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

我相信min(upper(A)，upper(B))>=max(lower(A)，lower(B))将是一个很好的解决方案，不仅因为它的简单性，而且因为它超越了两个范围的可扩展性。

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);
};

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)。

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

这很容易扭曲正常人的大脑，所以我找到了一个更容易理解的视觉方法:

勒解释

如果两个范围“太胖”，无法放入正好是两者宽度之和的槽中，那么它们就会重叠。

对于范围[a1, a2]和[b1, b2]，这将是:

/**
 * we are testing for:
 *     max point - min point < w1 + w2    
 **/
if max(a2, b2) - min(a1, b1) < (a2 - a1) + (b2 - b1) {
  // too fat -- they overlap!
}