+15.5·m+0.5·m+k = 61m+k.
To obtain an upper bound on the cost of the quasi-tour, we observe that the tour has at
most ν + 1 components (one for each bi-wheel and one containing s). The lemma follows.
5.3 Assignment from Tour
In this section, we are going prove the other direction of our reduction. Given a tour in
GS , we are going to define an assignment to the variables of the associated instance of the
Hybrid problem and give the proof of the following lemma.
Lemma 2. If there is a tour in GS with cost 61m + k − 2, then, there is an assignment to the
variables of the corresponding instance of the Hybrid problem that leaves at most k equations
unsatisfied.
12
Again, let us give a useful definition. Consider a quasi-tour ET and a set V ′ ⊆ V (G). Let
conT (V ′ ) be the number of connected components induced by ET which are fully contained
in V ′ . Then, the full local cost of the set V ′ is defined as cFT (V ′ ) = cT (V ′ ) + 2conT (V ′ ). By
the definition, the full local cost of V (G) is equal to the cost of the quasi-tour (plus 2).
Intuitively, cFT (V ′ ) captures the cost of the quasi-tour restricted to V ′ : it includes the
cost of edges and the cost of added connected components. Note that now for two disjoint
sets V1 , V2 we have cFT (V1 ∪V2 ) ≥ cFT (V1 )+cFT (V2 ) since V1 ∪V2 could contain more connected
components than V1 , V2 together. If we know that the total cost of the quasi-tour is small,
then cFT (V ) is small (less than 61m + k). We can use this to infer that the sum of the local
full costs of all gadgets is small.
The high-level idea of the proof is the following: we will use roughly the same partition
of V (G) into sets as in the proof of Lemma 1. For each set, we will give a lower bound on
its full local cost for any quasi-tour, which will be equal to what the tour we constructed in
Lemma 1 pays. If a given quasi-tour behaves differently its local cost will be higher. The
difference between the actual local cost and the lower bound is called the credit of that
part of the graph. We construct an assignment for I2 and show that the total sum of credits
is higher that the number of unsatisfied equations. But using the reasoning of the previous
paragraph, the total sum of credits will be at most k.
Proof. We are going to prove a slightly stronger statement and show that if there exists a
quasi-tour in GS with cost 61m + k − 2, then, there exists an assignment leaving at most
k equations unsatisfied. Recall that the existence of a tour in GS with cost C implies the
existence of a quasi-tour in GS with cost at most C.
We may assume that simple edges are contained only once in ET due to the following
preprocessing step: If ET contains two copies of the same simple edge, we remove them
without increasing the cost, since the number of components can only increase by one.
In the following, given a quasi-tour ET in GS , we are going to define an assignment φT
and analyze the number of satisfied equations by φT compared to the cost of the quasi-tour.
The general idea is that each vertex of GS that corresponds to a variable of I2 has exactly
two forced and exactly two simple edges incident to it. If the forced edges are used once
each, the variable is called honest. We set it to 1 if the simple edges are both used once
and to 0 otherwise. It is not hard to see that, because simple cycle edges connect vertices
that represent the variables, this procedure will satisfy all cycle equations involving honest
variables. We then argue that if other equations are unsatisfied the tour is also paying
extra, and the same happens if a variable is dishonest.
Let us give more details. First, we concentrate on the assignment for checker variables.
Assignment for Checker Variables
Let us consider the following equations with two variables xui−1 ⊕ xui = 0, xui ⊕ xui+1 = 0,
xnj−1 ⊕ xnj = 0, xnj ⊕ xnj+1 = 0 and xui ⊕ xnj = 1. We are going to analyze the cost of a quasitour traversing the gadget displayed in Figure 2 (a) and define an assignment according
13
to ET . Let us first assume that our quasi-tour is honest, that is, the underlying quasi-tour
traverses forced edges only once.
Honest tours: For x ∈ {xui , xnj }, we set x = 1 if the quasi-tour traverses both simple
edges incident on x and x = 0, otherwise. Since we removed all copies of the same
simple edge, we may assume that cycle equations are always satisfied. If the tour uses
xui−1 − xui −F xnj −F xui − xui+1 , we get xui−1 = xui+1 = 1, xnj−1 = xnj+1 = 0 and 5 satisfied
equations. Given xnj−1 − xnj −F xui −F xnj − xnj+1 , we obtain 5 satisfied equations as well.
Let us define Vip := {xui , xnj }. Notice that in both cases, we have local cost cFT (Vip ) = 5.
We claim that cFT (Vip ) ≥ 5 for a valid quasi-tour. In order to obtain a valid quasi-tour, we
need to traverse both forced edges in Gpi and use at least two simple edges, as otherwise,
it implies cFT (Gpi ) ≥ 6. Given a quasi-tour ET , we introduce a local credit function defined
by crT (Vip ) = cFT (Vip ) − 5. If xui −F xnj −F xui forms a connected component, we get 4
satisfied equations and crT (Vip ) = 1, which is sufficient to pay for the unsatisfied equation
xui ⊕ xnj = 1. On the other hand, assuming xui−1 = xui+1 = 1 and xnj−1 = xnj+1 = 1, we get
crT (Vip ) = 1 and 1 unsatisfied equation.
Dishonest tours: We are going to analyze quasi-tours, which are using one of the forced
edges twice. By setting xui 6= xnj , we are able to find an assignment that always satisfies
xui ⊕ xnj = 1 and two other equations out of the five that involve these dishonest variables.
The local cost in this case is at least 7. Hence, the credit crT (Vip ) = 2 is sufficient to pay for
the two unsatisfied equations.
Assignment for Contact Variables
Again, we will distinguish between honest tours (which use forced edges exactly once) and
dishonest tours. This time we are interested in seven equations: the size-three equation
x ⊕ y ⊕ z = 0 and the six cycle equations containing the three contacts.
Observe that the local cost of Vj3S := {xr , xl , x, y r , y l, y, z r , z l , z, erj , elj } is at least 15.5.
The local edge cost of any quasi-tour is 9.5 for the forced edges. For each component
{γ, γ l , γ r } with γ ∈ {x, y, z}, we need to pay at least 2 more because there are two vertices
with odd degree (γ l , γ r ) and we also need to connect the component to the rest of the
graph (otherwise the component already costs 2 more). Let us define the credit of Vj3S
with respect to ET by crT = cFT (Vj3S ) − 15.5.
Honest tours: For each γ ∈ {x, y, z}, we set γ = 1 if the tour uses both simple
edges incident on γ and 0, otherwise. Notice that in the case (x + y + z = b) with b ∈ {0, 2},
this satisfies all seven equations and the tour has local cost at least cFT (Vj3S ) = 15.5.
Case (x = y = z = 1) : The assignment now failed to satisfy the size-three equation,
so we need to prove that the quasi-tour has local cost at least 16.5. Since all vertices are
balanced with respect to ET , the quasi-tour has to use at least one edge incident on erj and
elj besides {s, erj }F and {s, elj }F . If the quasi-tour takes {eαj , γ α } for a γ ∈ {x, y, z} and all
α ∈ {r, l}, since all simple edges incident on x, y, z are used, we get at total cost of at least
16.5, which gives a credit of 1.
Case (x + y + z = 1) : Without loss of generality, we assume that x = y = 0 6= z
14
holds. Again, only the size-three equation is unsatisfied, so we must show that the local
cost is at least 16.5. We will discuss two subcases. (i) There is a connected component
δ −F δ r − δ l −F δ for some δ ∈ {x, y}. We obtain that cFT ({δ, δ l , δ r }) ≥ 6 and therefore, a
lower bound on the total cost of 16.5. (ii) Since we may assume that xr , xl , y r and y l are
balanced with respect to ET , we have that {eαj , γ α } ∈ ET for all α ∈ {r, l} and γ ∈ {x, y}.
Because eαj are also balanced, we obtain {eαj , z α } ∈ ET for all α ∈ {r, l}, which implies a
total cost of 16.5.
Dishonest tours: Let us assume that the quasi-tour uses both of the forced edges {γ r , γ}
and {γ l , γ} for some γ ∈ {x, z, y} twice. We delete both copies and add {γ r , γ l } instead
which reduces the cost of the quasi-tour. Hence, we may assume that only one of the two
incident forced edges is used twice.
First, observe that if all forced edges were used once, then there would be eight vertices
in the gadget with odd degree: xr , xl , y r , y l , z r , z l , erj , elj . If exactly one forced edge is used
twice, then seven of these vertices have odd degree. Thus, it is impossible for the tour to
make the degrees of all seven even using only the simple edges that connect them. We can
therefore assume that if a forced edge is used twice, there exists another forced edge used
twice.
We will now take cases, depending on how many of the vertices x, y, z are incident on
forced edges used twice. Note that if one of the forced edges incident on x is used twice,
then exactly one of the simple edges incident on x is used once. So, first suppose all three
of x, y, z have forced edges used twice. The local cost from forced edges is at least 14.
Furthermore, there are three vertices of the form γ α , for γ ∈ {x, y, z} and α ∈ {l, r} with
odd degree. These have no simple edges connecting them, thus the quasi-tour will use
three simple edges to balance their degrees. Finally, the used simple edges incident on
x, y, z each contribute 0.5 to the local cost. Thus, the total local cost is at least 18.5, giving
us a credit of 3. It is not hard to see that there is always an assignment satisfying four out
of the seven affected equations, so this case is done.
Second, suppose exactly two of x, y, z have incident forced edges used twice, say, x, y.
For z, we select the honest assignment (1 if the incident simple edges are used, 0 otherwise) and this satisfies the cycle equations for this variable. We can select assignments for
x, y that satisfy three of the remaining five equations, so we need to show that the cost in
this case is at least 17.5. The cost of forced edges is at least 12.5, and the cost of simple
edges incident on x, y adds 1 to the local cost. One of the vertices xl , xr and one of y l , y r
have odd degree, therefore the cost uses two simple edges to balance them. Finally, the
vertices z l , z r have odd degree. If two simple edges incident to them are used, we have a
total local cost of 17.5. If the edge connecting them is used, then the two simple edges
incident on z must be used, again pushing the local cost to 17.5.
Finally, suppose only x has an incident forced edge used twice. By the parity argument
given above, this means that one of the forced edges incident on s is used twice. We can
satisfy the cycle equations for y, z by giving them their honest assignment, and out of the
three remaining equations some assignment to x satisfies two. Therefore, we need to
15
show that the cost is at least 16.5. The local cost from forced edges is 11.25 and the simple
edge incident on x contributes 0.5. Also, at least one simple edge incident on xl or xr is
used, since one of them has odd degree. For y l , y r , either two simple edges are used, or if
the edge connecting them is used the simple edges incident on y contribute 1 more. With
similar reasoning for z l , z r , we get that the total local cost is at least 16.75.
Let us now conclude our analysis. Consider the following partition of V : we have a singleton set {s}, 9m sets of size 2 containing the matching edge gadgets and m sets of size
11 containing the gadgets for size-three equations (except s). The sum of their local costs
is at most cFT (V ) ≤ 61m
P + k. But the sum of their local costs is (using the preceding analysis) equal to 61m + crT (Vi ). Thus, the sum of all credits is at most k. Since we have
already argued that the sum of all credits is enough to cover all equations unsatisfied by
our assignment, this concludes the proof.
We are ready to give the proof of Theorem 4.
Proof of Theorem 4. We are given an instance I1 of the MAX-E3LIN2 problem with ν variables and m equations. For all δ > 0, there exists a k such that if we repeat each equa(k)
tion k time we get an instance I1 with m′ = km equations and ν variables such that
2(ν + 1)/m′ ≤ δ.
(k)
Then, from I1 , we generate an instance I2 of the Hybrid problem and the corresponding graph GS . Due to Lemmata 1, 2 and Theorem 3, we know that for all ǫ > 0, it is NP hard to tell whether there is a tour with cost at most 61m′ +2ν +2+ǫ·m′ ≤ 61·m′ +(δ +ǫ)m′
or all tours have cost at least 61m′ + (0.5 − ǫ)m′ − 2 ≥ 61.5 · m′ − ǫ · m′ − δ · m′ . The ratio
between these two cases can get arbitrarily close to 123/122 by appropriate choices for ǫ, δ.
6
ATSP
In this section, we prove the following theorem.
Theorem 5. It is NP -hard to approximate the ATSP to within any constant approximation
ratio less than 75/74.
6.1 Construction
Let us describe the construction that encodes an instance I2 of the Hybrid problem into
an instance of the ATSP. Again, it will be useful to have the ability to force some edges to
be used, that is, we would like to have bidirected forced edges. A bidirected forced edge
of weight w between two vertices x and y will be created in a similar way as undirected
forced edges in the previous section: construct L − 1 new vertices and connect x to y
through these new vertices, making a bidirected path with all edges having weight w/L.
16
It is not hard to see that without loss of generality we may assume that all edges of the
path are used in at least one direction, though we should note that the direction is not
prescribed. In the remainder, we denote a directed forced edge consisting of vertices x and
y by (x, y)F , or x →F y.
Let I2 consist of the collection {Wi }νi=1 of bi-wheels. Recall that the bi-wheel consists of
two cycles and a perfect matching between their checkers. Let {xui , xni }zi=1 be the associated
set of variables of Wp . We write u(i) to denote the function which, given the index of a
checker variable xui returns the index j of the checker variable xnj to which it is matched
(that is, the function u is a permutation function encoding the matching). We write n(i) to
denote the inverse function u−1 (i).
Now, for each bi-wheel Wp , we are going to construct the corresponding directed
graph GpA as follows. First, construct a vertex for each checker variable of the wheel.
For each matching equation xui ⊕ xnj = 1, we create a bidirected forced edge {xui , xnj }F with
w({xui , xnj }F ) = 2.
For each contact variable xk , we create two corresponding vertices xrk and xlk , which
are joined by the bidirected forced edge {xrk , xlk }F with w({xrk , xlk }F ) = 1.
Next, we will construct two directed cycles Cup and Cnp . Note that we are doing arithmetic on the cycle indices here, so the index z + 1 should be read as equal to 1. For Cup , for
any two consecutive checker vertices xui , xui+1 on the un-negated side of the bi-wheel, we
add a simple directed edge xnu(i) → xui+1 . If the checker xui is followed by a contact xui+1 in
ul
u
the cycle, then we add two simple directed edges xnu(i) → xur
i+1 and xi+1 → xi+2 . Observe
that by traversing the simple edges we have just added, the forced matching edges in the
direction xui →F xnu(i) and the forced contact edges for the un-negated part in the direction
ul
xur
i →F xi we obtain a cycle that covers all checkers and all the contacts of the un-negated
part.
We now add simple edges to create a second cycle Cnp . This cycle will require using
the forced matching edges in the opposite direction and, thus, truth assignments will be
encoded by the direction of traversal of these edges. First, for any two consecutive checker
vertices xni , xni+1 on the un-negated side of the bi-wheel, we add the simple directed edge
xun(i) → xni+1 . Then, if the checker xni is followed by a contact xni+1 in the cycle then we add
nl
n
the simple directed edges xun(i) → xnr
i+1 and xi+1 → xi+2 . Now by traversing the edges we
have just added, the forced matching edges in the direction xni →F xun(i) and the forced
contact edges for the negated part in the direction xnr
→F xnl
i
i , we obtain a cycle that
covers all checkers and all the contacts of the negated part, that is, a cycle of direction
opposite to Cup .
What is left is to encode the equations of size three. Again, we have a central vertex s
that is connected to gadgets simulating equations with three variables. For every equation
with three variables, we create the gadget displayed in Figure 1 (b), which is a variant
of the gadget used by Papadimitriou and Vempala [PV00]. Let us assume that the j-th
equation with three variables in I3 is of the form x ⊕ y ⊕ z = 1. This equation is simulated
α
by G3A
j . The vertices used are the contact vertices γ , γ ∈ {x, y, z}, α ∈ {r, l}, which
we have already introduced, as well as the vertices {sj , tj , eij | i ∈ [3]}. For notational
17
simplicity, we define Vj3A = sj , tj , eij , γ α | i ∈ [3], γ ∈ {x, y, z}, α ∈ {r, l} . All directed
non-forced edges are simple. The vertices sj and tj are connected to s by forced edges with
w((s, sj )F ) = w((tj , s)F ) = λ, where λ > 0 is a small fixed constant. To simplify things, we
also force them to be used in the displayed direction by deleting the edges that make up
the path of the opposite direction. This is the whole description of the graph GA .
6.2 Assignment to Tour
We are going to construct a tour in GA given an assignment to the variables of I2 and prove
the following lemma.
Lemma 3. Given an instance I2 of the Hybrid problem with ν bi-wheels and an assignment
that leaves k equations in I2 unsatisfied, then, there exists a tour in GA with cost at most
37m + 5ν + 2mλ + 2νλ + k.
Before we proceed, let us again give a definition for a local edge cost function. Let
G be an edge-weighted digraph and ET a multi-set of edges of E(G) that defines a tour.
Consider a set V ′ ⊆ V (G). The local edge cost of the set V ′ is then defined as
X X
cT (V ′ ) =
w (u, v)
u∈V ′ (u,v)∈ET
In words, for each vertex in V ′ we count the total weight of its outgoing edges used in
the quasi-tour (including multiplicities). Thus, that this sum contains the full weight for
edges with their source in V ′ , regardless of where their other endpoint is. Also note that
again for two sets V1 , VP
2 we have cT (V1 ∪ V2 ) ≤ cT (V1 ) + cT (V2 ) (with equality for disjoint
sets) and that cT (V ) = e∈ET w(e).
Proof of Lemma 3. Let Wp be a bi-wheel with variables {xui , xni }zi=1 . Given an assignment
to the variables of I2 , due to Theorem 3, we may assume that either xui = 1 6= xnj for all
i, j ∈ [z] or xui = 0 6= xnj for all i, j ∈ [z]. We traverse the cycle Cup if xu1 = 1 and the cycle Cnp
otherwise. This creates ν strongly connected components. Each contains all the checkers
of a bi-wheel and the contacts from one side.
For each matching edge gadget, the local edge cost is 3. We pay two for the forced
edge and 1 for the outgoing simple edge. We will account for the cost of edges incident on
contacts when we analyze the size-three equation gadget below.
Let us describe the part of the tour traversing the graph G3A
j , which simulates x⊕y⊕z =
1. Recall that if x is set to true in the assignment we have traversed the bi-wheel gadgets
in such a way that the forced edge xr →F xl is used, and the simple edge coming out of xl
is used.
According to the assignment to x, y and z, we traverse G3A
j as follows:
Case (x + y + z = 1): Let us assume that z = y = 0 6= x holds. Then, we use
s →F sj → e2j → y l →F y r → e3j → z l →F z r → e1j → tj →F s. The cost is 3 + λ for the
forced edges, 6 for the simple edges inside the gadget, plus 1 for the simple edge going out
of xl . Total local edge cost cost: cT (Vj3A ) = 10 + λ.
18
Case (x + y + z = 3): Then, we use s →F sj → e2j → e1j → e3j → tj →F s. Again we pay
3 + λ for the forced edges, 4 for the simple edges inside the gadget and 3 for the outgoing
edges incident on xl , y l , z l . Total local edge cost: cT (Vj3A ) = λ + 10.
Case (x + y + z = 2): Let us assume that x = y = 1 6= z holds. Then, we use
s →F sj → e3j → z l →F z r → e1j → e3j → e2j → tj →F s with total local edge cost
cT (Vj3A ) = λ + 11.
Case (x + y + z = 0): We use s →F sj → e2j → y l →F y r → e3j → z l →F z r → e1j →
xl →F xr → e2j → tj →F s with cT (Vj3A ) = λ + 11.
The total edge cost of the quasi-tour we constructed is 3 · 9m + (10 + 2λ)m + k =
37m + 2λm + k. We have at most ν + 1 strongly connected components: one for each
bi-wheel and one containing s. A component representing a bi-wheel can be connected
to s as follows: let xl , xr be two contact vertices in the component. Add one copy of each
edge from the cycle s →F sj → e1j → xl →F xr → e2j → tj →F s. This increases the cost by
5 + 2λ but decreases the number of components by one.
6.3 Tour to Assignment
In this section, we are going to prove the other direction of the reduction.
Lemma 4. If there is a tour with cost 37 · m + k + 2λ · m, then, there is an assignment that
leaves at most k equations unsatisfied.
Proof. Given a tour ET in GA , we are going to define an assignment to checker and contact
variables. As in Lemma 2, we will show that any tour must locally spend on each gadget
at least the same amount as the tour we constructed in Lemma 3. If the tour spends more,
we use that credit to satisfy possible unsatisfied equations.
Assignment for Checker Variables
Let us consider the following equations with two variables xui ⊕ xui+1 = 0, xui−1 ⊕ xui = 0,
xui ⊕ xnj = 1, xnj ⊕ xnj+1 = 0, xnj−1 ⊕ xnj = 0 and the corresponding situation displayed in
Figure 2 (b). Since ET is a valid tour in GA , we know that {xui , xnj }F is traversed and due
to the degree condition, for each x ∈ {xui , xnj }, the tour uses another incident edge e on x
with w(e) ≥ 1. Therefore, we have that cT ({xui , xnj }) ≥ 3. The credit assigned to a gadget
is defined as crT ({xui , xnj }) = cT ({xui , xnj }) − 3.
Let us define the assignment for xui and xnj . A variable xui is honestly traversed if either
both the simple edge going into xui is used and the simple edge coming out of xnj is used,
or neither of these two edges is used. In the first case, we set xui to 1, otherwise to 0.
Similarly, xnj is honest if both the edge going into xnj and the edge out of xui are used, and
we set it to 1 in the first case and 0 otherwise.
Honest tours: First, suppose that both xui and xnj are honest. We need to show that the
credit is at least as high as the number of unsatisfied equations out of the five equations
19
that contain them. It is not hard to see that if we have set xui 6= xnj all equations are
satisfied. If we have set both to 1, then the forced edge must be used twice, making the
local edge cost at least 6, giving a credit of 3, which is more than sufficient.
Dishonest tours: If both xui and xnj are dishonest the tour must be using the forced
edge in both directions. Thus, the local cost is 5 or more, giving a credit of 2. There is
always an assignment that satisfies three out of the five equations, so this case is done. If
one of them is dishonest, the other must be set to 1 to ensure strong connectivity. Thus,
there are two simple edges used leaving the gadget, making the local cost 4 (perhaps the
same edge is used twice). We can set the honest variable to 1 (satisfying its two cycle
equations), and the other to 0, leaving at most one equation unsatisfied.
Assignment for Contact Variables
First, we note that for any valid tour, we have cT (Vj3A ) ≥ 10 + λ. This is because the two
forced edges of weight λ must be used, and there exist 10 vertices in the gadget for which
all outgoing edges have weight 1. Let us define the credit crT (Vj3A ) = cT (Vj3A ) − (10 + λ).
Honest Traversals: We assume that the underlying tour is honest, that is, forced edges
are traversed only in one direction. We set x to 1 if the forced edge is used in the direction
xr →F xl and 0 otherwise. In the first case we know that the simple edges going into xr
and out of xl are used. In the second, the edges e1j → xl and xr → e2j are used. We do
similarly for y, z.
We are interested in the equation x ⊕ y ⊕ z = 1 and the six cycle equations involving
x, y, z. The assignment we pick for honest variables satisfies the cycle equations, so if it
also satisfies the size-three equation we are done. If not, we have to prove that the tour
pays at least 11 + λ.
Case (x = y = z = 0): Due to our assumption, we know that e2j → y l →F y r → e3j →
z l →F z r → e1j → xl →F xr → e2j is a part of the tour. Since ET is a tour, there exists
a vertex in Vj3A \{sj , tj } that is visited twice and we get cT (Vj3A ) ≥ 11 + λ. Thus, we can
spend the credit crT (Vj3A ) ≥ 1 on the unsatisfied equation x ⊕ y ⊕ z = 1.
Case (x + y + z = 2): Without loss of generality, let us assume that x = y = 1 6= z
holds. Then, we know that e3j → z l →F z r → e1j is a part of the tour. But, this implies that
3A
there is a vertex in V (G3A
j ) that is visited twice. Hence, we have that crT (Vj ) ≥ 1.
Dishonest Traversals: Consider the situation, in which some forced edges {γ r , γ l }F
are traversed in both directions for some variables γ ∈ {x, y, z}. For the honest variables,
we set them to the appropriate value as before, and this satisfies their cycle equations.
Observe now that if a forced edge γ l →F γ r is also used in the opposite direction, then
there must be another edge used to leave the set {γ l , γ r }. Thus the local edge cost of this
set is at least 3. It follows that the credit we have for the gadget is at least as large as the
number of dishonest variables. We can give appropriate values to them so each satisfies
one cycle equation and the size-three equation is satisfied. Thus, the number of unsatisfied
equations is not larger than our credit.
In summary, for every tour ET in GA , we can find an assignment to the variables of I2
such that all unsatisfied equations are paid by the credit induced by ET .
20
We are ready to give the proof of Theorem 5.
Proof of Theorem 5. We are again given an instance I1 of the MAX-E3LIN2 problem with
ν variables and m equations. For all δ > 0, there exists a k such that if we repeat each
(k)
equation k time we get an instance I1 with m′ = km equations and ν variables such that
ν/m′ ≤ δ.
(k)
Then, from I1 , we generate an instance I2 of the Hybrid problem and the corresponding directed graph GA . Due to Lemmata 3, 4 and Theorem 3, we know that for all ǫ > 0, it
is NP -hard to tell whether there is a tour with cost at most 37m′ + 5ν + 2m(ν + λ) + ǫ · m′ ≤
37 · m′ + ǫ′ m′ or all tours have cost at least 37m′ + (0.5 − ǫ)m′ ≥ 37.5 · m′ − ǫ′ · m′ , for some
ǫ′ depending only on ǫ, δ, λ. The ratio between these two cases can get arbitrarily close to
75/74 by appropriate choices for ǫ, δ, λ.
7
Concluding Remarks
In this paper, we proved that it is hard to approximate the ATSP and the TSP within any
constant factor less than 75/74 and 123/122, respectively. The proof method required essentially new ideas and constructions from the ones used before in that context. Since the
best known upper bound on the approximability is O(log n/ log log n) for ATSP and 3/2 for
TSP, there is certainly room for improvements. Especially, in the asymmetric version of the
TSP, there is a large gap between the approximation lower and upper bound, and it remains a major open problem on the existence of an efficient constant factor approximation
algorithm for that problem. Furthermore, it would be nice to investigate if some of the
ideas of this paper, and in particular the bi-wheel amplifiers, can be used to offer improved
hardness results for other optimization problems, such as the Steiner Tree problem.
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Richard Schmied