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Research ArticleGames Based Study of Nonblind Confrontation
Yixian Yang,1,2,3 Xinxin Niu,1,2,3 and Haipeng Peng2,3
1Guizhou Provincial Key Laboratory of Public Big Data, Guizhou University, Guiyang 550025, China2Information Security Center, State Key Laboratory of Networking and Switching Technology,Beijing University of Posts and Telecommunications, Beijing 100876, China3National Engineering Laboratory for Disaster Backup and Recovery, Beijing University of Posts and Telecommunications,Beijing 100876, China
Correspondence should be addressed to Haipeng Peng; [email protected]
Received 4 January 2017; Accepted 20 March 2017; Published 19 April 2017
Academic Editor: Liu Yuhong
Copyright ยฉ 2017 Yixian Yang et al. This is an open access article distributed under the Creative Commons Attribution License,which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Security confrontation is the second cornerstone of the GeneralTheory of Security. And it can be divided into two categories: blindconfrontation and nonblind confrontation between attackers and defenders. In this paper, we study the nonblind confrontation bysome well-known games. We show the probability of winning and losing between the attackers and defenders from the perspectiveof channel capacity. We establish channel models and find that the attacker or the defender wining one time is equivalent to one bittransmitted successfully in the channel. This paper also gives unified solutions for all the nonblind confrontations.
1. Introduction
The core of all security issues represented by cyberspacesecurity [1], economic security, and territorial security isconfrontation. Network confrontation [2], especially in bigdata era [3], has been widely studied in the field of cyberspacesecurity. There are two strategies in network confrontation:blind confrontation and nonblind confrontation. The so-called โblind confrontationโ is the confrontation in whichboth the attacker and defender are only aware of their self-assessment results and know nothing about the enemyโsassessment results after each round of confrontation. Thesuperpower rivalry, battlefield fight, network attack anddefense, espionage war, and other brutal confrontations,usually belong to the blind confrontation. The so-calledโnonblind confrontationโ is the confrontation in which boththe attacker and defender know the consistent result aftereach round.The games studied in this paper are all belongingto the nonblind confrontation.
โSecurity meridianโ is the first cornerstone of the GeneralTheory of Security which has been well established [4, 5].Security confrontation is the second cornerstone of theGeneral Theory of Security, where we have studied the blindconfrontation and gave the precise limitation of hacker attack
ability (honker defense ability) [4, 5]. Comparing with theblind confrontation, the winning or losing rules of nonblindconfrontation are more complex and not easy to study. Inthis paper, based on the Shannon InformationTheory [6], westudy several well-known games of the nonblind confronta-tion: โrock-paper-scissorsโ [7], โcoin tossingโ [8], โpalm orback,โ โdraw boxing,โ and โfinger guessingโ [9], from anovel point of view.The famous game, โrock-paper-scissors,โhas been played for thousands of years. However, there arefew related analyses on it. The interdisciplinary team ofZhejiang University, Chinese Academy of Sciences, and otherinstitutions, in cooperation with more than three hundredvolunteers, spent four years playing โrock-paper-scissorsโand giving corresponding analysis of game. And the findingswere awarded as โBest of 2014: MIT technology review.โ
We obtain some significant results. The contributions ofthis paper are as follows:
(i) Channel models of all the above three games areestablished.
(ii) The conclusion that the attacker or the defenderwining one time is equivalent to one bit transmittedsuccessfully in the channel is found.
HindawiMathematical Problems in EngineeringVolume 2017, Article ID 8679079, 11 pageshttps://doi.org/10.1155/2017/8679079
2 Mathematical Problems in Engineering
(iii) Unified solutions for all the nonblind confrontationsare given.
The rest of the paper is organized as follows. The modelof rock-paper-scissors is introduced in Section 2, models ofcoin tossing and palm or back are introduced in Section 3,models of finger guessing and drawing boxing are introducedin Section 4, unified model of linear separable nonblind con-frontation is introduced in Section 5, and Section 6 concludesthis paper.
2. Model of Rock-Paper-Scissors
2.1. Channel Modeling. Suppose ๐ด and ๐ต play โrock-paper-scissors,โ whose states can be, respectively, represented byrandom variables๐ and ๐:๐ = 0, ๐ = 1, and ๐ = 2 denote the โscissors,โ โrock,โand โpaperโ of ๐ด, respectively;๐ = 0, ๐ = 1, and ๐ = 2 denote the โscissors,โ โrock,โand โpaperโ of ๐ต, respectively.
Law of Large Numbers indicates that the limit of thefrequency tends to probability; thus the choice habits of ๐ดand ๐ต can be represented as the probability distribution ofrandom variables๐ and ๐:
Pr(๐ = 0) = ๐means the probability of ๐ด for โscissorsโ;Pr(๐ = 1) = ๐means the probability of ๐ด for โrockโ;Pr(๐ = 2) = 1 โ ๐ โ ๐ means the probability of ๐ด for
โpaperโ, where 0 < ๐, ๐ and ๐ + ๐ < 1;Pr(๐ = 0) = ๐means the probability of ๐ต for โscissorsโ;Pr(๐ = 1) = ๐ means the probability of ๐ต for โrockโ;Pr(๐ = 2) = 1 โ ๐ โ ๐ means the probability of ๐ต for
โpaper,โ where 0 < ๐, ๐ and ๐ + ๐ < 1.Similarly, the joint probability distribution of two-
dimensional random variables (๐, ๐) can be listed as follows:Pr(๐ = 0, ๐ = 0) = ๐ means the probability of ๐ด for
โscissorsโ and ๐ต for โscissorsโ;Pr(๐ = 0, ๐ = 1) = ๐ means the probability of ๐ด for
โscissorsโ and ๐ต for โrockโ;Pr(๐ = 0, ๐ = 2) = ๐ โ ๐ โ ๐means the probability of ๐ด
for โscissorsโ and ๐ต for โpaper,โ where 0 < ๐, ๐, and ๐+๐ < ๐;Pr(๐ = 1, ๐ = 0) = ๐ means the probability of ๐ด for
โrockโ and ๐ต for โscissorsโ;Pr(๐ = 1, ๐ = 1) = ๐ means the probability of ๐ด for
โrockโ and ๐ต for โrockโ;Pr(๐ = 1, ๐ = 2) = ๐ โ ๐ โ ๐means the probability of ๐ด
for โrockโ and ๐ต for โpaper,โ where 0 < ๐, ๐, and ๐ + ๐ < ๐;Pr(๐ = 2, ๐ = 0) = ๐ means the probability of ๐ด for
โpaperโ and ๐ต for โscissorsโ;Pr(๐ = 2, ๐ = 1) = โ means the probability of ๐ด for
โpaperโ and ๐ต for โrockโ;Pr(๐ = 2, ๐ = 2) = 1โ๐โ๐โ๐โโmeans the probability
of๐ด for โpaperโ and๐ต for โpaper,โ where 0 < ๐, ๐, and ๐+๐ <1 โ ๐ โ ๐.Construct another random variable ๐ = [2(1 + ๐ +๐)] mod 3 from๐ and ๐. Because any two random variables
can form a communication channel, we get a communicationchannel (๐; ๐)with๐ as the input and๐ as the output, whichis called โChannel ๐ด,โ which is shown in Figure 1.
Channel A
Channel B
X
Y
Z
Z
Figure 1: Block diagram of the channel model.
If ๐ด wins, then there are only three cases.
Case 1. โ๐ด chooses scissors, ๐ต chooses paperโ; namely, โ๐ =0, ๐ = 2.โ This is also equivalent to โ๐ = 0, ๐ = 0โ; namely,the input of โChannel ๐ดโ is equal to the output.Case 2. โ๐ด chooses stone, ๐ต chooses scissorsโ; namely, โ๐ =1, ๐ = 0.โ This is also equivalent to โ๐ = 1, ๐ = 1โ; namely,the input of โChannel ๐ดโ is equal to the output.Case 3. โ๐ด chooses cloth, ๐ต chooses stoneโ; namely, โ๐ = 2,๐ = 1.โ This is also equivalent to โ๐ = 2, ๐ = 2โ; namely, theinput of โChannel ๐ดโ is equal to the output.
In contrast, if โChannel ๐ดโ sends one bit from the senderto the receiver successfully, then there are only three possiblecases.
Case 1. The input and the output equal 0; namely, โ๐ = 0,๐ = 0.โ This is also equivalent to โ๐ = 0, ๐ = 2โ; namely, โ๐ดchooses scissors, ๐ต chooses paperโ; ๐ด wins.
Case 2. The input and the output equal 1; namely, โ๐ = 1,๐ = 1.โ This is also equivalent to โ๐ = 1, ๐ = 0โ; namely, โ๐ดchooses rock, ๐ต chooses scissorsโ; ๐ด wins.
Case 3. The input and the output equal 2; namely, โ๐ = 2,๐ = 2.โ This is also equivalent to โ๐ = 2, ๐ = 1โ; namely, โ๐ดchooses paper, ๐ต chooses rockโ; ๐ด wins.
Based on the above six cases, we get an important lemma.
Lemma 1. ๐ด wins once if and only if โChannel ๐ดโ sends onebit from the sender to the receiver successfully.
Now we can construct another channel (๐; ๐) by usingrandom variables ๐ and ๐ with ๐ as the input and ๐ as theoutput, which is called โChannel ๐ต.โ Then similarly, we canget the following lemma.
Lemma 2. ๐ต wins once if and only if โChannel ๐ตโ sends onebit from the sender to the receiver successfully.
Thus, the winning and losing problem of โrock-paper-scissorsโ played by ๐ด and ๐ต converts to the problem ofwhether the information bits can be transmitted successfullyby โChannel ๐ดโ and โChannel ๐ต.โ According to Shannonโssecond theorem [3], we know that channel capacity is equalto the maximal number of bits that the channel can transmit
Mathematical Problems in Engineering 3
successfully. Therefore, the problem is transformed into thechannel capacity problem. More accurately, we have thefollowing theorem.
Theorem 3 (โrock-paper-scissorsโ theorem). If one does notconsider the case that both ๐ด and ๐ต have the same state; then
(1) for ๐ด, there must be some skills (corresponding to theShannon coding) for any ๐/๐ โค ๐ถ, such that ๐ด wins ๐times in ๐๐ถ rounds of the game; if ๐ด wins ๐ข times in๐ rounds of the game, then ๐ข โค ๐๐ถ, where ๐ถ is thecapacity of โChannel ๐ดโ;
(2) for ๐ต, there must be some skills (corresponding to theShannon coding) for any ๐/๐ โค ๐ท, such that ๐ต wins ๐times in ๐๐ท rounds of the game; if ๐ต wins ๐ข times in๐ rounds of the game, then ๐ข โค ๐๐ท, where ๐ท is thecapacity of โChannel ๐ตโ;
(3) statistically, if ๐ถ < ๐ท, ๐ต will win; if ๐ถ > ๐ท, ๐ด will win;if ๐ถ = ๐ท, ๐ด and ๐ต are evenly matched.
Here, we calculate the channel capacity of โChannel ๐ดโand โChannel ๐ตโ as follows.
For channel (๐; ๐) of ๐ด: ๐ denotes its transition proba-bility matrix with 3 โ 3 order,๐ (0, 0) = Pr (๐ = 0 | ๐ = 0) = (๐ โ ๐ โ ๐)๐ ,๐ (0, 1) = Pr (๐ = 1 | ๐ = 0) = ๐๐ ,๐ (0, 2) = Pr (๐ = 2 | ๐ = 0) = ๐๐ ,๐ (1, 0) = Pr (๐ = 0 | ๐ = 1) = ๐๐ ,๐ (1, 1) = Pr (๐ = 1 | ๐ = 1) = ๐๐ ,๐ (1, 2) = Pr (๐ = 2 | ๐ = 1) = (๐ โ ๐ โ ๐)๐ ,๐ (2, 0) = Pr (๐ = 0 | ๐ = 2) = ๐(1 โ ๐ โ ๐) ,๐ (2, 1) = Pr (๐ = 1 | ๐ = 2) = (1 โ ๐ โ ๐ โ ๐ โ โ)(1 โ ๐ โ ๐) ,๐ (2, 2) = Pr (๐ = 2 | ๐ = 2) = โ(1 โ ๐ โ ๐) .
(1)
The channel transfer probability matrix is used to calculatethe channel capacity: solve the equations ๐๐ = ๐, where๐ isthe column vector:
๐ = (๐0, ๐1, ๐2)๐ ,๐ = ( 2โ
๐=0
๐ (0, ๐) log2๐ (0, ๐) , 2โ๐=0
๐ (1, ๐) log2๐ (1, ๐) ,2โ๐=0
๐ (2, ๐) log2๐ (2, ๐)) .(2)
Consider the transition probability matrix ๐.
(1) If ๐ is reversible, there is a unique solution; that is,๐ = ๐โ1๐; then ๐ถ = log2(โ2๐=0 2๐๐).According to the formula ๐๐ง(๐) = 2๐๐โ๐ถ, ๐๐ง(๐) =โ2๐=0 ๐๐ฅ(๐)๐(๐, ๐), ๐, ๐ = 0, 1, 2, where ๐๐ง(๐) is the probability
distribution of ๐.And the probability distribution of ๐ is obtained. If๐๐(๐) โฅ 0, ๐ = 0, 1, 2, the channel capacity can be confirmed
as ๐ถ.(2) If๐ is irreversible, the equation hasmultiple solutions.
Repeat the above steps; then we can get multiple ๐ถ and thecorresponding ๐๐(๐). If ๐๐(๐) does not satisfy ๐๐(๐) โฅ 0, ๐ =0, 1, 2, we delete the corresponding ๐ถ.
For channel (๐; ๐) of ๐ต:๐ denotes its transition probabil-ity matrix with 3 โ 3 order,
๐ (0, 0) = Pr (๐ = 0 | ๐ = 0) = ๐๐ ,๐ (0, 1) = Pr (๐ = 1 | ๐ = 0) = ๐๐ ,๐ (0, 2) = Pr (๐ = 2 | ๐ = 0) = (๐ โ ๐ โ ๐)๐ ,๐ (1, 0) = Pr (๐ = 0 | ๐ = 1) = ๐๐ ,๐ (1, 1) = Pr (๐ = 1 | ๐ = 1) = ๐๐ ,๐ (1, 2) = Pr (๐ = 2 | ๐ = 1) = (๐ โ ๐ โ ๐)๐ ,๐ (2, 0) = Pr (๐ = 0 | ๐ = 2) = (1 โ ๐ โ ๐)(1 โ ๐ โ ๐ ) ,๐ (2, 1) = Pr (๐ = 1 | ๐ = 2) = (1 โ ๐ โ โ)(1 โ ๐ โ ๐ ) ,๐ (2, 2) = Pr (๐ = 2 | ๐ = 2) = (1 โ ๐ โ ๐)(1 โ ๐ โ ๐ ) .
(3)
The channel transfer probability matrix ๐ is used tocalculate the channel capacity ๐ต.
Solution equation group ๐๐ค = ๐ข, where ๐ค, ๐ข are thecolumn vector:
๐ค = (๐ค0, ๐ค1, ๐ค2)๐ ,๐ข = ( 2โ
๐=0
๐ (0, ๐) log2๐ (0, ๐) , 2โ๐=0
๐ (1, ๐) log2๐ (1, ๐) ,2โ๐=0
๐ (2, ๐) log2๐ (2, ๐)) .(4)
Consider the transition probability matrix ๐.(1) If ๐ is reversible, there is a unique solution; that is,๐ค = ๐โ1๐ข; then๐ท = log2(โ2๐=0 2๐ค๐).According to the formula ๐๐ง(๐) = 2๐ค๐โ๐ท, ๐๐ง(๐) =โ2๐=0 ๐๐ฆ(๐)๐(๐, ๐), ๐, ๐ = 0, 1, 2.And the probability distribution of ๐ is obtained. If๐๐(๐) โฅ 0, ๐ = 0, 1, 2, the channel capacity can be confirmed
as๐ท.
4 Mathematical Problems in Engineering
(2) If๐ is irreversible, the equation hasmultiple solutions.Repeat the above steps, then we can get multiple ๐ท and thecorresponding ๐๐(๐). If ๐๐(๐) does not satisfy ๐๐(๐) โฅ 0, ๐ =0, 1, 2, we delete the corresponding๐ท.
In the above analysis, the problem of โrock-paper-scissorsโ game has been solved perfectly, but the correspond-ing analysis is complex. Here, we give a more abstract andsimple solution.
Law of Large Numbers indicates that the limit of thefrequency tends to probability; thus the choice habits of ๐ดand ๐ต can be represented as the probability distribution ofrandom variables๐ and ๐:0 < Pr (๐ = ๐ฅ) = ๐๐ฅ < 1,๐ฅ = 0, 1, 2, ๐0 + ๐1 + ๐2 = 1;0 < Pr (๐ = ๐ฆ) = ๐๐ฆ < 1,
๐ฆ = 0, 1, 2, ๐0 + ๐1 + ๐2 = 1;0 < Pr (๐ = ๐ฅ, ๐ = ๐ฆ) = ๐ก๐ฅ๐ฆ < 1,
๐ฅ, ๐ฆ = 0, 1, 2, โ0โค๐ฅ,๐ฆโค2
๐ก๐ฅ๐ฆ = 1;๐๐ฅ = โ0โค๐ฆโค2
๐ก๐ฅ๐ฆ, ๐ฅ = 0, 1, 2;๐๐ฆ = โ0โค๐ฆโค2
๐ก๐ฅ๐ฆ, ๐ฆ = 0, 1, 2.
(5)
The winning and losing rule of the game is if ๐ = ๐ฅ, ๐ =๐ฆ, then the necessary and sufficient condition of the winningof ๐ด(๐) is (๐ฆ โ ๐ฅ) mod 3 = 2.
Now construct another random variable ๐น = (๐ โ2) mod 3. Considering a channel (๐; ๐น) consisting of ๐ and๐น, that is, a channel with ๐ as an input and ๐น as an output,then, there are the following event equations.
If๐ด(๐) wins in a certain round, then (๐โ๐) mod 3 = 2,so ๐น = (๐ โ 2) mod 3 = [(2 + ๐) โ ๐] mod 3 = ๐. That is,the input (๐) of the channel (๐; ๐น) always equals its output(๐น). In other words, one bit is successfully transmitted fromthe sender to the receiver in the channel.
Conversely, if โone bit is successfully transmitted from thesender to the receiver in the channel,โ it means that the input(๐) of the channel (๐; ๐น) always equals its output (๐น).That is,๐น = (๐ โ 2) mod 3 = ๐, which is exactly the necessary andsufficient conditions for๐ winning.
Based on the above discussions, ๐ด(๐) winning oncemeans that the channel (๐; ๐น) sends one bit from the senderto the receiver successfully and vice versa. Therefore, thechannel (๐; ๐น) can also play the role of โChannel ๐ดโ in thethird section.
Similarly, if the random variable๐บ = (๐โ2) mod 3, thenthe channel (๐; ๐บ) can play the role of the above โChannel ๐ต.โ
And now the form of channel capacity for channel (๐; ๐น)and channel (๐; ๐บ) will be simpler. We have๐ถ(๐; ๐น) = max๐ [๐ผ(๐, ๐น)] = max๐ [๐ผ(๐, (๐ โ 2) mod3)] = max๐ [๐ผ(๐, ๐)] = max๐ [โ ๐ก๐ฅ๐ฆlog(๐ก๐ฅ๐ฆ/(๐๐ฅ๐๐ฆ))]. Themaximal value here is taken for all possible ๐ก๐ฅ๐ฆ and ๐๐ฅ. So,๐ถ(๐; ๐น) is actually the function of ๐0, ๐1, and ๐2.
Similarly, ๐ถ(๐; ๐บ) = max๐ [๐ผ(๐, ๐บ)] = max๐ [๐ผ(๐, (๐โ2) mod 3)] = max๐ [I(๐, ๐)] = max๐ [โ ๐ก๐ฅ๐ฆlog(๐ก๐ฅ๐ฆ/(๐๐ฅ๐๐ฆ))].The maximal value here is taken for all possible ๐ก๐ฅ๐ฆ and ๐๐ฆ.So, ๐ถ(๐; ๐บ) is actually the function of ๐0, ๐1, and ๐2.2.2. The Strategy of Win. According to Theorem 3, if theprobability of a specific action is determined, the victory ofboth parties in the โrock-paper-scissorsโ game is determinedas well. In order to obtain the victory with higher probability,one must adjust his strategy.
2.2.1.The Game between Two Fools. The so-called โtwo foolsโmeans that ๐ด and ๐ต entrench their habits; that is, theychoose their actions in accordance with the established habitsno matter who won in the past. According to Theorem 3,statistically, if ๐ถ < ๐ท, ๐ด will lose; if ๐ถ > ๐ท, then ๐ด will win;and if ๐ถ = ๐ท, then both parties are well-matched in strength.
2.2.2. The Game between a Fool and a Sage. If ๐ด is a fool,he still insists on his inherent habit; then after confronting asufficient number of times, ๐ต can calculate the distributionprobabilities ๐ and ๐ of random variable ๐ correspondingto ๐ด. And ๐ต can get the channel capacity of ๐ด by somerelated conditional probability distribution at last, and thenby adjusting their own habits (i.e., the probability distributionof the random variable ๐ and the corresponding conditionalprobability distribution, etc.); then ๐ต enlarges his own chan-nel capacity to make the rest of game more beneficial tohimself; moreover, the channel capacity of ๐ต is larger enough,๐ถ(๐ต) > ๐ถ(๐ด); then ๐ต win the success at last.
2.2.3. The Game between Two Sages. If both๐ด and ๐ต get usedto summarizing the habits of each other at any time, andadjust their habits, enlarge their channel capacity. At last, thetwo parties can get the equal value of channel capacities; thatis, the competition between them will tend to a balance, adynamically stable state.
3. Models of (Coin Tossing) and(Palm or Back)
3.1. The Channel Capacity of โCoin Tossingโ Game. โCointossingโ game: โbankerโ covers a coin under his hand on thetable, and โplayerโ guesses the head or tail of the coin. Theโplayerโ will win when he guesses correctly.
Obviously, this game is a kind of โnonblind confronta-tion.โ We will use the method of channel capacity to analyzethe winning or losing of the game.
Based on the Law of Large Numbers in the probabilitytheory, the frequency tends to probability. Thus, accordingto the habits of โbankerโ and โplayer,โ that is, the statisticalregularities of their actions in the past, we can give theprobability distribution of their actions.
We use the random variable ๐ to denote the state of theโbanker.โ ๐ = 0 (๐ = 1) means the coin is head (tail).So the habit of โbankerโ can be described by the probabilitydistribution of ๐; that is, Pr(๐ = 0) = ๐, Pr(๐ = 1) = 1 โ ๐,where 0 โค ๐ โค 1.
We use the random variable ๐ to denote the state of theโplayer.โ ๐ = 0 (๐ = 1) means that he guesses head (tail).
Mathematical Problems in Engineering 5
So the habit of โplayerโ can be described by the probabilitydistribution of ๐; that is, Pr(๐ = 0) = ๐, Pr(๐ = 1) =1 โ ๐, where 0 โค ๐ โค 1. Similarly, according to the paststates of โbankerโ and โplayer,โ we have the joint probabilitydistribution of random variables (๐,๐); namely,
Pr (๐ = 0, ๐ = 0) = ๐;Pr (๐ = 0, ๐ = 1) = ๐;Pr (๐ = 1, ๐ = 0) = ๐;Pr (๐ = 1, ๐ = 1) = ๐,
(6)
where 0 โค ๐, ๐, ๐, ๐, ๐, ๐ โค 1 and๐ + ๐ + ๐ + ๐ = 1;
๐ = Pr (๐ = 0)= Pr (๐ = 0, ๐ = 0) + Pr (๐ = 0, ๐ = 1)= ๐ + ๐;
๐ = Pr (๐ = 0)= Pr (๐ = 0, ๐ = 0) + Pr (๐ = 1, ๐ = 0)= ๐ + ๐.
(7)
Taking ๐ as the input and ๐ as the output, we obtain thechannel (๐;๐) which is called โChannel๐โ in this paper.
Because ๐ guesses correctly = {๐ = 0, ๐ = 0} โช {๐ =1, ๐ = 1} = one bit is successfully transmitted from the sender๐ to the receiver ๐ in โChannel ๐,โ โ๐ wins one timeโ isequivalent to transmitting one bit of information successfullyin โChannel๐.โ
Based on the channel coding theorem of Shannonโs Infor-mation Theory, if the capacity of โChannel ๐โ is ๐ถ, for anytransmission rate ๐/๐ โค ๐ถ, we can receive ๐ bits successfullyby sending ๐ bits with an arbitrarily small probability ofdecoding error. Conversely, if โChannel๐โ can transmit ๐ bitsto the receiver by sending ๐ bits without error, there must be๐ โค ๐๐ถ. In a word, we have the following theorem.
Theorem 4 (banker theorem). Suppose that the channelcapacity of โChannel ๐โ composed of the random variable(๐;๐) is ๐ถ. Then one has the following: (1) if ๐ wants to win๐ times, he must have a certain skill (corresponding to theShannon coding) to achieve the goal by any probability closeto 1 in the ๐/๐ถ rounds; conversely, (2) if ๐ wins ๐ times in ๐rounds, there must be ๐ โค ๐๐ถ.
According to Theorem 3, we only need to figure out thechannel capacity ๐ถ of โChannel ๐โ; then the limitation oftimes that โ๐ winsโ is determined. So we can calculate thetransition probability matrix ๐ด = [๐ด(๐, ๐)], ๐, ๐ = 0, 1 ofโChannel๐โ:
๐ด (0, 0) = Pr (๐ = 0 | ๐ = 0) = Pr (๐ = 0,๐ = 0)Pr (๐ = 0)
= ๐๐ ;
๐ด (0, 1) = Pr (๐ = 1 | ๐ = 0) = Pr (๐ = 1,๐ = 0)Pr (๐ = 0)
= ๐๐ = 1 โ ๐๐ ;๐ด (1, 0) = Pr (๐ = 0 | ๐ = 1) = Pr (๐ = 0,๐ = 1)
Pr (๐ = 1)= ๐(1 โ ๐) = (๐ โ ๐)(1 โ ๐) ;
๐ด (1, 1) = Pr (๐ = 1 | ๐ = 1) = Pr (๐ = 1,๐ = 1)Pr (๐ = 1)
= ๐(1 โ ๐) = 1 โ (๐ โ ๐)(1 โ ๐) .(8)
Thus, the mutual information ๐ผ(๐, ๐) of๐ and ๐ equals
๐ผ (๐, ๐) = โ๐
โ๐
๐ (๐, ๐) log( ๐ (๐, ๐)[๐ (๐) ๐ (๐)])= ๐ log[ ๐(๐๐)] + ๐ log[ ๐[๐ (1 โ ๐)]]+ ๐ log[ ๐[(1 โ ๐) ๐]]+ ๐ log[ ๐[(1 โ ๐) (1 โ ๐)]]
= ๐ log[ ๐(๐๐)] + (๐ โ ๐) log[ (๐ โ ๐)[๐ (1 โ ๐)]]+ (๐ โ ๐) log[ (๐ โ ๐)[(1 โ ๐) ๐]]+ (1 + ๐ โ ๐ โ ๐) log[ (1 + ๐ โ ๐ โ ๐)[(1 โ ๐) (1 โ ๐)]] .
(9)
Thus, the channel capacity ๐ถ of โChannel ๐โ is equalto max[๐ผ(๐, ๐)] (the maximal value here is taken fromall possible binary random variables ๐). In a word, ๐ถ =max[๐ผ(๐, ๐)] 0 < ๐, ๐ < 1 (where ๐ผ(๐, ๐) is the mutualinformation above).Thus, the channel capacity๐ถ of โChannel๐โ is a function of ๐, which is defined as ๐ถ(๐).
Suppose the random variable ๐ = (๐+ 1) mod 2. Taking๐ as the input and ๐ as the output, we obtain the channel(๐; ๐) which is called โChannel ๐โ in this paper.Because {๐wins} = {๐ = 0,๐ = 1} โช {๐ = 1,๐ = 0} ={๐ = 0, ๐ = 0} โช {๐ = 1, ๐ = 1} = {one bit is successfully
transmitted from the sender ๐ to the receiver ๐ in theโChannelYโ}, โ๐wins one timeโ is equivalent to transmittingone bit of information successfully in โChannel ๐.โ
Based on theChannel coding theoremof Shannonโs Infor-mation Theory, if the capacity of โChannel ๐โ is ๐ท, for anytransmission rate ๐/๐ โค ๐ท, we can receive ๐ bits successfullyby sending ๐ bits with an arbitrarily small probability of
6 Mathematical Problems in Engineering
decoding error. Conversely, if โChannel๐โ can transmit ๐ bitsto the receiver by sending ๐ bits without error, there must be๐ โค ๐๐ท. In a word, we have the following theorem.
Theorem5 (player theorem). Suppose that the channel capac-ity of โChannel ๐โ composed of the random variable (๐;๐) is๐ท. Then one has the following: (1) if ๐ wants to win ๐ times,he must have a certain skill (corresponding to the Shannoncoding) to achieve the goal by any probability close to 1 in the๐/๐ถ rounds; conversely, (2) if ๐ wins ๐ times in the n rounds,there must be ๐ โค ๐๐ท.
According to Theorem 4, we can determine the winninglimitation of๐ as long as we know the channel capacity๐ท ofโChannel ๐.โ
Similarly, we can get the channel capacity ๐ท =max [๐ผ(๐, ๐)], 0 < ๐, ๐ < 1, of โChannel๐.โThus, the channelcapacity๐ท of โChannel๐โ is a function of๐, which is denotedas๐ท(๐).๐ผ (๐, ๐) = โ
๐
โ๐
๐ (๐, ๐) log( ๐ (๐, ๐)[๐ (๐) ๐ (๐)])= ๐ log[ ๐(๐๐)] + (๐ โ ๐) log[ (๐ โ ๐)[๐ (1 โ ๐)]]+ (๐ โ ๐) log[ (๐ โ ๐)[(1 โ ๐) ๐]]+ (1 + ๐ โ ๐ โ ๐) log[ (1 + ๐ โ ๐ โ ๐)[(1 โ ๐) (1 โ ๐)]] .
(10)
From Theorems 3 and 4, we can obtain the quantitativeresults of โthe statistical results of winning and losingโ andโthe game skills of banker and player.โ
Theorem 6 (strength theorem). In the game of โcoin tossing,โif the channel capacities of โChannel ๐โ and โChannel ๐โ are๐ถ(๐) and๐ท(๐), respectively, one has the following.Case 1. If both๐ and๐ do not try to adjust their habits in theprocess of game, that is, ๐ and ๐ are constant, statistically, if๐ถ(๐) > ๐ท(๐), ๐ will win; if ๐ถ(๐) < ๐ท(๐), ๐ will win; and if๐ถ(๐) = ๐ท(๐), the final result of the game is a โdraw.โ
Case 2. If๐ implicitly adjusts his habit and๐ does not, that is,change the probability distribution ๐ of the random variable๐ to enlarge the๐ท(๐) of โChannel๐โ such that๐ท(๐) > ๐ถ(๐),statistically,๐will win.On the contrary, if๐ implicitly adjustshis habit and๐ does not, that is,๐ท(๐) < ๐ถ(๐), ๐ will win.
Case 3. If both ๐ and ๐ continuously adjust their habits andmake ๐ถ(๐) and ๐ท(๐) grow simultaneously, they will achievea dynamic balance when ๐ = ๐ = 0.5, and there is no winneror loser in this case.
3.2. The Channel Capacity of โPalm or Backโ Game. Theโpalm or backโ game: three participants (๐ด, ๐ต, and๐ถ) choosetheir actions of โpalmโ or โbackโ at the same time; if one of theparticipants choose the opposite action to the others (e.g., the
others choose โpalmโ when he chooses โbackโ), he will winthis round.
Obviously, this game is also a kind of โnonblind con-frontation.โ We will use the method of channel capacity toanalyze the winning or losing of the game.
Based on the Law of Large Numbers in the probabilitytheory, the frequency tends to probability.Thus, according tothe habits of ๐ด, ๐ต, and ๐ถ, that is, the statistical regularities oftheir actions in the past, we have the probability distributionof their actions.
We use the random variable ๐ to denote the state of ๐ด.๐ = 0 (๐ = 1) means that he chooses โpalm (back).โ Thus,the habit of๐ด can be described as the probability distributionof ๐; that is, Pr(๐ = 0) = ๐, Pr(๐ = 1) = 1 โ ๐, where0 โค ๐ โค 1.
We use random variable ๐ to denote the state of ๐ต. ๐ =0 (๐ = 1) means that he chooses โpalm (back)โ. Thus, thehabit of ๐ต can be described as the probability distribution of๐, that is, Pr(๐ = 0) = ๐, Pr(๐ = 1) = 1 โ ๐, where 0 โค ๐ โค 1.
We use the random variable ๐ to denote the state of ๐ถ.๐ = 0 (๐ = 1) means that he chooses โpalm (back).โ Thus,the habit of๐ถ can be described as the probability distributionof๐; that is, Pr(๐ = 0) = ๐, Pr(๐ = 1) = 1โ๐, where 0 โค ๐ โค 1.
Similarly, according to the Law of Large Numbers, wecan obtain the joint probability distributions of the randomvariables (๐, ๐, ๐) from the records of their game results aftersome rounds; namely,
Pr (๐ด for palm, ๐ต for palm, ๐ถ for palm)= Pr (๐ = 0 ๐ = 0 ๐ = 0) = ๐;
Pr (๐ด for palm, ๐ต for palm, ๐ถ for back)= Pr (๐ = 0 ๐ = 0 ๐ = 1) = ๐;
Pr (๐ด for palm, ๐ต for back, ๐ถ for palm)= Pr (๐ = 0 ๐ = 1 ๐ = 0) = ๐;
Pr (๐ด for palm, ๐ต for back, ๐ถ for back)= Pr (๐ = 0 ๐ = 1 ๐ = 1) = ๐;
Pr (๐ด for back, ๐ต for palm, ๐ถ for palm)= Pr (๐ = 1 ๐ = 0 ๐ = 0) = ๐;
Pr (๐ด for back, ๐ต for palm, ๐ถ for back)= Pr (๐ = 1 ๐ = 0 ๐ = 1) = ๐;
Pr (๐ด for back, ๐ต for back, ๐ถ for palm)= Pr (๐ = 1 ๐ = 1 ๐ = 0) = ๐;
Pr (๐ด for back, ๐ต for back, ๐ถ for back)= Pr (๐ = 1 ๐ = 1 ๐ = 1) = โ,
(11)
where 0 โค ๐, ๐, ๐, ๐, ๐, ๐, ๐, ๐, ๐, ๐, โ โค 1 and๐ + ๐ + ๐ + ๐ + ๐ + ๐ + ๐ + โ = 1;๐ = Pr (๐ด for palm) = Pr (๐ = 0) = ๐ + ๐ + ๐ + ๐;๐ = Pr (๐ต for palm) = Pr (๐ = 0) = ๐ + ๐ + ๐ + ๐;๐ = Pr (๐ถ for palm) = Pr (๐ = 0) = ๐ + ๐ + ๐ + ๐.
(12)
Mathematical Problems in Engineering 7
Suppose the random variable๐ = (๐ + ๐ + ๐) mod 2;then the probability distribution of๐ is
Pr (๐ = 0) = Pr (๐ = 0, ๐ = 0, ๐ = 0)+ Pr (๐ = 0, ๐ = 1, ๐ = 1)+ Pr (๐ = 1, ๐ = 1, ๐ = 0)+ Pr (๐ = 1, ๐ = 0, ๐ = 1)
= ๐ + ๐ + ๐ + ๐,Pr (๐ = 1) = Pr (๐ = 0, ๐ = 0, ๐ = 1)
+ Pr (๐ = 0, ๐ = 1, ๐ = 0)+ Pr (๐ = 1, ๐ = 0, ๐ = 0)+ Pr (๐ = 1, ๐ = 1, ๐ = 1)
= ๐ + ๐ + ๐ + โ.
(13)
Taking๐ as the input and๐ as the output, we obtain thechannel (๐;๐) which is called โChannel ๐ดโ in this paper.
After removing the situations in which three participantschoose the same actions, we have the following equation:{๐ด wins} = {๐ด for palm, ๐ต for back, ๐ถ for back} โช{๐ด for back, ๐ต for palm, ๐ถ for palm} = {๐ = 0, ๐ = 1, ๐ =1}โช{๐ = 1, ๐ = 0, ๐ = 0} = {๐ = 0,๐ = 0}โช{๐ = 1,๐ = 1}= {one bit is successfully transmitted from the sender (๐) tothe receiver (๐) in the โChannel Aโ}.
Conversely, after removing the situations that three par-ticipants choose the same actions, if {one bit is successfullytransmitted from sender (๐) to the receiver (๐) in theโChannel Aโ}, there is {๐ = 0,๐ = 0} โช {๐ = 1,๐ = 1} ={๐ = 0, ๐ = 1, ๐ = 1} โช {๐ = 1, ๐ = 0, ๐ = 0} = {๐ด forpalm, ๐ต for back, ๐ถ for back} โช {๐ด for back, ๐ต for palm, ๐ถfor palm} = {๐ด wins}. Thus, โ๐ด wins one timeโ is equivalentto transmitting one bit successfully from the sender ๐ to thereceiver ๐ in the โChannel ๐ด.โ From the channel codingtheorem of Shannonโs Information Theory, if the capacity ofthe โChannel ๐ดโ is ๐ธ, for any transmission rate ๐/๐ โค ๐ธ,we can receive ๐ bits successfully by sending ๐ bits with anarbitrarily small probability of decoding error. Conversely, ifthe โChannel๐ดโ can transmit ๐ bits to the receiver by sending๐ bits without error, there must be ๐ โค ๐๐ธ. In a word, we havethe following theorem.
Theorem7. Suppose that the channel capacity of the โChannel๐ดโ composed of the random variable (๐;๐) is ๐ธ. Then, afterremoving the situations in which three participants choose thesame actions, one has the following: (1) if ๐ด wants to win๐ times, he must have a certain skill (corresponding to theShannon coding theory) to achieve the goal by any probabilityclose to 1 in the ๐/๐ธ rounds; conversely, (2) if ๐ด wins ๐ times inthe ๐ rounds, there must be ๐ โค ๐๐ธ.
In order to calculate the channel capacity of the channel(๐;๐), we should first calculate the joint probability distri-bution of the random variable (๐,๐):
Pr (๐ = 0,๐ = 0) = Pr (๐ = 0, ๐ = 0, ๐ = 0)+ Pr (๐ = 0, ๐ = 1, ๐ = 1)
= ๐ + ๐;
Pr (๐ = 0,๐ = 1) = Pr (๐ = 0, ๐ = 1, ๐ = 0)+ Pr (๐ = 0, ๐ = 0, ๐ = 1)
= ๐ + ๐;Pr (๐ = 1,๐ = 0) = Pr (๐ = 1, ๐ = 1, ๐ = 0)
+ Pr (๐ = 1, ๐ = 0, ๐ = 1)= ๐ + ๐;
Pr (๐ = 1,๐ = 1) = Pr (๐ = 1, ๐ = 0, ๐ = 0)+ Pr (๐ = 1, ๐ = 1, ๐ = 1)
= ๐ + โ.(14)
Therefore, the mutual information between ๐ and ๐equals
๐ผ (๐,๐) = (๐ + ๐) log[ (๐ + ๐)[๐ (๐ + ๐ + ๐ + ๐)]]+ (๐ + ๐) log[ (๐ + ๐)[(1 โ ๐) (๐ + ๐ + ๐ + ๐)]]+ (๐ + ๐) log[ (๐ + ๐)[๐ (๐ + ๐ + ๐ + โ)]] + (๐ + โ)โ log[ (๐ + โ)[(1 โ ๐) (๐ + ๐ + ๐ + โ)]] = (๐ + ๐)โ log[ (๐ + ๐)[๐ (๐ + ๐ + ๐ + ๐)]] + (๐ + ๐)โ log[ (๐ + ๐)[(1 โ ๐) (๐ + ๐ + ๐ + ๐)]] + (๐ โ ๐ โ ๐)โ log[ (๐ โ ๐ โ ๐)[๐ (1 โ (๐ + ๐ + ๐ + ๐))]]+ (1 โ (๐ + ๐ + ๐))โ log[ (1 โ (๐ + ๐ + ๐))[(1 โ ๐) (1 โ (๐ + ๐ + ๐ + ๐))]] .
(15)
Thus, the channel capacity of โchannel ๐ดโ is equal to ๐ธ =max[๐ผ(๐,๐)] and it is a function of ๐ and ๐, which is definedas ๐ธ(๐, ๐).
Taking ๐ as the input and ๐ as the output, we obtainthe channel (๐,๐)which is called โChannel ๐ต.โ Similarly, wehave the following.
Theorem8. Suppose that the channel capacity of the โChannel๐ตโ composed of the random variable (๐;๐) is ๐น. Then, afterremoving the situation in which the three participants choosethe same action, one has the following: (1) if ๐ต wants to win๐ times, he must have a certain skill (corresponding to theShannon coding) to achieve the goal by any probability close
8 Mathematical Problems in Engineering
to 1 in the ๐/๐น rounds; conversely, (2) if ๐ต wins ๐ times in the nrounds, there must be ๐ โค ๐๐น.
The channel capacity ๐น can be calculated as the sameway of calculating ๐ธ. Here, the capacity of โChannel ๐ตโ is afunction of ๐ and ๐, which can be defined as ๐น(๐, ๐).
Similarly, taking ๐ as the input and๐ as the output, weobtain the channel (๐,๐)which is called โChannel๐ถ.โ So wehave the following.
Theorem9. Suppose that the channel capacity of the โChannel๐ถโ composed of the random variable (๐;๐) is ๐บ. Then, afterremoving the situations in which three participants choose thesame actions, one has the following: (1) if ๐ถ wants to win๐ times, he must have a certain skill (corresponding to theShannon coding theory) to achieve the goal by any probabilityclose to 1 in the ๐/๐น rounds; conversely, (2) if ๐ถ wins ๐ times inthe ๐ rounds, there must be ๐ โค ๐๐บ.
The channel capacity ๐บ can be calculated by the sameway of calculating ๐ธ. Now the capacity of โChannel ๐ถโ is afunction of ๐ and ๐, which can be defined as ๐บ(๐, ๐).
FromTheorems 6, 7, and 8, we can qualitatively describethe winning or losing situations of๐ด, ๐ต, and ๐ถ in the palm orback game.
Theorem 10. If the channel capacities of โChannel๐ด,โ โChan-nel ๐ต,โ and โChannel ๐ถโ are ๐ธ, ๐น, and ๐บ, respectively, thestatistical results of winning or losing depend on the values of๐ธ, ๐น, and ๐บ. The one who has the largest channel capacity willgain the priority.We can know that the three channel capacitiescannot be adjusted only by one participant individually. It isdifficult to change the final results by adjusting oneโs habit(namely, only change one of the ๐, ๐, and ๐ separately), unlesstwo of them cooperate secretly.
4. Models of (Finger Guessing) and(Draw Boxing)
4.1. Model of โFinger Guessingโ. โFinger guessingโ is a gamebetween the host and guest in the banquet. The rules of thegame are the following. The host and the guest, respectively,choose one of the following four gestures at the same timein a round: bug, rooster, tiger, and stick. Then they decidethe winner by the following regulations: โBugโ is inferior toโroosterโ; โroosterโ is inferior to โtigerโ; โtigerโ is inferior toโstickโ; and โstickโ is inferior to โbugโ. Beyond that, the gameis ended in a draw and nobody will be punished.
The โhost ๐ดโ and โguest ๐ตโ will play the โfinger guessinggameโ again after the complete end of this round.Themathe-matical expression of โfinger guessing gameโ is as follows:suppose ๐ด and ๐ต are denoted by random variables ๐ and ๐,respectively; there are 4 possible values of them. Specifically,
๐ = 0 (or ๐ = 0) when ๐ด (or ๐ต) shows โbugโ;๐ = 1 (or ๐ = 1) when ๐ด (or ๐ต) shows โcockโ;๐ = 2 (or ๐ = 2) when ๐ด (or ๐ต) shows โtigerโ;๐ = 3 (or ๐ = 3) when ๐ด (or ๐ต) shows โstickโ.If ๐ด shows ๐ฅ (namely, ๐ = ๐ฅ, 0 โค ๐ฅ โค 3) and ๐ต shows๐ฆ (namely, ๐ = ๐ฆ, 0 โค ๐ฆ โค 3) in a round, the necessary and
sufficient condition of ๐ด wins in this round is (๐ฅ โ ๐ฆ) mod4 = 1. The necessary and sufficient condition of ๐ต wins inthis round is (๐ฆ โ ๐ฅ) mod 4 = 1. Otherwise, this round endsin a draw and proceeds to the next round of the game.
Obviously, the โfinger guessingโ game is a kind of โnon-blind confrontation.โWho is thewinner and howmany timesthe winner wins? How can they make themselves win more?We will use the โchannel capacity methodโ of the โGeneralTheory of Securityโ to answer these questions.
Based on the Law of Large Numbers in the probabilitytheory, the frequency tends to probability.Thus, according tothe habits of โhost (๐)โ and โguest (๐),โ that is, the statisticalregularities of their actions in the past (if they meet for thefirst time, we can require them to play a โwarm-up gameโ andrecord their habits), we can give the probability distribution of๐,๐ and the joint probability distribution of (๐, ๐), respectively:0 < Pr (๐ = ๐) = ๐๐ < 1,๐ = 0, 1, 2, 3; ๐0 + ๐1 + ๐2 + ๐3 = 1;0 < Pr (๐ = ๐) = ๐๐ < 1,๐ = 0, 1, 2, 3; ๐0 + ๐1 + ๐2 + ๐3 = 1;0 < Pr (๐ = ๐, ๐ = ๐) = ๐ก๐๐ < 1,
๐, ๐ = 0, 1, 2, 3; โ0โค๐,๐โค3
๐ก๐๐ = 1.๐๐ฅ = โ0โค๐ฆโค3
๐ก๐ฅ๐ฆ, ๐ฅ = 0, 1, 2, 3;๐๐ฆ = โ0โค๐ฅโค3
๐ก๐ฅ๐ฆ, ๐ฆ = 0, 1, 2, 3.
(16)
In order to analyze the winning situation of ๐ด, weconstruct a random variable ๐ = (๐ + 1) mod 4. Then weuse the random variables ๐ and ๐ to form a channel (๐; ๐),which is called โchannel ๐ดโ; namely, the channel takes ๐as the input and ๐ as the output. Then we analyze someequations of the events. If ๐ด shows ๐ฅ (namely, ๐ = ๐ฅ, 0 โค๐ฅ โค 3) and ๐ต shows ๐ฆ (namely, ๐ = ๐ฆ, 0 โค ๐ฆ โค 3) in a round,one has the following.
If ๐ด wins in this round, there is (๐ฅ โ ๐ฆ) mod 4 = 1; thatis, ๐ฆ = (๐ฅ โ 1) mod 4, so we have ๐ง = (๐ฆ + 1) mod 4 = [(๐ฅ โ1) + 1] mod 4 = ๐ฅ mod 4 = ๐ฅ. In other words, the output ๐is always equal to the input ๐ in the channel ๐ด at this time.That is, a โbitโ is successfully transmitted from the input๐ toits output ๐.
In contrast, if a โbitโ is successfully transmitted from theinput ๐ to the output ๐ in the โchannel ๐ด,โ โthe output ๐ง isalways equal to its input ๐ฅ; namely, ๐ง = ๐ฅโ is true at this time.Then there is (๐ฅ โ ๐ฆ) mod 4 = (๐ง โ ๐ฆ) mod 4 = [(๐ฆ + 1) โ๐ฆ] mod 4 = 1 mod 4 = 1. Hence, we can judge that โ๐ด winsโaccording to the rules of this game.
Combining with the situations above, one has the follow-ing.
Lemma 11. In the โfinger guessingโ game, โ๐ดwins one timeโ isequivalent to โa โbitโ is successfully transmitted from the inputto its output in the โchannel A.โโ Combine Lemma 1 with theโchannel coding theoremโ of Shannonโs Information Theory; ifthe capacity of the โchannel ๐ดโ is ๐ถ, for any transmission rate๐/๐ โค ๐ถ, we can receive ๐ bits successfully by sending ๐ bits with
Mathematical Problems in Engineering 9
an arbitrarily small probability of decoding error. Conversely, ifthe โchannel ๐ดโ can transmit ๐ bits to the receiver by sending ๐bits without error, there must be ๐ โค ๐๐ถ. In a word, we havethe following theorem.
Theorem 12. Suppose that the channel capacity of the โchannel๐ดโ composed of the random variable (๐; ๐) is ๐ถ. Then afterremoving the situation of โdraw,โ one has the following: (1)if ๐ด wants to win ๐ times, he must have a certain skill(corresponding to the Shannon coding) to achieve the goal byany probability close to 1 in the ๐/๐ถ rounds; conversely, (2) if๐ดwins ๐ times in the ๐ rounds, there must be ๐ โค ๐๐ถ.
According to Theorem 12, we only need to figure out thechannel capacity ๐ถ of the โchannel ๐ด,โ then the limitation ofthe times of โ๐ด winsโ is determined. So we can calculate thechannel capacity ๐ถ: first, the joint probability distribution of(๐, ๐) is Pr(๐ = ๐, ๐ = ๐) = Pr(๐ = ๐, (๐ + 1) mod 4 = ๐) =Pr(๐ = ๐, ๐ = (๐ โ 1) mod 4) = ๐ก๐(๐โ1) mod 4, ๐, ๐ = 0, 1, 2, 3, 4.
Therefore, the channel capacity of the channel ๐ด(๐;๐) is๐ถ = max [๐ผ (๐, ๐)]= max
{{{ โ0โค๐,๐โค3
[๐ก๐(๐โ1) mod 4] log [๐ก๐(๐โ1) mod 4](๐๐๐๐)}}} .
(17)
The max in the equation is the maximal value taken from thereal numbers which satisfy the following conditions: 0 < ๐๐,๐ก๐๐ < 1, ๐, ๐ = 0, 1, 2, 3;๐0+๐1+๐2+๐3 = 1;โ0โค๐,๐โค3 ๐ก๐๐ = 1;๐๐ฅ =โ0โค๐ฆโค3 ๐ก๐ฅ๐ฆ. Thus, the capacity๐ถ is actually the function of thepositive real variables which satisfy the following conditions๐0 + ๐1 + ๐2 + ๐3 = 1 and 0 < ๐๐ < 1, ๐ = 0, 1, 2, 3; namely, itcan be written as ๐ถ(๐0, ๐1, ๐2, ๐3), where ๐0 + ๐1 + ๐2 + ๐3 = 1.
Similarly, we can analyze the situation of โ๐ตwins.โWe cansee that the times of โ๐ดwinsโ (๐ถ(๐0, ๐1, ๐2, ๐3)) depend on thehabit of ๐ต(๐0, ๐1, ๐2, ๐3). If both ๐ด and ๐ต stick to their habits,their winning or losing situation is determined; if either ๐ด or๐ต adjusts his habit, he can win statistically when his channelcapacity is larger; if both ๐ด and ๐ต adjust their habits, theirsituations will eventually reach a dynamic balance.
4.2. Model of โDraw Boxingโ. โDraw boxingโ is more compli-cated than โfinger guessing,โ and it is also a game between thehost and guest in the banquet.The rule of the game is that thehost (๐ด) and the guest (๐ต) independently show one of the sixgestures from 0 to 5 and shout one of eleven numbers from0 to 10. That is, in each round, โhost ๐ดโ is a two-dimensionalrandom variable ๐ด = (๐, ๐), where 0 โค ๐ โค 5 is the gestureshowed by the โhostโ and 0 โค ๐ โค 10 is the number shoutedby him. Similarly, โguest๐ตโ is also a two-dimensional randomvariable ๐ต = (๐น, ๐บ), where 0 โค ๐น โค 5 is the gesture showed bythe โguestโ and 0 โค ๐บ โค 10 is the number shouted by him. If๐ด and ๐ต are denoted by (๐ฅ, ๐ฆ) and (๐, ๐) in a certain round,respectively, the rules of the โdraw boxingโ game are
If ๐ฅ + ๐ = ๐ฆ, ๐ด wins;If ๐ฅ + ๐ = ๐, ๐ต wins.
If the above two cases donot occur, the result of this roundis a โdraw,โ and๐ด and๐ต continue the next round. Specifically,when the numbers shouted by both sides are the same
(namely, ๐ = ๐ฆ), the result of this round is a โdraw.โ However,the numbers shouted by both sides are different and the ges-tures showed by them are not equal to โthe number shoutedby any sideโ; the result of this round also comes to a โdraw.โ
Obviously, the โdraw boxingโ game is a kind of โnonblindconfrontation.โ Who is the winner and how many times thewinner wins? How can they make themselves win more? Wewill use the channel capacity method of the โGeneral Theoryof Securityโ to answer these questions.
Based on the Law of Large Numbers in the probabilitytheory, the frequency tends to probability.Thus, according tothe habits of โhost (๐ด)โ and โguest (๐ต)โ, that is the statisticalregularities of their actions in the past (if they meet for thefirst time, we can require them to play a โwarm-up gameโ andrecord their habits), we can give the probability distributionof ๐ด, ๐ต and their components ๐, ๐, ๐น, and ๐บ, and the jointprobability distribution of (๐, ๐, ๐น, ๐บ), respectively.
The probability of โ๐ด shows ๐ฅโ:0 < Pr(๐ = ๐ฅ) = ๐๐ฅ < 1, 0 โค ๐ฅ โค 5; ๐ฅ0 + ๐ฅ1 + ๐ฅ2 +๐ฅ3 + ๐ฅ4 + ๐ฅ5 = 1;
The probability of โ๐ต shows ๐โ:0 < Pr(๐น = ๐) = ๐๐ < 1, 0 โค ๐ โค 5; ๐0 + ๐1 + ๐2 +๐3 + ๐4 + ๐5 = 1;
The probability of โ๐ด shouts ๐ฆโ:0 < Pr(๐ = ๐ฆ) = ๐๐ฆ < 1, 0 โค ๐ฆ โค 10; โ0โค๐ฆโค10 ๐๐ฆ = 1;The probability of โ๐ต shouts ๐โ:0 < Pr(๐บ = ๐) = ๐ ๐ < 1, 0 โค ๐ โค 10; โ0โค๐โค10 ๐ ๐ = 1;The probability of โ๐ด shows ๐ฅ and shouts ๐ฆโ:0 < Pr[๐ด = (๐ฅ, ๐ฆ)] = Pr(๐ = ๐ฅ, ๐ = ๐ฆ) = ๐๐ฅ๐ฆ < 1,0 โค ๐ฆ โค 10, 0 โค ๐ฅ โค 5, โ0โค๐ฆโค10,0โค๐ฅโค5 ๐๐ฅ๐ฆ = 1;The probability of โ๐ต shows ๐ and shouts ๐โ:0 < Pr[๐ต = (๐, ๐)] = Pr(๐น = ๐, ๐บ = ๐) = โ๐๐ < 1,0 โค ๐ โค 10, 0 โค ๐ โค 5, โ0โค๐โค10,0โค๐โค5 โ๐๐ = 1;The probability of โ๐ด shows๐ฅ and shouts๐ฆโ and โ๐ต shows๐ and shouts ๐โ at the same time:0 < Pr[๐ด = (๐ฅ, ๐ฆ), ๐ต = (๐, ๐)] = Pr(๐ = ๐ฅ, ๐ =๐ฆ, ๐น = ๐, ๐บ = ๐) = ๐ก๐ฅ๐ฆ๐๐ < 1, where 0 โค ๐ฆ, ๐ โค 10,0 โค ๐ฅ, ๐ โค 5, โ0โค๐ฆ,๐โค10,0โค๐ฅ,๐โค5 ๐ก๐ฅ๐ฆ๐๐ = 1.
In order to analyze the situation of ๐ด wins, we constructa two-dimensional random variable๐ = (๐,๐) = (๐๐ฟ (๐บ โ ๐) ,๐ + ๐น) . (18)The function ๐ฟ is defined as ๐ฟ(0) = 0; ๐ฟ(๐ฅ) = 1 when ๐ฅ ฬธ= 0.Therefore,Pr [๐ = (๐ข, V)] = โ
๐ฅ+๐=V,๐ฅ๐ฟ(๐โ๐ฆ)=๐ข๐ก๐ฅ๐ฆ๐๐ ลก ๐๐ขV,
where 0 โค V โค 10, 0 โค ๐ข โค 5. (19)
Then, we use the random variables๐ด and๐ to form a channel(๐ด; ๐), which is called โchannel ๐ดโ and takes ๐ด as the inputand ๐ as the output.
Then we analyze some equations. In a certain round, ๐ดshows ๐ฅ (i.e.,๐ = ๐ฅ, 0 โค ๐ฅ โค 5) and shouts ๐ฆ (i.e., ๐ = ๐ฆ, 0 โค๐ฆ โค 10); meanwhile, ๐ต shows ๐ (i.e., ๐น = ๐, 0 โค ๐ โค 5) and
10 Mathematical Problems in Engineering
shouts๐ (i.e.,๐บ = ๐, 0 โค ๐ โค 10). According to the evaluationrules, we have the following: if๐ดwins in this around, we have๐ฅ + ๐ = ๐ฆ and ๐ฆ ฬธ= ๐. Thus, ๐ฟ(๐ โ ๐ฆ) = 1 and ๐ = (๐ข, V) =(๐ฅ๐ฟ(๐ โ ๐ฆ), ๐ฅ + ๐) = (๐ฅ, ๐ฆ) = ๐ด. In other words, the output ๐of the โchannel ๐ดโ is always equal to its input ๐ด at this time;that is to say, a โbitโ is sent successfully from the input ๐ด toits output ๐.
In contrast, if one bit is successfully sent from the input๐ดto the output ๐ in the โchannel ๐ด,โ โthe output ๐ง = (๐ข, V) =(๐ฅ๐ฟ(๐ โ ๐ฆ)๐ฅ + ๐)โ is always equal to the โinput (๐ฅ, ๐ฆ)โ at thistime; also there is ๐ฅ๐ฟ(๐ โ ๐ฆ) = ๐ฅ when ๐ฅ + ๐ = ๐ฆ; that is,๐ฆ ฬธ= ๐ and ๐ฅ + ๐ = ๐ฆ. Thus, ๐ด wins this round according tothe evaluation rules.
Combining with the cases above, we have the following.
Lemma 13. In a โdraw boxingโ game, โ๐ด wins one timeโ isequivalent to one bit is successfully sent from the input ofโchannel ๐ดโ to its output.
Combining Lemma 13with the โchannel coding theoremโof Shannonโs Information Theory, if the capacity of theโchannel ๐ดโ is ๐ท, for any transmission rate ๐/๐ โค ๐ท, wecan receive ๐ bits successfully by sending ๐ bits with anarbitrarily small probability of decoding error. Conversely, ifthe โchannel ๐ดโ can transmit ๐ bits to the receiver by sending๐ bits without error, there must be ๐ โค ๐๐ท. In a word, we havethe following theorem.
Theorem 14. Suppose that the channel capacity of the โchannel๐ดโ composed of the random variable (๐ด; ๐) is ๐ท. Then afterremoving the situation of โdraw,โ one has the following: (1)if ๐ด wants to win ๐ times, he must have a certain skill(corresponding to the Shannon coding) to achieve the goal byany probability close to 1 in the ๐/๐ท rounds; conversely, (2) if๐ด wins ๐ times in the ๐ rounds, there must be ๐ โค ๐๐ท.
According to Theorem 4, we only need to figure out thechannel capacity๐ท of the โchannel ๐ดโ; then the limitation oftimes that โ๐ด winsโ is determined. So we can calculate thechannel capacity๐ท:
๐ท = max [๐ผ (๐ด, ๐)] = max{โ๐,๐ง
Pr (๐, ๐ง)โ log [ Pr (๐, ๐ง)[Pr (๐)Pr (๐ง)]]}= max
{{{ โ๐ฅ,๐ฆ,๐,๐
Pr (๐ฅ, ๐ฆ, ๐ฅ๐ฟ (๐ โ ๐ฆ) , ๐ฅ + ๐)
โ log[ Pr (๐ฅ, ๐ฆ, ๐ฅ๐ฟ (๐ โ ๐ฆ) , ๐ฅ + ๐)[Pr (๐ฅ, ๐ฆ)Pr (๐ฅ๐ฟ (๐ โ ๐ฆ) , ๐ฅ + ๐)]]}}}= max
{{{ โ๐ฅ,๐ฆ,๐,๐
๐ก๐ฅ,๐ฆ,๐ฅ๐ฟ(๐โ๐ฆ),๐ฅ+๐โ log[ ๐ก๐ฅ,๐ฆ,๐ฅ๐ฟ(๐โ๐ฆ),๐ฅ+๐[๐๐ฅ๐ฆ๐๐ฅ๐ฟ(๐โ๐ฆ),๐ฅ+๐]]
}}} .
(20)
Themaximal value in the equation is a real number whichsatisfies the following conditions:
โ0โค๐ฆโค10
๐๐ฆ = 1; 0 โค ๐ฆ โค 10;โ
0โค๐ฆโค10,0โค๐ฅโค5
๐๐ฅ๐ฆ = 1; 0 โค ๐ฆ โค 10, 0 โค ๐ฅ โค 5,โ
0โค๐โค10,0โค๐โค5
โ๐๐ = 1; 0 โค ๐ โค 10, 0 โค ๐ โค 5.(21)
Thus, the capacity ๐ท is actually the function of ๐๐, ๐๐,which satisfies the following conditions: 0 โค ๐ โค 5; ๐0 + ๐1 +๐2 + ๐3 + ๐4 + ๐5 = 1; 0 โค ๐ โค 10; โ0โค๐โค10 ๐ ๐ = 1, where0 โค ๐ โค 5 and 0 โค ๐ โค 10.
Similarly, we can analyze the situation of โ๐ต wins.โ Wecan see that the times of โ๐ด winsโ (๐ท(๐๐, ๐๐)) depend on thehabit of ๐ต(๐๐, ๐๐). If both ๐ด and ๐ต stick to their habits, theirwinning or losing is determined; if either ๐ด or ๐ต adjusts hishabit, he can win statistically when his channel capacity islarger; if both๐ด and ๐ต adjust their habits, their situations willeventually reach a dynamic balance.
5. Unified Model of Linear Separable(Nonblind Confrontation)
Suppose that the hacker (๐) has n methods of attack; that is,the random variable ๐ has ๐ values which can be denoted as{๐ฅ0, ๐ฅ1, . . . , ๐ฅ๐โ1} = {0, 1, 2, . . . , ๐ โ 1}. These ๐methods makeup the entire โarsenalโ of the hacker.
Suppose that the honker (๐) has ๐ methods of defense;that is, the random variable ๐ has ๐ values, which can bedenoted as {๐ฆ0, ๐ฆ1, ๐ฆ๐โ1} = {0, 1, 2, . . . , ๐ โ 1}. These ๐methods make up the entire โarsenalโ of the honker.
Remark 15. In the following, we will equivalently transformbetween โthe methods ๐ฅ๐, ๐ฆ๐โ and โthe numbers ๐, ๐โ asneeded; that is, ๐ฅ๐ = ๐ and ๐ฆ๐ = ๐. By the transformation, wecan make the problem clear in the interpretation and simplein the form.
In the nonblind confrontation, there is a rule of winningor losing between each hackerโs method ๐ฅ๐ (๐ = 0, 1, . . . , ๐ โ1) and each honkerโs method ๐ฆ๐ (๐ = 0, 1, . . . , ๐ โ 1). Sothere must exist a subset of the two-dimensional number set{(๐, ๐), 0 โค ๐ โค ๐ โ 1, 0 โค ๐ โค ๐ โ 1}, which makesโ๐ฅ๐ is superior to ๐ฆ๐โ true if and only if (๐, ๐) โ ๐ป. If thestructure of the subset๐ป is simple, we can construct a certainchannel to make โthe hacker wins one timeโ equivalentto โone bit is successfully transmitted from the sender tothe receiver.โ Then, we analyze it using Shannonโs โchannelcoding theorem.โ For example,
in the game of โrock-paper-scissors,โ๐ป = (๐, ๐) : 0 โค๐, ๐ โค 2(๐ โ ๐) mod 3 = 2;in the game of โcoin tossing,โ๐ป = (๐, ๐) : 0 โค ๐ = ๐ โค1;in the game of โpalm or back,โ๐ป = (๐, ๐, ๐) : 0 โค ๐ ฬธ=๐ = ๐ โค 1;
Mathematical Problems in Engineering 11
in the game of โfinger guessing,โ๐ป = (๐, ๐) : 0 โค ๐, ๐ โค3(๐ โ ๐) mod 4 = 1;in the game of โdraw boxing,โ ๐ป = (๐ฅ, ๐ฆ, ๐, ๐) : 0 โค๐ฅ, ๐ โค 50 โค ๐ ฬธ= ๐ฆ โค 10๐ฅ + ๐ = ๐ฆ.
We have constructed corresponding communicationchannels for each ๐ป above in this paper. However, it isdifficult to construct such a communication channel for ageneral ๐ป. But if the above set ๐ป can be decomposed into๐ป = {(๐, ๐) : ๐ = ๐(๐), 0 โค ๐ โค ๐ โ 1, 0 โค ๐ โค๐ โ 1} (namely, the first component ๐ of ๐ป is a function ofits second component), we can construct a random variable๐ = ๐(๐). Then considering the channel (๐; ๐), we can givethe following equations.
If the โhacker๐โ attacks with themethod ๐ฅ๐, and โhonker๐โ defends with the method ๐ฆ๐ in a certain round, then if โ๐wins,โ that is, ๐ = ๐(๐), the output of the channel (๐; ๐) is๐ = ๐(๐ฆ๐) = ๐(๐) = ๐ = ๐ฅ๐. So the output of the channelis the same as its input now; that is, one bit is successfullytransmitted from the input of the channel (๐; ๐) to its output.Conversely, if โone bit is successfully transmitted from theinput of the channel (๐; ๐) to its output,โ there is โinput =outputโ; that is, โ๐ = ๐(๐)โ, which means โ๐ wins.โ
Combining the cases above, we obtain the followingtheorem.
Theorem 16 (the limitation theorem of linear nonblindconfrontation). In the โnonblind confrontationโ, supposethe hacker ๐ has n attack methods {๐ฅ0, ๐ฅ1, . . . , ๐ฅ๐โ1} ={0, 1, 2, . . . , ๐ โ 1} and the honker ๐ has m defense methods{๐ฆ0, ๐ฆ1, ๐ฆ๐โ1} = {0, 1, 2, . . . , ๐ โ 1}, and both sides complywith the rule of winning or losing: โ๐ฅ๐ is superior to ๐ฆ๐โ if andonly if (๐, ๐) โ ๐ป, where ๐ป is a subset of the rectangular set{(๐, ๐), 0 โค ๐ โค ๐ โ 1, 0 โค ๐ โค ๐ โ 1}.
For๐, if๐ป is linear and can be written as๐ป = {(๐, ๐) : ๐ =๐(๐), 0 โค ๐ โค ๐ โ 1, 0 โค ๐ โค ๐ โ 1} (i.e., the first component๐ of ๐ป is a certain function ๐(โ ) of its second component ๐),we can construct a channel (๐; ๐) with ๐ = ๐(๐) to get that,if ๐ถ is the channel capacity of channel (๐; ๐), we have thefollowing.
(1) If๐ wants to win ๐ times, he must have a certain skill(corresponding to the Shannon coding) to achieve the goalby any probability close to 1 in the ๐/๐ถ rounds.
(2) If๐wins ๐ times in ๐ rounds, there must exist ๐ โค ๐๐ถ.For๐, if๐ป is linear and can be written as๐ป = {(๐, ๐) : ๐ =๐(๐), 0 โค ๐ โค ๐โ1, 0 โค ๐ โค ๐โ1} (i.e., the second component๐ of ๐ป is a certain function ๐(โ ) of its first component ๐), we
can construct a channel (๐; ๐บ) with ๐บ = ๐(๐) to get that,if ๐ท is the channel capacity of channel (๐; ๐บ), we have thefollowing.
(3) If ๐ wants to win ๐ times, he must have a certain skill(corresponding to the Shannon coding) to achieve the goalby any probability close to 1 in the ๐/๐ท rounds.
(4) If๐wins ๐ times in ๐ rounds, there must exist ๐ โค ๐๐ท.6. Conclusion
It seems that these games of nonblind confrontation aredifferent. However, we use an unified method to get the
distinctive conclusion; that is, we establish a channel modelwhich can transform โthe attacker or the defender wins onetimeโ to โone bit is transmitted successfully in the channel.โThus, โthe confrontation between attacker and defenderโ istransformed to โthe calculation of channel capacitiesโ by theShannon coding theorem [6]. We find that the winning orlosing rules sets of these games are linearly separable. Forlinearly inseparable case, it is still an open problem. Thesewinning or losing strategies can be applied in big data field,which provides a new perspective for the study of the big dataprivacy protection.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This paper is supported by the National Key Research andDevelopment Programof China (Grant nos. 2016YFB0800602,2016YFB0800604), the National Natural Science Founda-tion of China (Grant nos. 61573067, 61472045), the BeijingCity Board of Education Science and technology project(Grant no. KM201510015009), and the Beijing City Board ofEducation Science and Technology Key Project (Grant no.KZ201510015015).
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