Minimizing Tracks in VLSI Design with Graph Coloring, Study notes of Mathematics

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A Programmable Logic Array Example

Suppose we want the following Gate connections

< 1 ; 3 > < 1 ; 3 > < 2 ; 3 > < 2 ; 4 > < 2 ; 5 > < 4 ; 5 > This layout ì realizes î the above connections using 5 tracks

G 1

G 2

G 3

G 4

G 5

track

Interval Graph

I (^) Interval Graph : a graph G with a oneñtoñone correspondence between its vertices and a collection of intervals on the line such that two vertices of G are adjacent when the corresponding intervals overlap.

I (^) Example applications: competition graph (used in ecology; species compete for survival), VLSI routing problems (PLA folding).

I (^) Given the VLSI design problem of connecting the rows of gates: (1, 3) (1, 3) (2, 3) (2, 4) (2, 5) (4, 5)

we can model the problem of determining the minimum number of tracks by nding the Chromatic Number of a related interval graph.

I (^) Let vertices be the connection pairs, and consider them as intervals, for example: (1, 3) → [1..3]

I (^) Let edges join intervals without overlap.

I (^) The minimum number of tracks will be the Chromatic Number of the graph since intervals can share a track only if they do not overlap.

Another Example

I (^) The largest complete graph in the gure below is a triangle, therefore it requires 3 colors (and we cannot color it in 2 colors)

I (^) Thus we need only 3 tracks when the rows are rearranged.

(D, E)

(C, E)

(B, C)

(A, B)

(C, D)

(A, B)

Sec. 2.4óColoring Theorems

I (^) Polygon : a planar graph consisting of a single circuit with edges drawn as straight lines.

I (^) Triangulation of a Polygon : the process of adding a set of straightñline chords between pairs of vertices of the polygon so that all interior regions are bounded by a triangle.

I (^) Note: Chords cannot cross each other nor the sides of the polygon.

Triangulation Theorem

Theorem 1. The vertices in a triangulation of a polygon can be 3ñcolored.

Proof is by induction on n, the number of edges in the polygon.

BC. Let n = 3. Then the polygon is a triangle, and clearly can be 3ñcolored.

IH. (Strong induction) Assume any triangulated polygon with 4  k < n boundary edges can be 3ñcolored for some arbitrary n  4.

IS. Show a triangulated polygon, T , with n boundary edges can be 3ñcolored.

Pick some chord edge e = < vi ; vj >, which must exist since T has been triangulated.

Since all chord edges connect vertices of the polygon, the chord edge e splits T into two smaller triangulated polygons, each of which can be 3ñcolored by the IH.

In each coloring, vi will have some color, and vj will have some other color.

Then the two subgraphs can be combined to yield a 3ñcoloring of the original polygon since, if need be, the coloring of one of the smaller polygons can be modied. Note: this 3ñcoloring is unique.

Fisk's Corollary

Corollary : the Art Gallery Problem with n walls requires at most b n 3 c guards (where b c is the oor function.)

Proof: Let the n walls form a polygon P with triangulation T. 3ñcolor T and note each triangle will have a corner of each color. Pick one color, c, and place a guard at each corner colored c (1 in each triangle). Hence the sides (and thus all walls) of every triangle will be watched.

A polygon with n walls has n corners. If there are n corners and 3 colors, some color is used on b n 3 c or fewer corners.

Other Coloring Theorems

Notes: I (^) If a graph is bipartite, it is 2ñcolorable (and viceñversa)

I (^) A graph is 2ñcolorable IFF all circuits have even length (this doesn't require the graph to be connected)

I (^) Let (G ) denote the chromatic number of G.

Theorem 2. If the graph G is not an odd circuit or a complete graph, then (G )  d where d is the maximum degree of a vertex in G. (This gives a usually poor upper bound on (G ))

Theorem 5. Every planar graph can be 5ñcolored

Note : in Section 1.4, exercise 16, the reader was asked to prove Any connected planar graph has a vertex of degree at most 5.

Theorem 5 Proof, by induction on the number of vertices.

BC. Let 1  n  5. Trivially, any such n vertex graph can be 5ñcolored.

IH. Assume for some arbitrary n  1, that connected planar graphs with n 1 vertices can be 5ñcolored.

IS. Show a graph with n vertices can be 5ñcolored.

By the note, G must have a vertex, x, of degree at most 5. Delete x from G to obtain a graph, G ′, with n 1 vertices.

By the IH , G ′^ is 5ñcolorable.

Now, reconnect x to the graph and try to properly color x.

If x has degree  4, then simply assign a color to x which is different from any of its neighbors. The same coloring works if the degree of x is 5 and 2 or more of its neighbors has the same color.

There remains the case of how to color x if all 5 neighbors have different colors.

Case 1. If there is no path between them (other than through a), a may be recolored with color 3, all vertices adjacent to a which are 3 can be assigned 1, and so on.

This re-coloring will not affect c since there is no path from a to c, and furthermore, will only affect vertices reachable from a which are colored 1 or 3.

After the re-coloring, x may be colored 1.

Case 2. There exists a path from a to c. This path either encompasses b, or it encompasses d, but not both, since G is planar.

Thus there can be no path between b and d (other than through a), so the same type of re-coloring may be applied to b (color 2), using d's color (4).

Thus allowing x to be colored with 2.

Hence, every planar graph can be 5ñcolored.