Let us now start proving the Gibbard-Sutherbrook theorem. As we have mentioned in our previous module, it says that if you have three or more alternatives, then every social choice function that is onto and strategy proof implies and is implied by the fact that this social choice function is dictatorial. And we have also seen a corollary in the previous module it says that under strategy proofness, all these notions of parity efficiency, unanimity and ontoness are the same. So this is this corollary here. And therefore, you can restate the Gibbard-Sutherbrook theorem replacing this ontoness with parity efficiency or unanimity. So in some texts you will find similar definitions and they are all the same. Now before moving on to the proof, let us look at some of the points that we can note from this statement which is implied and which are not implied by Gibbard-Sutherbrook theorem. So the first observation that we can make is that this number of alternatives needs to be at least three or more. If you had two alternatives, then we know that there exists some certain kind of onto and strategy proof mechanisms, strategy proof social choice functions for instance, plurality or any kind of scoring rule-based mechanisms that we have discussed before. They are all non-dictatorial social choice functions yet they are onto and strategy proof. So that is one interesting point to note. The Gibbard-Sutherbrook result does not hold when you have two alternatives. The second observation that we are going to make is that this domain of this social choice function is the entire set of all possible strict preferences. So script P essentially denotes that it is the permutation of all possible alternatives. So if you have M-ordinates script P lists all possible M-factorial permutations of these alternatives which can potentially be the preferences of these agents. So this actually gives so one way to visualize or build an intuition about this result is that the domain of the social choice function is too rich and the voter has many options to misreport and that essentially gives rise to such kind of an impossibility result. Later on we will see that if your domain was limited that is not all possible strict preferences over these alternatives were feasible in your social choice function then GSTRM may not hold and we will see examples of that. The third observation is about indifference. We have so far discussed about strict preferences. You can actually generalize or include more things about indifference in preferences. So you can either expand P because indifference could actually lead to a superset of script P that we have discussed. If that happens then you already know the type of constructions that we are going to do already exist there. So in that case GSTRM will continue to hold but if you have removed some of these preferences, some of the strict preferences that we are going to consider then possibly GSTRM won't hold. So it all depends on the domain. So just remember that our social choice function was a mapping script P to the power n that means all the players have these the preferences which lives in script P and we are picking one alternative. So this is the social choice setup that we are picking one alternative out of from this preference profile. If you have this domain to be very much restricted which we will discuss very soon then GSTRM can be bypassed and we will see such examples. And the cardinalization, the last point is about cardinalization. Here we are only talking about the ordinal preferences of these agents but if there are cardinal preferences which also includes note that cardinalization is just another way of representing or maybe a richer way of representing the preferences. If the numbers are equal, the cardinal numbers are equal then you can assume that they are indifferent. If they are unequal then you can have a complete ordering between those alternatives. If by cardinalization you are not ruling out all the possible P's all the strict preferences that we are going to use in the proof then GSTRM will continue to hold. So in some sense we can say that cardinalization that admits all those strict preferences for all these agents then in that domain also we can say that an equivalent GSTRM will hold. So it does not really matter on cardinalization or ordinalization, it just matters that how these alternatives are positioned, what are the possible ways you can order them. Okay so with that let us start the proof and for this proof we are going to follow a very direct approach and this is due to SEN 2001. So at the end I will give the complete reference for this for this paper and what we are going to do in this module because the proof is essentially a little long and quite laborious to follow. So therefore we will just look at the simpler setting of two players, two voters and then apply induction. So this proof essentially uses proofs it for two agents and then applies induction. We will not discuss this induction step because that will be too involved but the two agent case is quite easy to follow and that is what we are going to do in this module to build the intuition. So in order to prove it let us go in steps. So the first sub result the lemma is that if we have three or three or more alternatives and you have two agents one and two and the fact that F is on point strategy proof then for every preference profile what we can conclude is that this outcome will either be the top most alternative of player one or the top most alternative of player two. So it cannot be any other alternative apart from these two things. So this is what it means. If it happens that these two alternatives are the same then we already know and this is something that we are going to use over and over again even though we have stated on-to-ness and strategy proofness in the statement of the of this theorem we all also know from the previous corollary that it is equivalent to say this is Pareto efficient or unanimous. That is because under strategy proofness all these results are the same and we are going to in our proof we are going to use all of these results interchangeably. So we have already made use of one of those results. So here we are assuming that F is unanimous as well. So if that is unanimous then and this two alternatives are one and the same then the theorem holds the lemma holds because then unanimity will actually imply that it is it should be the same outcome the outcome of the social choice function should be the should be the same because the both the players are agreeing on the on their top alternatives. So therefore let us assume for the general case these two alternatives are not the same. The top most alternative of one is A and the top most alternative of B of two is B. Now let us assume for contradiction that the outcome the final outcome is neither A nor B. So we are actually negating this statement. So it is not neither A nor B and notice that this requires you to use three alternatives at least three alternatives. And now we are going to look at a as before we are going to look at certain transitions. So we are going to construct certain preference profiles. We are given this particular preference profile P1 and P2 and we know that the outcome here is C by our assumption which is neither the top alternative of one neither the top alternative of two. So now what we are going to do so we are looking at this first alternative here so the top most alternative of player one and we are putting that A in this in this way. Similarly we are also constructing another preference profile where the the second so B that is the top alternative of player two has been pushed to the top to the second top and the other alternative remains the other preference profile remains the same. Similarly here it was the first place preference profile was kept the same and for the second place preference so A was maybe somewhere down below here it has been pushed to the second top. Now what can we say so we can conclude for this particular preference profile here that the outcome should be either A or B and why is that that is because of the Pareto efficiency and here we are actually using using this fact that F is also Pareto efficient so that means it will not pick something which is Pareto dominated and what is Pareto dominated so anything except B and A is Pareto dominated by both these players with respect to A so A Pareto dominates all other alternatives here so therefore none of those can be an outcome in this preference profile if because F is Pareto efficient so it has the only option of of either A or B that that should be the outcome. Now let us notice that if this outcome was B for instance then what player two gets if you look at it so here the here the outcome is B and here the outcome is C which is strictly worse than B for player two so and because player one's preference is not changing then player two will misreport from P2 to P2 prime and that will be a beneficial deviation so since A phase strategy proof as well we cannot have this alternative so B is ruled out the only option therefore is under this alternative is A so it should have A as its outcome so similar arguments you can apply here as well and look at the transition from here to here and apply the same argument for player one and you can see that here the outcome is definitely going to be B it cannot have A so B will be the outcome in this case it's a very symmetric argument now we are in a nice position to make a transition so you notice that this is P2 prime and this preference profile is nothing but collecting together this P1 prime and this P2 prime for these two pairs and what we can observe is if you trans if you transition from from this preference profile to here what is happening is that for I mean the position of B is remaining the same in this case so if you apply the equivalent definition of strategy proofness which is monotonicity then you can conclude that here the outcome would have been B but if you apply the same argument from here to here of course so what is happening is that here the the relative position of B was same here the relative position of A remains same right so here also A's relative position does not change you can again apply monotonicity and you can conclude that this outcome will be A and since A is not equal to B this is a contradiction I mean you arrive at a contradiction from one transition you are getting an alternative which is A the other transition is telling the alternative will be the outcome will be B and that cannot happen so what we have assumed that here the outcome is C which is neither A nor B then that is that is not true so we have actually proved this this lemma all right so with that lemma in our hand and now we can actually go and prove the give us other word result for two alternatives so what do we have here so we have for two alternatives we have at least three so two agents we have at least three alternatives and we know that this is an ongoing strategy proof so strategy proof means that this is equivalent to monotonicity we have used it already and on to this means that it can also be better efficient or unanimous so just to keep in mind that this this is what it means and let us assume that the top most alternative of player one and player two are not the same so this is A and this is B and let us look at a different preference profile of both these players where the top most alternative is C for player one and the top most alternative is D for player two and there are two conclusions that if the outcome in the first preference profile is the top alternative of player one that is it is A then it should continue to be the top alternative of player one so that will be C and similarly if the top alternative is of B of player two which is B then in the new preference profile the top alternative will be the same the top alternative of the same player which is D and we will we will just prove this one part because the the proof of the second part will be quite symmetric so without loss of generality we will just prove this so just to mention that this actually proves the dictatorship for two players and and then later on this this direct proof that we have mentioned actually uses an induction for more than two players now one thing that we have already observed and we have also mentioned earlier that if C and D are same so in this case we are not putting any constraints about C and D so if C and D are same then unanimity implies that this this conclusion will be true so the lemma is trivial in that case so for all all the non-trivial cases we will assume that C is not equal to D so that is what we are going to assume now let us list down all possible exhaustive cases and there are six cases where C is not equal to D and and since we are considering only only this part so the first implication of of this lemma we are going to assume so even even when we say case one you can also create an equivalent statement which involves the the second agent's case so it is it can be the same argument can be interchangeably used for player one and player two to show that if you flip the names so here this is player one this is player two but if you flip their identities make player one player two and player two player one then the similar argument will follow and we can get the similar conclusion so we are not going to prove that part in the in this proof but whenever we are using some of these cases some of these later cases here and we are referring to case one we are going to assume that that that this other counterpart is also true because sometimes we will need those cases to be used okay so so let us first list down all the cases so here the so the first case is where the C and B is equal to A and B then you keep the second alternative is fixed and the first alternative is neither in or B then these the other cases where it is the first alternative of this player is remaining the same and the second or the second player's top alternative is neither B not equal to B and similarly you can you can list out all possible cases so just check that this actually covers all the possible cases that can potentially happen when C is not equal to D and we have listed it in this way because that this will give us some advantage in proving we will use some of this the proofs of some of these cases in the later cases as well okay so let us go over these cases one by one so the first case is where C is equal to A and D is equal to B so when C is equal to A and D is equal to B so this is the the modified profile so this is the original profile and we are going to assume that the outcome here was A so the outcome here is A now we are we are we are supposed to say what is the out what is going to be the outcome here and C that is this the top most alternative of player one in this in this new profile is A and that for player two is B now we are going to construct a different preference profile and that is very carefully chosen what we have done is we have pushed B onto the second top position and similarly for player two A to the second top position now we already know by the previous lemma the lemma that we proved in the beginning of this module that the in this preference profile the outcome will be either A or B that is all that is something that we already know now our conclusion is that it should be the outcome will be A and now let us for contradiction contradiction of the current lemma let us assume that this outcome is B so that is even though the in the first preference profile the top most alternative of player one was chosen say in the second preference profile the top most alternative for player two was chosen so let us do that then let us see what is the contradiction then what you can say here is from if you go from this preference profile so from this preference profile to this preference profile that we have constructed we see that the A's relative position has weakly increased so A was somehow down below in this case and it has moved to the second top position because B was on the top in this previous one A's relative position has weakly improved so we can apply monotonicity onto it and because A was the outcome here A will continue to be the outcome according to monotonicity but now we can apply the same thing between these two preference profiles because here B was somewhere down below and B has been pushed to the second top position for player one using the same argument and same monotonicity we can conclude that the outcome will be B this case so that is essentially a contradiction so in one of these transitions we conclude that this is B and the other transition we conclude that is A which is not which is not possible so we cannot have this this condition B to be true so this has to be A so this this question mark is essentially A so we have approved case one in this case so let us look at case two where C is neither A nor B and the second alternative is B now we know that this C this is a different alternative neither A nor B and we also know that this top most alternative in this this two top most alternatives one of them will be the outcome and again as before we are going to assume for contradiction this is the outcome then again we we can construct a P1 hat and notice that P2 we are going to keep the same so this is the same as this P2 now what what did we do in this P1 hat we have actually we kept A somewhere here and we have pushed A onto the top now again if you apply the condition from P1 P2 right so P1 P2 to this condition that we already know that using this case one so what was case one so the top most alternatives for both these players are remaining identical maybe the position of the other alternatives are changing but their top most positions are remaining the same therefore using case one we can conclude because and this is we are using the same case one for player player two there we were we were proving it for player one but player two we have we have said that a similar argument will hold even for player two so we are using that so here the top most alternative of player B player two that is B was the outcome so therefore that will continue to be the outcome here that was that was our case one so because this using case one we have an outcome so here the outcome is going to be B now what one can observe is that here the outcome was A and for player one B was leaving somewhere below so because A was the so we know that C is neither A nor B A is in the second position so B will leave somewhere below it's lower in preference than both C and A for player one and the outcome is going to be B so player one might as well be tempted to misreport its preference when it is when its true preference profile is B one and B two in that world player one will misreport to B one because the outcome is going to be A which it strictly prefers over B which is the current outcome so we have actually constructed an example that this is actually manipulable this social choice function is actually manipulable which is a contradiction so therefore we cannot have this this condition to be true the outcome will certainly be C in this case that proves our case two as well so for case three we have to use this case two in a repeated manner so we know what was what was case three we have C which is neither A nor B and D which is not equal to B so we know that D this is not equal to B and C as before neither A nor B right so now we can construct this preference profile as as we have done in in in the previous cases so P1 hat P2 hat and their top most alternate is the C and B right so we are keeping the second player's top most alternative to be the same and the first player's top most alternative has moved to C which is neither A nor B now as before let us assume for contradiction that this outcome here even though the outcome here is A the outcome here is let's say D then if we look at this transition from P prime from this preference profile to this one then what we can see using case two so remember what was case two so case two was the the situation where the so player wants so one of the players top alternative does not change which is which is happening in the case of player one so its top most alternative is not changing the other agents top most alternative is changing and that player's top most alternative was the outcome here so if that holds so using case two we have seen that if you change so whichever players outcome was here and that the top most alternative was chosen as the outcome that players top most alternative will also be chosen here so therefore we can conclude from this transition that the outcome here will be B but similarly we can use the same case two between these two transition between the transition between these two alternatives preference profiles so here because the the top outcome was A which was the top alternative of player one then it should continue to be the top alternative of the of the same player so which is C now we know that C is not equal to B so therefore this is a contradiction and the rest of the proof is essentially trying to argue about such kind of proof by contradiction we are going to assume something which is which is contrary to the conclusion of this lemma and we'll construct certain preference profiles where we can show that that it leads to a contradiction so the the case four is when the the top alternative of player one is remaining the same both of them are A and the second alternative the top alternative of the second player is neither A nor B so it is moving to some D and now we can look at this the same different constructed preference profile the top alternative of player one and two are the same as the original ones so now we can apply from for this transition from here to here we can apply case one which will conclude that this is going to be A and from this this case to this case you can use case two because here the one of the player's top alternatives is remaining fixed the other alternative has actually changed and for contradiction we have assumed that this outcome is D therefore this will be equal to D and because D is not equal to A by assumption here then this is a contradiction so we have actually proved case one two three four so case four so case five is essentially the case where you have a kind of a flip so the top alternative of player two has become the top alternative of player one in this case and D is as before neither equal to A neither not equal to B and we can apply a very similar so we can actually construct a preference profile for for these two players so here this will be A so the top most alternative of player one is A and that for player two is D so this is the same as this D and this is the same as the original preferences preference of player one we don't really care about the rest of the alternatives so now if you look at this transition from P hat P prime to P hat then what we can observe again using case case two essentially this will be case two because one of the player's top alternative is not changing the other one's top alternative is changing and we have actually used the fact that so we are assuming that this is the this is the top alternative the outcome here so in that case in the new preference profile also the outcome would be D and if you are looking at the transition from the original preference profile to this this last preference profile and then what we can observe is that this agent so this is coming from this case four so that the outcome was A and whatever we are changing here is essentially we are looking at that agent whose top alternative is not changing so essentially this will also be so the first case was also for case four so this was also a conclusion from the case four because here the the top most the outcome was the top most alternative of that agent whose top alternative is not changing the other agent's alternative is changing and that is exactly what we have proved in the in case four and then by using this transition from P1 P2 to P P hat we can conclude that this outcome is going to be A and that is because these two are not the same we actually hit a contradiction here and the final case that is case six is the following when you have a C to be equal to B and D to be equal to A so the top alternatives of these two players have actually flipped so what what can we do in this context is that so since the outcome here is the top alternative of player one and let us assume for contradiction that the top alternative the outcome here is the top alternative of player two and let us assume another alternative which is neither N or B we can pick such an alternative because there are three or more alternatives in this in the assumption itself so now we have created some new preference profiles as you can see so you have pushed that so X was somewhere below so we have constructed a preference profile where B is on the top and that is this that is the same B as the top alternative of player two in the original preference and we have pushed X to the second highest position and in the second one we are keeping this preference so here all these preferences are the same for player two and we are pushing X to the top so in one case X is in the second top position and in the last one it is on the top most position now let us look at the transitions and we will be using all the different cases that we have already proved here so let us look at this transition from P prime to to this transition this preference profile transition we already know that here the top alternative is D and because the top most alternatives are not changing so here B and A and here also they are B and A so we can use the first case that is case one that the top alternatives are not changing and therefore if the outcome was D in this case which was which was the top alternative of player two then it will continue to be the top alternative of player two so we will have A here so that is the that is the conclusion similarly if you look at the transition from this preference profile to the last preference profile here what is happening there we can we can see that this is this is falling under the condition of case three and because A was the outcome here which was the top alternative of player one and that should be the outcome here so the outcome here will be X now if you if you look at these two alternatives here so you can already begin to see that for player one there is a profitable deviation in in these two cases so suppose it was currently looking at this preference profile so player once true preference profile was P1 at and in that case the A was both below B and X because X is not neither A nor B and that is going to be the outcome here right but if if it transitions to and misreports its preference to P1 tilde then at least it knows that the outcome will be X which is more preferred than A in in its original preference profile so player one finds it beneficial to deviate and misreport its preference and which is going to lead to the fact that this F is not a strategy proof mechanism so that cannot happen so therefore what we have assumed this is not true so it must be the case that the outcome is C which is equal to P in this case okay so that actually concludes the proof of all the exhaustive cases and we have proved the keyboard side of the board result for for two agents so this is the the main intuition how it is it is done the the actual proof with the with more than two agents essentially uses induction on this on this number of agents and I am not going to prove that if you are interested you can look at this the this paper which is titled a direct proof of GS theorem and that will give you the complete details it's just you'll have to do a lot more accounting than than what we have done in this case


A lot of things went into it. I think first and foremost, my wife, Emily, and I really love it here. We love the city, we love the energy, we love Yankee Stadium, we love going to Yankee Stadium and playing at that park every day. So that was really the most important thing. And then the team, you look at the team and the New York Yankees are a team that this organization has franchised that has consistently put winning products on the field.


Today, I have the pleasure of speaking with Rob Cook from the CSE or the Canadian Securities Exchange. How are you this morning? I'm just great. Thanks, Tracy. So hockey stick performance in the market, let's talk about the CSE. You have outperformed pretty much everybody. Can you give us some of the highlights, Rob? Well, the highlights have gone on for many years, but it's been the last couple of years that we've seen the hockey stick that we were always hoping would come about. And we had a lot of confidence that it would, because we just win over one person at a time to our exchange, to our market model, to the way that we do business. And it's been very gratifying to see the recent results. The number of companies that are listing, the amount of money that they're raising in financing, it's just been tremendous. So let's talk to the numbers, and many investors out there are a little more short-term memory and they're looking forward. So with the CSE, your market cap has increased from X to X in the last year. It's increased from a little over $4 billion to close to $14 billion. In addition to that, how many new companies have listed in the last year? I could tell you on 12 months, but I do know on nine months, I think we're up 126 companies, which is going to already exceed our largest year before. I've always perceived it to be a fallacy that your volume of trading is lower than other exchanges. Can you correct that fallacy? Yes, quite easily. One of the things that we look at is a turnover ratio, which measures the value of the trading divided by the market cap of the stocks. It's about the fact that our market cap has been rising. So is the trading volume. So we routinely, we just look at it on a monthly basis. Using that measure, the liquidity on the CSE has exceeded that of the TSX and the venture exchange for well over two years. We don't tell people that that means that we have a more liquid market. We tell people that that means we have a no less liquid market. What we have today, and for the recent past, is we have stocks that investors are really interested in trading. So it's extremely active. I was recently speaking at an investment conference, and I was stating that one of the areas we are following are the Americans that have started to trade in Canada and list specifically on the CSE. Are they all cannabis stocks, or are you actually getting other types of stories right now, Rob? We're getting a few other stories. They're because of the increased interest in the cannabis sector, and the fact that we have had several from the United States that have come to us already. It's broadened the awareness, but we've always had a few. When we launched our exchange in 2003, we only had three companies, and one of those was America. That company raised about $100 million while it was listed on the CSE. So there's always been some interest from the United States. Laterally, it's been the cannabis sector. We're the only stock exchange in North America that will list a company that has cannabis assets in the United States. We have approximately 40 of them at the moment. Most of those are Canadian companies, however, that have gone south, set up operations or acquired interests mostly in the medical marijuana space, and we've also seen more and more large cap companies come north of the border, not just to get a listing, but to get financed, because it's a very growing industry, and it's a very competitive industry, and they want to raise capital so that they can grow fast. Of course, we both know that with the public markets, these companies are trying to build shareholder value. They are trying to secure financings. Can you tell me what the competitive advantage is for listing on the CSE? Well, the competitive advantage has always been what our mission statement is, and that's reduce the cost of capital for emerging companies, and how do we do that? We charge low fees, and we have what we call a streamlined regulation program that allows the company to focus on their business, reduces the regulatory burden, without dispensing with rules. They have rules like other stock exchanges, but the companies have to honor those rules, but they don't have to come to us every time they have a transaction and ask us for approval. That enables them to move quicker and focus on their business. We call it a reduced friction cost, and the companies tell us that our low fees are nice, and they don't want us to change that, but they appreciate more the reduced friction cost. So Rob, as the market development head for the CSE or the Canadian Securities Exchange, what should we expect maybe in the next six months? Can you give us, share anything with us? What are you focused on? Well, we've recently, our CEO and our head of listings development have spent some time in Israel, and they've met with a lot of technology companies there. So we're seeing increasing interest. We already have applications on hand from I think three of them, and we're expecting several more. We have a new business development person who's currently in Singapore, and he's meeting with companies there. He's actually being accompanied by people from the OTC Markets Group in the United States because they have a couple of regulated boards called the QB and the QX, and any of the CSE companies that want to get traded in the United States can easily access that market, and that works out very well for them. So let me just reconfirm, you went from four billion to 14 billion in one year? Yes. So what should we expect this next year? Are you going to try and repeat that? Well, we don't have a crystal ball. That kind of growth, you can never promise. We do know that there are over 100 companies that have applied for listening that we have yet to complete the reviews, but we are working very fast on that, and we've staffed up, so that's been going very well. And some of those will be worth hundreds of millions of dollars, so the market cap could increase. Then again, it's a market, markets are volatile, prices drop, the market cap drops. So we don't know, but we do know that there's a growing number of companies coming that adds to our market cap, and I think there's a growing percentage that are going to have a market cap of well over $100 million. For our investor intel audience, our theme here is we focus on what works. Basically the CFC is working. So thank you for joining us today, Rob. It's always a pleasure. Right. Thank you very much.


What's up guys, my name is Michael and welcome to my YouTube channel today. We are going to go over this problem called what's next. Basically you're given three numbers a one a two and a three and you just have to determine whether is an arithmetic progression or a geometric progression. And it's guaranteed that's one or the other.  So that's, that's when you break through the loop. Okay. And, um, yeah. And then basically that's it. Just print whether it's an arithmetic progression or geometric progression. And then the next, uh, next thing you're printing is basically the, the next member of the sequence. So the next, the next value of the sequence. So in this, in this case, we have four, seven, 10. We know this is an arithmetic progression based on the definition of arithmetic progressions and, uh, because we're adding three every time. So four plus three is seven, seven plus three is 10, 10 plus three is 13. And that's why the output is 13. In the other one, we have two, six and 10, uh, two, six and 18. We know this is a geometric progression because we're multiplying three every time. Two times three is six, six times three is 18, 18 times three is 54. Right. And that's why the next pro next value is 54 because 18 times three is 54. Okay. And that's basically the gist of the problem. You just have to determine if it's an arithmetic progression or not. So if we want to write this down, let's say I have a one, a two, a three, we just have to determine a four, right? And if there's, if it's an arithmetic progression, we're adding by some number X every single time. So this would be an arithmetic progression. A four will just be a three plus X. And if it's a geometric progression, you're multiplying by some number. You're multiplying by some X. So we would multiply by some X value and multiply by some X value. And then a four is multiplying is a three multiplied by an X value. So yeah, that's, that's basically the definition of arithmetic progression and geometric progression. And I'm going to explain the code. So guys, I have not actually done this problem before with, uh, in C plus plus. So I'll do it with you guys. Okay. So we know in arithmetic progression is if they, each of the sums, if you add a value to the first value, it equals the second one. And if you add the same value to the second one equals a third value. So there are differences between them that have to be the same. So if a one, um, eight, if a two minus a one is equal to a three minus a two. Yeah. If a two minus a one is equal to a three minus a two, then we're going to see out arithmetic progression and then a space. And then we're, we're going to do is we are going to print the sum, the difference. So the difference is a three minus a two, right? And we're just going to take this value and then add it with the last value, which is a three and then we print in. Um, so if it's not an arithmetic progression, that means it's a geometric progression. So we're going to see out GP. Else see out GP, we're going to print out a new space. And then we're going to do is we're going to take the third value, right? The third value divided by the second value. So that'll give us the common ratio and then multiply by the third value to get our fourth value. So third value divided by second value. And then we're going to multiply by the third value again to get our. So this gives us our common ratio and this is going to multi, uh, give us the next value. And it's the same thing as arithmetic progression. We get the common difference, right? And then we add by the next value. And then, yeah, because they said the last value, the last, all three numbers are all zero in the end. That's why I read in the three numbers first and while they are not all equal to zero, then I do this and then I read in the three numbers again, so then by the next iteration, it would check if they are all equal to zero or not. Okay. And yeah, let's see if this works. I'm not a hundred percent sure because I did this in Java and I didn't do this in C plus plus before. So yes, uh, let's see. GCC, let's see if this works. It could be a problem with absolute value. Hope there isn't gives me a wrong answer. Okay. What, what is the problem? Um, let's see. What is the problem? Uh, a two minus a one is equal to a three minus a two. Um, let me check my Java solution real quick. It should be right. Hold up. Let me just find my previous Java solution because this, uh, a three minus a two is equal to a two minus a one. Yeah, that is right. Um, a three plus a two minus, I think that's right. Wait a three plus a two, a two minus a one. This is right. If a two minus, let's put parentheses a two minus a one is equal to a three minus a two, a three plus a three minus a two, this should be right. I mean, we could do a two minus a one. It's still the same thing because they should have a common, common number. So let's, let's try it again. But yeah, that was my solution before. So I don't know why don't know why it's not working now. Let's see. Let's see if it works now. Still wrong answer. Okay. Let's see. Let's see. Let's see. What is the problem? A three minus a two plus. Are you sure this is right? Okay. Let's, let's, let's go through this again. Four seven 10 should give us, let's print this out. Okay. Let's do this real quick. Four seven 10 AP 13. Yeah. That's right. Two six 18 GP 54. It's right. Zero zero zero. Yep. Okay. So that's really strange why my thing is not working. Five zero zero should be zero. Oh, wait, the whoops. I did not do five zero zero. Hold up. Let's see. Ah, okay. So this loop should actually be an or sign. Yeah. Yeah. So if you, I should have put a parentheses on not that. So then yeah, now it should work. So the reason why you have to put or is because if they're not all zeros, then yeah, and that's a problem, but should work now. Yeah. Now it got accepted. Okay. Yeah. So yeah, if they're not, so basically the reason why this didn't work is they have to all be zeros and then it would break. So I could have done this way while they're, let's say they're all equal to zero. Instead of putting or kind of this way, put parentheses on all of them and then put a not so while they're not all equal to zero. So this would actually distribute the, then the not equal to zero into all of these and it would give the same answer. So yeah, that's why the or you had to put an or if you're going to distribute the not to each of these. But yeah, this is basically the solution to the problem. Ray coms subscribe. I'll check you guys later. This is a pretty basic problem, but yeah, Ray coms subscribe. I'll check you guys later. Peace.


Hi everyone, Joe for Jaspeyscasebreaks.com coming at you with our last filler pack to unlock that noir break. Now there's a few things happening here, there's going to be three different dice rolls in this break. The first dice roll, we pulled six spots out of this pack to give them away, so we only sold 24 spots. So the first dice roll will be giving away those six spots, the 24 people, 24 spots. The second dice roll will be for the break itself, we'll do names and teams and we'll rip open this pack. And the third and final dice roll will be randomizing all 30 spots and then seeing who's going to get into that noir break which is going to be in a different video. So with all that being said, let's get on with it. Here's the first dice roll. We're going to give away some spots within this break. Three and a six, so top six after nine. One. Two. Three. Four. Five. Five. Six. Seven. Eight. And ninth and final time. So after nine, top six, Cal, Robert, Anthony, Sean, Robert, and Kym, you get an extra spot. So you have slightly better odds to maybe get those noir spots.  You can still win those noir spots, which is the whole point of this break here. All right, so we'll put little rooftops next to your name just so you know that you won, those are your winning spots right there. So now let's do the break itself. This is the second dice roll. There's all the teams, all 30 teams are in. Let's roll it and randomize this. Three and a six. Again, nine times for names and teams. One. Two. Three. Four. Six. Seven. Eight. And ninth and final time. Three and a six. Nine times for the teams. One. Two. Three. Four. Five. Six. Seven. Eight. And ninth and final time. After nine, we've got the mavericks down to the pelicans. All right, so here's the first half of the list right here. And there's the second half of the list right there. And no trade windows. Let's just go ahead and print this out, rip that pack. There'll be some surprises in there. Let's see what we got. And here is the final printout, hot off the presses. Thanks everybody for getting in, taking the risk on this break. Appreciate it. John Collins Silver, Dennis Rodman Green, Keon Martin, Wendell Carter Jr., and something in the bonus pack here, Red, White and Blue, Julius Randall, Jared Allen, and Avery Bradley. Nothing too crazy there. Let's give away those noir spots. That's the crazy part. From Sean down to Robert and everybody in between, third and final dice roll. All the names in. Let's roll it. Let's randomize it. Two and a three. Five times. It's going to be top six after five. Good luck. One. Two. Three. Four. And fifth and final time after five. First off. It's only the top six. So from seven on down, sad times, but I appreciate everybody taking a shot at this. Thank you. Much happier times for the top six because you're in the next noir break coming up in a separate video. Kym's happy. So is Cal, Anthony, Robert, Brett, and Gregory. Congrats to the top six. You're in that noir break. We'll see you for that next video. Jaspyscasebreaks.com.


When you are performing a header it's extremely important that you keep your eyes on the ball. Most players close their eyes and hope the ball hits them, rather than focusing on exactly where they want to meet the ball with their head. It's near impossible to keep your eyes open at the point of contact, but do your best to focus on the ball until it meets your head. Make contact with your forehead, not the top of your head and definitely not on your nose. You want the ball to come flush off the center of your forehead. To produce power, you need to engage your whole body. An athletic stance and slight bend in your knees will allow you to have more balance and generate power, but the real power comes from using your total upper body. Bend your body back and then forward through the ball. You are not just using your head and neck muscles to create power, you are using your whole upper body snapping back and forward through the ball. To produce accuracy you need to guide the ball with your follow-through. Push the ball towards your target with your head and upper body. This is another reason it's so important to keep your eyes open when performing headers. Jumping headers follow the same principles. Keep your eyes on the ball, make contact with your forehead, snap your body back and through the ball, push the ball towards your target with your follow-through. Jumping headers are all about timing. You want to attack the ball and meet it at your highest point. If your timing is off, you'll either catch the ball too low or too high and you won't be able to direct the ball where you want it to go. Focus on your footwork. You want to be moving towards the ball rather than jumping from a standstill. If you need to take one or two steps back then a few quick steps forward as you leap in the air and attack the ball. Learning to read the flight of the ball takes practice, but after a while you will understand when to start your jump. Just think about attacking the ball rather than letting it come to you and jumping forward if possible, meeting the ball in the air at your highest point. If you're performing a defensive header especially in your own defensive third, you just want to think about getting that ball as far and as wide as possible. You don't want to clear it into the center of the field if possible. You want to get it wide. If you can hit a teammate that's great, but really you're trying to clear your lines, get distance on the ball, get height on the ball, and make sure that you get it out of trouble so your team can get out the field. So if he's throwing me the ball, again, throw a little lower please. Hey, what's going on? This is Dylan Tubi of Progressive Soccer Training. I hope you enjoyed that video. This was actually a look into the new program I'm releasing, The Match Winner Method. Become a game changer, improve your performances, and maximize your match stats in 21 days. This program will be released on December 21st. Please subscribe to my channel for updates and related information.


Good afternoon everyone Okay We're gonna start So good afternoon everyone out. My name is Eli Kapilovsky. I I work for Melnox Product manager for the cloud market and today we're gonna talk about Melnox interconnect So Melnox is a leading provider for storage and server We've been around for 13 years now based off California and Yocanam Israel we're about 1200 employees Worldwide our revenue last year was 500 million and growing and Before I start with our integration solution for open stack I would like to give you a quick introduction of our portfolio and what we're doing in the space so Melnox is building their own Silicon adapter cards getway switches cables to connect the interconnect Software to accelerate application and software to manage the interconnect We support both internet and infini men technologies With a concept we call VPI virtual protocol interconnect. This concept is really unique It allows us to communicate either through internet or in Fini band with the same fabric with no Change of cables hardware software driver. We now can support both internet and infinity band So again a very unique capability with a very fast interconnect So we support up to 40 gigi internet today and up to 56 gigabit per second infinity band or as we call it FDR 14 data rate so this is the products that support our fast interconnect and We're gonna show you this slide shows you a little bit about our customers Melnox is becoming I will say the secret sauce of many many public cloud providers out there You can see that some of the names out there some known ones some less known but The message here is that if you want to create the public cloud Service and offer your users with with a more efficient Infrastructure as a service Melanus Melanox have the ingredients to support that infrastructure to make your infrastructure less complicated more efficient and at the end of the day provide your end user with a better performance and lower cost So what is our added value we provide a variety of advantages from the application to Storage access How we simplified and integrated solution with open stack or with other? cloud management In higher infrastructure efficiency, so just by using Melanox interconnect when you're using 40 e Internet adapter Or 56 gigabit per second infinity band you could support more virtual machines per a physical server You could off load the hypervisor CPU And you could create an unlimited scalable infrastructure That will cost less when you design your architecture So a few examples on how we accelerate application. We use a very unique technology called our DMA remote direct memory access This technology is really Fascinating it started from the infinity band wall With high performance computing and now it's expanded to other areas Such as cloud RDMA allows you to bypass the TCP overhead and to copy memory from one application to another and by that reduce latency increase Performance and bandwidth and reduce the pu overhead. So three examples starting from storage. We're able to Provide a much faster storage access Over six acts performance improvement when using RDMA on top of ice Kazi, for example We can provide a much faster migration If you compare it with a 10 gig solution, we're talking about Almost four acts faster migration time. In matter of fact, we have public cloud providers that are using this as one of the advantages to show that when you migrate or when you create a new VM the downtime is zero And it's very unique and it's only because of the The fast interconnect and thirdly you have the VM VM to VM latency performance And this is a very unique technology that we now supports all multiple hypervisors So VM to VM latency using Melonox interconnect is now two micro seconds Just to compare that with other solution out there We're talking about 40 micro seconds latency when using power virtualization mode if using Melonox SRRV technology you could get 20 acts faster Late or a faster communication between virtual machine one virtual machine to the other And it's working on both KVM and ESX Now I'm going to talk a little bit about the integration we have with open stack and that's really some of the uniqueness that we have We showing at our booth today. So if you're interested to learn more, please visit our booth and see the demo that we have on this We believe there's two ways to integrate with with open flow with sorry with open stack either the you know Falsum or Grizzly We support both options So we create our own quantum API our quantum plug-in, sorry and it talks with our adapter card So our adapter card knows How to communicate seamless integration with open stack? dashboard We could do this directly through a technology we called embedded switch Which is a new revolutionary way To set a virtualized switch layer to switch not on the software, but actually on the hardware so all the capabilities that I'm going to show you is Done by some hardware offloads on the adapter card, which allows us to create a more efficient Better performance communication to set different rules and provisioning on our adapter card So one of the approach is a direct communication with the embedded switch again a technology that we saw now support Along with SRV it allows you to get a hardware security configuration quality of service We can introduce in Fini band as the underlying technology While the user is still experiencing internet tools and provisioning capabilities and That the other option is using SDN controller in this case. We're using floodlight big-switch controller To communicate between the open stack Folsom and Melonox card so we implemented a Open-flow agent that runs on top of our nick And we believe this to be very unique to Melonox and we're very proud with the collaboration. We have with week switch on this one It allows or it benefit at the end user with a lot of new exciting capabilities So we have automated nick provisioning we could do counters statistics ACLs Configuration Mac anti spoofing and many other features that are now done on the actual nick This expose us to a much scalable Infrastructure if you think about it things that start from VLAN configuration, right when you have a villain configuration on a software It's limited When you use it on Melonox interconnect on our adapter card with the e-switch configure Technology and the open-flow agent We now get unlimited resources and villain configuration and all of that is done seamlessly Through our embedded switch technology So in other words the IT manager will configure his open stack dashboard a CLI the traditional way Underneath the hood we will provision it through the hardware make it more scalable and with a better performance So that's one of the uniqueness of the solution Here's a quick look of how it looks from the inside So we have an open-flow agent that talks with the our embedded switch and we we expose the VM with RDMA Either through SRV or power virtualization Providing a much faster simpler as the end connectivity Another exciting way to leverage your Existing open-stack storage access is by using Melonox interconnect with projects such as Cinder We just we announced it last week and we have a lot of I Will say partners that working with us on this solution Where we showing that by using Melonox adapter cards and it's not necessarily meaning Fini band even using Melonox 10 gigi adapter card By using iSCSI over RDMA We were able to get 5x the performance improvement versus just standard iSCSI over TCP So again just by applying a patch one specific patch To the Cinder project or to the driver the iSCSI driver you now benefit from 5x performance improvement We find this to be very Beneficial for our customers or anyone that using Cinder With open stack so to summarize the benefits of what we provide our customers in an open-stack deployment We're talking about best performance So again Cinder about 5x performance improvement as to storage access SRV allows us to give you full bare metal performance on a virtualized nick When we say bare metal performance we melt at Melonox We mean 40 gig internet with PCIe Gen 3 So really taking the full advantage of what SRV technology could give us We could provide you with the most cost-effective solution We have ways to consolidate the network on one fabric it Migration storage management One adapter one cable could do it all and not just that it could accelerate the performance off of each of those aspects With a 40 gigi you also get more virtual machines per server without affecting the SLA You know one unique story for that will be let's assume you have 80 Virtual machines running on one physical server and let's assume that at the peak time Half of your users reaching about 500 megabit per second for each VM With one 10 gigi adapter card, you're gonna affect your SLA 40 gigi connectivity Ensure you that you're gonna get the SLA even at peak time and that's what we offer our customers Simpler to manage it through standard API is we are just introduced the project We're doing with Cinder the projects we're doing with the quantum API and the SDN approach where we showing how we can communicate To the adapter card using open-flow agent running talking to big switch controller We're gonna add more as the end controller moving forward And it's available today. It's running today and Doing all of this by using the e-switch technology on our adapter card. We could truly ensure our customers that we could show hardware Isolation and filtering that are more much more efficient and better in scaling Thank you very much. If you have any question, please come and visit us at our booth We would love to show you our demo of the open stack as the integration with Melanox interconnect. Thank you