A Polynomial Algorithm for the Two-Variable Integer
Programming Problem
RAVINDRAN KANNAN
Cornell University, Ithaca, N e w Y o r k
ABSTRACT A polynomial time algorithm is presented for solving the following two-variable integer programming
problem
maximize ClXl + c2x2
subject to a, lxl + a,2x2 =< b,, I = 1, 2,
and x~, x2 => O, integers,
, n,
where a,j, cj, and b, are assumed to be nonnegattve integers This generahzes a result of Htrschberg and Wong,
who developed a polynomial algorithm for the same problem with only one constraint (l e, where n = 1)
However, the techniques used here are quite different
KEY WORDS AND PHRASES integer programming, knapsack problem, polynomial algorithm, coefficient size,
feasible region decomposition
CR CATEGORIES 3 15, 5 25, 5 30, 5 40
Introduction
We consider the following integer programming problem:
(1)
maximize
subject to
and
clxa + czx2
a a x l + a,2xz ~ b,,
xl, x2 => 0,
a= 1,2 ....
n,
integers,
where au, G, and b, are assumed to be nonnegative integers. We first show that the solution
to (1) can be obtained easily from the solutions to at most n problems, each of which is of
the form
(2)
maximize
subject to
c l x l + c2x2
a l x i + a2x2 =< b,
f ~ xl =< g,
and xl, x2 => 0, integers,
(2a)
(2b)
(2c)
where all constants involved are positive integers; further they can all be computed in time
no greater than a fixed polynomial of the size of the input which is assumed to be
represented in binary encoding. It is also assumed that g _-< [ b / a i J . If not, we can replace
g by [ b / a i J . (Using different techniques, Hirschberg and Wong [l] have developed a
polynomial algorithm to solve (2) where constraint (2b) is deleted.)
Permission to copy without fee all or part of this material is granted provided that the copies are not made or
distributed for direct commercial advantage, the ACM copyright notice and the title of the publication and its
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This research was supported by Sonderforschungsberelch 21 (DFG), Institut fur Okonometne und Operations
Research, OniversttM Bonn, Bonn, West Germany, and the National Science Foundation under Grant ENG 7609936 to Cornell University, Ithaca, New York
Author's present address Department of Electrical Engmeenng and Computer Science, Division of Computer
Science, University of California at Berkeley, Berkeley, CA 94720
© 1980 ACM 0004-5411/80/0100--0118 $00 75
Journalofthe AsSoclaUonforComputingMachinery,Vol 27,No I, January 1980,p 118-122
A Polynomtal Algorithm f o r the Two. Variable Integer Programming Problem
119
1. Decomposition
THEOREM 1
(DECOMPOSITIONTHEOREM). The set S defined as S = {(x~, x2)]a, lxl +
a,2x2 ~ b,, i = 1. . . . .
n; Xl, x2 ~ 0} where a u and b, are positive integers can be expressed as
UP-iS,, where m =<n and each S, is o f the f o r m
S, = {(xl, x2)[fi =<x~ _-<g,; z,x~ + z,2x~ =<s,; xl, x2 _->0}.
Here z,~, z,2, s, are all posmve integers, fi and g, are positive rationals, and all these constants
can all be computed in polynomial time f r o m a,j and b,.
Of course we note that intuitively the theorem is quite o b v i o u s - - o n e can draw the
straight lines a , x l + a,2x2 = b, and by looking at the graph, it would be easy to determine
the "tightest" constraint in each o f several regions. Since there are at most (~) points o f
intersection of these hnes in the first quadrant, there would be at most (~) of these regions.
The fact that this number can be reduced to n is also easy to see. W h a t follows is a rigorous
proof of the theorem.
Definitton. A constraint a,o,~x~ +ato2X2~ b,o is said to be binding f o r S in the region
f_-< Xl =< g if for f - <_ Xl ~ g, a,oxXl +a~o2x2 ~ b, o implies a,lxx + a,2x2 _-< b , for each i = 1,
2, . . . , n .
LEMMA 1. I f a constraint is binding f o r S in the two regions f =< xl ~ g, f ' _-< x l ~ g'
where f _ <
- g _-<f ' _-<g', then it is binding f o r S in the region f_-< x~ _-<g'.
PROOf. Since S is convex, the lemma is obvious.
[]
Thus in the decomposition o f S into S,, we can assume that each o f the n constraints o f
S is a binding constraint for at most one region. Hence, there are at most n regions, i.e., at
most n S,'s.
PROOF OF THEOREM 1. The proof is by reduction on n. F o r n = 1, $1 = S. Note that
we could take f i = 0 and gt = b~/an. Now suppose for S with n constraints, we could
decompose S into {S,},m=l, m _--<n. Let g = S N {.(xl, X2)lan+l,lXl + an+l,2X2 ~ bn+l) (we
have added one more constraint). To decompose S as m the theorem, we find the point of
intersection (u,, vJ of an+uX~ + a,+~,2x2 = b,+~ with z,~xl + z,2x2 = s, for each l, t = 1, 2,
. . . . m. (We are using the notauon of Theorem 1 for S and S,.) Then for each i, i = 1, 2,
. . . . m, we do the following: Iffi =< u, _-<g,, then the (n + l)th constraint is binding for S for
a segment of [fi, g,] and the constraint z,~x~ + z,2xe _-<s, for the rest of [fi, g,]. It is easy to
determine these segments. (See Figure 1.) Here the (n + l)th constraint is binding for the
segment [fl, u,] and the other constraint for the segment [u,, g,].
If, however, u, ¢ [fl, g,], then either a,+a,xxi + an+l,2x2 _-< b,+~ or z,ax~ + z,2x2 ~ s, IS
binding for the whole of [fi, g,] for S. It is again easy to determine which constraint is
binding~ Thus we have decomposed S. By Lemma 1, there are at most (n + 1) regions into
which S has been decomposed. It is also clear that this algorithm runs in time O(n2). []
Thus to solve (1) we only need to find the maximum of caxl + c2x2 over the integer
points in each set in the decomposition and then take the maximum among all these
maxima.
",K-'-'Zil X l + Z i 2
X2 = Si
FIG |.
120
2.
RAVINDRAN KANNAN
The Algorithm
We now need to solve a problem of the form
(3)
maximize
subject to
ClX] + C2X2
alxl + azxz =<b,
and
L <~ xl ~_ U,
Xl, x2 _->0, integers,
and
U
(3a)
(3b)
(3c)
where
(4)
a~, as,
L,
are nonnegativeintegers.
If any one of the following three conditions is met, then (3) is obviously trivially solved:
(5)
Cl (or c2) ~ 0:xl
(6)
(or x2)
al
can be set to its lowest permissible value and the
resulting one-variable problem solved,
or
a2=0
or
b-<_0,
L = u.
(7)
Our algorithm successively reduces (3) to other problems of the same form until one of the
conditions (5), (6), or (7) is met. Assume to start with that none of these three conditions
is satisfied. We can assume, without loss of generality, that al U =< b. If not, replace U by
[b/alJ.
The set of x~, x2 satisfying (3a), (3b), and (3c) is the set of integer points inside the
triangle T2 and the rectangle T~ shown in Figure 2. Maximizing a linear function over the
integer points in the rectangle T~ is trivial. So, we need only to consider the triangle T2. It
is clear that with the change of variables,
(8)
x] = xl - L,
[ b - axU]
x~ ffi xz - /
az
/'
x] and x~ are nonnegative for every integer point in T2. Thus, we can always reduce
problem (3) to the following one:
(9)
maximize
clx] + c2x~
subject to
a~x~ + a2x~ ~ b - [b
and
x~, x~ ~ 0,
-
-
axU]
- a l l = b' (say)
integers,
(9b)
where al, a2, Cl, c2 > 0, integers.
T 2 ~ i + a2x z = b
x2
T,
f
U
FIG 2
(9a)
"'4" X I
A Polynomial Algorithm for the Two-Variable Integer Programming Problem
121
If again b' <= 0, the problem is trivial. So assume b' > 0. After renumbering variables, if
necessary, assume that aa ~ a2. Let k -- [a~/a2J and let al = ka2 + p. (Note that as # 0.)
Then since ca, c2 > 0, we may add the constraint x[ ~ [(b' - alx])/a2J to (9) without
altering the set of optimal solutions. But
a2
J
l_
=
a2
~
_-> ~ - x~
- x~ k
~
(in the context of (9a) and (9b))
(again in the same context).
Therefore, the set of optimal solution to (9) is the same as that of (10) below:
(10)
maximize
s u b j e c t to
and
c~x~ + c2x~
a~x~ + a2xi ~ b',
(10a)
x~, x~ ~ O, integers,
(10b)
x2' =
>
(10c)
-x
k.
Replacing x[ ~ 0 by x~ ~ [b'/aaJ (in the context of (10c)) and substituting
(11)
x~' = x l -
- x~ k
into (10), we get
(12)
maximize
(cl -c2(k))x~ + c2x~' + c2(k)
subject to p x l + a2x~' ~ b' - (k)
x~, x~' ~ 0,
Lb,]
a2,
integers,
We now have a problem of the same form as (3), and it is clear that using (11) we can
convert an optimal solution of (12) into one o f (9). If now (12) satisfies (5), (6), or (7), we
are done. Otherwise, we iterate the same process on (12). Note that to obtain (12) from (3),
we need to perform a constant number of multiplications and divisions each o f which
takes at most O(d log d log log d) time [2], where d is the length of the input. Further, we
claim that after two iterations the size of the lesser coefficient in the knapsack constraint
is cut down by a factor of at least 2. This is because in (9) either p = (al - [alia2] a2) =<
a2/2, in which case already in (12) the lesser coefficient is at most ½ of that in (3), o r p >
a2/2. But m this latter case a further iteration on (12) yields a problem with coefficients p
and a2 - [a2/p] p = a2 - p < a2/2. Thus at most 2 log as iterations are required before
condition (6) is satisfied. Thus, the problem is solved in O(d 2 log d log log d) time.
To summarize the above discussion we present the algorithm to solve a problem of the
form (3). The procedure below assumes that the input sausfies (4). We will also assume
that either the problem given is infeasible (in which case by convention we set the value of
the optimal solution to -oo) or it has a finite optimal solution value, The procedure tells
which of the two cases holds and in the latter case gives an optimal solution (xa, x2) as a
vector.
procedure KP(cl, c2, al, a2, b, L, U )
begin
if (5), (6), or (7) holds t h e n r e t u r n the solutton after a tnvtal calculation,
if ( U ~ oo) t h e n U ~-- ram(U, lb/alJ,
C o m m e n t W e n o w r e d u c e (3) to (9),
if (L ~ 0 or U ~ oo) t h e n r e t u r n the better o f the two solutwns
(U, l(b - azU)/a2J) and
(KP(cl, c2, al, a2, b - [(b - alU)/a2] a~ - alL, O, ~ ) + (L, [(b - alU)/a2])),
122
RAVINDRANKANNAN
CommentNow L = 0 and U = oo. After ensuringthat a~ ~ a2, we recurstvelysolve(12),
0 I
0);
if(a'<a2) thenreturnKP(c2'cl'a2'a"b'O'°°)(!
k ~ [aa/a~], p ~ al - kay;
KP(cl - c2k, c2, p, az, b - k i b/ad az, O, [ b/ad)
return
0
!
+ (0, [b/aiJk),
end
3. C o n c l u s i o n s
In [1] it was conjectured that there is a polynominal algorithm that solves the general kvariable knapsack provided k is assumed to be a constant. The algorithm given here in
Section 2 for the two-variable case seems to be simpler than the one in [1], and hence is
probably more amenable to efforts at generalizing tt to any fixed n u m b e r of variables It
has, however, the same time complexity as the latter.
It is not clear that this algorithm is different from that given by Hirschberg and Wong,
although it was arrived at in a rather different way. The referee suggests that ~t would be
of interest to see if in fact the algorithm given here could not be proved to be identical to
or a simple modification of that of Hlrschberg and Wong.
REFERENCES
!
HIRSCHBERG, D
S., AND WONG, C K A polynomialalgorithmfor the knapsackproblemin two variables J
A C M 23, 1 (Jan 1976), 147-154
2 AHO,A V, HOPCROFT,J E, ANDULLMAN,J D The Design and Analysts of Computer Algorithms AddisonWesley, Reading,Mass, 1974
RECEIVED JULY 1977, REVISED FEBRUARY 1979
Journal of the Assooatlon for Computing Machinery, Vol 27 No 1, January 1980