Lab 5: Inheritance

Simulate the inheritance of blood types for each member of a family.

$ ./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

Background

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.

Accepting this Lab

  1. Accept this lab via GitHub Classroom.
  2. After about a minute, refresh the page and ensure you see “You’re ready to go!”.

Getting Started

Log into code.cs50.io, click on your terminal window, and execute cd by itself. You should find that your terminal window’s prompt resembles the below:

$

Next execute

get50 inheritance

in order to download a directory called inheritance into your codespace.

Then execute

cd inheritance

in order to change into that directory. Your prompt should now resemble the below:

inheritance/ $

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

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

Understanding

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').

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. 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). We then call print_family to print out each of those family members and their blood types. Finally, the function calls free_family to free any memory that was previously allocated with malloc.

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

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.

We’ve divided the create_family function into a few TODOs for you to complete.

  • 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.
  • Next, we’ve included a condition to check if 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.
    • Otherwise (if generations == 1), then 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.
  • Finally, your function should return a pointer for the person that was allocated.

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.

Hints

  • 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.
  • Remember, to allocate memory for a particular person, we can use malloc(n), which takes a size as argument and will allocate n bytes of memory.
  • Remember, to access a variable via a pointer, we 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].

How to Test Your Code

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

Execute the below to evaluate the correctness of your code using check50. But be sure to compile and test it yourself as well!

check50 cs50/labs/2021/fall/inheritance

Execute the below to evaluate the style of your code using style50.

style50 inheritance.c

How to Submit

In your terminal, execute the below to submit your work.

submit50 inheritance

To confirm your submission, go to github.com/classroom50/inheritance-USERNAME where USERNAME is your GitHub username. You should see the code from your latest submission.

Labs are assessed only on whether you’ve submitted an honest attempt.

Want to see the staff's solution?
// Simulate genetic inheritance of blood type

#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>

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

const int GENERATIONS = 3;
const int INDENT_LENGTH = 4;

person *create_family(int generations);
void print_family(person *p, int generation);
void free_family(person *p);
char random_allele();

int main(void)
{
    // Seed random number generator
    srand(time(0));

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

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

    // Free memory
    free_family(p);
}

// Create a new individual with `generations`
person *create_family(int generations)
{
    // Allocate memory for new person
    person *new_person = malloc(sizeof(person));

    // If there are still generations left to create
    if (generations > 1)
    {
        // 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];
    }

    // If there are no generations left to create
    else
    {
        // 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();
    }

    // Return newly created person
    return new_person;
}

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

    // Free parents recursively
    free_family(p->parents[0]);
    free_family(p->parents[1]);

    // Free child
    free(p);
}

// Print each family member and their alleles.
void print_family(person *p, int generation)
{
    // Handle base case
    if (p == NULL)
    {
        return;
    }

    // Print indentation
    for (int i = 0; i < generation * INDENT_LENGTH; i++)
    {
        printf(" ");
    }

    // Print person
    if (generation == 0)
    {
        printf("Child (Generation %i): blood type %c%c\n", generation, p->alleles[0], p->alleles[1]);
    }
    else if (generation == 1)
    {
        printf("Parent (Generation %i): blood type %c%c\n", generation, p->alleles[0], p->alleles[1]);
    }
    else
    {
        for (int i = 0; i < generation - 2; i++)
        {
            printf("Great-");
        }
        printf("Grandparent (Generation %i): blood type %c%c\n", generation, p->alleles[0], p->alleles[1]);
    }

    // Print parents of current generation
    print_family(p->parents[0], generation + 1);
    print_family(p->parents[1], generation + 1);
}

// Randomly chooses a blood type allele.
char random_allele()
{
    int r = rand() % 3;
    if (r == 0)
    {
        return 'A';
    }
    else if (r == 1)
    {
        return 'B';
    }
    else
    {
        return 'O';
    }
}