return x2 >= y1 && x1 <= y2;

为什么这样做有效: 范围不重叠的唯一情况是当一个范围的结束在另一个范围的开始之前。所以我们想要!(x2 < y1 || x1 > y2)它等价于上面。

如果你正在处理，给定两个范围[x1:x2]和[y1:y2]，自然/反自然顺序范围同时存在:

自然顺序:x1 <= x2 && y1 <= y2或 反自然顺序:x1 >= x2 && y1 >= y2

然后你可能想用这个来检查:

它们重叠<=> (y2 - x1) * (x2 - y1) >= 0

其中只涉及四个操作:

2倍 一个乘法 一个比较

我想问题是关于最快的代码，而不是最短的代码。最快的版本必须避免分支，所以我们可以这样写:

对于简单的例子:

static inline bool check_ov1(int x1, int x2, int y1, int y2){
    // insetead of x1 < y2 && y1 < x2
    return (bool)(((unsigned int)((y1-x2)&(x1-y2))) >> (sizeof(int)*8-1));
};

或者，对于这种情况:

static inline bool check_ov2(int x1, int x2, int y1, int y2){
    // insetead of x1 <= y2 && y1 <= x2
    return (bool)((((unsigned int)((x2-y1)|(y2-x1))) >> (sizeof(int)*8-1))^1);
};

什么新东西。只是可读性更强。

def overlap(event_1, event_2):

    start_time_1 = event_1[0]
    end_time_1 = event_1[1]

    start_time_2 = event_2[0]
    end_time_2 = event_2[1]

    start_late = max(start_time_1, start_time_2)
    end_early = min(end_time_1, end_time_2)


    # The event that starts late should only be after the event ending early.
    if start_late > end_early:
        print("Absoloutly No overlap!")
    else:
        print("Events do overlap!")

重叠(X, Y):= if (X1 <= Y1) then (Y1 <= X2) else (X1 <= Y2)。

证明:

考虑X在Y之前或与Y左对齐的情况，即X1 <= Y1。那么Y要么在X内部开始，要么在X的末尾开始，即Y1 <= X2;或者Y远离x，第一个条件是重叠;第二个，不是。

在互补的情况下，Y在X之前，同样的逻辑适用于交换的实体。

So,

重叠(X, Y):= if (X1 <= Y) then (Y1 <= X2) else重叠(Y, X)。

但这似乎并不完全正确。在递归调用中，第一个测试是多余的，因为我们已经从第一个调用的第一个测试中知道了实体的相对位置。因此，我们实际上只需要测试第二个条件，即交换后(X1 <= Y2)。所以,

重叠(X, Y):= if (X1 <= Y1) then (Y1 <= X2) else (X1 <= Y2)。

QED.

Ada的实现:

   type Range_T is array (1 .. 2) of Integer;

   function Overlap (X, Y: Range_T) return Boolean is
     (if X(1) <= Y(1) then Y(1) <= X(2) else X(1) <= Y(2));

测试程序:

with Ada.Text_IO; use Ada.Text_IO;

procedure Main is

   type Range_T is array (1 .. 2) of Integer;

   function Overlap (X, Y: Range_T) return Boolean is
     (if X(1) <= Y(1) then Y(1) <= X(2) else X(1) <= Y(2));

   function Img (X: Range_T) return String is
     (" [" & X(1)'Img & X(2)'Img & " ] ");

   procedure Test (X, Y: Range_T; Expect: Boolean) is
      B: Boolean := Overlap (X, Y);
   begin
      Put_Line
        (Img (X) & " and " & Img (Y) &
         (if B then " overlap .......... "
               else " do not overlap ... ") &
         (if B = Expect then "PASS" else "FAIL"));
   end;
         
begin
   Test ( (1, 2), (2, 3), True);  --  chained
   Test ( (2, 3), (1, 2), True);

   Test ( (4, 9), (5, 7), True);  --  inside
   Test ( (5, 7), (4, 9), True);

   Test ( (1, 5), (3, 7), True);  --  proper overlap
   Test ( (3, 7), (1, 5), True);

   Test ( (1, 2), (3, 4), False);  -- back to back
   Test ( (3, 4), (1, 2), False);

   Test ( (1, 2), (5, 7), False);  -- disjoint
   Test ( (5, 7), (1, 2), False);
end;

以上程序输出:

 [ 1 2 ]  and  [ 2 3 ]  overlap .......... PASS
 [ 2 3 ]  and  [ 1 2 ]  overlap .......... PASS
 [ 4 9 ]  and  [ 5 7 ]  overlap .......... PASS
 [ 5 7 ]  and  [ 4 9 ]  overlap .......... PASS
 [ 1 5 ]  and  [ 3 7 ]  overlap .......... PASS
 [ 3 7 ]  and  [ 1 5 ]  overlap .......... PASS
 [ 1 2 ]  and  [ 3 4 ]  do not overlap ... PASS
 [ 3 4 ]  and  [ 1 2 ]  do not overlap ... PASS
 [ 1 2 ]  and  [ 5 7 ]  do not overlap ... PASS
 [ 5 7 ]  and  [ 1 2 ]  do not overlap ... PASS

反过来思考:如何使这两个范围不重叠?给定[x1, x2]，则[y1, y2]应在[x1, x2]之外，即y1 < y2 < x1或x2 < y1 < y2，这等价于y2 < x1或x2 < y1。

因此，使两个范围重叠的条件是:不(y2 < x1或x2 < y1)，这相当于y2 >= x1和x2 >= y1(与Simon接受的答案相同)。