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
