Mechanism Design - Enumeration of Kinema Episode 1 Part 1 potx

Appendix A Solving m Linear Equations in n Unknowns In this appendix, we develop a procedure for solving a system of m linear equations in n variables subject to a constraint that all the variables are nonnegative integers. We first discuss a method for solving one equation in n unknowns. Then, we extend the method to solving a system of m equations in n unknowns. A.1 Solving One Equation in n Unknowns Consider the following linear equation: c 1 x 1 + c 2 x 2 + c 3 x 3 +···+c n x n = k, (A.1) where x i s are the variables and c i s are the coefficients. The c i s are nonnegative integers and k is a positive integer. We wish to solve for x i , for i = 1, 2, ,n subject to a constraint that all x i s must be nonnegative integers. In addition, the following constraint may be imposed: x i ≤ q i (constant) . (A.2) Since there are n unknowns in one equation, we may choose n − 1 number of unknowns arbitrarily and solve Equation (A.1) for the remaining unknown, provided that all the solutions are nonnegative integers. This can be accomplished by a com- puter program using a nested-do loops algorithm to vary the value of each x i and check for the validity of the solutions. A more rigorous procedure for solving one linear equation in two unknowns can be found in [1]. © 2001 by CRC Press LLC Table A.1 A Nested-do Loops Algorithm for Solving One Linear Equation in n Unknowns. FOR I 1 =0TOq 1 x(1) = I 1 FOR I 2 =0TOq 2 x(2) = I 2 FOR I 3 =0TOq 3 x(3) = I 3 . . . FOR I n−1 =0TOq n−1 x(n − 1) = I n−1 x(n) = (k − c(1) ∗ x(1) − −c(n − 1) ∗ x(n − 1))/c(n). IF x(n) < 0, DISCARD THE SOLUTION. IF x(n) ≥ 0, SAVE THE SOLUTION. NEXT I n−1 . . . NEXT I 3 NEXT I 2 NEXT I 1 A.2 Solving m Equations in n Unknowns Next, we consider a system of m linear equations in n unknowns: c 11 x 1 + c 12 x 2 +···+c 1n x n = k 1 , c 21 x 1 + c 22 x 2 +···+c 2n x n = k 2 , . . . (A.3) c m1 x 1 + c m2 x 2 +···+c mn x n = k m , where the coefficients c ij s are nonnegative integers, the constants k i s are positive integers, and n>m. Furthermore, the solutions to the system of equations are subject to a constraint that all thevariables x i , for i = 1, 2, ,n, must benonnegative integers. Writing Equation (A.3) in matrix form yields C x = 0 , (A.4) © 2001 by CRC Press LLC where C =      c 11 c 12 c 13 c 1n −k 1 c 21 c 22 c 23 c 2n −k 2 . . . . . . . . . . . . . . . . . . c m1 c m2 c n3 c mn −k m      , x = [ x 1 ,x 2 , ,x n , 1 ] T . Applying Gauss elimination, Equation (A.4) can be reduced to an upper triangular form: G x = 0 , (A.5) where G =      g 11 g 12 g 13 g 1n p 1 0 g 22 g 23 g 2n p 2 00 . . . . . . . . . . . . 000g m,m g mn p m      . The last row in Equation (A.5) contains n − m + 1 unknowns which can be solved by the procedure outlined in the preceding section. Once x m , ,x n are known, the remaining unknowns are found by back substitution. Example A.1 We wish to solve the following two equations for all possible nonneg- ative integers of x i s x 1 + x 2 + x 3 = 6 , (A.6) 2x 1 + 3x 2 + 4x 3 = 16 . (A.7) Subtracting Equation (A.6)×2 from Equation (A.7), we obtain x 2 + 2x 3 = 4 . (A.8) Equation (A.8) contains two unknowns. Since both 2x 3 and 4 are even numbers, x 2 must be an even number. Let x 2 assume the values of 0, 2, and 4, one at a time, and solve Equation (A.8) for x 3 . Once x 3 is found, we solve Equation (A.6) for x 1 .Asa result, we obtain the three solutions: Solution x 1 x 2 x 3 1 402 2 321. 3 240 © 2001 by CRC Press LLC References [1] Gelfond, A.O., 1981, Solving Equations in Integers, Translated by Sheinin, O.B., Mir Publisher, Moscow. © 2001 by CRC Press LLC . I 3 NEXT I 2 NEXT I 1 A.2 Solving m Equations in n Unknowns Next, we consider a system of m linear equations in n unknowns: c 11 x 1 + c 12 x 2 +···+c 1n x n = k 1 , c 21 x 1 + c 22 x 2 +···+c 2n x n =. I 3 . . . FOR I n 1 =0TOq n 1 x(n − 1) = I n 1 x(n) = (k − c (1) ∗ x (1) − −c(n − 1) ∗ x(n − 1) )/c(n). IF x(n) < 0, DISCARD THE SOLUTION. IF x(n) ≥ 0, SAVE THE SOLUTION. NEXT I n 1 . . . NEXT I 3 NEXT. x i , for i = 1, 2, ,n, must benonnegative integers. Writing Equation (A.3) in matrix form yields C x = 0 , (A.4) © 20 01 by CRC Press LLC where C =      c 11 c 12 c 13 c 1n −k 1 c 21 c 22 c 23