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191014K02 | Day 1 Lecture 1

191014K02: Day 1 Tutorial

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Definitions

ccc-approximate mincut

A ccc-approximate mincut is a set of at most crcrcr edges if rrr is the number of edges in a mincut.

min kkk-way cut

A minimum k-cut is a smallest set of edges whose removal would partition the graph to at least kkk connected components

G(n,p)G(n,p)G(n,p) graphs

The G(n,p)G(n, p)G(n,p) model, due to Erdös and Rényi, has two parameters, nnn and ppp. Here nnn is the number of vertices of the graph and ppp is the edge probability. For each pair of distinct vertices, vvv and w,pw, pw,p is the probability that the edge (v,w)(v, w)(v,w) is present. The presence of each edge is statistically independent of all other edges. The graph-valued random variable with these parameters is denoted by G(n,p)G(n, p)G(n,p). When we refer to “the graph G(n,p)G(n, p)G(n,p)”, we mean one realization of the random variable.

Problems

  1. Generalize the mincut argument to ccc-approximate mincuts.

  2. Generalize the mincut argument to min kkk-way cut.

  3. Prove #min kkk-cuts is at most nO(k)n^{O(k)}nO(k).

  4. Show that G(n,1/2)G(n, 1/2)G(n,1/2) graphs have:

    • many cliques of size 2log⁡n−o(log⁡n)2 \log n-o(\log n)2logn−o(logn) in expectation, and
    • no cliques of size 2log⁡n+o(log⁡n)2 \log n+o(\log n)2logn+o(logn) in expectation (and with high probability).
  5. Consider the following algorithm for finding a minimum cut. Assign a random score to each edge, and compute a minimum spanning tree. Removing the heaviest edge in the tree breaks it into two pieces. Argue that with probability ω(1/n2)\omega(1/n^2)ω(1/n2), those pieces will be the two sides of a minimum cut. Hint: relate this algorithm to the contraction algorithm we did in the class. Also think about Kruskal’s algorithm.

  6. Show that for every n≥4n \geq 4n≥4, there is a simple graph GnG_nGn​ on nnn vertices that has at least (n2){n \choose 2}(2n​) distinct minimum cuts.

  7. Show that for every n≥3n \geq 3n≥3, there is a simple graph GnG_nGn​ on nnn vertices such that the value of ILPOPT of the vertex cover ILP associated with GnG_nGn​ is at least one less than twice the value of LPOPT of the vertex cover LP associated with GnG_nGn​, i.e:

ILPOPT(Gn)≥2⋅LPOPT(Gn)−1.\text{ILPOPT}(G_n) \geq 2\cdot \text{LPOPT}(G_n) - 1.ILPOPT(Gn​)≥2⋅LPOPT(Gn​)−1.

  1. Consider the Set Cover instance shown in the figure below.

    • Show that all-half is the unique LPOPT for this instance.
    • Show that if you include every set in F′\mathcal{F}^\primeF′ with probability xsx_sxs​, then the probability that F′\mathcal{F}^\primeF′ covers UUU is at most 2−Ω(n)2^{-\Omega(n)}2−Ω(n).


© 2022 • Neeldhara Misra • Credits •

 

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