crypto thought crackme #6 [Archive] - RCE Messageboard's Regroupment

View Full Version : crypto thought crackme #6

mike

May 20th, 2006, 11:58

Not quite crypto, but still a fun little puzzle. Start with s=2^n. Then pick the first fraction f in the list such that sf is an integer. Repeat until there is no such fraction. What's the result?

Code:



33   5  13  1  2  10  7

--, --, --, -, -, --, -

20  11  10  5  3   7  2

Admiral

May 20th, 2006, 14:42

Hey

I'm not sure if I totally get this game. My first answer was n=0, 'cause the list contains no integers. That seems a little silly, so I guess we have to start at n=1 (which forces us to pick the last fraction).

The only problem is that you've missed out the 'inductive step'. What exactly does 'repeat' mean? if it means 'increment n' as I'd have guessed then we may be here for a long time, as (7/2)*2^n = 7*2^(n-1) which is integral for every n >= 1.

Maybe somebody can clear this up for me.

Regards
Admiral

mike

May 20th, 2006, 17:29

The assignment should be s[I]=s[i-1]*f[i-1] at each step, where f[i-1] is the first fraction in the list such that s[i-1]*f[i-1] is an integer.

Assume n>4 for now. So the first steps go
s=2^n, f=7/2
s=7*2^{n-1}, f=10/7
s=5*2^n, f=33/20
s=3*11*2^{n-2}, f=5/11
s=3*5*2^{n-2}, f=33/20
s=3^2*11*2^{n-4}, f=5/11
s=3^2*5*2^{n-4}, etc.

P.S. If you have an answer, remember to ROT13 so others can still play!

LLXX

May 20th, 2006, 20:05

Numbers don't rot13 well... let's use base64 to encode the solution instead

Code:

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I think that's the first part of the solution...

wtbw

May 20th, 2006, 20:23

I'll follow LLXX's example...

Code:

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Edit: Uh.. guess it would be nicer spread out!

Will

gabri3l

May 21st, 2006, 03:38

Cool little math puzzle.

Heres my proof:
I am think it is correct so I use "proof" very lightly.

Code:

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I must warn you I say divide a few times when talking about the numerators and denominators but I mean multiply by the fractions, just didn't feel like going back and fixing it.

mike

May 22nd, 2006, 08:00

LLXX: You got the first three terms right. What's the pattern?
wtbw: You got the first three terms, but messed up on n=4

Code:



s=2^n, f=7/2

s=7*2^{n-1}, f=10/7

s=5*2^n, f=33/20 

s=3*11*2^{n-2}, f=5/11

s=3*5*2^{n-2}, f=33/20

s=3^2*11*2^{n-4}

s=3^2*11, f=5/11 

s=3^2*5, f=1/5

s=3^2*11, f=2/3 <-- here, should be 3^2, not 3^2*11

To avoid typos, why don't you write up some code to follow the algorithm, and then see what happens?

gabri3l:

Quote:

[Originally Posted by gabri3l]Fascinating. It plays on the parts of the fractions.

Yep.

Quote:

[Originally Posted by gabri3l]So we can predict that the next fraction we choose is 10/7.
So our next equation is:
(10/7)(7*2^(n-1))
which is equal to
10*2^(n-1)

Yes, though it would probably be more useful to write 5*2^n.

Quote:

[Originally Posted by gabri3l]Here we need to find a divisor for the product of 10 and 2.
Our options have to be products of primes 2 and 5.

2,5,10,20 are all options for the denominator.

Here is where the options vary, we are going to examine each one to its conclusion.

No, you have no options! There is only one fraction that you're allowed to pick at each stage. The only option you have is choosing n.

Quote:

First we will pick the simple 1/5

You can only pick 1/5 if 33/20 doesn't work. And that always works if n>1.

wtbw

May 22nd, 2006, 09:11

Ah, what a stupid mistake, sorry Mike.

I see that there is a little more fun in this yet.

Code:

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Cheers,

Will

Edit: fixed a typo

gabri3l

May 22nd, 2006, 10:37

ahh! I missed this part in the instructions:

Quote:

first fraction f in the list

Man that would have made all that writing much easier. lol

Thank god this wasn't a test because seems that little fact now changes everything. I would have failed.

mike

May 22nd, 2006, 11:17

wtbw: You got it! The pseudocode itself is a proof, so you're good there. Here's some more info.

Code:

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gabri3l: Yes, that rule makes all the difference, doesn't it?

Next challenge for everyone:

Code:

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wtbw

May 22nd, 2006, 11:45

More more more :-)

Code:

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bGwNCnJlcGVhdCBhcyBkZXNpcmVkLg==

Edit is explained inside!

Will

mike

May 23rd, 2006, 16:56

wtbw:

Code:

SXQgbG9va3MgbGlrZSB5b3UncmUgdHJ5aW5

nIHRvIHNvbHZlIDNebiAyXm0gLT4gNV5IVyhuIHhvciBtKSwgDQp3

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aGUgZnJhY3Rpb24h

JMI

May 25th, 2006, 19:31

My apologies to wtbw for having overwritten his Post four times with my efforts at creating a second forum with the current database to practice in setting up the OllyForum, which will be appearing here shortly.

I thought I had solved all the problem, but still had one last glitch which caused the database to, again, revert to the 05.23.06 backup and then be brought up again to the 05.24.06 backup. Unfortunately, when that backup was made, wtbw had not reposted his contribution, which he hopefully can post one last time without me screwing it up again.

Our test forum is up and running with the 05.24.06 current forum backup and we are about to attempt importing the OllyForum, which Kayaker worked dilegently on to convert, then convert the conversion, and finally import into a working vBulletin forum. Now we attempt to import into a working copy of the current Forum and see how that works. If all goes smoothly, we will have it open very soon.

Regards,

wtbw

May 26th, 2006, 01:34

What a mission :-)

Code:

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Will

mike

May 26th, 2006, 20:44

wtbw: Great! I hope you can see that the system is universal for computation. Can you come up with a systematic way to call a subroutine? That is, given your program, how could you use it in another program that may use those primes for something else?

wtbw

May 27th, 2006, 07:54

Yeah, it's a nice thing indeed, thank you for sharing it

I haven't coded anything up for this, because it's just a first idea and I think it would be horrible to code (and I'm exhausted from doing the T2 challenge).. but how about using two primes above whatever the function uses as a "store"...

One prime (let's say 7) will be a count of bits used per prime, and the other (say 11) will be huge, being a concatted sequence of binary numbers.

eg: before the call we have 2*2*2*3*5*5*5*5*7*11*11*13

In our call we use 2,3,5. 7 and 11 are used for this. 13 we can leave alone.

So we store as follows before we do any of our call:
Maximum number of bits needed is 3 (to store four 5s). So we have 7*7*7.

then we have three bits for each of 2,3,5,7,11:
2: 3 - 011
3: 1 - 001
5: 4 - 100
7: 1 - 001
11: 2 - 010

concat them all: 011 001 100 001 010 = 0x330A = 13066decimal.

So we create 11^13066. This may seem a bit excessive (it is, I'm sure), but it leaves all the information intact so as long as it's possible to do the encoding.... now I think about it, maybe that's the difficulty here, not representing it... how do you start the sequence off without losing anything, if you can't assume that any particular prime is free? How do you know that whatever possible "first" step is done was not the state beforehand, if you see what I mean... maybe the key is to always dedicate one/two/whatever as free and available at the start of calls? You tell me mike!

I've left this reply cleartext because I think it suits more of a discussion now... and if people are scrolling down this far anyway they probably deserve it

But someone feel free to edit me if you disagree.

Will

dELTA

May 28th, 2006, 10:48

Yeah, please cleartext from now on guys, the base64 posting is starting to get a bit old...

mike

May 28th, 2006, 21:03

I was just thinking something along the lines of shifting all the primes. That is, if we let p[I] be the ith prime (i is the index on the prime), then if I've already used primes with indices 1 to n, then if I add n to all the prime indices in the subroutine I'm guaranteed that there won't be any collisions.

By the way, this system is called Fractran and was invented by John Conway.

wtbw

May 29th, 2006, 00:40

Oh I see, I thought you meant doing it dynamically, without any knowledge about the use of primes in what you were inserting it to.

Cool

Will