##
Lovász's reduction of

COLORABILITY to 3-COLORABILITY

(Lecture notes written by Vašek Chvátal)

### A problem and an infinite sequence of problems.

Consider the problem
COLORABILITY

Input: A graph G and a positive integer k.

Question: Is G k-colorable?

and, for each positive integer k, consider the problem
k-COLORABILITY

Input: A graph G.

Question: Is G k-colorable?

### Two observations.

Trivially, for each positive integer k,
k-COLORABILITY reduces to COLORABILITY.

In addition, for each positive integer k,
k-COLORABILITY reduces to (k+1)-COLORABILITY:

if G' denotes the graph arising from G by adding a new vertex and making this
new vertex adjacent to all the old vertices, then G is k-colorable if and
only if G' is (k+1)-colorable.

### A theorem of Lovász

László Lovász (Coverings and coloring of
hypergraphs,* Proceedings of the Fourth Southeastern Conference on
Combinatorics, Graph Theory, and Computing,* Utilitas Math.,
Winnipeg, Man., 1973, pp. 3--12.) proved that
COLORABILITY reduces to 3-COLORABILITY,
and so all the problems
3-COLORABILITY,

4-COLORABILITY,

5-COLORABILITY,

...

COLORABILITY

are equivalent.

### A plan for the reduction

Given a graph G and a positive integer k, we are required to construct
quickly a graph H such that
H is 3-colorable if and only if G is k-colorable.

The idea is to build H around a rectangular *screen*, an array with
k rows that correspond to the colors for use in G and with n columns that
correspond to the n vertices of G. To this screen, we will attach a number
of *peripherals* whose function is to coerce every
3-coloring of H into having the following properties:
- only two of the three colors appear on the screen

and that these two colors can be named "green" and "red" in such a way
that
- in each column (1,u), (2,u), ..., (k,u) of the screen,

at least one vertex (i,u) is colored green and
- for each pair v,w of adjacent vertices and for each color i,

at least one of the two vertices (i,v) and (i,w) is colored red.

If we manage to accomplish this plan, then every 3-coloring of H will
point out a k-coloring f of G: specifically, f(u) is any i such that
vertex (i,u) is colored green.

### A peripheral to enforce property (i)

We begin the construction of H with the vertices of the screen and no
edges at all; to enforce property (i), we add a new vertex b and make it
adjacent to all the vertices of the screen. Since we are free to name
the three colors in any 3-coloring of H whatever we want, we might as well
call the color of b "blue" and to call the two other colors "green" and "red".
Now property (i) is forced.

### Peripherals to enforce property (ii)

There is no loss of generality in assuming that k is odd. (If k
happens to be even, then we may replace k by k+1 and replace G by G'
that consists of G and a new vertex adjacent to all the vertices of G:
now G is k-colorable if and only if G' is (k+1)-colorable.) Under this
assumption, we may enforce property (ii) by adding to H one odd cycle
on k new vertices for each of the n columns of the screen, then matching
the k new vertices with the k vertices of the column, and making each
vertex adjacent to its mate: if all the vertices of the column were
colored by the same color, then the k vertices of the odd cycle would
have to be colored by the remaining two colors, which is impossible.

### Peripherals to enforce property (iii)

So far "green" and "red" have been interchangeable. We will now break
the symmetry by adding to H a new vertex named g and making g adjacent to
b: the color of g cannot be blue and we might just as well call it green.
To enforce property (iii), we add to H three vertices and six edges
for each choice of i, v, and w such that i is one of the k colors for use in G
and v,w are two adjacent vertices of G. The three new vertices, x, y, and z,
form a triangle; the six new edges are the three edges of the triangle plus
the three edges
(x,(i,v)), (y,(i,w)), (z,g).

If both (i,v) and (i,w) were colored green, then all three vertices x,
y, z of the triangle would have to be colored red or blue, which is
impossible.

### From a k-coloring of G to a 3-coloring of H.

Now the construction of H is complete. It follows from what we have
said so far that every 3-coloring of H has properties (i), (ii), (iii),
and so it yields a k-coloring f of G by letting f(u) be any i such that
(i,u) is colored green. Conversely, every k-coloring f of G yields a
3-coloring of H: first, each vertex (i,u) of the screen can be colored
green if f(u)=i,

red
otherwise

and then this 2-coloring of the screen can be extended into a
3-coloring of the entire H as follows. The special vertex b is colored
blue and the special vertex g is colored green. As for the n odd
cycles that enforce property (ii), each of them corresponds to one of
the n columns of the screen; its k vertices are matched with the k
vertices of the column;of these k vertices, k-1 are red and one is
green. The mate of the green vertex on the cycle gets colored red; the
colors on the remaining k-1 vertices of the cycle alternate between
green and blue. As for the various triangles xyz that enforce property
(iii), its vertex z always gets colored red; one of the vertices
(i,v), (i,w) is colored red and its neighbor in the triangle gets
colored green; the other one of the vertices (i,v), (i,w) is colored
red or green and its neighbor in the triangle gets colored blue.

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