有關三元數列的探討

Abstract

長度為n的三元數列(0, 1, 2)，探討(一)0為偶數個1為偶數個，或(二)0為偶數個1為奇數個，或(三)0為奇數個1為偶數個，或(四)0為奇數個1為奇數個的方法數時，就離散的傳統上來說是用遞迴關係去求解。本文將建構一對一函數，利用一對一函數的特性去求此問題的解，與以前的方法比較起來僅需要了解一對一函數的特性即可求解，易懂且不需要用到比較複雜的遞迴觀念。

The problem of the number of ternary sequences of length n with :\n(a) 0 is even, 1 is even, \n(b) 0 is even, 1 is odd,\n(c) 0 is odd, 1 is even, \n(d) 0 is odd, 1 is odd, \nhas been solved by recurrence relations before. We solve the problem by constructingone-to-one functions, and use the characteristics of one-to-one functions to solve this problem. Our method is simpler than those methods which have been done before.