环形5城市中用n次行程返回起点城市的路径数求解

Problem Statement #

We have 5 cities arranged in a ring: c1-c2, c2-c3, c3-c4, c4-c5, c5-c1. Starting from c1, each move between adjacent cities counts as 1 trip. We need to find the total number of valid paths that start at c1, use exactly n trips, and return to c1.

Approach: Dynamic Programming & Recurrence Relation #

Alright, let's break this down step by step. This is a classic state-transition problem, and we can simplify it using symmetry and dynamic programming to derive an efficient solution.

Step 1: Define State Variables

To leverage the ring's symmetry, we split the problem into three state categories:

• a[k]: Number of paths ending at c1 after k trips
• b[k]: Number of paths ending at either c2 or c5 (each directly adjacent to c1) after k trips
• c[k]: Number of paths ending at either c3 or c4 (two steps away from c1) after k trips

Step 2: Initial Conditions

Let's set our base cases based on the problem's starting point:

• a[0] = 1: 0 trips means we're still at the starting city c1
• b[0] = 0: No moves, so we can't be at c2 or c5
• c[0] = 0: No moves, so we can't be at c3 or c4
• a[1] = 0: 1 trip can only take us to c2 or c5—no way to get back to c1
• b[1] = 1: Each of c2 and c5 has exactly 1 path from c1 in 1 trip
• c[1] = 0: 1 trip isn't enough to reach c3 or c4

Step 3: State Transitions

Using the ring's adjacency rules, we can derive how paths flow between states:

1. To end at c1 (a[k]), you must come from either c2 or c5:
   a[k] = 2 * b[k-1]
2. To end at c2 (or c5, b[k]), you can come from c1 or c3 (or c4 for c5):
   b[k] = a[k-1] + c[k-1]
3. To end at c3 (or c4, c[k]), you can come from c2 or c4 (or c3 for c4):
   c[k] = b[k-1] + c[k-1]

Step 4: Simplify to a Single-Variable Recurrence

By substituting and eliminating b[k] and c[k], we end up with a clean recurrence relation for a[k]:

a[k] = a[k-1] + 3*a[k-2] - 2*a[k-3]

With fixed initial values:

• a[0] = 1
• a[1] = 0
• a[2] = 2

Efficient Calculation Methods #

1. Iterative DP (For Small n)

If n is small (e.g., ≤ 10^5), we can compute the result iteratively in O(n) time with this simple Python function:

def count_circular_paths(n):
    if n == 0:
        return 1
    if n == 1:
        return 0
    # Initialize base cases
    a0, a1, a2 = 1, 0, 2
    if n == 2:
        return a2
    # Iterate up to n
    for k in range(3, n + 1):
        a_next = a2 + 3 * a1 - 2 * a0
        a0, a1, a2 = a1, a2, a_next
    return a2

2. Matrix Exponentiation (For Large n)

For very large n (like 10^9), iterative DP will be too slow. Instead, use matrix exponentiation to get the result in O(log n) time. The transformation matrix for our recurrence is:

[ 1  3  -2 ]
[ 1  0   0 ]
[ 0  1   0 ]

Raise this matrix to the power of n-2, multiply it by the initial state vector [a2, a1, a0] = [2, 0, 1], and the first element of the resulting vector is a[n].

3. Closed-Form Formula

We can also derive a direct formula using the recurrence's characteristic equation. This is great for one-off calculations, though you'll need to handle floating-point precision carefully (or use exact integer arithmetic):

a[k] = (1/5)*2^k + (2/5)*[ ((-1+√5)/2)^k + ((-1-√5)/2)^k ]

Example Verifications #

Let's test our solutions with small values to confirm correctness:

• n=0: 1 path (stay at c1) ✔️
• n=1: 0 paths (can't return in 1 move) ✔️
• n=2: 2 paths (c1→c2→c1, c1→c5→c1) ✔️
• n=5: 2 paths (c1→c2→c3→c4→c5→c1, c1→c5→c4→c3→c2→c1) ✔️

内容的提问来源于stack exchange，提问作者Manveer Singh