奇想西卡

[手機] HTC diamond on Android 2.1

HTC diamond 這手機買了快兩年 ,, 刷了不少大大小小的ROM

最後沉迷在 gpower 威能底下

不過 gpower 也幾乎不更新了 ,

最後一個 ROM 畫面

基本上完全符合需求了 , 可以撥接電話、出門在外可以開 papago 導航、開個 wifi 上上網看看東西

真的沒WIFI也可以開通威寶單日3G上網

不過當機不斷也鈍鈍的

升級不能、更新無望身邊的人也一個一個跳 iphone 跟 android

也害我想換手機了

去年就看到有人用外掛的android在 dimond上執行

但是 wifi 不能、 gPS不能、相機不能 ,, 所以就放棄了

直到最近發現新版的 android 2.1 成熟多了

參考 : http://www.mobile01.com/topicdetail.php?f=224&t=1525030&last=21001545

原始網址 http://forum.xda-developers.com/showthread.php?t=621498

基本上幾乎完美了，

安裝蠻簡單的 , 基本上抓下 cab檔後一直跑就好了

不過執行很慢要有耐心 ... 一度還以為死當了

用了幾天感覺很順手, market更是方便

缺點

1. 太耗電 , 使用中幾乎就是燙電力也是直線的消耗 ...

2. 有點頓 (雖然習慣了 diamond 跑M2 也是盾阿...)

3. 相機還不能用

除此以外已經可以完全取代 wm6.1 (or 6.5) , 害我好心動想去敗一隻原生 android 2.2 的來用

>__________ <"

http://forum.xda-developers.com/showthread.php?t=711692

[prj][XDANDROID][2.2][AOSP][18.7.10]RC1 RELEASED

android 2.2 也出了 ,, 等成熟一點在裝上去試試看

Known Issues
Many things are still not working properly, and the developers are fully aware of 99% of them.

SDCard - Most SDCards are working, but if you the 'waiting for SDCard' message, you should try one of these fixes (and please let us know if it works)
Fix 1: Format your card with this tool http://www.sdcard.org/consumers/formatter/
Fix 2: Edit your startup.txt file and add these entries to your cmdline:
set cmdline "lcd.density=240 msmts_calib=0x395.0x3c9.0x79.0x77 clock-7x00.a11=500 msmvkeyb_toggle=off pmem.extra=1 gsensor_axis=-2,-1,3
msmsdcc_1bit msmsdcc_fmax=14000000 msmsdcc_nopwrsave"
Fix 3: Tell your wife / girlfriend that you are about to make some modifications to her phone... change the ringtone and steal her old Class 2 SD Card to use with Android. She will never discover what you have done, and if she does, you'll be able to use Android to send texts to your former girlfriends ;-)
Camera - Not working
GPS - Should be working, but it's only just been enabled (so could be buggy)
Battery Status - Highly problematic for most users (don't trust the capacity Android tells you, it's simply up and down a lot, and highly inaccurate)
WiFi - Working for most people, suspected to cause slight heat, definitely drains the battery faster than WinMo. Some users have reported freezes when switching WiFi off
Audio - Speaker working for most people, microphone can have issues
Deep Sleep - Working for most people, but can have issues waking up (sometimes, it will not wake up)
GPRS - Working
3G - Working for most
PIN Code - Not working
GSensor - Working (Screen only handles 0 and 270 degrees)
Bluetooth - Experimentally working (don't expect too much, but it's working for a number of users)
OpenGL ES - Not working
USB - Supported, but will NOT operate in mass storage mode. You can charge via USB and you can debug, but forget accessing your files by it.
Headphone Socket - Semi working (see FAQ for details on how to make it work)
Application Errors: It's very common to see 'Application (xxxxxx) is not responding: Force Close or Wait'. Click wait. I do not know the precise reason for this, but from what I can gather, it COULD be related to the slow speed of the SD Card causing the app response times to JUST be a little too slow for Android's liking, thus causing the message. This seems to happen more after bootup or wakeup when the device is busy doing stuff in the background.
Cannot choose ringtones: This is believed to be caused by the media scanner failing (due to discovering some filenames with none English characters). The best option of all is to use a CLEAN SD Card. But, if you insist on hosting other files on the same SD Card, you MUST make sure they have English character filenames.

[文件] 電子交易安全期中考 -2

之前版本有幾個錯誤先修改了，就重寄一份希望不會打擾到大家 ^^;;

檔案中紅字為修改過部分。

第一題：補上第二個式子的證明

第二題：第二個式子筆記沒抄到不知道有哪為好心的人士幫忙補上 ,,

第五題： Z3 的三個表應該是寫錯了，再mod 3 之下應該不會有大於三的數 (感謝士軒的提醒)

1. 寫出並證明費碼定理

Fermat 定理，當 (a,p)=1 (a跟p互質)

a. a^p-1 = 1 mod p

證明：在Zp 之中 a、p互質

[1*2*3..*(p-1)]modp={(1*a)(2a)(3a)...[a(p-1)]}modp <-同成a係數只是重排

={a^(p-1)*[1*2*3..*(p-1)]}modp <-右邊將a提出成為a的p-1次方

因為 a、p互質所以存在反元素將兩邊同乘反元素 [1*2*3..*(p-1)]^-1 消掉

所以 1modp = a^(p-1)modp -> a^p-1 = 1 mod p

b. a^p = a mod p

證明：假設 (a,p)≠1 則 a為p的倍數並可寫成a^p≠a mod p

a^p≠a mod p

a^p –a ≠ (a-a) mod p (兩邊同減a)

a(a^p-1 –1) ≠ 0 mod p (左邊將a提出)

∵a(a^p-1 –1) ≠ 0 mod p

∴ p不整除 a 且p不整除 (a^p-1 –1)

與假設不符合所以得證 a^p=a mod p

2. 寫出並證明尤拉定理

尤拉定理，當 (a,p)=1 (a跟p互質)

a. a^Φ(n)＝1 mod n

[X1*X2*X3..XΦ(n)]mod n =[(X1modn)(X2modn)(X3modn)…[XΦ(n)]modn] modn

([(X1modn)(X2modn)(X3modn)…[XΦ(n)]modn] modn 同乘a等於係數重排)

= [aX1modn aX2modn aX3modn …. [aXΦ(n)]modn ] modn (右邊係數同乘a)

=a^Φ(n) [X1*X2*X3..XΦ(n)]modn (右邊將a提出)

1 mod n= a^Φ(n)(左右同乘反元素[X1*X2*X3..XΦ(n)]^-1)

得證：a^Φ(n)＝1 mod n

b. 當 (a,n)≠1　且(a<n) ，a^Φ(n^＋¹⁾＝a mod n

3. 寫出並證明RSA所用的基本定理

RAS基本定理：m代表一個訊息，n=p*q ， m^Φ(n)+1＝m mod n

a.當 (m,n) = 1 根據尤拉定理 a^Φ(n) ＝ 1 mod n

m^Φ(n)+1＝m mod n(兩邊同乘m後得證 )

b.當 (m,n)≠1 所以　b-1 m = C1p b-2 m = C2q

b-1 m = C1 * p , (m , q) = 1 根據尤拉定理可寫為　

m^Φ(q) ＝ 1 mod q

[m^Φ(q)]^Φ(p)＝ 1^Φ(p)mod q ( 兩邊同乘Φ(p) 次方)

m^Φ(n) = 1 mod q (因為 n =p*q，1的任何次方還是1 )

m^Φ(n) = 1+kq (存在一整數k，1modq = 1+kq)

m^Φ(n+1) =m + kqm (兩邊同乘m)

m^Φ(n+1) = m + kqC1p (m =C1*p代入)

m^Φ(n+1) = m + kC1n (其中 n=p*q代入)

m^Φ(n+1) = m mod n (m+kC1n等於 m mod n)

b-2 m = C2*q 同 b-1 可證 m^Φ(n+1) = m mod n

4. 證明 (a+b) mod n = [(a mod n) + (b mod n)] mod n

假設 (a mod n)= Ra (b mod n) = Rb , 則 a = Ra + j n ; b = Rb + kn

(a+b) mod n = [(Ra+jn )+(Rb+kn) ]mod n

= [Ra+Rb+ (j+k)n ] mod n

=(Ra+Rb) mod n <- 根據假設 (a mod n)= Ra (b mod n) = Rb帶入

= [(a mod n )+ (b mod n )] mod n

5. 何謂一個體, 做 Z₃ 的三個表

體：除了0之外，有兩個交換環(加法和乘法)，加法和乘分具分配律，並有不具有0的真因子，除0之外皆有反元素

加法	0	1	2
0	0	1	2
1	1	2	0
2	2	0	1

乘法	1	2
0	0	0
1	1	2
2	2	1

W	-w	w^-1
0	0	0
1	2	1
2	1	2

6. 若 Z=k．Φ(n)+q 則 a^z = ? mod n

a^z =a ^k^．^Φ(n)+q = [a^Φ(n)]^k * a^q

= {[(a^Φ(n))^k modn ][a^qmodn]}modn (根據尤拉 a^Φ(n)＝1modn )

= {(1modn)(a^qmodn)}modn

=a^q modn

7. 說明 discrete logarithm problem

離散對數依照方程式： y=g^x mod p

給定g x p 可以很快算出 y ，但是只給 y g p 則很難算出 x的值。

8. 用 Euclid algorithm 求 GCD(86,32)

GCD(86,32)

86=2*32+22 GCD(32,22)

32=1*22+10 GCD(22,10)

22=2*10+2 GCD(10,2)

10=2*5+0 得 GCD(86,32)=2

[文件] 電子交易安全期中考

1、2 題不確定是否完整，

就是第二個部份的證明不知道要不要寫，

如果需要就要再補上去，

不確定是否有打錯或是筆記操錯的地方有錯請指正，thanks