MIT-SCIENCE-Lectures-hand_out2_listof

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.

An important and easy-to-neglect issue in building the frame is the weight of the water it will need to support.One cubic foot of water contains about 71/2 gallons and weighs approximately 63 pounds [2]. For a pool of any reasonable size, this quickly adds up; for large volumes of water, careful structural analysis will be important to ensure that the frame will be able to hold the weight, especially if the pool is being recessed into a platform. One simple way to minimize the amount of water used and thus the amount of weight that must be supported is to make the pool less deep; this is often a very reasonable option since the actual depth of the pool will always not be obvious to the audience.The frame should then be covered with plywood (given the loads it will be sup-porting, something substantial like 3/4 ′′ plywood should be used. It then needs to be covered with a pond liner so that it can be filled with water. These are readily available from garden stores as waterproof plastic sheets, which can be cut to size and fit into the pool.The pool can then be filled with water. Once filled, it should be emptied or changed regularly — the last thing anyone needs on the stage is a place for bacteria and other unpleasant things to grow, and a stagnant pool of water fits that bill all too well. A pool treatment can also be useful here, though ones containing chlorine are not ideal due to their distinctive smell.3 Building the waterfallMany of the same techniques from building a pool of water apply to building a waterfall. Again, we will need to use a waterproof pond liner around a frame and support structure. The difference, of course, is that the waterfall will be a stream (albeit one that sometimes flows horizontally and sometimes flows vertically), so it will not have as much depth.This allows the support structure to be more lightweight.A common means for building streams is to construct a set of curved ribs, then lay a thin layer of plywood or lauan over it that conforms to the curve of the ribs. This should then be covered with the pond liner.The water should flow through the stream and down the waterfall into the pool, which serves as a reservoir. It then needs to be returned to the top of the waterfall; for this, a pump is required. A household sump pump, available from the Home Depot or similar hardware store in the $100 range, is probably the easiest option.ℵ0-2For especially large installations, a common pump may not suffice. One option is to obtain a larger pump; McMaster-Carr carries sump pumps up to 3/4 hp and effluent pumps up to 2 hp, though the latter may be a bit expensive for this purpose at over $700[3]. A simple alternative is to use multiple smaller pumps. Submersible pumps are best for this application: other types of pumps may allow the pump to be placed further away from the pool, reducing noise, but adding a drain to the pool requires creating and sealing a hole in the liner, which only adds complexity. Sump pumps may pump into a standard garden hose, which is certainly easy to find and easy to work with. However, the pump will be able to operate more efficiently if PVC pipe (usually or 11/2 ′′ , though it depends on the pump) is used.11/4 ′′The pressure created by a pump, especially a large one, may be rather high. This could lead to the unusual effect of the water spewing forcefully upward or in a powerful stream — which, while useful for a fountain, isn’t very realistic in a stream effect.The solution here is to increase the diameter of the pipe by using couplers with successively larger diameters. This causes the same amount of water to flow through a larger area, and thus with less force.Waterfalls and streams are generally rather noisy. The sound of water flowing down the stream and into the pool can be a rather pleasant one, and can be a helpful part of the sound design for creating the outdoor atmosphere. If, however, that doesn’t fit with the sound designer’s vision, it becomes the technical designer’s problem to fix.Reducing the velocity of the flowing water is an obvious solution with an obvious drawback.Adding rocks and similar objects to the stream will help reduce the volume of water that needs to flow, reducing the noise created (and also allowing a smaller, quieter pump to be used). It is also quieter to not have water fall directly from the waterfall into the pool; instead, it should flow onto a surface that slopes into the pool.References[1] M. Powers, “Pools, waterfalls, etc.” available at /howto/water/pools.htm.[2] P. Carter, Backstage Handbook, 3rd ed. Louisville, Kentucky: Broadway Press,1994.ℵ0-3[3] McMaster-Carr, Inc. /.ℵ0-4。

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.。