Python 如何检查周期范围的重叠（重叠的年度重复周期）

我试图找到一个优雅的算法来检查两年周期是否重叠。这一时期与年份无关，但可以预期一年总是闰年

例如，周期A=（3月1日至5月1日）和周期B=（4月1日至9月1日）重叠。 此外，A期=（10月1日至2月1日）和B期=（1月1日至3月1日）重叠

然而，我发现这比我预期的要困难得多。复杂性来自于跨越年底的时期

我已经找到了一个有效的解决方案（请参见下面的

doesOverlap（a，B）

方法），但我发现它很难实现

# for the rest of the MWE context code, see further
# WORKING, but a bit convulted
def doesOverlap(A, B):  
    '''returns True if yearly period A and B have overlapping dates'''
    # list to track if day in year is part of a period A
    # (this could probably be done a bit cheaper with a dictionary of tuples, but not relevant for my question)
    yeardayCovered = [False for x in range(366)]  # leap year

    # mark the days of A
    for d in range(A.start, A.start + A.length):
        yeardayCovered[d % 366] = True

    # now check each of the days in B with A
    for d in range(B.start, B.start + B.length):
        if yeardayCovered[d % 366]:
            return True
    return False

我相信这应该是可能的，用更少的检查和更优雅。我尝试将其中一个开始日期设置为零偏移，应用一些模运算符，然后进行常规（非循环）范围重叠检查（）。但我并没有在所有的测试用例中都使用它

#NOT WORKING CORRECTLY!!
def doesOverlap(A, B):   
    '''determines if two yearly periods have overlapping dates'''
    Astart = A.start
    Astop = A.stop
    Bstart = B.start
    Bstop = B.stop

    # start day counting at Astart, at 0
    offset = Astart
    Astart = 0
    Astop = (Astop - offset) % 366
    Bstart = (Bstart - offset) % 366
    Bstop = (Bstop - offset) % 366

    # overlap?
    # https://stackoverflow.com/a/13513973
    return (Astart <= Bstop and Bstart <= Astop)

---

您可以将跨年底的时段划分为两个时段，然后对这两个时段进行比较。这将产生一个非常简单的递归函数，不需要任何特定于Python的特性：

def overlap(start_a, stop_a, start_b, stop_b):
    if start_a > stop_a:
        return overlap(start_a, 365, start_b, stop_b) or overlap(0, stop_a, start_b, stop_b)
    elif start_b > stop_b:
        return overlap(start_a, stop_a, start_b, 365) or overlap(start_a, stop_a, 0, stop_b)
    else:
        return start_a <= stop_b and start_b <= stop_a

def doesOverlap(A, B):
    return overlap(A.start, A.stop, B.start, B.stop)

def重叠（启动a、停止a、启动b、停止b）：
如果开始>停止：
返回重叠（开始a、365、开始b、停止b）或重叠（0、停止a、开始b、停止b）
elif开始>停止：
返回重叠（开始a、停止a、开始b、365）或重叠（开始a、停止a、0、停止b）
其他：
返回开始位置a
def重叠（a0、a1、b0、b1）：
#首先，我们从每年的“循环”中“提升”间隔
#在以下情况下，通过调整结束日期调整时间“线”
#在开始日期之前结束。。。
如果a1b0且a0b0和a0+365b0和a0-365
啊，这就是我要找的！我已经完成了从循环到线性的提升，但是向前/向后移动是我没有想到的。很好的解决方案，尤其是递归
def overlap(start_a, stop_a, start_b, stop_b):
    if start_a > stop_a:
        return overlap(start_a, 365, start_b, stop_b) or overlap(0, stop_a, start_b, stop_b)
    elif start_b > stop_b:
        return overlap(start_a, stop_a, start_b, 365) or overlap(start_a, stop_a, 0, stop_b)
    else:
        return start_a <= stop_b and start_b <= stop_a

def doesOverlap(A, B):
    return overlap(A.start, A.stop, B.start, B.stop)

def overlap(a0, a1, b0, b1):
    # First we "lift" the intervals from the yearly "circle"
    # to the time "line" by adjusting the ending date if
    # ends up before starting date...
    if a1 < a0: a1 += 365
    if b1 < b0: b1 += 365

    # There is an intersection either if the two intervals intersect ...
    if a1 > b0 and a0 < b1: return True

    # ... or if they do after moving A forward or backward one year
    if a1+365 > b0 and a0+365 < b1: return True
    if a1-365 > b0 and a0-365 < b1: return True

    # otherwise there's no intersection
    return False