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MA:SSACUSETTS.' INSTI TUTE
<at;V;;7
OF ECHNOLG
PARAMETRIC SHORTEST PATH ALGORITHMS
WITH AN APPLICATION TO CYCLIC STAFFING
by
Richard M. Karp
and
James B. Orlin
OR 103-80
October 1980
Research supported, in part, by the U.S. Army Research Office under
Contract No. DAAG29-80-C-0061.
Parametric Shortest Path Algorithms
with an Application to Cyclic Staffing
by
Richard M. Karp and James B. Orlin
Abstract
Let
vertex
G = (V,E) be a digraph with
vertices including a special
s. Let E' C E be a designated subset of edges.
there is an associated real number
f 2(e)
The length of edge
fl(e).
For each
eE E
Furthermore, let
1 if
e E E'
:
0 if
e
e is flpe)-Af2(e),
takes on real values.
edge of
n
where
E-E'
X is a parameter that
Thus the length varies additively in X for each
E'.
We shall present two algorithms for computing the shortest path from s
to each vertex
running times
v E V parametrically in the parameter
O(n3 ) and
O(nlE l logn).
X, with respective
For dense digraphs the running
time of the former algorithm is comparable to the fastest (non-parametric)
shortest path algorithm known.
This work generalizes the results of Karp [2] concerning the minimum
cycle mean of a digraph, which reduces to the case that
E' = E. Furthermore,
the second parametric algorithm may be used in conjunction with a transformation given by Bartholdi, Orlin, and Ratliff [1] to give an
algorithm for the cyclic staffing problem.
-- -
O(n21logn)
Parametric Shortest Path Algorithms
with an Application to Cyclic Staffing
by
Richard M. Karp
and
James B. Orlin
Let G = (V,E) be a digraph with
n vertices including a special vertex
s. Let E' c E be a designated subset of edges. For each
an associated real number
fl(e).
Furthermore, let
I
f2(e)
The length of edge
e E E there is
if
e E'
eE'
O .'if e E-E'
e is fl(e)- Xf2(e),
where
X is a parameter that
takes on real values. Thus the length varies additively in X for each edge
of E'.
We shall present two algorithms for computing the shortest path from
to each vertex
running times
v
E
s
V parametrically in the parameter X, with respective
O(n3 )
and
O(nlEI log n).
For dense digraphs the running time
of the former algorithm is comparable to the fastest (non-parametric) shortest
path algorithm known.
This work generalizes.the results of Karp [2] concerning the minimum
cycle mean of a digraph, which reduces to the case that
E' = E. Furthermore,
the second parametric algorithm may be used in conjunction with a transformation given by Bartholdi, Orlin, and Ratliff [1] to give an
O(n21ogn)
algorithm for the cyclic staffing problem.
In the following a walk is a directed edge progression from an initial
vertex to a terminal vertex.
A cycle is a walk in which the initial vertex
1
2
is equal to the terminal vertex.
A path is a walk in which no vertex is
repeated.
For each walk
P and for each
·Q(P,X)= :
L(P,X)
Thus
is the length of walk
X E R define
(fl(e)-Xf 2(e))
k and.each
P. For each integer
XE R
let
I f2 (e) = k}
gk(v,A) = {in £(P,): P is a walk :from s to v and e2E
Then
gk(vX)
is a linear function of
)X which measures the shortest length
gk.
path from s to v subject to the additional constraint that the path has
E'.
k edges of
exactly
s to
In the following we assume that there is a path from
v for each
vertex v E V and that there is no cycle of negative length composed solely
Thus for all
of edges in E-E'.
is a minimum length path from
both be checked in O(nJlEl)
Let
value
F(v,X)
X, where
v E V and for sufficiently small
s to
X there
v. Of course, these assumptions may
stens.
be the minimum length of a walk from
s to
v- for fixed
F(Y,X) = -~ if there is a sequence of walks with length
unbounded from below.
min gk(v,X) for all v E V and for all
O<k<n-1
for which the resulting digraph has no cycles of negative length.
Lemma 1. F(v,)
Proof.
'=
XE R
Since there are no negative length cycles, there is always a
n-l
minimum length walk which is a path and hence has at most
I
edges in E'.
O
The following result is a direct extension of a theorem proved by Karp
in [2].
Let X* be the maximum value of X for which there are no cycles of
negative length.
Let V(k) = {veV: Igk(V)
}.
Let G(X) = min max (g(v,)-gk(v,x)).
veV Okn-
vE V(k)
Theorem 2. Either
for which G(X)
Proof.
X* =
or else X* is the unique value of- X
0.
Note first that, if G(X) is finite (i.e., there is some
such that gn(v,X) f +),
then
v
G(X) is a strictly decreasing function, and
thus there is a unique value of X for which G(X) = 0.
For X = X* we have that all cycles have non-negative length and hence
by Lemma l, for all
v
V
gn(V,*) > F(v,X*) =
n
O<k<n-l
9k('X*)
Hence, G(X*) > 0.
Conversely, for value
at least one edge in
E'.
X* there must be a cycle
Else we could increase
C of length
0 with
X* without creating a
negative length cycle.
Let v
from
be a vertex on cycle
C and let
P be a minimum length path
s to v for value X*. P followed by any number of repetitions of
C is also a minimum length walk from
s to
of such a walk is a minimum length walk from
sufficient repetitions of
v. Hence, any initial part
s to its endpoint. After
C there will be an initial part
n edges in E' from s to w. Then
P' with exactly
4
(P,X*) = g (w,X*) =
G(X*)
Thus
_0.
Thus
gk(,X*)
min
O<k<n-1
O.
G(X*) =
Let d(u,v) be the minimum length of a path from u to v in G
consisting solely of edges of
Let
E-E'.
be the minimum length of
g(v,A)
k edges in E' and such that the last
a walk from s to v with exactly
edge of the walk is in E'. Then we may calculate g and g' by the following recursive scheme:
gk1(u
) + f 1 (u,v) - X)
gi(vx)-
umin
(u,v)EE·
-9k(vX) =
min (g(u,X)+d(u,v)) for
and
k =,...,n
subject to the initial conditions
g(V,X) =' d(s,v)
In order to compute
O(n3 )
F(v,X)
for all
X < X*, we may calculate
steps from Theorem 2 and the functions
F(v,X)
late
of g'
and
from Lemma 1 in another
g take
0(n3 )
O(n3 )
gk(v,X),
steps.
X* in
and we may calcu-
Since the calculations
steps, the entire algorithm requires
0(n3 )
steps.
An O(nlE log n) Algorithm
The previous algorithm is not typical for'parametric programming
algorithms in that the solutions are calculated simultaneously for all values
of X.
The following is a more standard approach in that optimal solutions
are calculated for one value of
X at a time as
X increases from
-.
5
In the following a tree will refer to a connected subgraph of
exactly
n-l
edges, with one directed into each vertex of
tree is really an arborescence with root
s to
v on
tree T and for each vertex v
distance from
TX
s to v on T. Thus
E' on the path from
s to
TX such that the
is a path of minimum distance.
V, let
(Thus a
s.)
It is well known that for X < X* there is a tree
unique path from
V-{s}.
G with
dl(T,v) - Xd2(T,v)
d 2 (T,v)
For each
denote the
is the number of edges of
v in T.
A neighbor of tree T is a tree which differs from T in exactly one
edge. Let T be any tree and let e = (u,v) E E. Then we let
denote the subgraph obtained from T by adding edge
unique edge of T directed into
N(T,e)
e and deleting the
v. The following is an immediate conse-
quence of this construction.
Lemma 3. Let T be a tree and let
e = (u,v).
tree if and only if v is not on the path from
If e = (u,v),
s to
Then
N(T,e)
is a
u in T.
0
let
Ai(T,e) = d i(T,u)+f i(e)-d i(T,v) for
i = 1,2
and let
P(T,v) = {wEV such that v is a vertex on the path from s to w in T}
Lemma 4. Suppose
T' = N(T,e)
is a tree.
J d (T,w) + A (T,e)
di(T,w)
and if A2(T,e) s 0 then X' = A(T,e)/A2(T,e)
Then for
if
w
if w
i = 1,2
P(T,v)
P(T,v)
is the unique value of
for which
I
X
6
dl(T,w)- Xd2 (T,w) = dl(T',w) -d
Proof.
w
P(T,v),
=
A'
and
w EV .
weP(T,v),
di(T',v)-di(T,v) = i(T,e)
d i(T',w) = d i(T,w).
follows from the fact that
for all
i = 1,2
It is easy to verify that for
di(T',w)-di(T,w)
And for
2(T,w)
The last statement in the theorem
is the unique value of
X for which
A l(T,e)- XA2(T,e) : 0.
0
Lemma 5. If T' = N(T,e)
A(T,e)/A2(T,e)
.clircuit of length
is not a tree and
is the unique value of
-
0.
Proof. Suppose that
.
subpaths in T from s to v and from
edge
(u,v).
X for which the digraph T' has a
T' is not a tree.
in T' by deleting the subpath
A 2(T,e) f 0, then
s
Let Pv
to
and
Pu
denote the
u. Then 'acycle is formed
Pv from the subpath
Pu and then adding
The length of this circuit is thus:
d1(T,u) - d(T,v) + fl (e)- X(d2 (T,u)-d2(T,v)+f 2(e)) = A (T,e) - XA2 (T,e)
L
The following algorithm will calculate a sequence of trees
and a sequence of real numbers
the tree
X=
T
-: = XO'Xl''A),
is a tree of minimum distances for
T1,...,Tm
= A* in such a way that
X E [il,Ai]
or X = -I in which case we do not define the trees.
except for
The trees will
only be stored implicitly via appropriate data structures; otherwise, to
-3
store all the trees. would take S2(n ) space.
The algorithm requires a data structure to store and retrieve entries
of the form
(e,c(e)),
where
e E E and
c(e) E R. The operations performed
on the structure are the insertion of an entry, the deletion of the unique
7
entry
(e,c(e)) associated with a given edge e, and the retrieval of an
entry
(e,c(e)) for which c(e) is minimum. If a balanced tree [3] is used
for this purpose, then each of the basic operations can be performed in
O(log IEI)
steps.
Algorithm 2
Step 0 (Initialize)
0.1
Let M
=
1+eEElfl(e)l.
Let X = -C.
-M. (Use a
0.2 Let T be the tree of minimum distances for X
standard O(nJEI) shortest path algorithm to determine this tree.)
0.3
Compute di(Tl,v) for i = 1,2 and v E V.
0.4 For each edge e E E let
if A2 (T ,e)< 0
Ai(T1,e)
c(e)=
2(Tl,e)
if A2(Tl,e) > 0
0.5 Store the values c(e) for e E E in a balanced tree B.
0.6 Let i = 1.
Step 1 (Compute the new tree)
1.1
Let X i = c(e') = min c(e) (by finding the minimum element of the
eEE
balanced tree B).
If
=
iX
then quit.
Let (u,v) = e'.
1.2 Ti+ l = nTi,e')
1.3 Let Pi
If u
=
P(Tiv).
P then quit with X* = Xi; else, let Aj = Aj(Ti,e') for
j = 1,2.
1.4 For each w E Pi and for j = 1,2
d (T
W) =
i+
dj(Tiw)+Aj
if
wE P
dj(Tiw)
if
W
P
;
8
Step 2 (Calculate the next change of trees)
2.1
Let E be the subset of edges of E with at least one endpoint in P.
2.2
For e EE delete the value c(e) from B and instead let
Go
if A2 (Ti+
1,e) <
2Tic(e)
= i if A 2 (Ti+le) > 0
and store this new value of c(e) in the balanced tree B.
2.3 Let i = i+l and return to step 1.
Theorem 6. For each
X E [ii-l'Xi
Proof.
i = l,...,m,
is optimal for all
with X finite.
Suppose the theorem is false.
there is a value
X' E
Choose the minimum
i so that
[XilXi] such that the result fails at X'.
can happen for one of two reasons:
for
the tree T i
X = X' and a vertex
either there is a tree
v so that the distance from
less than the corresponding distance in T i for
fails because there is a negative cycle for
T that is optimal
s to
X = X';
X = X'.
This
v in T is
else the theorem
We first investigate
the former case.
By our choice of
where
for
M, for X < -M it follows that
k is the minimum index for which gk(v,)
X E (--,-M] and thus
Furthermore
have that Ti_ 1
X'
Xi_.1
F(v,X) = gk(v,X)
I a. Thus Tl is optimal
X' > -M.
To see this note that by our choice of
is optimal for value Xi_ 1.
Lemma 4, the distances in T i and
By our choice of
Ti_ 1 are the same for
X=
Xi
i-l
T i is optimal at Xi-l
For each
l
w E V let
h(w,;) = d l (T,w) - d(T i ,w) - A(d2 (T,w)- d 2 (Ti,w)) .
i we
and by
Thus
9
Then, by the optimality of T at X = ',
h(s,X') = 0. Since
Ti
h(v,X') < 0 and
that
precedes
for all
w. Also,
is not optimal at X = X',
there is a v such
h(u,X') = 0 for the vertex
u which immediately
v on the path from
of T directed into
hw,X') <
s to
v in T. Let
e be the edge
(u,v)
v. Then
h(v,)
= A(T i ,e)- XA2(T i ,e)
Furthermore,
h(v,X') < 0 < h(v,Xi
l)
This can only happen if
A2(Ti,e.)
>0
i
and
Al(Ti,e)
ki_<
A2i,e)<
But this would contradict our choice of
tree
X'
X i in step 1, thus proving that
T cannot exist.
Suppose instead that there is a negative cycle for
l' > Xi
l,
length
0 for
as
Ti
is optimal in [X -1'l
Ai1 < A* < Xi
X* with
E' is in C. Let C = Vl,...,Vk,Vl
vj_ 1 does not precede
vj
.
X = '.
As before
Let there be a cycle
C of
and such that at least one edge of
and let
on the path from
vj
s to
be a vertexsuch that
vj
in T i.
Let
e = (vj_lvj).
We now claim that A(Ti,e)/A2(Ti,e) = X* and
contradicting our choice of
a minimum length walk from
walk.
Let
Thus
vj_ 1
e = (Vjl,vj).
the above that
precedes
Since
A2(Ti,e) > 0 thus
Xi. To see this we note first that for
s to
vj
v followed by
X = X*
C is also a minimum length
on some minimum length walk from
s to
vj.
T i is a minimum distance tree, it follows from
10
dl(Tv
j)
- X*d2(Ti,vj) = dlTi,vj_l)+fl(e)
*(d 2(Ti,vjl) + f 2 (e))
Thus
Al(Ti ,e)-X*
Let a(w)
be the number of sets
2(Ti,e) =
Pi
0
0.
as defined in Step 1.3 such that
w E Pi' In order to determine the number of steps taken by the algorithm,
we will use the following lemma.
Lemma 7. For each w E V,
Proof. If w E Pi
d2 (Ti,w) < n-l
for all
(w) < n-l.
d 2 (Ti ,w) > (d2 (Ti_ 1 ,w)-
then
Since
0
i it follows that a(w) < n-l.
We now can see that Algorithm 2 takes
(1) By Lemma 7, dj(Ti,w)
0(nEi log n) steps as follows:
is updated at most
n-l
w E V. Thus this updating may be carried out in 0(n2)
(2) By Lemma 7, c(e)
for each endpoint.
1>0.
is updated at most
2n-2
times for each
steps overall.
times,
Thus the total number of updates of
c(e)
n-l
times
is 0(nlEI).
At each update we must both delete and insert the value into a balanced tree.
These operations may be carried out in 0(log n) steps.
Furthermore, we may
have to choose the minimum element in the balanced tree as many as
n 2 times.
Thus the total computation count for these steps is 0(nlEl log n).
An Application to Cyclic Staffing
Vector y is said to be circular if each component has value 0 or 1
and if the components with value 1 occur consecutively, where the first and
last components are considered consecutive. Matrix
(resp. row) circular if each column (resp. row) of
A is said to be column
A is circular.
11
In [1], a certain class of cyclic staffing problems was represented as
an integer programming problem
minimize
,
(3)
subject to xl+"'+xn =
Ax > b
x > 0 and integer
where A is column circular.
An example of the problem is to find the
minimum number of persons needed to staff a weekly schedule so as to satisfy
demands that vary from day to day within the week but repeat weekly and so
that each person receives two consecutive days off per week.
This problem may
be represented as the integer program (3)where
A
'1 001 11 1
11 001 11
1 '1 10 0 1 1
1 1 1 10 01
1 1'I 1100
011 1 1 10
0011 1 1 1
The followinig s pro,:ed in [1]:
is considered as a parameter,
If
then every basic solution to (3) is integral whenever
pJ is integral.
solve (3) it suffices to determine the minimum integral value of vl
which (3) is feasible.
To
for
Furthermore, this may be accomplished by transform-
ing this parametric problem into an equivalent parametric shortest path
problem of the type solved in this paper.
The number of edges in this problem
is linear in n and thus the entire problem takes
0(n2 logn)
steps.
We briefly describe the problem transformation given in [1].
loss of generality, we may assume that there is no pair
column indices such that
aij > aik
for all
j,
Without
k of distinct
i; for if such a pair did
12
exist, then column
k could be eliminated.
We may also assume that the
columns of A are in lexicographic order; i.e., if j < k then there
exists some q such that
aij = aik, i < q and
aiq > ajq
It follows
that A is row circular as well as column circular; i.e., in each row of A,
either all the 's are consecutive or all the O's are consecutive.
We next make a change of variables that will transform our problem to
a parametric shortest-path problem.
yj =
X,= -1
-
+ x 2x+
-- + xj;
For
in particular, YO
-(xl+x 2 + --+xn).
j = O,l,...,n-l,
let
is identically zero.
The nonnegativity constraints on the
YO <Yl' Yl <Y2""Yn 2 Yn-l' Yn-1
Suppose row i has consecutive l's, occurring in columns
Also, let
xi
become
-X
j+l,...,j+k.
Then
this row represents the inequality
xj+l +x
j +2
which can be rewritten yj < k- bi
+
+ xk
> bi
Suppose row i does not have consecu-
tive l's but has consecutive O's, occurring in columns
j+l,...,k, where
k < n. Then this row represents the inequality
(X1 X2+ .+Xn)-
(Xj+l+-.-+Xk) > bi
which can be rewritten
Yk
<
Yj -bi-X
The cyclic staffing problem is thus transformed to the maximization of
X
subject to linear inequalities, each of which is of the form
Yv
<y
+be - XB
where
(4)
Be E {0,1}
;
I
13
Let N be a network with vertex set
inequality of the form (4), an edge
{O,l,...,n-l}
e from u to
and, for each
v of parametric cost
b e - XBe . Then the cyclic staffing problem is reduced to the computation of
A* for the network N. Once X* is determined, yv is simply the cost of
a shortest path in N from
0 to
v, at
X = X*.
The transformation just given, coupled with Algorithm 3, yields an
algorithm to solve the cyclic staffing problem in time
2
2 1 og n).
O(n
An
experimental computer program that executes this algorithm was run on several
examples in which the columns of the constraint matrix consisted of all
rotations of an n-vector of
k consecutive l's, and the components of the
right hand side were drawn from a uniform distribution.
balanced tree operations never exceeded
time of the algorithm is O(nlog n).
2n,
The number of
suggesting that the expected
14
References
1.
Bartholdi,J.J., Orlin, J.B., and Ratliff, H.D., "Cyclic scheduling via
integer programs with circular ones," to appear in Operations Research.
2.
Karp, R.M., "A characterization of the minimum cycle mean in a digraph,"
Discrete Mathematics 23 (1978) 309-311.
3.
Knuth, D.E., The Art of Computer Programming. Vol. 3. Sorting and
Searching, Addison-Wesley, Reading, Mass., 1973.
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