This message was too old and has been purged

[minerpeabody]

Sorry guys, I had a lot of hassle due to my paper deadline and some other ongoing projects.
I have just tried to eval the formula in Mathematica 9.1, however it does not yield to a valid result?
Any Ideas what might be wrong?

A huge problem are the "Null" exponents abviously.
Something somehow cannot be evaluated analytically.

http://imagize[Suspicious link removed]ageshack.us/v2/1280x1024q90/545/5f8u.png

http://imagize[Suspicious link removed]ageshack.us/v2/1280x1024q90/833/81py.png

Did you try the latest version of the program that uses -1 as a base instead of 0?  Also, I have access to Mathematica, so perhaps I can see what you are up against...

MP

[BurtW]

Yes, I think he may need to try the new version here:

https://bitcointalk.org/index.php?topic=427712.msg4769094#msg4769094

Not any previous version.

[Evil-Knievel]

This message was too old and has been purged

[minerpeabody]

Sorry guys, I had a lot of hassle due to my paper deadline and some other ongoing projects.
I have just tried to eval the formula in Mathematica 9.1, however it does not yield to a valid result?
Any Ideas what might be wrong?

A huge problem are the "Null" exponents abviously.
Something somehow cannot be evaluated analytically.

[ This is the Mathematica compatible version that I sent to EK via PM]

I have figured out what you were doing wrong.  Since you are using Mathematica, the modulus (mod) operator is not '%' as it is in Perl and in Python.  I have since rewritten the formula to be Mathematica compatible.  The following version was tested to run under Mathematica 7:

((1+((-1)^(((1-(-1)^(Mod[a*2^(32-0),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-0),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-0),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-0),(2^32-1)]))/2))))/(-2))/2)*2^0+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-1),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-1),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-1),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-1),(2^32-1)]))/2))))/(-2))/2)*2^1+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-2),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-2),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-2),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-2),(2^32-1)]))/2))))/(-2))/2)*2^2+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-3),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-3),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-3),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-3),(2^32-1)]))/2))))/(-2))/2)*2^3+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-4),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-4),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-4),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-4),(2^32-1)]))/2))))/(-2))/2)*2^4+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-5),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-5),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-5),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-5),(2^32-1)]))/2))))/(-2))/2)*2^5+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-6),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-6),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-6),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-6),(2^32-1)]))/2))))/(-2))/2)*2^6+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-7),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-7),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-7),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-7),(2^32-1)]))/2))))/(-2))/2)*2^7+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-8),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-8),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-8),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-8),(2^32-1)]))/2))))/(-2))/2)*2^8+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-9),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-9),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-9),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-9),(2^32-1)]))/2))))/(-2))/2)*2^9+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-10),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-10),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-10),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-10),(2^32-1)]))/2))))/(-2))/2)*2^10+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-11),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-11),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-11),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-11),(2^32-1)]))/2))))/(-2))/2)*2^11+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-12),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-12),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-12),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-12),(2^32-1)]))/2))))/(-2))/2)*2^12+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-13),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-13),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-13),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-13),(2^32-1)]))/2))))/(-2))/2)*2^13+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-14),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-14),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-14),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-14),(2^32-1)]))/2))))/(-2))/2)*2^14+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-15),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-15),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-15),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-15),(2^32-1)]))/2))))/(-2))/2)*2^15+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-16),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-16),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-16),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-16),(2^32-1)]))/2))))/(-2))/2)*2^16+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-17),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-17),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-17),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-17),(2^32-1)]))/2))))/(-2))/2)*2^17+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-18),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-18),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-18),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-18),(2^32-1)]))/2))))/(-2))/2)*2^18+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-19),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-19),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-19),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-19),(2^32-1)]))/2))))/(-2))/2)*2^19+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-20),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-20),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-20),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-20),(2^32-1)]))/2))))/(-2))/2)*2^20+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-21),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-21),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-21),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-21),(2^32-1)]))/2))))/(-2))/2)*2^21+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-22),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-22),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-22),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-22),(2^32-1)]))/2))))/(-2))/2)*2^22+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-23),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-23),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-23),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-23),(2^32-1)]))/2))))/(-2))/2)*2^23+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-24),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-24),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-24),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-24),(2^32-1)]))/2))))/(-2))/2)*2^24+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-25),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-25),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-25),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-25),(2^32-1)]))/2))))/(-2))/2)*2^25+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-26),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-26),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-26),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-26),(2^32-1)]))/2))))/(-2))/2)*2^26+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-27),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-27),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-27),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-27),(2^32-1)]))/2))))/(-2))/2)*2^27+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-28),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-28),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-28),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-28),(2^32-1)]))/2))))/(-2))/2)*2^28+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-29),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-29),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-29),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-29),(2^32-1)]))/2))))/(-2))/2)*2^29+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-30),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-30),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-30),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-30),(2^32-1)]))/2))))/(-2))/2)*2^30+
((1+((-1)^(((1-(-1)^(Mod[a*2^(32-31),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-31),(2^32-1)]))/2))+(-1)^(2-(((1-(-1)^(Mod[a*2^(32-31),(2^32-1)]))/2)+((1-(-1)^(Mod[b*2^(32-31),(2^32-1)]))/2))))/(-2))/2)*2^31

MP

[Mitchell]

My brain hurts after reading this topic.

[abacus]

My brain hurts after reading this topic.

Mine too.
And I have to admit I also skipped the 3^rd page to jump directly to the end...

[minerpeabody]

Sorry guys. 

Many years after writing my first computer program in BASIC, I still enjoy programming.  In RL, my programming is more mundane... stuff like managing email aliases, creating accounts, monitoring log files, etc.  This was the most interesting challenge I've considered in a little while, and really the only one where the "how it was done" was more interesting than the "what was done."

The OP, kinda presented the challenge like he might have some interest in how it was done.  Others answered the call, no one seemed to have the answer.  I thought it might be interesting to share the answer and how I came up with it because the principles involved actually have value for bitcoin/altcoin mining... Why? Because if you can describe logical operations mathematically, you can investigate SHA256 and sCrypt mathematically.  Who knows where that leads?  If you could simplify the mathematical expression of sCrypt for instance (and look at it), you might be able to derive an algorithm that can hash sCrypt hashes even faster by saving your compute platform redundant/superfluous processing cycles. You never know.

I didn't mean to bore anyone; my intentions were good... and then of course, there was the $200 :-)

The ironic thing? I got into computers because I wasn't particularly good at math...

MP

PS.: I'll be putting my ASIC Erupter Blade V2 configuration/monitoring tools out in source code form and GPL'd for FREE soon. I have a few mining boards, like them a lot, but realized that I would like them more if I could apportion/manage/restart/reconfigure them automatically under program control. When ready (probably I'll start with a 0.5 release), look on the forums for where to get them if you have and like the Erupter Blade hardware....

[minerpeabody]

Evil-Knievel:

Could we get an update?

MP

[BurtW]

We are all very interested to learn your intentions regarding the promised bounty.  As one who attempted to solve the problem and claim the bounty I submit to you that minerpeabody's solution does in fact work and does satisfy the criteria set forth in your original post and subsequent clarifications.

1) Did he solve it?  If not then what specifically is lacking?
2) If solved then when can he expect payment?

[FiatKiller]

The answer reminds me of why I made a parentheses counting tool to make sure longa$$ stuff like this has an even number of open and closed parentheses without going blind.  lol   I only have to check stuff if it says it's NG and it tells me whether it's a left or right parenthesis missing...

[deepceleron]

This stuff reminds me why I don't do any design work or enter any "contests" here unless the prize or bounty is held in escrow.

[abacus]

This stuff reminds me why I don't do any design work or enter any "contests" here unless the prize or bounty is held in escrow.

Yeah, good to know...

[minerpeabody]

This stuff reminds me why I don't do any design work or enter any "contests" here unless the prize or bounty is held in escrow.

Hi,

I'm obviously new here.  While I know what escrow is, how does escrow work here?  How do you identify services that you can trust?  Is there a thread that covers this?

MP
(still unpaid)

[BurtW]

I have tried to get him to come here and comment.  He has not.  It appears he is too busy to care about his reputation.

Any person trusted by both parties can be an escrow agent in situations like this.

[Evil-Knievel]

This message was too old and has been purged

[Evil-Knievel]

This message was too old and has been purged

[minerpeabody]

The Bounty:

minerpeabody, I have just checked your solution and it indeed meets all requirements in the original posting.
So it looks like you have perfectly succeeded the task and thus qualified to claim the bounty.

Current Mt.Gox BTC Price: 1 BTC = 915 US$
If I calculate correctly, 200 US$ = (200/915)*1BTC = 0.2185... BTC - I will round it up to 0.22.

All you have to do, is provide me your BTC address.

Further, although you have perfectly completed this task I would like to dig into it a bit deeper. As most modular operations are in the exponent now, they cannot be pulled out of the equation that easily. So I hope we can brainstorm a bit more and think about how we could get rid of the nested exponents or at least of the modulus in those. This way Mathematica cannot reduce/simplify equations - that consist of many XOR terms - at all.

I hope you guys still have fun thinking about this and maybe we can work towards a pure mathematical representation of SHA256 (even if it is just for a round reduced one - e.g. the first 4 or 8 rounds).  

I will post my thoughts (and Mathematica worksheets) here soon.

Thank you Evil-Knievel,

You can pay me at this bitcoin address:

1KUhB2S8Xwp2eE3Tsn9NRtd9HWNxvp3Dx2

As I contemplate your response, I think that you might be able to get somewhere using fuzzy logic.  If one substitute (-1) for (0), ie: {-1, 1} as opposed to {0, 1}, a single XOR can be expressed as:

(-1)^((a+b)^2/4)

It is difficult to escape the modulo math because without it, it's difficult to isolate individual bits. Using negative vs positive numbers (as opposed to individual numbers) might work, but that is just the first thought off the top of my head.

MP

[Evil-Knievel]

This message was too old and has been purged

[Evil-Knievel]

This message was too old and has been purged

[abacus]

I still have to well understand how Trust rating works, but if I click on your trust link, I can't see any negative feedback for you.
Maybe the user deleted it, I don't know.

PS: I am using the default trust list.