各位大大好
剛寫了台科考古但手邊沒答案
在這邊po上我的答案
如果有寫錯的拜託各位多多指教 ((跪

題目:
https://imgur.com/w84LKMj
https://imgur.com/ZxDTFi5
https://imgur.com/QDiOI2W
https://imgur.com/cWCRoLW



我的答案：

1.F T F F T

2.(a) O(1) < O(logn) < O(n) < O(nlogn) < O(n^2) < O(2^n) < O(n!)
(b) Theta(n^2) 不是很確定

3.(a) //-de*+ab-cdf
(b) de-ab+cd-*/f/

4.(a) worse case: O(n^2)
best case: O(nlogn)
(b) 若是要將數列由小到大排序，worse case的情況是pivot都選到最小或最大值
best case 的情況是pivot都選到中位數
不是很確定是不是這樣寫就好

5.假設每個resource都只有一個=>RAG畫出來有circular wait的現象=>有deadlock

6.(a)log_2(16)+log_2(2048) = 4+11 = 15 bits
(b)log_2(16) = 4 bits
(c)log_2(4)+log_2(2048) = 2+11 = 13 bits

7.8 times 有查過前面的文章，應該是對的

8.(a)code sequence 1: 1+1+8 = 10 => code sequence 1 執行比較多的指令
code sequence 2: 2+1+1 = 4

(b)code sequence 1: 1x7 + 1x3 + 8x1 = 18 cycles => 一樣快
code sequence 2: 2x7 + 1x4 + 1x1 = 18 cycles

(c)code sequence 1: 18/10 = 1.8
code sequence 2: 18/4 = 4.5

(d)code from compilier 1: 10^9 x( 3x7 + 1x3 + 1x1)/( 5x10^9 ) = 5 seconds
code from compilier 2: 10^9 x( 1x7 + 3x3 + 4x1)/( 5x10^9 ) = 4 seconds

=> code form compilier 2 is faster

(e)code from compilier 1: 10^9 x(3+1+1)/(5x10^6) = 1000 mips
code from compilier 2: 10^9 x(1+3+4)/(4x10^6) = 2000 mips


=>code from compilier 2 is faster



9.如圖 https://imgur.com/MjbWiXu

10.(a) 2 no-ops between 16 and 20
2 no-ops between 20 and 24

(b) 假設branch是在EX stage決定的
total cycle = 3+1+3+(5-1) = 10 ctcles



雖然台科相較台清交容易，但還是怕手殘有算錯
還請各位幫忙找錯，感激不盡！

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對欸! 感謝b大
對!我手殘寫成8個，感謝
※ 編輯: brilliantl (111.249.1.196), 01/17/2018 00:32:47