=Paper= {{Paper |id=Vol-2098/paper14 |storemode=property |title=On Bounded Diameter MST Problem on Random Instances |pdfUrl=https://ceur-ws.org/Vol-2098/paper14.pdf |volume=Vol-2098 |authors=Edward Kh. Gimadi,Alexey M. Istomin,Ekaterina Yu. Shin }} ==On Bounded Diameter MST Problem on Random Instances== https://ceur-ws.org/Vol-2098/paper14.pdf
         On Bounded Diameter MST Problem on
                  Random Instances

       Edward Kh. Gimadi1,2 , Alexey M. Istomin1 , and Ekaterina Yu. Shin2
                            1
                          Sobolev Institute of Mathematics,
                4 Acad. Koptyug avenue, 630090, Novosibirsk, Russia
                          2
                            Novosibirsk State University,
                    2 Pirogova Str. 630090, Novosibirsk, Russia
    gimadi@math.nsc.ru, alexeyistomin@gmail.com, katherine15963@gmail.com



         Abstract. We give an approximation deterministic algorithm for solv-
         ing the Random bounded diameter minimum spanning tree (BDMST)
         problem on an undirected graph. The algorithm has a quadratic time
         complexity. A probabilistic analysis was performed under conditions that
         edge weights of given graph are identically independent uniformly dis-
         tributed random variables on an interval (an ; bn ). Conditions of asymp-
         totic optimality are presented.

         Keywords: Graph · Bounded diameter minimum spanning tree · Min-
         imum spanning tree · Asymptotically optimal algorithm · Probabilistic
         analysis · Performance guarantees · Random inputs · Uniform




1     Introduction

     The Minimum Spanning Tree (MST) problem is a one of the classic discrete
optimization problems. Given undirected weighted graph G = (V, E), MST is to
find a spanning tree of a minimal weight. MST is polynomially solvable, there are
classic algorithms by Boruvka (1926), Kruskal (1956) and Prim (1957). These
algorithms have complexity O(n2 ) and O(M log n) where M = |E| and n = |V |.
    In current paper a modification of the classical MST is studied. We study
a bounded diameter minimum spanning tree problem (BDMST). The goal is
to find in the graph Gn a spanning tree Tn of minimal total weight having its
diameter limited by given number d. The diameter of a tree is the number of
edges on the longest path between two leaves in the tree. This problem is N P -
hard in the common case [10].
    The Bounded Diameter Minimum Spanning Tree Problem has many practical
applications in various fields such as telecommunication networks and linear

     Copyright c by the paper’s authors. Copying permitted for private and academic purposes.
    In: S. Belim et al. (eds.): OPTA-SCL 2018, Omsk, Russia, published at http://ceur-ws.org
160    E. Kh. Gimadi et al.

light wave network design [5], bit compression for information retrieval [7] and
distributed mutual exclusion [28, 30, 31, 34].
    A good example of usage is the Distributed Mutual Exclusion algorithms.
Here we have a computer network of k computers and the internal communi-
cation is done by sending messages between computers along a tree. Only One
Computer is Allowed to Enter a Critical Section. If Some Computer Wants the
Right to Enter a Critical Section It must request it by sending a message to
the computer which currently has this right. The time of this request depends
on the number of edges in the path to the computer with the right. The goal
is to build a communication tree with the minimal cost and bounded time of
communication. And the solution is exactly the Bounded Diameter Minimum
Spanning Tree.
    Techniques for solving the BDMST problem may be classified into tree cate-
gories: exact methods, heuristic methods with experimentally measured perfor-
mance ratio and algorithms with guaranteed performance ratio.
    There are exact approaches for solving the BDMST problem based on mixed
linear integer programming [2], [17] and 0-1 integer linear programming based
branch and cut approaches [18]. But, these approaches could only be used to
solve small problem instances, like complete graphs with less than 100 nodes.
    As for the heuristic methods with experimentally measured performance ra-
tio, there was presented [1] a greedy heuristic algorithm - the One Time Tree
Construction (OTTC) for solving the BDMST problem followed by its mod-
ification [27], called Randomized Greedy Heuristics (RGH). Later it was also
studied and extended in [32] and [6]. Genetic algorithms for solving BDMST
problems were considered as well [26], [21], [22], [23]. Local search approaches
were considered in [19], [20].
    There were not really much attempts to solve BDMST by algorithms with
guaranteed performance ratio. A study was done in [3], however the proof in not
really easy to follow. In this paper we give the first approximation deterministic
polynomial time algorithm for solving the Random DBMST on an undirected
graph.
    In the papers [15, 14] this problem was studied with a graph diameter bounded
from below. In the current paper we consider the problem with a graph diame-
ter bounded from above. We introduce a polynomial-time algorithm to solve this
problem and provide conditions for this algorithm to be asymptotically optimal.
A probabilistic analysis was performed under conditions that edges weights of
given graph are identically independent distributed random variables.
    By FA (I) and OP T (I) we denote respectively the approximate (obtained by
some approximation algorithm A) and the optimum value of the objective func-
tion of the problem on the  input I. An algorithm A is said to have performance
guarantees εA (n), δA (n) on the set of random inputs of the problem of the size
n, if                  n                               o
                                             
                    Pr FA (I) > 1 + εA (n) OP T (I) ≤ δA (n),                  (1)

where εA (n) is an estimation of the relative error of the solution obtained by
algorithm A, δA (n) is an estimation of the failure probability of the algorithm,
                                                        On BDMST Problem          161

which is equal to the proportion of cases when the algorithm does not hold the
relative error εA (n) or does not produce any answer at all.
    Following by [13], we say that an algorithm A is called asymptotically optimal
on the class of instances of the problem, if there are exist such performance
guarantees that εA (n) → 0 and δA (n) → 0 as n → ∞. Apparently, judging by the
review article [33], the first examples of asymptotically optimal algorithms were
presented in the works [11, 12] for the traveling salesman problem on random
input data.
    Let’s denote UNI(an ; bn ) a class of complete graphs with n vertices where
edge weights are independent identically distributed random variables with uni-
form distribution on an interval (an ; bn ).
    Frieze shown that the mathematical expectation of weight of classic MST on
a random graph can be unexpectedly small. So for example on a complete graph
with weights of edges from class U N I(0; 1), the weight of a MST w.h.p. (with
high probability) is close to the constant 2.02 ... [9].
    As it was said in the papers [14, 15] the MST was studied with a graph diam-
eter bounded from below. In [14] presented an asymptotically optimal algorithm
Ae with time-complexity O(n2 ) for graphs which belong to UNI(an ; bn )-class. On
the first stage algorithm A  e build a d-vertex path P , using the greedy strategy
”Go to the nearest unexplored vertex” , starting from an arbitrary vertex. On
the second stage in a graph G with edge weights equal to an for all e ∈ P , by
means of Prim’s algorithm [25], a spanning tree of a minimal weight is built.
This tree is taken as a solution.
    On graphs which belong to UNI(an ; bn ) class algorithm A   e has the following
performance guarantees:
                                b /a 
                                  n    n           −0.25(n−d)
                      n = O          n−1 , δn = e            .
                                n/ ln n−d

Thus, the sufficient conditions for the asymptotic optimality of the algorithm A
                                                                               e
are
                                        n
                               bn /an ≤    , d = o(n).
                                      ln n
   Next, let’s proceed to the description of the algorithm for solving the BDMST
problem.


2    An Algorithm A for Finding a Bounded Diameter MST

    Let d be a parameter exceeding the tree diameter.
   Stage 1. Arbitrary select d vertices subset from V , let’s denote selected
subset V1 . Using the Prim’s algorithm [25] construct in the graph G[d] induced
by these d vertices a minimum spanning tree T0 : edge by edge grow up tree by
edges e1 , . . . , ed−1 . Obviously, its diameter is smaller than the parameter d. Put
V2 = V \ V1 .
162      E. Kh. Gimadi et al.

   Stage 2. Every vertex u ∈ V2 is connected by the shortest possible edge
with a vertex v ∈ V1 . As a result we obtain an n-vertex tree TA which is an
approximate solution of the problem.
   A comment. If the constructed tree T0 has the form of a chain, then the
nearest vertices v are selected from the set V1 \ v 0 , where v 0 is one of the two end
vertices of the chain.
   Finally we built the spanning tree TA with a diameter smaller than the
parameter d.
   Further, we denote by W (G0 ) the weight of the subgraph G0 of the given
graph G, by WA the weight of the solution built by algorithm A. Also by EX
we will denote an expectation of a random variable X and by VarX its variance.


3     Analysis of Algorithm A

     The algorithm has polynomial complexity O(n2 ), since the construction of
the tree T0 in Stage 1 is done by the Prim’s algorithm [25] in time O((n − d)2 ),
and in the Stage 2 it takes about d(n − d) comparison operations.
    A probabilistic analysis we perform under conditions that graph edges weights
are identically independent distributed random variables with uniform distribu-
tion on a set (an , bn ), 0 < an ≤ bn < ∞. Further we suppose that the parameter
d is defined on the set of values d in the range ln n ≤ d < n.
    Statement 1. The spanning tree TA is restricted by a diameter not exceeding
d, since on the second Stage the diameter of the tree T0 can increase by no more
than 1.
    Statement 2. W (TA ) = W (T0 ) + S, where S = u∈V2 φdu , φdu is a random
                                                       P
variable equal to minimum from d identically independent distributed random
variables from with uniform distribution on an interval [an , bn ], an > 0.
    Also, according to step 1 of the algorithm A, weight of selected edge ei is a
random variable equal to minimum from i identically independent distributed
random variables from with uniform distribution on an interval [an , bn ], an > 0.
    So,
                                            d−1
                                            X
                                  W (T0 ) =     φdei .
                                                i=1

                                     φd
                                      u −an
Or using variables ξud , where ξud = bn −an , distributed on [0, 1], we get

                                         d−1
                                         X
      W (T0 ) = (d − 1)an + (bn − an )         ξedi = (d − 1)an + (bn − an )W (T0 )0 .
                                         i=1

                    Pd−1
We denote W (T0 )0 = i=1 ξedi .
  Denote S 0 = u∈V2 ξud . Obviously,
               P


                           S = (n − d)an + S 0 (bn − an ).
                                                          On BDMST Problem        163

ξud is a random variable equal to minimum from d identically independent dis-
tributed random variables with uniform distribution on an interval [0, 1].
     We have
                W (TA ) = (n − 1)an + (bn − an )(W (T0 )0 + S 0 ).

   Statement 3. We can estimate the expectation of W (T0 )0 by the sum
                              1 1         1
                               + + . . . + ≤ ln d
                              2 3         d
for the expectation of the length ECh ≤ ln d, were Ch is a chain in G[d], obtained
by the greedy procedure ”Go to the nearest city”. So we have W (T0 )0 ≤ h ln d
w.h.p., where the constant h is large 1.
    Let λn be a positive constant. Denote

                                bn (h ln d + (1 + λn )ES 0 )
                         εn =                                .                    (2)
                                an          n−1

   Statement 4.
                         
                       Pr WA ≤ (1 + εn )OP T ≥ 1 − δn ,                           (3)

where
                          δn = Pr S 0 > (1 + λn )ES 0 .
                                 
                                                                                  (4)

Proof.

     WA     (n − 1)an + (bn − an )(W (T0 )0 + S 0 )        bn (h ln d + S 0 )
          ≤                                         =≤ 1 +                    .
     OP T                (n − 1)an                         an     n−1

   By virtue of formulas (2) and (3) the inequality can be continued with the
probability 1 − δn .


    WA       bn (h ln d + S 0 )     bn (h ln d + (1 + λn )ES 0 )
         ≤1+                    ≤1+                              = 1 + εn .
    OP T     an     n−1             an          n−1
The Statement 4 is proved.

   Statement 5.
                                            n−d
                                   ES 0 =       .
                                             d
Proof. S 0 is equal to the sum of n−d random independent identically distributed
variables each of them equal to minimum over d − 1 uniformly distributed on a
segment [0, 1] variables.
   Using the statement 5 we have the following expression for the εn
                                                                  
             bn (h ln d + (1 + λn )(n − d)/d)   bn h ln d 1 + λn
       εn =                                   ≤          +           .            (5)
             an             n−1                 an n − 1       d
164         E. Kh. Gimadi et al.

      Next for the probabilistic analysis of Algorithm A we need the following
   Petrov’s Theorem [24]. Consider independent random variables X1 , . . . , Xn .
Let there be positive constants g1 , . . . , gn and T such that for all 1 ≤ k ≤ n and
0≤t≤T
                                                 n g t2 o
                                                    k
                              EetXk ≤ exp                .                         (6)
                                                    2
             Pn                  Pn
   Put S = k=1 Xk and G = k=1 gk . Then
                                            x2
                                         exp −2G  , for 0 ≤ x ≤ GT,
                     Pr{S > x} ≤
                                           exp − T2x , if x ≥ GT.


      Theorem 1. Let the parameter d be defined so that

                                         ln n ≤ d < n.                           (7)

Then Algorithm A solves the problem asymptotically optimal w.h.p.

Proof. We introduce a proof for two cases for a values of the parameter d: ln n ≤
d < n/2 and n/2 ≤ d < n − 1.

                               Case 1:        ln n ≤ d < n/2.

               q
      Put λn = 4 ln  n
                    n .
      According to the formula (5)
                                                                
                                   bn        h ln d 1 + λn
                              εn ≤                 +                 .
                                   an        n−1       d

We see that εn → 0 under condition

                                         bn
                                            = o(dn ).
                                         an

      Now using Petrov’s Theorem estimate the fault probability

                          δn = Pr S 0 > (1 + λn )ES 0 ,
                                 


      Put
                                                   d
                                             T =     ;
                                                   2
                                               n−d
                                         G=        ;
                                                d2
                                                         n−d
                                   x = λn ES 0 = λn          .
                                                          d
                                                               On BDMST Problem   165

   The inequality T G > x is satisfied. Indeed from

                              1n−d                    n−d
                    TG =           > x = λn ES 0 = λn
                              2 d                      d
it follows that                                r
                                 1                 4 ln n
                                   > λn =                 .
                                 2                    n
    According to Petrov’s Theorem, we have an estimate for the failure proba-
bility of the algorithm A:
                                                       n       x2 o
                       δn = Pr{Ŝ 0 > x} ≤ exp             −       .
                                                               2G
   Now show that
                                      x2
                                         ≥ ln n.
                                      2G
   Indeed, since n − d ≥ n2 , according to inequality (7), we get
                           2
           x 2     λn (n−d)
                        d            (n − d) 2   n − d  4 ln n 
              =                  =          λn =                  ≥ ln n.
           2G       2 (n−d)
                        d2
                                        2          2       n

   From this it follows that
                                      n       x2 o                1
           δn = Pr{S 0 > x} ≤ exp         −        ≤ exp(− ln n) = → 0,
                                              2G                  n
as n → ∞. So in the Case 1 Algorithm A solves the problem asymptotically
optimal.

                          Case 2 : n/2 ≤ d < n − 1.


   Put λn = ln n.
   According to the formula (5)
                                                              
                               bn         h ln d 1 + λn
                          εn ≤                  +                  .
                               an         n−1       d

We see that within the values of the parameter d fot the case
                                                           2, the  expression
                                                           bn ln n
in parentheses reaches a maximum at d = n/2 .So εn = O an n and εn → 0
under condition
                                bn      n 
                                    =o       .
                                an      ln n
   Now using Petrov’s Theorem estimate the fault probability

                       δn = Pr S 0 > (1 + λn )ES 0 ,
                              
166         E. Kh. Gimadi et al.

      Put
                                                d
                                          T =     ;
                                                2
                                          (n − d)
                                       G=         ;
                                            d2
                                                  n−d
                             x = λn ES 0 = ln n       .
                                                    d
      The inequality T G < x is satisfied.

                                   1 (n − d)            (n − d)
                           TG =              < x = ln n         .
                                   2    d                  d
    According to Petrov’s Theorem, we have an estimate for the failure proba-
bility of the algorithm A:
                                                      n       Txo
                            δn = Pr{S 0 > x} ≤ exp        −      .
                                                               2
      Now
                             Tx    d   (n − d)
                                = ln n         ≥ ln n.
                              2    2      d
      From this it follows that
                                     n Txo                  1
             δn = Pr{Ŝ 0 > x} ≤ exp −       ≤ exp(− ln n) = → 0,
                                        2                   n
as n → ∞. So in the Case 2 Algorithm A solves the problem asymptotically
optimal as well.
      Theorem 1 is completely proved.


4      Conclusion

It would be interesting to investigate (a) the Random BDMST problem on input
data with infinite support like exponential or truncated-normal distribution, (b)
the problem of finding several edge-disjointed spanning trees with a bounded
diameter.

Acknowledgement. Authors are supported by the Russian Foundation for Ba-
sic Research (project 16-07-00168), and by the program of fundamental scientific
researches of the SB RAS I.5.1.


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