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Topic: Crypto Compression Concept Worth Big Money - I Did It! - page 9. (Read 13895 times)

legendary
Activity: 1176
Merit: 1011
I do not understand your propositions , both of them . Am I allow to try to .rar whatever you send me?
Sure, go ahead. I haven't tried it but I'm pretty sure Rar won't reduce them to 950K or less (in fact, I think Rar won't even reduce 1 bit off them).

See link above for 1MB test files Smiley
legendary
Activity: 1176
Merit: 1011
Here is a zip containing 3 test files of exactly 1MB (1048576 bytes) each:

http://www.mediafire.com/download/z2sniezt41c8hi0 or mega mirror

Just to be sure you're getting the correct data, the MD5 checksums of the 3 included files are:

statistics-2b.ocx = d027872ea8e2b5b7db33cf6c2da23080
example.mp5 = e4d85efed99135c8d55ff23dc5b10c34
delta.iff = cb44279ee8c9e7a6859498f5ebd8427e

I'm curious to see if B(asic)Miner or anyone else can successfully compress any of these files to 950K or smaller. What the heck, even 990K (1013760 bytes) would already be very impressive.




legendary
Activity: 2128
Merit: 1073
OP is simply delusional and people keep feeding the troll. He keeps saying his yada yada yadas with no useful information. Reminds me of Herbalife; "you can be RICH like the actor in this video. You just have to buy a ton of our expensive overpriced inventory...". The focus is on how rich you can be instead of the product you'll try to sell.

Even if OP could break mathematical laws, he should look for angel investors, showing them his "discovery". Google how to patent an idea instead of wasting people's time. See how long this thread is? OP is just looking for attention or trying to scam somebody without knowledge about compression.
C'mon, don't get paranoid. This isn't any sort of scam, it is just a harmless crackpottery. Nobody's going to lose any money or health or life or anything really valuable after reading this thread. Remember that Bitcoin a currency backed by gold, commedy gold.
full member
Activity: 193
Merit: 100
OP is simply delusional and people keep feeding the troll. He keeps saying his yada yada yadas with no useful information. Reminds me of Herbalife; "you can be RICH like the actor in this video. You just have to buy a ton of our expensive overpriced inventory...". The focus is on how rich you can be instead of the product you'll try to sell.

Even if OP could break mathematical laws, he should look for angel investors, showing them his "discovery". Google how to patent an idea instead of wasting people's time. See how long this thread is? OP is just looking for attention or trying to scam somebody without knowledge about compression.
legendary
Activity: 1240
Merit: 1001
Thank God I'm an atheist
I just invented a beautiful  compression software: it's called md5

It can compress files of any size to just 128 bit

Still  developing decompression tool... Smiley
sr. member
Activity: 476
Merit: 250
If anyone else is still reading at this point, and wants to check my working, I think I've demonstrated clearly that this is not a reversible compression scheme.

The two letter word "Pi" gives an index of 148.
The following 156 two letter words also give the same index:
(This is restricted to just printable characters, there could be far more matches using the full 256 combinations for each byte)

String: 0i (0011000001101001) Index: 148
String: 0j (0011000001101010) Index: 148
String: 0l (0011000001101100) Index: 148
String: 1I (0011000101001001) Index: 148
String: 1J (0011000101001010) Index: 148
String: 1L (0011000101001100) Index: 148
String: 8I (0011100001001001) Index: 148
String: 8J (0011100001001010) Index: 148
String: 8L (0011100001001100) Index: 148
String: 9A (0011100101000001) Index: 148
String: 9B (0011100101000010) Index: 148
String: 9D (0011100101000100) Index: 148
String: :A (0011101001000001) Index: 148
String: :B (0011101001000010) Index: 148
String: Cheesy (0011101001000100) Index: 148
String: String: String: String: @; (0100000000111011) Index: 148
String: @= (0100000000111101) Index: 148
String: @> (0100000000111110) Index: 148
String: @[ (0100000001011011) Index: 148
String: @] (0100000001011101) Index: 148
String: @^ (0100000001011110) Index: 148
String: @k (0100000001101011) Index: 148
String: @m (0100000001101101) Index: 148
String: @n (0100000001101110) Index: 148
String: @y (0100000001111001) Index: 148
String: A9 (0100000100111001) Index: 148
String: A: (0100000100111010) Index: 148
String: A< (0100000100111100) Index: 148
String: AK (0100000101001011) Index: 148
String: AM (0100000101001101) Index: 148
String: AN (0100000101001110) Index: 148
String: AY (0100000101011001) Index: 148
String: AZ (0100000101011010) Index: 148
String: A\ (0100000101011100) Index: 148
String: Ai (0100000101101001) Index: 148
String: Aj (0100000101101010) Index: 148
String: Al (0100000101101100) Index: 148
String: B9 (0100001000111001) Index: 148
String: B: (0100001000111010) Index: 148
String: B< (0100001000111100) Index: 148
String: BK (0100001001001011) Index: 148
String: BM (0100001001001101) Index: 148
String: BN (0100001001001110) Index: 148
String: BY (0100001001011001) Index: 148
String: BZ (0100001001011010) Index: 148
String: B\ (0100001001011100) Index: 148
String: Bi (0100001001101001) Index: 148
String: Bj (0100001001101010) Index: 148
String: Bl (0100001001101100) Index: 148
String: CI (0100001101001001) Index: 148
String: CJ (0100001101001010) Index: 148
String: CL (0100001101001100) Index: 148
String: D9 (0100010000111001) Index: 148
String: D: (0100010000111010) Index: 148
String: D< (0100010000111100) Index: 148
String: DK (0100010001001011) Index: 148
String: DM (0100010001001101) Index: 148
String: DN (0100010001001110) Index: 148
String: DY (0100010001011001) Index: 148
String: DZ (0100010001011010) Index: 148
String: D\ (0100010001011100) Index: 148
String: Di (0100010001101001) Index: 148
String: Dj (0100010001101010) Index: 148
String: Dl (0100010001101100) Index: 148
String: EI (0100010101001001) Index: 148
String: EJ (0100010101001010) Index: 148
String: EL (0100010101001100) Index: 148
String: GA (0100011101000001) Index: 148
String: GB (0100011101000010) Index: 148
String: GD (0100011101000100) Index: 148
String: H9 (0100100000111001) Index: 148
String: H: (0100100000111010) Index: 148
String: H< (0100100000111100) Index: 148
String: HK (0100100001001011) Index: 148
String: HM (0100100001001101) Index: 148
String: HN (0100100001001110) Index: 148
String: HY (0100100001011001) Index: 148
String: HZ (0100100001011010) Index: 148
String: H\ (0100100001011100) Index: 148
String: Hi (0100100001101001) Index: 148
String: Hj (0100100001101010) Index: 148
String: Hl (0100100001101100) Index: 148
String: MA (0100110101000001) Index: 148
String: MB (0100110101000010) Index: 148
String: MD (0100110101000100) Index: 148
String: P9 (0101000000111001) Index: 148
String: P: (0101000000111010) Index: 148
String: P< (0101000000111100) Index: 148
String: PK (0101000001001011) Index: 148
String: PM (0101000001001101) Index: 148
String: PN (0101000001001110) Index: 148
String: PY (0101000001011001) Index: 148
String: PZ (0101000001011010) Index: 148
String: P\ (0101000001011100) Index: 148
String: Pi (0101000001101001) Index: 148
String: Pj (0101000001101010) Index: 148
String: Pl (0101000001101100) Index: 148
String: QI (0101000101001001) Index: 148
String: QJ (0101000101001010) Index: 148
String: QL (0101000101001100) Index: 148
String: SA (0101001101000001) Index: 148
String: SB (0101001101000010) Index: 148
String: SD (0101001101000100) Index: 148
String: UA (0101010101000001) Index: 148
String: UB (0101010101000010) Index: 148
String: UD (0101010101000100) Index: 148
String: YA (0101100101000001) Index: 148
String: YB (0101100101000010) Index: 148
String: YD (0101100101000100) Index: 148
String: ZA (0101101001000001) Index: 148
String: ZB (0101101001000010) Index: 148
String: ZD (0101101001000100) Index: 148
String: \A (0101110001000001) Index: 148
String: \B (0101110001000010) Index: 148
String: \D (0101110001000100) Index: 148
String: `9 (0110000000111001) Index: 148
String: `: (0110000000111010) Index: 148
String: `< (0110000000111100) Index: 148
String: `K (0110000001001011) Index: 148
String: `M (0110000001001101) Index: 148
String: `N (0110000001001110) Index: 148
String: `Y (0110000001011001) Index: 148
String: `Z (0110000001011010) Index: 148
String: `\ (0110000001011100) Index: 148
String: `i (0110000001101001) Index: 148
String: `j (0110000001101010) Index: 148
String: `l (0110000001101100) Index: 148
String: aI (0110000101001001) Index: 148
String: aJ (0110000101001010) Index: 148
String: aL (0110000101001100) Index: 148
String: cA (0110001101000001) Index: 148
String: cB (0110001101000010) Index: 148
String: cD (0110001101000100) Index: 148
String: eA (0110010101000001) Index: 148
String: eB (0110010101000010) Index: 148
String: eD (0110010101000100) Index: 148
String: iA (0110100101000001) Index: 148
String: iB (0110100101000010) Index: 148
String: iD (0110100101000100) Index: 148
String: jA (0110101001000001) Index: 148
String: jB (0110101001000010) Index: 148
String: jD (0110101001000100) Index: 148
String: lA (0110110001000001) Index: 148
String: lB (0110110001000010) Index: 148
String: lD (0110110001000100) Index: 148
String: qA (0111000101000001) Index: 148
String: qB (0111000101000010) Index: 148
String: qD (0111000101000100) Index: 148
String: rA (0111001001000001) Index: 148
String: rB (0111001001000010) Index: 148
String: rD (0111001001000100) Index: 148
String: tA (0111010001000001) Index: 148
String: tB (0111010001000010) Index: 148
String: tD (0111010001000100) Index: 148

In fact, even with the single character 'P', there are three other printable characters with the same index:
String: B (01000010) Index: 74
String: D (01000100) Index: 74
String: P (01010000) Index: 74
String: ` (01100000) Index: 74


legendary
Activity: 2646
Merit: 1137
All paid signature campaigns should be banned.
I am not trying to get the system, as described, to "work" but out of mathematical curiosity I wonder if there is a starting search digit that would create unique ending positions for all 16 of the 4 bit combinations.  We would have to try starting at 0, then 1, then 2, ... up to 9.

If there is one of more starting locations that do in fact lead to unique ending locations for all 16 of the four bit combinations then we could see if any of those could be extended to 8 bits.

If there is even one starting digit that creates a unique map of the 4 bit values or the 8 bit values then it could be used as an esoteric code mapping.

This could also be tried on all 16 starting locations of a hexidecimal representation of pi.

Pretty simple program.  If I get time I might try it for fun.
legendary
Activity: 1039
Merit: 1005
The biggest question is not whether a computer can do the "compression" but whether it is possible to reconstruct the file.

Here are the chains of digit positions for just 4 bits:
Pi:    314159265358979323846264338327950288419716939937510
0000     4                4   4            4
0001     4                4   4       5
0010     4                4           5                5
0011     4                4           5         6
0100     4 5   5 5
0101     4 5   5           6
0110     4 5  6            6
0111     4 5  6     7
1000       5   5 5                    5
1001       5   5 5         6
1010       5   5           6 6
1011       5   5           6        7
1100       5  6            6 6
1101       5  6            6        7
1110       5  6     7               7
1111       5  6     7    8

As you can see, the bit combinations 0101, 0110 and 1001 all lead to the same ending digit position. The same is true for 0001/1000, 1010/1100 and 1011/1101/1110. This means that the algorithm's output of a file starting with one of the bit patterns in such a group is indistinguishable from the output if the file had started with one of the other bit patterns in the group.

It follows that it is impossible to uniquely decompress the output.

Don't take it personally, but your scheme is not usable.

Onkel Paul
sr. member
Activity: 364
Merit: 253
On average, the index will be at least as large as the actual data itself, and very likely (in 99.999999% cases) even larger.

Not true at all.  If the idea worked, and it does not, but if it did the idea would produce just three small numbers, something like this:

Code:
File size:                52,428,800 bytes
Number of ones found:    209,715,200
Ending location in pi: 5,242,878,123

This is far less than the input of 50 megabytes.

Now the next question is if this is possible, can a consumer computer do a compression within few seconds if not minutes for average file sizes?
sr. member
Activity: 434
Merit: 250

 I am still waiting for 1mb file mail Cheesy Send it to me , I will try to compress it , what can I lose but time Cheesy
legendary
Activity: 2646
Merit: 1137
All paid signature campaigns should be banned.
On average, the index will be at least as large as the actual data itself, and very likely (in 99.999999% cases) even larger.

Not true at all.  If the idea worked, and it does not, but if it did the idea would produce just three small numbers, something like this:

Code:
File size:                52,428,800 bytes
Number of ones found:    209,715,200
Ending location in pi: 5,242,878,123

This is far less than the input of 50 megabytes.
legendary
Activity: 2646
Merit: 1137
All paid signature campaigns should be banned.
B(asic)Miner questions and specific answers for you:

For every 1 byte of data, I need to index 100 - 150 indexes into Pi.
Where did you get this 100-150 number?  

II am simply moving forward according to some rules I figured out for creating a unique pathway through Pi.  
We would need to see all the rules in order to help you figure out if this idea is theoretically sound.

So I would need (100*50,000) or 500,000 indexes into Pi at least.  Let's say that the last bit of data I find is at index location 501,500 into Pi.  
As pointed out previously a 50 megabyte file contains 50*1024*1024 = 52,428,800 bytes.  So using your number of 100 digits /byte you would need at least 5,242,880,000 digits of pi.

Well, here is my crypto key:
Please stop using this term.  It has a meaning and this is not it.  Use "here is my metadata"

For 0s, we hop from the first 4 in Pi to every other 4 in Pi only.  But if we encounter a 1 in our binary data, then the Hunt Value increments +1, so now we are only hopping on every 5 in Pi.  This is what keeps the record of our original bits.  All the other numbers in Pi are useless as we are encoding.  Here is an example:   001      We would be looking for the first 4 in Pi then the 2nd 4 in Pi and now we must increment +1 and hop to the next 5 in Pi.  We keep hopping on 5s as long as our data is 0s, but if we encounter another 1, we increment and begin hopping along 6s.
Is this the full description?

In this way, our pathway is totally unique
I don't think this is true.

to the data we are feeding into it, and thus our arrival point (end point) can let us figure out the path taken to reach itself by knowing how many 1s were used and then attempting to solve backwards to the decimal point using the precise number of steps it took to encode it, the original file size recorded in bits.
Even if the path was unique, which I don't think it is, this process sounds very very hard, possibly impossibly hard.

Also, I'm not sure I should be saying 8 bits per byte, but my friend taught me 7 bits per byte when working with Ascii Binary, was I misinformed on that?
A byte is always 8 bits by definition.  How that byte and those 8 bits are used to encode data varies.  (Simple) ASCII maps all the various printed symbols into these 8 bits.  8 bits per character.

Remember, the theory works by looking at the data in a file as characters in a book, in Ascii format, and thus will need to encode precisely the same number of bits for every character.  But if you look at a hex and decimal and binary converter online, if you type in just one letter, you only get 3 bits or 4 bits or 2 bits ... some erratic bit size.  Every character must have the same bit size, so I want to translate the data into Ascii Binary.
Don't worry too much about these details.  Just describe the idea itself as well as you can and let the geeks and engineers worry about the details - like whether it is possible or not.
sr. member
Activity: 288
Merit: 251
The OP is side stepping the information part though, he's using a huge number that theoretically contains all possible data combinations and indexing it that instead of attempting to compress the raw data. Possibly with other numbers with pi's quality (illogical number or something, nothing recursive in it) the indexing needed will be reduced.
Makes no difference. There is no way to 'cheat' the problem at hand. On average, the index will be at least as large as the actual data itself, and very likely (in 99.999999% cases) even larger.

Quote
Personally I can't see how the indexing would be any different that the formula or fractional method I'd posted, for ideal numbers the data could be represented with a very short formula/fraction etc. but the majority of possible data wouldn't have a good solution so would have little or no compression, not worked it out for successive approximation indexing yet though
Besides some exceptional (very rare) 'lucky' numbers, the vast majority will require more data to store the index.

Consider this: even if it would be possible to compress (or 'encode' or 'represent' or 'index') just a tiny bit off any file, you could repeat the same process over and over, and eventually reducing every file to a few or even one bit. I think you'll intuitively feel that's not gonna work.

Quote
Pi is even a transcendental number (as opposed to e.g. √2 which is also irrational, but algebraic).
sr. member
Activity: 476
Merit: 250
I changed my program to use a precalculated file of Pi to 1000000 digits, rather than compute the digits on the fly.

When I apply it to the same 1000000 digits file, it runs out of Pi digits at only 12500 bytes of the input file (out of 1000000 bytes total).

'Compression' stats at that point:
After byte 12505, pi position is 999428
After byte 12506, pi position is 999525
After byte 12507, pi position is 999561
After byte 12508, pi position is 999628
After byte 12509, pi position is 999729
After byte 12510, pi position is 999795
After byte 12511, pi position is 999856
After byte 12512, pi position is 999950
legendary
Activity: 2646
Merit: 1137
All paid signature campaigns should be banned.
OOPS, reading back through the previous posts I see this has already been shown here:

https://bitcointalksearch.org/topic/m.3120822

Oh well, might as well post my work here anyway....

-----------------------------------------------------------------

Thanks for writing the program!  Cool, I was just about to do the same.

Here is a quick example showing that the paths are not unique for the idea as described.  We may not have the full description and there may be more to it than we have been told but using the start at 4, jump to next 4 as long as the input is zero and increment when the input is one idea:

The following two sequences must have different paths:

00001000
00010000

Assume what if we are in an area of pi with the following sequence of digits:

   4 ... 4 ... 4 ... 5 ... 4 ... 5 ... 5 ... 5 ... 5  (where ... means digits that do not matter in this example)

If my understanding of the idea is correct both inputs land at the same end location.

00001000 would map to 4 4 4 4 5 5 5 5
00010000 would map to 4 4 4 5 5 5 5 5

but with the totally possible distribution given we end up at the same ending location.

Is our understanding of your idea complete and correct or is there more to it than described so far?
sr. member
Activity: 476
Merit: 250
Listen, if I sat down with a programmer who was able to listen to what I'm saying and not throw up objections that have no relevance to my method, and we could sit and work on this until it fits my theory exactly, then I know it would work.  You are right, I cannot do it myself, but I have been very clear about that from the very first post.  I have asked for help.  But it could be done soon.  Very very soon.  A few days for the encoding portion, it's just a file lines really.  The decoding portion would take a lot of brain work, research, testing, modification, etc ...  and we'd have to make our choices about how to resolve the solution of this software by seeing the results of those tests.  If I had the money to hire a coder myself, I would not have even come here at all.  Again, not asking for money now, but I am asking for someone to just TRY to do this with me.  How much time could it take to prove me wrong?  Maybe less than the time it's taking you to actually type out these responses to belittle me so you can go on believing everything is impossible.

Ok, to go the extra mile, I have written a toy program which does what I think you are describing:
- Set the first Pi search digit to 4
- Start at the beginning of the digits of Pi, and the beginning of an input file
- Repeatedly:
  - Take the next byte of the input file and convert it into binary
  - For each binary bit:
    - if it is is 1 increment the Pi search digit, wrapping round to 0 as necessary
    - find the next digit of Pi which matches the search digit
- Print the resulting index digit of Pi where the last binary bit of the last byte of the input file was found.

The program is extremely slow, so I'm not going to run it on a huge file.
For the first just over 1.5KB of its own source file, the program produces:

After byte 1610, pi position is 130339
After byte 1611, pi position is 130407
After byte 1612, pi position is 130468
After byte 1613, pi position is 130531
After byte 1614, pi position is 130666
After byte 1615, pi position is 130747
After byte 1616, pi position is 130857
After byte 1617, pi position is 130915
After byte 1618, pi position is 130995

I hope you can see that your 50MB file is never going to have a final index position of 500,000, if I have already reached 131,000 with only 1.5KB.

[Note that you haven't demonstrated any decompression routine, and this 'compression' routine is not proved to be reversible, so that two input files may give the same index value, which means that decompression is not possible.]
legendary
Activity: 2646
Merit: 1137
All paid signature campaigns should be banned.
That is why I called it a "very large out of band reproducible random data set (like pi)"

That is why it is an encoding/decoding system, not a compression system.

Very large amounts of information can be encoded in very small metadata:  ISBN numbers, URLs, etc.

He is not talking about data compression at all.  He just does/did not know the proper terminology is all.
legendary
Activity: 1176
Merit: 1011
I for one am very glad you thought this up and have spent so much time and effort on it and want to personally thank you for bringing it to my attention.  It is fun to dream up new things and share them.  You could have saved yourself a lot of grief and flames by just publishing your idea in your original post but I know from personal experience that is very hard to do when you think you have really got something new and valuable.
Although I highly appreciate creativity and thinking out of the box, this whole idea is fundamentally flawed, as could have been proven in advance, regardless of the exact implementation details.

Not because of 'stubborn thinking' or refusing to look beyond known methods, but simply because it is mathematically impossible to create a compression (or 'encoding' or whatever you wanna call it) algorithm that reduces any file to less than its own size. No matter if you use Pi, or black magic, or voodoo, or whatever.

Let's repeat the same argument one more time: there are N possible files of 1MB. There are M possible 4K crypto keys. N is much, much, MUCH larger than M. So, by definition, the vast majority of 1MB files (that is, way over 99.9999999999999999999999%) can NOT be represented by a 4K crypto key. Period.

Even if you would create a HUGE (de)compression program that contains a vast library of ALL possible 1MB files, and when compressing a file it just stores some index to specify the exact 1MB file you need, it's still impossible. There's much more possible input files, than the number of unique index keys (or 'compressed files') you can output.

What riddles me, is why people who are obviously not stupid, keep believing in fairy tales and ignoring plain simple logic.
legendary
Activity: 2646
Merit: 1137
All paid signature campaigns should be banned.
Everyone:

I also know how much fun it can be to bait, flame and troll some of the losers and scammers who post here.  I think this guy is simply trying to find out if his idea will work.  I know I could be wrong and he may just be having fun or trying to pull a scam - it is so hard to tell sometimes but so far he looks genuine to me.

Carry on...

full member
Activity: 193
Merit: 100
Oh God, please just give OP a bitch-slap!
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