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Our final task in gen_aux is to examine every element of r If the element is bracketed, we have to expand it into a sequence of words; otherwise, we append it to ret What may seem like magic on first reading is that this processing is exactly what we are doing in gen_aux and therefore, we can call gen_aux to do it! Such a call is called recursive, and it is one of those techniques that looks like it can't possibly work until you've tried it a few times To convince yourself that this function works, begin by noting that the function obviously works if word is not bracketed Next, assume that word is bracketed, but its rule's right-hand side has no bracketed words of its own It should still be easy to see that the program will work in this case, because when it makes each recursive call, the gen_aux that it calls will immediately see that the word is not bracketed Therefore, it will append the word to ret and return The next step is to assume that word refers to a slightly more complicated rule one that uses bracketed words in its right-hand side, but only words that refer to rules with no bracketed words of their own When you encounter a recursive call to gen_aux, do not try to figure out what it does Instead, remember that you have already convinced yourself that it works in this case, because you know that at worst, its argument is a category that does not lead to any further bracketed words Eventually, you will see that the function works in all cases, because each recursive call simplifies the argument We do not know any sure way to explain recursion Our experience is that people stare at recursive programs for a long time without understanding how they work Then, one day, they suddenly get it and they don't understand why they ever thought it was difficult Evidently, the key to understanding recursion is to begin by understanding recursion The rest is easy Having written gen_sentence, read_grammar, and the associated auxiliary functions, we'll want to use them:
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int main() { // generate the sentence vector<string> sentence = gen_sentence(read_grammar(cin)); // write the first word, if any vector<string>::const_iterator it = sentencebegin(); if (!sentenceempty()) { cout << *it; ++it; } // write the rest of the words, each preceded by a space while (it != sentenceend()) { cout << " " << *it; ++it; } cout << endl; return 0; }
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We read the grammar, generate a sentence from it, and then write the sentence a word at a time The only even minor complexity is that we put a space in front of the second and subsequent words of the sentence
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It is now time to write nrand We begin by noting that the standard library includes a function named rand (defined in <cstdlib>) That function takes no arguments, and returns a random integer in the range [0, RAND_MAX], where RAND_MAX is a large integer that is also defined in <cstdlib> Our job is to reduce the range [0, RAND_MAX], which includes both 0 and RAND_MAX, to [0, n), which includes 0 but not n, with the understanding that n <= RAND_MAX
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You might think that it would suffice to compute rand() % n, which is the remainder when dividing the random integer by n In practice, this technique fails for two reasons The most important reason is pragmatic: rand() really returns only pseudo-random numbers Many C++ implementations' pseudo-random-number generators give remainders that aren't very random when the quotients are small integers For example, it is not uncommon for successive results of rand() to be alternately even and odd In that case, if n is 2, successive results of rand() % n will alternate between 0 and 1 There is another, more subtle reason to avoid using rand() % n: If the value of n is large, and RAND_MAX is not evenly divisible by n, some remainders will appear more often than others For example, suppose that RAND_MAX is 32767 (the smallest permissible value of RAND_MAX for any implementation) and n is 20000 In that case, there would be two distinct values of rand() that would cause rand() % n to be 10000 (namely, 10000 and 30000), but only one value of rand() that would cause rand() % n to be 15000 (namely, 15000) Therefore, the naive implementation of nrand would yield 10000 as a value of nrand(20000) twice as often as it would yield 15000 To avoid these pitfalls, we'll use a different strategy, by dividing the range of available random numbers into buckets of exactly equal size Then we can compute a random number and return the number of the corresponding bucket Because the buckets are of equal size, some random numbers may not fall into any bucket at all In that case, we keep asking for random numbers until we get one that fits The function is easier to write than to describe:
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// return a random integer in the range [0, n) int nrand(int n) { if (n <= 0 || n > RAND_MAX) throw domain_error("Argument to nrand is out of range"); const int bucket_size = RAND_MAX / n; int r; do r = rand() / bucket_size; while (r >= n); return r; }
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The definition of bucket_size relies on the fact that integer division truncates its result This property implies that RAND_MAX / n is the largest integer that is less than or equal to the exact quotient As a consequence, bucket_size is the largest integer with the property that n * bucket_size <= RAND_MAX The next statement is a do while statement A do while is like a while statement, except that it always executes the body at least once, and tests the condition at the end If that condition yields true, then the loop repeats, executing the push_back until the while fails In this case, the body of the loop sets r to a bucket number Bucket 0 will correspond to values of rand() in the range [0, bucket_size), bucket 1 will correspond to values in the range [bucket_size, bucket_size * 2), and so on If the value of rand() is so large that r >= n, the program will continue trying random numbers until it finds one that it likes, at which point it returns the corresponding value of r For example, let's assume that RAND_MAX is 32767 and n is 20000 Then bucket_size will be 1, and nrand will work by discarding random numbers until it finds one less than 20000 As another example, assume that n is 3 Then bucket_size will be 10922 In this case, values of rand() in the range [0, 10922) will yield 0, values in the range [10922, 21844) will yield 1, values in the range [21844, 32766) will yield 2, and values of 32766 or 32767 will be discarded
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