如何从两端迭代序列？

我想“从两端”（到（该部分的）中间，此处）迭代Python序列（的一部分）。
在 C/C++/Java 中，我可能会尝试

int start = 2, beyond = 7;
for (int forward = start, back = beyond ; forward <= --back ; forward++)
    printf("%d %d\n", forward, back);

我似乎无法决定我最不喜欢以下哪种变体：

seq = tuple(chr(c) for c in range(ord('a'), ord('z')+1))
start, beyond = 2, 7

print("subtract")
for forward in range(start, beyond):
    back = start + beyond - 1 - forward
    if (back < forward):
        break
    print(seq[forward], seq[back])
#    if (back <= forward): XXX ERROR, as YvesgereY pointed out
#        break
print("reversed")
for forward, back in zip(range(start, beyond),
                         reversed(range(start, beyond))):
    if (back < forward):
        break
    print(seq[forward], seq[back])
print("step -1")
for forward, back in zip(range(start, beyond), range(beyond-1, start-1, -1)):
    if (back < forward):
        break
    print(seq[forward], seq[back])
print("naked code")
forward, back = start, beyond-1 # thanks again, YvesgereY
while (forward <= back):
    print(seq[forward], seq[back])
    forward += 1
    back -= 1

有没有一种令人愉快/理智/惯用的方法来处理这个问题？

Why not zip与自身相反的列表？

mid = (len(lst) + 1) // 2   
for x, y in zip(lst[:mid], lst[::-1]):
     print((x, y))

(1, 5)
(2, 4)
(3, 3)

zip自动处理不同长度的列表，只需确保正确设置中点（这取决于列表大小是否均匀）。