If an egg is dropped from above that floor it will break.
1 egg 100 floors.
That leaves us with just floor n 1 to try with the second egg.
Second thought is to try xth floor then 2xth floor till 100th in this case worst case time will be 100 x x 1.
2 eggs 100 floors worst case solution.
With this method the maximum tries is 51.
It occurs when the egg survives 98 floors.
A building has 100 floors.
Then we have to test floors 1 13 with the 2nd egg to see where the egg breaks again and the egg will not break on any of those floors.
What strategy should be used to drop eggs such that total number of drops in worst.
The solution for the worst case in this scenario occurs when the threshold floor is floor number 14 because we will drop the first egg on floor 14 and it will break.
Worst case will be when egg 1 breaks at 100th floor then we have to try egg 2 from 100 x th to 99th floors.
Instead of taking equal intervals we can decrease the.
The following is a description of the instance of this famous puzzle involving 2 eggs and a building with 100 floors.
If it is dropped from that.
Now for the ultimate method.
In 100 x x 1 equation with increase in x 100 x decreases while x 1 increases thus we can minimize it when 100.