现在的分叉距离是多少

分叉距离 = 两边各自独有的提交数之和。运行下面这段程序：

def ancestors(g, c):
    seen = set()
    st = [c]
    while st:
        x = st.pop()
        if x in seen:
            continue
        seen.add(x)
        st.extend(g[x])
    return seen

def only_on(g, a, b):
    return ancestors(g, a) - ancestors(g, b)

g = {'3f2a91c': [],
     '7b4e2d0': ['3f2a91c'],
     'a1c5f83': ['7b4e2d0'],
     'e90d417': ['7b4e2d0'],
     '5c8b206': ['e90d417']}
d = len(only_on(g, 'a1c5f83', '5c8b206')) + \
    len(only_on(g, '5c8b206', 'a1c5f83'))
print(d)