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Graph Theory: The Stable Marriage Problem
After writing about DP for a while, I came back to graph theory. Today we will look at a simple but interesting problem. Stable marriage problem is a type of bipartite matching problem, but it does not need to know any other algorithm to learn it.
Suppose there are n boys and n girls. Now they have to get married in such a way that the marriage is “stable”. It is possible to marry everyone with everyone, but every boy and girl has some preferences, everyone wants to marry the person of their choice. If the names of 3 boys are Tom,Bob,Peter and 3 girls are named Alice,Mary,Lucy then the list of choices for boys
The list is in order of most to least preferred. As Tom likes Alice more, likes Lucy less.
Again the list of girls’ preferences
Now how to get married will be stable marriage? First we need to understand what is meant by stable. Suppose a marriage is given as follows:
This matching/beta is not stable, because Tom prefers Mary to Lucy, and Mary prefers Tom to Peter. So Tom and Mary can break up and move in together. If A,B are boys and C,D are girls and A-C and B-D are married then the marriage will not be stable if both of the following statements are true:
1. If A prefers D to C.
2. If D prefers A to B.
If 2 statements are true, A and D will break up the marriage! But if any of the statements are false, the marriage will be stable.
In 1962, David Gale and Lloyd Shapley proved that there is a solution to the problem of simultaneous stable marriages for equal numbers of boys and girls. They invented a very simple algorithm to solve the problem. The algorithm is like this:
1. First every single guy will propose to his most favorite girl whom he hasn’t proposed yet, no problem if the girl is already engaged, more than one guy can propose to a girl. A boy will never propose to a girl twice.
2. Now each girl will choose the one she likes most from those who have sent her proposals, discarding all others. If the girl likes someone before, she will also cancel him.
3. If no one is still single, step 1 will be repeated.
Why does the algorithm work? Suppose A-C and B-D are married. Then the marriage will break if A prefers D and if D prefers A more. But in the above algorithm it is possible. Reason:
If A likes D more then he will send proposal to D first, if D refuses or quits then only send proposal to C.
If D likes A more then he will leave anyone else and marry A.
And if D does not marry A and marries someone else, it means that he prefers someone else more, in this case there is no possibility of breaking the marriage.
This algorithm will give stable matching but may not give optimal results. The results will be optimal for each boy, but may not be optimal for girls, i.e. there may be stable matching such that one of the girls could have married a more desirable guy. That is, the result will be optimal for the proposal that sends.
The algorithm is very easy to implement in code. The preference list will be input to you. You can easily write the code by keeping this information in an array, who has sent a proposal to whom, and who is currently engaged with whom. The sudocode given on wikipedia is as follows:
function stableMatching {
Initialize all m ∈ M and w ∈ W to free
while ∃ free man m who still has a woman w to propose to {
w = m’s highest ranked such woman to whom he has not yet proposed
if w is free
(m, w) become engaged
else some pair (m’, w) already exists
if w prefers m to m’
(m, w) become engaged
m’ becomes free
else
(m’, w) remain engaged
}
}
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Coin Change + Rock Climbing
If you have read the previous chapters, you have now learned the basics of dynamic programming, and the more you solve problems, the better your skills will be. Since there is no specific algorithm in DP, we have to think a lot, each DP problem is different, but if you know the solutions of some classic DP problems like knapsack, coin change, then you will understand how to move your thinking forward, how to determine the state of DP. Should be done, then you will be able to solve new type of DP problem. I have already discussed the nCr determination and the DP solution of the 0-1 knapsack, and will discuss some more classic or standard problems later.
Coin Change:
Now we will look at the coin change problem. In fact, after learning knapsack, you can change coins, but I am writing to make things clear.
You have some coins worth 5,8,11,15,18
dollar Each coin has an infinite number of times, you can take any coin as many times as you want. So your coin array could look like this:
coin[] = {5, 8, 11, 15, 18};
Now you have to create some value with these coins. Let the number be make
. If make=18 we can make 5+5+8 thus 18. You have to say whether the value can be created with the coins or not.
First we think about what will be the state in DP. We will continue to try to make the number one coin at a time. Then you have to keep the state of which coins you are taking at the moment, and the total value of the coins you have taken before. If the function name is call , the prototype will be:
int call(int i,int amount)
After that, it’s the same as before. First I will try to get coin number i:
if(amount+coin[i]<=make)
ret1=call(i,amount+coin[i]);
else
ret1=0;
Here i am calling i instead of calling i+1 because one coin can be taken many times. If a coin could not be taken multiple times, I would call i+1. If amount+coin[i] is greater than make then the coin is likely to be taken. If the coin doesn’t exist then we move to the next coin:
ret2=call(i+1,amount);
Make can be created even if one of ret1 and ret2 is true. So finally I will write:
return ret1|ret2;
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