Stable Marriage Algorithm in C - c

Question

I am trying my best to get this code done. It's been well over 15 hours now at least over the course of a couple days, and I am still stuck. I am programming in C, and it is my first language, so I am a bit of a noob I'm sorry. Any kind of help would be very helpful.
A quick first question:
while (manStatus[i] == -1 || womanStatus[i] == -1)
What would be the proper syntax or way to say that I want to continuously check the elements of my array to see if any value inside there is "-1" (or in reality, that the men and women inside those arrays are single, so keep attempting to pair them).
I have more, but this one has stumped me for days. I'll keep struggling with the rest, but would love some help with this for starters if anyone has a moment.

Answer 1

In C one cannot simply expect manStatus[i] == -1 to check all the values within the array and return true if any value within matches -1.
Rather you need to write an algorithm which manually does the meticulous checking of each element in the array. It's actually very simple and uses a loop. I've implemented it here as a function:
bool checkArray(int array[],int arrayLength, int val)
{
int i;
for(i = 0; i < arrayLength; i++)
{
if(array[i] == val)
return true;
}
return false;
}
Now you can call this function in your statement:
while (checkArray(manStatus, LENGTH_OF_ARRAY, -1) || checkArray(womenStatus, LENGTH_OF_ARRAY, -1))

Answer 2

Well, there is lsearch that could be handy, but that is a non-standard function.
Define a helper function to check if a value is in an array.
int containsInt(int *array, size_t length, int value) {
size_t i;
for(i = 0; i < length; i++) {
if(array[i] == value) {
return 1;
}
}
return 0;
}
Then your check becomes,
while(!containsInt(manStatus, numMen, -1) && !containsInt(womanStatus, numWomen, -1))

Answer 3

If you want to check that any value in manStatus array is -1 you can do like this:
#define NON_WITH_MINUS_ONE 0
#define SOME_WITH_MINUS_ONE 1
int checkMarriage(int *manStatus,int N)
{
for(i=0;i<N;i++)
if(manStatus[i]==-1)
return SOME_WITH_MINUS_ONE
return NON_WITH_MINUS_ONE
}
where N is number of elements in the array. Then you can call this function e.g. by
if(checkMarriage(manStatus, 20)==1)
{//do something}
else
{ //do something else}

Answer 4

With the way you have done it, if you run out of either men or women before the other array has run out, then you are stuck in an infinite loop.
The best way to break out of the loop once there can be no more matches is:
while (manStatus[i] == -1` && `womanStatus[i] == -1)

Answer 5

Let me first answer on a math point of view. From my understanding, you have two finite sets (happily the number of human beings on earth is finite ... Hugh). The first one, which I will denote Sw is the set of women (ladies first guys !). The second is the men's one : Sm.
Now, you want to pair women from Sw with men from Sm. The cardinal number of each set is not guaranteed to be the same. I will denote them i = |Si|, where i ∈ {w, m}.
Suppose we have w > m (heh heh), then all you have to do is to choose a subset Sw,m of m women amongst the w that are in Sw (there are w! / ((w - m)!.m!) such subsets) and then set up a bijection between Sm and Sw,m. There are m! such bijections. Then you extend the bijection by setting the other women to the single status (which is -1 if I got it right).
From now on, you have two possibilities :
Generate one such "extended bijection" p and bind man i with women p(i) (the leftover poor little women to status -1).
You are given the two sets partially "paired" and you want to pair remaining singles as much as possible.
Before I go further, can you tell me which approach is the one that interests you ?