CS50 PSet3: Tideman

A guide to the ‘Tideman’ problem in CS50 Week 3.

Goal: To write functions to determine the winner of a tideman election and to print the name of the winner. Before reading further it is important to understand how a tideman voting system works, which is explained in the problem description.

The vote function must take arguments rank, name, and ranks. If name is a match for the name of a valid candidate, then the ranks array should be updated to indicate that the voter has the candidate as their rank preference (where 0 is the first preference, 1 is the second preference, etc.) The function should return true if the rank was successfully recorded, and false otherwise (if, for instance, name is not the name of one of the candidates).

The record_preferences function is called once for each voter, and takes as argument the ranks array. The function should update the global preferences array to add the current voter’s preferences.

The add_pairs function should add all pairs of candidates where one candidate is preferred to the pairs array. A pair of candidates who are tied (one is not preferred over the other) should not be added to the array.
The function should update the global variable pair_count to be the number of pairs of candidates.

The sort_pairs function should sort the pairs array in decreasing order of strength of victory, where strength of victory is defined to be the number of voters who prefer the preferred candidate. If multiple pairs have the same strength of victory, you may assume that the order does not matter.

The lock_pairs function should create the locked graph, adding all edges in decreasing order of victory strength so long as the edge would not create a cycle.

The print_winner function should print out the name of the candidate who is the source of the graph. You may assume there will not be more than one source.

Just like with plurality, the main function is already written for us, and our objective is to populate the functions called within main. The correct behaviour of the program is shown below, taking as command line arguments the names of the candidates in the election before prompting the user for number of voters followed by each voter’s ranking of the candidates. Finally the winner of the election is printed.


The vote function is called once for every voter (i) and ranking (j) in a nested loop within the main function. For example, if there were three candidates the function would be called three times per voter to check each voter’s first, second and third ranking candidate choices. Just like in plurality, if it returns false then “Invalid vote.” is printed and the program ends.

The goal of the function is thus to first determine if the candidate name in each ranking position is valid and secondly to update each voter’s ranks array with that voter’s rank preferences.

The ranks array is a 1D array which defines the ranking order of the candidates for each voter. The candidate index integer (the order in which they were defined in the command line arguments) is what is stored in the ranks array. For example, if there are three candidates Alice, Bob and Charlie, then Alice will be [0], Bob will be [1] and Charlie will be [2]. If the voter votes for Bob in rank 1, Charlie in rank 2 and Alice in rank 3 then the outputted ranks array will be [1, 2, 0] for that voter.

The function itself is similar to the vote function in the plurality problem. It should loop through each candidate, checking using strcmp each time if the candidate name matches the name argument. If it does, add that candidate’s index to the current position of the ranks array and return true since the name is valid. If this condition isn’t met after checking all candidates against the name argument, return false.


This function is also called once for each voter, inside the same loop as the vote function. It takes one argument, the ranks array that was populated in the vote function. The objective here is to update the preferences array, which is a 2D array that indicates the number of voters who prefer candidate i to candidate j. In other words, if there are three candidates, preferences is a 3x3 table.

The code for this one is short, but the logic can be challenging. A nested loop must be used, with both the inner and outer loops looping up to the number of candidates. The inner loop however should start from i + 1, as we only want to add to the preferences array when j comes after i in the ranks array. Once this condition is included, the preferences count can be increased by 1 at the position where the candidate in ranks[i] is being compared to the candidate in ranks[j].


This function must add every pair of candidates where one is preferred over the other to the pairs array. This is a 1D array of pair structs, which have both a winner and loser integer value. The global variable pair_count must also be incremented for every pair added.

Once again this can be achieved using a nested loop up to the amount of candidates, with the inner loop incrementally increasing from i + 1. This is done to prevent repeat checks of the preferences array. Again taking the example of three candidates, candidate [0] will be checked against [1] and [2], then candidate [1] will be checked against candidate [2] only. This fulfils every combination exactly once, as candidates should not be checked against themselves.

Within the inner loop, each combination can be checked to see if candidate i or candidate j is the winner using separate conditional statements. To check if i is the winner, we check if the number of people who prefer candidate i to candidate j is greater than the number of people who prefer j to i, reversing this to check if j is the winner. If there is a draw, both conditions pass and the loop continues.

With a winner determined, the pair struct in the pair_count position of the pairs array can be updated with a winner and loser. Recall that pair_count is initially equal to zero, as defined within the main function. pair_count should then be incremented, which allows the next pair to be inserted at the next position in the array.


With the pairs array defined, the purpose of this function is to sort them in order of decreasing strength of victory, where the strength of victory is defined to be the number of voters who prefer the preferred candidate. In terms of our program, it is equal to position [pairs.winner][pairs.loser] of the preferences array, which gives the number of people who prefer the winner to the loser.

Any sorting algorithm, as taught in the week 3 lecture, can be used for this function however I have opted for bubble sort. To do this I have once again used, yep you guessed it, a nested loop. An important thing to note here is that i is incremented downwards. This is because each pass of the loop will require less elements to be sorted as those to the right of i will already be sorted.

For each adjacent couple of pair structs in the pairs array, the strength of victory is compared. If the pair at position j + 1 has a greater strength of victory than the pair at position j, then they are swapped using a temporary variable.


The purpose of this function is to populate the locked 2D array, which is composed of boolean values and is initially set to all false. A pair can be ‘locked’ and the value set to true if locking that pair does not create a cycle back to the winning candidate. This is shown graphically in the below image. In this example Alice was preferred to Bob, and Charlie was preferred to Alice. If Bob was preferred to Charlie in one of the sorted pairs then a cycle would be created and there would not be a winner of the election. In this case however, Charlie can be declared the winner as he has no arrows pointing to him. In terms of the array, all values are false pointing towards him (more on that later).

It’s important that you fully understand what you want the function to do before starting with the syntax, as it can take a bit of time to get your head around. Personally I found it really clicked after reading this post. It’s important to realise that the sole purpose of the function is to populate the locked array based on the information given in the pairs array, nothing more nothing less.

Now on to the syntax itself. There are many ways to approach this with some using recursion, some not. My approach was to define a separate cycle function that takes as argument two candidates and checks if a cycle is created.

This is a recursive function, so a base case must first be defined that will exit the function if a cycle is created. This occurs when the function is being called for two identical candidate values, meaning a candidate points at themselves, thus creating a cycle.

The recursive case loops through each candidate and performs two checks. This is perhaps easier to understand with the knowledge that when this function is being called, end will initially be the loser and cycle_start will be the winner of each pair being checked. Firstly, if there is an arrow going from end to another candidate then there is a potential for a cycle. If not, there cannot be a cycle and the function returns false. If the first check returns true then cycle is called again, this time using the current candidate as the end argument. This process repeats until false is returned, meaning the chain ends and there is no cycle, or the base case is triggered as the chain has looped back round to the start, thus creating a cycle and returning true.

Within the lock_pairs function, I have looped through each pair and called the cycle function for each loser and winner. If cycle returns false, then the locked array can be set to true for the pair.

If this is still not making sense I recommend sitting down with a sheet of paper and walking through a few example pairs arrays. This was definitely the most challenging problem in CS50 thus far and it took me a while to get my head around.


The final function must print the winner of the election, given the information defined in the locked array. In terms of the array, the winner is the candidate with no true values in their column. In graphical terms, this means there are no arrows pointing towards them.

To determine this, we can loop through each candidate, declaring an integer to count the number of false values in that candidate’s column within locked. We then iterate through each candidate’s column, iterating false_count if a false value is found. If false_count is equal to candidate_count then the entire column is false and that candidate can be declared as a source of the graph and their name printed.


The difficulty definitely escalated significantly for this PSet as we had our first taste of recursion and sorting algorithms. Very satisfying to finally see a successful check50 for this one.