Inheritance
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.
Demo
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 cs50.dev, click on your terminal window, and execute cd
by itself. You should find that your terminal window’s prompt resembles the below:
$
Next execute
wget https://cdn.cs50.net/2023/fall/psets/5/inheritance.zip
in order to download a ZIP called inheritance.zip
into your codespace.
Then execute
unzip inheritance.zip
to create a folder called inheritance
. You no longer need the ZIP file, so you can execute
rm inheritance.zip
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 viamalloc
) oneperson
for each member of the family of that number of generations, returning a pointer to theperson
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 havealleles
assigned to them. The oldest generation should have alleles randomly chosen (as by calling therandom_allele
function), and younger generations should inherit one allele (chosen at random) from each parent. - Each
person
should haveparents
assigned to them. The oldest generation should have bothparents
set toNULL
, and younger generations should haveparents
be an array of two pointers, each pointing to a different parent.
- For example,
Hints
Click the below toggles to read some advice!
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];
}
person;
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
srand(time(0));
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
free_family(p);
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 byp->parents[0]
. - You might find the
rand()
function useful for randomly assigning alleles. This function returns an integer between0
andRAND_MAX
, or32767
. In particular, to generate a pseudorandom number that is either0
or1
, you can use the expressionrand() % 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 beforefree
ing 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)
{
return;
}
// Free parents recursively
free_family(p->parents[0]);
free_family(p->parents[1]);
// Free child
free(p);
}
Walkthrough
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
Correctness
check50 cs50/problems/2024/x/inheritance
Style
style50 inheritance.c
How to Submit
submit50 cs50/problems/2024/x/inheritance