Colored Necklace Bisection Victor Kostyuk Advisor: Michael Capalbo

Problem setup Consider a necklace with 2n beads of k colors. There are 2a i beads of color i, and the beads are arranged on the necklace arbitrarily.

Color (i) Number of beads (2a i )

Goal - efficient cutting alg. Is there an O(n c ) algorithm for making the least number of cuts between beads such that the resulting bead strings can be partitioned into two groups, with a i beads of color i per group?

Group 1:Group 2:

Existing results (1) Goldberg and West (1985) proved that such a partition is always possible with k+1 cuts. Fewer cuts, however, can often be used. Right: only 4 cuts

Existing results (2) There is an essentially brute force O(n k-2 ) algorithm for finding least number of cuts. However no algorithm is known which is polynomial time in both n and k. Such an algorithm we are trying to find.

Investigation approaches (1) Can be formulated as an integer program To the right: x ij = 0 means bead i is in same group as j. e ij =1 iff beads i and j are adjacent.

Investigation approaches (2) But, solving integer programs is hard Random algorithm: needs O(k log k) cuts and the resulting bisection is not precise.

Investigation approaches (3) Graph bisection: division of the vertex set of a graph with an even number of vertices into two equal, non-intersecting subsets which minimizes number edges between vertices belonging to different subsets General graph bisection is NP-complete, but bisections of some graphs can be done in polynomial time

Investigation approaches (4) Conjectures rejected: –There always exists a smallest bisection in which some two cuts are n beads apart –Given a division of beads into two equal strings, there is always an exchange of some r > 1 bead substrings which improves color distribution

Prospects and possibilities Relaxations of the original problem –O(k) cuts –particular bead configurations Perhaps the problem is NP-complete

Work continues...

Any questions?