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**bob farey****Guest**

ok my last post was a 196, smiles.... and this is..........

**Devantè****Real Member**- Registered: 2006-07-14
- Posts: 6,400

bobbym wrote:

Hi;

That too is interesting. Does a Cambridge graduate rate higher than a Tutorial Fellow in Mathematics at Wadham College and author of a couple of books? Possibly.

Hodge's classic work could be an exaggeration of the story.

If I recall correctly his information was also based on Simon Singh - but the best source would be someone who was actually alive at the time.

@Bob Farey - is this your code?

fhmxvwzwqcdicnxiyzwtyelazwczsyewdevyzmxhicuhvmhssl

qvvhwewjzzchvcsjcylenucxijsvlezicjzeyykplhvztjhzcz

uetvwzoecjhooujbxpysczgwtwfkwgsvfwvikxhxcecvlfxwne

iweujtjwywjtrczewcmrdidejeikourfzwjyiwfcocypwjishs

ecfehsdziyescqfwthwclwpgszvxwxjgvwzmzefykwgizhbijc

zeeiwsqlvyviwjxoxelejvhzjkwcvryehyewzjmewwhsajzjuz

ggmepidujviamzeavwwyzcwdyywwcfmufyzshmwwwvgipzwhbm

hnzcsclaledziwlehwswzzsizzjjdlveujvwszfsgjiaeimhvx

ejibyvtedycwfhwpcuiwczfhhfvavjdkzxsvfexsvjmwjwdumq

kzzhpzwiwxlfcvxyyczwhixwjwpzzrhfjwvwqsxwhiwzjhszee

So we know that you use a 26-letter wheel with letters arranged in a random (by a human or a computer - what method was used to pick the numbers?) order, and that the order in which the letters are read could be significant. Solvable by a computer? Eventually, yes. Solvable by a human? Most probably not.

Interesting.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,730

bob farey wrote:

It does not start or end with "Heil Moderators" every time and the message itself sets the "rotors".

I am the only moderator conversing with you.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**bob farey****Guest**

the letter order was arranged by a computer then I asked my wife to give a few pairs of numbers between 1 and 26 and swopped them, smiles......so a female brain is involved.....

The only way you can be certain of ordering the message correctly is to look at every permutation, ie factorial 900

Then for each one of these, guess how many letters are fillers , say between 100 and 800

For each of those 700 filler possibilities, guess how I inserted the dummies, eg 1 then 2 then three then one etc or 7 then 10 then 3

or 4 then 9 then 2 then 6 in other words many trillions

so when you do factorial 900 times 700 times many trillions operations FOR EACH OF THOSE YOU NEED TO CARRY OUT 26 TO THE POWER (your guess of how many letters in message)

I give you an infinitely fast computer, so you got some sensible words - a lot of them tell me what the message is then...

**Devantè****Real Member**- Registered: 2006-07-14
- Posts: 6,400

...and twenty-thousand years later, when code-breakers are anxiously awaiting the final permutation of Bob Farey's code, it is cracked. The message reads:

rofl i juts wast3d ur ti3m lols!!!!!!!!11

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**bob farey****Guest**

another clue, the message is in plain english, no l8trs, m8t

**bob farey****Guest**

So you could not crack it, well here is the hidden message, c an you find where the message is in the text I have given you.

I could have given you 2000 letters instead of 900 without affecting the recipient's ability to transcribe it in 2 minutes.

Purely from the number of permutations to try, it is uncrackable, because the code letters are not consecutive in the grid I gave you...

M E E T M E A T T H E B U L L S H E A D T U E S D A Y S E V E N P M

**BOB FAREY****Guest**

So does anybody dispute that this method of encryption is uncrackable?

**calccrypto****Member**- Registered: 2010-03-06
- Posts: 96

Kerckhoffs' principle

In cryptography, Kerckhoffs' principle (also called Kerckhoffs' assumption, axiom or law) was stated by Auguste Kerckhoffs in the 19th century:

a cryptosystem should be secure even if everything about the system, except the key, is public knowledge.Kerckhoffs' principle was reformulated (perhaps independently) by Claude Shannon as "The enemy knows the system." In that form, it is called Shannon's maxim. In contrast to "security through obscurity," it is widely embraced by cryptographers.

Show the algorithm but not the key. if it can be reconstructed without the key, it is breakable. i doubt that i can do it, but a really good cryptanalyst can probably do it

Visit calccrypto.wikidot.com for detailed descriptions of algorithms and other crypto related stuff (not much yet, so help would be appreciated).

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,730

Hi;

The point I made to him on another thread. Need to see the algorithm to prove it is difficult to crack. The output is no help at all.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**calccrypto****Member**- Registered: 2010-03-06
- Posts: 96

sorry for the repeat

Visit calccrypto.wikidot.com for detailed descriptions of algorithms and other crypto related stuff (not much yet, so help would be appreciated).

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,730

Hi calccrypto;

sorry for the repeat

No need, you did nothing wrong. Just stating the fact why no progress can be made with his algorithm.

Been to your site, wished that the tabs on the hash 2x pages were working for me.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**calccrypto****Member**- Registered: 2010-03-06
- Posts: 96

thanks for visiting. sorry my site isnt fully compatible with your computer

*Last edited by calccrypto (2010-03-07 06:20:40)*

Visit calccrypto.wikidot.com for detailed descriptions of algorithms and other crypto related stuff (not much yet, so help would be appreciated).

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,730

Hi;

Then it is a problem with my browser, I will retry thanks.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**bob farey****Guest**

I am not familiar with the terms, especially key. Hidden within the code is instructions on where to start, which direction to go, and details of the gaps between encrypted letters. Only the recipient knows which of the 900 letters in the code give these instructions. For example, a particular letter in the coded text would tell you which ROW to start in, so A=1, B=2 etc etc. Where to start in the row is more cunningly coded. The gaps between encoded letters are also defined in particular cells of the matrix.

I am not being difficult, folks, but you have to take any letter, ie 900 choices, then any other, ie 899 and so on.

So for a 20 word message that is 900 x 899 x 898..................x 880 =X

Then for EACH of these strings, you have to try all 26 letters of the alphabet.

So the total number of tries is X multiplied by 20 to the power 26

Within your tries you will find many many many words that string together.

I could have encoded the words in a different order to make it more difficult.......

I do not know how to calculate the total number of permutations, but 20 to the power 26 is pretty big......in itself.

**S.G. Shredmaster****Member**- Registered: 2010-03-19
- Posts: 290

codes make my head hurt.........but theyre good for me.

When you're old and your eyes are dim, There ain't no old shep gonna happen again

We'll still go walking down country lanes, I'll sing the same old song

Hear me call your name

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**LordDannyl****Member**- Registered: 2010-07-29
- Posts: 1

This all depends on the amount of "waffle". The way I would do it normally would be to take the first, 27th, 53rd etc and do a frequential analysis of those to determine the substitution for those, then repeat that for the 2nd, 28th, 54th and so on. However, the "waffle" would throw this analysis off; a little would probably still work because it would only affect the proportions marginally. However lots would seriously mess up the proportions (though it would still be breakable- just harder to get frequency analysis right. The most common would still correspond to "e", but then there's the trouble of working out what's waffle).

If you want a tough code, try this one:

ÍÐÇÆ|vr¸ºin||vd²»¿it¥ÎÄ²gboµr{xÍtwiÁÂÉÂÍÆ»if{{}eµº¾¼qtmÈÄÄ®µµÄ§`Woo¡ªß¦xd´opt»º³lt}«`\|Ã´»»xqqÇy£rrpyÊÖ¨¨}}reµº¾¼»»ÈÄÄ®µµÄ¹¼¹ÍÒÄ¹¸¥}¢Á|noµrmI¢¢ÎÃxyn»³pn}}vg½ÏÏÍ

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**Leornardo da vinci****Guest**

you guys try to decipher this code by using the keyword ER this isnt a hard cryptic code here it is : JZFXAJUJSAZQZJZFXAJUJSAZQZPABOMIYBWMHVHTFHPWNJQWPLZAMEBBVLFAZHKAWTHQUAMEBWYAPVRMF

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,730

Hi Leornardo da vinci;

Welcome to the forum. Did you leave something out?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**YZF****Member**- Registered: 2012-04-24
- Posts: 6

I wrote a code generator a while back in Delphi.

The following code when deciphered would read ABCDEF

81012191411911941194119411941131815181419141181319013101813110

The only drawback with this is that if you wrote a whole sentence, it would be quite a long string, can anyone say how I have done this?

Jason

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