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
