Discussion 2C Notes (Week 5, February 4) TA: Brian Choi Section Webpage:

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1 Discussion 2C Notes (Week 5, February 4) TA: Brian Choi Section Webpage: Recursion A recursion is a function-writing technique where the function refers to itself. Assuming you have learned the concept of recursion sufficiently in the lecture, we ll focus on the tricks in writing a recursive function. The Way to Think About Recursion Writing a recursive function involves the division of a problem into subproblems. A good example is the factorial function that computes n! = n x (n-1) x (n-2) x... x 2 x 1 for some arbitrary n 0. Here is the iterative way of writing factorial: int factorial(int n) int temp = 1; for (int i = 1; i <= n; i++) temp *= i; return temp; Iterative functions are more intuitive than recursive functions because we can see the whole process -- you see how factorial goes from 1 to n and multiply them up in that for loop. When writing the recursive version of the same function, you need not worry about going from 1 to n. Instead, we only worry about going from n-1 to n, believing factorial(n-1) will return the right value. In factorial s case, if you know the value of factorial(n-1), it is very easy to compute factorial(n) -- we only need to multiply n to the value of factorial(n-1) (because n * factorial(n-1) = n * (n-1)*(n-2)*...*2*1 = factorial(n)). This is what s meant by going from n-1 to n. Translating this into a C++ code, we get the following function: int factorial(int n) int temp = n * factorial(n - 1); return temp; My factorial function on the left simply assumes that factorial(n-1) returns what we want, and computes the value of factorial(n) based on this assumption. When factorial(n-1) is called, then it makes the same assumption and uses the value of factorial(n-2) to compute its value,..., and so on. For example, if factorial(5) is called, it relies on the value factorial(4) returns. When factorial(4) is called, it relies on the value factorial(3) returns. When factorial(3) is called, it relies on the value factorial(2)... Wait, there is something wrong here. We seem to be stuck in an infinite recursion. We need to stop this loop at some point. This is the role of the base case. In this example, factorial(0) becomes the base case. (Use your common sense when picking your base case -- we stop at 0 because factorial(n) is not defined for a negative n.) If we make factorial(0) explicitly return the right value, which is 1, then factorial(1) can compute its value based on this. Now that because factorial(1) works, factorial(2) can compute its value based on what factorial(1) returns. factorial(3) can use factorial(2) s value to compute its value, and so on. Copyright 2011 Brian Choi Week 5, Page 1/6

2 int factorial(int n) if (n == 0) // base case return 1; int temp = n * factorial(n - 1); return temp; This is like a domino effect, with the first domino block being factorial(0). Once we make sure the first case that works for sure, and define the relationship between two blocks, then it should work for the whole sequence. Use the following guideline when designing a recursive function: 1. Find the relationship between the current step and the previous step. Can you use some pre-computed value (e.g., factorial(n-1)) in the computation your current value (e.g., factorial(n))? 2. Find the base case. What is the smallest value that makes sense? (e.g., negative n doesn t make much sense in the factorial case). When should this chain of recursive calls stop? Make sure you re returning the right value for each of these base cases. If you know anything about proof by induction, you should find this concept very similar. Once you figured out 1 and 2 above, write a function this way: function_header take care of all the base cases, returning appropriate values write out the relationship between the previous step and the current step Don t forget to make sure it always hits some base case (i.e., no infinite recursion)! Practice helps you understand this concept. Let s do some practice. Recursion Exercises 1. Write the function average, that takes n doubles in an array arr and returns the average of all values in arr recursively (do not use while or for loops). double average(const double arr[], int n) // n > 0 if (n == 1) return arr[0]; double prevavg = average(arr, n - 1); return ((n - 1) * prevavg + arr[n - 1]) / n; Copyright 2011 Brian Choi Week 5, Page 2/6

3 2. Write the function sumofdigits, that takes in a positive integer n, and computes the sum of digits of n. For example, n=123 returns = 6. int sumofdigits(const int n) // n >= 0 if (n < 10) return n; return n % 10 + sumofdigits(n / 10); 3. Write the function deletechar, which takes in a string and a character, and return the string with all instances of the character removed. string deletechar(const string &s, const char c) if (s.empty()) return s; if (s[0] == c) return deletechar(s.substr(1), c); return s[0] + deletechar(s.substr(1), c); Copyright 2011 Brian Choi Week 5, Page 3/6

4 4. Consider the following sequence of numbers. 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, This sequence is known as the Fibonacci numbers, and is defined by the following recurrence relation: Write the function fibo that correctly returns the F(n) for a nonnegative n. int fibo(const int n) // n >= 0 if (n == 0) return 0; if (n == 1) return 1; F(0) = 0 F(1) = 1 F(n) = F(n-1)+F(n-2), n 2 return fibo(n - 1) + fibo(n - 2); 5. A palindrome is a phrase whose reverse is the same as itself. Examples include eye, racecar, and deed. Write the recursive function palindrome, which will check if a string is a palindrome or not. Assume there s no space in the string. bool palindrome(const string &s) int size = s.size(); if (size <= 1) return true; if (s[0]!= s[size - 1]) return false; return palindrome(s.substr(1, size - 2)); Copyright 2011 Brian Choi Week 5, Page 4/6

5 6. Given a set of integers, represented by an array s, write a function called exactsum that checks if the elements of some subset of s sums up to exactly target. Examples) If s[] = 1, 2, 3, then exactsum(s, 3, 6) returns true. If s[] = -3, 5, 2, then exactsum(s, 3, 2) returns true. If s[] = -3, 5, 2, then exactsum(s, 3, 8) returns false. If s[] =, then exactsum(s, 0, 10) returns false. If s[] = -3, 5, 2, -10, then exactsum(s, 3, 2) returns true. bool exactsum(int a[], int n, int targetsum) if (n == 0) return false; if (a[0] == targetsum) return true; return exactsum(a + 1, n - 1, targetsum - a[0]) exactsum(a + 1, n - 1, targetsum); 7. Write reverse(), which takes in a string and reverses the string recursively. string reverse(const string &s) if (s.empty()) return s; return reverse(s.substr(1)) + s[0]; // Could also do: // // return s[s.size() - 1] + reverse(s.substr(0, s.size() - 1)); Copyright 2011 Brian Choi Week 5, Page 5/6

6 8. To help remember phone numbers, people use mnemonic phone numbers, such as UCLA-CSD (spurious number, don t call this one). We will generate a function that, given an ordered sequence of digits, prints all possible mnemonics on the screen (one in each line). To help ease the implementation efforts, here is a digit-to-letters mapping function. string digittoletters(char digit) switch (digit) case '0': return "0"; case '1': return "1"; case '2': return "ABC"; case '3': return "DEF"; case '4': return "GHI"; case '5': return "JKL"; case '6': return "MNO"; case '7': return "PQRS"; case '8': return "TUV"; case '9': return "WYZ"; default: cout << "ERROR" << endl; abort(); Using the helper function above, implement mnemonics() below. It takes in prefix (e.g ) and digits (e.g ) as strings, and print all possible mnemonics. For this function, you are allowed to use a for loop. void mnemonics(const string &prefix, const string &digits) if (digits.empty()) cout << prefix << endl; return; string letters = digittoletters(digits[0]); for (int i = 0; i < letters.size(); i++) mnemonics(prefix + letters[i], digits.substr(1)); For example, mnemonics("", "723") prints PAD, PAE, PAF, PBD,..., QAD, QAE, QAF, etc. mnemonics("1-800-", "723") should print PAD, PAE, PAF, etc. The values do not have to be printed in any specific order, however. Copyright 2011 Brian Choi Week 5, Page 6/6