Date: Jan 28, 2013 1:28 AM Author: Subject: Re: MUSATOV´Ś PRIME (_time_) polynomial MACHINE > 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