blocks|key|3712369|text|就我记忆中的布尔代数而言，你可以将它们作为普通数字相乘。因此(x2+%2B+x1x3x5)(x5+%2B+x6)将是x2x5+%2B+x2x6+%2B+x1x3x5x5+%2B+x1x3x5x6。但就我记忆所及，这只适用于"AND“，而不适用于"OR”|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3712370|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

As far as I remember the boolean algebra you can just multiply them as ordinary numbers.
So (x2 + x1x3x5)(x5 + x6) will be x2x5 + x2x6 + x1x3x5x5 + x1x3x5x6. But again as far as I remember this applied only to "AND" not to "OR"

blocks|key|5180058|text|这看起来更像是一道数学题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5180059|但如果我没弄错你想要什么，它应该是这样的|5180060|x2x4x5%2Bx2x4x6%2Bx2x3x5x6x5%2Bx2x3x5x6x6%2Bx1x3x5x4+ext.+|code-block|syntax|javascript|5180061|基本(a1%2Ba2)a3=a1a3%2Ba2a3+(a1%2Ba2)(a3%2Ba4)=a1a3%2Ba1a4%2Ba2a3%2Ba3a4|5180062|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Well this looks more like a math question.

But if I understand what you want correctly it would look like this

<pre><code>x2x4x5+x2x4x6+x2x3x5x6x5+x2x3x5x6x6+x1x3x5x4 ext. 
</code></pre>

basically 
(a1+a2)a3=a1a3+a2a3 
(a1+a2)(a3+a4)=a1a3+a1a4+a2a3+a3a4

blocks|key|5180092|text|这就是了：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5180093|(x2x4x5+%2B+x1x3x5)(x4x6)+%2B+(x2x5+%2B+x1x3x5)(x3x5x6)
(x2x4x5x6+%2B+x1x3x4x5x6)+%2B+(x2x5+%2B+x1x3x5)(x3x5x6)
x2x4x5x6+%2B+x1x3x4x5x6+%2B+x2x3x5x6+%2B+x1x3x5x6
x2x4x5x6+%2B+x1x3(x4x5x6+%2B+x5x6)+%2B+x2x3x5x6
x2x5x6(x4%2B1)+%2B+x1x3(x4x5x6+%2B+x5x6)
x2x4x5x6+%2B+x1x3(x5x6(x4%2B1))
x2x4x5x6+%2B+x1x3(x4x5x6)
x2x4x5x6+%2B+x1x3x4x5x6
(x2%2Bx1x3)x4x5x6|code-block|syntax|javascript|5180094|我可能在什么地方弄错了，所以你应该先测试一下。|5180095|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Here you go:

<pre><code>(x2x4x5 + x1x3x5)(x4x6) + (x2x5 + x1x3x5)(x3x5x6)
(x2x4x5x6 + x1x3x4x5x6) + (x2x5 + x1x3x5)(x3x5x6)
x2x4x5x6 + x1x3x4x5x6 + x2x3x5x6 + x1x3x5x6
x2x4x5x6 + x1x3(x4x5x6 + x5x6) + x2x3x5x6
x2x5x6(x4+1) + x1x3(x4x5x6 + x5x6)
x2x4x5x6 + x1x3(x5x6(x4+1))
x2x4x5x6 + x1x3(x4x5x6)
x2x4x5x6 + x1x3x4x5x6
(x2+x1x3)x4x5x6
</code></pre>

I probably made a mistake somewhere, so you should test it first.

blocks|key|1294039|text|你的原件：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1294040|f(x)+=+(x2+%2B+x1x3x5)(x4+%2B+x3x5x6)(x5+%2B+x6)|code-block|syntax|javascript|1294041|现在使用一些简单的数学方法：|1294042|f(x)+=+(x2x4+%2B+x2x3x5x6+%2B+x1x3x5x4+%2B+x1x3x5x3x5x6)(x5+%2B+x6)

f(x)+=+x2x4x5+%2B+x2x3x5x6x5+%2B+x1x3x5x4x5+%2B+x1x3x5x3x5x6x5+%2B+x2x4x6+%2B+x2x3x5x6x6+%2B+x1x3x5x4x6+%2B+x1x3x5x3x5x6x6|1294043|简化给我们带来了答案(虽然不一定更短)，|1294044|f(x)+=+x2x4x5+%2B+x2x3x5x6+%2B+x1x3x5x4x5+%2B+x1x3x5x6+%2B+x2x4x6+%2B+x2x3x5x6+%2B+x1x3x5x4x6+%2B+x1x3x5x6|1294045|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Your original: 

<pre><code>f(x) = (x2 + x1x3x5)(x4 + x3x5x6)(x5 + x6)
</code></pre>

Now using some simple maths:

<pre><code>f(x) = (x2x4 + x2x3x5x6 + x1x3x5x4 + x1x3x5x3x5x6)(x5 + x6)

f(x) = x2x4x5 + x2x3x5x6x5 + x1x3x5x4x5 + x1x3x5x3x5x6x5 + x2x4x6 + x2x3x5x6x6 + x1x3x5x4x6 + x1x3x5x3x5x6x6
</code></pre>

Simplifying gets us the answer (though not necessarily shorter),

<pre><code>f(x) = x2x4x5 + x2x3x5x6 + x1x3x5x4x5 + x1x3x5x6 + x2x4x6 + x2x3x5x6 + x1x3x5x4x6 + x1x3x5x6
</code></pre>

I have the following function:

<pre><code>f(x) = (x2 + x1x3x5)(x4 + x3x5x6)(x5 + x6)
</code></pre>

How can I make the expression like:

<pre><code>f(x) = x1x2x3 + x2x3x4 + ...
</code></pre>

out of this? Is there any method?

I'm not sure if SO is the right place to post this...I guess it's not, but still, I found the tag and around 100 posts with it, so here I am :P

How to shorten the boolean function

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有以下功能：f(x) = (x2 + x1x3x5)(x4 + x3x5x6)(x5 + x6)怎样才能使表达式像这样：f(x) = x1x2x3 + x2x3x4 + ...走出这个世界？有什么方法吗？我不确定这是否是发布这篇文章的正确位置……我猜不是，但我仍然找到了标签和大约100篇帖子，所以我在这里:P

问如何缩短布尔函数
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