Προηγμένη Τεχνητή Νοημοσύνη (2022-11-21-20:10:16) / Μέρος 3 / Προγραμματισμένη Μετάδοση μαθήματος - Σταματόπουλος Παναγιώτης (ΕΘΝΙΚΟ ΚΑΙ ΚΑΠΟΔΙΣΤΡΙΑΚΟ ΠΑΝΕΠΙΣΤΗΜΙΟ ΑΘΗΝΩΝ, 2022)
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Let's name these transmitters X11, X12, X13, X17, X21, X22, X23, X27, X31, X37 So I have 21 transmitters I want to go a bit further and I want to express the limit problems in the same way as before So, some limits that are very easy to express, 5, 6 and 7 are only limits, 5 says 2 things, 4 means X14 So X14 must be A and X17 must be N These are limits that can be broken into 2 limits and we can easily reduce the area of the transmitter Other transmitters with limits 5, 6 and 7 are only 2 limits Limit 4, not M after A says that in each continuous box of a transmitter the second must be M and the first must be A This is a series of limits for each continuous transmitter So X11 and X12 must not be such that X11 must be A and X12 must be M This can be expressed with the following limit If X11 is A, then X12 will be different from M The same limit must be expressed between X12 and X13 between X13 and X14, X14 and X15, X17 and X11 So I must have 7 limits for Jim, 7 for Jane and 7 for John to express limit 4 The number 2 means that the 3 transmitters must be all different X11 must be different from X21, X21 from X31 and X21 from X31 This is a very useful limit in many applications All these systems have a version of the so-called All-Different Limit which takes transmitters to which we want to go at different rates all different from X11, X21, X31 and vice versa from X12, X22, X32, X32 and every day we have a different limit for the 2 Now, what is a little more complicated is the 3, which says that each transmitter will have at least 2 R in each continuous from day to day since it will have at least 2R, it will have 2R, 3R, 4R, 5R, 6R or 7R In theory, we do not know now if there is a solution with more than 3R but most likely it will have 3R but this is a knowledge that we knew from the work we did so we can not pay attention to it and not take it into account and for the most part, as you can see in the problem here and simply the system then can go through the solutions and see that there is no solution that someone will take 4R or 5R or 6R or 7R but these solutions can be written, they can be combined and what does this mean for the gym? We can write such a solution for the gym, for the gym and for the gym let's take the gym it means that the gym must have at least 2R all the time that is, the 2R must be Sunday and Monday that is, X11 must be R and X12 must be R and the rest of the numbers must be different from R don't worry about the rest now, I will say it a little later or X12 and X13 must be R that is, Monday and Tuesday must be R and all the rest must be different from R or X13 and X14 must be R and all the rest must be different from R so that we can cover all the cases with 2 continuous numbers or 3 continuous numbers X11, X12 and X13 must be R and all the rest must be different from R and so on 4, 5, 6 and 7 a very simple solution, but it can be impressed with all these problem solutions if we put them in some system of optimization as a system of programming solutions it should find the solution that we have found without having to do what is called an imitation that is, what should I put there? Should I put M or A? We will need to make an imitation in the end because there are two solutions, you remember these and with a lot of exaggerations the solution, one of the oranges is correct and the other orange is correct but there you will not be able to write anything more so I will say it now if Jim takes A, what happens? if Jim takes A, Jane will take M and John will take R there is a solution but if Jim takes M, Jane will take R and John will take A there are two solutions so there it will be enough to make an imitation but until then, all the others will be lost deterministically because of the algorithm we have again so I will turn around here and see how we will treat first of all the Convon-convergence and the Convon-convergence with some simple algorithms Convon-convergence is like this we first define a continuous optimization of the algorithm it is not consistent it is consistent with the Convon when it is not inconsistent yet that is, when for each converter for all x values in the DDI field it is true that the bias that consists only of x belongs to the limitation that is applied to this converter that is, I do not have in the field a converter in x values which violate the limitation that exists for this converter if it is not a problem to satisfy the limitations not consistent I can do it not consistent with this procedure that says that for each converter in the DDI field for each x value in the DDI field if the bias that consists only of x does not belong to the limitation does not allow the limitation that is applied to the converter then remove the x value from the DDI field and this with the new DDI and when this is done it means that our problem is not consistent and we do not have to deal with the limitations as there is the idea of not consistency there is also the idea of non-consistency we say that a continuous to satisfy the limitations is consistent with the non-consistent limits when every non-consistent limit in the DDI field the DDI field was not directed the fields had no direction for each non-consistent limit as a problem to satisfy the limits there is also a directed field to satisfy the limits that in each of the initial fields we have put the two arcs the arc in one direction towards the other direction so indeed in a non-consistent limit we have arcs in each combination of two components which is a limit from the second component and from the second component to the first component so the definition of a problem to satisfy the limits is arc-consistent when every directed field in the DDI field is consistent and what does this mean? when for every non-consistent limit in the DDI field in the first peak of the limit so that the DDI field is consistent in the VJ field so that the Psi field is consistent in the DDI field in other words this definition says that a problem is arc-consistent when every directed field in the DDI field is arc-consistent and when a directed field is arc-consistent when every field has support in the DDI field having support means that there is at least one field in the DDI field so that both of these limits are consistent based on the combination of these two fields it is important that we speak about consistent directed fields because look here a very big problem with two peaks V1 and V2 the V1 field is R the V2 field is RG and in the DDI field the V1 field must be different from the V2 field in this very simple problem the arc V1-V2 this arc is consistent why? because for every peak in the V1 field there is a consistent peak based on the V2 field the V2-V1 is not consistent the V2-V1 is not consistent because it is not true that for every peak in the V2 field for example the arc there is a consistent peak in the V1 field it does not exist so while the V1-V2 is consistent the V2-V1 is not consistent and it is a problem to have consistency in one direction and to have consistency in the other direction so now we want to see how we will apply consistent peak-to-peak to a peak that is not consistent in an arc that is not consistent what will say that it is consistent in the arc? as we said it will say that for the arc VI-VJ there is a peak in the VI field so that there is no consistent peak with it in the VI field if there are such peaks in the VI field, we should remove them I have no reason to give there is no value I do not gain anything to give to the VI the value A when the VJ does not have consistent peak with the value A of the VI there is a revised process revised by VI-VJ which says the following it uses a boolean to translate this deleted so what it uses is that the revised we will continue and we are interested to know if the revised made at least one signature we need this information so we have this boolean to translate and if the deleted is taken by Dimitru as you say the deleted returns if it takes Dimitru it means that it has made at least one signature if it takes the value false it means that it has made no signature that is why the deleted becomes false and if it is true the deleted will take Dimitru and it will remain true so what does the revised say after making the deleted true for each X in the VI field of the peak which is the beginning of the peak for each X in the VI field if there is no peak in the VI field so that the X pair is acceptable from the limitation between VI and VJ then you remove the peak from the VI field the peak has no value in the VI field to be given some value in the VI field because it will be condemned to not give a final solution and if it has made one signature it will remain true and it will return true if I pass and all the values in the VI field have the same value in the VI field there will be no signature and the deleted will remain false and it will return false so this revised refers to a value but be careful it is not enough to write a limitation that is invalid in the valid values to go and make valid the values that are invalid and to say that I have finished let me give an example take the following problem the multipliers X1, X2, X3 with a field D1, D2, D3 with a field D1, D2, D3 and my limitation is X1, X2 and X2, X3 let me do the following let me write the fields here separately D1 is 1, 2, 3, 4 D2 is 1, 2, 3, 4 D3 is 1, 2, 3, 4 so I have these three multipliers I have these two fields if I want to write the definition it is X1 X2 and X3 there is a definition of X1, X2 and there is a definition of X2, X3 this is the definition if I want to talk about arcs the arcs are this and this this this and this and I take what happens with the arcs if the arcs are consistent and if they are not I can revise them and make them consistent ok let's say I will examine the arcs in this order X1, X2 X2, X1 X2, X3 X3, X2 ok and I will revise X1, X2 what will revise X1, X2 say? let's see if each arc in the X1 field has a consistent value based on the X1, X2 definition in X2 field X1 has a consistent value X2, X3, X4 X2 has a consistent value X3 has a consistent value X4 has a consistent value X4 has a consistent value in X2 field X4 has a consistent value in X2 field therefore X4 in X1 field will be removed when I revise X1, X2 when I revise X4 it should no longer belong to X1 I have described it let's revise X2, X1 there are values in the X2 field so that they do not have a consistent value in the X1 field based on the definition between X1 and X2 the value 1 in the X2 field does not have a consistent value in the X1 field because there is no value in the X1 field smaller than this one therefore this must be removed therefore 1 cannot belong to X2 value 2 has a consistent value value 2 in X2 has a consistent value in X1 field value 3 has a consistent value value 4 has a consistent value therefore let's revise X2, X3 let's revise X2, X3 there are values in the X2 field that do not have consistent values in the X3 field value 2 has value 1 has been removed value 2 has been removed from the previous one value 2 has value 3 has value 4 does not therefore 4 cannot belong to X2 and let's revise X3 X2 there are values in the X3 field that do not have consistent values in the X2 field value 2 has value 1 has value 2 has value 1 has no smaller value than this one in the X2 field likewise value 2 has no smaller value than this one in the X1 field must be does not belong to X3 and also value 2 does not belong to X3 let's revise and I have revised all the values is my problem now inconsistent? it is not, because with the divisions that were made in sequence X2 is not anymore therefore X1 and X2 have the value 3 which does not have consistent value in the X2 field there is no value in the X2 field greater than the value 3 in the X1 field so, to revise all the values at one time after another it resolves that my problem will have consistent value some divisions that I have revised due to some divisions that were made in sequence at some point you can see here by checking X2, X3 by revising X2, X3 the 4 came out this was removed but the X2 field was used to revise X1, X2 now that I have an additional value in the X2 field I have to do re-revise X1, X2 so now who can be an algorithm that will correctly revise the division that will ensure that my problem will have consistent value look it is the AC1 algorithm it is the algorithm proposed by Magworth in the paper that was inspired by Wolf's work it is the first attempt to systematize this algorithm and it is not like this I have a problem that I want to do consistently so what I do I start and I do it non-consistent to disperse with the nodes and in a Q structure a single Q I go and put all the values of the problem I put the value VIVJ for each VIVJ that I have and I go and put the value VJVI for each value of the problem I go and put both the N arc and the D arc and I do this repeat until that you see here which is controlled by a logic called change in each repeat until it is initialized in false and after a revise if if the revise returns true the revise that was before the revise that was before returns true if the revise returns true because of the or that is here the change will be true so the until not change means that if the change is true the condition does not change the not change will be false and the repeat will be true the idea of this method is this I do a first pass and I do a revise if I have done even one pass again from the beginning I do a revise if I have done even one pass I do a revise only when I pass all the passes only when I pass all the passes only when I pass all the passes and all the revise that I will do will return false so no pass will be true then the change will remain false so the repeat will be true it is a method that goes what is called on the safe side before when we tested this did I do a pass of all the passes once were they rewritten from the pass again from the beginning if you do not need to re-write you will not do rewrites if I do a re-write with the second pass again a re-write when I do a re-write with all the passes and there is no re-write I know that I do not need to re-write but if I do the re-write there will be no re-write this is the basic idea of this method it is for me that it works of course it does not have an abstract multiplicity so the space is nd nd because we need to protect essentially the area of ​​transmitters we have n transmitters we have nd values in each field so we need nd space to be able to protect the fields in the sense that to protect the information if a transmitter continues and continues in the field what we are interested in is to keep the active fields of the transmitters at any time for each for each transmitter to have a boolean table that is as big as the number of values in the field and the 1 means that the corresponding value is expressed in the field and the 0 means that it has a description something like that now because the space, the time is nd look, the revise as I said before has multiplicity nd if it is the revise that is going to do the revise for each x in the field of Vi for each x in the field of Vi it goes to see if there is a corresponding value in the field of Vj that is, this if here, it hides this if does not exist, it hides because it does not see the representation which is not the same as the one that is here it needs a double representation in the field of Vi and in the field of Vj in the field of Vi and in the field of Vj this in the field of Vi has a multiplicity and this in the field of Vj has a multiplicity we need the same time to do this revise now, the revise is hidden sorry, here the revise is hidden in a repeat until the point is what is the worst case for this repeat until that is, how can we find what is the worst that can happen from the point of reviewing this repeat look, the worst that can happen is to become continuous diagrams and we need to do them from the beginning and each time to become the few diagrams that can become that is, one each time I make one diagram if each time I make one diagram this will become the most so many times that are values in the fields how many values are there in the fields? there are nd so this repeat until will become the most nd times that is the worst case that is, in each revise only one value is written of course, this nd has in it a foreach that is a criterion so for each of these nd in the worst case we need to do a revise or times or therefore nd because of the repeat until or nd d square because of the revise or nd times this is the complexity of color of my method of ACM so, you can understand you can understand that this method is not the best to think of a person because every time a person makes a diagram again everything from the beginning because everything from the beginning and not what is needed because I previously showed you you saw that while I was doing a revise x1 and x2 were not consistent after the first revise because the field of x2 had been affected due to some acme due to the x2 and x3 the field of x2 had been affected due to the x1 and x2 the field of x1 had been affected due to the x1 and x2 yes, can I somehow take advantage of this information and not do everything from the beginning this is the method that follows but we will discuss it in the next lesson it is the AC3 you will ask why did we go from P1 to AC1 to AC3 because P1 left AC2 for the Woltz algorithm somewhat smarter than this nonsense that we are doing here but not as much as AC3 suggested by Magworth so he left a gap in the method of arc consistency AC is from arc consistency to estimate if it is for the Woltz algorithm so I would suggest to stop here for today we will continue on Thursday from this point on we will discuss AC3 yes, on Thursday we will go and collect the last lesson was to pay for it the last lesson was 13 weeks but the lesson along with the presentations was designed for 12 weeks so we have a margin if we paid the lesson due to the strike that was on 9th of November due to the absence of me that was on 2nd of November these two we paid the other one we will pay it is free you can use it the program that is called PDF is called Draw for PDF it is free it is very nice You can write whatever you want. Whenever I discover something, I can stop writing on the table. I have the pencil here. Do you think the pencil is programmed? No, it's not a pencil. It's free, independent of the pencil. The pencil works like a mouse, it's not a pencil. It's a mouse, the computer can see. Why do you need a pencil? Because it's very useful. Yes, you can write on a PDF, not just for teaching, you can write notes on it. But it's a bit weird. That's what you have on your desk, and you can write on your PDF. It's very useful. Thank you very much. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. You're welcome. 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