https://ameblo.jp/fpq-cipher/ Something new about FPQ-Cipher is written. ja [$a.apn]
　This means that "a" is a prime number.
[$b@c.Lpn]
　This means that a larger "b" prime number than "c"
[$d&@e.Lpn]
　This means that "d" and "e" are two prime numbers
　and "d" is larger than "e."

#Explanation_FPQ Cipher Basis(FPQCB)
1.Alice makes "F" which is $x.apn+1.
2.She chooses Q.apn which can be made $Q+F.apn.
3.She chooses $P@Q.Lpn which can be made $PQ+F.apn.
4.She calculates "e" and "d" with two expression
　to do it to the version she'll use.
　[@A] e(d+F) mod (P-1){(Q+F)-1}=1
　[@B] e(d+F) mod {(F-1)-1}{(PQ+F)-1}=1
　[Common Ver.]@A
　Public Key(PK)=(e,P)　Secret Key(SK)=(d+F)
　[1st Normal Ver.]@B
　PK=(e,PQ+F)　SK=(d+F)
　[2nd Normal Ver.]@A
　PK=(e,Q+F)　SK=(d+F)
　[3rd Normal Ver.]@A
　PK=(d+F,P)　SK=(e)
5.She opens PK she'll use to the public.
6.Bob makes $P'&@Q'.Lpn and calculates A'～F' with these ways.
　A'={(P'^2)-(Q'^2)} mod P'
　B'={(P'^2)-(Q'^2)-A'} / P'
　C'=P'-B'={A+(Q'^2} / P'
　D'=B'C'
　E'=(P'^2) mod Q'
　F'=A'+(Q'^2*E')
7.He ciphers M which
he wanna tell Alice a message with A' , D' and F'.
　{A'M+(Q'^2*E')} / D'=[C] rest [D]
　But in case of Common Ver, M is Bob's F(→FPQCB1).
　(I'll teach to you about Common Ver. in detail.)
8.He ciphers A' with Alice's PK.
　{A'^(Left of PK)} mod (Right of PK)=[C']
　And he sends Alice D', F', [C], [D] and [C'].
9.She disciphers [C'] with her SK.
　{[C']^(SK)} mod (Right of PK)=A'
10.She disciphers [C] and [D] with A', D' and F':
　M={D'[C]+[D]-(Q'^2*E')} / A'
　;and gets M.

#The detailed Explanation of Common Ver.
1.Both Alice and Bob do to FPQCB7.
2.They calculates [E] and [F].
　F(Q+F)=[E]
　Q(d+F)=[F]
3.They do FPQCB8 and send [E] and [F] each other.
4.They do FPQCB9 and 10.
5.They get Partner's "Q" and "d" with these expression.
　But the things which are used are made by Partner.
　([E]/F)-F=Q
　([F]/Q)-F=d
6.Finish!

Make a comment if you don't understand.

See you again. Bye.

(Visit "phampicky blog.")
]]> https://ameblo.jp/fpq-cipher/entry-11445821067.html Thu, 10 Jan 2013 01:47:51 +0900