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.
1 egg 100 floors.
Now for the ultimate method.
One of the floors is the highest floor an egg can be dropped from without breaking.
Suppose that we wish to know which stories in a 100 story building are safe to drop eggs from and which will cause the eggs to break on landing.
What strategy should be used to drop eggs such that total number of drops in worst.
2 eggs 100 floors worst case solution.
It will take 50 tries to reach floor 100 and one more egg to try on the 99th floor so the total is 51 tries.
Second thought is to try xth floor then 2xth floor till 100th in this case worst case time will be 100 x x 1.
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.
That leaves us with just floor n 1 to try with the second egg.
If it is dropped from that.
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.
It occurs when the egg survives 98 floors.
With this method the maximum tries is 51.
In 100 x x 1 equation with increase in x 100 x decreases while x 1 increases thus we can minimize it when 100.