Stable marriage problem Puns

I have studied the Gale-Shapley algorithm. I am a new guy to the field but what I get basically is that the algorithm provides a stable matching for a set of n boys and n girls. I have seen this question in mit ocw

Prove or disprove the following claim: for some n ≥ 3 (n boys and n girls, for a total of 2n people), there exists a set of boys’ and girls’ preferences such that every dating arrangement is stable.

The solutions tell me that the claim in the question is false but the Gale-Shapley algorithm must provide a solution right? Where am I getting the thing wrong?

m1: w4, w1, w3, w2

m2: w3, w1, w4, w2

m3: w1, w2, w3, w4

m4: w2, w3, w4, w1

w1: m2, m3, m1, m4

w2: m4, m1, m3, m2

w3: m1, m2, m3, m4

w4: m3, m1, m2, m4

Find a stable matching using the Gale-Shapley algorithm with MEN proposing?

Each woman would be receiving one proposal so there is only one round because there will be no rejections and those are stable matches?

i.e. m1/w4, m2/w3, m3/w1, m4/w2

Same for the Women proposing? Or am I missing something?

Both Akshay and Ajay were big time casanovas in 90s. Even after getting married, news about their extramarital affairs with different actresses were always in headlines.

Yet somehow they have managed to be in a stable marriage with kids. They aren't married to complete nobodies either. Twinkle was daughter of the Superstar Rajesh Khanna. Kajol has her own identity. She was a Superstar herself in 90s.

Yet they never thought of leaving their cheating husbands. On the other hand we have Aamir and Hrithik who are divorcees now. They were also into extra-marital affairs but their reputation was not as bad as Akshay and Ajay.

What would be the reason Akshay and Ajay managed to have stable marriages while Aamir and Hrithik could not?

Edit: Man that grammatically wrong title is going to invite some jokes.

This stuff was figured out over 50 years ago...

I want a clean and fair debate in the comments. Why is it that colleges have not switched over to a system like this where all students are the "men" and all colleges are the "women"? What are the disadvantages of this system on both sides?

https://en.wikipedia.org/wiki/Stable_marriage_problem

These past two years have been hard, even before the pandemic. I started intensive treatment for my MDD and CPTSD in January 2020, and I’ve been striving to get my mind well enough to perform my duties in my marriage while my husband works hard. I’ve learned a lot about myself and have accepted that I can’t work and need to remain a stay at home wife. I’ve finally managed to keep my house clean and consistently cook every day for long periods, so that way my husband can keep getting promoted and we can earn more income. My medication has stabilized my depression and has allowed me to be able to think and feel like a normal adult again.

However, I still have a crippling coping mechanism hurting me. I binge eat and can’t stop. I spent my entire childhood underweight and not eating for days at a time, and now I’m obese with no appetite control. I’m scared to go outside, and exercising has me worried I won’t get something done, even though I have plenty of time. I’m scared any change to my routine will rock the boat and destroy my husband’s last chance of going Army Ranger, after his last two attempts were thwarted by me having an acute mental health hospitalization. He’s getting so close, and I’m scared to do anything for fear something bad will happen.

If he ends up not fulfilling his dream and we end up not being able to stay in the army, I will likely spiral and need to be hospitalized again…..we wouldn’t be able to afford it without Tricare. But he is also really worried about my health and the ticking clock as I age and my family history of heart issues.

After an extensive literature survey (i.e. Googling for fifteen minutes), I wasn't able to find a solution to the problem I am about to propose. The closest I was able to find was the Stable Marriage/Roommate Problem but I am not sure if the solution can be applied to this problem.

Introduction:

In the Pokémon video game, there is a mechanic where Pokémon can breed to make eggs that hatch into new Pokémon. Much like how selective breeding works on our world, Pokémon are bred to pass on certain desirable qualities to an offspring. However, unlike real animals, different species of Pokémon can breed with each other as long as they belong to the same "Egg Group." A Pokémon that belong to two Egg Groups are particularly useful because they can transfer attributes from one Egg Group to another. There are 13 (12 after modification) Egg Groups that are connected and are depicted by this chart. (An observant Pokémon player might notice a few differences from this original post. Please see my post reply describing modifications and the reasons for making them.)

Question:

• How many ways can the egg groups be paired with the available connections? And what are the egg group pairs?

• Is there a more generalized solution?

• Can the chart be redrawn with fewer line intersections?

I was able to find one solution randomly making pairs but I was wondering if there were more solutions. Thank you!

Hi everyone, thank you for the support. I'm taking this post down because it's starting to grow really fast and I don't want it all over the internet. I'll take the advice to talk it out really well and continue my tendency to respond with a slap on the butt, pull of the nose or ear, or similar act. Thank you all for taking the time to read, and rest assured I will certainly seek out help if things escalate. I fortunately have a lot of people I can turn to.

If you are very curious, DM me and I will send the original text.

Idk if you guys are interested in such things, but in greek mythology Hades and Persephone are actually in love with each other, none of theme ever cheated, they live a happy life in the underworld and on top of that, hades is the god of wealth. Also they never had kids (presumably hades is sterile).coincidence? I think not!

One of my assignments is the stable marriage problem. I've written out all the checks I think i need to the best of my knowledge. I'm just not sure of what I'm doing. I'm sure I messed up at the looping part. I tried using goto statements, which are a mess. I don't think I'm properly breaking out of the loops.

I used this video: https://www.youtube.com/watch?v=Qcv1IqHWAzg as a frame of reference for my logical structure.

It's should be easier for me as I don't have to worry about inputs. I am given data to work.

My main steps are:

-Start with a man, make sure he is single

-Check through to the first non-negative preference(I used negative 1 to indicate that he has been rejected by a woman)

-Check if that woman is single, if she is great, reset

-If she isn't. check if she prefers this man over her current partner, if she does, great! The man who lost his partner should be marked rejected by that partner.

if not, go back and ask the next woman. (I'm not sure if I should mark this as rejection though, my main criteria for rejection is if a woman has left a man she partnered up with previously. should I actually be checking the other men first?)

-keep cycling through all the men until no one is single.

This is a system where the man proposes.

Here's a link to my main.cpp so that you could take a look: https://drive.google.com/open?id=1_SKoXhf2u-iczvhKiquKlptIv87kkzz-

I removed all the goto start functions on that cpp because I'm just not sure where they should be anymore.

I'm trying to create a simple matching algorithm within Excel. Ideally it'd be as complex as the stable marriage, but along only two variables and at most 50 total pairs. Here are the two groups I want to pair and the variables to pair along.

Intern:

-Team preference (i.e. Pref 1 = X team, Pref 2 = Y team, Pref 3 = Z team)

-Skillsets: this would be a list of 5-6 skills with a self identified competence scale (no experience, moderate, expert)

Manager:

-Team that the manager sits on (the intern list will match the available teams from every available manager)

-Required and helpful skills (same skills list for the interns)

Honestly, anything that will help automate this very manual matching process would be helpful and appreciated. It need not be a Stable Marriage. Thanks in advance!

I have written this java code to the Stable marriage problem. Is my code correct ? In my solution there are a number of persons that choose some choices. Unlike the Stable marriage problem choices don't prioritize the persons.

I am by no means a math wiz and even less familiar with some of the algorithms out there. My problem is a resource allocation problem.

I have 2 subsets: lets call them boys and girls. The boys are trying to match with the girls and vice versa. The caveats to the standard stable matching problem are as follows.

1. The boys can choose up to 2 girls (and can be split up whatever percentage worked. E.g. 30% of the boy to one girl, 70% to another)

2. The girls can match with any number of boys (and partial boys as described above), but cannot have more than three distinct heights of boys. They also have a max weight that all the boys (and partial boys) they match with to get close to without going over.

I'm struggling to find an algorithm that factors in the ability for more than one match both ways between boys and girls and allows for caveats like described above. Not looking for a silver bullet, just something to get me closer to a solution. Any help would be greatly appreciated.

I (28F) dont come from a very stable life/childhood. I moved homes, schools multiple times. Moved countries 3 times. My parents are divorced, quite a lot of death in the family, some drug problems. Basically my life was always a bit chaotic until I met my current partner that I am due to marry in June next year.

My partner(28M) comes from a very stable family with no divorce history in his whole immediate and extended family, they are all very close, amazing, happy people. My life is very stable now, I am very in love with my partner, he is in live with me and treats me amazing. We got a dog, building a house, planning childreb I got a promotion at work. Yet I am soo scared of this, I'm not used it and been getting anxious recently as if am waiting for something bad to happen. I think it is important to mention I do haveI an anxiety disorder.

I really feel like I am scared of stability it is just such a foreign feeling. It is very weird, I am scared of something that most people strive for.

This is a classic problem in computer science:

In a town there are n boys and n girls who will be paired up for marriage. Each boy has a list where he ranks the girls in order of preference, and each girl has a similar list of boys.

Once they become of age, the boys begin to court the girls using the traditional dating algorithm. Each boy approaches his favorite girl and asks for her hand in marriage. If a girl only has one suitor, she marries him. Otherwise, she chooses the guy she likes best among her suitors and marries him.

Once a boy has been rejected, he will propose to his second favorite girl, regardless of whether or not she is already married. If she likes him better than her current husband, she will elope with the suitor, and her ex-husband will be thrown back into the dating pool. Otherwise, she will remain faithful and reject the proposal. A rejected boy will then propose marriage to his next favorite girl, and so on until he finally finds a wife.

Once everyone is married, the traditional dating algorithm terminates.

1. Define a stable marriage to be a marriage where neither partner can find anyone else of the opposite gender in town whom they prefer to their current spouse and who would also accept their marriage proposal. In other words, if a marriage between a boy and a girl is stable, then for every other girl in town, either the boy does not prefer the other girl to his wife or the other girl does not prefer the boy to her current husband.

Show that the traditional dating algorithm places everybody in stable marriages.

2. For each boy, define their optimal girl to be the highest-ranked girl on their list that it is possible to be in a stable marriage with. Similarly, define their pessimal girl to be the lowest-ranked girl that he could end up in a stable marriage with. Optimal and pessimal boys are defined similarly.

Show that the stable algorithm produces pairings where every boy ends up with his optimal girl and each girl ends up with her pessimal boy.

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I wrote a stable marriage demonstrator program for the purpose of helping student in my college:

I have tested the program myself and am going to continually testing it, however I would like some opinion on the code, I tried to make the naming of the function user friendly but sometimes the coder himself forgets how unintuitive is it for others to read the code.

Of course let me know if there are any bugs.

Thanks

https://gist.github.com/anonymous/a2b6844138035be38f9e

Here is the algorithm:

Every Morning:

Each man goes to the first woman on his list not yet crossed off and proposes to her.

Every Afternoon:

Each woman says “maybe, come back tomorrow” to the man she likes best among the proposals (she now has him on a string) and “never” to all the rest.

Every Evening:

Each rejected suitor crosses off the woman who rejected him from his list.

The above loop is repeated each successive day until there are no more rejected suitors. On this day, each woman marries the man she has on a string.

Here is the datastructure representation:

Male preference = {male_id: [[preference_list], [current_preference_list], [woman_on_string]]}

Female preference = {female_id : [[preference_list], [man_on_string]]}

The preference list is sorted highest preference from right to left

[low,.....,high]

m1: w4, w1, w3, w2

m2: w3, w1, w4, w2

m3: w1, w2, w3, w4

m4: w2, w3, w4, w1

w1: m2, m3, m1, m4

w2: m4, m1, m3, m2

w3: m1, m2, m3, m4

w4: m3, m1, m2, m4

Find a stable matching using the Gale-Shapley algorithm with MEN proposing?

Each woman would be receiving one proposal so there is only one round because there will be no rejections and those are stable matches?

i.e. m1/w4, m2/w3, m3/w1, m4/w2

Same for the Women proposing? Or am I missing something?