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2011-07-28
, 10:17
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Posts: 226 |
Thanked: 195 times |
Joined on Nov 2009
@ Malaysia
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#2802
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Just to update you guys, there's been invites going around to a "preview" of the N9 in Malaysia sometime soon - within the next couple of weeks. Judging from the amount of time it took for Nokia Malaysia to organize something like this for the N900 (6 months after official launch), I would assume that the N9 can't be that far off - right on track for the late August/early September launch date that's been going around on the boards for awhile.
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2011-07-28
, 10:27
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Posts: 52 |
Thanked: 74 times |
Joined on Jun 2011
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#2803
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ElGamal Example (explained easier)
This scheme requires several variables.
a is a secret key less than the value of p. We will assume it is 304 since p must be greater than it.
g is the generator variable. Nokia has told us its value is 10.
p is a large prime number that Nokia have told us is >304.
For our situation, p is the most important. It is a prime number that must have 'g' as a primitive root.
For fun, we will select 2789.
m is our encrypted message (if there is one) that has to be less than p. For fun we'll choose 832 (from taxi)
k is a prime number used for encrypting the message. It also has the condition it must be less than p and gcd(k, p − 1) = 1. For fun, we will choose 15 (from truck).
So let's do it.
Generate Public Key
p = 2789; g = 10; a = 304
g^a % p = 10^304 (% 2789) ≡ 1161
Now we have what is known as a 'public key' (p, g, g^a%p).
The public key is P = (p ← 2789, g ← 10, g^a%p ← 1161), the secret key is a (304).
Encrypt Message
Next, we encrypt a message m = 832 as follows:
k = 15
γ = 10^15 (% 2789) ≡ 1063
δ ≡ 832 x 1161^15 (% 2789) ≡ 181
c = (γ ← 1063, δ ← 181)
The ciphertext, c, is then sent to the recipient.
Decrypt Message
The recipient can then calculate m from this. The message is calculated by:
m ≡ 1063^(2789-1-304) x 181 ≡ (1063^2484) x 181 (% 2789) ≡ 832
In summary -- to get your 6 digits for the competition:
Generating Public Key
<p><10><10^a % p> (6 digits)
In this situation, p is most likely 3 digits. Also, a or y is likely 15 (where else would it be used?)
This means you can get p from the list below:
313, 337, 367, 379, 383, 389, 419, 433, 461, 487, 491, 499, 503, 509, 541, 571, 577, 593, 619, 647, 659, 701, 709, 727, 743, 811, 821, 823, 857, 863, 887, 937, 941, 953, 971, 977, 983.
Generating Cipher Key
<10^k % p><m*(10^a%p)^k %p> (6 digits)
In this situation, p is most likely 4 digits.
This means Nokia has given us the p from the 'colors on the wall'.
Last edited by recluse; 2011-07-28 at 12:43.
This scheme requires several variables.
a is a secret key less than the value of p. We will assume it is 304 since p must be greater than it.
g is the generator variable. Nokia has told us its value is 10.
p is a large prime number that Nokia have told us is >304.
For our situation, p is the most important. It is a prime number that must have 'g' as a primitive root.
For fun, we will select 2789.
m is our encrypted message (if there is one) that has to be less than p. For fun we'll choose 832 (from taxi)
k is a prime number used for encrypting the message. It also has the condition it must be less than p and gcd(k, p − 1) = 1. For fun, we will choose 15 (from truck).
So let's do it.
Generate Public Key
p = 2789; g = 10; a = 304
g^a % p = 10^304 (% 2789) ≡ 1161
Now we have what is known as a 'public key' (p, g, g^a%p).
The public key is P = (p ← 2789, g ← 10, g^a%p ← 1161), the secret key is a (304).
Encrypt Message
Next, we encrypt a message m = 832 as follows:
k = 15
γ = 10^15 (% 2789) ≡ 1063
δ ≡ 832 x 1161^15 (% 2789) ≡ 181
c = (γ ← 1063, δ ← 181)
The ciphertext, c, is then sent to the recipient.
Decrypt Message
The recipient can then calculate m from this. The message is calculated by:
m ≡ 1063^(2789-1-304) x 181 ≡ (1063^2484) x 181 (% 2789) ≡ 832
In summary -- to get your 6 digits for the competition:
Generating Public Key
<p><10><10^a % p> (6 digits)
In this situation, p is most likely 3 digits. Also, a or y is likely 15 (where else would it be used?)
This means you can get p from the list below:
313, 337, 367, 379, 383, 389, 419, 433, 461, 487, 491, 499, 503, 509, 541, 571, 577, 593, 619, 647, 659, 701, 709, 727, 743, 811, 821, 823, 857, 863, 887, 937, 941, 953, 971, 977, 983.
Generating Cipher Key
<10^k % p><m*(10^a%p)^k %p> (6 digits)
In this situation, p is most likely 4 digits.
This means Nokia has given us the p from the 'colors on the wall'.
Last edited by recluse; 2011-07-28 at 12:43.
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2011-07-28
, 10:29
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Posts: 66 |
Thanked: 27 times |
Joined on Jul 2011
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#2804
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Originally Posted by recluse
Damn man, looks like you're solving the world hunger!
3.5 Example
Generation of a key: p and a are randomly generated, g is a primitive root of p appropriate.
p = 2357
g = 2
a = 1751
g^a mod p = 2^1751 (mod 2357) ≡1185
The public key is now P = (p ← 2357, g ← 2, g^a ← 1185), the secret key is a.
Encrypt a message m = 2035 is as follows (k is randomly chosen):
k = 1520
γ = 2^1520 (mod 2357) ≡ 1430
δ ≡ 2035 x 1185^1520 (mod 2357) ≡ 697
c = (γ ← 1430, δ ← 697)
The ciphertext, c, is then sent to the recipient. We can then calculate m from this. the message is calculated by:
m ≡ 1430^(2357-1-1751) x 697 ≡ 872 x 697 (mod 2357) ≡ 2035
__________________
Free yourself from nokia's morlock cave, discover the future, today: iPhone!
Free yourself from nokia's morlock cave, discover the future, today: iPhone!
| The Following 4 Users Say Thank You to Cod3rror For This Useful Post: | ||
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2011-07-28
, 10:43
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Posts: 2 |
Thanked: 0 times |
Joined on Jul 2011
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#2805
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Originally Posted by aikipy
Where is the truck? This may be dumb..
I know that nowbut last night, before I watched the movie again, I thought it was a wall, not a truck, so the idea kinda got stuck in my mind
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2011-07-28
, 10:48
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Posts: 54 |
Thanked: 11 times |
Joined on Jul 2011
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#2806
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Originally Posted by biatch0
@biatch0
Just to update you guys, there's been invites going around to a "preview" of the N9 in Malaysia sometime soon - within the next couple of weeks. Judging from the amount of time it took for Nokia Malaysia to organize something like this for the N900 (6 months after official launch), I would assume that the N9 can't be that far off - right on track for the late August/early September launch date that's been going around on the boards for awhile.
im fr msia too. where/when's the n9 preview? btw is the invites private oni?
thx..
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2011-07-28
, 10:54
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Posts: 130 |
Thanked: 57 times |
Joined on Jul 2010
@ UK
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#2807
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Originally Posted by recluse
ElGamal Example
p is a prime, g is a primitive root of p, a is a random secret number.
p = 2789; g = 10; a = 2351
g^a % p = 10^2351 (% 2789) ≡1980
Now we have what is known as a 'public key' (p, g, g^a%p).
The public key is P = (p ← 2789, g ← 10, g^a%p ← 1980), the secret key is a (2351).
Next, we encrypt a message m = 832 as follows (k is randomly chosen, m has to be less than p):
k = 1520
γ = 10^1520 (% 2789) ≡ 2300
δ ≡ 832 x 1980^1520 (% 2789) ≡ 2271
c = (γ ← 2300, δ ← 2271)
The ciphertext, c, is then sent to the recipient.
The recipient can then calculate m from this. The message is calculated by:
m ≡ 2300^(2789-1-2351) x 2271 ≡ (2300^437) x 2271 (% 2789) ≡ 832
<head ache>
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2011-07-28
, 10:54
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Posts: 30 |
Thanked: 14 times |
Joined on Mar 2011
@ Romania
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#2808
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Originally Posted by notGeekyEnough
in the second video. it almost blends in with the wall
Where is the truck? This may be dumb..
| The Following 2 Users Say Thank You to aikipy For This Useful Post: | ||
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2011-07-28
, 11:02
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Posts: 52 |
Thanked: 74 times |
Joined on Jun 2011
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#2809
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When looking for 6 digit numbers, it will either be the public key, the cipher text or, possibly, p.
Possible primes, p > 304, with a prime root of 10:
313, 337, 367, 379, 383, 389, 419, 433, 461, 487, 491, 499, 503, 509, 541, 571, 577 ...
Last edited by recluse; 2011-07-28 at 11:13.
Possible primes, p > 304, with a prime root of 10:
313, 337, 367, 379, 383, 389, 419, 433, 461, 487, 491, 499, 503, 509, 541, 571, 577 ...
Last edited by recluse; 2011-07-28 at 11:13.
| The Following User Says Thank You to recluse For This Useful Post: | ||
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2011-07-28
, 11:11
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Posts: 2 |
Thanked: 0 times |
Joined on Jul 2011
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#2810
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bye bye #n9seconds.. My head is aching. thanks to you. I have watched the video zillion times. Did not complete my office deadlines. Have a deliverable tomorrow.
this has become too geeky now..
Thank god, there were only 6 ads and each for 9 seconds...
this has become too geeky now..
Thank god, there were only 6 ads and each for 9 seconds...
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| disapoint, eflop, epic win!, laggy interface, n9 rox, so much win, wateriswet, who cares, whyyyyy?????? |
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All times are GMT. The time now is 17:08.
to answer your question, yes, still on el gamal, because it's f***ing awesome to try to solve something like this! it's not about the n9 anymore (at least, not for me), but about the thrill! but that's something you'll never understand, I guess...