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
