Problem to Solve

A person’s blood type is determined by two alleles (i.e., different forms of a gene). The three possible alleles are A, B, and O, of which each person has two (possibly the same, possibly different). Each of a child’s parents randomly passes one of their two blood type alleles to their child. The possible blood type combinations, then, are: OO, OA, OB, AO, AA, AB, BO, BA, and BB.

For example, if one parent has blood type AO and the other parent has blood type BB, then the child’s possible blood types would be AB and OB, depending on which allele is received from each parent. Similarly, if one parent has blood type AO and the other OB, then the child’s possible blood types would be AO, OB, AB, and OO.

In a file called inheritance.c in a folder called inheritance, simulate the inheritance of blood types for each member of a family.


Distribution Code

For this problem, you’ll extend the functionality of code provided to you by CS50’s staff.

Download the distribution code

Log into, click on your terminal window, and execute cd by itself. You should find that your terminal window’s prompt resembles the below:


Next execute


in order to download a ZIP called into your codespace.

Then execute


to create a folder called inheritance. You no longer need the ZIP file, so you can execute


and respond with “y” followed by Enter at the prompt to remove the ZIP file you downloaded.

Now type

cd inheritance

followed by Enter to move yourself into (i.e., open) that directory. Your prompt should now resemble the below.

inheritance/ $

Execute ls by itself, and you should see and see a file named inheritance.c.

If you run into any trouble, follow these same steps again and see if you can determine where you went wrong!

Implementation Details

Complete the implementation of inheritance.c, such that it creates a family of a specified generation size and assigns blood type alleles to each family member. The oldest generation will have alleles assigned randomly to them.

  • The create_family function takes an integer (generations) as input and should allocate (as via malloc) one person for each member of the family of that number of generations, returning a pointer to the person in the youngest generation.
    • For example, create_family(3) should return a pointer to a person with two parents, where each parent also has two parents.
    • Each person should have alleles assigned to them. The oldest generation should have alleles randomly chosen (as by calling the random_allele function), and younger generations should inherit one allele (chosen at random) from each parent.
    • Each person should have parents assigned to them. The oldest generation should have both parents set to NULL, and younger generations should have parents be an array of two pointers, each pointing to a different parent.


Understand the code in inheritance.c

Take a look at the distribution code in inheritance.c.

Notice the definition of a type called person. Each person has an array of two parents, each of which is a pointer to another person struct. Each person also has an array of two alleles, each of which is a char (either 'A', 'B', or 'O').

// Each person has two parents and two alleles
typedef struct person
    struct person *parents[2];
    char alleles[2];

Now, take a look at the main function. The function begins by “seeding” (i.e., providing some initial input to) a random number generator, which we’ll use later to generate random alleles.

// Seed random number generator

The main function then calls the create_family function to simulate the creation of person structs for a family of 3 generations (i.e. a person, their parents, and their grandparents).

// Create a new family with three generations
person *p = create_family(GENERATIONS);

We then call print_family to print out each of those family members and their blood types.

// Print family tree of blood types
print_family(p, 0);

Finally, the function calls free_family to free any memory that was previously allocated with malloc.

// Free memory

The create_family and free_family functions are left to you to write!

Complete the create_family function

The create_family function should return a pointer to a person who has inherited their blood type from the number of generations given as input.

  • Notice first that this problem poses a good opportunity for recursion.
    • To determine the present person’s blood type, you need to first determine their parents’ blood types.
    • To determine those parents’ blood types, you must first determine their parents’ blood types. And so on until you reach the last generation you wish to simulate.

To solve this problem, you’ll find several TODOs in the distribution code.

First, you should allocate memory for a new person. Recall that you can use malloc to allocate memory, and sizeof(person) to get the number of bytes to allocate.

// Allocate memory for new person
person *new_person = malloc(sizeof(person));

Next, you should check if there are still generations left to create: that is, whether generations > 1.

If generations > 1, then there are more generations that still need to be allocated. We’ve already created two new parents, parent0 and parent1, by recursively calling create_family. Your create_family function should then set the parent pointers of the new person you created. Finally, assign both alleles for the new person by randomly choosing one allele from each parent.

  • Remember, to access a variable via a pointer, you can use arrow notation. For example, if p is a pointer to a person, then a pointer to this person’s first parent can be accessed by p->parents[0].
  • You might find the rand() function useful for randomly assigning alleles. This function returns an integer between 0 and RAND_MAX, or 32767. In particular, to generate a pseudorandom number that is either 0 or 1, you can use the expression rand() % 2.
// Create two new parents for current person by recursively calling create_family
person *parent0 = create_family(generations - 1);
person *parent1 = create_family(generations - 1);

// Set parent pointers for current person
new_person->parents[0] = parent0;
new_person->parents[1] = parent1;

// Randomly assign current person's alleles based on the alleles of their parents
new_person->alleles[0] = parent0->alleles[rand() % 2];
new_person->alleles[1] = parent1->alleles[rand() % 2];

Let’s say there are no more generations left to simulate. That is, generations == 1. If so, there will be no parent data for this person. Both parents of your new person should be set to NULL, and each allele should be generated randomly.

// Set parent pointers to NULL
new_person->parents[0] = NULL;
new_person->parents[1] = NULL;

// Randomly assign alleles
new_person->alleles[0] = random_allele();
new_person->alleles[1] = random_allele();

Finally, your function should return a pointer for the person that was allocated.

// Return newly created person
return new_person;
Complete the free_family function

The free_family function should accept as input a pointer to a person, free memory for that person, and then recursively free memory for all of their ancestors.

  • Since this is a recursive function, you should first handle the base case. If the input to the function is NULL, then there’s nothing to free, so your function can return immediately.
  • Otherwise, you should recursively free both of the person’s parents before freeing the child.

The below is quite the hint, but here’s how to do just that!

// Free `p` and all ancestors of `p`.
void free_family(person *p)
    // Handle base case
    if (p == NULL)

    // Free parents recursively

    // Free child


Not sure how to solve?

How to Test

Upon running ./inheritance, your program should adhere to the rules described in the background. The child should have two alleles, one from each parent. The parents should each have two alleles, one from each of their parents.

For example, in the example below, the child in Generation 0 received an O allele from both Generation 1 parents. The first parent received an A from the first grandparent and a O from the second grandparent. Similarly, the second parent received an O and a B from their grandparents.

$ ./inheritance
Child (Generation 0): blood type OO
    Parent (Generation 1): blood type AO
        Grandparent (Generation 2): blood type OA
        Grandparent (Generation 2): blood type BO
    Parent (Generation 1): blood type OB
        Grandparent (Generation 2): blood type AO
        Grandparent (Generation 2): blood type BO


check50 cs50/problems/2024/spring/inheritance


style50 inheritance.c

How to Submit

  1. Download your inheritance.c file by control-clicking or right-clicking on the file in your codespace’s file browser and choosing Download.
  2. Go to CS50’s Gradescope page.
  3. Click Problem Set 5: Inheritance.
  4. Drag and drop your inheritance.c file to the area that says Drag & Drop. Be sure it has that exact filename! If you upload a file with a different name, the autograder likely will fail when trying to run it. Ensuring you have uploaded files with the correct filename is your responsibility!
  5. Click Upload.

You should see a message that says “Problem Set 5: Inheritance submitted successfully!”