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Groups > sci.crypt > #49977 > unrolled thread
| Started by | austin obyrne <ao402468@gmail.com> |
|---|---|
| First post | 2021-09-17 09:07 -0700 |
| Last post | 2021-09-18 15:20 +0100 |
| Articles | 20 on this page of 55 — 9 participants |
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Triangulation Cryptography - Statistical Cryptanalysis Attack. austin obyrne <ao402468@gmail.com> - 2021-09-17 09:07 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Leo <usenet@gkbrk.com> - 2021-09-17 16:17 +0000
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Richard Heathfield <rjh@cpax.org.uk> - 2021-09-17 17:34 +0100
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. MM <mrvmurray@gmail.com> - 2021-09-18 02:28 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. austin obyrne <ao402468@gmail.com> - 2021-09-17 09:36 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. MM <mrvmurray@gmail.com> - 2021-09-18 09:14 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. 711 Spooky Mart <711@spooky.mart> - 2021-09-24 15:14 -0500
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Rich <rich@example.invalid> - 2021-09-24 21:57 +0000
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Richard Heathfield <rjh@cpax.org.uk> - 2021-09-17 17:23 +0100
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Max <maxturv26@gmx.net> - 2021-09-17 18:44 +0200
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. austin obyrne <ao402468@gmail.com> - 2021-09-17 10:11 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Max <maxturv26@gmx.net> - 2021-09-17 19:32 +0200
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Richard Heathfield <rjh@cpax.org.uk> - 2021-09-17 18:43 +0100
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. austin obyrne <ao402468@gmail.com> - 2021-09-17 11:04 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Max <maxturv26@gmx.net> - 2021-09-17 20:14 +0200
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. austin obyrne <ao402468@gmail.com> - 2021-09-17 11:35 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Max <maxturv26@gmx.net> - 2021-09-17 20:49 +0200
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. austin obyrne <ao402468@gmail.com> - 2021-09-17 12:09 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Max <maxturv26@gmx.net> - 2021-09-17 21:30 +0200
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2021-09-17 13:39 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Max <maxturv26@gmx.net> - 2021-09-17 22:44 +0200
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2021-09-17 13:55 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. 711 Spooky Mart <711@spooky.mart> - 2021-09-24 15:33 -0500
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Richard Heathfield <rjh@cpax.org.uk> - 2021-09-17 20:42 +0100
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Max <maxturv26@gmx.net> - 2021-09-17 22:25 +0200
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. wizzofozz <oxxxxxxxxxxxs@gmail.com> - 2021-09-17 22:56 +0200
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Max <maxturv26@gmx.net> - 2021-09-17 23:11 +0200
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. wizzofozz <oxxxxxxxxxxxs@gmail.com> - 2021-09-17 23:53 +0200
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Max <maxturv26@gmx.net> - 2021-09-18 00:16 +0200
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. austin obyrne <ao402468@gmail.com> - 2021-09-18 00:12 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. MM <mrvmurray@gmail.com> - 2021-09-18 01:56 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Richard Heathfield <rjh@cpax.org.uk> - 2021-09-18 10:09 +0100
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. MM <mrvmurray@gmail.com> - 2021-09-18 02:25 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Rich <rich@example.invalid> - 2021-09-17 20:57 +0000
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. wizzofozz <oxxxxxxxxxxxs@gmail.com> - 2021-09-17 23:03 +0200
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. wizzofozz <oxxxxxxxxxxxs@gmail.com> - 2021-09-17 22:47 +0200
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Rich <rich@example.invalid> - 2021-09-17 20:54 +0000
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. austin obyrne <ao402468@gmail.com> - 2021-09-17 11:23 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Rich <rich@example.invalid> - 2021-09-17 20:50 +0000
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Rich <rich@example.invalid> - 2021-09-17 20:46 +0000
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2021-09-17 13:56 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. MM <mrvmurray@gmail.com> - 2021-09-18 02:41 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Richard Heathfield <rjh@cpax.org.uk> - 2021-09-18 10:55 +0100
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. MM <mrvmurray@gmail.com> - 2021-09-18 03:34 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Rich <rich@example.invalid> - 2021-09-18 12:26 +0000
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Richard Heathfield <rjh@cpax.org.uk> - 2021-09-18 14:02 +0100
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. MM <mrvmurray@gmail.com> - 2021-09-18 06:17 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Richard Heathfield <rjh@cpax.org.uk> - 2021-09-18 14:28 +0100
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Richard Heathfield <rjh@cpax.org.uk> - 2021-09-18 14:51 +0100
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Rich <rich@example.invalid> - 2021-09-18 16:20 +0000
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Rich <rich@example.invalid> - 2021-09-18 16:18 +0000
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. MM <mrvmurray@gmail.com> - 2021-09-18 06:24 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Richard Heathfield <rjh@cpax.org.uk> - 2021-09-18 14:31 +0100
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. MM <mrvmurray@gmail.com> - 2021-09-18 06:44 -0700
Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. Richard Heathfield <rjh@cpax.org.uk> - 2021-09-18 15:20 +0100
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| From | austin obyrne <ao402468@gmail.com> |
|---|---|
| Date | 2021-09-17 09:07 -0700 |
| Subject | Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <438ad408-35cb-4e67-968d-9c0b67861b31n@googlegroups.com> |
Given that the ciphertext is comprised of groups of 3 x seven-digit integers. It follows that the sample space is of the order of (9 factorial)^21 in any attack. This is 362880^21- an enormous number that can even be made emuch larger if needs be. The expectation of repeats in a string of 10000 items of vector ciphertext has been explored and no repeat was found. Collecting enough repeats to submit to the sample space of 9!^21 hoping to get some meaningful frequency and some assocoiated probability that could assigned to some character in ASCII would be a useless exercise. The probability index would be useless. It wouldn't even come on scale. It would in fact be tantamount to another brute force program that would fail for the same reason as before. Comment: Trying out old defunct methods of scalar cryptography as some readers are doing won't work in vector cryptography. Keep rowing. Austin O’Byrne.
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| From | Leo <usenet@gkbrk.com> |
|---|---|
| Date | 2021-09-17 16:17 +0000 |
| Message-ID | <si2f25$is1$1@dont-email.me> |
| In reply to | #49977 |
On Fri, 17 Sep 2021 09:07:33 -0700, austin obyrne wrote: > Given that the ciphertext is comprised of groups of 3 x seven-digit > integers. It follows that the sample space is of the order of (9 > factorial)^21 in any attack. This is 362880^21- an enormous number that > can even be made emuch larger if needs be. > > The expectation of repeats in a string of 10000 items > of vector ciphertext has been explored and no repeat > was found. > > Collecting enough repeats to submit to the sample space of 9!^21 hoping > to get some meaningful frequency and some assocoiated probability that > could assigned to > some character in ASCII would be a useless exercise. > > The probability index would be useless. It wouldn't even come on scale. > It would in fact be tantamount to another > brute force program that would fail for the same reason as before. > > Comment: > > Trying out old defunct methods of scalar cryptography as some readers > are doing won't work in vector cryptography. > > Keep rowing. > > Austin O’Byrne. Is there a public implementation of this cipher available, so we can generate some plaintext / ciphertext pairs to analyze it better? Saying traditional methods won't work is one thing, but letting people try and fail is how you make claims. -- Leo
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| From | Richard Heathfield <rjh@cpax.org.uk> |
|---|---|
| Date | 2021-09-17 17:34 +0100 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <si2g35$49a$1@dont-email.me> |
| In reply to | #49978 |
On 17/09/2021 17:17, Leo wrote: > On Fri, 17 Sep 2021 09:07:33 -0700, austin obyrne wrote: > >> Given that the ciphertext is comprised of groups of 3 x seven-digit >> integers. It follows that the sample space is of the order of (9 >> factorial)^21 in any attack. This is 362880^21- an enormous number that >> can even be made emuch larger if needs be. >> >> The expectation of repeats in a string of 10000 items >> of vector ciphertext has been explored and no repeat >> was found. >> >> Collecting enough repeats to submit to the sample space of 9!^21 hoping >> to get some meaningful frequency and some assocoiated probability that >> could assigned to >> some character in ASCII would be a useless exercise. >> >> The probability index would be useless. It wouldn't even come on scale. >> It would in fact be tantamount to another >> brute force program that would fail for the same reason as before. >> >> Comment: >> >> Trying out old defunct methods of scalar cryptography as some readers >> are doing won't work in vector cryptography. >> >> Keep rowing. >> >> Austin O’Byrne. > > Is there a public implementation of this cipher available, so we can > generate some plaintext / ciphertext pairs to analyze it better? This raised a smile. You are now in broadly the same place I was a few years ago. MM tried to save me some effort by saying he'd already broken AOB's stuff. He hadn't kept hold of the break, so although of course I believed him I decided to try to break it for myself, which I did in full view of sci.crypt. But... I haven't bothered to keep hold of the break. And then along came you. When you break AOB's stuff, keep hold of the break. :-) -- Richard Heathfield Email: rjh at cpax dot org dot uk "Usenet is a strange place" - dmr 29 July 1999 Sig line 4 vacant - apply within
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| From | MM <mrvmurray@gmail.com> |
|---|---|
| Date | 2021-09-18 02:28 -0700 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <d05e9f2a-d5b7-4245-b048-c61b0c782a33n@googlegroups.com> |
| In reply to | #49980 |
On Friday, 17 September 2021 at 17:34:51 UTC+1, Richard Heathfield wrote: > And then along came you. When you break AOB's stuff, keep hold of the > break. :-) How about this: https://groups.google.com/g/sci.crypt/c/GmJrmCW-dTI/m/6mboBBcvBQAJ M --
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| From | austin obyrne <ao402468@gmail.com> |
|---|---|
| Date | 2021-09-17 09:36 -0700 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <4ff67537-31ef-4b96-891d-45c9f885140cn@googlegroups.com> |
| In reply to | #49978 |
On Friday, 17 September 2021 at 17:17:12 UTC+1, Leo wrote: > On Fri, 17 Sep 2021 09:07:33 -0700, austin obyrne wrote: > > > Given that the ciphertext is comprised of groups of 3 x seven-digit > > integers. It follows that the sample space is of the order of (9 > > factorial)^21 in any attack. This is 362880^21- an enormous number that > > can even be made emuch larger if needs be. > > > > The expectation of repeats in a string of 10000 items > > of vector ciphertext has been explored and no repeat > > was found. > > > > Collecting enough repeats to submit to the sample space of 9!^21 hoping > > to get some meaningful frequency and some assocoiated probability that > > could assigned to > > some character in ASCII would be a useless exercise. > > > > The probability index would be useless. It wouldn't even come on scale. > > It would in fact be tantamount to another > > brute force program that would fail for the same reason as before. > > > > Comment: > > > > Trying out old defunct methods of scalar cryptography as some readers > > are doing won't work in vector cryptography. > > > > Keep rowing. > > > > Austin O’Byrne. > Is there a public implementation of this cipher available, so we can > generate some plaintext / ciphertext pairs to analyze it better? > > Saying traditional methods won't work is one thing, but letting people > try and fail is how you make claims. > > -- > Leo Hi Leo, No unfortunately , I have no had any exposure whatever. At this stage I need some support beacuse although I have great confidence in this cipher it needs backing from the establishment. I baulk at submitting anything to the AMS (MM has worked on a sample but I discontinued with it because I saw my stuff being carved up. (as a professional standard submission needs to be) and feared having it trivialised by the establishment). No fault of MM. I'm even becoming rusty on my own programming work and must start refreshing my Ada-95 source code . AOB
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| From | MM <mrvmurray@gmail.com> |
|---|---|
| Date | 2021-09-18 09:14 -0700 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <b9dd1c9a-25c2-44dc-9bdf-d27bbde58bf4n@googlegroups.com> |
| In reply to | #49981 |
On Friday, 17 September 2021 at 17:36:03 UTC+1, austin obyrne wrote: > I'm even becoming rusty on my own programming work > and must start refreshing my Ada-95 source code . If you return to that, remember the input of comp.lang.ada, amongst many others. Find someone who has experience, and solicit their help. LOSE YOUR ASCII FIXATION! Computer data is a sequence of numbers in the range 0..255, that is all. What they mean, ASCII, GIF, MPEG, audio, whatever, is of no concern. If the result of decryption is the same set of numbers as in the original file(s), then you have succeeded. Doing this actually simplifies your programs! Your programs are /terrible/ - they obfuscate the algorithm, they are usually WAY too verbose and they crash with consumate ease. They require editing where a proper design would accept input. They provide Eve with numerous clues as to how to attack your ciphers. (Hint: if you try to decrypt with a wrong key you should get nonsense, not a helpful (to Eve) exception fault). M --
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| From | 711 Spooky Mart <711@spooky.mart> |
|---|---|
| Date | 2021-09-24 15:14 -0500 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <silbic$1fsp$1@gioia.aioe.org> |
| In reply to | #49981 |
On 9/17/21 11:36 AM, austin obyrne wrote: > I'm even becoming rusty on my own programming work > and must start refreshing my Ada-95 source code . Ada is a obscure and niche specific language. It is not ideal for collaboration and exploration of simple cryptographic primitives if you want many eyes on it. If you want people to read your code and comment on it use C or Python or even JavaScript. -- ███████████████████████████████████ █░░░░░░░░░░░█░░░░░░░░███░░░░░░░░███ █░░███████░░█░░████░░███░░████░░███ [chan] 711 █░░░░░░░██░░█░░░░██░░███░░░░██░░███ spooky mart ██████░░██░░███░░██░░█████░░██░░███ always open ██████░░██░░███░░██░░█████░░██░░███ stay spooky ██████░░██░░█░░░░██░░░░█░░░░██░░░░█ https://bitmessage.org ██████░░██░░█░░██████░░█░░██████░░█ ██████░░░░░░█░░░░░░░░░░█░░░░░░░░░░█ ███████████████████████████████████
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| From | Rich <rich@example.invalid> |
|---|---|
| Date | 2021-09-24 21:57 +0000 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <silhkq$71g$1@dont-email.me> |
| In reply to | #50103 |
711 Spooky Mart <711@spooky.mart> wrote: > On 9/17/21 11:36 AM, austin obyrne wrote: > > >> I'm even becoming rusty on my own programming work and must start >> refreshing my Ada-95 source code . > > Ada is a obscure and niche specific language. It is not ideal for > collaboration and exploration of simple cryptographic primitives if > you want many eyes on it. If you want people to read your code and > comment on it use C or Python or even JavaScript. Yes, Ada is not 'mainstream' by any definition now. However, Ada is *all* AOB knows -- and he only knows a very limited subset of Ada at that, and then of Ada-95, which is a very old version at this point in time. So he is unlikely to switch to C or Python or Javascript, as to him the effort needed to make the switch is likely not worth it in his limited world view.
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| From | Richard Heathfield <rjh@cpax.org.uk> |
|---|---|
| Date | 2021-09-17 17:23 +0100 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <si2feh$qea$1@dont-email.me> |
| In reply to | #49977 |
On 17/09/2021 17:07, austin obyrne wrote: > Given that the ciphertext is comprised of groups of > 3 x seven-digit integers. It follows that the sample > space is of the order of (9 factorial)^21 in any attack. > This is 362880^21- an enormous number that can even > be made emuch larger if needs be. Don't be too impressed by big numbers. We polished off bigger numbers than that in Shuttlepads. > The expectation of repeats in a string of 10000 items > of vector ciphertext has been explored and no repeat > was found. > > Collecting enough repeats to submit to the sample > space of 9!^21 hoping to get some meaningful frequency > and some assocoiated probability that could assigned to > some character in ASCII would be a useless exercise. > > The probability index would be useless. It wouldn't even > come on scale. It would in fact be tantamount to another > brute force program that would fail for the same reason as before. You said that sort of thing about ShuttlePads. Which I seem to recall fell rather easily. > > Comment: > > Trying out old defunct methods of scalar cryptography as > some readers are doing won't work in vector cryptography. They have already been shown to work. > Keep rowing. Until you learn from previous breaks, you're just going to keep on rowing the same old river. -- Richard Heathfield Email: rjh at cpax dot org dot uk "Usenet is a strange place" - dmr 29 July 1999 Sig line 4 vacant - apply within
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| From | Max <maxturv26@gmx.net> |
|---|---|
| Date | 2021-09-17 18:44 +0200 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <si2glu$3nn$1@gioia.aioe.org> |
| In reply to | #49977 |
On 17.09.21 18:07, austin obyrne wrote: > Given that the ciphertext is comprised of groups of > 3 x seven-digit integers. It follows that the sample > space is of the order of (9 factorial)^21 in any attack. > This is 362880^21- an enormous number that can even > be made emuch larger if needs be. 9 factorial? I'd like to see that math. Is there any point in your cipher where each number from 1 to 9 has to appear exactly once? I hope all values are independent from one another. > > The expectation of repeats in a string of 10000 items > of vector ciphertext has been explored and no repeat > was found. > > Collecting enough repeats to submit to the sample > space of 9!^21 hoping to get some meaningful frequency > and some assocoiated probability that could assigned to > some character in ASCII would be a useless exercise. > > The probability index would be useless. It wouldn't even > come on scale. It would in fact be tantamount to another > brute force program that would fail for the same reason as before. > > Comment: > > Trying out old defunct methods of scalar cryptography as > some readers are doing won't work in vector cryptography. This has nothing to do with cryptography. As long as you work with integers (which you do), you can't escape combinatorics. > > Keep rowing. > > Austin O’Byrne. >
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| From | austin obyrne <ao402468@gmail.com> |
|---|---|
| Date | 2021-09-17 10:11 -0700 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <d4b3bdb9-21b1-465e-ae4c-9afcd319be7dn@googlegroups.com> |
| In reply to | #49982 |
On Friday, 17 September 2021 at 17:44:50 UTC+1, Max wrote: > On 17.09.21 18:07, austin obyrne wrote: > > Given that the ciphertext is comprised of groups of > > 3 x seven-digit integers. It follows that the sample > > space is of the order of (9 factorial)^21 in any attack. > > This is 362880^21- an enormous number that can even > > be made emuch larger if needs be. > 9 factorial? I'd like to see that math. Is there any point in your > cipher where each number from 1 to 9 has to appear exactly once? I hope > all values are independent from one another. > > > > The expectation of repeats in a string of 10000 items > > of vector ciphertext has been explored and no repeat > > was found. > > > > Collecting enough repeats to submit to the sample > > space of 9!^21 hoping to get some meaningful frequency > > and some assocoiated probability that could assigned to > > some character in ASCII would be a useless exercise. > > > > The probability index would be useless. It wouldn't even > > come on scale. It would in fact be tantamount to another > > brute force program that would fail for the same reason as before. > > > > Comment: > > > > Trying out old defunct methods of scalar cryptography as > > some readers are doing won't work in vector cryptography. > This has nothing to do with cryptography. As long as you work with > integers (which you do), you can't escape combinatorics. > > > > Keep rowing. > > > > Austin O’Byrne. > > Hi Max, As I see I don't think combinatorics is valid as a method in vector cryptography .- OK in scalar data that have magnitude but not with vectors ( that have direction as well as magnitude). AOB
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| From | Max <maxturv26@gmx.net> |
|---|---|
| Date | 2021-09-17 19:32 +0200 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <si2jeu$1fbv$1@gioia.aioe.org> |
| In reply to | #49983 |
On 17.09.21 19:11, austin obyrne wrote: [..] > > As I see I don't think combinatorics is valid as a method in vector cryptography > .- OK in scalar data that have magnitude but not with vectors ( that have direction as well as magnitude). You don't use "directions". They are, at best, an "emergent property" of your vectors. Your vectors are sufficiently defined by their x, y and z-values. As these values are integers, I can easily map them / encode them to a single integer (143, 752, 12) simply becomes 143752012. Still wondering, where do the 9! come from? I think, this is wrong. > > AOB > Cheers, Max
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| From | Richard Heathfield <rjh@cpax.org.uk> |
|---|---|
| Date | 2021-09-17 18:43 +0100 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <si2k4o$23a$1@dont-email.me> |
| In reply to | #49984 |
On 17/09/2021 18:32, Max wrote: > On 17.09.21 19:11, austin obyrne wrote: > [..] >> >> As I see I don't think combinatorics is valid as a method in vector >> cryptography >> .- OK in scalar data that have magnitude but not with vectors ( that >> have direction as well as magnitude). > > You don't use "directions". They are, at best, an "emergent property" of > your vectors. You are of course correct, but you will never convince AOB of this. It's "Eve thinking", and AOB doesn't understand Eve thinking. -- Richard Heathfield Email: rjh at cpax dot org dot uk "Usenet is a strange place" - dmr 29 July 1999 Sig line 4 vacant - apply within
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| From | austin obyrne <ao402468@gmail.com> |
|---|---|
| Date | 2021-09-17 11:04 -0700 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <9e90a0c8-df42-4883-b2a4-9ddd0199f70cn@googlegroups.com> |
| In reply to | #49984 |
On Friday, 17 September 2021 at 18:32:18 UTC+1, Max wrote: > On 17.09.21 19:11, austin obyrne wrote: > [..] > > > > As I see I don't think combinatorics is valid as a method in vector cryptography > > .- OK in scalar data that have magnitude but not with vectors ( that have direction as well as magnitude). > You don't use "directions". They are, at best, an "emergent property" of > your vectors. Your vectors are sufficiently defined by their x, y and > z-values. As these values are integers, I can easily map them / encode > them to a single integer (143, 752, 12) simply becomes 143752012. > > Still wondering, where do the 9! come from? I think, this is wrong. > > > > > > AOB > > > > Cheers, > > Max Each column (of the 7) can be filled in 9! ways i.e. nine factorial ways. (The possibility space must consider filling all 9 places in every possible way) There are 7 'columns' => 9^7 There are 3 sets of 7 integers as coefficients of i, j, k => (9!^7) ^3 = 9! ^ 21 in words - nine factorial to the power of twenty one. AOB
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| From | Max <maxturv26@gmx.net> |
|---|---|
| Date | 2021-09-17 20:14 +0200 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <si2lv0$kr0$1@gioia.aioe.org> |
| In reply to | #49986 |
On 17.09.21 20:04, austin obyrne wrote: > On Friday, 17 September 2021 at 18:32:18 UTC+1, Max wrote: >> On 17.09.21 19:11, austin obyrne wrote: >> [..] >>> >>> As I see I don't think combinatorics is valid as a method in vector cryptography >>> .- OK in scalar data that have magnitude but not with vectors ( that have direction as well as magnitude). >> You don't use "directions". They are, at best, an "emergent property" of >> your vectors. Your vectors are sufficiently defined by their x, y and >> z-values. As these values are integers, I can easily map them / encode >> them to a single integer (143, 752, 12) simply becomes 143752012. >> >> Still wondering, where do the 9! come from? I think, this is wrong. >> >> >>> >>> AOB >>> >> >> Cheers, >> >> Max > Each column (of the 7) can be filled in 9! ways i.e. nine factorial ways. > (The possibility space must consider filling all 9 places in every possible way) > There are 7 'columns' => 9^7 > There are 3 sets of 7 integers as coefficients of i, j, k => (9!^7) ^3 = 9! ^ 21 > in words - nine factorial to the power of twenty one. I asked, where the 9! comes from, not the 21. Why do you think, each column can be filled in 9! ways? If we're talking digits, then each spot in a 7-digit-number can be filled in 10 ways. So, there are 10^7 possible combinations. > > AOB >
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| From | austin obyrne <ao402468@gmail.com> |
|---|---|
| Date | 2021-09-17 11:35 -0700 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <22751d32-7d5e-43ce-9060-6b0fa3ba4bbdn@googlegroups.com> |
| In reply to | #49987 |
On Friday, 17 September 2021 at 19:15:02 UTC+1, Max wrote: > On 17.09.21 20:04, austin obyrne wrote: > > On Friday, 17 September 2021 at 18:32:18 UTC+1, Max wrote: > >> On 17.09.21 19:11, austin obyrne wrote: > >> [..] > >>> > >>> As I see I don't think combinatorics is valid as a method in vector cryptography > >>> .- OK in scalar data that have magnitude but not with vectors ( that have direction as well as magnitude). > >> You don't use "directions". They are, at best, an "emergent property" of > >> your vectors. Your vectors are sufficiently defined by their x, y and > >> z-values. As these values are integers, I can easily map them / encode > >> them to a single integer (143, 752, 12) simply becomes 143752012. > >> > >> Still wondering, where do the 9! come from? I think, this is wrong. > >> > >> > >>> > >>> AOB > >>> > >> > >> Cheers, > >> > >> Max > > Each column (of the 7) can be filled in 9! ways i.e. nine factorial ways. > > (The possibility space must consider filling all 9 places in every possible way) > > There are 7 'columns' => 9^7 > > There are 3 sets of 7 integers as coefficients of i, j, k => (9!^7) ^3 = 9! ^ 21 > > in words - nine factorial to the power of twenty one. > I asked, where the 9! comes from, not the 21. Why do you think, each > column can be filled in 9! ways? > If we're talking digits, then each spot in a 7-digit-number can be > filled in 10 ways. So, there are 10^7 possible combinations. > > > > > AOB > > > So, there are 10^7 possible combinations. Zero (0) doesn't count anywhere n the first column= > 9 only ways - Doesn't happen in my work but you may have a point to argue in other cases .- Better settle for 9 all round - hence 9! - agreed ? AOB
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| From | Max <maxturv26@gmx.net> |
|---|---|
| Date | 2021-09-17 20:49 +0200 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <si2nut$1i95$1@gioia.aioe.org> |
| In reply to | #49989 |
On 17.09.21 20:35, austin obyrne wrote: > On Friday, 17 September 2021 at 19:15:02 UTC+1, Max wrote: >> On 17.09.21 20:04, austin obyrne wrote: >>> On Friday, 17 September 2021 at 18:32:18 UTC+1, Max wrote: >>>> On 17.09.21 19:11, austin obyrne wrote: >>>> [..] >>>>> >>>>> As I see I don't think combinatorics is valid as a method in vector cryptography >>>>> .- OK in scalar data that have magnitude but not with vectors ( that have direction as well as magnitude). >>>> You don't use "directions". They are, at best, an "emergent property" of >>>> your vectors. Your vectors are sufficiently defined by their x, y and >>>> z-values. As these values are integers, I can easily map them / encode >>>> them to a single integer (143, 752, 12) simply becomes 143752012. >>>> >>>> Still wondering, where do the 9! come from? I think, this is wrong. >>>> >>>> >>>>> >>>>> AOB >>>>> >>>> >>>> Cheers, >>>> >>>> Max >>> Each column (of the 7) can be filled in 9! ways i.e. nine factorial ways. >>> (The possibility space must consider filling all 9 places in every possible way) >>> There are 7 'columns' => 9^7 >>> There are 3 sets of 7 integers as coefficients of i, j, k => (9!^7) ^3 = 9! ^ 21 >>> in words - nine factorial to the power of twenty one. >> I asked, where the 9! comes from, not the 21. Why do you think, each >> column can be filled in 9! ways? >> If we're talking digits, then each spot in a 7-digit-number can be >> filled in 10 ways. So, there are 10^7 possible combinations. >> >>> >>> AOB >>> >> So, there are 10^7 possible combinations. > Zero (0) doesn't count anywhere n the first column= > 9 only ways Yes, it does. Or do you only allow for vector coefficients greater or equal to 100? 7 would be 007. Same with all the following digits. 700 is a valid value, too. > - Doesn't happen in my work but you may have a point to argue in other cases > .- Better settle for 9 all round - hence 9! - agreed ? No. Even if there were 9 possible values per digit, that would be 9*9*9*...*9 aka 9^n. 9! would be 9*8*7*6*5*4*3*2*1, only applicable, if each digit in a 9-digit-number would have to be unique and also not allowed to be 0. > > AOB >
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| From | austin obyrne <ao402468@gmail.com> |
|---|---|
| Date | 2021-09-17 12:09 -0700 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <15d4e9a7-59f7-47ea-9840-aa266dc7f781n@googlegroups.com> |
| In reply to | #49990 |
On Friday, 17 September 2021 at 19:49:04 UTC+1, Max wrote: > On 17.09.21 20:35, austin obyrne wrote: > > On Friday, 17 September 2021 at 19:15:02 UTC+1, Max wrote: > >> On 17.09.21 20:04, austin obyrne wrote: > >>> On Friday, 17 September 2021 at 18:32:18 UTC+1, Max wrote: > >>>> On 17.09.21 19:11, austin obyrne wrote: > >>>> [..] > >>>>> > >>>>> As I see I don't think combinatorics is valid as a method in vector cryptography > >>>>> .- OK in scalar data that have magnitude but not with vectors ( that have direction as well as magnitude). > >>>> You don't use "directions". They are, at best, an "emergent property" of > >>>> your vectors. Your vectors are sufficiently defined by their x, y and > >>>> z-values. As these values are integers, I can easily map them / encode > >>>> them to a single integer (143, 752, 12) simply becomes 143752012. > >>>> > >>>> Still wondering, where do the 9! come from? I think, this is wrong. > >>>> > >>>> > >>>>> > >>>>> AOB > >>>>> > >>>> > >>>> Cheers, > >>>> > >>>> Max > >>> Each column (of the 7) can be filled in 9! ways i.e. nine factorial ways. > >>> (The possibility space must consider filling all 9 places in every possible way) > >>> There are 7 'columns' => 9^7 > >>> There are 3 sets of 7 integers as coefficients of i, j, k => (9!^7) ^3 = 9! ^ 21 > >>> in words - nine factorial to the power of twenty one. > >> I asked, where the 9! comes from, not the 21. Why do you think, each > >> column can be filled in 9! ways? > >> If we're talking digits, then each spot in a 7-digit-number can be > >> filled in 10 ways. So, there are 10^7 possible combinations. > >> > >>> > >>> AOB > >>> > >> So, there are 10^7 possible combinations. > > Zero (0) doesn't count anywhere n the first column= > 9 only ways > Yes, it does. Or do you only allow for vector coefficients greater or > equal to 100? 7 would be 007. Same with all the following digits. 700 is > a valid value, too. > > - Doesn't happen in my work but you may have a point to argue in other cases > > .- Better settle for 9 all round - hence 9! - agreed ? > No. Even if there were 9 possible values per digit, that would be > 9*9*9*...*9 aka 9^n. 9! would be 9*8*7*6*5*4*3*2*1, only applicable, if > each digit in a 9-digit-number would have to be unique and also not > allowed to be 0. > > > > > > AOB > > This is beyond argument - it is a standard result => n places can be filled in n! ways - I've been using this on trust for many years In our case these is no point in filling a place in the first column.with zero - okay in all the others of course => total = ( 9! + 10!^6) ^3 in a single ciphertext item comprised of 3 x 7digit integers. - please check. Aob
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| From | Max <maxturv26@gmx.net> |
|---|---|
| Date | 2021-09-17 21:30 +0200 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <si2qd0$ln9$1@gioia.aioe.org> |
| In reply to | #49991 |
On 17.09.21 21:09, austin obyrne wrote: > On Friday, 17 September 2021 at 19:49:04 UTC+1, Max wrote: >> On 17.09.21 20:35, austin obyrne wrote: >>> On Friday, 17 September 2021 at 19:15:02 UTC+1, Max wrote: >>>> On 17.09.21 20:04, austin obyrne wrote: >>>>> On Friday, 17 September 2021 at 18:32:18 UTC+1, Max wrote: >>>>>> On 17.09.21 19:11, austin obyrne wrote: >>>>>> [..] >>>>>>> >>>>>>> As I see I don't think combinatorics is valid as a method in vector cryptography >>>>>>> .- OK in scalar data that have magnitude but not with vectors ( that have direction as well as magnitude). >>>>>> You don't use "directions". They are, at best, an "emergent property" of >>>>>> your vectors. Your vectors are sufficiently defined by their x, y and >>>>>> z-values. As these values are integers, I can easily map them / encode >>>>>> them to a single integer (143, 752, 12) simply becomes 143752012. >>>>>> >>>>>> Still wondering, where do the 9! come from? I think, this is wrong. >>>>>> >>>>>> >>>>>>> >>>>>>> AOB >>>>>>> >>>>>> >>>>>> Cheers, >>>>>> >>>>>> Max >>>>> Each column (of the 7) can be filled in 9! ways i.e. nine factorial ways. >>>>> (The possibility space must consider filling all 9 places in every possible way) >>>>> There are 7 'columns' => 9^7 >>>>> There are 3 sets of 7 integers as coefficients of i, j, k => (9!^7) ^3 = 9! ^ 21 >>>>> in words - nine factorial to the power of twenty one. >>>> I asked, where the 9! comes from, not the 21. Why do you think, each >>>> column can be filled in 9! ways? >>>> If we're talking digits, then each spot in a 7-digit-number can be >>>> filled in 10 ways. So, there are 10^7 possible combinations. >>>> >>>>> >>>>> AOB >>>>> >>>> So, there are 10^7 possible combinations. >>> Zero (0) doesn't count anywhere n the first column= > 9 only ways >> Yes, it does. Or do you only allow for vector coefficients greater or >> equal to 100? 7 would be 007. Same with all the following digits. 700 is >> a valid value, too. >>> - Doesn't happen in my work but you may have a point to argue in other cases >>> .- Better settle for 9 all round - hence 9! - agreed ? >> No. Even if there were 9 possible values per digit, that would be >> 9*9*9*...*9 aka 9^n. 9! would be 9*8*7*6*5*4*3*2*1, only applicable, if >> each digit in a 9-digit-number would have to be unique and also not >> allowed to be 0. >> >> >>> >>> AOB >>> > This is beyond argument - it is a standard result => n places can be filled in n! ways - I've been using this on trust for many years > > In our case these is no point in filling a place in the first column.with zero - okay in all the others of course => total = ( 9! + 10!^6) ^3 in a single ciphertext item comprised of 3 x 7digit integers. - please check. Yes, there is. If you want to express a k digit number with n digits and n > k > 0 , you have to fill up the left n-k digits with zeros. There is no other way to express 7 as a 3-digit-number than 007. > > Aob >
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| From | "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> |
|---|---|
| Date | 2021-09-17 13:39 -0700 |
| Subject | Re: Triangulation Cryptography - Statistical Cryptanalysis Attack. |
| Message-ID | <si2udf$94r$2@gioia.aioe.org> |
| In reply to | #49992 |
On 9/17/2021 12:30 PM, Max wrote: > On 17.09.21 21:09, austin obyrne wrote: >> On Friday, 17 September 2021 at 19:49:04 UTC+1, Max wrote: >>> On 17.09.21 20:35, austin obyrne wrote: >>>> On Friday, 17 September 2021 at 19:15:02 UTC+1, Max wrote: >>>>> On 17.09.21 20:04, austin obyrne wrote: >>>>>> On Friday, 17 September 2021 at 18:32:18 UTC+1, Max wrote: >>>>>>> On 17.09.21 19:11, austin obyrne wrote: >>>>>>> [..] >>>>>>>> >>>>>>>> As I see I don't think combinatorics is valid as a method in >>>>>>>> vector cryptography >>>>>>>> .- OK in scalar data that have magnitude but not with vectors ( >>>>>>>> that have direction as well as magnitude). >>>>>>> You don't use "directions". They are, at best, an "emergent >>>>>>> property" of >>>>>>> your vectors. Your vectors are sufficiently defined by their x, y >>>>>>> and >>>>>>> z-values. As these values are integers, I can easily map them / >>>>>>> encode >>>>>>> them to a single integer (143, 752, 12) simply becomes 143752012. >>>>>>> >>>>>>> Still wondering, where do the 9! come from? I think, this is wrong. >>>>>>> >>>>>>> >>>>>>>> >>>>>>>> AOB >>>>>>>> >>>>>>> >>>>>>> Cheers, >>>>>>> >>>>>>> Max >>>>>> Each column (of the 7) can be filled in 9! ways i.e. nine >>>>>> factorial ways. >>>>>> (The possibility space must consider filling all 9 places in every >>>>>> possible way) >>>>>> There are 7 'columns' => 9^7 >>>>>> There are 3 sets of 7 integers as coefficients of i, j, k => >>>>>> (9!^7) ^3 = 9! ^ 21 >>>>>> in words - nine factorial to the power of twenty one. >>>>> I asked, where the 9! comes from, not the 21. Why do you think, each >>>>> column can be filled in 9! ways? >>>>> If we're talking digits, then each spot in a 7-digit-number can be >>>>> filled in 10 ways. So, there are 10^7 possible combinations. >>>>> >>>>>> >>>>>> AOB >>>>>> >>>>> So, there are 10^7 possible combinations. >>>> Zero (0) doesn't count anywhere n the first column= > 9 only ways >>> Yes, it does. Or do you only allow for vector coefficients greater or >>> equal to 100? 7 would be 007. Same with all the following digits. 700 is >>> a valid value, too. >>>> - Doesn't happen in my work but you may have a point to argue in >>>> other cases >>>> .- Better settle for 9 all round - hence 9! - agreed ? >>> No. Even if there were 9 possible values per digit, that would be >>> 9*9*9*...*9 aka 9^n. 9! would be 9*8*7*6*5*4*3*2*1, only applicable, if >>> each digit in a 9-digit-number would have to be unique and also not >>> allowed to be 0. >>> >>> >>>> >>>> AOB >>>> >> This is beyond argument - it is a standard result => n places can be >> filled in n! ways - I've been using this on trust for many years >> >> In our case these is no point in filling a place in the first >> column.with zero - okay in all the others of course => total = ( 9! + >> 10!^6) ^3 in a single ciphertext item comprised of 3 x 7digit >> integers. - please check. > > Yes, there is. If you want to express a k digit number with n digits and > n > k > 0 , you have to fill up the left n-k digits with zeros. There is > no other way to express 7 as a 3-digit-number than 007. Sorry for butting in, but what about: 124 = 1 + 2 + 4 = 7 ;^)
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