设一个三角形的三条边分别为 Z1、Z2、Z3，而且 Z1+Z2+Z3=15。请问有多少个不 同的等腰三角形？采用回溯法给出解答过程，用Python写出来

要解决这个问题，我们可以使用回溯算法遍历所有可能的边长组合，因为题目提到的是等腰三角形，所以只有两条边长度相同，另一条边长度不同。由于总和固定且是15，我们只需要找到所有可以构成等腰三角形的有效组合。

首先，我们将15拆分成三部分，然后递归地检查每种组合是否满足三角形条件（任意两边之和大于第三边）。下面是用Python实现的回溯算法：

def count_triangles(Z1, Z2, current_sum, z3, solutions):
    # Base case: If we've used all the sum and have a valid triangle
    if current_sum == 15 and Z1 <= z3 and Z2 <= z3:
        solutions.append([Z1, Z2, z3])
        return
    
    # Check all possible combinations with Z3 as the largest side
    if current_sum + z3 <= 15:
        count_triangles(Z1, Z2, current_sum + z3, z3, solutions)

    # Check for an equal side when Z1 or Z2 is the largest
    if current_sum + Z1 == 15:
        count_triangles(Z2, Z3, 0, Z1, solutions)
    if current_sum + Z2 == 15:
        count_triangles(Z1, Z3, 0, Z2, solutions)

# Start the recursion
count_triangles(1, 1, 0, 7, [])  # Since the minimum length is 1, we start from here
solutions = [sorted(triangle) for triangle in solutions]  # Sort the sides to eliminate duplicates
unique_solutions = list(set(tuple(solution) for solution in solutions))  # Remove duplicates

print(f"Total different non-degenerate equilateral triangles: {len(unique_solutions)}")
print("Unique solutions:")
for sol in unique_solutions:
    print(sol)

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