IT討論區(209) - 水退？！
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做左兩年土木頂唔順個環境想轉行
想請教下咁多位師兄,如果靠自學儲skills,之後將啲做過既project塞入profolio有無機會入到行? 暫時睇過Freecodecamp, hyperskill, udemy, codecademy呢幾個平台, 打算揀一個浸返幾個月~半年再去搵IT工, 如果入到行做多排再走去補master
但就有啲驚HR見我無相關degree或者讀過bootcamp就直接filter走我
本身有個唔關事既Mech Degree,叫讀過下IT既intro course可以自學到
想請教下咁多位師兄,如果靠自學儲skills,之後將啲做過既project塞入profolio有無機會入到行? 暫時睇過Freecodecamp, hyperskill, udemy, codecademy呢幾個平台, 打算揀一個浸返幾個月~半年再去搵IT工, 如果入到行做多排再走去補master
但就有啲驚HR見我無相關degree或者讀過bootcamp就直接filter走我
本身有個唔關事既Mech Degree,叫讀過下IT既intro course可以自學到
又咪喺一個伏
…
…
太多古代人
本身做緊Sub仔既Engineer位,但真係做到對呢行無哂信心,加上有部分實在啲既原因(E.g. 地盤環境太惡劣,安全問題,做細公司要駐site行到邊都充滿煙味), 呢啲我諗至少轉返做純文書工作可以解決到
我嗰部真係用唔到nfc
IT行業水退緊
Opening 愈黎愈少 （globally + locally)
然後多得班大公司+agent 吹奏入行既人愈黎愈多
唔計bootscam/自學 你覺得點解你白紙讀幾個月就可以係一堆IT related degree 仔搶飯碗
然後讀master 唔覺cs master 依家收non related bachelor grad
Opening 愈黎愈少 （globally + locally)
然後多得班大公司+agent 吹奏入行既人愈黎愈多
唔計bootscam/自學 你覺得點解你白紙讀幾個月就可以係一堆IT related degree 仔搶飯碗
然後讀master 唔覺cs master 依家收non related bachelor grad
你有追下post應該知bootcamp撚係會比人screen 走多過有認受性
老細根本唔知你做緊乜
點升
點升
出席多啲會 +
見老闆多過打曲
見老闆多過打曲
你對IT邊樣有興趣
純為錢 不如nursing /老師
仲香啦
最緊要係老左都高人工
純為錢 不如nursing /老師
仲香啦
最緊要係老左都高人工
個ring network algorithm咁催眠。。。左迴轉，右迴轉自己認大佬。。。
1. 屌你都係冇問題
「醫生 我有病呀」
「你邊到唔舒服呀」
「我病到返唔撚到工呀」
2. 你做啲咩要ring network
「醫生 我有病呀」
「你邊到唔舒服呀」
「我病到返唔撚到工呀」
2. 你做啲咩要ring network
定係去上返個bootcamp<<(寫落cv認受性應該高啲?)
你夠膽寫落CV 手巴夠膽碎左你封job application
之前你唔係咁講㗎喎
答得到啲題目 最多笑下你on9哂咗幾皮嘢
答得到啲題目 最多笑下你on9哂咗幾皮嘢
Suppose all processes are
identical (same sets of states, transition
functions, etc.).
Then it’s impossible to elect a leader,
even under the best assumptions
(bidirectional communication, ring size n
known to all).
By contradiction. Assume an algorithm that solves the
problem.
• Assume WLOG that each process has exactly one start
state (could choose same one for all).
• Then there is exactly one execution
C0, M1, N1, C1, M2, N2, C2
,…
• Prove by induction on the number r of completed rounds
that all processes are in identical states after r rounds.
– Generate same messages, to corresponding neighbors.
– Receive same messages.
– Make the same state changes.
• Since the algorithm solves the leader election problem,
someone eventually gets elected.
• Then everyone gets elected, contradiction.
睇唔明個prove
identical (same sets of states, transition
functions, etc.).
Then it’s impossible to elect a leader,
even under the best assumptions
(bidirectional communication, ring size n
known to all).
By contradiction. Assume an algorithm that solves the
problem.
• Assume WLOG that each process has exactly one start
state (could choose same one for all).
• Then there is exactly one execution
C0, M1, N1, C1, M2, N2, C2
,…
• Prove by induction on the number r of completed rounds
that all processes are in identical states after r rounds.
– Generate same messages, to corresponding neighbors.
– Receive same messages.
– Make the same state changes.
• Since the algorithm solves the leader election problem,
someone eventually gets elected.
• Then everyone gets elected, contradiction.
睇唔明個prove
碎完痴返Re:2
個course lecture 1 教咪學下
個course lecture 1 教咪學下