Lin–Kernighan heuristic explained

In combinatorial optimization, Lin–Kernighan is one of the best heuristics for solving the symmetric travelling salesman problem. It belongs to the class of local search algorithms, which take a tour (Hamiltonian cycle) as part of the input and attempt to improve it by searching in the neighbourhood of the given tour for one that is shorter, and upon finding one repeats the process from that new one, until encountering a local minimum. As in the case of the related 2-opt and 3-opt algorithms, the relevant measure of "distance" between two tours is the number of edges which are in one but not the other; new tours are built by reassembling pieces of the old tour in a different order, sometimes changing the direction in which a sub-tour is traversed. Lin–Kernighan is adaptive and has no fixed number of edges to replace at a step, but favours small numbers such as 2 or 3.

Derivation

For a given instance

(G,c)

of the travelling salesman problem, tours are uniquely determined by their sets of edges, so we may as well encode them as such. In the main loop of the local search, we have a current tour

T\subsetE(G)

and are looking for new tour

T'\subsetE(G)

such that the symmetric difference

F=Tn{\triangle}T'

is not too large and the length

\sum_ec(e)

of the new tour is less than the length

\sum_ec(e)

of the current tour. Since

is typically much smaller than

and

, it is convenient to consider the quantity

g(F)=\sum_ec(e)-\sum_ec(e)

— the gain of using

F\subseteqE(G)

when switching from

—since

g(Tn{\triangle}T')=\sum_ec(e)-\sum_ec(e)

: how much longer the current tour

is than the new tour

. Naively

-opt can be regarded as examining all

F\subseteqE(G)

with exactly

elements (

but not in

, and another

but not in

) such that

Tn{\triangle}F

is again a tour, looking for such a set which has

g(F)>0

. It is however easier to do those tests in the opposite order: first search for plausible

with positive gain, and only second check if

Tn{\triangle}F

is in fact a tour.

Define a trail in

to be alternating (with respect to

) if its edges are alternatingly in

and not in

, respectively. Because the subgraphs

l(V(G),Tr)

and

l(V(G),T'r)

are

-regular, the subgraph

G[Tn{\triangle}T']=l(V(G),Tn{\triangle}T'r)

will have vertices of degree

, and

only, and at each vertex there are as many incident edges from

as there are from

. Hence (essentially by Hierholzer's algorithm for finding Eulerian circuits) the graph

G[Tn{\triangle}T']

decomposes into closed alternating trails. Sets

F\subseteqE(G)

that may satisfy

F=Tn{\triangle}T'

for some tour

may thus be found by enumerating closed alternating trails in

, even if not every closed alternating trail

makes

Tn{\triangle}F

into a tour; it could alternatively turn out to be a disconnected

-regular subgraph.

Key idea

Alternating trails (closed or open) are built by extending a shorter alternating trail, so when exploring the neighbourhood of the current tour

, one is exploring a search tree of alternating trails. The key idea of the Lin–Kernighan algorithm is to remove from this tree all alternating trails which have gain

\leq0

. This does not prevent finding every closed trail with positive gain, thanks to the following lemma.

Lemma. If

a_0,...c,a_n-1

are numbers such that

	n-1
\sum
	i=0

a_i>0

, then there is a cyclic permutation of these numbers such that all partial sums are positive as well, i.e., there is some

such that

	r
\sum
	i=0

a_(k+i)>0

for all

r=0,1,...c,n-1

For a closed alternating trail

F=e₀e₁...e_n-1

, one may define

a_i=c(e_i)

e_i\inT

and

a_i=-c(e_i)

e_i\notinT

; the sum

	n-1
\sum\nolimits
	i=0

a_i

is then the gain

g(F)

. Here the lemma implies that there for every closed alternating trail with positive gain exists at least one starting vertex

v₀

for which all the gains of the partial trails are positive as well, so

will be found when the search explores the branch of alternating trails starting at

v₀

. (Prior to that the search may have considered other subtrails of

starting at other vertices but backed out because some subtrail failed the positive gain constraint.) Reducing the number of branches to explore translates directly to a reduction in runtime, and the sooner a branch can be pruned, the better.

This yields the following algorithm for finding all closed, positive gain alternating trails in the graph.

State: a stack of triples

(u,i,g)

, where

u\inV(G)

is a vertex,

i\geq0

is the current number of edges in the trail, and

is the current trail gain.

For all

u\inV(G)

, push

(u,0,0)

onto the stack.

While the stack is nonempty:
1. Pop

(u,i,g)

off the stack and let

v_i:=u

. The current alternating trail is now

F=\{v₀v_1,v₁v_2,...c,v_i-1v_i\}

1. If

is even then:

For each

u\inV(G)

such that

v_iu\inT\setminus\{v₀v_1,v₁v_2,...c,v_i-1v_i\}

(there are at most two of these), push

l(u,i+1,g+c(v_iu)r)

onto the stack.

1. If instead

is odd then:

1. 1. If

g>c(v_iv₀₎

then report

\{v₀v_1,v₁v_2,...c,v_i-1v_i,v_iv₀\}

as a closed alternating trail with gain

g-c(v_iv₀₎>0

1. 1. For each

u\inV(G)

such that

g>c(v_iu)

and

v_iu\notinT\cup\{v₀v_1,v₁v_2,...c,v_i-1v_i\}

(there may be

O(n)

of these, or there could be none), push

l(u,i+1,g-c(v_iu)r)

onto the stack.

Stop

As an enumeration algorithm this is slightly flawed, because it may report the same trail multiple times, with different starting points, but Lin–Kernighan does not care because it mostly aborts the enumeration after finding the first hit. It should however be remarked that:

Lin–Kernighan is not satisfied with just having found a closed alternating trail

of positive gain, it additionally requires that

Tn{\triangle}F

is a tour.

Lin–Kernighan also restricts the search in various ways, most obviously regarding the search depth (but not only in that way). The above unrestricted search still terminates because at

i=2n

there is no longer any unpicked edge remaining in

, but that is far beyond what is practical to explore.

In most iterations one wishes to quickly find a better tour

, so rather than actually listing all siblings in the search tree before exploring the first of them, one may wish to generate these siblings lazily.

Basic Lin–Kernighan algorithm

The basic form of the Lin–Kernighan algorithm not only does a local search counterpart of the above enumeration, but it also introduces two parameters that narrow the search.

The backtracking depth

p₁

is an upper bound on the length of the alternating trail after backtracking; beyond this depth, the algorithm explores at most one way of extending the alternating trail. Standard value is that
p₁=5

.

The infeasibility depth

p₂

is an alternating path length beyond which it begins to be required that closing the current trail (regardless of the gain of doing so) yields an exchange to a new tour. Standard value is that

p₂=2

.Because there are

	\lfloorp_1/2\rfloor
n

)

alternating trails of length

p₁

, and the final round of the algorithm may have to check all of them before concluding that the current tour is locally optimal, we get

\lfloorp_1/2\rfloor

(standard value

) as a lower bound on the exponent of the algorithm complexity. Lin & Kernighan report

2.2

as an empirical exponent of

in the average overall running time for their algorithm, but other implementors have had trouble reproducing that result.^[1] It appears unlikely that the worst-case running time is polynomial.^[2]

In terms of a stack as above, the algorithm is:

Input: an instance

(G,c)

of the travelling salesman problem, and a tour

T\subsetE(G)

Output: a locally optimal tour

Variables:

a stack of triples

(u,i,g)

, where

u\inV(G)

is a vertex,

i\geq0

is the current number of edges in the trail, and

is the current trail gain,

the sequence

v_0,v_1,...c

of vertices in the current alternating trail,

the best set

of exchange edges found for current tour, and its corresponding gain

g^*

Initialise the stack to being empty.

Repeat

Set

g^*:=0

and

F:=\varnothing

For all

u\inV(G)

, push

(u,0,0)

onto the stack.

While the stack is nonempty:

Pop

(u,i,g)

off the stack and let

v_i:=u

is even then

for each

u\inV(G)

such that

v_iu\inT\setminus\{v₀v_1,v₁v_2,...c,v_i-1v_i\}

push

l(u,i+1,g+c(v_iu)r)

onto the stack if:

i\leqp₂

, or

uv₀\notinT\cup\{v₀v_1,v₁v_2,...c,v_i-1v_i,v_iu\}

and

Tn{\triangle} \{v₀v_1,v₁v_2,...c,v_i-1v_i,v_iu,uv₀\}

is a tour (Hamiltonicity check)

else (

is odd):

g>c(v_iv₀₎

g-c(v_iv₀₎>g^*

, and

Tn{\triangle} \{v₀v_1,v₁v_2,...c,v_i-1v_i,v_iv₀\}

is a tour (Hamiltonicity check) then let

F:=\{v₀v_1,v₁v_2,...c,v_i-1v_i,v_iv₀\}

and

g^*:=g-c(v_iv₀₎

For each

u\inV(G)

such that

g>c(v_iu)

and

v_iu\notinT\cup\{v₀v_1,v₁v_2,...c,v_i-1v_i\}

, push

l(u,i+1,g-c(v_iu)r)

onto the stack.

End if.

Let

(u,j,g)

be the top element on the stack (peek, not pop). If

i\leqj

then

g^*>0

then

set

T:=Tn{\triangle}F

(update current tour) and clear the stack.

else if

i>p₁

then

pop all elements

(u,j,g)

off the stack that have

j>p₁

end if

end while

until

g^*=0

Return

The length of the alternating trails considered are thus not explicitly bounded, but beyond the backtracking depth

p₁

no more than one way of extending the current trail is considered, which in principle stops those explorations from raising the exponent in the runtime complexity.

Limitations

The closed alternating trails found by the above method are all connected, but the symmetric difference

Tn{\triangle}T'

of two tours need not be, so in general this method of alternating trails cannot explore the full neighbourhood of a trail

. The literature on the Lin–Kernighan heuristic uses the term sequential exchanges for those that are described by a single alternating trail. The smallest non-sequential exchange would however replace 4 edges and consist of two cycles of 4 edges each (2 edges added, 2 removed), so it is long compared to the typical Lin–Kernighan exchange, and there are few of these compared to the full set of 4-edge exchanges.

In at least one implementation by Lin & Kernighan there was an extra final step considering such non-sequential exchanges of 4 edges before declaring a tour locally optimal, which would mean the tours produced are 4-opt unless one introduces further constraints on the search (which Lin and Kernighan in fact did). The literature is vague on exactly what is included in the Lin–Kernighan heuristic proper, and what constitutes further refinements.

For the asymmetric TSP, the idea of using positive gain alternating trails to find favourable exchanges is less useful, because there are fewer ways in which pieces of a tour can be rearranged to yield new tours when one may not reverse the orientation of a piece. Two pieces can only be patched together to reproduce the original tour. Three pieces can be patched together to form a different tour in one way only, and the corresponding alternating trail does not extend to a closed trail for rearranging four pieces into a new tour. To rearrange four pieces, one needs a non-sequential exchange.

Checking Hamiltonicity

The Lin–Kernighan heuristic checks the validity of tour candidates

Tn{\triangle}F

at two points: obviously when deciding whether a better tour has been found, but also as a constraint to descending in the search tree, as controlled via the infeasibility depth

p₂

. Concretely, at larger depths in the search a vertex

v_2k+1

is only appended to the alternating trail if

Tn{\triangle}\{v₀v_1,v₁v_2,...c,v_2kv_2k+1,v_2k+1v₀\}

is a tour. By design that set of edges constitutes a 2-factor in

, so what needs to be determined is whether that 2-factor consists of a single Hamiltonian cycle, or instead is made up of several cycles.

If naively posing this subproblem as giving a subroutine the set of

edges as input, one ends up with

O(n)

as the time complexity for this check, since it is necessary to walk around the full tour before being able to determine that it is in fact a Hamiltonian cycle. That is too slow for the second usage of this test, which gets carried out for every alternating trail with more than

edges from

. If keeping track of more information, the test can instead be carried out in constant time.

A useful degree of freedom here is that one may choose the order in which step 2.3.2 iterates over all vertices; in particular, one may follow the known tour

. After picking

edges from

, the remaining subgraph

l(V(G),T\setminus\{v₀v_1,...c,v_2k-2v_2k-1\}r)

consists of

paths. The outcome of the Hamiltonicity test done when considering the

(k+1)

th edge

v_2kv_2k+1

depends only on in which of these paths that

v_2k

resides and whether

v_2k+1

is before or after

v_2k

. Hence it would be sufficient to examine

different cases as part of performing step 2.3.2 for

v_2k-1

; as far as

v_2k+1

is concerned, the outcome of this test can be inherited information rather than something that has to be computed fresh.

References

10.1287/opre.21.2.498. Shen. Lin. Shen Lin. B. W.. Kernighan. Brian Kernighan. 1973 . An Effective Heuristic Algorithm for the Traveling-Salesman Problem . Operations Research . 21. 2 . 498–516.
10.1016/S0377-2217(99)00284-2. K. Helsgaun . An Effective Implementation of the Lin-Kernighan Traveling Salesman Heuristic . European Journal of Operational Research . 126 . 1 . 2000 . 106–130 . 10.1.1.180.1798 .
Book: http://www.research.att.com/~dsj/papers/TSPchapter.pdf. The Traveling Salesman Problem: A Case Study in Local Optimization. David S.. Johnson. Lyle A.. McGeoch. Local Search in Combinatorial Optimization . E. H. L. Aarts . J. K. Lenstra . Jan Karel Lenstra . John Wiley and Sons. London. 1997. 215–310.

External links

Notes and References

Melamed . I. I. . Sergeev . S. I. . Sigal . I. Kh. . The traveling salesman problem. Approximate algorithms . Avtomatika i Telemekhanika . 1989 . 11 . 3–26.
Papadimitriou . C. H. . The complexity of the Lin–Kernighan heuristic for the travelling salesman problem . SIAM Journal on Computing . 1992 . 21 . 3 . 450–465. 10.1137/0221030 .