两边独有的加起来是多少

这个数叫"分叉距离"——两条分支各自往前走了多远。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']}
a = len(only_on(g, 'a1c5f83', '5c8b206'))
b = len(only_on(g, '5c8b206', 'a1c5f83'))
print(str(a) + "+" + str(b) + "=" + str(a + b))