I only know that I was in similar positions - and I was right and sincere - till I just *learnt* not to claim what I want to keep as "trade secrets" anyway.
That's simply maturing. When you open your mouth you gotta prove that, simple. Otherwise accept to look like someone who is "spandimerding" (an old demo scene word which means you're claiming false things).
I cover/balance my lack of maturity of not wanting to share the most important things, with my maturity to keep my mouth closed.
But at the end I believe that what has to speak for itself is the final result.. the EXE.
For that one has to be concrete and productive - i.e. not waste his time posting 20 hours a day, but rather code. ;)

It might seem strange, but after 23 years of coding I've not the strength to code such a program. I cannot follow nested rollbacks any more. I am not that old, but I start to see some limits. At least I see them :)

Giovanni
Posted on 2002-05-07 03:21:28 by sch.jnn
If you don't feel you have the strength to code something like this, I'm pretty sure 'someone' *cough* would be willing to help you *cough*. :grin: :grin:

The reason I gave up was because it seemed a lot of work. Of course I had a bit of a crush on that prof so maybe that factored into it too :(. Really cute girl :grin: . Besides, I thought that the 'endings' solution was too simple so maybe someone else would come up with it, with a better result. Or find a better solution, one more elegant... you for example.

My thinking is, if you're factoring, the least number of factors you'll get is 1 (for a prime). The second least is having two prime factors. In RSA the public key is the product of two prime factors so it's not enough of a 'challenge' :rolleyes: . So if the number was the product of three or n prime factors, it'll be more challenging. As if I could actually solve any of it...
Posted on 2002-05-07 04:14:06 by AmkG

If you don't feel you have the strength to code something like this, I'm pretty sure 'someone' *cough* would be willing to help you *cough*. :grin: :grin:

My thinking is, if you're factoring, the least number of factors you'll get is 1 (for a prime). The second least is having two prime factors. In RSA the public key is the product of two prime factors so it's not enough of a 'challenge' :rolleyes: . So if the number was the product of three or n prime factors, it'll be more challenging. As if I could actually solve any of it...

You went out of road. If you had more than 2 factors, my algo would not work.

17 * 11
-- -------
17
017
---------
187

Hint: The solution is merely into what you see than what it makes, WYSIWYG ;)

The initial zero has to be ignored. It's only a graphical adjustment because the message parser does a left trim.

Giovanni

PS: Thanks for your offer. Actually I don't like even to think about how to code an EXE, but I am sure as soon as somebody follows our little message exchange, there will be many of them ready to go. The main question is and will be, how to make me talk more :grin:

Posted on 2002-05-07 08:07:47 by sch.jnn
sch.jnn,

TOO COOL!!!!!

I think he may be on to something!!!

what you see, will allow you to reduce the possible answers

B
Posted on 2002-05-07 08:30:27 by Brad

sch.jnn,

TOO COOL!!!!!

I think he may be on to something!!!

what you see, will allow you to reduce the possible answers
B

Yep! Think easy!

Giovanni
Posted on 2002-05-07 08:56:22 by sch.jnn
This Is Scarry!!! :grin:
Posted on 2002-05-07 08:57:44 by Brad
Originally posted by sch.jnn
``````17 * 11
---------
17
17
---------
187``````
Hint: The solution is merely into what you see than what it makes, WYSIWYG ;)
How do you know the 7 in 187 comes from 1*7 rather than 3*9?
``````A * B = Xn

| 1 3 7 9 = A
-----------
1 | 1 7 3 9 \
3 | 3 1 9 7  \_B
7 | 7 9 1 3  /
9 | 9 3 7 1 /
n``````
The 11 is a little too deceptive for me? ;)
``````  17
x 13
----
17
51
----
221``````
Now what is it that I'm looking for?

Edit: So, you get a decision tree that is exponential
with respect to the number of digits in the product?
Posted on 2002-05-07 08:58:28 by bitRAKE
the question is:

Is least significant: least or more

B
Posted on 2002-05-07 09:02:34 by Brad

Now what is it that I'm looking for?

Please try to use primary numbers which single additions don't produce carries (for now). You'll see better.

Anyway, 11 *is* malicious :grin:

Giovanni
Posted on 2002-05-07 09:07:22 by sch.jnn

the question is:

Is least significant: least or more

B

For example?
Posted on 2002-05-07 09:11:33 by sch.jnn
quote:
"...The main question is and will be, how to make me talk more.." :grin:

I'm trying....give me a minute

B
Posted on 2002-05-07 09:16:48 by Brad

Please try to use primary numbers which single additions don't produce carries (for now). You'll see better.
Once you have your guess, you can calculate the carries. It is still almost binary expansion on 'guess' space as the number of digits increase.
Posted on 2002-05-07 09:25:05 by bitRAKE
"....It is still almost binary expansion on 'guess' space...."

I don't think so....try to only carry a 1

B

edited!
Posted on 2002-05-07 09:28:34 by Brad

"....It is still almost binary expansion on 'guess' space...."

I don't think so....try to only carry a 1
I don't understand you - the carry has nothing to do with the 'guess' space. :tongue: Maybe, I'm hard to understand, too. :tongue:
Posted on 2002-05-07 09:39:58 by bitRAKE
bitRAKE,

The guesses should resolve as you step through the carries....I think!

B
Posted on 2002-05-07 09:45:06 by Brad

bitRAKE,

The guesses should resolve as you step through the carries....I think!

B
Try it with bigger numbers - this doesn't happen completely - there is some pruning. The probagation of the carries actually prevents you from determining which guess is correct.
Posted on 2002-05-07 09:48:08 by bitRAKE
VERY INTERESTING!

Compliments!

Giovanni
Posted on 2002-05-07 10:14:40 by sch.jnn
ARRRGH!!!!

Got to go to work!!!

Hey, please everyone, stop working on this until I get back :grin:

anyway, I believe we should only look at the addition carries....not the multiplication carries....

Good Luck!!!!

B
Posted on 2002-05-07 10:24:23 by Brad
I've been silently watching this tread, and doing some math on my own.

There is one interesting point not being said yet.

(( bitRake, was surprised you made a simular lookup table ;) ))

The problem: Key == Prime1 (P1) * Prime2 (P2)

Thus Key Is a product of two primes. (no biggy here).

Now, from the table, and the last digit of the key, you can deduce a good estimated choice of what the last digists of the two primes will be.

When visually multiplied as suggested to do so, you can see that the first 'stage' product is interestingly enough, a product of two primes.

``````  17
13
----
51  <- Prod of two primes (3 and 17)
17
====
221

Thus Key = (P1 * P2) = P1 * (P2' + 10*P2")

Where P1 = 17 (unknown)
p2 = 13 (unknown)
P3 = 3 (known Guess)
P2' = Last dig. P2 MUST be 7 (known)``````

The last intersting point, is that at BEST, the intermittant product of two primes is one digit LESS than the total. But we still dont know '51'. Its like we're starting all over again, but one digit back!.

To me this is a Deeeeeeeeply rucursive solution. And through rucursive concultions, you can expand outward to the final two "large" primes. Since if i was able to get '51', i would know that P1 == 17. And then i can subtract this from the 221 key value and know the result is x * 10 * P1. All of which is solveable. Just need to 'recurse' to it.

My problem with this idea, is i cant figure out your big-number library. I try to run it in VB and it just 'flashes' quickly a few numbers and the window is gone before i realize it. (( Im too lazy to learn VB... :( ))

Anywho this is what i see in this problem.
NaN
Posted on 2002-05-07 10:31:42 by NaN

anyway, I believe we should only look at the addition carries....not the multiplication carries....

It's not the point yet to think much about carries. It's all much easier. But it took me a very long time to see (about 6 months) what I had to see. From then on it took only less than a month to work it all out. But carries *are* important, later.

Good work, I'm off this evening :grin:

Giovanni
Posted on 2002-05-07 10:36:36 by sch.jnn