Problem Description
There are two types of soup: type A and type B. Initially, we have n ml of each type of soup. There are four kinds of operations: serve 100 ml of soup A and 0 ml of soup B, serve 75 ml of soup A and 25 ml of soup B, serve 50 ml of soup A and 50 ml of soup B, and serve 25 ml of soup A and 75 ml of soup B. Each turn, we choose from the four operations with equal probability. If the remaining volume of soup is not enough to complete the operation, we will serve as much as possible. We stop once we no longer have some quantity of both types of soup. Return the probability that soup A will be empty first, plus half the probability that A and B become empty at the same time. Answers within 10^-5 of the actual answer will be accepted.
Key Insights
- The problem can be approached using dynamic programming and probability theory.
- The operations affect the volumes of soup A and B differently, creating various paths to emptying the soups.
- The states can be represented as a grid or a recursive function with memoization to avoid recalculating probabilities for the same state.
- The recursive nature of the problem allows us to build up the probabilities based on previous states.
Space and Time Complexity
Time Complexity: O(n^2), where n is the initial volume of the soups. The recursion can create a state space proportional to n. Space Complexity: O(n^2) for memoization of previously calculated states.
Solution
The solution involves a recursive function that computes the probability of soup A being empty first from the current state of the volumes of soup A and B. The four operations are considered in the function, and the results are combined based on their probabilities. The recursive function is memoized to improve efficiency, avoiding recalculation of the same states.