By Pan V.

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**Extra resources for Algorithm for approximating complex polynomial zeros (1998)**

**Example text**

There is a source s that feeds all the lines. The capacity of each arc (s, S) emanating from the source equals x(S) · c(S). There is a sink t that is fed by all the rectangles. The capacity of every arc (u, t) entering the sink equals 1. Observation 1. There is a one-to-one correspondence between vectors y such that (x, y) is a partial cover and ﬂows f in Nx . The correspondence y ↔ fy satisﬁes fy (u, t) = S|u∈S y(S, u), for every rectangle u ∈ U, and fy (s, S) = u∈S y(S, u), for every line S ∈ S.

6: for each rank ρ in DR ∩ [c, . . , c + ci1 ] do 7: Let be the line stored at A2 [ρ − c]. 8: Update DI to record the pair ( 1 , ) as the intersection with rank ρ. } 10: Let c := c + ci1 . 11: i1 := i1 + 1. } 13: i2 := i2 + 1. 14: end if 15: end for algorithm CountAndRecord is an iterative algorithm that computes, during the i-th iteration the element with rank i in the ﬁnal sorted order (Line 3 of Algorithm 2). Thus we only need to be able to ﬁnd out whether the line that will be the i-th element in sorted order is an element involved in an intersection.

We denote the LP-relaxation by lp-soft. The integrality gap of both lp-hard and lp-soft is at least 2 − o(1) even in the one-dimensional case. Consider an instance that contains k + 1 rectangles and two lines of capacity k that intersect all the rectangles. A fractional optimal solution is x∗ (S) = (k + 1)/(2k) for each line S and y ∗ (S, u) = 1/2 for every line S and rectangle u. This means that the value of the fractional minimum is 1 + k1 , while the integral optimum is 2. The following deﬁnitions apply to both lp-hard and lp-soft.

### Algorithm for approximating complex polynomial zeros (1998) by Pan V.

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