Game Theory [letter] entry notes _SG competing game theory Nash equilibrium function _

background

It has long been hovering in the short board to make up for the polynomial / write string cool title / yard cancer data structure / look at the math book this four strange things, so the level is food.

Previous exposure to a little simple game theory, but that it is too simple, is against search (Min-Max).

Recent training, so that the title game on a SAMParent-Tree after doubling, SAM code all over, a look at that game, more and more panic, they simply will not find: far more than not, they fail to get a point.

He began catching game theory, find it wonderful, but very good understanding.

This article because the author was working captured, there is no complete contents of the Nash Equilibrium.

update: study found that road led me to the subject of game theory is false title track, I can see the concrete content of the anti-NIM in

some documents

oi-wiki basic game theory can be a beginning

SG function entry and examples in this article conscience

Getting Started with anti-NIM question is very short, anti-NIM recommends that after they get to know the SG as a better understanding of the application

Nash equilibrium is an example of initial presentation Nash equilibrium in OI in usage

Nash Equilibrium / scribing method this article I get out, just to illustrate streak method (beep -) very, very

Nash Equilibrium / reliable calculation personally think is more reliable Nash equilibrium calculation method (crazy diss streak method)

Nash Equilibrium / mixed strategies (FBI Warning) This stuff is too immortal, individual plans if there is time to find a book on game theory to study this piece of content, but the reason OI which probably will not happen this kind of thing (do a Nash equilibrium out pretty good).

Nash Equilibrium / vivid example (fog This is up Lord my grandfather's grandfather concern [doge] manually funny

I believe we saw it, I do not want to write Nash equilibrium, so a lot of data collection.

Equality combination game

Some definitions, conventions and basic laws

• Equality combination play (loose)
  • Two people involved, and a win and one loss
  • Each person may choose to face all situations are the same
  • Game always ends that can be abstracted into the status of all points, select the abstract to the edge, the game will become \ (DAG \)
  • Generally not lose operator (an operator can not win problem is anti-NIM)
• N-position&P-position
  • For a state, if the current state of the person can get win, it is called N-position (Now)
  • For a state, if a person is transferred from the previous state to win, it is called P-position (Pre)
• If a state is the N-position, then he will be transferred to a P-position
• If a state is a P-position, the transfer of all of his points are certain N-position

Min-Max search

Role and Profile

Equality is a combination search method most violent game, you can use this method results in almost all of these issuesExponential violencealgorithm

Very basic, so it is nothing to talk about.

The general process

First abstract state, and then write the given initial state violence from the title, with each state to determine basic facts and boundary conditions is N or P.

Therefore, memory search is commonly used.

Such a problem can be combined to understand this process:

• Alice and Bob are doing the game, this game is a (n \ n \ times) \ carried out on the board, the board in each grid has a certain gold coins, when Alice and Bob alternate handling an initial in the upper left corner of the board pieces, per person moving a grid, the grid and the target collect coins (repeated after repeated collection) after one round Alice Bob mobile moves, a total movement \ (m \) wheel.
• Because Alice is mhw, so all she owned all the gold coins, and Bob will be nothing. Naturally, Alice want to maximize the number of coins collected, while Bob is desirable to minimize this value.
• Several rounds gives the board size, the number of coins for each of the grid, made, seeking the ultimate Alice how much gold can get.

(Original title, but perhaps more classic, there will be a similar topic in any place)

NIM problem

Brief introduction

This is almost the classic game theory problem inside. SG function as well as a basic understanding of the problem of anti-NIM after its text should be used as the basis.

description

Now you have \ (n \) heap of stones, the \ (i \) heap of stones there \ (a_i \) months. Two people take turns for the game, each can take any number of stones from any pile in, but can not take people's actions can not lose. He asked the upper hand to win or FLAC.

in conclusion

The upper hand to win, if and only if \ (A_1 \) ^ \ (A_2 \) ^ \ (A_3 \) ^ ... ^ \ (A_N \ neq0 \)

In other words, a state of the exclusive OR is not 0 and N-position, and a state of the exclusive OR is 0 to P-position

prove

• The final winner of the last step, XOR and reflect on, is to become a non-0 from 0.

• Subsequently, as long as there is a method to prove that the non-exclusive or 0 and iso after subtracting a number or 0 and can become.

• And current is provided to the exclusive-OR \ (K \) , its most significant bit are binary \ (D \) position, it certainly has \ (a_i \) of \ (D \) bit is one. We call this \ (a_i \) becomes \ (^ {a_i \} = a_i Prime \) ^ \ (K \)
• Considering the \ (K \) and \ (a_i \) at the \ (D \) the bits are 1, so \ (a_i ^ {\ prime} \) at the \ (D \) on the bit is 0, so \ (^ {a_i \} Prime <a_i \) , the movement of legitimate.

• While the other, no matter how moving after got a XOR and zero status, or move completely different and certainly not 0

We give a constructive proof.

SG function

Definitions and conventions

• We assume that a game model: a \ (DAG \) is selected on the \ (n \) points \ (A_1, A_2, A_3, ..., A_N \) , each operator can select a point, along this (DAG \) \ taking a side step, people can not go lose, or win the upper hand and asked flip wins.
• \ (MEX (S) = \ min \ {n-limits_ \ in N, n-\ {n-notin S}} \) , is not in the set \ (S \) the smallest non-negative integers.
• \ (SG (X) = MEX \ left (\ left \ {SG (Y) | X \ rightarrow Y \ right \} \ right) \) , where \ (x \ rightarrow y \) This expression may be irregular, it is status \ (X \) can be transferred to the state \ (Y \) means.
• We call a game ({, ..., a_n \ a_1 , a_2} A = \ \) \ this initial state \ (SG (A) \) of \ (sg (a_i) \) of the exclusive OR and .

theorem

A state \ (A \) is the N-position (upper hand wins), if and only if \ (SG (A) \ neq0 \)

prove

We found that this thing is really like and NIM problem.

• Consider \ (DAG \) is not a point on the edge, it \ (sg \) certain to 0, and a person loses, that he got a state where all the points are not out of the side. At this time, therefore this state \ (SG = 0 \) , a P-position.
• Consider a point \ (A \) , if the \ (SG (A) = K \) , then he can be transferred to state (b), there must be \ (sg (b) = 0 , sg (b) = 1 ,. ..sg (b) = k-1 \) , which is much like the problem with the NIM, because NIM, we can also put a bunch of \ (k \) a pebble becomes \ (0,1,2, ..., k-1 \) any number of one.
• So, when we encounter a \ (SG (A) \ neq0 \) state, we can \ (sg \) the size of this value is seen as a bunch of a few stones, to borrow above problems NIM method, the state \ (a_i | highbit (SG (a)) = highbit (SG (a)) \) , becomes a \ (SG \) is \ (SG (a) \) ^ \ (SG (a) \ ) state \ (a_i ^ {\ prime} \) can.
• Similarly, a \ (SG (A) = 0 \) state, a one-step change, \ (SG (A ^ {\ Prime}) \) is certainly not equal to zero.

anti-NIM

I discussion below is immature, he takes issue only for NIM stones effectively, or to prove that any (sg = 0 \) \ state or not a side, or even have to \ (sg \) value of 1 the edge points.

In fact, that prompted me to study game theory topics I mentioned in the background because this thing is fake.

Proof of the problem in writing this article is found to be perfect after update

description

The same model that question the other conditions stipulated in the SG function, but we can not take a step, said people are the winners.

theorem

For a state \ (A \) :

• If \ (\ max a_i \ Leq. 1 \) , which is the N-position if and only if \ (SG (A) = 0 \)
• If \ (\ max a_i>. 1 \) , it is when the N-position and only if \ (SG (A) \ neq0 \)

prove

• \(\max a_i>1\)。
  • If \ (SG (A) \ neq0 \)
    • When there are multiple (a_i> 1 \) \ when the \ (SG \) function in the proof of the same, we are always able to \ (SG (A ') \ ) becomes 0
    • When only one \ (a_i> 1 \) , we found that the modified result \ (\ max a_i \ leq1 \) , so that we should \ (SG (A ') \ ) is not zero. Suppose we follow the original method, to obtain \ (a_i \) strain \ (a_i '\) , then we should actually becomes the \ (a_i' '= a_i' \) ^ \ (. 1 \) . At this time, easy to know \ (a_i '' \ in \ {0,1 \} \) , so we successfully transferred to a P-position, and \ (a_i \ rightarrow a_i '' \) is legitimate transfer
  • If \ (SG (A) = 0 \)
    • At this point there must be more than 1 \ (a_i> 1 \) , therefore \ (SG \) function where the discussion is still applicable.
• \(\max a_i\leq1\)
  • If \ (SG (A) = 0 \)
    • Over, proved wrong, this fear is not a false conclusion

Complete information static game

Nash Equilibrium

definition

Calculation