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1F→:第二小題的CLA 應該是 , g1,p1 ... g15,p15 可以同步11/14 01:08
2F→:計算,所以只花 1 gate delay , 然後 c0 要先被計算出來11/14 01:08
3F→:(否則無法算C1,C2,C3,C4) , 這需要 2 gate delays11/14 01:09
4F→:(Sorry c0 那邊應該是 G0,P0,....G4,P4 , 且 G0...G4需11/14 01:13
5F→:要花 2 gate delays 才對)11/14 01:13
6F→:最後才是 C1,C2,C3,C4 , 這四個同步計算,也是花 2 gate11/14 01:14
7F→:delay , 所以 1+2+2=5 (我是這樣看的@@)11/14 01:15
1F→:推!11/14 01:34
5F→:可以請教樓上, 4 disks in protection group 應該被解11/10 19:44
6F→:讀成 "4+1個disks"儲存all data block與 parity block11/10 19:45
7F→:嗎?我是指交大那一題 , 3q11/10 19:46
1F→:因為原題意要求,恰好一人被每一個人love, BUT 前半段只11/08 17:28
2F→:說明 "存在一人" (一人以上) everyone loves , 所以要11/08 17:29
3F→:消弭存在2人 everyone loves 的情況11/08 17:29
1F→:第一題是 O(n),它會check n/2 次 if statment11/08 17:53
5F→:YES11/03 17:51
3F→:用列運算把A轉成上三角矩陣的過程:第一步,第一列n個11/01 23:57
4F→:entry會執行 n-1次,因為要把 a21,a31...an1削掉11/02 00:00
5F→:所以第一步是 n(n-1),依此類推列運算第二步 (n-1)(n-2)11/02 00:02
6F→:, (n-2)(n-3) .... 2*1 , 當A變成上三角矩陣後,det(A)11/02 00:05
7F→:相當於上三角矩陣對角項相乘 , 所以有N-1次乘法11/02 00:06
8F→:so 全部需要的乘法次數:11/02 00:07
9F→:n(n-1)+(n-1)(n-2)+...+3*2+2*1+ (n-1)11/02 00:07
10F→:ㄟ等等我發現答案跟你的不太一樣囧,我先收回我的解法xd11/02 00:09
2F→:找TLB與page table是看virtual address的page number欄11/01 23:37
3F→:位,在page table中 page number會對應到一個physical11/01 23:39
4F→:number , physical number+page offset就形成 physical11/01 23:40
5F→:address , 此時再抓 physial address 的 index欄位與11/01 23:41
6F→:tage欄位到cache 找data block (就是cache Hit/miss)11/01 23:42
1F推:題目意思是求x (length=8的bit string) , 將x帶入狀態11/01 23:25
2F→:轉換函數(initial state = s0)之後,輸出的狀態還是 s011/01 23:27
3F→:你把 對應的FSM畫出來之後 trace 2,3次會發現11/01 23:28
4F→:length=8 的 bit string之中 有 8個1`s , 8個0`s 還有11/01 23:30
5F→:4個0's+4個1's 作不全相異排列11/01 23:31
1F→:我算的跟你一樣10/29 00:05