逻辑： is ( A && !(B || C)) || ( B || C ) 与 ( A || B || C ) 相同吗？

发布于 2024-10-12 23:50:54 字数 482 浏览 4 评论 0原文

我遇到过一些 obj-c 代码，我想知道是否有办法简化它：

#if ( A && !(B || C)) || ( B || C )

这是否相同？

#if ( A || B || C )

如果没有，是否有另一种更容易阅读的方式来表述它？

[编辑] 在提出问题之前，我尝试了真值表，但我认为我一定遗漏了一些东西，因为我怀疑 Foundation.framework/Foundation.h 会采用这种更复杂的形式。有充分的理由吗？

这是原始代码（来自 Foundation.h）：

#if (TARGET_OS_MAC && !(TARGET_OS_EMBEDDED || TARGET_OS_IPHONE)) || (TARGET_OS_EMBEDDED || TARGET_OS_IPHONE)

原文

I've encountered some obj-c code and I'm wondering if there's a way to simplify it:

#if ( A && !(B || C)) || ( B || C )

is this the same as?

#if ( A || B || C )

If not, is there another way to formulate it that would be easier to read?

[edit]
I tried the truth table before asking the question, but thought I had to be missing something because I doubted that Foundation.framework/Foundation.h would employ this more complex form. Is there a good reason for it?

Here's the original code (from Foundation.h):

#if (TARGET_OS_MAC && !(TARGET_OS_EMBEDDED || TARGET_OS_IPHONE)) || (TARGET_OS_EMBEDDED || TARGET_OS_IPHONE)

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ˇ宁静的妩媚 2024-10-19 23:50:54

是的。就像其他人说的，你可以用真值表来表示。德摩根规则也能有所帮助。

但是，我认为最好的选择是使用卡诺地图。学习需要几分钟，但卡诺图可以让您始终如一地找到布尔逻辑的最小表达式。真值表可以验证最小化，但他们无法将其提供给您。

我是这样得到的：

首先，表格布局：

         AB
     00   01   11   10
  0|    |    |    |    |
C 1|    |    |    |    |

现在，考虑你的方程，B || C总是会导致一个事实：

         AB
     00   01   11   10
  0|    |  T |  T |    |
C 1|  T |  T |  T |  T |

这样就只剩下两种情况了。无论哪种情况，右侧的计算结果都是 false。对于 000，左侧的计算结果也为 false（0 && !(whatever) 为 false）。对于 100, 1 && !(0 ||| 0) 计算结果为 true。因此，该陈述是正确的。填写：

         AB
     00   01   11   10
  0|  F |  T |  T |  T |
C 1|  T |  T |  T |  T |

现在，我们只需“掩盖”所有真相即可。 “C”将覆盖底行。 “B”将覆盖中间的方块（四个值）。因此，“B || C”覆盖了除右上角方块之外的所有区域。现在，“A”将覆盖右侧的四格正方形。没关系，这是多余的。因此，“A || B || C”覆盖了所有正确的方格并忽略了唯一错误的方格。

Yes. Like others said, you can truth table it. The De Morgan rules can also help.

However, I think the best option is to use a Karnaugh Map. It takes a few minutes to learn, but Karnaugh Maps allow you to consistently find the most minimal expression for boolean logic. Truth tables can verify a minimization, but they can't give it to you.

Here's how I got it:

First, the table layout:

         AB
     00   01   11   10
  0|    |    |    |    |
C 1|    |    |    |    |

Now, considering your equation, B || C will always cause a truth:

         AB
     00   01   11   10
  0|    |  T |  T |    |
C 1|  T |  T |  T |  T |

This leaves only two cases. In either case, the right side evaluates to false. For 000, the left side also evaluates to false (0 && !(whatever) is false). For 100, 1 && !(0 ||| 0) evaluates to true. Thus, the statement is true. Filling in:

         AB
     00   01   11   10
  0|  F |  T |  T |  T |
C 1|  T |  T |  T |  T |

Now, we only need to "cover" all the truths. "C" will cover the bottom row. "B" will cover the middle square (of four values). Thus, "B || C" covers all but the top right square. Now, "A" will cover the right four-space square. It's OK that this is redundant. Thus, "A || B || C" covers all the true squares and omits the only false one.

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凉城 2024-10-19 23:50:54

拿起笔+纸+尝试一下，只有8种可能的输入

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迷离° 2024-10-19 23:50:54

他们是一样的。您可以使用真值表生成器来测试它。这两个表达式仅在一种情况下给出 false，即 A、B 和 C 为 false< /代码>。

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半枫 2024-10-19 23:50:54

A | B | C | (B || C) | (!(B || C)) | (A && !(B || C)) | (A && (!(B || C)) || (B || C) | (A || B || C)
------------------------------------------------------------------------------------------------------
T | T | T |     T    |       F     |         F        |                 T             |         T      
T | T | F |     T    |       F     |         F        |                 T             |         T 
T | F | T |     T    |       F     |         F        |                 T             |         T 
T | F | F |     F    |       T     |         T        |                 T             |         T 
F | T | T |     T    |       F     |         F        |                 T             |         T 
F | T | F |     T    |       F     |         F        |                 T             |         T 
F | F | T |     T    |       F     |         F        |                 T             |         T 
F | F | F |     F    |       T     |         F        |                 F             |         F

根据最后两列，我会说是的。

A | B | C | (B || C) | (!(B || C)) | (A && !(B || C)) | (A && (!(B || C)) || (B || C) | (A || B || C)
------------------------------------------------------------------------------------------------------
T | T | T |     T    |       F     |         F        |                 T             |         T      
T | T | F |     T    |       F     |         F        |                 T             |         T 
T | F | T |     T    |       F     |         F        |                 T             |         T 
T | F | F |     F    |       T     |         T        |                 T             |         T 
F | T | T |     T    |       F     |         F        |                 T             |         T 
F | T | F |     T    |       F     |         F        |                 T             |         T 
F | F | T |     T    |       F     |         F        |                 T             |         T 
F | F | F |     F    |       T     |         F        |                 F             |         F

Based on the last two columns, I would say yes.

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