Ch2 Nash Equilibrium
双醛淀粉_苯胺希夫碱的合成
Abstract： Dialdehyde starchaniline Schiffbase （ DASAN ） was synthesized， with dialdehyde starch （ DAS ） and aniline as reactants and acetic acid as catalyst，in dimethyl sulfoxide （ DMSO ） AN were studied and the homogeneous system in N2 atmosphere． The synthesis conditions of DASoptimum conditions were as follows： the mass concentration of DAS is 100 g / L in DMSO， the dosage of acetic acid is 3% （ volume fraction） ， reaction temperature is 50 ℃ ， the molar ratio of aldehyde group to amino group is 2∶1 ， and reaction time is 12 h． Under these optimum conditions， the utilization ratio of aniline is 94% with DAS5 （ the dialdehyde molar fraction is 5% ） as reactant， and 55% with DAS20 （ the dialdehyde molar fraction is 20% ） as reactant． The structure of DASAN is characterized by means of FTIR， NMR and XRD． Key words： dialdehyde starch； aniline； homogeneous system； Schiffbase； starch chemicals Foundation item： Supported by the State Key Program of National Natural Science Foundation of China （ 20836001 ） 双醛淀粉（ DAS ） 是一类重要的氧化淀粉， 是以 高碘酸钠或高碘酸为氧化剂， 使淀粉葡萄糖环中的 C2 ～ C3 键断裂形成醛基而得到的产物。 双醛淀粉 可以和其他化合物 中的醛基具有很高的反应活性， 上的羟基或氨基发生反应， 生成相应的缩醛、 半缩醛 或希夫碱衍生物。 双醛淀粉可与含氨基的化合物反应生成希夫 ， 碱 通过 CN 双键将双醛淀粉与伯胺化合物连接 起来。能与双醛淀粉发生反应生成希夫碱的含氨基
第二章纳什均衡(理论-策略性博弈)
3BeerQuiche——啤酒与热狗,信号博弈
NTU IO(I):Classnote03Meng-Yu Liang March,2009 Kohlberg and Mertens(Econometrica1986)We will use the term”(game)tree”for the extensive form of a game with perfect recall (i.e.,where every player remembers whatever he knew previously,including his past actions).The agent normal form(Selten)of a tree is the normal form of the game between agents, obtained by letting each information set be manned by a different agent,and by giving any agent of the same player that player’s payoff.A behavioral strategy of a player in a tree is a list of(mixed)strategies,one for each of his agents.Kuhn(1953)has shown that every mixed strategy of a player in a tree is equivalent to some behavioral strategy,in the sense that both give the same probability distribution on the endpoints whatever be the strategies of all opponents.A sequential equilibrium(Kreps-Wilson)of an n-player tree is an n-tuple of behavioral strategies which is the limit of a sequence(σm)of completely mixed(i.e.,strictly positive) behavioral strategies,such that every agent maximizes his expected payoffgiven the strategies of all other agents and given the limiting conditional probability distribution on his information set implied by(σm).Anε-perfect equilibrium of a normal form game(Selten)is a completely mixed strategy vector,such that any pure strategy which is not a best reply has weight less than E.Anε-proper equilibrium of a normal form game(Myerson)is a completely mixed strategy vector,such that whenever some pure strategy s1,is a worse reply than some other pure strategy s2,,the weight on s1,is smaller thanεtimes the weight on s2,.A perfect(proper)equilibrium of a normal form game is a limit(ε→0)ofε-perfect (proper)eqilibria.A perfect(proper)equilibrium of a tree is a perfect(proper)equilibrium of its agent normal form.It is evident that”proper”is a stronger requirement than”perfect.”It is also easy to verify that a perfect equilibrium of a tree is sequential(Kreps-Wilson).Existence theorems have been proved for all the above concepts(Kreps-Wilson,Myerson, Selten).Proposition1For generic extensive-form games,the set of Nash equilibrium outcomes is finite and all Nash equilibrium within a given connected component induce the same outcome. Proposition2Every game has at least one stable component.1Beer QuicheConsider the following version of the Beer Quiche Game.First Nature selects whether player 1,the sender,will be of the strong(s)or weak(w)type.The probability with which the strong type is selected equals910,and this value is common knowledge between the two players. Player1then sends a signal,whether to have Beer of Quiche for breakfast.Upon seeing this signal,Player2,the receiver,then decides whether or not to challenge player1to a duel.Player1knows that the strong type prefers beer,and that the weak type prefers quiche. More precisely,each type gets a utility of+1from having its most preferred breakfast.Player 1also know that each type does not like to duel.Not having to duel yields each sender type an additional utility of+2.Player2profits from challenging if and only if he faces the weak type.Thus Player2gets a utility of+1if he faces the weak type and duels,and a utility of−1if he faces the strong type and duels.The payoffs to this signalling game are summarized in the table below:d n s1,−13,0 w0,12,0Bd n s0,−12,0 w1,13,0QThe attached picture draws the extensive form of this signalling game.We can also compute the normal form of this two-player game,as follows.A strategy for the sender is a pair of signals from the set{B,Q},thefirst one describing the choice of the strong sender type,and the second one describing the choice of the weak sender type.A strategy for the receiver is a pair of actions from the set{d,n},thefirst one describing his choice when he sees the sender selected beer,and the second one describing his choice when he sees the sender selected quiche.Note that we are treating the sender as a single player,so the normal form computes the expected payoffof player1and2,conditional on a pure strategy profile.We then obtain:dd dn nd nnBB910,−45910,−452910,02910,0BQ1,−4565,−910145,1103,0QB0,−4595,11015,−9102,0QQ110,−452110,0110,−452110,02Nash Equilibria of the Normal FormWe start by computing the Nash equilibria for the normal form.Observe that the strategy dd for the receiver is strictly dominated by the strategy nn.Hence the above game is strategically equivalent to the game in which dd is eliminated as a receiver strategy:dn nd nnBB910,−452910,02910,0BQ65,−910145,1103,0QB95,11015,−9102,0QQ2110,0110,−452110,0We now claim that in the remaining game the strategy QB for the sender is strictly dominated by the strategy17BB+67QQ.Indeed,the latter strategy yields an expected payoffvector of(2714,12,3114),which strictly dominates the payoffvector from using QB,(95,15,2)that resultsfrom using QB.Hence we may again reduce the game by eliminating the strategy QB for the sender.This results in the3×3game:dn(p1)nd(p2)nn(p3)BB(q1)910,−452910,02910,0BQ(q2)65,−910145,1103,0QQ(q3)2110,0110,−452110,0Note that in this game no remaining strategy for the sender or the receiver is strictly dominated.1However,the strategy dn is weakly dominated for the receiver(by the strategy nn).Let us nowfind all the Nash equilibria of this3×3game.Let p1denote the probability with which the receiver selects dn,p2the probability with which the receiver selects nd,and p3=1−p1−p2the probability with which the receiver selects nn.Then the expected payoffof the sender equalsPure Strategy Expected Payoffto SenderBB910p1+2910p2+2910(1−p1−p2)=2910−2p1BQ65p1+145p2+3(1−p1−p2)=3−95p1−15p2QQ2110p1+110p2+2110(1−p1−p2)=2110−2p2Therefore the sender prefersBB to BQ⇐⇒p1≤p2−12BB to QQ⇐⇒p1≤p2+25BQ to QQ⇐⇒p1≤p2+12Therefore BB is a best response for the sender if and only if p1≤p2−12,BQ is a best response if and only if p2−12≤p1≤p2+12,and QQ is a best response if and only if p1≥p2+12.Next,let q1denote the probability with which the sender selects BB,q2the probability with which the sender selects BQ,and q3=1−q1−q2the probability with which the sender selects QQ.Then the receiver’s expected payoffequals:dn−45q1−910q2nd110q2−45(1−q1−q2)=−45+910q2+45q1nn0Therefore the receiver prefersdn to nd⇐⇒q1≤12−98q2dn to nn⇐⇒q1=q2=0nd to nn⇐⇒q1≥1−98q2Therefore dn is a best response for the receiver if and only if q1=q2=0,nd is a best response if and only if q1≥1−98q2,and nn is a best response if and only if q1≤1−98q2.To determine the Nash equilibria,we consider three cases:1BB is a best response to nd,BQ is a best response to nn,and QQ is a best response to dn.Also,dn is a best response to QQ,nd is a best response to BQ,and nn is a best response to QQ.1.q 1=q 2=0,i.e.the sender selects QQ with probability one.Then the receiver can respond with any mixture between dn and nn ,i.e.p 2=0.Given this receiver response,QQ will be a best response if and only if p 1≥12.2.q 1>0or q 2>0.Then from the receiver’s best response we obtain p 1=0.But when p 1=0,the sender will select QQ with probability zero,i.e.we must have q 1+q 2=1.(a)q 1>1−98q 2.Given that q 1+q 2=1this inequality is equivalent to q 1<1.When q 1>1−98q 2,the receiver must use nd with probability 1,i.e.we have p 2=1.Now with p 2=1the sender must select BB with probability 1,i.e.we have q 1=1,a contradiction.(b)q 1<1−98q 2.Given that q 1+q 2=1this inequality is equivalent to q 1>1,a contradiction.(c)q 1=1−98q 2.Given that q 1+q 2=1this equality is equivalent to q 1=1,i.e.thesender selects BB for sure.In this case,the receiver can mix between nd and nn ,but we must have p 1=0.The mixture between nd and nn must be selected so that the sender prefers BB to BQ ,i.e.p 2≥12.We conclude that it is a Nash equilibrium for the receiver to mix with between nd and nn ,with nd being selected with probability of at least 12,and for the sender to select BB with probability 1.We summarize this discussion as follows:Lemma 1In any Nash equilibrium the sender selects QB with probability zero,and the re-ceiver selects dd with probability zero .Let q 1denote the probability with which the sender selects BB ,and q 2the probability with which the sender selects BQ (with the remaining prob-ability allocated to QQ ).Also let p 1denote the probability with which the receiver selects dn ,and p 2the probability with which the receiver selects nd (with the remaining probability allocated to nn ).Then there are two components of Nash equilibria to the Beer Quiche game,given by:(i)q 1=q 2=0,p 1≥12and p 2=0;(ii)q 1=1,p 2≥12and p 1=0.Consider the sender’s expected payoffin the Nash equilibria associated with the first component,in which both sender types select Q .We have u ∗1=2110.Meanwhile,the sender’s expected utility from selecting the strategy BB in these equilibria equals 910p 1+2910(1−p 1)=2910−2p 1≤1910,and his expected payofffrom selecting BQ equals 65p 1+3(1−p 1)=3−95p 1≤2110.Thus only strategy BQ is ever a weak best response,and then only in the equilibrium where p 1=12.Consider the sender’s expected payoffin the Nash equilibria associated with the second component,in which the sender selects BB and the receiver selects nd with probability p 2≥12and nn with probability 1−p 2.We have u ∗1=2910.Meanwhile,the sender’s expected utility from selecting the strategy BQ in these equilibria equals 145p 2+3(1−p 2)≤2910,and his expected payofffrom selecting QQ equals 110p 2+2110(1−p 2)≤2110.Thus only the strategy BQ is ever a weak best response,and then only in the equilibrium in which p 2=p 3=12.Now let us consider matters from the receiver’s perspective.Her equilibrium utility in the Nash equilibria associated with the first component of Nash equilibria in which the senderselects QQ equals u∗2=0.Therefore there are no weak best responses other than the equilibrium responses dn and nn.In the Nash equilibria associated with the second component,in which the sender selects BB,the receiver’s expected utility equals u∗2=0.Therefore there are no weak best responses other than the equilibrium responses nd and nn.3Stable SetsTofind the stable sets of this game,we make use of the following result,contained in Kohlberg-Mertens:Every stable set contains a stable set of the game in which a strategy that is weakly dominated or never a weak best response to any equilibrium in the set is eliminated.We restrict attention to stable sets contained in a single component of Nash equilibria.For equilibria associated with the component in which the sender uses QQ,the strategies BB and nd are not a weak best response to any equilibrium in the component.The game in which these strategies have been eliminated is as follows:dn nnBQ65,−9103,0QQ2110,02110,0Since the strategy dn is weakly dominated,it must be used with minimal probability in any equilibrium of a pertubed game.But then the sender will select BQ with maximal probability, i.e.there is no equilibrium of the perturbed game near this component.We conclude that this component contains no stable set.Next,consider equilibria associated with the component in which the sender uses BB. The strategies QQ and dn are not a weak best response to any equilibrium in the component, and after eliminating them,we obtain the following game:nd nnBB2910,02910,0BQ145,1103,0In this game,the strategy nn is weakly dominated,and hence will not be used in voluntarily any equilibrium of the perturbed game.But when the strategy nd is used with probability1, the sender’s best response is to select BB with maximal probability.The resulting equilibrium is near the component.However,this does not end the analysis.To actually prove that this component is stable, we need to exhibit Nash equilibria for perturbations of the full game.This will be is second place where the Cho-Sobel condition comes in.Note in particular that we cannot merely turn the equilibrium of the reduced game into an equilibrium of the full game by having all strategies other than BB and nd being selected with minimal probability:if the sender trembles sufficiently more on QQ than on BQ then the receiver would want to use the strategy nn with maximal probability.And in that case the sender can profitably deviate by selecting BQ with maximal probability.A more intricate construction is needed.We illustrate this construction below.3.1Perturbations that can be Stabilized without Voluntary ContributionLet εs =(εs BB ,εs BQ ,εs QB ,εs QQ )be a vector of trembles for the sender,i.e.a vector satisfying εs i >0for all i ∈S s ≡{BB,BQ,QB,QQ }and i ∈S s εs i <1.Similarly,let εr =(εr dd ,εr dn ,εr nd ,εr nn )be a vector of trembles for the sender,i.e.a vector satisfying εr j >0for all j ∈S r ≡{dd,dn,nd,nn }and j ∈S r εr j <1.Then the (εs ,εr )perturbed game is the (infinite)normal form game derived from the original game in which the sender is restrictedto using completely mixed strategies σs =(σs BB ,σs BQ ,σs QB ,σs QQ )satisfying σs i ≥εs i for all i ∈S s ,and the receiver is restricted to using completely mixed strategies satisfying σr =(σr dd ,σr dn ,σr nd ,σr nn )satisfying σr j ≥εr j for all j ∈S r .We start by a simple observation :in any Nash equilibrium of any sufficiently smallperturbed game near a component of equilibria of the unperturbed game,any strategy which is not a weak best response to any equilibrium in the component must be used with minimal probability.Thus in any such Nash equilibrium near the component in which the sender uses BB with probability one we must haveσs QB =εs QB ,σs QQ =εs QQ ,σr dd =εr dd ,σr dn =εr dn .First,we consider perturbations for which there are equilibria to the perturbed game near the component in which there are no voluntary contributions.That is,we consider equilibriain which σs BQ =εs BQ .For such equilibria to exist,the sender’s expected payofffrom using the strategy BB must exceed his payofffrom using BQ .The expected payofffrom using BB equals 910(εr dd +εr dn )+2910(1−εr dd −εr dn )for any allowed equilibrium strategy of the receiver.Meanwhile,the expected payofffrom using BQ equals εr dd +65εr dn +145σr nd +3σr nn ,where σr nd +σr nn =1−εr dd −εr dn .Substituting in the constraint yields an expected payofffrom using BQ equal to εr dd +65εr dn +145(1−εr dd −εr dn −σr nn )+3σr nn ,where εr nn ≤σr nn ≤1−εr dd −εr dn −εr nd .Sincethis payoffis increasing in σr nn ,BB will be an optimal strategy if and only if its expected payoffexcceds the expected payofffrom BQ when σr nn =εr nn .Thus it is necessary that 910(εr dd +εr dn )+2910(1−εr dd −εr dn )≥εr dd +65εr dn +145(1−εr dd −εr dn −εr nn )+3εr nn .Straightforwardalgebra reduces this inequality to12−εr dd −εr nn +27εr dn ≥0which always holds if the perturbation is sufficiently small.The second condition for such equilibria to exist,is that the receiver’s expected payofffrom using the strategy nd must exceed her expected payofffrom the strategy nn .The receiver’s expected payofffrom using the strategy nd equals 110εs BQ −910εs QB −45εs QQ .Meanwhile,her expected payofffrom the strategy nn equals zero.Hence we must haveεs BQ −9εs QB −8εs QQ ≥0.In particular,note that it is impossible for this inequality to be satisfied if 9εs QB ≥εs BQ ,as wepredicted above.We may summarize this discussion as follows:Lemma 2(No Voluntary Constributions)For sufficiently small perturbations,the (εs ,εr )perturbed game has a Nash equilibrium without voluntary contributions near the component of the original game in which the sender selects BB with probability one if and only if εs BQ −9εs QB −8εs QQ ≥0.One such equilibrium is given by σs =(1−εs BQ −εs QB −εs QQ ,εs BQ ,εs QB ,εs QQ )and σr =(εr dd ,εr dn ,1−εr dd −εr dn −εr nn ,εr nn ).3.2Equilibria with Voluntary ContributionsWhen εs BQ −9εs QB −8εs QQ <0only Nash equilibria with voluntary contribution can be nearthe component of the original game in which the sender selects BB with probability one.Here we document the construction of such equilibria.For an equilibrium with voluntary contribution to exist,it must be the case that in the perturbed game the sender’s expected payofffrom using the strategy BB equals his expected payofffrom using the strategy BQ .Since for sufficiently small perturbations in any equi-librium near the component the receiver mus use the strategies dd and dn with minimal probability,the sender’s expected payoffform BB equals 910(εr dd +εr dn )+2910(1−εr dd −εr dn ).Meanwhile,as we computed above,his expected payofffrom BQ equals εr dd +65εr dn +145(1−εr dd −εr dn −σr nn )+3σr nn ,where σr nn must satisfy the inequalities εr nn ≤σr nn ≤1−εr dd −εr dn −εr nd .We showed above that at σr nn =εr nn the inequality 910(εr dd +εr dn )+2910(1−εr dd −εr dn )>εr dd +65εr dn +145(1−εr dd −εr dn −σr nn )+3σr nn holds when the perturbations are sufficiently small.Meanwhile,at σr nn ≤1−εr dd −εr dn −εr nd ,the inequality 910(εr dd +εr dn )+2910(1−εr dd −εr dn )<εr dd +65εr dn +145(1−εr dd −εr dn −σr nn )+3σr nn holds when the perturbations are sufficiently small.Thus there exists a unique value of σr nn ∈[εr nn ,1−εr dd −εr dn −εr nd ]such that the sender is indifferent between using BB and using BQ .This value can be calculated to be:σr nn =12(1−2εr dd +54εr dn )Note in particular that σr nn →12as the perturbation vanishes,i.e.we pick up the weak Nash equilibrium in the component!This is because it is the only equilibrium in the component ofthe unperturbed game for which the sender is indifferent between sending BB and BQ .The second condition for such an equilibrium to exist is that the receiver is indifferent between sending nd and nn in the Nash equilibrium of the perturbed game.Since the receiver’s expected payofffrom using the strategy nd equals 110σs BQ −910εs QB −45εs QQ ,and her expected payofffrom using nn equals zero,this requires thatσs BQ =9εs QB +4εs QQ .We summarize this discussion as follows:Lemma 3(Voluntary Constributions)For sufficiently small perturbations,the (εs ,εr )per-turbed game has a Nash equilibrium with voluntary contributions near the component of theoriginal game in which the sender selects BB with probability when εs BQ −9εs QB −8εs QQ <0.This equilibrium is given by σs =(1−10εs QB −5εs QQ ,9εs QB +4εs QQ ,εs QB ,εs QQ )and σr =(εr dd ,εr dn ,12−28εr dn ,12−εr dd +27εr dn ).。
第三讲 纳什均衡：案例分析
Nash equilibrium： illustrations
一、Cournot’s model of oligopoly：
duopoly with constant unit cost and linear inverse demand function
Players：two firms； Actions：Each firm’s set of actions is the set of its possible outputs（nonnegative numbers）； Preferences：Each firm’s preferences are represented by its profit.
基本假定
Constant unit cost：
Linear demand function：
Assume：
pifp D( p) 0ifp
Profit functions
Best response functions
Best response functions
j j
Conclusion
Revenue equivalent theorem
Firm 1’s best response function：
Firm 2’s best response function：
1 ( c q2 )ifq1 c b2 (q1 ) 2 0ifq1 c
Best response functions
Conclusion
二、Bertrand’s model of oligopoly：
duopoly with constant unit and linear demand function
二硫化四乙基秋兰姆液相方法
二硫化四乙基秋兰姆液相方法
二硫化四乙基秋兰姆，也被称为促进剂TETD，是一种有机化合物，化学式为C10H20N2S4。
这种化合物具有白色或淡黄色结晶性固体，熔点为70-74℃，沸点为117℃，相对密度为1.27。
二硫化四乙基秋兰姆是一种橡胶硫化促进剂，属于秋兰姆类橡胶硫化促进剂。
这种促进剂焦烧快，硫化速度也快，属于超速硫化促进剂。
硫化平坦性虽小，但能得到抗张强度、定伸强度和硬度高的硫化橡胶。
因此，它可用作二烯类橡胶的助硫化促进剂或无硫磺硫化剂，也可作为低不饱和橡胶的硫化促进剂。
此外，二硫化四乙基秋兰姆还有其他用途。
例如，它可以作为天然橡胶、丁苯橡胶、丁腈橡胶、丁基橡胶、顺丁橡胶及胶乳的超促进剂和硫化剂，还可作为杀菌剂和杀虫剂使用。
在医疗领域，它也常被用作抗酒精中毒的药物。
然而，二硫化四乙基秋兰姆也具有一定的危险性。
根据世界卫生组织国际癌症研究机构的致癌物清单初步整理参考，该物质被归类为3类致癌物。
此外，应避免直接接触这种化合物，以防止可能对皮肤造成的刺激和伤害。
二硫化四乙基秋兰姆是一种重要的化学物质，在多个领域都有广泛的应用。
然而，由于其潜在的危险性，应谨慎处理和使用这种物质。
如需更多关于二硫化四乙基秋兰姆液相
方法的详细信息，建议咨询化学领域专业人士或查阅相关文献资料。
博弈论1：纯策略
如何制定规则来约束
• 世界上有两种符合集体利益的正义： 实质正义与规则正义。实质正义必须通过 规则正义来实现，否则将被虚置。因此规 则正义最重要。
• 分蛋糕或分粥故事： 20个人的小集体， 每天吃粥。如果所有人抡吃，导致纳什悖 论：每人早早来等粥，成本极高，对谁都 不利。
• 纳什均衡：在完全信息静态 博弈中各方都实现个利益最 大化的策略组合状态
• 纯策略：假定每个参加者每 次只选择一个策略进行博弈
• 混合策略：每个参加者每次 按照各种几率选择不同策略 进行博弈，如监督博弈
二、博弈案例1：占优策略与劣策略
（dominant strategy and dominated strategy) 课本上译为“支配性策略”与“被支配策略”
完全信息静态博弈游戏各方在完全知道博弈过程的三大要素的情况下同时选择策略完全信息动态博弈走象棋围棋游戏各方的完全知识博弈过程的三大要素的情况下相继作出策略选择不完全信息动态博弈国际市场竞争10矩阵的每个位置显示每个参加者的每个策略组合此位置即一博弈格局每个博弈格局上列出每个参加者的利益函数于是得到payoffmatrix例如
• Since then game theory has be taken as the Foundation for Understanding Complex Economic Issues
一、博弈论的基本理论框架
1/博弈论的理论意义与特征
▪ 古典经济学：根据资源稀缺性配置资源 的科学，价格是资源稀缺性的信号，由 此对经济行为进行成本收益分析，MR＝ MC实现利润最大化 忽视了他人决策对彼此经济利益的影响
• 制度经济学：
科斯：对人之间的交易过程的成本收益分析 诺斯：对交易过程的制度环境进行分析，包括 制度约束与制度变迁与交易过程的关系
邻苯二甲醛缩乙二胺对称双希夫碱及其金属配合物的合成
邻苯二甲醛缩乙二胺对称双希夫碱及其金属配合物的合成作者：张漩卓一来源：《知音励志·教育版》2017年第05期摘要对称双希夫碱在医药、催化、分析等多个领域有重要应用。
因其结构中有较多的配位点，易与金属离子形成金属配合物。
本文合成了一种新型邻苯二甲醛双缩乙二胺配体L1，并将其与过渡金属盐配合得到金属配合物。
表征了配体L1和金属配合物的结构，并设计反应方案，优化了实验条件，验证了配体L1及其金属化合物的物理、化学性质，为进一步研究邻苯二甲醛类希夫碱提供了实验依据。
【关键词】邻苯二甲醛；乙二胺；双希夫碱；金属配合物1 引言希夫碱（Schiff-base）是醛或酮分子中的羰基与伯胺分子中的氨基发生亲核加成，脱水而成的一类含有C=N双键的化合物。
近年来希夫碱及其配合物广泛应用于光致变色、催化、医药、分析检测等众多领域。
希夫碱种类繁多，有单希夫碱、双希夫碱、大环类希夫碱等多种类型。
对称双希夫碱因其结构的特殊性使其在配位化学中占有重要的地位。
对称希夫碱具有多个配位点，能与多种过渡金属盐配位，较易与金属离子形成配合物。
乙二胺是常见的试剂，广泛用于有机合成及制药工业。
同时，乙二胺可作分析试剂、有机试剂等，也是蛋白质和纤维蛋白质的溶剂，目前对水杨醛缩乙二胺希夫碱的合成研究较多。
本文采用邻苯二甲醛双缩乙二胺得到一种新型双希夫碱，该双希夫碱结构不仅对称，还有单双键交替共轭大环结构，以及氮原子配位点。
实验表明，邻苯二甲醛缩乙二胺双希夫碱容易与过渡金属配合得到结构和性质稳定的金属配合物。
本文首先合成了邻苯二甲醛缩乙二胺配体L1，将配体L1与Cu（Ⅱ）、Zn（Ⅱ）盐进行配合，得到2种过渡金属配合物；并且设计不同实验方案，优化了反应条件，并用红外对其进行了表征分析。
对于了邻苯二甲醛缩乙二胺配体L1的研究，为进一步研究邻苯二甲醛类希夫碱提供一些基础理论数据。
2 实验步骤2.1 实验试剂及设备过渡金属盐有Cu（NO3）2·3H2O、Zn（CH3COO）2·2H2O，甲醇，无水乙醇，乙醚，邻苯二甲醛，乙二胺，薄层层析硅胶均为分析纯。
纳什均衡 求解 matlab代码
纳什均衡求解 matlab代码纳什均衡（Nash Equilibrium）是一种策略组合，其中每个参与者都选择最优策略，使得其他参与者的策略不变。
下面是一个简单的MATLAB代码示例，用于求解一个简单的两人零和博弈的纳什均衡。
```matlabfunction [strategy1, strategy2] = nash_equilibrium(payoff1, payoff2) % payoff1和payoff2是两个矩阵，分别表示两个参与者的收益。
% 矩阵的每个元素(i,j)表示当第i个参与者选择策略1，第j个参与者选择策略2时的收益。
n = size(payoff1, 1); % 策略空间的大小% 初始化每个参与者的策略向量strategy1 = zeros(1, n);strategy2 = zeros(1, n);% 遍历每个参与者的策略空间for i = 1:nmax_payoff = -inf; % 初始化最大收益为负无穷大for j = 1:n% 计算当第i个参与者选择策略i时，第j个参与者的最大收益 max_payoff = max(max_payoff, max(payoff2(j, :)));end% 如果第i个参与者的最大收益大于0，则选择该策略if max_payoff > 0strategy1(i) = 1; % 选择策略1endendfor j = 1:nmax_payoff = -inf; % 初始化最大收益为负无穷大for i = 1:n% 计算当第j个参与者选择策略j时，第i个参与者的最大收益 max_payoff = max(max_payoff, max(payoff1(i, :)));end% 如果第j个参与者的最大收益大于0，则选择该策略if max_payoff > 0strategy2(j) = 1; % 选择策略2endendend```这个函数接受两个参数：`payoff1`和`payoff2`，分别表示两个参与者的收益矩阵。