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    <p>Since you haven’t specified what your problem is I will just go through what you need to know.</p> <p><strong>Closed under union</strong> - what does it mean? <br> If we have a language <code>L</code> and a language <code>S</code> and both are context-free we know that the union of the languages is also context free. </p> <blockquote> <p>L ∪ S = A Context-free Language</p> </blockquote> <p>See <a href="http://en.wikipedia.org/wiki/Context-free_language#Closure_properties" rel="nofollow noreferrer">this</a> for more closure properties of CFLs and you can find proofs of this on the web if you are interested (or you can try to make your own proof). </p> <p><strong>How can you use this to solve your problem?</strong> <br> You have this language specification (let’s call it <code>L</code>): </p> <blockquote> <p>L = {a<sup>m</sup> b<sup>n</sup> c<sup>p</sup> d<sup>q</sup> : n=q or m &lt;= p or m+n=p+q }</p> </blockquote> <p>You can split this language into 3 other languages:</p> <blockquote> <p>A = {a<sup>m</sup> b<sup>n</sup> c<sup>p</sup> d<sup>q</sup> : n = q } <br> B = {a<sup>m</sup> b<sup>n</sup> c<sup>p</sup> d<sup>q</sup> : m &lt;= p }<br> C = {a<sup>m</sup> b<sup>n</sup> c<sup>p</sup> d<sup>q</sup> : m+n = p+q }</p> </blockquote> <p>It is easy to see that <code>A ∪ B ∪ C = L</code>.</p> <p>If you can prove that A, B and C are context-free you can conclude that L is also context-free.</p> <p><strong>Solution</strong><br> To determine whether a language is context-free, see <a href="https://stackoverflow.com/a/3510165/1401257">this answer</a>. To quote the answer:</p> <blockquote> <p>First, you should attempt to build a <a href="http://en.wikipedia.org/wiki/Context-free_grammar" rel="nofollow noreferrer">context-free grammar</a> that forms the language in subject. A grammar is context-free if left-hand sides of all productions contain exactly one non-terminal symbol. By definition, if one exists, then the language is context-free.</p> <p>An equivalent construct would be a <a href="http://en.wikipedia.org/wiki/Pushdown_automaton" rel="nofollow noreferrer">pushdown automaton</a>. It's the same as DFA, but with a stack available. It may be easier to build than a grammar.</p> </blockquote> <p>So if you can build a CFG (or a PDA) for each of the languages A, B and C, you have solved your problem.</p> <p>Let’s take language <code>A</code>: We need a grammar that generates a string of the form <code>a..b..c..d..</code> where we must have an equal amount of <code>b</code>s and <code>d</code>s.</p> <pre><code>S -&gt; AE A -&gt; Aa | ε E -&gt; bEd | C C -&gt; Cc | ε </code></pre> <p><code>S</code> is the start variable (or non-terminal), <code>ε</code> is the empty string (some people use null symbol <code>^</code> but I have always been told to use <code>ε</code>). This grammar should be able to generate the language <code>A</code> and therefore <code>A</code> is a context-free. (Please let me know if anyone detects an error. I haven't created CFGs in a while).</p> <p>Since this is an exercise I will let you solve the remaining parts, but you should have gotten an idea of how to do it. </p>
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    1. COhuge thanks for such a great text!
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