**Moderator:** Good morning, and welcome to this press briefing. Today, we are here to discuss a fascinating mathematical puzzle that has captured the public's imagination and found relevance in fields ranging from computer science to economics. The subject is the problem of repeatedly withdrawing money in denominations of 1, 2, and 5 yuan. Our purpose is to provide an objective and accurate overview of this problem, its mathematical underpinnings, its computational significance, and its broader implications. We will now open the floor for a detailed presentation. *** The core problem can be stated as follows: "In how many distinct ways can one withdraw a total of *n* yuan by using an unlimited supply of 1-yuan, 2-yuan, and 5-yuan notes?" This seemingly simple question is a gateway to a rich area of mathematics known as combinatorics, and more specifically, to the study of partition and composition problems with constrained parts. It is a classic example that demonstrates how a straightforward scenario can model complex systems. **1. Defining the Problem and Its Mathematical Framework** To analyze this problem accurately, we must first define its parameters. We are dealing with combinations of the three denominations {1, 2, 5} that sum to a target value *n*. The "ways" are considered distinct if the sequences of denominations used are different, even if the total number of bills is the same. For instance, to make 3 yuan, there are three distinct ways: 1. 1 + 1 + 1 2. 1 + 2 3. 2 + 1 Note that "1 + 2" and "2 + 1" are counted as two different ways. If the problem were to count combinations where order does not matter, it would be a different, though related, problem. Our focus today is on the ordered sequences, often referred to as "compositions." This problem can be elegantly modeled using a recurrence relation. Let a_n denote the number of ways to form the amount *n*. We can reach the amount *n* by first forming an amount that is one step away and then adding the final bill. Specifically: - We can arrive at *n* by adding a 1-yuan bill to any sequence that forms *n-1*. There are a_{n-1} such sequences. - We can arrive at *n* by adding a 2-yuan bill to any sequence that forms *n-2*. There are a_{n-2} such sequences. - We can arrive at *n* by adding a 5-yuan bill to any sequence that forms *n-5*. There are a_{n-5} such sequences. Since these paths are mutually exclusive, the total number of sequences is the sum of these three possibilities. This gives us the linear recurrence relation: **a_n = a_{n-1} + a_{n-2} + a_{n-5}** This recurrence holds for n >= 5. To use it, we must first establish the initial conditions, the base cases for small values of *n*: - a_0 = 1 (There is one way to form 0 yuan: using no bills). - a_1 = 1 (Only one way: a single 1-yuan bill). - a_2 = 2 (1+1, or a single 2). - a_3 = 3 (1+1+1, 1+2, 2+1). - a_4 = 5 (1+1+1+1, 1+1+2, 1+2+1, 2+1+1, 2+2). From a_5 onwards, the recurrence takes over: a_5 = a_4 + a_3 + a_0 = 5 + 3 + 1 = 9. One can verify this by listing the sequences, though the recurrence provides a far more efficient computational method. **2. The Generating Function Approach** A more powerful and general tool for solving such problems is the generating function. A generating function is a formal power series whose coefficients encode the sequence we are interested in. For this problem, the generating function F(x) is constructed as follows: Each 1-yuan bill contributes a factor of x^1. Each 2-yuan bill contributes a factor of x^2. Each 5-yuan bill contributes a factor of x^5. Since we can use any number of each bill, the choices for each denomination are represented by an infinite series: - For 1-yuan: 1 + x^1 + x^2 + x^3 + ... = 1/(1-x) - For 2-yuan: 1 + x^2 + x^4 + x^6 + ... = 1/(1-x^2) - For 5-yuan: 1 + x^5 + x^10 + x^15 + ... = 1/(1-x^5) The overall generating function, which counts all possible combinations, is the product of these three: **F(x) = 1 / [ (1 - x)(1 - x^2)(1 - x^5) ]** The coefficient of x^n in the expansion of F(x) is precisely the number a_n, the number of ways to form *n* yuan. This method is exceptionally versatile. For example, if we were to change the problem—perhaps by limiting the number of 5-yuan bills that can be used—we would simply modify the corresponding factor in the generating function. This approach connects the problem to deep areas of mathematics, including partition theory and analytic number theory. **3. Computational Significance and Algorithmic Solutions** From a computer science perspective, this problem is a canonical example used to teach fundamental algorithmic techniques. It is a special case of the more general "coin change problem," which is a staple in competitive programming and technical interviews. Several standard algorithms can be applied: - **Recursion with Memoization:** A direct implementation of the recurrence relation a_n = a_{n-1} + a_{n-2} + a_{n-5}. A naive recursive implementation would be wildly inefficient due to repeated calculations. Memoization, or caching results of subproblems, transforms this into an efficient O(n) time and O(n) space solution. - **Dynamic Programming (Iterative):** This is often the preferred method. We initialize an array `dp` where `dp[i]` represents the number of ways to form amount `i`. We set `dp[0] = 1` and then iteratively build the solution for larger amounts by adding the contributions from `dp[i-1]`, `dp[i-2]`, and `dp[i-5]`. This bottom-up approach is robust and clearly demonstrates the principle of optimal substructure. - **Matrix Exponentiation:** For extremely large values of *n* (e.g., n = 10^18), even an O(n) dynamic programming solution is too slow. The linear recurrence relation can be represented as a matrix transformation. By using fast exponentiation of this transformation matrix, the n-th term of the sequence can be computed in O(log n) time, a dramatic improvement for massive inputs. The problem thus serves as a perfect pedagogical bridge from basic recursion to advanced algorithmic paradigms. **4. Broader Implications and Real-World Analogies** The "1, 2, 5 Yuan" problem is not merely an abstract puzzle. Its mathematical structure is analogous to numerous real-world phenomena. - **Economics and Finance:** The problem can model savings plans or investment strategies where contributions can be made in fixed amounts at regular intervals. The count a_n could represent the number of distinct contribution sequences to reach a savings goal. Furthermore, the denominations {1, 2, 5} mirror a standard currency system, making the problem a simplified version of cash management and transaction analysis. - **Computer Science and Systems Design:** In task scheduling or resource allocation, a large task of size *n* may need to be broken down into smaller, fixed-size subtasks. The number of ways to decompose the task, if the subtask sizes are 1, 2, and 5 units, is directly given by this model. This can be relevant for parallel processing and load distribution. - **Linguistics and Bioinformatics:** In linguistics, the problem can be seen as counting the number of ways a sentence of *n* syllables can be formed from words of 1, 2, and 5 syllables. In bioinformatics, it could analogously represent forming a polymer chain from monomer units of specific lengths. The recurrence relation itself, a_n = a_{n-1} + a_{n-2} + a_{n-5}, is a generalization of the famous Fibonacci recurrence (which would be a_n = a_{n-1} + a_{n-2} if only 1 and 2 yuan bills were available). This highlights how small changes in the parameters of a system can lead to significantly different and more complex behaviors. **5. Variations and Extensions** The core problem can be extended in many directions, each leading to a new set of mathematical challenges: - **The "Number of Coins" Variation:** Instead of counting sequences, what is the minimum number of bills required to form *n* yuan? This is the classic "Minimum Coin Change" problem, solvable with dynamic programming and central to greedy algorithm analysis (the {1, 2, 5} system is "canonical," meaning the greedy algorithm
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