如何以編程方式解決2玻璃球難題？

Question

關於一座100層高的建築物和兩個玻璃球，有一種流行的puzzle。我看到了解決方案，現在我想知道我是否可以通過編程的方式解決難題。

微不足道的編程解決方案是一個完整的搜索（我相信我可以使用回溯代碼）。有沒有更好的編程解決方案？我可以使用動態編程來解決這個難題嗎？

Answer 1

這與卵滴拼圖類似。我會給你在動態編程的基本策略。將此與您的問題緊密聯繫起來。

# include <stdio.h> 
# include <limits.h> 

// A utility function to get maximum of two integers 
int max(int a, int b) { return (a > b)? a: b; } 

/* Function to get minimum number of trails needed in worst 
case with n eggs and k floors */ 
int eggDrop(int n, int k) 
{ 
/* A 2D table where entery eggFloor[i][j] will represent minimum 
    number of trials needed for i eggs and j floors. */ 
int eggFloor[n+1][k+1]; 
int res; 
int i, j, x; 

// We need one trial for one floor and0 trials for 0 floors 
for (i = 1; i <= n; i++) 
{ 
    eggFloor[i][1] = 1; 
    eggFloor[i][0] = 0; 
} 

// We always need j trials for one egg and j floors. 
for (j = 1; j <= k; j++) 
    eggFloor[1][j] = j; 

// Fill rest of the entries in table using optimal substructure 
// property 
for (i = 2; i <= n; i++) 
{ 
    for (j = 2; j <= k; j++) 
    { 
     eggFloor[i][j] = INT_MAX; 
     for (x = 1; x <= j; x++) 
     { 
      res = 1 + max(eggFloor[i-1][x-1], eggFloor[i][j-x]); 
      if (res < eggFloor[i][j]) 
       eggFloor[i][j] = res; 
     } 
    } 
} 

// eggFloor[n][k] holds the result 
return eggFloor[n][k]; 
} 

/* Driver program to test to pront printDups*/ 
int main() 
{ 
    int n = 2, k = 36; 
    printf ("\nMinimum number of trials in worst case with %d eggs and %d floors is %d \n", n, k, eggDrop(n, k)); 
    return 0; 
} 

輸出： 在最壞的情況下試驗的最小數量與2個雞蛋和36層爲8

Answer 2

我很抱歉，但以前的答案只是不能妥善處理這個問題。最好的時間是N的sqrt。有人會試圖用logn來反駁這一點，但我很抱歉，其實並非如此。

// my prentend breaking function 
function itBreaks(level) { 
    return level > 36; 
} 

function search(maxLevel) { 
    var sqrtN = Math.floor(Math.sqrt(maxLevel)); 
    var i = 0; 
    for (;i < maxLevel; i += sqrtN) { 
     if (itBreaks(i)) { 
      break; 
     } 
    } 

    for (i -= sqrtN; i < maxLevel; i++) { 
     if (itBreaks(i)) { 
      return i - 1; 
     } 
    } 
}