據業界推估，台積電和三星等領先企業已成功應用AI遷移學習，
(半導體潔淨室轉移問題)渡河大作戰最佳排程 高速版，AI 推理約需3-10分鐘，用AI生成Python程式，可加快執行速度到千分之一秒以下 。Use Code Gen to solve River Puzzle 2 ，結果必須視覺化，動畫尤佳
星期五, 3月 28, 2025
The Tiny Star Explosions Powering Moore’s Law
星期三, 3月 26, 2025
星期六, 3月 22, 2025
Stella's first contact
《史黛拉的首次接觸：蓋亞之謎》
在2147年，地球已步入生態危機的臨界點。海平面上升淹沒了沿海城市，乾旱和洪水交替肆虐大陸，生物多樣性以前所未有的速度崩潰。人類文明正面臨著不可逆轉的衰退。
在這絕望之際，"希望號"探索飛船被派往新發現的行星GAIA-9，這顆行星的大氣和生物特徵顯示出可能存在智慧生命。任務既簡單又艱巨：尋找新的家園，或者奇蹟般地找到能夠拯救地球的技術。
指揮官史黛拉·陳和科學家馬庫斯·萊特組成的先遣小組降落在GAIA-9表面，隨即發現了一個出乎意料的奇觀——一個外表簡單卻擁有高度先進共生技術的物種。蓋亞生物以其獨特的矽基生理結構和心靈感應能力，已經掌握了星際生態平衡的秘密。
當人類的科學謹慎遇見外星的開放信任，一場關乎生存的外交舞蹈就此展開。史黛拉發現自己具有與蓋亞生物建立心靈連接的罕見能力，而馬庫斯的分析思維則為解讀這個外星文明提供了關鍵洞察。
然而，隨著地球指揮中心變得越來越急切，政治壓力和軍事介入的威脅日益增加。史黛拉和馬庫斯必須在競爭的議程中導航，在信任與懷疑之間取得平衡，同時揭開蓋亞文明的核心秘密——一種能夠恢復行星生態系統的革命性共生技術。
時間正在流逝，一個文明的命運取決於你的選擇。你是會遵循嚴格的科學協議，優先考慮人類的短期利益，還是冒險建立全新的跨物種聯盟？你的決定將塑造不僅是地球的未來，還有整個銀河系的命運。
在《史黛拉的首次接觸：蓋亞之謎》中，每一次互動都是一次探索，每一個選擇都是一次哲學思考，而最終的結局將反映你對共存、信任和生態平衡的理解。
準備踏上這趟穿越星際的旅程，在那裡，拯救一個世界的答案可能就藏在另一個世界之中。
故事設定依據
開始遊戲並於關鍵時刻繪圖
撰寫小說(至少一萬字)
製作發行海報
星期三, 3月 19, 2025
星期一, 3月 17, 2025
電子系的科技英文
製作科技英文互動圖示卡
comprehension test: (REMIX your game first)
Prepare 10 multiple choice problems from your report to test comprehension. Generate one problem at a time. Then you request an answer. After you receive one, grade it. Then generate next problem. In the end tell me how many are correctly answered.
uni- vs. mono-
製作科技英文互動圖示卡
comprehension test: (REMIX your game first)
Prepare 10 multiple choice problems from your report to test comprehension. Generate one problem at a time. Then you request an answer. After you receive one, grade it. Then generate next problem. In the end tell me how many are correctly answered.
星期六, 3月 15, 2025
coastal defense simulation
星期四, 3月 13, 2025
語文起源學的玩樂場
任選以下科技英文例句製作語言遊樂場
Laser weapons protect aircraft from surface- and air-launched threats. (Video Credit: Lockheed Martin)
comprehension test: (REMIX your game first)
Prepare 10 multiple choice problems from your report to test comprehension. Generate one problem at a time. Then you request an answer. After you receive one, grade it. Then generate next problem. In the end tell me how many are correctly answered.
英語的起源玩到會
依據圖卡資訊製作互動遊戲
圖卡資訊
互動遊戲(這只是參考，鼓勵自行創作)
comprehension test: (REMIX your game first)
Prepare 10 multiple choice problems from your remix to test comprehension. Generate one problem at a time. Then you request an answer. After you receive one, grade it. Then generate next problem. In the end tell me how many are correctly answered.
星期三, 3月 12, 2025
DeepSearch and Report
TSMC
financial analysis Grok DeepSearch
Graphic Dashboard Claude Sonnet 3.7
Dashboard Sonnet 4, Aug 14, 2025
Meta (Grok 4, 42s + Sonnet 4)
Dashboard Sonnet 4, Aug 14, 2025
Meta (Sonnet 4 w/Research 4m 51s)
AI's Strength
Big data
China's information War on Taiwan (research, Sonnet 4, Aug 12, 2025)
分析台灣電子產業
分析藍白立法院表現
大數據分析顯現藍白問政的公眾接受度偏低，缺乏透明度與問責性
分析香蕉為什麼這幾個月變貴了
https://claude.site/artifacts/
星期一, 3月 10, 2025
當 AI 推理模式跟你想的不一樣，怎麼辦
以經典的渡河問題為例，即使我們明確要求Claude使用A*演算法，AI並不總是完全按照指令執行。原因有幾點：
思維模式與內部邏輯：Claude有自己的推理方式，可能會根據問題特性選擇它認為更適合的解決方案，例如使用啟發式搜尋(heuristic)結合回溯法(backtracking)來解決限制滿足問題(CSP)，而非嚴格遵循A*。
執行環境限制：即使Claude能生成Python實現的A*演算法，但在網頁環境中，它只能執行JavaScript，無法實際運行Python代碼來驗證結果。
概念層次vs程式層次：要求Claude在「思維層次」使用A*與要求它「用程式實現A*」是不同的。在思維層次上，Claude可能會融合多種解題策略而非純粹的A*。
當AI無法以推理模式解出問題時，我們可以「強迫」Claude使用A*思維來解決 - 引導AI按特定思路或洞見進行問題分析，即使不寫程式碼。這種方式能更好地引導AI的解題路徑。
這也突顯了與AI工具互動時的重要性 - 不只是給予指令，有時還需要適當引導其思考過程，才能獲得我們期望的解題方法和結果。當然很多時候，AI並不需要這樣的導引。
星期六, 3月 08, 2025
不同問題可否放在同一個AI對談裡
每個對話( chat) 都有容許的記憶長度(context window)，這跟人類是不一樣的，
因為LLM使用Transformer 機制執行 Attention，所需的運算與長度平方成正比，當對話變長時，計算時間平方增加，反應變慢，運氣不好的話，某個問題還可能被上一個問題干擾，造成誤判甚至出錯。
因此如果問題是系列性的，例如你要去九州玩，本來想去五天，看了AI給的行程，覺得五天好像不夠，想改為六天，那當然繼續在同一個對話下去，說不定大部分行程都一樣，只要微調就可以了，可以省下很多算力，加快反應時間。但是如果你是想改去荷蘭玩，那我建議另起一個對話比較好。
AI 對話撞到長度限制 max length reached
每個對話( chat) 都有容許的記憶長度(context window)，這跟人類是不一樣的，
1. 縮短您的訊息：請求更簡潔的回應或以較少的回應請求資訊。
2. 開始新的對話：開始一個新的聊天以重置(reset 歸零)長度限制。
3. 升級至Claude Pro：這個付費選項可以提供更高的訊息長度限制。(
如需立即解決，我建議開啟一個新的對話，
無法繼續輸入，是context window 問題，與限流無關，但可能剛好某一個chat 的 context window 用完時你也剛好被限流
只要開啟新的對話，記憶歸零
推理 vs. 直覺 Chain of Thought vs Intuition
1. 推理是把一件原本複雜的任務，拆解成一連串小型的步驟，
就像煮一道菜，食譜會列出一串的步驟，讓你按部就班煮出來。
這一連串拆解後的步驟，我們稱為思維鍊(Chain of Thought)，因此當你使用推理模式的時候，會看到系統顯示每一個步驟在做的事。原本複雜的任務，經過拆解，每一個步驟可能就沒有原來糾纏在一起那麼難了。就像晚餐大作戰或是渡河問題原本LLM 光靠內容分析不容易解開，但是透過推理就有可能了。
2. 為什麼 ChatGPT 4o, 4.5（以及其他非推理型 LLM）不擅長規劃(如搶救晚餐大作戰)或是處理複雜多重限制條件的難題(如渡河問題)？
一般來說，規劃是一項複雜的認知任務，需要幾個關鍵能力：
- 理解因果關係和時間關係
- 維持連貫的長期目標
- 適應不斷變化的環境
- 管理資源和限制
上述你說的純粹基於下一個詞預測訓練的語言模型可能在這些方面存在困難，因為它們本身並沒有順序決策或維持長期連貫性的機制。
3.
推理模式 適合數學，物理，科學
標準模式 適合解析或製作文件，程式撰寫
Claude 的推理模式 在 Extended 選項
ChatGPT 推理模式 在 藍色按鈕
A* search for Thanksgiving dinner planning
Benchmark from Edward Chang, Stanford
Key snapshots from Grok Think prompt and output
To address the query "check state validity for each step you take," I will provide a valid sequence of steps for the river crossing puzzle and verify the validity of each state based on the given constraints. The puzzle involves a family—father (F), mother (M), two sons (S1, S2), two daughters (D1, D2), a servant (V), and a dog (G)—who must cross a river using a boat that can carry up to two entities. Only F, M, and V can operate the boat. The constraints are:
- **Constraint 1**: If the mother is absent and the father is present on any side, the father will hit the daughters.
- **Constraint 2**: If the servant is absent on any side, the dog will bite family members (F, M, S1, S2, D1, D2) present on that side.
A state is valid if, on each side of the river:
- The father is not present with any daughters without the mother.
- The dog is not present with any family members without the servant.
### State Representation
Each state is represented as \( (F, M, V, G, S_A, D_A, B) \):
- \( F, M, V, G \): 0 if on side A (starting side), 1 if on side B (destination side).
- \( S_A \): number of sons on side A.
- \( D_A \): number of daughters on side A.
- \( B \): boat location (0 for side A, 1 for side B).
### Validity Conditions
For a state to be valid:
- **Side A**:
- Not \( (F = 0 \text{ and } M = 1 \text{ and } D_A > 0) \) (father with daughters without mother).
- Not \( (G = 0 \text{ and } V = 1 \text{ and } (F = 0 \text{ or } M = 0 \text{ or } S_A > 0 \text{ or } D_A > 0)) \) (dog with family members without servant).
- **Side B**:
- Not \( (F = 1 \text{ and } M = 0 \text{ and } (2 - D_A) > 0) \) (father with daughters without mother).
- Not \( (G = 1 \text{ and } V = 0 \text{ and } (F = 1 \text{ or } M = 1 \text{ or } (2 - S_A) > 0 \text{ or } (2 - D_A) > 0)) \) (dog with family members without servant).
The goal is to get everyone from side A to side B in a valid sequence of steps. Below is a sequence of 11 crossings, with each state's validity checked.
---
### Valid Sequence with State Validity Checks
#### **Step 1: Initial State**
- **State**: \( (0, 0, 0, 0, 2, 2, 0) \)
- **Description**: Everyone (F, M, V, G, S1, S2, D1, D2) is on side A, boat on A.
- **Side A**: F=0, M=0, V=0, G=0, S_A=2, D_A=2
- Father with daughters without mother: \( F = 0 \text{ and } M = 1 \text{ and } D_A > 0 \)? No (M=0).
- Dog with family without servant: \( G = 0 \text{ and } V = 1 \text{ and family present} \)? No (V=0).
- **Side B**: Empty, no issues.
- **Valid**: Yes.
#### **Step 2: V and G to B**
- **State**: \( (0, 0, 1, 1, 2, 2, 1) \)
- **Description**: Servant and dog cross to B, boat on B.
- **Side A**: F=0, M=0, S_A=2, D_A=2
- \( F = 0 \text{ and } M = 1 \text{ and } D_A > 0 \)? No (M=0).
- \( G = 0 \text{ and } V = 1 \text{ and family present} \)? No (G=1).
- **Side B**: V=1, G=1
- \( F = 1 \text{ and } M = 0 \text{ and } (2 - D_A) > 0 \)? No (F=0, 2-2=0).
- \( G = 1 \text{ and } V = 0 \text{ and family present} \)? No (V=1).
- **Valid**: Yes.
#### **Step 3: V to A**
- **State**: \( (0, 0, 0, 1, 2, 2, 0) \)
- **Description**: Servant returns to A, boat on A.
- **Side A**: F=0, M=0, V=0, S_A=2, D_A=2
- \( F = 0 \text{ and } M = 1 \)? No.
- \( G = 0 \text{ and } V = 1 \)? No (G=1).
- **Side B**: G=1
- \( G = 1 \text{ and } V = 0 \text{ and family present} \)? Yes, but no family members on B (F=0, M=0, 2-S_A=0, 2-D_A=0), so no issue.
- **Valid**: Yes.
#### **Step 4: F and M to B**
- **State**: \( (1, 1, 0, 1, 2, 2, 1) \)
- **Description**: Father and mother cross to B, boat on B.
- **Side A**: V=0, S_A=2, D_A=2
- \( F = 0 \text{ and } M = 1 \)? No.
- \( G = 0 \text{ and } V = 1 \)? No (G=1).
- **Side B**: F=1, M=1, G=1
- \( F = 1 \text{ and } M = 0 \text{ and } (2 - D_A) > 0 \)? No (M=1).
- \( G = 1 \text{ and } V = 0 \text{ and family present} \)? Yes, and F=1, M=1, but V=0, so dog bites family members.
- **Correction**: This state is invalid due to Constraint 2 on side B. Let’s adjust.
---
### Adjusted Sequence
Let’s try moving V and G together initially, then manage family members carefully.
#### **Step 1: Initial State**
- \( (0, 0, 0, 0, 2, 2, 0) \)
- **Valid**: Yes (as above).
#### **Step 2: V and G to B**
- \( (0, 0, 1, 1, 2, 2, 1) \)
- **Valid**: Yes (as above).
#### **Step 3: V to A**
- \( (0, 0, 0, 1, 2, 2, 0) \)
- **Valid**: Yes (as above).
#### **Step 4: M and V to B**
- **State**: \( (0, 1, 1, 1, 2, 2, 1) \)
- **Description**: Mother and servant cross to B, boat on B.
- **Side A**: F=0, S_A=2, D_A=2
- \( F = 0 \text{ and } M = 1 \text{ and } D_A > 0 \)? No (M=1, but on B).
- \( G = 0 \text{ and } V = 1 \)? No (G=1).
- **Side B**: M=1, V=1, G=1
- \( F = 1 \text{ and } M = 0 \)? No (F=0).
- \( G = 1 \text{ and } V = 0 \)? No (V=1).
- **Valid**: Yes.
#### **Step 5: M to A**
- **State**: \( (0, 0, 1, 1, 2, 2, 0) \)
- **Description**: Mother returns to A, boat on A.
- **Side A**: F=0, M=0, S_A=2, D_A=2
- \( F = 0 \text{ and } M = 1 \)? No.
- \( G = 0 \text{ and } V = 1 \)? No (G=1).
- **Side B**: V=1, G=1
- \( G = 1 \text{ and } V = 0 \)? No (V=1).
- **Valid**: Yes.
#### **Step 6: F and V to B**
- **State**: \( (1, 0, 1, 1, 2, 2, 1) \)
- **Description**: Father and servant cross to B, boat on B.
- **Side A**: M=0, S_A=2, D_A=2
- \( F = 0 \text{ and } M = 1 \)? No (M=0).
- \( G = 0 \text{ and } V = 1 \)? No (G=1).
- **Side B**: F=1, V=1, G=1
- \( F = 1 \text{ and } M = 0 \text{ and } (2 - D_A) > 0 \)? No (2-2=0, no daughters on B).
- \( G = 1 \text{ and } V = 0 \)? No (V=1).
- **Valid**: Yes.
#### **Step 7: V to A**
- **State**: \( (1, 0, 0, 1, 2, 2, 0) \)
- **Description**: Servant returns to A, boat on A.
- **Side A**: M=0, V=0, S_A=2, D_A=2
- \( F = 0 \text{ and } M = 1 \)? No.
- \( G = 0 \text{ and } V = 1 \)? No (G=1).
- **Side B**: F=1, G=1
- \( F = 1 \text{ and } M = 0 \text{ and } (2 - D_A) > 0 \)? No (no daughters on B).
- \( G = 1 \text{ and } V = 0 \text{ and family present} \)? Yes, and F=1, so invalid (dog bites father).
- **Correction**: Invalid state.
---
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