Approximately 2^(n^2)-2^(2n) - 2n as n increases?
On Mon, Aug 27, 2012 at 8:16 PM, Jim Bromer <[email protected]> wrote: > Damn. > Approximately 2^(n^2)-2^(2n) - n as n increases? > > > On Mon, Aug 27, 2012 at 8:14 PM, Jim Bromer <[email protected]> wrote: >> Approximately 2^(n^2)-2^(2n)-log(2)n as n increases? >> >> >> On Mon, Aug 27, 2012 at 8:11 PM, Jim Bromer <[email protected]> wrote: >>> 2^(n^2)-2^(2n)-log(e)n as n increases approximately? (That is just a >>> guess. Something about the proportion of primes since the primes won't >>> be detected.) >>> Maybe. >>> >>> On Mon, Aug 27, 2012 at 8:07 PM, Jim Bromer <[email protected]> wrote: >>>> That last messaged contained another error. >>>> There are 2^(n^2) possible ways to fill the n^2 bits of the partial >>>> products but there are 2^(2n) possible ways to do it with a >>>> well-formed set (given a nXn bit multiplier). So the worse case is >>>> that I might have to go through would be 2^(n^2)-2^(2n) trials before >>>> I found a system that represented a number that could be factored. So >>>> the conclusion is the same. >>>> I think that is right - or at least it is getting closer to the ballpark. >>>> Jim Bromer >>>> >>>> On Mon, Aug 27, 2012 at 7:44 PM, Jim Bromer <[email protected]> wrote: >>>>> Ok the wheels are slipping but train is slowly getting up to steam. >>>>> >>>>> How many possible ways are there to set up the n^2 bits of the partial >>>>> products? 2^(n^2). Only n^2 are well-formed. (This is for a nXn >>>>> multiplication algorithm). So even if I had the dreamed of polynomial >>>>> time solution to Boolean Satisfiabiity the worse case is that I might >>>>> have to go through 2^(n^2)-n^2 trials before I got a factorization. >>>>> >>>>> So even though you do not have to make the multiplication algorithm >>>>> reversible in order to derive a Boolean Formula that will detect a >>>>> non-prime number (given a set of possible partial products), if the >>>>> goal is to use the method with a polynomial time solution to Boolean >>>>> Satisfiabilty then you are going to have to make it reversible (or >>>>> something like that) if you are want to use it to find a factorization >>>>> solution to any given number in polynomial time. >>>>> >>>>> Jim Bromer >>>>> (I might be wrong...It has happened. I admit it.) >>>>> >>>>> >>>>> >>>>> On Mon, Aug 27, 2012 at 7:21 PM, Jim Bromer <[email protected]> wrote: >>>>>> Unfortunately I was wrong about this, but the method does have a flaw. >>>>>> >>>>>> The partial products are fully specified in the sense that they detail >>>>>> the relationships between each bit as it is used in the cross-products >>>>>> in a consistent manner based on the bits of the multiplicand. So if >>>>>> you had a Boolean Solver then the Boolean formula of the cross >>>>>> products (derived from a standard multiplication algorithm) could be >>>>>> used to determine if there was a solution for a given system of >>>>>> cross-products. >>>>>> >>>>>> However, if the system detected that there was no solution to a >>>>>> problem (given a system of partial products) it would not definitely >>>>>> represent a prime number since the given system could be poorly >>>>>> formed. On the other hand it should be able to detect a >>>>>> factorization. If it found that a system could be factored based on >>>>>> the fact that there was a solution to the Boolean Formula, then the >>>>>> given system of partial products could be added to detect the number >>>>>> that can be factored. >>>>>> >>>>>> Jim Bromer >>>>>> >>>>>> >>>>>> >>>>>> >>>>>> >>>>>> >>>>>> On Mon, Aug 27, 2012 at 6:48 PM, Jim Bromer <[email protected]> wrote: >>>>>>> I was really surprised when I discovered that converting the >>>>>>> cross-products, used to determine the partial products, of a binary >>>>>>> multiplication into the form of Boolean logic was logically simple. >>>>>>> It turns out that the methodology of the multiplication problem, when >>>>>>> you are given the multiplicands is simple just as the multiplication >>>>>>> algorithm is simple. However, if you were to try to work backwards to >>>>>>> use the partial products to try to determine the multiplicands the >>>>>>> problem is much more difficult. It is simple as long as poorly formed >>>>>>> partial products have been filtered out beforehand. The formalization >>>>>>> of the problem, (expressed in pure Boolean form), is not quite so >>>>>>> simple if the algorithm is supposed to detect or avoid poorly formed >>>>>>> partial products. >>>>>>> >>>>>>> It took me a long time to figure this out so I won't be calling any of >>>>>>> the other boys in this group slow anytime soon. So anyway, if you >>>>>>> wanted to use the multiplication algorithm as a factorization >>>>>>> algorithm you would have to work the algorithm backwards in order to >>>>>>> determine the multiplicands given the product, or to determine the >>>>>>> multiplicands given a system of partial products. The use of the >>>>>>> partial products as the given isn't simple because if a bit in a >>>>>>> partial product = 0 it could be because the corresponding bits in one >>>>>>> or both of the multiplicands that produced the cross product were 0. >>>>>>> That shows that there are three ways to explain a bit in a cross >>>>>>> product that equals 0. On the other hand if the bit in the cross >>>>>>> product was 1 then we would know that both bits were 1, which shows >>>>>>> that there should be a reasonably "easy" way to define the Boolean >>>>>>> Logic of the partial products so that it would only allow correctly >>>>>>> formed partial products to get past the Boolean Formula. >>>>>>> >>>>>>> This reasoning shows how complicated turning an algorithm into a true >>>>>>> Boolean Formula can be. First we start off with the standard >>>>>>> multiplication algorithm, ok. But if you want to use that algorithm in >>>>>>> an unconventional way, you have to make sure that the implicit >>>>>>> relations are well-formed in the sense that the unconventional usage >>>>>>> will not present the given values in an unacceptable poorly-formed >>>>>>> way. So even if you want to consider the problem in purely abstract >>>>>>> way -using only Boolean variables- you have to be prepared for >>>>>>> unexpected effects when your claim jumps the conventional rails. >>>>>>> >>>>>>> If you want to take the conversion of a multiplication function into a >>>>>>> pure (variable) Boolean form and then use it to detect a factorization >>>>>>> given a solution to the derived Boolean Formula, the multiplication >>>>>>> method would first have to be expressed as a reversible function. >>>>>>> Jim Bromer ------------------------------------------- AGI Archives: https://www.listbox.com/member/archive/303/=now RSS Feed: https://www.listbox.com/member/archive/rss/303/21088071-c97d2393 Modify Your Subscription: https://www.listbox.com/member/?member_id=21088071&id_secret=21088071-2484a968 Powered by Listbox: http://www.listbox.com
