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

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...

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...

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:

Try some more, please.

sch.jnn,

TOO COOL!!!!!

I think he may be on to something!!!

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

B

TOO COOL!!!!!

I think he may be on to something!!!

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

B

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

This Is Scarry!!! :grin:

*Originally posted by sch.jnn*

```
17 * 11
```

---------

17

17

---------

187

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

Can we do 13*17 instead?
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?

the question is:

Is least significant: least or more

B

Is least significant: least or more

B

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

the question is:

Is least significant: least or more

B

For example?

quote:

"...The main question is and will be, how to make me talk more.." :grin:

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

B

"...The main question is and will be, how to make me talk more.." :grin:

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

B

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

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

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

B

edited!

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

B

edited!

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

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

bitRAKE,

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

B

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

B

bitRAKE,

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

B

VERY INTERESTING!

Compliments!

Giovanni

Compliments!

Giovanni

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

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

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 ;) ))

Thus Key

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

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

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

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