## Generating EC parameters

… is not as easy as generating random numbers.

P1363 Section 1.9.5 mention that

The most difficult part of generating EC parameters is finding a base point of prime order

So the next things to do is finding a random point in an elliptic curve (prime case A.11.1/binary case A.11.2), and use A.2.5 to find a square root modulo p and use A.2.1 to calculate modular exponentiation.

In the text book, algorithm for elliptic curve key pair generation is only 5 lines. But implementing one line requires many hours understanding P1363.

• #### Budi Rahardjo 5:29 am on October 30, 2009 Permalink | Reply

implementing one line requires many hours of understanding

no kidding. and then after that, many more hours of coding time.
i am in the middle of it right now.

• #### CG 10:20 am on October 30, 2009 Permalink | Reply

so true.

that’s why we really glad you joined in 😉

## El-Gamal with Pari

Encrypt – decrypt successful.

• #### Budi Rahardjo 10:16 pm on October 21, 2009 Permalink | Reply

Congratulation! Good stuff 🙂

• #### CG 11:11 am on May 19, 2010 Permalink | Reply

thank you 🙂

• #### romi 3:45 am on May 17, 2011 Permalink | Reply

what its mean: gen P(l): to automatically generate the parameter p, where |logp|2  l bits, and set the appropriate generator g.
jpj2  can be interpreted as:
| jp|2 = l, or
| jp|2 = l – 1, or
|jp|2 = l + 1:

help me pliss

• #### romi 3:46 am on May 17, 2011 Permalink

this thing is in elgamal
i am not understand what the task want

• #### oman 7:47 pm on May 18, 2010 Permalink | Reply

can i know how u did this? any coding involved? thanks 😉

## Paper – Bali

Another paper,
more here.

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