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D&C, DP

D&C, DP. Divide and Conquer Dynamic Programming. Recursive Method…. Big Problem!!!. Big Problem!!!. Sub-Problem. Sub-Problem. Sub-Problem. Three major steps. Big Problem!!!. Solved!!!. Conquer. Divide. Merge. Sub-Problem. Solved. Sub-Problem. Solved. Sub-Problem. Solved.

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D&C, DP

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  1. D&C, DP Divide and Conquer Dynamic Programming

  2. Recursive Method… Big Problem!!! Big Problem!!! Sub-Problem Sub-Problem Sub-Problem

  3. Three major steps Big Problem!!! Solved!!! Conquer Divide Merge Sub-Problem Solved Sub-Problem Solved Sub-Problem Solved Solved Solved Solved

  4. Stairs Climbing 每次可踏上一階或踏上兩階 眼前有四階樓梯， 每次可踏上一階或踏上兩階， 那麼爬完四階共有幾種踏法？ = = + + = + f(3) 3 = f(2) 2 f(1) f(4) = f(3) f(2) + 1 + 5 3 2 f(4) = f(3) + f(2)

  5. Tiling Problem with L-shaped Tile

  6. Fast Exponentiation 7^13=96889010407 7^13=?

  7. 7^13=7^7 * 7^6 • 7^7= 7^4 * 7^3 • 7^6=7^3 * 7^3 • 7^4= 7^2 * 7^2 7^3=7^2 * 7^1 • 7^2= 7^1 * 7^1 • 7^1= 7

  8. Exception(Prune and search) Kth smallest(related to Quick sort) Binary Search

  9. Dynamic Programming(動態規劃) Programming  最佳化(Optimization) DP  運用”動態”的方式求最佳解 Dynamic 動態

  10. DP可視為 D&C 的延伸 f(4) f(2) f(3) f(0) f(1) f(1) f(2) f(0) f(1) Overlapping

  11. 讀取、計算、儲存 f(4) f(2) f(3) f(0) f(1) f(1) f(2) f(3) = f(2) + f(1) f(4) = f(3) + f(2) f(0) f(1) f(2) = f(0) + f(1)

  12. 讀取、計算、儲存 f(4) f(2) f(3) f(0) f(1) f(1) f(2) f(3) = 3 f(4) = 5 f(1) = 1 f(0) f(1) f(2) = 2 f(0) = 1

  13. DP的時間、空間複雜度 時間複雜度:O(1) 時間複雜度:O(n) f(n) = f(n-1) + f(n-2) 時間複雜度:O(s) 時間複雜度:O(n*s*1)

  14. DP的時間、空間複雜度 空間複雜度:O(n) 空間複雜度:以子問題的出現期間決定。 空間複雜度:O(1)

  15. Note on DP 2.決定問題的計算順序。 3.確認初始值、問題的計算範圍。 1.發現相似的求解方式，紀錄重複的子問題資訊 。

  16. Practice in Dp 2.Bottom-up, Iterative way. 1.Top-down, Recursive way.

  17. Stairs Climbing(初步思考) f(n) = { 1 , if n = 1 { 2 , if n = 2 { f(n-1) + f(n-2) , if n >= 3

  18. Stairs Climbing(美觀) f(n) = { 0 , if n < 0 [Exterior] { 1 , if n = 0 [Initial] { 1 , if n = 1 [Initial] { f(n-1) + f(n-2) , if n >= 2 and n <= 5 [Compute] { 0 , if n > 5 [Exterior]

  19. Stairs Climbing(不好的示範) int f(int n) { if (n == 0 || n == 1) return1; else return f(n-1) + f(n-2); }

  20. Stairs Climbing(Two ways) f(4) f(2) f(3) f(0) f(1) f(1) f(2) f(0) f(1) 2. Bottom-up, Iterative. 1.Top-down, Recursive.

  21. Stairs Climbing(Top-down) inttable[5];// 表格，儲存全部問題的答案。 boolsolve[5];// 記錄問題是否已計算 int f(int n) { if (n == 0 || n == 1) return1; if (solve[n]) return table[n];     table[n] = f(n-1) + f(n-2); // 將答案存入表格 solve[n] = true;            // 紀錄已計算 return table[n]; }

  22. Stairs Climbing(Top-down) f(n) = f(n-1) + f(n-2) 1.好處是不必斤斤計較計算順序。 2.只計算必要的問題，而不必計算整個陣列。

  23. Stairs Climbing(Top-down) f(4) f(2) f(3) f(n) = f(n-1) + f(n-2) f(0) f(1) f(1) f(2) f(0) f(1)  1.壞處是不斷呼叫函式，執行效率較差。 2.無法自由地控制計算順序，無法妥善運用記憶體。

  24. Stairs Climbing(Bottom-up) int table[5]; voiddp() { // initial for (inti=0; i<=4;i++)         table[i] = 0; table[0] = 1;     table[1] = 1; // compute for (inti=2; i<=4;i++)         table[i] = table[i-1] + table[i-2]; } Int table[5]; table[0] = 1; table[ 1] = 1;

  25. Stairs Climbing(Bottom-up) for (int i=0; i<=5; i++) table[i] = 0; table[2] = ? table[3] = ? for (int i=2; i<=5; i++)         table[i] = table[i-1] + table[i-2]; table[4] = ? 訂定一個計算順序，然後由最小的問題開始計算。 其特色是程式碼通常只有幾個迴圈。 這個方式的好處與壞處恰與前一個方式互補。

  26. The Summary of D&C, DP 2.可考慮使用表格紀錄出現頻繁的子問題數值。 3.To iterate is man, to recurse is divine. +智慧 靈感 = 經驗 1.找出結構相似之處(recursion)，設計一套公式(recurrence)。

  27. UVA DP 623 900 11069 10446 495 D&C 11038 620 10230 920 374

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