entityMap|blocks|key|1dlbm|text|有很多正确的方法可以做到这一点。我将只介绍其中一个，我希望这是解决这些问题的一种有用的方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3bkrm|首先，我们注意到三个段之间的共性n。我们将划掉每个部分的符号，一次一个，以确保它们具有正确的关系。首先，我们可以验证a和b之间的关系是正确的。然后，我们可以检查a和c之间的关系。如果这两个都是正确的，我们就完成了。|fql4p|首先，让我们去掉a中令人讨厌的"%2B1“，这意味着不管n是什么，我们至少有一个a。现在，剩余的输入应该有a的n个实例，b的2n个实例和c的3n个实例。如果我们没有a，我们可以立即停止-拒绝；如果我们没有至少一个，我们就不能有a的n%2B1实例。|5940t|现在，如果有更多的a实例，通过在其位置上写A来划掉它，通过在b的位置上写B来划掉b的两个实例。然后，返回并查找a的更多实例，来回跳转，直到不再有a的实例。然后，验证是否不再有b的实例；如果是，则b的实例是a的两倍，并且前两个部分具有正确的关系。如果在任何时候你没有足够的b来删除，或者如果你用完了a，你仍然有b，那么你可以安全地在这一点上停止-reject。|3o3ah|接下来，您可以对A和c的实例执行相同的操作，只需划掉c的三个实例，而不是两个。如果我们发现a和c同时被耗尽，我们就完成并停止接受。|fdc8h|转换可能如下所示：|fdodi|Q++++T++++Q'++++T'++++d++++++++comment
q0+++a++++q1++++X+++++right++++account+for+%2B1

q1+++a++++q2++++A+++++right++++n>0+case,+continue
q1+++#++++hA++++#+++++same+++++n=0+case,+accept

q2+++a++++q2++++a+++++right++++skip+uncrossed+a
q2+++B++++q2++++B+++++right++++skip+crossed+b
q2+++b++++q3++++B+++++right++++find+first+uncrossed+b,+cross+it

q3+++b++++q4++++B+++++left+++++find+next+b,+cross+it

q4+++B++++q4++++B+++++left+++++find+last+uncrossed+a
q4+++a++++q2++++A+++++right++++cross+it
q4+++A++++q4++++A+++++left+++++skip+crossed+a,+if+any
q4+++X++++q5++++X+++++right++++ran+out+of+a+to+cross

q5+++A++++q5++++A+++++right++++skip+crossed+a
q5+++B++++q5++++B+++++right++++skip+crossed+b
q5+++c++++q6++++c+++++left+++++verify+existence+of+some+c

q6+++C++++q6++++C+++++left+++++skip+crossed+c
q6+++B++++q6++++B+++++left+++++skip+crossed+b
q6+++A++++q7++++a+++++right++++find+last+crossed+a,+uncross
q6+++X++++q10+++X+++++right++++ran+out+of+crossed+a

q7+++a++++q7++++a+++++right++++skip+uncrossed+a
q7+++B++++q7++++B+++++right++++skip+crossed+b
q7+++C++++q7++++C+++++right++++skip+crossed+c
q7+++c++++q8++++C+++++right++++find+first+uncrossed+c,+cross
q8+++c++++q9++++C+++++right++++cross+2nd+uncrossed+c
q9+++c++++q6++++C+++++left+++++cross+3rd+uncrossed+c

q10++a++++q10+++a+++++right++++skip+uncrossed+a
q10++B++++q10+++B+++++right++++skip+crossed+b
q10++C++++q10+++C+++++right++++skip+crossed+c
q10++#++++hA++++#+++++same+++++accept+if+no+leftover+symbols+until+end|code-block|syntax|javascript|9ukvv^0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|S|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|T|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|U|9|@]|A|@]|B|$]]|$2|G|4|H|6|7|8|V|9|@]|A|@]|B|$]]|$2|I|4|J|6|7|8|W|9|@]|A|@]|B|$]]|$2|K|4|L|6|7|8|X|9|@]|A|@]|B|$]]|$2|M|4|N|6|O|8|Y|9|@]|A|@]|B|$P|Q]]|$2|R|4|-4|6|7|8|Z|9|@]|A|@]|B|$]]]]

There are a lot of correct ways to do this. I will just walk through one of them I hope illustrates a useful way to attack these problems.

First, we notice the commonality n between the three segments. We will cross off symbols from each section, one at a time, to make sure they have the right relationships. First, we can verify that the relationship between a and b is right. Then, we can check the relationship between a and c. If those are both right, we are done.

First, let's get rid of the pesky "+1" from the a. This means we have at least one a regardless of what n is. So, we can begin by changing a to X. Now, the remaining input should have n instances of a, 2n instances of b and 3n instances of c. If we don't have the one a, we can halt-reject immediately; we can't have n+1 instances of a if we don't have at least one.

Now, if there are more instances of a, cross it off by writing A in its place, and go cross off two instances of b by writing B in their places. Then, go back and look for more instances of a, bouncing back and forth until there are no more instances of a. Then, verify there are no more instances of b; if so, there were twice as many instances of b as there were of a, and the first two sections have the right relationship. If at any point you didn't have enough b to cross off, or if after you ran out of a you still had b, then you can safely halt-reject at this point.

Next, you can do the same thing for instances of A and c, just cross off three instances of c instead of two. If we find the A get exhausted at the same time as the c do, we are done and halt-accept.

The transitions might look like this:

<pre><code>Q T Q' T' d comment
q0 a q1 X right account for +1

q1 a q2 A right n&gt;0 case, continue
q1 # hA # same n=0 case, accept

q2 a q2 a right skip uncrossed a
q2 B q2 B right skip crossed b
q2 b q3 B right find first uncrossed b, cross it

q3 b q4 B left find next b, cross it

q4 B q4 B left find last uncrossed a
q4 a q2 A right cross it
q4 A q4 A left skip crossed a, if any
q4 X q5 X right ran out of a to cross

q5 A q5 A right skip crossed a
q5 B q5 B right skip crossed b
q5 c q6 c left verify existence of some c

q6 C q6 C left skip crossed c
q6 B q6 B left skip crossed b
q6 A q7 a right find last crossed a, uncross
q6 X q10 X right ran out of crossed a

q7 a q7 a right skip uncrossed a
q7 B q7 B right skip crossed b
q7 C q7 C right skip crossed c
q7 c q8 C right find first uncrossed c, cross
q8 c q9 C right cross 2nd uncrossed c
q9 c q6 C left cross 3rd uncrossed c

q10 a q10 a right skip uncrossed a
q10 B q10 B right skip crossed b
q10 C q10 C right skip crossed c
q10 # hA # same accept if no leftover symbols until end
</code></pre>

entityMap|0|type|IMAGE|mutability|IMMUTABLE|data|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/3dfeab5b2b043bd4249f042dc1b19e44.png|imageAlt|blocks|key|ddh8o|text|由于我们不应该解决您的作业:)，我在JFLAP上解决了以下语言，您可以针对您的语言稍作修改。逻辑是相同的，您需要添加几个状态。L=+{a%5En%2B1+b%5E2n+:n+>=+0}|unstyled|depth|inlineStyleRanges|entityRanges|l1q9|​|4o79j|?|atomic|offset|length|ek86h|dttto^0|0|0|0|1|0|0|0^^$0|$1|$2|3|4|5|6|$7|8|9|-4]]]|A|@$B|C|D|E|2|F|G|S|H|@]|I|@]|6|$]]|$B|J|D|K|2|F|G|T|H|@]|I|@]|6|$]]|$B|L|D|M|2|N|G|U|H|@]|I|@$O|V|P|W|B|X]]|6|$]]|$B|Q|D|K|2|F|G|Y|H|@]|I|@]|6|$]]|$B|R|D|-4|2|F|G|Z|H|@]|I|@]|6|$]]]]

Since we are not supposed to solve your homework :), I solved the following language on JFLAP and you can change it a little bit for your language. The logic is the same and you'd need to add a couple of states. 
L = {a^n+1 b^2n : n >= 0}

<a href="https://i.stack.imgur.com/ltc7c.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ltc7c.png" alt="the solution on JFLAP"></a>

I need some help with designing Turing machine that accepts language 
L= {a^n+1 b^2n c^3n: n>=0}

Design a turing machine that accepts the language L= {a^n+1 b^2n c^3n: n>=0}

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。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

 我需要一些帮助来设计图灵机，它接受语言L= {a^n+1 b^2n c^3n: n>=0} 

问设计一个接受语言L= {a^n+1 b^2n c^3n: n>=0}的图灵机
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问设计一个接受语言L= {a^n+1 b^2n c^3n: n>=0}的图灵机EN