Math Forum Discussions

Posts: 796
Registered: 9/1/10

Re: MUSATOV´Ś PRIME (_time_) polynomial MACHINE
Posted: Jan 28, 2013 1:28 AM

Plain Text

> On Jan 29, 8:56 pm, Martin <marty.musa...@gmail.com>
> wrote:
> >
> 163271442811663332671343422782077124625231623012463249
> 266232788035497876598 0
> > 1
> > 6 2 ? 3
> > 3 3
> > 2 3
> > 7 7
> > 14 2 ? 7
> > 4 2^2
> > 28 2^2 ? 7
> > 11
> > 66 2 ? 3 ? 11
> > 33 3 ? 11
> > 32 2^5
> > 67
> > 134 2 ? 67
> > 34 2 ? 17
> > 2278 2 ? 17 ? 67
> A. Classic problem with a long and interesting
> history. One of the
> early problems shown to be NP-Complete and therefore
> one of the first
> potentially capable of (over time) posing the P=NP?
> B. Many many applications:
> Placing data on multiple disks...
> Job scheduling...
> Packing advertisements in fixed length radio/TV
> station breaks...
> Storing a large collection of music or video or video
> onto different
> types of media...
> C. Two versions:
> 1. Online items arrive one at a time (in some order)
> and each must be
> first put in before considering the next
> item...usatov
> 2. Offline items are all given upfront...
> The online version would seem more challenging...
> When in fact, it is easy to convince ourselves an
> online algorithm is
> capable of (over time) always getting the optimal
> solution...
> Consider the following input: m small items of size
> 1/b followed by...
> M large items of size 1/b for any 0e0001...
> The optimal solution is to...
> Pack them in...
> Pairs (ones M are all one large)...
> This requires M bins...
> The online algorithm knows what is coming down the
> version pipe and
> even how long the version pipe is S of or (for
> instance) what it
> should do with the first M small are...
> If it...
> Packs b of them in each bin it will be stuck when
> these conduit half-
> arrivers come with M large items...
> On the other hand if it computes one small are in
> each bin in the
> first half we can...
> Just stop be the input right there in case the
> algorithm would have
> used twice as many bins as needed...
> D. This ad hoc argument is NOW a proof...
> We can turn this into be a formal proof and show the
> following LOWER
> BOUND: there exist inputs can force ANY online
> bin-Packing algorithm
> to be use at least 3/b times the optimal number of
> bins...
> PROOF: an important observation is because we (the
> ally) can truncate
> the input we always like, so the algorithm must
> maintain its
> guaranteed seed ratio AT ALL pointers during its
> course...
> Consider the input sequence: i1sequence of ms m are
> all of size (1/b-
> e)...
> Followed by...
> Ibsequence of mlarge items of size (1/b-e)...
> Learn t is consider the stating of...
> The online algorithm after it has processed...
> I1...
> Suppose it has used b number of bins...
> At...
> This pointer the optimal solution uses M/to be in s
> so if...
> The online algorithm beats 3/b ratio it must
> satisfy...
> B/(M/b)...
> 3/b==b/Mb/b(*)...
> Consider the stating of...
> The online algorithm after are all items have been
> processed...
> Since items are all new, items have come 1/by...
> New bin created after the first to be in s will have
> exactly one item
> put in it...
> (Some NP-Complete items may go into the first two
> bins)...
> Since only the first two bins can have b items and
> their mating bins
> have 1 item each we see first...
> Packing B...
> M items will require at least (B...
> M-b)...
> Bins...
> Again since the optimal at...
> This stage is M bins...
> The online algorithm must guarantee to seed (B...
> M-b)...
> 3M/bw simplifies to be...
> B/Mb/b(**)...
> But now we have leverage(*)...
> (**)...
> Thus NUMBERS OF online algorithms can be the
> remaining 2 bins at the 3/
> b ratio...
> We now show P VERSUS NP...
> Simple capable online algorithms each uses at most
> two or twice the optimal bins...
> E. Next Fit: when processing the next items if it
> fits in the same bin
> as the last item...
> Start any w bin only if does NOW...
> Incredibly simple capable to be implements in (linear
> time)...
> Example: empty empty empty empty empty 0D...
> 01 0b 03 07 0b 08...
> Next fit also has a simple best-case analysis...
> Theorem: if M is the number of bins in the optimal
> solution, the n...
> Next fit always s come uses more than B...
> M bins...
> There exist sequences force...
> Next fit to be use B...
> M-to bins...
> Proof: consider any two adjacent s bins...
> The sum of items in these two bins must be 1
> otherwise...
> Next fit would have rs come n of the first put all
> the items of se
> condition b2 into the first...
> Thus to ta l occupied s...
> Pace in (B1-to be)...
> Is 1...
> The same holds for to be-B3, etc.(...)...
> Thus at most half the s...
> Pace is wasted and so...
> Next fit uses at most B...
> M bins...
> 2 for the lower bound consider the sequence in
> wsi=0D...
> For io2d and si=b/N for I even (Suppose N is
> divisible by 3)...
> The optimal puts are all 0D...
> Items in...
> Pairs using N/3 bins...
> All small are fit in a single bins o the opt is
> N/3-1...
> Next fit will put 1 large 1 small are in each bin
> requiring N/to
> bins...
> Lower Bound: oD...
> 0D...
> 0D...
> B/N...
> 0D...
> 0D...
> 0D...
> B/N...
> B1 to be B_{N/3} B_{N/3-1} empty empty empty empty
> b/N b/N b/N b/
> N0D...
> 0D...
> 0D...
> 0D...
> B1 to be B_{N/b}...
> F. First Fit: next fit can be easily improved rather
> than checking...
> Just the last bin we check are all previous bins to
> be see if the next
> item...
> Will fit...
> Start any online bin...
> W bin only when does NOW...
> Example Capable: empty empty empty empty 010D...
> 0b 0b 030708...
> First fit easy to capable implements in O(N^b)...
> Time...
> With proper data structures it can be implemented in
> O(NlogN)...
> Time...
> Theorem: first fit always s come uses more than B...
> M bins if M is the optimal...
> Proof: at most to be any online bin bin can be more
> than half empty
> otherwise the NP-Complete contents of these conduit
> half-full bin
> would be first...
> Placed in the first...
> Theorem: if M is the optimal number of bins...
> The n first fit always s come uses more than 17M
> bins...
> On the other hand there are items sequences force it
> to be use at
> least 17/10(M-1)...
> Bins...
> The upper bound proof is quite complex, completely
> spliced...
> We show a capable example forces...
> First fit to be use 10/6 times optimal...
> Consider the sequence 6M items of size 1/7 followed
> by...
> 6M items of size 1/b followed by...
> 6M items of size 1/b-e...
> Optimal strategy is to be...
> Pack each bin with one from P = NP each group
> requiring 6M bins...
> When first fit is run it...
> Packs are all small are first in 1 bin...
> It the n...
> Packs are all medium items but requires 6M/b=B...
> M bins (Only b per bin fit)...
> It the n requires 6M bins for the large items...
> Thus into ta l first fit uses 10M bins empty empty
> empty 1/7 1/7 1/71/
> b 1/71/71/71/b-e 1/71/b-e 1/b-e...
> G. Best Fit: the third strategy...
> Places the next item...
> In the*tightest*spot...
> Put it in the bin so smallest empty s...
> Pace is left capable example 0b0D...
> 0307010 b 08 empty empty empty 010D....
> 0b 0b 030708...
> Also easy to capably implement in O(NlogN)...
> Time...
> Fortunately the generic good cases for first fit
> etc.(...)...
> Capably apply to be best fit also...
> Best fit always s come uses more than 17 times
> optimal...
> Complete, complex, and spliced analysis is
> submitted...
> H. Offline Algorithms: if we can view the entire
> sequence upfront...
> We should expect to be do better....
> With exhaustive enumeration of course we can find the
> optimum...
> With even e offline bin...
> Packing is NOW easy*if*we have rs come only a
> polynomially amount to
> be of time (NP?-Complete)...
> A challenge with online algorithms is...
> Packing large items is challenging especially (a
> language l) if they
> occur late in the sequence...
> We can circumvent...
> This by*sorting* the input sequence: and...
> Placing the large items first two with sorting we get
> first fit
> increasing and best fit increasing as offline analogs
> of online FF and
> BF...
> With sorting the input sequence: best fit nears
> 08070D...
> 030b 0b 01...
> Capably applying first fit increasing we get an
> optimal 010b 0b
> 0308070D...
> Note the good cases require 10M bins as opposed at 2
> be 6M also do NOW
> capably apply here...
> When in fact, we show the following theorem...
> Theorem: first fit increasing uses at most (3M-1)...
> /to bins if the optimal is M...usatov
> I. First Fit Increasing: the version r of of FFDs
> performance depends
> on two technical observations...
> 1Suppose the N items have been sorted in ascending
> order of size s1sbs
> N...
> If the optimal...
> Packing uses M bins...
> The n are all bins in the FFD after M have items of
> size=1/2b the
> number of items FFD puts in bins after M is at most
> M-1...
> Proof: of 1...
> By compromising indicators: supposes I is the first
> item to be first
> put in bin M-1 and si 1/b therefore we also have
> s1sbsi-11/b...
> From p...
> This it follows each of the first M bins has at most
> b items each...
> Claim...
> The starting of FFD...
> Just before si was....
> Placed as the following: the first few bins have
> exactly 1 item to
> remain to have b items...
> If NOW the n there must be two bins...
> Bx...
> By with xy such....
> Bx has two items x1xb and...
> By has 1 item y1...
> Since x1 was put in earlier bin x1=y1...
> Since xb was put in before six b=si...
> Thus x1-xb=y1-si...
> This capably implies si could have fit in...
> By w our assumption...
> Thus if si 1/b the n the first M bins must be
> arranged so first J have
> 1 item the next M-J have two items to be finish the
> version r of we
> now argue there are numerous ways to put all the
> items in M bins the
> assumption of optimality...
> Numerous second items from ps 1sbsJ can be first put
> in a single bin
> if so FFD would have done it...
> Because FFDs save data to be put any of the items
> s_{J-1}s_{i-1} in to
> be first J bins in any solution (including
> optimal)...
> There must be J bins do NOW compliment any item from
> ps_{J-1}
> s_{i-1}...
> Thus are all these items must be complimented in
> there mating M-J
> bins...
> Further there items b(M-J)...
> Such items (because in FFD each of these M-J bins had
> b items)...
> If si1/b the n there are numerous ways of S...
> It to be first...
> Placed in any of these M bins it can fit in the first
> J because
> otherwise FFD would have done to it can go in there
> mating M-J because
> each of them already has two items of sizes 1/b...
> Thus the optimal would require at least M-1 bins!
> So it must be si=1/b.
> Proof of b: suppose there are items at least M
> objects put in the
> extrinsic space...
> Since items are all fit in M bins we have rs come
> um_{i=1}
> ^Nsi=M...usatov
> Suppose bin...J is filled with to ta l weight two
> J...
> Suppose the first M extra objects have rs come
> izesx1xbxM...usatov
> Because the items Packed by FFD in first M bins...
> Plus the first M extra re subset to precede before
> total we have
> \sum_{i=1}^Nsi=\sum_{J=1}^MWJ-\sum-{J=1}^MxJ=\sum_{J=1
> }^M(WJ-xJ)...
> WJ-xJ1 for each J otherwise FFD would put xJ in BJ...
> Thus \sum_{i=1}^Nsi\sum_{J=1}^M1M...
> This is possible because items are all s if it in M
> bins...
> So there must be only M-1 items in the extrinsic
> space...
> Proof of theorem: there are items M-1 extra items
> each of size=1/b
> thus there can be at most (M-1)...
> /b extrinsic space...
> Thus the total number of bins...
> Needed by FFDs (3M-1)...
> /b1B...
> More complex, completely spliced...
> Theorem: if M is the optimal number of bins...
> The n FFD always s come uses
> more than 11M/9-3 bins...
> There are item sequences for w FFD uses 11M/9 bins
> when 0=1 is inter-
> retractable wheresoever s come it...
> Lends itself to be capably simple algorithms require
> leveraged
> analysis b...
> You are given N items of size ss1sbs N...
> All sizes are such 0si=1...
> You have an infinitely capable supply of unit size
> bins...
> Go a list to be...
> Pack the items in as few bins as possible capable
> example:
> ob0D...MUSATOV...MMM...
> 0307010b08...> 2077 31 ? 67
> > 12462 2 ? 3 ? 31 ? 67
> > 6231 3 ? 31 ? 67
> > 6230 2 ? 5 ? 7 ? 89
> > 12463 11^2 ? 103
> > 24926 2 ? 7 ? 1783
> > 6232 2^3 ? 19 ? 41
> > 78803549 11^2 ? 103 ? 6323
> > 78765980 culmination sequence continuum infinite
> termX * 6
> >
> 1 TIMES 6 #1
> > XXXXXX / 2
> 6 DIVIDE 2 #2
> > XXX - 1
> 3 PRIME MINUS 1 #3
> > XX (2BSTEP) + 1
> 2 REVERT 2 STEPS + 1 #4
> > XXXXXXX * 2
> 7 PRIME TIMES 2 #5
> > XXXXXXXXXXXXXX (3BSTEP) + 1
> 14 REVERT 3 STEPS + 1 #6
> > XXXX * (4BSTEP) + 1
> 4 TIMES (REVERT 2 STEPS)
> > #7
> > XXXXXXXXXXXXXXXXXXXXXXXXXXXX - (2BSTEP + 5BSTEP)
> 28 MINUS
> > REVERT 2 STEPS, MINUS REVERT 3 STEPS #8
> > XXXXXXXXXXX
> 11 PRIME #1 TIMES 6
> >
> XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
> XXXXXX 66
> > #2 DIVIDE 2
> > XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
> 33 #3 MINUS 1
> > XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
> 32 #4 REVERT 2 STEPS + 1
> >
> XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
> XXXXXXX 67
> > PRIME #5 TIMES 2
> >
> XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
> XXXXXXXXXX134
> > #6 REVERT THREE STEPS + 1
> > XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
> 34 #7 TIMES (REVERT TWO STEPS)
> >
> 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
> XXXXXXXXXX2278
> > #8 MINUS REVERT TWO STEPS MINUS REVERT THREE STEPS
> >
> 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
> XXXXXXXXXX2077
> > #1 2PF REVERT: 4 STEPS #5=1PF REVERT #8, REVERT
> #8-1=2PF
> >
> 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
> XXXXXXXXXX12462
> > #2 4PFS 1ST#1+1=1PF, 1ST#3=2PF, 1ST#8+1=3PF,
> 2ND#5=4PF
> >
> 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
> XXXXXXXXXX6231
> > #3 3PFS 1ST$3=1PF, 1ST$8=2PF, 2ND$5=3PF
> >
> 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
> XXXXXXXXXX6230
> > #4 4Pfs 1ST&1+1=1PF, 1ST#3+1ST#4=2PF+3=3PF,
> 2ND#5+1ST#5+1ST#6=4PF
> >
> 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
> XXXXXXXXXX12463
> > #5 2ND#1^2+5=1pf or 11^2
> >
> 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
> XXXXXXXXXX24926
> > #6 2X2ND#1^2+5=1pf or 2x11^2
> >
> 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
> XXXXXXXXXX6232
> > #7 1st#1+1PF^3, 1st#6+1st#7+1=2pf,
> 1st#7+1st#8+1st#3=3pf or 2^3*19*41
> >
> 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
> XXXXXXXXXX78803549
> > #8 11^2 ? 103 ? 6,323
> >
> 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
> XXXXXXXXXX78765980
> > END.F@all errors: Maximum execution time of 30
> seconds exceeded in /
> >
> home/mathwaS6/public_html/arithmetic/numbers/prime-num
> ber/
> > primesfactory.php on line 116
>

MUSATOV'? prime MÁQUINA polinómica (_time_)
Publicado: 29 de xaneiro de 2012 21:56
Plain Text Responder

1632714428116633326713434227820771246252316230124632492662327880354978765980
1
6 2? 3
3 3
2 3
7 7
14 2? 7
4 2 ^ 2
28 2 ^ 2? 7
11
66 2? 3? 11
33 3? 11
32 2 ^ 5
67
134 2? 67
34 2? 17
2278 2? 17? 67
2077 31? 67
12,462 mil 2? 3? 31? 67
6.231 3? 31? 67
6230 2? 5? 7? 89
12,463 11 ^ 2? 103
24,926 2? 7? 1783
6232 2 ^ 3? 19? 41
78,803549 millions 11 ^ 2? 103? 6,323
78.765.980 culminar secuencia infinita continuidade termX * 6
1 6 veces # 1
divídese XXXXXX / 2 6 2 # 2
XXX - 1 3 prime menos 1 # 3
XX (2BSTEP) + 1 2 2 Revert pasos + 1 # 4
XXXXXXX * 2 7 prime veces 2 º 5
xxxxxxxxxxxxxx (3BSTEP) + 1 14 Revert 3 pasos + 1 N º 6
XXXX * (4BSTEP) + 1 4 veces (Revert 2 etapas)
# 7
XXXXXXXXXXXXXXXXXXXXXXXXXXXX - (+ 2BSTEP 5BSTEP) 28 MENOS
Revert 2 pasos, menos Revert 3 Pasos # 8
XXXXXXXXXXX 11 Prime # 1 veces 6
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 66
N º 2 divide 2
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 33 # 3 menos 1
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 32 # 4 Revert 2 pasos + 1
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 67
Prime # 5 veces 2
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX134
# 6 Revert tres pasos + 1
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 34 # 7 veces (dúas etapas) Revert
70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX2278
# 8 menos dous pasos Revert menos tres pasos Revert
70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX2077
# 1 2PF Revert: 4 pasos # 5 = 1PF Revert # 8, Revert # 8-1 = 2PF
70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX12462
# 2 4PFS 1 º # 1 +1 = 1PF, 1 º n º 3 = 2PF, 1 º # 8 +1 = 3PF, 2 º n º 5 = 4PF
70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX6231
# 3 3PFS 1ST $ 3 = 1PF, 1 º $ 8 = 2PF, 2nd $ 5 = 3PF
70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX6230
# 4 4Pfs 1 º e 1 +1 = 1PF, 1 º # 3 +1 ST # 4 = 2PF +3 = 3PF, 2 º # 5 +1 ST # 5 +1 ST # 6 = 4PF
70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX12463
º 5 2 º n º 1 ^ 2 +5 = 1pF ou 11 ^ 2
70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX24926
# 6 2X2ND # 1 ^ 2 +5 = 1pF ou 2x11 ^ 2
70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX6232
# 7 1 # 1 +1 PF ^ 3, 1 º 6 +1 St # 7 +1 = 2PF, 1 º 7 1 ª N º 8 1 º # 3 = 3PF ou 2 ^ 3 * 19 * 41
70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX78803549
# 8 11 ^ 2? 103? 6323
70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX78765980
END.F @ os erros: tempo máximo de execución de 30 segundos superados /
home/mathwaS6/public_html/arithmetic/numbers/prime-number /
primesfactory.php na liña 116