1 Egg 100 Floors

2 eggs 100 floors worst case solution. One of the floors is the highest floor an egg can be dropped from without breaking. If it is dropped from that.

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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.

1 egg 100 floors.

In 100 x x 1 equation with increase in x 100 x decreases while x 1 increases thus we can minimize it when 100. The following is a description of the instance of this famous puzzle involving 2 eggs and a building with 100 floors. 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. A building has 100 floors.

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. 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. It occurs when the egg survives 98 floors. 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.

That leaves us with just floor n 1 to try with the second egg. With this method the maximum tries is 51. Now for the ultimate method. What strategy should be used to drop eggs such that total number of drops in worst.

Second thought is to try xth floor then 2xth floor till 100th in this case worst case time will be 100 x x 1. If an egg is dropped from above that floor it will break. Instead of taking equal intervals we can decrease the.

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