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By Pierluigi Crescenzi, Viggo Kann.

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5) we get z ∗ = Ψ(f (Ψ−1 (z ∗ ))). Hence, Ψ−1 (z ∗ ) = f (Ψ−1 (z ∗ )). This means, that x∗ := Ψ−1 (z ∗ ) is a fixed point of f satisfying x∗ ∈ K. 23 2 Brouwer’s Fixed Point Theorem Case 2. We assume that K has dimension m with m < n. , ym so that x = y1 · u(1) + ... + ym · u(m) . With U := . u(1) .. .. u(m) ∈ Rn×m we have x = U · y and U T · x = y. Setting  y1  .  m  and y :=   ..  ∈ R ym  ˜ := {y ∈ Rm : y = U T x with x ∈ K} K ˜ is bounded, convex, and closed with dimension m. For any we have that K ˜ we define y∈K f˜(y) := U T · f (U · y).

N. In 1974, Tamir published an algorithm for solving the NCP for the case that f is a so-called Z-function. Tamir’s algorithm is a generalization of Chandrasekaran’s algorithm which solves the linear complementarity problem for the case that the given matrix M is a so-called Z-matrix. Proving his results, Tamir used the iterative processes of Gauss-Seidel and of Jacobi. 4 we present a different proof where these iterative processes are not used. Instead, we use the least element theory and the Poincar´e-Miranda theorem.

Then due to f (x) = x ⇔ x · (x − 1) = 0 f has two fixed points x∗ = 0 and x∗∗ = 1, but neither belongs to K. 4. (concerning convexity) A set K in Rn is convex if x, y ∈ K ⇒ λ · x + (1 − λ) · y ∈ K for all λ ∈ [0, 1] and for all x, y ∈ K. Let K = {x = x1 x2 ∈ R2 : 1 ≤ 2 x21 + x22 ≤ 1}. K is just a disc with a hole and therefore not convex: x= 3 4 0 − ,y = 3 4 0 ∈K but 0 0 K = 1 1 · x + (1 − ) · y. 2 2 Let f : K → K be defined by rotating K by 45 degree to the left. Then the zero vector is the only fixed point of f .

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A Compendium of NP Optimization Problems by Pierluigi Crescenzi, Viggo Kann.

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