Eggs dropping puzzle (Binomial Coefficient and Binary Search Solution)

Given n eggs and k floors, find the minimum number of trials needed in worst case to find the floor below which all floors are safe. A floor is safe if dropping an egg from it does not break the egg. Please see n eggs and k floors. for complete statements

Input : n = 2, k = 10
Output : 4
We first try from 4-th floor. Two cases arise,
(1) If egg breaks, we have one egg left so we
    need three more trials.
(2) If egg does not break, we try next from 7-th
    floor. Again two cases arise.
We can notice that if we choose 4th floor as first
floor, 7-th as next floor and 9 as next of next floor,
we never exceed more than 4 trials.

Input : n = 2. k = 100
Output : 14


We have discussed the problem for 2 eggs and k floors. We have also discussed a dynamic programming solution to find the solution. The dynamic programming solution is based on below recursive nature of the problem. Let us look at the discussed recursive formula from a different perspective.

How many floors we can cover with x trials?
When we drop an egg, two cases arise.

  1. If egg breaks, then we are left with x-1 trials and n-1 eggs.
  2. If egg does not break, then we are left with x-1 trials and n eggs
Let maxFloors(x, n) be the maximum number of floors 
that we can cover with x trials and n eggs. From above 
two cases, we can write.

maxFloors(x, n) = maxFloors(x-1, n-1) + maxFloors(x-1, n) + 1
For all x >= 1 and n >= 1

Base cases : 
We can't cover any floor with 0 trials or 0 eggs
maxFloors(0, n) = 0
maxFloors(x, 0) = 0

Since we need to cover k floors, 
maxFloors(x, n) >= k           ----------(1)

The above recurrence simplifies to following,
Refer this for proof.

maxFloors(x, n) = ∑xCi
                  1 <= i <= n   ----------(2)
Here C represents Binomial Coefficient.

From above two equations, we can say.
&Sum;xCj  >= k
1 <= i <= n
Basically we need to find minimum value of x
that satisfies above inequality. We can find
such x using Binary Search.

C++

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// C++ program to find minimum 
// number of trials in worst case.
#include <bits/stdc++.h>
  
using namespace std;
  
// Find sum of binomial coefficients xCi
// (where i varies from 1 to n). If the sum
// becomes more than K
int binomialCoeff(int x, int n, int k)
{
    int sum = 0, term = 1;
    for (int i = 1; i <= n && sum < k; ++i) {
        term *= x - i + 1;
        term /= i;
        sum += term;
    }
    return sum;
}
  
// Do binary search to find minimum 
// number of trials in worst case.
int minTrials(int n, int k)
{
    // Initialize low and high as 1st 
    // and last floors
    int low = 1, high = k;
  
    // Do binary search, for every mid, 
    // find sum of binomial coefficients and 
    // check if the sum is greater than k or not.
    while (low < high) {
        int mid = (low + high) / 2;
        if (binomialCoeff(mid, n, k) < k)
            low = mid + 1;
        else
            high = mid;
    }
  
    return low;
}
  
/* Drier program to test above function*/
int main()
{
    cout << minTrials(2, 10);
    return 0;
}

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Java














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// Java program to find minimum  


// number of trials in worst case. 


class Geeks { 


  


// Find sum of binomial coefficients xCi 


// (where i varies from 1 to n). If the sum 


// becomes more than K 


static int binomialCoeff(int x, int n, int k) 


{ 


    int sum = 0, term = 1; 


    for (int i = 1; i <= n && sum < k; ++i) { 


        term *= x - i + 1; 


        term /= i; 


        sum += term; 


    } 


    return sum; 


} 


  


// Do binary search to find minimum  


// number of trials in worst case. 


static int minTrials(int n, int k) 


{ 


    // Initialize low and high as 1st  


    //and last floors 


    int low = 1, high = k; 


  


    // Do binary search, for every mid,  


    // find sum of binomial coefficients and  


    // check if the sum is greater than k or not. 


    while (low < high) { 


        int mid = (low + high) / 2; 


        if (binomialCoeff(mid, n, k) < k) 


            low = mid + 1; 


        else


            high = mid; 


    } 


  


    return low; 


} 


  


/* Driver program to test above function*/


public static void main(String args[]) 


{ 


    System.out.println(minTrials(2, 10)); 


} 


} 


  


// This code is contributed by ankita_saini 



















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Python3





# Python3 program to find minimum

# number of trials in worst case.


# Find sum of binomial coefficients

# xCi (where i varies from 1 to n).

# If the sum becomes more than K

def binomialCoeff(x, n, k):


	sum = 0;

	term = 1;

	i = 1;

	while(i <= n and sum < k): 
		term *= x - i + 1;
		term /= i;
		sum += term;
		i += 1;
	return sum;

# Do binary search to find minimum 
# number of trials in worst case.
def minTrials(n, k):

	# Initialize low and high as 
	# 1st and last floors
	low = 1;
	high = k;

	# Do binary search, for every 
	# mid, find sum of binomial 
	# coefficients and check if 
	# the sum is greater than k or not.
	while (low < high):

		mid = (low + high) / 2;
		if (binomialCoeff(mid, n, k) < k):
			low = mid + 1;
		else:
			high = mid;

	return int(low);

# Driver Code
print(minTrials(2, 10));

# This code is contributed 
# by mits



[tabby title="C#"]

                











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// C# program to find minimum  


// number of trials in worst case. 


using System; 


  


class GFG  


{ 


  


// Find sum of binomial coefficients 


// xCi (where i varies from 1 to n).  


// If the sum becomes more than K 


static int binomialCoeff(int x,  


                         int n, int k) 


{ 


    int sum = 0, term = 1; 


    for (int i = 1;  


             i <= n && sum < k; ++i)  


    { 


        term *= x - i + 1; 


        term /= i; 


        sum += term; 


    } 


    return sum; 


} 


  


// Do binary search to find minimum  


// number of trials in worst case. 


static int minTrials(int n, int k) 


{ 


    // Initialize low and high 


    // as 1st and last floors 


    int low = 1, high = k; 


  


    // Do binary search, for every  


    // mid, find sum of binomial  


    // coefficients and check if the 


    // sum is greater than k or not. 


    while (low < high)  


    { 


        int mid = (low + high) / 2; 


        if (binomialCoeff(mid, n, k) < k) 


            low = mid + 1; 


        else


            high = mid; 


    } 


  


    return low; 


} 


  


// Driver Code 


public static void Main() 


{ 


    Console.WriteLine(minTrials(2, 10)); 


} 


} 


  


// This code is contributed 


// by Akanksha Rai(Abby_akku) 



















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PHP














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<?php 


// PHP program to find minimum  


// number of trials in worst case. 


  


// Find sum of binomial coefficients  


// xCi (where i varies from 1 to n).  


// If the sum becomes more than K 


function binomialCoeff($x, $n, $k) 


{ 


    $sum = 0; $term = 1; 


    for ($i = 1; $i <= $n &&  


         $sum < $k; ++$i)  


    { 


        $term *= $x - $i + 1; 


        $term /= $i; 


        $sum += $term; 


    } 


    return $sum; 


} 


  


// Do binary search to find minimum  


// number of trials in worst case. 


function minTrials($n, $k) 


{ 


    // Initialize low and high as  


    // 1st and last floors 


    $low = 1; $high = $k; 


  


    // Do binary search, for every  


    // mid, find sum of binomial  


    // coefficients and check if  


    // the sum is greater than k or not. 


    while ($low < $high)  


    { 


        $mid = ($low + $high) / 2; 


        if (binomialCoeff($mid, $n, $k) < $k) 


            $low = $mid + 1; 


        else


            $high = $mid; 


    } 


  


    return (int)$low; 


} 


  


// Driver Code 


echo minTrials(2, 10); 


  


// This code is contributed  


// by Akanksha Rai(Abby_akku) 


?> 



















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Output:


4





Time Complexity : O(n Log k)



          
          
          
          

            

    

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