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9/1/10


Re: MUSATOV´Ś PRIME (_time_) polynomial MACHINE
Posted:
Mar 5, 2012 1:37 AM


On Jan 29, 8:56 pm, Martin <marty.musa...@gmail.com> wrote: > 163271442811663332671343422782077124625231623012463249266232788035497876598 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 NPComplete 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 binPacking 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/be)... 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 NPComplete 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... Mb)... Bins... Again since the optimal at... This stage is M bins... The online algorithm must guarantee to seed (B... Mb)... 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 bestcase 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... Mto 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 (B1to be)... Is 1... The same holds for to beB3, 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/31... 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/31} 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 NPComplete contents of these conduit halffull 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(M1)... 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/be... 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/be 1/71/be 1/be... 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 (3M1)... /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 M1... Proof: of 1... By compromising indicators: supposes I is the first item to be first put in bin M1 and si 1/b therefore we also have s1sbsi11/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 x1xb=y1si... 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 MJ 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_{J1}s_{i1} in to be first J bins in any solution (including optimal)... There must be J bins do NOW compliment any item from ps_{J1} s_{i1}... Thus are all these items must be complimented in there mating MJ bins... Further there items b(MJ)... Such items (because in FFD each of these MJ 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 MJ because each of them already has two items of sizes 1/b... Thus the optimal would require at least M1 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(WJxJ)... WJxJ1 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 M1 items in the extrinsic space... Proof of theorem: there are items M1 extra items each of size=1/b thus there can be at most (M1)... /b extrinsic space... Thus the total number of bins... Needed by FFDs (3M1)... /b1B... More complex, completely spliced... Theorem: if M is the optimal number of bins... The n FFD always s come uses more than 11M/93 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 > XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 66 > #2 DIVIDE 2 > XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 33 #3 MINUS 1 > XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 32 #4 REVERT 2 STEPS + 1 > XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 67 > PRIME #5 TIMES 2 > XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX134 > #6 REVERT THREE STEPS + 1 > XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 34 #7 TIMES (REVERT TWO STEPS) > 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX2278 > #8 MINUS REVERT TWO STEPS MINUS REVERT THREE STEPS > 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX2077 > #1 2PF REVERT: 4 STEPS #5=1PF REVERT #8, REVERT #81=2PF > 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX12462 > #2 4PFS 1ST#1+1=1PF, 1ST#3=2PF, 1ST#8+1=3PF, 2ND#5=4PF > 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX6231 > #3 3PFS 1ST$3=1PF, 1ST$8=2PF, 2ND$5=3PF > 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX6230 > #4 4Pfs 1ST&1+1=1PF, 1ST#3+1ST#4=2PF+3=3PF, 2ND#5+1ST#5+1ST#6=4PF > 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX12463 > #5 2ND#1^2+5=1pf or 11^2 > 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX24926 > #6 2X2ND#1^2+5=1pf or 2x11^2 > 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX6232 > #7 1st#1+1PF^3, 1st#6+1st#7+1=2pf, 1st#7+1st#8+1st#3=3pf or 2^3*19*41 > 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX78803549 > #8 11^2 ? 103 ? 6,323 > 70XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX78765980 > END.F@all errors: Maximum execution time of 30 seconds exceeded in / > home/mathwaS6/public_html/arithmetic/numbers/primenumber/ > primesfactory.php on line 116


Date

Subject

Author

1/29/12



1/29/12


adamk

3/5/12



3/5/12


adamk

1/28/13




