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Frobenius Number for Three Numbers

Frobenius Number for Three Numbers. By Arash Farahmand MATH 870 Spring 2007. Coin Exchange Problem. What is the largest amount that cannot be changed? First tackled by G. Frobenius (1849-1917) and J. Sylvester (1814-1897). N = 2. With two coins Formula: g ( n , m ) = nm – n – m.

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Frobenius Number for Three Numbers

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  1. Frobenius Numberfor Three Numbers By Arash Farahmand MATH 870 Spring 2007

  2. Coin Exchange Problem • What is the largest amount that cannot be changed? • First tackled by G. Frobenius (1849-1917) and J. Sylvester (1814-1897)

  3. N = 2 • With two coins • Formula: g (n, m) = nm – n – m

  4. Complexity for N > 2 • Frobenius numbers by lattice point enumeration • Polynomial time on average for fixed N • Relatively fast algorithm for lattice reduction applied to find Frobenius number when N = 3

  5. Algorithm • Step 1. Form the homogeneous basis and then use lattice reduction to obtain a reduced basis V • Step 2. Make use of V and the ILP methods to determine the two axial protoelbows (x1, y1) and (x2, y2)

  6. Algorithm (continued) • Step 3. If y1 or x2 is zero then the elbow set is {(x1, 0), (0, y2)}; otherwise it is {(x1, 0), (0, y2), (x1+ x2, y1 + y2)} • Step 4. In all cases the Frobenius number is max [{(x1, y1 + y2), (x1+ x2, y2} B] - ΣA

  7. References • M. Beck and S. Robins, Computing the Continuous Discretely, 2006, Springer • S. Wagon, D. Einstein, D. Lichtblau, and A. Strzenbonski, Frobenius Numbers by Lattice Point Enumeration, http://stanwagon.com/public/FrobeniusByLatticePoints.pdf , Revised Aug 1, 2006, last visited February 15, 2007

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