We consider several algorithmic problems concerning geodesics in finitely
generated groups. We show that the three geodesic problems considered by
Miasnikov et al [arXiv:0807.1032] are polynomial-time reducible to each other.
We study two new geodesic problems which arise in a previous paper of the
authors and Fusy [arXiv:0902.0202] .Comment: 6 page

Elder, Murray

Rechnitzer, Andrew

English

arXiv

Murray Elder

Andrew Rechnitzer

Crossref

Groups – Complexity – Cryptology

Some geodesic problems in groups

We consider several algorithmic problems concerning geodesics in finitely generated groups. We show that the three geodesic problems considered by Miasnikov et al. are polynomial-time reducible to each other. We study two new geodesic problems which arise in a previous paper of the authors and Fusy

University of Queensland eSpace

Groups Complex. Cryptol. 2 (2010), 223–229
DOI 10.1515/GCC.2010.014 © de Gruyter 2010
Some geodesic problems in groups
Murray Elder and Andrew Rechnitzer
Abstract. We consider several algorithmic problems concerning geodesics in finitely gen-
erated groups. We show that the three geodesic problems considered by Miasnikov et al.
are polynomial-time reducible to each other. We study two new geodesic problems which
arise in a previous paper of the authors and Fusy.
Keywords. Word problem, geodesic problems.
2010 Mathematics Subject Classification. 20F10, 20F65.
1 Introduction
The study of algorithmic problems in group theory goes back to Dehn [3], and has
been a key theme in much of combinatorial and geometric group theory since. In
a recent paper Miasnikov et al. [11] considered three algorithmic problems con-
cerning geodesics in finitely generated groups. In this article we show that these
three problems are in fact equivalent, in the sense that each is polynomial time
reducible to the others. We consider two more related problems which arose in
work of the authors with Éric Fusy, in computing the growth and geodesic growth
rates of groups [6]. We show that these new problems have efficient polynomial
time solutions for a large class of groups.
The two problems concern how adding a generator to a geodesic effects length.
Let hG j Ri be a presentation for a group G, with G finite, and `.u/ the length of
a geodesic representative for a word u 2 .G ˙1/.
Problem 1. Given a geodesic u and generator x 2 G , decide whether `.ux/ 
`.u/ D 0; 1 or 1.
Note that if all the relators in R have even length, then `.ux/ D `.u/ would
imply a word of odd length equal to the identity, so `.ux/  `.u/ D 0 is not an
option. So the problem turns into the decision problem:
Problem 2. Given a geodesic u and generator x 2 G , decide if `.ux/ > `.u/.
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224 M. Elder and A. Rechnitzer
In [6] we use the fact that Thompson’s group F has an efficient (time
O.n2 log n/ and space O.n log n/) solution to Problem 1, together with a tech-
nique from [10], to compute the number of geodesics and elements up to length
22. We remark that the same procedure will work efficiently for any group and
generating set that has an efficient (polynomial space and time) solution to Prob-
lem 1.
This article is organised as follows. In Section 2 we consider the geodesic
problems of Miasnikov et al. [11]. We prove that they are polynomially reducible
to each other, and that Problems 1 and 2 polynomially reduce to them. In Section 3
we prove that if R is countably enumerable, the problems of Miasnikov et al.
reduce to Problems 1 and 2, this time not necessarily in polynomial time and space.
This implies a solvable word problem, and so we have examples (from [2] for
instance) for which Problems 1 and 2 are unsolvable. In Section 4 we describe
groups which have polynomial time solutions to Problems 1 and 2 and give some
open problems.
2 Geodesic problems of Miasnikov et al.
In [11] Miasnikov et al. consider the following algorithmic and decision problems
for a group G with finite generating set G .
Problem 3 (Geodesic problem). Given a word in G ˙1, find a geodesic representa-
tive for it.
Problem 4 (Geodesic length problem). Given a word in G ˙1, find the length of a
geodesic representative.
Problem 5 (Bounded geodesic length problem). Given a word in G ˙1 and an in-
teger k, decide if a geodesic representative has length  k.
They show that for free metabelian groups (with standard generating sets), Prob-
lem 5 is NP-complete. They also show that a polynomial time solution to any of
these problems implies a polynomial time solution to the next, and each implies a
polynomial time solution to the word problem for the group.
Proposition 1. If G is a group with finite generating set G , then Problems 3–5 of
Miasnikov et al. are polynomial time and space reducible to each other.
Proof. It is clear that a solution to Problem 3 solves Problem 4, and a solution to
Problem 4 solves Problem 5. Suppose we can solve Problem 4 in time f .n/ and
space g.n/. Then consider the following solution to Problem 3. Given a word u
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Some geodesic problems in groups 225
of length n in the finite (inverse closed) generating set G , with jG j D k, apply the
solution to Problem 4 to u. If it returns n, then u is a geodesic. Else the output is
m < n. Pick a generator x and run Problem 4 on ux. If the output is not m  1,
then pick another x. If the output is m  1, set x1 D x. For i D 2; : : : ; m, pick a
generator x, and run Problem 4 on ux1 : : : xi1. If output is m i then set xi D x,
else pick again. After m iterations we have ux1x2 : : : xm has length 0 so is equal
to the identity, so x1m : : : x12 x11 is a geodesic for u. Each iteration takes at most
kf .n C i/ time, and i  m < n, so in total this algorithm takes nkf .2n/ time,
and space g.n C m/.
Next, suppose we can solve the Problem 5 in time f .n/ and space g.n/. Then
consider the following solution to Problem 4. Given a word u of length n, run
the solution to Problem 5 on the pair .u; n  1/. If the output is No, then u is a
geodesic, so output n. While the answer is Yes, run Problem 5 on .u; n  i/ for
i D 2; 3; : : : until the answer is No, and thus the length of a geodesic for u is n i .
The total time is at most nf .n/ and space g.n/.
This answers Problem 5.3 in [11].
Proposition 2. A polynomial time and space solution to Problem 5 implies a poly-
nomial time and space solution to Problem 1.
Proof. Given a geodesic word u D u1 : : : u`.u/ and a generator x 2 G , run Prob-
lem 5 on .ux; `.u/  1/. If it returns Yes, then `.ux/  `.u/ D 1. If not, then
run Problem 5 on .ux; `.u//. If this returns Yes, `.ux/  `.u/ D 0. If not, then
`.ux/  `.u/ D 1.
The converse of this proposition, that a polynomial time and space solution to
Problem 2 implies a polynomial solution to Problems 3–5, is not obvious. In the
next section we prove that the solvability of Problem 2 implies some solution to
Problems 3–5, but not preserving time and space complexity.
3 Word problem
If G D hG j Ri is recursively presented, by which we mean R is countably enu-
merable, and has a solution to Problem 1 or 2, then a very brute force procedure
which runs through all possible words can solve the geodesic problem (Problem 3).
It follows that Problems 4 and 5, and the word problem, are solvable. Since there
are many groups with unsolvable word problem, including finitely presented ex-
amples [2], then there are groups for which Problems 1 and 2 are unsolvable. This
also shows that a solution to Problem 2 implies some solution to Problem 1 for
any recursively presented group.
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226 M. Elder and A. Rechnitzer
The following proof was described to us by Volker Diekert and Andrew Duncan.
The notation u DG v means that u and v represent the same group element.
Proposition 3. If G D hG j Ri with G finite and R countably enumerable, and
has a solution to Problem 2, then Problem 3 is solvable.
Proof. We proceed by induction on the length of the input word u to Problem 3.
If juj D 0 then clearly u is geodesic. Assume for all words of length n  0 we
can find a geodesic representative, and consider the word w of length nC1. Write
w D ux where x is a generator. So we can find a geodesic representative v for u
by inductive assumption.
Input v; x into Problem 2. If it returns `.vx/ > `.x/, then vx is a geodesic
for w. If not, then we know that vx DG z for some word z of length  n. So
1 DG z.vx/1 DG c1c2 : : : ck , for some k, where each ci is a conjugate of an
element of R, and so z DG c1c2 : : : ck.vx/.
Since R is countably enumerable, so too is the set of conjugates, as is the set of
products of conjugates. If the list of products of conjugates is p1; p2; : : : then run-
ning through p1.vx/; p2.vx/; : : : , freely reducing each one, we must eventually
find a word of freely reduced length  n equal to z DG vx.
If this word has length n  1 then it must be a geodesic for vx (since v was
geodesic of length n). Otherwise it has length n, so applying the inductive as-
sumption we can find a geodesic for it.
4 Examples
We end by considering groups that have efficient solutions to Problems 1 and 2.
As noted in the introduction, Thompson’s group F has an efficient solution to
Problem 4 (and thus Problems 1–5). This result is essentially due to Fordham [8],
who first introduced a simple technique for computing the geodesic length of an
element represented as a pair of rooted binary trees. The implementation used in
[6] which gives the O.n2 log n/ time and O.n log n/ space solution is the version
of Fordham’s technique due to Belk and Brown [1]. So what other groups have
polynomial time and space solutions to Problems 1 and 2?
Proposition 4. If the full set of geodesics for a group with a finite generating set
forms a regular language, and one can construct the corresponding finite state
automaton, then Problem 2 is solvable in linear time and constant space.
Proof. Given a word u and a generator x, read ux into the finite state automaton.
ux is accepted if and only if `.ux/ > `.u/. The space required is proportional to
the number of states in the automaton.
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Some geodesic problems in groups 227
It follows that all groups satisfying the falsification by fellow traveler prop-
erty [12] have a linear time solution to Problem 2, provided the fellow traveling
constant is known. Examples include hyperbolic groups, abelian groups, Coxeter
groups, and virtually abelian groups with some generating sets [12].
An automatic structure on a group is strongly geodesic if it includes the set of
all geodesics [7]. In this case our argument to solve Problem 2 may not work
– if ux is rejected by the acceptor automaton then we know `.ux/  `.u/, but
non-geodesic words may also be accepted. More generally, we may ask:
Open Question 1. Does every automatic group have a polynomial time and space
solution to Problems 1 and 2?
In [5] the first author gives an algorithm to compute a geodesic representative
for a word in the solvable Baumslag–Solitar groups ha; t j tat1 D ani in linear
time and space, thus solving Problem 3. By [9] the full set of geodesics for these
groups (with this generating set) fails to be regular. An intriguing open problem is
the following:
Open Question 2. For the non-solvable Baumslag–Solitar groups ha; t jtamt1 D
ani with jmj; jnj > 1, is there an algorithm to solve any of Problems 1–5 in
polynomial time and space?
In the case m D ˙n the group is isomorphic to Fjnj  Z so the answer is
yes (such groups enjoy the falsification by fellow traveler property, for example).
Recently Volker Diekert and Jürn Laun partially answered this question – they
give a clever geodesic normal form which yields a polynomial solution for the
cases where n divides m [4].
As noted, for presentations which have only even length relators, Problems 1
and 2 are solved by the same algorithm. For a presentation with odd length relators,
it is not clear that a solution to Problem 2 solves Problem 1 in the same time and
space complexity.
Open Question 3. Is there an example with odd length relators where Problems 1
and 2 do not have the same time and space complexities?
And lastly
Open Question 4. Is there an example where Problems 1 and/or 2 have polynomial
time (and space) solutions, but problems 3–5 are superpolynomial?
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228 M. Elder and A. Rechnitzer
Acknowledgments. We thank Volker Diekert and Andrew Duncan for explain-
ing how Problems 1 and 2 imply a solvable word problem and geodesic problem
for recursively presented groups. Thanks to Alexei Miasnikov, Vladimir Shpil-
rain and Sasha Ushakov for many fruitful discussions about this work, and to the
anonymous reviewer for their careful proofreading and suggestions.
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Internat. J. Algebra Comput. 15 (2005), 815–850.
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[3] M. Dehn. Über unendliche diskontinuierliche Gruppen. (German) Math. Ann. 71
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[5] M. Elder. A linear time algorithm to compute geodesics in solvable Baumslag–
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[10] E. J. Janse van Rensburg and A. Rechnitzer. Generalised Atmospheric Rosenbluth
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Received June 11, 2010.
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Some geodesic problems in groups 229
Author information
Murray Elder, Mathematics, University of Queensland, Brisbane, Australia.
E-mail: murrayelder@gmail.com
Andrew Rechnitzer, Department of Mathematics, University of British Columbia,
Vancouver, Canada.
E-mail: andrewr@math.ubc.ca
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We consider several algorithmic problems concerning geodesics in finitely generated groups. We show that the three geodesic problems considered by Miasnikov et al. are polynomial-time reducible to each other. We study two new geodesic problems which arise in a previous paper of the authors and Fusy. © de Gruyter 2010

Elder, M

Rechnitzer, A

OPUS - University of Technology Sydney

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Some geodesic problems in groups

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