Section 3. Recursion and Intro to Backtracking

Section materials curated by Kylie and Nick, drawing upon materials from previous quarters.

This week’s section exercises continue our exploration of recursion to tackle even more challenging and interesting problems. In particular, many of this week's section problems center around recursive backtracking, a very powerful and versatile problem-solving technique.

Remember that every week we will also be releasing a Qt Creator project containing starter code and testing infrastructure for that week's section problems. When a problem name is followed by the name of a .cpp file, that means you can practice writing the code for that problem in the named file of the Qt Creator project. Here is the zip of the section starter code:

📦 Starter code

1) Recursion Mystery Part 2

Topics: recursive function calls, output tracing

void recursionMystery2(int x, int y) {
    if (y == 1) { 
        cout << x;
    } else {
        cout << (x * y) << ", ";
        recursionMystery2(x, y - 1);
        cout << ", " << (x * y);

For each call to the above recursive function, write the output that would be produced, as it would appear on the console.

Call                                   Output
recursionMystery2(4, 1);        ___________________________________
recursionMystery2(4, 2);        ___________________________________
recursionMystery2(8, 2);        ___________________________________
recursionMystery2(4, 3);        ___________________________________
recursionMystery2(3, 4);        ___________________________________
8, 4, 8
16, 8, 16
12, 8, 4, 8, 12
12, 9, 6, 3, 6, 9, 12

2) Recursion Tracing

Topics: Recursion, strings, recursion tracing

Below is a recursive function to reverse a string:

string reverseOf(string s) {
    if (s.empty()) {
        return "";
    } else {
        return reverseOf(s.substr(1)) + s[0];

Trace through the execution of reverseOf("stop") along the lines of what we did in lecture, showing recursive call information for each call that’s made and how the final value gets computed.

Our initial call to reverseOf("stop") fires off a call to reverseOf("top"). This call fires off a call to reverseOf("op"). This in turn calls reverseOf("p"). This in turn calls reverseOf(""). This triggers the base case and returns the empty string. (Notice that the reverse of the empty string "" is indeed the empty string ""). We now append p to return "p". We now append o to return "po". We append t to return "pot". And finally we append s to return "pots" back to whoever called us. Yay!

3) Random Shuffling (shuffle.cpp)

How might the computer shuffle a deck of cards? This problem is a bit more complex than it might seem, and while it's easy to come up with algorithms that randomize the order of the cards, only a few algorithms will do so in a way that ends up generating a uniformly-random reordering of the cards.

One simple algorithm for shuffling a deck of cards is based on the following idea:

  • Choose a random card from the deck and remove it.
  • Shuffle the rest of the deck.
  • Place the randomly-chosen card on top of the deck. Assuming that we choose the card that we put on top uniformly at random from the deck, this ends up producing a random shuffle of the deck.

Write a function

	   		   string randomShuffle(string input)

that accepts as input a string, then returns a random permutation of the elements of the string using the above algorithm. Your algorithm should be recursive and not use any loops (for, while, etc.).

The header file "random.h" includes a function

                           int randomInteger(int low, int high);

that takes as input a pair of integers low and high, then returns an integer greater than or equal to low and less than or equal to high. Feel free to use that here.

Interesting note: This shuffling algorithm is a variant of the Fisher-Yates Shuffle. For more information on why it works correctly, take CS109!

Here is one possible solution:

string randomShuffle(string input) {
    /* Base case: There is only one possible permutation of a string
    * with no characters in it.
    if (input.empty()) {
        return input;
    } else {
        /* Choose a random index in the string. */
        int i = randomInteger(0, input.length() - 1);
        /* Pull that character to the front, then permute the rest of
        * the string.
        return input[i] + randomShuffle(input.substr(0, i) + input.substr(i + 1));

This function is based on the recursive observation that there is only one possible random shuffle of the empty string (namely, itself), and then using the algorithm specified in the handout for the recursive step.

Here is another possible solution (using a modified function header) which shows how to implement this function using references and no return value. Shoutout to one of our awesome SLs, Rachel Gardner, for this alternate solution!

void randomShuffle(string &input){
    if (input == "") return;
    int rand = randomInteger(0, input.length() - 1);
    char chosen = input[rand];
    input.erase(rand, 1);
    input = chosen + input;

4) Zig Zag (zigzag.cpp)

Topics: Recursion, printing output to console

Write a recursive function named zigzag that returns a string of n characters as follows. The middle character (or middle two characters if n is even) is an asterisk (*). All characters before the asterisks are '<'. All characters after are '>'. Report an error if n is not positive.

Call            Output
zigzag(1)       * 
zigzag(4)       <**> 
zigzag(9)       <<<<*>>>>
string zigzag(int n) {
    if (n < 1) {
        error("The value of n was negative");
    } else if (n == 1) {
        return "*";
    } else if (n == 2) {
        return "**";
    } else {
        return "<" + zigzag(n-2) + ">";

5) Double Stack (double.cpp)

Topics: Recursion, Stacks

Write a recursive function named doubleStack that takes a reference to a stack of ints and replaces each integer with two copies of that integer. For example, if s stores {1, 2, 3}, then doubleStack(s) changes it to {1, 1, 2, 2, 3, 3}.

void doubleStack(Stack<int>& s) 
    if (!s.isEmpty()) {
        int n = s.pop();

6) String Subsequences (subsequence.cpp)

Topics: Recursion, verifying properties

Write a recursive function named isSubsequence that takes two strings and returns true if the second string is a subsequence of the first string. A string is a subsequence of another if it contains the same letters in the same order, but not necessarily consecutively. You can assume both strings are already lower-cased.

Call                                    Output
isSubsequence("computer", "core")       false 
isSubsequence("computer", "cope")       true 
isSubsequence("computer", "computer")   true
bool isSubsequence(string big, string small) 
    if (small.empty()) {
        return true;
    } else if (big.empty()) {
        return false;
    } else {
        if (big[0] == small[0]) {
            return isSubsequence(big.substr(1), small.substr(1));
        } else {
            return isSubsequence(big.substr(1), small);

7) Recursion and Time Complexity and Exponents, (Big-)Oh My

Topics: recursion, time complexity, big O, algorithm comparison

Below is a simple function that computes the value of m^n when n is a nonnegative integer:

int raiseToPower(int m, int n) {
    int result = 1;
    for (int i = 0; i < n; i++) {
        result *= m;
    return result;

1) What is the big-O complexity of the above function, written in terms of m and n? You can assume that it takes time O(1) to multiply two numbers.

2) If it takes 1 microsecond (μs) to compute raiseToPower(100, 100), approximately how long will it take to compute raiseToPower(200, 10000)?

Below is a recursive function that computes the value of m^n when n is a nonnegative integer:

int raiseToPower(int m, int n) {
    if (n == 0) return 1;
    return m * raiseToPower(m, n - 1);

3) What is the big-O complexity of the above function, written in terms of m and n? You can assume that it takes time O(1) to multiply two numbers.

4) If it takes 1μs to compute raiseToPower(100, 100), approximately how long will it take to compute raiseToPower(200, 10000)?

Here’s an alternative recursive function for computing m^n that uses a technique called exponentiation by squaring. The idea is to modify the recursive step as follows:

  • If n is an even number, then we can write n as n = 2k. Then m^n = m^(2k) = (m^k)^2.
  • If n is an odd number, then we can write n as n = 2k + 1. Then m^n = m^(2k+1) = m * m^(2k) = m * (m^k)^2.

Based on this observation, we can write this recursive function:

int raiseToPower(int m, int n) {
    if (n == 0) {
        return 1;
    } else if (n % 2 == 0) {
        int halfPower = raiseToPower(m, n / 2);
        return halfPower * halfPower;
    } else {
        int halfPower = raiseToPower(m, n / 2);
        return m * halfPower * halfPower;

5) What is the big-O complexity of the above function, written in terms of m and n? You can assume that it takes time O(1) to multiply two numbers.

6) If it takes 1μs to compute raiseToPower(100, 100), approximately how long will it take to compute raiseToPower(200, 10000)?

  1. This function runs in time O(n). It runs the loop n times, at each step doing O(1) work. There is no dependence on m in the runtime.

  2. We know that this code runs in time O(n), so it scales roughly linearly with the size of n. Therefore, if it took 1μs to compute a value when n = 100, it will take roughly 100 times longer when we plug in n = 10000. As a result, we’d expect this code would take about 100μs to complete.

  3. If we trace through the recursion, we’ll see that we make a total of n recursive calls, each of which is only doing O(1) work. Adding up all the work done by these recursive calls gives us a total of O(n) work, as before.

  4. As before, this should take about 100μs.

  5. Notice that each recursive call does O(1) work (there are no loops anywhere here), then calls itself on a problem that’s half as big as the original one. This means that only O(log n) recursive calls will happen (remember that repeatedly dividing by two is the hallmark of a logarithm), so the total work done here is O(log n).

  6. We know that the runtime when n = 100 is roughly 1μs. Notice that 100^2 = 10,000, so we’re essentially asking for the runtime of this function when we square the size of the input. Also notice that via properties of logarithms that log n^2 = 2 log n. Therefore, since we know the runtime grows roughly logarithmically and we’ve squared the value of n, this should take about twice as long as before, roughly 2μs.

Note: The topics for the last two problems on this handout will not be covered until lecture on Friday, July 16. We provide them here not with the intent that you will cover them in section, but instead with the expectation that you can use them as extra practice when reviewing topics from Friday's lecture.

8) Win some, lose sum (sum.cpp)

Topics: recursive backtracking

Write a recursive function named canMakeSum that takes a reference to a Vector<int> and an int target value and returns true if it is possible to have some selection of values from the Vector that sum to the target value. In particular, you should be implementing a function with the following declaration:

bool canMakeSum(Vector<int>& values, int target)

For example, let's say that we executed the following code:

Vector<int> nums = {1,1,2,3,5};
canMakeSum(nums, 9)

We should expect that the call to canMakeSum should return true. Given the values specified in nums, we can select 1, 3, and 5 such that 5 + 3 + 1 = 9.

However, let's say that we executed the following code instead:

Vector<int> nums = {1,4,5,6};
canMakeSum(nums, 8);

We should expect that the call to canMakeSum in this case should return false, since there is no possible combination of values from the vector that sum up to the target value of 8.

bool canMakeSumHelper(Vector<int>& v, int target, int cur, int index) {
	if (index >= v.size()) {
		return cur == target;
	return canMakeSumHelper(v, target, cur + v[index], index + 1) ||
		   canMakeSumHelper(v, target, cur, index + 1);

bool canMakeSum(Vector<int>& v, int target) {
	return canMakeSumHelper(v, target, 0, 0);

9) Weights and Balances (weights.cpp)

Topics: recursive backtracking

I am the only child of parents who weighed,
measured, and priced everything; for whom
what could not be weighed, measured, and
priced had no existence.
—Charles Dickens, Little Dorrit, 1857

In Dickens’s time, merchants measured many commodities using weights and a two-pan balance – a practice that continues in many parts of the world today. If you are using a limited set of weights, however, you can only measure certain quantities accurately.

For example, suppose that you have only two weights: a 1-ounce weight and a 3-ounce weight. With these you can easily measure out 4 ounces, as shown below:

two-pan balance with unmarked weight on the left side and two weights on the right side, marked 1 and 3. The balance is level.

It’s more interesting to discover that you can also measure out 2 ounces by shifting the 1-ounce weight to the other side, as follows below:

two-pan balance with weight marked 1 and unmarked weight on the left side and one weight on the right side, marked 3. The balance is level.

Write a recursive function

bool isMeasurable(int target, Vector<int>& weights)

that determines whether it is possible to measure out the desired target amount with a given set of weights, which is stored in the vector weights.

As an example, the function call

Vector<int> weights = {1, 3};
isMeasurable(2, weights);

should return true because it is possible to measure out two ounces using the sample weight set as illustrated in the preceding diagram. On the other hand, calling

Vector<int> weights = {1, 3};
isMeasurable(5, weights);

should return false because it is impossible to use the 1- and 3-ounce weights to add up to 5 ounces. However, the call

Vector<int> weights = {1, 3, 7};
isMeasurable(6, weights);

should return true: you can measure the six-ounce weight by placing it and the one-ounce weight on one side of the scale and by placing the seven-ounce weight on the other.

Here’s a function question to ponder: let’s say that you get to choose n weights. Which ones would you pick to give yourself the best range of weights that you’d be capable of measuring?

Imagine that we start off by putting the amount to be measured (call it n) on the right side of the balance. This makes the target of measurement equal to n. Imagine that there is some way to reach this target n. If we put the weights on the scale one at a time, we can look at where we put that first weight (let’s suppose it weighs w). It must either

  • go on the left side, making the target weight on the scale n - w,
  • go on the right side, making the target weight on the scale n + w, or
  • not get used at all, leaving the target weight on the scale n.

If it is indeed truly possible to measure n, then one of these three options has to be the way to do it, even if we don’t know which one it is. The question we then have to ask is whether it’s then possible to measure the new net imbalance using the weights that remain – which we can determine recursively! On the other hand, if it’s not possible to measure n, then no matter which option we choose, we’ll find that there’s no way to use the remaining weights to make everything balanced!

If we’re proceeding recursively, which we are here, we need to think about our base case. There are many options we can choose from. One simple one is the following: imagine that we don’t have any weights at all, that we’re asked to see whether some weight is measurable using no weights. In what circumstances can we do that? Well, if what we’re weighing has a nonzero weight, we can’t possibly measure it – placing it on the scale will tip it to some side, but that doesn’t tell us how much it weighs. On the other hand, if what we’re weighing is completely weightless, then putting it on the scale won’t cause it to tip, convincing us that, indeed, it is weightless! So as our base case, we’ll say that when we’re down to no remaining weights, we can measure n precisely if n = 0. With that in mind, here’s our code:

bool isMeasurable(int target, Vector<int>& weights) {
	if (weights.isEmpty()) {
		return target == 0; // base case; no weights left to place
	} else {
		int last = weights[weights.size() - 1]; //using last index is fastest
		weights.remove(weights.size() - 1);

		// "choose and explore" all of the three possibilities
		bool result = isMeasurable(target + last, weights) ||
			   isMeasurable(target - last, weights) ||
			   isMeasurable(target, weights);
		// un-choose
		return result;