Summary: Two soups A and B are presented with n mL each. In each turn, one of four operations is randomly selected with probability 0.25: serve 100 mL of A and 0 mL of B; serve 75 mL of A and 25 mL of B; serve 50 mL of A and 50 mL of B; serve 25 mL of A and 75 mL of B. The process ends when A or B is emptied. If an operation requests more than the available amount, only what remains is served. The goal is to calculate the probability that A empties before B plus half the probability that both empty in the same step.
Intuition: Each turn reduces both quantities and the result depends on which container runs out first or if both run out simultaneously. The four operations generate a probabilistic tree structure and the problem consists of calculating the expected value by traversing that tree. It is natural to use memoization to store results of repeated subproblems. For large n, specifically above 4500 mL, the result converges very close to 1.0, so 1 can be returned as a shortcut.
Simplification: To reduce the state space, work with units of 25 mL. That is, convert n to m equal to the ceiling of n divided by 25. This way each operation is represented by integer decrements in those units and the number of states to calculate is limited.
Approach: Define a function p(a, b) that returns the desired probability starting from a units of A and b units of B. Base cases: if a is less than or equal to zero and b is less than or equal to zero return 0.5; if a is less than or equal to zero return 1; if b is less than or equal to zero return 0. For a general state p(a, b) equals 0.25 times the sum of the four recursive calls with the corresponding decrements. Memoization is used with a two-dimensional array or a dictionary to cache results and avoid recalculations.
Pseudocode: function p of a comma b: if a is less than or equal to zero and b is less than or equal to zero return 0.5. If a is less than or equal to zero return 1. If b is less than or equal to zero return 0. If result exists in memo return result. Calculate result equal to 0.25 multiplied by the sum of p with a minus 4 comma b; p with a minus 3 comma b minus 1; p with a minus 2 comma b minus 2; p with a minus 1 comma b minus 3. Save result in memo and return result. Main function soupServings with n: if n greater than 4500 return 1. Calculate m equal to the ceiling of n divided by 25. Return p of m comma m.
Versions and implementation notes: In C++ you can use a double array size 300 by 300 and the recursive function p with base cases and memoization. In JavaScript it is practical to use a Map to memoize concatenated keys and a recursive dp function. In Python it is convenient to use functools lru cache to simplify memoization. In all three implementations, the conversion to 25 mL units and the shortcut for n greater than 4500 improve performance.
Time and space complexity: The number of unique states is on the order of m squared, where m is the ceiling of n divided by 25. In the worst case for n equal to 4500 that corresponds approximately to 300 by 300 which gives around 90000 states. The space complexity is of the same order due to the use of the two-dimensional memoization table.
Final reflection: This exercise combines probability and recursive dynamic programming. The key elements are grouping quantities into 25 mL units to limit the state space, using memoization to avoid recomputation, and leveraging convergence for large n. It is a recommended problem for practicing memoization, probabilistic reasoning, and state space optimization.
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