考試

106成大資聯離散[參考解答]

1. In how many ways can 36 identical robots be assigned to five assembly lines with

(a) at least four robots assigned to each line? (5%)
(b) at least four, but no more than ten, assigned to each line? (5%)

參考解答:
(a) $\binom{20}{4}$
(b) $\binom{20}{16} – \binom{5}{1}\binom{13}{9} + \binom{5}{2}\binom{6}{2}$


2. Let $
D =
\begin{bmatrix}
2 & -1\\
-1 & 2 & -1 & & 0\\
& -1 & 2 & \ddots\\
& & -1 & \ddots & -1\\
& 0 & & \ddots & 2 & -1\\
& & & & -1 & 2
\end{bmatrix}
, \, i.e., \,
\begin{equation}
D(i,j) =
\begin{cases}
2, & \text{if $i = j.$}\\
-1, & \text{if $|i – j| = 1.$}\\
0, & \text{elsewhere.}
\end{cases}
\end{equation}
$

(a) Use recurrence relation to express determinant of $D, \, i.e., \, |D|.$ (5%)
(b) Find the general solution for $|D_n|.$ (5%)
(c) $
|D_1| = 2, |D_2| =
\begin{vmatrix}
2 & -1\\
1 & 2
\end{vmatrix}
= 3.
$ Find $|D_{100}|.$ (5%)

參考解答:
(a) $D_n = 2D_{n-1} – D_{n-2}, \, \forall n \geq 3. \ D_1 = 2, \, D_2 = 3.$
(b) $|D_n| = n + 1$
(c) $|D_{100}| = 101$


3. Solve the following recurrence relations: $6a_n – 5a_{n-1} + a_{n-2} = \sin(n \pi)$ with $a_0 = 1, \, a_{-1} = a_{-2} = 0.$ (10%)

參考解答:$a_n = 3(\dfrac{1}{2})^n + (-2)(\dfrac{1}{3})^n + \dfrac{1}{12} \sin(n \pi), \, \forall n \geq 0.$


4. Let $(Q,\oplus,\otimes)$ denote the field, where $\oplus$ and $\otimes$ are defined by $a \oplus b = a + b – k, \, a \otimes b = a + b – (ab/m),$ for fixed elements $k,m (\neq 0)$ of $Q.$ Determine the values for $k$ and $m$ in each of the following:

(a) The zero element for the field is 5. (5%)
(b) The additive inverse of the element 8 is -7. (5%)
(c) The multiplicative inverse of 3 is 1/6. (5%)

參考解答:
(a) $k = 5, \, m = 5.$
(b) $k = \dfrac{1}{2}, \, m = \dfrac{1}{2}.$
(c) 略


試題(pdf):連結

有任何問題,或想要完整解釋的,歡迎在下方留言唷!

7 留言

  1. ting

    大大安安,想問地2題的(2)答案是怎麼算的 謝謝!!

  2. ting

    *更正:是第一題的(2) 想請問怎麼算,謝謝大大

    • 文章作者的留言

      mt

      第一題的(b)是全部assembly lines at least four robots的情況 – 一個assembly lines大於10 robots其他四個lines at least four robots的情況 + 兩個assembly lines大於10 robots其他三個lines at least four robots的情況,因為沒有大於三個assembly lines大於10的情況,所以不用再繼續算。

      • ting

        終於搞懂了,非常感謝您!!!

      • ting

        不好意思再追問一下,假設有”大於三個assembly lines大於10的情況”,算法我了解了但正負號的部分該如何決定呢,是按照第一個情況是”-“,接續著”+”、”-“、”+”下去嗎?

回覆留言對象 取消回覆