In Facebook, and in Xiaonei. People sometimes do a survey game(Chinese is a popular点名游戏) something like this

Rules: Once you’ve been tagged, you are supposed to write a note with answering the following 10 questions. Change one of those question into a question you want to ask, and tag 10 other people, so they will do a survey with your question.

1. Who is your favorite person
....

Why 10? why only change 1 question, why tag 10 other people? Can these number be something else that's BETTER?

First we have to think of a criteria for a good game
1. The person who just answered a survey, might get tagged by someone else, and it would suck if most question are the same.
2. Each person wants to maximize the amount of people answer his own questions.
Suppose the probability of not having to do the same question is P, and the expected amount of people to answer someone's question is R.
The larger the RP the better.

Suppose there are n survey questions each survey. Each time one answer the survey can remove m problems and add m of their own. The removal of problem is completely random. and no one will ever propose the same problem as someone else. Then each person have to name k people randomly from their friends to do their survey.
Now, if there are y people, each have f friends(a is b's friend means b is a's friend). What n,k,m will maximize RP?

This helps to form a more scientific survey game.

We can add further restrictions later.

It sucks as for a y, some f is not possible.
Hell. I'm still stuck in the first part.
How many distinct graphs can have y vertexes and f edges each. and what is probability to come up with a specific graph?

Fun. going to think more about it.
and I lied about not posting before leave China.

There is a game called Numbrosia. The game is a 5*5 board with each block have a integer value. There are few operations, add 1 to every single block in a row or column. Shift every block in a row left or right. Shift every block in a column up or down. One want to minimize the amount of operation to get all blocks to 0.
We only consider Numbrosia with positive numbers. (if there are negative integers, just add the same amount of number on each row until all numbers are positive)

Now, create a generalized Numbrosia. Where it is n*n board instead of 5*5.
First question. Is all configuration solvable.
Answer. No. When the sum of all numbers is not divisible by n. It is not solvable.
If, the sum is divisible by n, is it solvable. Yeah. There is a simple yet inefficient algorithm to make sure you go solve it. Taking way more than the amount of optimal steps.
(a,b) = the position, = value at that position.
First, let's suppose it's we have a positive integers in each block. Find the lowest positive number, reduce it with the same row until it's 0. Do the same thing to other numbers, except if there is 0 on the same row, move it away until all numbers on the same row are positive. Do it until there is only one row left. Just to make everything easier, suppose that row is to and when i<=j. It's ok to assume since it's possible to alter the blocks into any arrangement if it's not in our desired formation.

Now, there is a simple algorithm to solve the game in this condition.
1. Do addition on row 2 so all number on row 2 = 1.
2. Move 1s from row 2 to 0s in row 1. If there is not enough 1s left, after all 1s are moved to row 1, go to step 1.
3. subtract row 1 again. Now there are 0s in row 1. If there is only 0s in row 1, end.
4. go to step 2.

So clearly, it is possible to solve a n*n Numbrosia when the sum of the numbers is divisible by n.

I assume the minimal number of step to solve it is 2an, where a is the largest number in the game.

It was my cosine's birthday.
Total of 5 people. Me, my Cosine, Mai(麦倩乔), Gu Yang(or just Yang) and Fang.
We went to Chinese pizza hat.
Where there is a photo of Yang with lamest pose thing I think of.

Waiting for seats

It is FUCKING EXPENSIVE.
So I did some calculations on the napkin on how to spend the money to maximize the amount of food we get.

Except I suck at calculations and my cosine already brought the pizza.

While waiting, Mai smile at the camera. Actually, she smile at the camera the exact same way in all photos. Creepy.

Cute shrimp.

The Pizza.

Where I realized the pizza is actually... square. I shall redo all my calculations.
Or I would just...FEAST

Yeah. Coffee

Then we go and play games. Where in this game. My face gets shown on the screen and my cosine punches a punching bag and the impact will reflect on my face on the screen.

We depart.
At night. I played some very excessively exhausting sport(badminton) and understand how weak I am physically.
Next day. I left Guangzhou.
Back to math.

So today I have been inspired by Kay's status message's commenter... and made(stole) some lame puns.

What do you call many people kill themselves due to math?
A math suicide!

Where is MIT?
In Mathachusetts.

What do you call a huge amount of calculus homeworks?
A mathive amount of homework.

I play MMOs, you know mathively multiplayer online games

I think I need a full body mathage after posting these.

For AIME, there are always questions ask m+n or mn where m and n and relatively prime integers. It's easy to infer the only way to m or n = 0 is the other one is 1. The answer of AIME are always integers from 0 to 999.
Suppose m and n are distributed evenly as long as m+n<1000 and mn<1000. How is the distribution of m+n and mn? Because if m and n can't be found easily, an educated guess can increase the chance.

Case 1: For the m+n case, the amount of (m,n) pairs for x are
Case 2: For the mn case, the amount of (m,n) pairs for x are
Where is the Euler's totient function, is the number of unique prime factors for x, I can't find any named function with that definition.

So for m+n one, it's better to chose a larger number, even a prime number.
For mn case, it's better to chose numbers with a lot of unique prime factors.

Case 3: It happens sometimes when m and n doesn't have any other requirement except m+n<1000. Then we have a linear relationship. The amount of times m+n = x is x-1.

Case 4: Sometimes there are Squarefree requirement for n, then it is still approximately linear.

A nice plot. Green = Case 4, Brown = Case 3, Purple = Case 2, Blue = Case 1 * 50
Then one just have to have a random number generator, and then convert it into a pick in the plot.