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Topic: N^19 + 10N ^19 = XN000000000-XN infinatum-- CASE AGAINST

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Re: N^19 + 10N ^19 = @HTTP://WWW.MEAMI.ORG/-XN infinatum-- CASE

Posted: Dec 18, 2009 8:51 PM
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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

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

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

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

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

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

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

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

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"
<> wrote:
> ec 18, 6:53 am, "Dik T. Winter" <> wrote:> In article <> Inverse 19 mathematics <> 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(;
>     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
> 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; 1 - 10 for .001 EQUALIZE inverse fractionation. (0.25 seconds)

> Inverse dose-rate effect for mutation induction by gamma-rays in
> 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 dose-rate effects on
> a cell line ..... ZEMAN, E. M. and BEDFORD, J. S., 1985, Dose
> fractionation effects
> ETD Search Results: sortinginitial:t... CPAP for 4 or more hours the
> first night strongly predicted 30-day 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 ...
> Physical Science Technologies - The University of UtahU-3812, Method
> of Obtaining a Strong Preferred [001] Texture in ..... U-3205, A
> Method of Channel Equalization and Diversity Combining .... U-2340,
> Inverse Tomographic Construction: A Non-Layered Construction ...
> U-1119, Optimized Field-Flow Fractionation System Based on Dual Stream
> Splitters, Giddings, John
> Initial Particle Form and Size on Change in Functional Specific
> 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 ...
> 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.
> Vol. 69 No. 5 May 1996 Section 4 Page 555by SA Amundson - 1996 - Cited
> by 27 - Related articles
> Inverse dose-rate 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
> Psychophysiological characteristics of narcissism during active
> 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 ...
> fulltextid=65862
> Protocol for X-ray Dosimetry and Exposure Arrangements Employed
> in J Zoetelief - 1985 - Cited by 1 - Related articles
> 109+001. 101 +001. 092+001. Mix D. 109_+001. 101 +001. 0-92+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
> MARCKSL1 - MARCKS-like 1F1 generation MLP-exposed offspring exhibited
> raised (P0<001) and this ... MLP-based filters in nonlinear channel
> equalization and inverse modeling tasks. .... Fractionation of HeLa
> cells has also revealed that 7-2/ MRP resides in the
> 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.,
> ?Field-Flow Fractionation: Analysis of Macromolecular,
>  1 2 3 4 5 6 7 8 9 10 Next
> mdik\/
> mkdir


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