MIT-SCIENCE-Lectures-sol8

24.00: Problems of PhilosophyProf. Sally HaslangerOctober 15, 2001Personal Identity IThere are several questions that might arise concerning personal identity. When we ask "Who am I?" we might be wondering about what "makes us tick", what we ultimately value, what matters to us. We might also be asking what sort of being we are, what our possibilities are, under what conditions "I" would continue to exist. We'll begin our discussion of personal identity with the latter set of questions.Consider a parallel set of questions:(Id) Under what conditions are baseball-events events in the same game? E.g., under what conditions are aparticular batter batting and a particular runner running parts of the same game?Break (I) into two questions. First, the question of synchronic identity:(SI) Under what conditions are simultaneous baseball-events events in the same game?Let a baseball "stage" consist of all the events at a given time that are part of a single game. Then we can formulate the question of diachronic identity as follows:(DI) Under what conditions are two baseball stages stages in the same game? E.g., what makes the beginning of the third inning and the end of the eighth inning stages of the very same baseball game?Restate the questions for persons:Synchronic identity: under what conditions are two simultaneous person-events events in the life of the same person? Diachronic identity: under what conditions are two person-stages stages in the life of a single person. In particular, what makes a particular person-stage a continuation of me as I am right now?Synchronic unity relation: the relation that unites simultaneous person-events into a stage of a single person.Diachronic unity relation: the relation that unites person-stages into a single temporally-extended person.Perry Dialogue, First Night:1. (Weirob) Is survival of death possible?ï What we want is survival of me (not the molecules of my body...).ï What survives must do justice to possibility of anticipation and memory.2. (Miller) Postulation of soul.Soul criterion: x is the same person as y iff x and y have the same soul.ï What is a soul? The subject of consciousness? A spiritual/disembodied self? The ghost in the machine?3. (Weirob) Practical applicability condition on adequate response:ï An adequate account of the survival/identity of persons must also be able to justify our practices of recognizing and identifying each other.4. (Weirob) The soul not a good basis for recognition.ï We can recognize each othersí bodies, but not each othersí souls.5. (Miller) Body-soul correlations provide a basis for recognition.ï We do observe the soul indirectly by observing the behavior of each others' bodies.6. (Weirob) But how do you know that there is a body-soul correlation? By observation?ï Chocolate example shows that correlation between body and soul untestable; if not testable, it must just be a matter of faith.7. (Miller) Personality-Soul correlations provide a basis for body-soul correlations.ï If the soul is the subject of mental states, the soul must be what explains the pattern of mental states, i.e., it must be what provides us with our personality or character. So there must be a correlation between soul and personality.8. (Weirob) But even if soul does account for personality, thereís no reason that one must have the same soul to exhibit the same personality.ï We can't infer from sameness/similarity of personality to sameness/similarity of soul.ï River analogy: rivers have the same appearance even though the water molecules are constantly changing; better example: Martians/God/mad scientist could make a duplicate of you with the same personality, but without the same soul. ï [Also, don't people sometimes undergo changes in their personality, e.g., after a conversion, a tragic experience, falling in love, etc.?]9. (Miller) But maybe the relevant correlations must be established by 1st person observations.ï I know of myself that there is a body-personality-soul correlation.10. (Weirob) But how do you know that even in your own case?ï All that you introspect is thoughts and feelings, but isnít this compatible with a sequence of souls passing through? How do you know that God doesnít give your body a new soul each morning? How would you tell if when you woke up there was a new soul responsible for your thoughts, or just the old one?11. (Weirob) Does the postulation of soul really help? Doesnít it push the question back?ï E.g., suppose I say that a flower blooming in my garden is the same one that was blooming yesterday; someone asks: on what basis do you say that? I say that it has the same flower-spirit as the previous one. Doesnít the question arise: what are the identity conditions for flower spirits? What does it take for a flower-spirit to continue to exist?12. Gretchenís bodily solution:ï Do I doubt my own soul? Not in the sense of doubting that Iím conscious. But my consciousness is a property of a body, and that body is me. But if I am my body, then the destruction of my body is the destruction of me. So:Body criterion: x is the same person as y iff x and y have the same living human body.In other words:Person-stage x is part of the same person as person-stage y iff x and y are person stages linked by bodily continuity (where bodily continuity is understood in terms of the continuity of a living human body).。

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where βn are arbitrary constants, corresponds to a function vanishing t > |T/2| ,that is, a time-limited function. The surprising import of (2.13) is that the Fourier transform of a time-limited function can be perfectly reconstructed from a knowledge of its values at the Fourier series frequencies alone. That means, in turn, that a knowledge of the countable infinity of Fourier coefficients can reconstruct the original function exactly. Putting it slightly differently, there is no purpose in computing a Fourier transform at frequency intervals closer than 1/T where T is either the
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ignoring all convergence issues. We thus have the nice result that a periodic function has a Fourier transform; it has the property of vanishing except precisely at the usual Fourier series frequencies where its value is a δ−function with amplitude equal to the complex Fourier series coefficient at that frequency.

Theory of Parallel Hardware April12,2004 Massachusetts Institute of Technology 6.896 Charles Leiserson,Michael Bender,Bradley Kuszmaul Handout14Solution Set6Due:In class on Wednesday,March31.Starred problems are optional.Problem6-1.A comparison network with inputs and comparators can be described as a list ,where for.The list represents a topological sort of the comparators,where each ordered pair stands for a comparison between the elements on lines and,with the minimum being output on line and the maximum on line.(a)Give an efficient algorithm for determining the depth of a comparison network so described.(Hint:Youcan’t just count the maximum number of comparators incident on a line.)Solution:We simply construct a dependency graph as follows.We assign create onevertex for each comparator,is a comparator.We add a directed edge between vertices and,with,if the two corresponding comparators have a line in common.That is,and or or or.For each,we assign a weight .Finally,we add a source vertex and a sink vertex with a0-weight edge from to every other vertex and a-weight edge from every other vertex to.Once we have the above dependency graph,we can just run any old shortest path algorithm tofind theshortest path between and.The negation of the path weight is depth of the comparison network.Thecorrectness of this algorithm is fairly obvious as our graph represents dependencies between comparators.The shortest path is the one with the most edges,or the most dependencies between comparators.Constructing our dependency graph requires time because we consider all pairs of compara-tors once.The graph is a dag since we only have edges if.Thus,we can runD AG-S HORTEST-P ATHS,which has a running time of Thus,we have a totalrunning time of.Define a comparison network as standard if for.(b)Give an algorithm to convert a comparison network with inputs and comparators into an equivalentstandard comparison network with inputs and comparators.Solution:We iterate over all the comparators in order.At a comparator,if,we do notperform any action and continue with the next comparator.If,then we need to standardize thecomparator by changing it to.Note that now the output that used to be on is on and vice-versa.Thus,we also need to reverse and in all subsequent comparators.That is to say,we considerall comparators for.If,then we change the comparator to be .If,then we change the comparator to be.We perform similar actions forand.We then continue on to the next comparator.The proof of correctness is just a simple induction argument over the number of comparators iterated over.To be equivalent,we simply need that afterflipping the comparator,all inputs to thecomparators are consistent with those of the original network.This fact is apparent in the construction.As for running time,we iterate over all comparators,and at each comparator we consider all the follow-ing comparators.Thus,we have a running time of.(c)Prove or give a counterexample:For any standard sorting network,if a comparator,where,isadded anywhere in the network,the network continues to sort.Solution:Simple counterexamples exist.Problem6-2.A comparison network is a transposition network if each comparator connects adjacent lines.Intu-itively,a transposition network represents the action over time of a linear systolic array making oblivious comparison exchanges between adjacent array elements.(a)Show that if a transposition network with inputs actually sorts,then it has comparators.Solution:For simplicity,let’s consider an even.If the network actually sorts,then it must sort theinput.Each input has to be moved across exactly lines.In a transposition network,each comparator moves two inputs by exactly line.Thus,each input must beinvolved in comparators.Since there are inputs,and each comparator can only move two inputs,we need at least comparators.The same argument works for odd.(b)Prove that a transposition network with inputs is a sorting network if and only if it sorts the sequence.Solution:Consider an input to a transposition network and the state of the networkafter the-th comparator.We denote the data held by line after the-th comparator by.Lemma1Consider the descending input to some-input transposition network.Consider also any other input.For any pairs of lines and with,if ,then.Proof.We use induction over the comparators in order.For a base case,consider the network before any comparators.For any and with,because the input is descending.Thus,we make no claim about any of the.Assume that the claim holds after the-th comparator.Consider the-th comparator.Because wehave a transposition network,we compare two adjacent lines and.We need to show that theclaim holds(for all pairs)of lines after the-th comparator.That is,consider any two lines and with .Suppose that.Then we need to show that.We have several cases:Case1Suppose that and.Note that because of how a comparator works, so.Case2Suppose that,,,and.Then lines and are not involved in the comparator.Thus,.By our inductive assumption,we have,which yields.Case3Suppose that,and.Then we have.By the inductive assumption,and.Thus,.Case4Suppose that,but.Then we have.If,then we have.Thus,we can apply the inductive as-sumption to get and.If instead,then.Thus,the inductive assump-tion gives us.Case5Suppose that but.Similar to case4.Case6Suppose that but.Similar to case3.Problem6-3.Give an algorithm for sorting elements on an mesh in time.Solution:We label each of the processors in the mesh according to coordinates.We sort such that smallest values are stored in the plane containing processors for all.The next smallest values are inthe plane,and so on.The smallest value is in the column,the next smallest are in the column ,and so on.Finally,the smallest value is in.Essentially,we are sorting lexicographically.For any two processors and,if,then the value stored in is less than that in.If but ,then the value stored in is less than that in.Similarly,if,,and,then the value stored in is less than that in.Our algorithm proceeds in several steps:1.Sort the values in every-plane(an-plane is the set of processors,withfixed,).2.Sort the values in every-plane(an-plane is the set of processors,withfixed,).3.Sort the values in every-plane,alternating the order between planes(a-plane is the set of processors withfixed,).4.Do two steps of the1-dimensional odd-even sort on each-row(an-row is the set of processors withfixedand,).5.Sort the values in every-plane.For each of the plane sorts in steps1,2,3,and5,we use the sort we discussed in class,which takes time.Step4only takes2time,so we have a total running time of.Now we need to argue correctness a bit.1We consider only0-1inputs as per the0-1lemma.After step1,on each -plane,the’s are moved towards the smallest value of.Since the values are sorted,there can be at most one dirty-column in each plane.Thus,every-column within a single-plane has(almost)the same number of’s(to within one).In step2,we sort-planes.Thus,we pull one-column from each of the-planes.Since each of the-columns in a given plane has within one the same number of’s,the-planes differ by at most’s.Thus,after we sort the values in each-plane,almost all the-planes are clean.In fact,we can only have two adjacent dirty-planes. More specifically,all the-planes above these two dirty-planes(all-planes with greater values of)contain only’s.Similarly,all the-planes below the two dirty planes contain only’s.In step3,we sort the-planes.In alternating planes,we move the’s towards the smallest values of or the largest values of.Thus,we still have at most two dirty-planes,but now the’s of one of these planes lines up with the ’s of the other.In step4,we try to consolidate the two dirty-planes into one.Suppose that between the two dirty planes are indicated by and.WLOG,suppose that the two planes collectively contain more’s than’s.Then all the’s in plane are below a in plane.Similarly,all the’s in plane are above a in plane.Thus,when we perform two steps of an odd-even sort in the-direction,we create a plane with all’s, leaving only one dirty plane.Going into step5,there is only one dirty plane.Now we can just sort it.1A more detailed proof can be found in Tom Leighton’s book,so it seems silly to try to reproduce one here.。

Carlos RenjifoDeutsch IV4 April, 2003Die Schachpartie aus der Perspektive von McConnerSchade, dass du die Schachpartie verpasst hast. Es war ein tolles Spiel zwischen zwei ziemliche guten Spielern. Der eine, Mirko Czentovic, ist zurzeit Weltschachmeister. Der andere kenne ich immer noch nicht. Ich weiß nur, dass er vor zwanzig Jahren keinen Schach gespielt haben soll.Was zuerst wie ein einseitiger Wettkampf aussah, war überraschend schnell quitt. Na ja. Der Weltschachmeister war auch nicht so gut, wie man erwartete. Vor zwei Tagen habe ich mit ein paar anderen gegen diesen Czentovic gespielt und obwohl wir zusammen die erste Partie verloren haben, haben wir die zweite Partie mit einem Remis beendet. Hätte ich gegen ihn allein gespielt, ohne von dem anderen gestört zu werden, hätte ich die Partie wahrscheinlich gewonnen. Aber genug damit und zurück zum Spiel.Also, zuerst war die Partie ein bisschen langweilig. Beide Spieler saßen vor einem Schachbrett und schoben ihre kleinen Holzfiguren auf das Brett herum. Aber nach ein paar Zügen wurde das Spiel viel interessanter.Ich mag es sehr, wenn ein Spiel plötzlich von dem Unterlegenen übernommen wird. Genau dieses ist in der Partie passiert. Nach zehn Zügen nahm der Unbekannte das Vorteil im Spiel. Jedes Mal, als Czentovic einer seine Figuren bewegte, reagierte der Unbekannte sofort. Nach diesen schnellen Gegenzügen musste Czentovic eine lange Überlegegungspause nehmen, um sich den folgenden Zug auszudenken. Ich musste lächeln jedes Mal, wenn Czentovics Zug automatischvon seinem Gegner entwertet wurde, wie ein geplatzter Scheck, der zerissen wird, weil es keinen Wert hat.Die Partie wurde immer weiter gespielt, bis der Unbekannte plötzlich schrie: "So! Erledigt!" Sobald ich diese Wörter gehört habe, ist mir das Herz fast aus der Brust herausgesprungen. Die Partie war zu Ende! Der Unbekannte hatte den Weltschachmeister, Mirko Czentovic, besiegt! Einfach unglaublich!Aber nach dieser kurzer Aufregung fiel ich in Schock, als ich hörte, dass Czentovic noch eine Partie spielen wollte. Weiß du, was ich denke? Ich glaube, er ist ein arroganter Mensch. Warum konnte er nicht zugeben, dass er verloren hatte? Wenn ich der Unbekannte wäre, hätte ich ihm keine zweite Chance gegeben. Aber, weil es nicht ich war, der spielte, wurde eine zweite Partie gespielt.Ich war von dieser zweiten Partie angewidert. Es war kein Gentlemanspiel mehr. Es ging jetzt nur noch ums Gesicht wahren. Czentovic wollte keinen wissen lassen, dass er gegen einen Amateur, einen “Niemanden”, verloren habe. Traurigerweise kriegte er seinen Wunsch. Die zweite Partie hat er gewonnen, aber nur durch betrüglisches Verhalten.Czentovic hat gemerkt, dass je länger der Unbekannte auf ihn warten musste, umso ungeduldiger wurde er. Deswegen hat er in der zweiten Partie immer seine ganze zehn minutenlange Überlegungspause benutzt. Er wusste, dass er mit dieser Technik die Konzentration des Unbekannten stören würde und auf diese Art und Weise gegen ihn gewinnen könne.Es dauerte nicht lange, bis dieser Plan seine Wirkung zeigte. Nach wenigen Zügen konnte man merken das etwas mit dem Unbekannten los war. Er fing an zu zittern und sein Gesichtsausdruck wurde immer mehr verkrampft. Ein paar Zügen später war das Spiel zu Ende.Der Unbekannte gab plötzlich auf, ist von seinem Stuhl aufgestanden und verließ den Rauchsalon, wo die Partie gespielt wurde.Damned Fool! Ich kann immer noch nicht glauben, dass er einfach aufgegeben hat. Auch wenn Czentovic geschummelt hat, gab es keinen Grund das Spiel aufzugeben. Wenn ich dieser Unbekannte wäre, hätte ich wenigstens die Schachpartie zu Ende gespielt! Schade, dass jetzt nichts mehr daran geändert werden kann.Aber sogar über dieses zweites Spiel muss ich zugeben, dass ich mich immer noch amüsiert habe. Es war ein gutes Spiel, obwohl es besser gewesen wäre, wenn Czentovic nicht geschummelt hätte und wenn der Unbekannte nicht so leicht aufgegeben hätte!。