We are playing the Guess Game. The game is as follows:
I pick a number from 1 to n. You have to guess which number I picked.
Every time you guess wrong, I'll tell you whether the number I picked is higher or lower.
However, when you guess a particular number x, and you guess wrong, you pay $x. You win the game when you guess the number I picked.
Example:
n = 10, I pick 8. First round: You guess 5, I tell you that it's higher. You pay $5. Second round: You guess 7, I tell you that it's higher. You pay $7. Third round: You guess 9, I tell you that it's lower. You pay $9.
Game over. 8 is the number I picked. You end up paying $5 + $7 + $9 = $21.
Given a particular n ≥ 1, find out how much money you need to have to guarantee a win.
Solution:
Core steps for Dynamic programming
(1) Define the subproblem
define DP[i][j] the money needs to pay to guarantee a win, so the answer is DP[1][n]
(2) Find the recursion
the number is in [1, n] , guess the number x (from 1 to n). if we pick up k, if k != x
then the worst case complexity to guarantee win is the get the maximum cost: k + max(dp[1,][k-1]), dp[k+1][n])
for all x from 1 to n, the the money needs to pay is dp[i][j] = min(dp[i][j], k + max(dp[1,][k-1]), dp[k+1][n]))
(3) Get the base case
the base case is dp[i][i] = 0 for i == j from 1 to n
1. Iterative method
1 dp = [[0] * (n+1) for i in range(0, n+1)] 2 print ("dp: ", dp) 3 4 for i in range(n-1, 0, -1): #from n-1 to 1 5 for j in range(i+1, n+1): #from i+1 to n 6 7 dp[i][j] = 2**64 8 for k in range(i, j): #from i to j-1 9 dp[i][j] = min(dp[i][j], k + max(dp[i][k-1], dp[k+1][j])) 10 #print ("dp ij ", i, j, k, dp[i][j]) 11 return dp[1][n] 12
2. Recursive method
1 def guesshelper(dp, i, j): 2 if i >= j: return 0 3 if dp[i][j] > 0: return dp[i][j] 4 5 dp[i][j] = 2**64 6 7 for k in range(i, j+1): 8 dp[i][j] = min(dp[i][j], k + max(guesshelper(dp, i, k-1), guesshelper(dp, k+1, j))) 9 10 return dp[i][j] 11 12 dp = [[0] * (n + 1) for i in range(n + 1)] 13 return guesshelper(dp, 1, n) 14