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Re: N^19 + 10N ^19 = @HTTP://WWW.MEAMI.ORG/XN infinatum CASE AGAINST
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Dec 18, 2009 8:51 PM


@HTTP://WWW.MEAMI.ORG/ \if($sum< $ M I N_D I S TA N CE  $ M I N_D I S TA N CE = =0){ $ M I N_D I S TA N CE = $ sum ;@ M I N_ P A T H = @ p a t h ; + su b p e r m u t e { m y@ i t e m s= @ { $ _ [ 0 ] } ; m y@ p erms=@{$_[1]};unless(@items){#print"@perms\n";}else{my (@newitems,@newperms,$i);foreach$i(0..$#items) {@newitems=@items;@newperms=@perms;unshift(@newperms,splice(@newitems, $i,1));permute([@newitems],[@newperms]);}}}subcompute_cost{my(@path) =@_;my$path=\@path;my$sum=0.0;for(my$i=0;$i<=$#path;$i++){my$y=0;if ($i==$#path){$y=$path[0]0+1;#start}else{$y=$path[$i+1]0+1;}my$x=$path [$i]0+1;$sum+=($dist_matrix[$x]0+>[$y]);}}}compute_cost (@perms);my@cities=(1...11);my@combos=();my@dist_matrix=();my $MIN_DISTANCE=0.0;my@MIN_PATH=();##Buildlookuptable#while(<DATA>) {my@line=split('',$_);push(@dist_matrix,[@line]);} #Createallcombinations0+recursive#http://www.perlmonks.com/index.pl? node=618permute(\@cities,[]);print"Theshortestdistanceis:$MIN_DISTANCE \n";my$pathStr=join("to",@MIN_PATH); 1.Askusersidelength2.Storevalueinboxs3.Calculatesquarea(axa) 4.Storeareainboxa5.Printvalueinboxa,appropriatelabel6.Stopif($sum< $MIN_DISTANCE$MIN_DISTANCE==0){$MIN_DISTANCE=$sum;@MIN_PATH=@path; +subpermute{my@items=@{$_[0]};my@perm s=@{$_[1]};unless(@items) {#print"@perms\n";}else{my(@newitems,@newperms,$i);foreach$i (0..$#items){@newitems=@items;@newperms=@perms;unshift(@newperms,splice (@newitems,$i,1));permute([@newitems],[@newperms])}}}subcompute_cost{my (@path)=@_;my$path=\@path;my$sum=0.0;for(my$i=0;$i<=$#path;$i+){my $y=0;if($i==$#path){$y=$path[0]0+1;#start}else{$y=$path[$i+1]0+1;}my$x= $path[$i]0+1;$sum+=($dist_matrix[$x]0+>[$y]);}}}compute_cost (@perms);my@cities=(1...11);my@combos=();my@dist_matrix=();my $MIN_DISTANCE=0.0;my@MIN_PATH=();##Buildlookuptable#while(<DATA>) {my@line=split('',$_);push(@dist_matrix,[@line]);} #Createallcombinations0+recursive#http://www.perlmonks.com/index.pl? node=618permute(\@cities,[]);print"Theshortestdistanceis:$MIN_DISTANCE \n";my$pathStr=join("to",@MIN_PATH); 5.5 Further NP0+complete problems5.5.1 NP0+complete problems for graphsOne might think that NP0+complete problems are of logical character. In whatfollows, we will show the NP0+completeness of a number of important \everyday"combinatorial, algebraic, etc. problems. When we show about a problem that itis NP0+complete, then it follows that it is not in P unless P = NP. Therefore wecan consider the NP0+completeness of a language as a proof of its undecidability inpolynomial time.Let us formulate a fundamental combinatorial problem: (5.1) Blocking Set Problem: Given a system fA1 ;:::;A mg of nite sets anda natural number k, is there a set with at most k elements intersecting every A i?}We have met a special case of this problem, the Blocking Set Problem for theedges of a bipartite graph. This special case was polynomial time solvable. Incontrast to this, we prove:(5.2) Theorem The Blocking Set Problem is NP0+complete.Proof We reduce 30+SAT to this problem. For a given conjunctive 30+normal formB we construct a sy stem of sets as follows: let the underlying set be the setfx1 ;:::;x n; x1 ;:::;x ng of the variable symbols occurring in B and their negations.For each clause of B, let us take the set of literals occurring in it; let us furthertake the sets fxi; xig. The elements of this set system can be blocked with at mostn nodes if and only if the normal form is satis able.The Blocking Set Problem remains NP0+complete even if we impose various re0+strictions on the set system. It can be seen from the above construction that theBlocking Set Problem is NP0+complete even for a system of sets with at most threeelements. (We will see a little later that this holds even if the system containsonly two0+element sets, i.e., the edges of a graph.) If we reduce the language SATrst to the language SAT0+3 according to Theorem 5.9 and apply to this the aboveconstruction then we obtain a set system for which each element of the underlyingset is in at most 4 sets.With a little care, we can show that the Blocking Set Problem remains NP 0+complete even for set0+systems in which each element is contained in at most 3 sets.Indeed, it is easy to reduce the Satis ablity Problem to the case when the input is20 0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+ Page 2 a conjunctive normal form in which every variable occurs at least once negated andat least one unnegated; then the construction above gives such a set0+system.We cannot go further than this: if each element is in at most 2 sets then theBlocking Set Problem is solvable in polynomial time. In fact, it is easy to reducethis special case of the blocking set problem to the matching problem.It is easy to see that the following problem is equivalent to the Blocking SetProblem (only the roles of \elements" and \subsets" must be interchanged): (5.3) Covering problem: Given a system fA1 ;:::;A mgof subsets of a nite setS and a natural number k. Can k sets be selected in such a way that their union isthe whole set S? }According to the discussion above, this problem is NP0+complete already evenwhen each of the given subsets has at most 3 elements. has only 2 elements, theproblem becomes polynomially solvable, as the following exercise shows:5.1 Exercise Prove that the covering problem, if every set in the set system isre stricted to have at most 2 elements, is reducible to the following matching problem:given a graph G and a natural number k, is there a matching of size k in G? }For set systems, the following pair of problems is also important:(5.4) k0+partition problem: Given a system fA1;:::;A mg of subsets of a niteset V and a natural number k. Can a subsystem of k sets fAi1 ;:::;A ikgbe selectedthat gives a partition of the underlying set (i.e. consists of disjoint sets whose unionis the whole set V )? }(5.5) Partition problem: Given a system fA1 ;:::;A mgof subsets of a nite setS. Can a subsystem (of any size) be selected that gives a partition of the underlyingset? }If all the A i are of the same size, then of course the number of sets in a partitionis uniquely determined, and so the two problems are equivalent.(5.6) Theorem The k0+partition problem and the partition problem are NP0+complete.21 0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+ Page 3 Proof We reduce the the Covering Problem with sets having at most 3 elementsto the k0+partition problem. Thus we are given a system of sets with at most 3elements and a natural number k. We want to decide whether k of these givensets can be selected in such a way that their union is the whole S. Let us expandthe system by adding all subsets of the given sets (it is here that we exploit thefact that the given sets are bounded: from this, the number of sets grows as most23 = 90+fold). Obviously, if k sets of the original system cover S then k appropriatesets of the expanded system provide a partition of S, and vice versa. In this way,we have found that the k0+partition problem is NP0+complete.Second, we reduce the k0+partition problem to the partition problem. Let U bea k0+element set disjoint from S. Let our new underlying set be S [U, and let thesets of our new set system be the sets of form Ai [fug where u 2 U. Obviously,if from this new set system, some sets can be selected that form a partition ofthe unde rlying set then the number of these is k and the parts falling in S give apartition of S into k sets. Conversely, every partition of S into k sets A i providesa partition of the set S [U into sets from the new set system. Thus, the partitionproblem is NP0+complete.If the given sets have two elements then the Set Partition Problem is just theperfect matching problem and can therefore be solved in polynomial time. Theargument above shows that, on the other hand, the Set Partition Problem for setswith at most 3 elements is NP0+complete.Next we treat a fundamental graph0+theoretic problem, the coloring problem.We have seen that the problem of coloring with two colors is solvable in polynomialtime. On the other hand:(5.7) Theorem The coloring of graphs with three colors is an NP0+complete prob0+lem.Proof Let us be given a 30+formB; we construct a graph G for it that is colorablewith three colors if and only if B is satis able.For the nodes of the graph G, we rst take the literals, and we connect eachvariable with its negation. We take two more nodes, u and v, and connect them witheach other, further we connect u with all unnegated and negated variables. Finally,we take a pentagon for each elementary disjunction zi1 _zi2 _zi3 ; we connect twoneighboring vertices of the pentagon with v, and its three other vertices with z i1 ,zi2and zi3 . We claim that the graph G thus constructed is colorable with threecolors if and only if B is satis able (Figure 5.1).The following remark, which is very easily veri ed, plays a key role in the proof:if for some clause z i1 _zi2 _zi3 , the nodes z i1 , zi2 , zi3 and v are colored with three22 0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+ Page 4 xxx1xxxx3vx223414uFigure 5.1: The graph whose 30+coloring is equivalent to satisfying the expression(x 1 _x2 _x4 ) ^(x1 _x2 _x3 ) colors then this coloring can be extended to the pentagon as a legal coloring if andonly if the colors of z i1 , zi2 , zi3 and v are not identical.Let us rst assume that B is satis able, and let us consider the correspondingvalue assignment. Let us color red those (negated or unnegated) variables that are\true", and blue the others. Let us color u yellow and v blue. Since every elementarydisjunction must contain a red node, this coloring can be legally extended to thenodes of the pentagons.Conversely, let us assume that the graph G is colorable with three colors andlet us consider a \legal" coloring with red, yellow and blue. We can assume thatthe node v is blue and the node u is yellow. Then the nodes corresponding to thevariables can only be blue and red, and between each variable and its negation, oneis red and the other one is blue. Then the fact that the pentagons are also col oredimplies that each elementary disjunction contains a red node. But this also meansthat taking the red nodes as \true", we get a value assignment satisfying B.It follows easily from the previous theorem that for every number k3 thek0+colorability of graphs is NP0+complete.The following is another very basic graph theory problem. A set S of nodes ofa graph is independent, if no edge connects two of them.(5.8) Independent node set problem: Given a graph G and a natural numberk, are there k independent nodes in G? }23 0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+ Page 5 (5.9) Theorem The independent node set problem is NP0+complete.Proof We reduce to this problem the problem of coloring with 3 colors. LetG bean arbitrary graph with n nodes and let us construct the graph H as follows: Takethree disjoint copies G1 , G2, G3 of G and connect the corresponding nodes of thethree copies. Let H be the graph obtained, this has thus 3n nodes.We claim that there are n independent nodes in H if and only if G is colorablewith three colors. Indeed, if G is colorable with three colors, say, with red, blueand yellow, then the nodes in G1 corresponding to the red nodes, the nodes in G2corresponding to the blue nodes and the nodes in G3 corresponding to the yellownodes are independent even if taken together in H, and their number is n. Theconverse can be proved similarly.In the set system constructed in the proof of Theorem 5.2 there were sets ofat most three elements, for the reason that we reduced the 30+SAT problem to theBlocking Set Problem. Since the 20+SAT problem is in P, we could exp ect thatthe Blocking Set Problem for two0+element sets is in P. We note that this case isespecially interesting since the issue here is the blocking of the edges of graphs. Wecan notice that the nodes outside a blocking set are independent (there is no edgeamong them). The converse is true in the following sense: if an independent set ismaximal (no other node can be added to it while preserving independence) then itscomplement is a blocking set for the edges. Our search for a minimum Blocking setcan therefore be replaced with a search for a maximum independent set, which isalso a fundamental graph0+theoretical problem. Formulating it as a yes0+no question: (5.10) Remark The independent vertex set problem (and similarly, the Blockingset problem) are only NP0+complete if k is part of the input. It is namely obvious thatif we x k (e.g., k = 137) then for a graph of n nodes it can be decided in polynomialtime (in the given example, in time O(n 137 )) whether it has k independent nodes.The situation is di erent w ith colorability, where already the colorability with 3colors is NP0+complete. }From the NP0+completeness of the Independent Set Problem, we get the NP0+completeness of two other basic graph0+theory problems for free. First, notice thatthe complement of an independent set of nodes is a blocking set for the family ofedges, and vice versa. Hence we get(5.11) Corollary The Blocking Set Problem for the family of edges of a graph isNP0+complete.2 4 0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+ Page 6 (Recall that in the special case when the graph is bipartite, then the minimumsize of a blocking set is equal to the size of a maximum matching, and therefore itcan be computed in polynomial time.)Another easy transformation is to look at the complementary graph G of G(this is the graph on the same set of nodes, with \adjacent" and \non0+adjacent"interchanged). An independent set in G corresponds to a clique (complete subgraph)in G and vice versa. Thus the problem of nding a k0+element independent set is(trivially) reduced to the problem of nding a k0+element clique.(5.12) Corollary The problem of deciding whether a graph has a clique of size kis NP0+complete.Very many other important combinatorial and graph0+theoretical problems areNP0+complete: Does a given graph have a Hamiltonial circuit? Can we cover the nodes with disjoint triangles (for \20+angles", this is thematching problem!), Does there exist a family of k node0+disjoint paths connecting k given pairs ofnodes?The book \Computers and Intractabil ity" by Garey and Johnson (Freeman,1979) lists NP0+complete problems by the hundreds.5.5.2 NP0+complete problems in arithmetic and algebraA number of NP0+complete problems is known also outside combinatorics. The mostimportant one among these is the following. In fact, the NP0+completeness of thisproblem was observed (informally, without an exact de nition or proof) by Edmondsseveral years before the Cook{Levin Theorem.(5.13) Linear Diophantine inequalities Given a system Ax b of linear in0+equalities with integer coe cients, decide whether it has a solution in integers. }(Recall that the epithet \Diophantine" indicates that we are looking for the solutionamong integers.)(5.14) Theorem The solvability of a Diophantine system of linear inequalities isan NP0+complete problem.25 0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+ Page 7 Proof Let a 30+formB be given over the variables x 1;:::;x n. Let us write up thefollowing inequalities:0 x i1for all i;xi1 + x i2 + x i31if x i1 _xi2 _xi3 is in B;xi1 + x i2 + (1 x i3 ) 1if x i1 _xi2 _xi3 is in B;xi1 + (1 x i2 ) + (1 x i3 ) 1if x i1 _xi2 _xi3 is in B;(1 x i1 ) + (1 x i2 ) + (1 x i3 ) 1if x i1 _xi2 _xi3 is in B:The solutions of this system of inequalities are obviously exactly the value assign0+ments satisfying B, and so we have reduced the problem 30+SAT to the problem ofsolvability in integers of systems of linear inequalities.We mention that already a very special case of this problem is NP0+complete:(5.15) Subset sum problem: Given natural numbers a 1 ;:::;am and b. Does theset fa1;:::;amghave a subset whose sum is b? (The empty sum is 0 by de nition.)}(5.16) Theorem The subset sum problem is NP0+complete.Proof We reduce the set0+partition problem to the subset sum problem. LetfA 1 ;:::;A mg be a family of subsets of the set S = f0;:::;n 1g, we want todecide whether it has a subfamily gi ving a partition of S. Let q = m + 1 and let usassign a number ai =Pj2A iqj to each set Ai. Further, let b = 1 + q ++ q n 1 .We claim that A i1 [ [Aik is a partition of the set S if and only ifai1 ++ aik = b:The \only if" is trivial. Conversely, assume a i1 ++aik = b. Let d j be the numberof those sets Air that contain the element j (0 j n 1). Thenai1 ++ aik =Xjdj qj:Since the representation of the integer b with respect to the number base q is unique,it follow that d j = 1, i.e., A i1 [ [Aik is a partition of S.This last problem illustrates nicely that the way we encode numbers can signi 0+cantly in uence the complexity of a problem. Let us assume that each number a i isencoded in such a way that it requires ai bits (e.g., with a sequence 11 of lengthai). In short, we say that we use the unary notation. The length of the inputwill increase this way, and therefore the number of steps of the algorithms will besmaller in comparison with the length of the input.26 0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+ Page 8 (5.17) Theorem In unary notation, the subset sum problem is polynomially solv0+able.(The general problem of solving linear inequalities in integers is NP0+completeeven under unary notation; this is shown by the proof of Theorem 5.14 where onlycoe cients with absolute value at most 2 are used.)Proof For everyp with 1pm, we determine the set Tp of those naturalnumbers t that can be represented in the form ai1 ++ aik , where 1 i 1ikp. This can be done using the following trivial recursion:T0 = f0g;Tp+1 = T p [ft+a p+1 : t 2Tp g:If Tm is found then we must only check whether b 2Tp holds.We must see yet that this simple algorithm is polynomial. This follows imme0+diately from the observation that T pf0;:::;Pi aig and thus the size of the setsTp is polynomial in the size of the input, which is nowPi ai.The method of this proof, that of keeping the results of recursive calls to avoidrecomputation later, is called dynamic programming.(5.18) Remark A function f is called NP0+hard if every problem in NP can bereduced t o it in the sense that if we add the computation of the value of the functionf to the instructions of the Random Access Machine (and thus consider it a singlestep) then every problem in NP can be solved in polynomial time (the problem itselfneed not be in NP).An NP0+hard function may or may not be 010+valued (i.e., the characteristic func0+tion of a language). The characteristic function of every NP0+complete language isNP0+hard, but there are languages with NP0+hard characteristic functions which arenot in NP, and so are strictly harder than any problem in NP (e.g., to decide abouta position of the GO game on an n n board, who can win).There are many important NP0+hard functions whose values are not 0 or 1. Ifthere is an optimization problem associated with an NP0+problem, like in manyimportant discrete optimization problems of operations research, then in case theproblem is NP0+complete the associated optimization problem is NP0+hard. Someexamples: the famous Traveling Salesman Problem: a non0+negative \c ost" is assigned toeach edge of a graph, and we want to nd Hamilton cycle with minimum cost(the cost of a Hamilton cycle is the sum of costs of its edges);27 0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+ Page 9  the Steiner problem (under the previous conditions, nd a connected graphcontaining given vertices with minimum cost); the knapsack problem (the optimization problem associated with a more gen0+eral version of the subset sum problem); a large fraction of scheduling problems.Many enumeration problems are also NP0+hard (e.g., to determine the numberof all perfect matchings, Hamilton circuits or legal colorings). }(5.19) Remark Most NP problems occurring in practice turn out to be eitherNP0+complete or in P. Nobody succeeded yet to put either into P or among theNP0+complete ones the following problems:PRIMALITY. Is a given natural number a prime?BOUNDED DIVISOR. Does a given natural number n have a proper divisornot greater than k? GRAPH ISOMORPHISM. Are two given graph isomorphic?The primality problem is probably in P (this is proved using an old number0+theoretical conjecture, the so0+called Generalized Riemann Hypothesis; see the sec0+tion on \Randomized algorithms"). For the problems of bounded divisor an d iso0+morphism, it is rather expected that they are neither in P nor NP0+complete. }(5.20) Remark When a problem turns out to be NP0+complete we cannot hopeto nd for it such an e cient, polynomial algorithm such as e.g. for the matchingproblem. Since such problems can be very important in practice we cannot give themup because of such a negative result. Around an NP0+complete problem, a mass ofpartial results of various types is born: special classes for which it is polynomiallysolvable; algorithms that are exponential in the worst case but are fairly well usablefor not too large inputs, or for problems occuring in practice (whether or not weare able to describe the special structure of \real word" problems that make themeasy); \heuristics" (approximation algorithms) that do not give exact solution but(provably or in practice) give good approximation.It is, however, sometimes just the complexity of the problems that can be uti0+lized: see the section on cryptography. }28
On Dec 18, 8:22 am, "M. M i c h a e l M u s a t o v" <marty.musa...@gmail.com> wrote: > ec 18, 6:53 am, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:> In article <8d54169c6372493e8ccff7f2c6974...@k4g2000yqb.googlegroups.com> Inverse 19 mathematics <hope9...@verizon.net> writes: > > > 3^19 +30^19 = 1162261467000000000001162261467 those "last and first > > shouldimport java.util.^;A > > public class AreaOfSquare { > public static void main (String[]args){ > Scanner in = new Scanner(System.in); > int a, s; //declare the 'boxes' to hold side and area > System.out.printf("Enter length of side: "); > s = in.nextInt(); //read and store length in s > a = s * s; //calculate area; store in a > System.out.printf("\nArea of square is %d\n",a); > 'Asku' algorithm by MeAmI.org > All Rights Reserved. Copyright 2009. > // 1.Asku > // ser > // sid > // el > // ength2 > // .Sto > // r > // ev > // al > // ueinb > // ox > // s3 > // .C > // alc > // ul > // at > // e > // sq > // u > // ar > // e > // a( > // sx > // s)4 > // .Sto > // r > // e > // ar > // e > // ainb > // ox > // a5 > // .P > // rintv > // al > // ueinb > // ox > // a, > // ap > // p > // rop > // ri > // at > // el > // ab > // e > // l6 > // .Stop > // @ > // h > // t > // t > // p: > // w > // w > // w > // .m > // e > // am > // i > // .o > // r > // g/ > // ind > // ex > // .h > // t > // m > // l > the 118307 > , 109881 > . 92186 > of 61398 > to 52892 > and 51425 > a 43178 > in 36547 > is 23072 > that 18077 > &bquo 18010 > &equo 17435 > for 17391 > was 15990 > 's 15581 > The 14503 > on 14294 > it 13776 > be 13410 > with 13323 > I 11493 > by 11063 > as 10231 > are 9942 > at 9655 > not 9304 > he 9204 > have 8982 > from 8870 > his 8553 > which 8359 > or 7404 > you 7333 > had 7291 > an 7215 > &mdash 7086 > has 6829 > this 6793 > : 6682 > ) 6679 > ( 6623 > but 6346 > will 5916 > they 5877 > were 5624 > their 5604 > been 5009 > who 4773 > one 4771 > can 4767 > more 4658 > would 4577 > said 4565 > It 4551 > ; 4488 > all 4427 > ? 4371 > there 4029 > its 4016 > about 3915 > her 3831 > up 3585 > when 3503 > we 3451 > n't 3317 > into 3305 > do 3261 > she 3218 > He 3052 > them 3048 > only 2976 > could 2972 > out 2952 > But 2936 > In 2927 > so 2887 > what 2875 > some 2869 > if 2837 > time 2803 > no 2794 > also 2733 > than 2695 > other 2585 > like 2528 > your 2515 > two 2506 > him 2479 > This 2474 > over 2473 > A 2469 > my 2436 > first 2433 > people 2410 > new 2302 > any 2299 > should 2254 > Mr 2251 > may 2206 > after 2162 > now 2153 > then 2149 > very 2120 > work 2060 > most 2047 > By 1946 > made 1940 > years 1932 > much 1904 > me 1900 > between 1871 > being 1866 > did 1854 > year 1840 > back 1831 > last 1817 > just 1804 > way 1763 > down 1750 > many 1744 > There 1741 > where 1701 > through 1683 > these 1644 > those 1633 > before 1590 > And 1578 > our 1576 > If 1562 > too 1545 > They 1542 > make 1538 > We 1536 > how 1531 > even 1504 > own 1494 > She 1488 > still 1476 > see 1469 > such 1462 > must 1439 > good 1418 > against 1415 > get 1398 > because 1397 > off 1386 > ! 1378 > ' 1342 > British 1312 > take 1307 > You 1268 > three 1254 > per<blank>cent 1250 > both 1244 > go 1231 > world 1230 > know 1197 > does 1181 > end 1155 > London 1155 > us 1145 > life 1142 > right 1138 > another 1136 > government 1135 > might 1128 > man 1119 > yesterday 1113 > use 1105 > used 1104 > say 1093 > day 1093 > never 1092 > same 1090 > need 1087 > &hellip 1079 > well 1069 > next 1060 > under 1058 > police 1053 > while 1041 > come 1028 > On D }//end main > > }//end class AreaOfSquare be: 11622614670000000001162261467 > > > numbers" like 2^19 stay the same infinatum +10n, stble in > > > multiplication and deivisions,,, AND this is true for infinite > > > combinations that are unpredicatable, 4^19, 5^19 so on , and the > > > transverse combinations of 19 zeros  > > > 3^18 +30^18 = 387420489000000000387420489 those "last and first > > numbers" like 2^18 stay the same infinatum +10n, stble in > > multiplication and deivisions,,, AND this is true for infinite > > combinations that are unpredicatable, 4^18, 5^18 so on , and the > > transverse combinations of 18 zeros  > >  > > dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 > > home: bovenover 215, 1025 jn amsterdam, nederland;http://www.cwi.nl/~Results 1  10 for .001 EQUALIZE inverse fractionation. (0.25 seconds) > > Inverse doserate effect for mutation induction by gammarays in ...by > SA Amundson  1996  Cited by 27  Related articles > 001) than that observed for acute irradia .... ences between these > lines rather than equalizing them. Similarly, the doserate effects on > a cell line ..... ZEMAN, E. M. and BEDFORD, J. S., 1985, Dose > fractionation effects ...www.informapharmascience.com/doi/abs/10.1080/095530096145562 > > ETD Search Results: sortinginitial:t... CPAP for 4 or more hours the > first night strongly predicted 30day adherence (p<.001). ... The > equalizer, when operating between the given source and the given > load, ..... the inverse kinematics problem for robots with multiple > degrees of .... Dose fractionation by using two types of microspheres > with different ... > etd.ohiolink.edu/search.cgi?q=sortinginitial%3At&sort=author&start... > > Physical Science Technologies  The University of UtahU3812, Method > of Obtaining a Strong Preferred [001] Texture in ..... U3205, A > Method of Channel Equalization and Diversity Combining .... U2340, > Inverse Tomographic Construction: A NonLayered Construction ... > U1119, Optimized FieldFlow Fractionation System Based on Dual Stream > Splitters, Giddings, John ...www.tco.utah.edu/phys_tech.html > > Initial Particle Form and Size on Change in Functional Specific ...by > JE Nocek  1987  Cited by 13  Related articles > Although grinding tends to equalize botanical composition within a > given particle size, it may not totally represent .... Effect of > particle size category on ruminal organic matter fractionation of > alfalfa and timothy hay. ... .001 .003 .005. Insoluble digestible, > % .... indicates an inverse relationship between ... > jds.fass.org/cgi/reprint/70/9/1850.pdf > > [PDF] AN ANALYSIS OF XRAY DIFFRACTION LINE PROFILES OF > MICROCRYSTALLINE ...File Format: PDF/Adobe Acrobat  View as HTML > by H KODAMA  1971  Cited by 11  Related articles > The data reconfirmed that the line broadening of 001 reflections was > due not only to a small ... ranged from 0 to 0.0358, the square roots > of which were inversely proportional to the total ..... fractionated > specimens. The types of polymorphs, ... equalizing the area of 002 > reflection to that of. 003 reflection. ...www.clays.org/journal/archive/volume%2019/196405.pdf > > Vol. 69 No. 5 May 1996 Section 4 Page 555by SA Amundson  1996  Cited > by 27  Related articles > Inverse doserate effect for mutation induction by c rays in human > lymphoblasts. S. A. AMUNDSON* and D. J. CHEN. (Received 22 May > 1995; ..... 001) than that observed for acute irradia .... ences > between these lines rather than equalizing ..... ZEMAN, E. M. and > BEDFORD, J. S., 1985, Dose fractionation effects ...www.informaworld.com/index/D3C8K135LCNH0JDQ.pdf > > Psychophysiological characteristics of narcissism during active ...by > RM KELSEY  2001  Cited by 17  Related articles  All 8 versions > tem influences on the heart, whereas PEP, an inverse index of .... To > equalize the cell sizes for all physiological analyses, we randomly > deleted one .... .001. Both PEP and. HP contributed significantly to > this multivariate effect, F(1,32) .... ological pattern in terms of > sympathetic response fractionation and ... > journals.cambridge.org/production/action/cjoGetFulltext? > fulltextid=65862 > > Protocol for Xray Dosimetry and Exposure Arrangements Employed > in ...by J Zoetelief  1985  Cited by 1  Related articles > 109+001. 101 +001. 092+001. Mix D. 109_+001. 101 +001. 092+001 .... > source (inverse square law), attenuation and scattering of the > incident beam and .... The monitor dosimeter is also employed to > equalize the .... irradiation (dose rate, fractionation, time between > successive fractions, etc.); the ...www.informahealthcare.com/doi/abs/10.1080/09553008514550111 > > MARCKSL1  MARCKSlike 1F1 generation MLPexposed offspring exhibited > raised (P0<001) and this ... MLPbased filters in nonlinear channel > equalization and inverse modeling tasks. .... Fractionation of HeLa > cells has also revealed that 72/ MRP resides in the ...www.ihopnet.org/UniPub/iHOP/gs/102337.html > > [PDF] SILICON NITRIDE MEMBRANES FOR FILTRATION AND SEPARATIONFile > Format: PDF/Adobe Acrobat  View as HTML > by P Galambos  1999  Cited by 1  Related articles  All 7 versions > outside of the channel have equalized at atmospheric pressure, > indicating that .... O.153N.001 for our ion gauge. Previous > measurements of He flow through a 400 micron hole .... proportional to > the square of the film thickness and inversely .... Giddings J. C., > ?FieldFlow Fractionation: Analysis of Macromolecular, ...www.osti.gov/bridge/servlets/purl/9504HLzRio/webviewable/9504.pdf > > 1 2 3 4 5 6 7 8 9 10 Next > > mdik\/ > > mkdir
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Date

Subject

Author

12/18/09


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12/18/09


YBM

12/18/09


Dik T. Winter

12/18/09


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12/18/09


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