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  1. #21
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    Quote Originally Posted by dasilvarsa View Post
    What is the fixed double word ?
    Can you explain further please.
    I see that every 4th byte is unencrypted.
    The encryption starts first byte after the part number.
    48 06 B3 00
    i called these 4 bytes - a double word. It is the same for "encrypting" all the file tail (starting at pos 312h = 786 decimal)

    "special_adding" is as follows:
    Let's take a byte @312h: namely 09
    Let's add the first syllable of 09 (0) with the the first syllable of 48 (4) - resulting in 4, LSbits of the sum will be the first syllable of "ecrypted" byte
    Let's add the second syllable of 09 (9) with the the second syllable of 48 (8) - resulting in 11, LSbits of the sum will be the second syllable of "ecrypted" byte
    So "encrypted" byte @312h is 41

    Let's take a byte @313h: namely 00
    Let's add the first syllable of 00 (0) with the the first syllable of 06 (0) - resulting in 0, LSbits of the sum will be the first syllable of "ecrypted" byte
    Let's add the second syllable of 00 (0) with the the second syllable of 06 (6) - resulting in 6, LSbits of the sum will be the second syllable of "ecrypted" byte
    So "encrypted" byte @313h is 06

    Let's take a byte @314h: namely 08
    Let's add the first syllable of 08 (0) with the the first syllable of B3 (B) - resulting in B, LSbits of the sum will be the first syllable of "ecrypted" byte
    Let's add the second syllable of 08 (8) with the the second syllable of B3 (3) - resulting in B, LSbits of the sum will be the second syllable of "ecrypted" byte
    So "encrypted" byte @314h is BB

    Let's take a byte @315h: namely - we don't care due to the corresponding byte of our dword is 00 - special_addition with 00 doesn't change the other argument
    So "encrypted" byte @315h is the sameas in the non-encrypted file (ahh.. 00 - we are lucky!)

    Let's take a byte @316h: namely 3A
    Let's add the first syllable of 3A (3) with the the first syllable of 48 (4) - resulting in 7, LSbits of the sum will be the first syllable of "ecrypted" byte
    Let's add the second syllable of 3A (A) with the the second syllable of 48 (8) - resulting in 12, LSbits of the sum will be the second syllable of "ecrypted" byte
    So "encrypted" byte @316h is 72

    and so on
    i can write one unique formula, but cannot decide which bit/byte-operation functions are well-known to the readers.
    In fact it is sufficient to use chr(), asc() and mod(,16)

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  3. #22
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    Quote Originally Posted by lolypop000 View Post
    Look here in decrypted file you find XX
    So what?
    i advised to look for <hr>XX - six bytes (in hex: 3C 68 72 3E 58 58), can't realize why you limit search string to two bytes only.
    Last edited by sbog; 28th June 2013 at 18:43.

  4. #23
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    Quote Originally Posted by catymag View Post
    I guess nuvi will see it like every unencrypted POIs, if unlocked.
    Caty is right. No need to encrypt it back after any changes.

  5. #24
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    Quote Originally Posted by sbog View Post
    So what?
    i advised to look for <hr>XX - six bytes (in hex: 3C 68 72 3E 58 58), can't realize why you limit search string to two bytes only.
    Becouse i search in text not hex
    Slovenian and Slovakian user now have working Fuel Types and Vehicle Charge Points.gpi thank to syzygy for decrypted file i hope user will give Rep to syzygy.

    [Only registered and activated users can see links. ]

    And i like to thank you again @sbog for tip for autochange becouse i forget this function.

    p.s.
    I am not programmer so i don't know all tips and tricks but i try to learn from you who know how to do it.
    My only goal is to help other users who don't know how to do it.
    So, respect and thank you both for all your help and tips.

    p.s.
    the same link for repaired gpi is also in post n18 on previous page.
    You have also in post #18 picture how is look before and after.
    Also in Caty post #2 you have link for download
    Last edited by lolypop000; 30th June 2013 at 18:30. Reason: typo, add more data.
    If you wish, you can [Only registered and activated users can see links. ]

  6. #25
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    Quote Originally Posted by sbog View Post
    48 06 B3 00
    i called these 4 bytes - a double word. It is the same for "encrypting" all the file tail (starting at pos 312h = 786 decimal)

    "special_adding" is as follows:
    ***Hidden content cannot be quoted.***
    i can write one unique formula, but cannot decide which bit/byte-operation functions are well-known to the readers.
    In fact it is sufficient to use chr(), asc() and mod(,16)
    @sbog
    Very good work, It works well for starting with a unencrypted file and creating a encrypted file.
    What I want to do is to start with a encrypted file and end up with a unencrypted file.
    Could you please do the explanation for decrypting.
    The difference between stupidity and genius is that genius has its limits.
    Albert Einstein.

  7. #26
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    Quote Originally Posted by dasilvarsa View Post
    @sbog
    Very good work, It works well for starting with a unencrypted file and creating a encrypted file.
    What I want to do is to start with a encrypted file and end up with a unencrypted file.
    Could you please do the explanation for decrypting.
    i think i'll have some gap this (upcoming) monday.
    it's a pity but i have no time right now although this looks to be easy enough.
    monday, ok?

  8. #27
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    Quote Originally Posted by sbog View Post
    "special_adding" is as follows:
    ***Hidden content cannot be quoted.***
    i can write one unique formula, but cannot decide which bit/byte-operation functions are well-known to the readers.
    In fact it is sufficient to use chr(), asc() and mod(,16)
    OK! You are genius! I have tried to do simple subtraction for each whole byte from key:
    48 06 B3 00
    but I always got some random wrong decrypted byte as say above. It had entered me to a screeching halt!!! The key was obvious after comparing of two simple and very small file that was made by GPICreator - one is encrypted and second is non-encrypted. Script for HEX editor worked fine. But always just a little broken result was on screen.
    It is pity what HEX editor is not operable with this algorithm. I am a deep stupid. So hope to you or may be syzygy give a ready for use program decision...
    Last edited by Giomen; 29th June 2013 at 02:17.
    Love your wife? Buy yourself a GPS for answering to her favorite question: "Dear, where are you?".

  9. #28
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    Quote Originally Posted by Giomen View Post
    OK! You are genius!
    No, i'm sbog.
    Quote Originally Posted by sbog View Post
    monday, ok?
    In fact i've done this.
    i wrote it in FoxPro, the "tool" is rather rough with the header analysis (no time to study its structure, maybe on monday) but it works for this "Fuel types" and for all "Cyclopes" from the parallel thread.

    By some reasons i shall not distribute full standalone executable, - only small one for VFP.9 (but user must install FoxPro libraries from elsewhere). If it is ok - i can do this.
    Anyway - here is the source:
    [hide]para _fnam1
    if type([_fnam1])#[C] or !file(_fnam1)
    ?[source (encrypted) file does not exist..]
    retu -10
    endi

    bb=fcrea([$dec_]+_fnam1)
    if bb<0
    ?[can't create output (decrypted) file..]
    retu -11
    endi

    aa=fope(_fnam1)
    if aa<0
    =fclo(bb)
    ?[can't open source (encrypted) file..]
    retu -12
    endi

    _o=0
    L0=frea(aa,4)
    do case
    case L0==chr(0)+chr(0)+chr(0)+chr(0) && not our case but often met
    ?[i'll think of it tomorrow. after RTFM. maybe..]
    _o=-13

    case L0==chr(0)+chr(0)+chr(8)+chr(0) && our case (Cyclopes & Fuel)
    L1=frea(aa,4)
    =fwri(bb,L0+L1)
    L0=doubval(L1)+65536*doubval(subs(L1,3))
    L1=frea(aa,4)
    =fwri(bb,L1)
    L1=doubval(L1)+65536*doubval(subs(L1,3))
    if L0-L1#13
    wait wind [unusual (not 13) difference: ]+allt(str(L0-L1))
    endi
    L0=frea(aa,L1)
    if L1>13
    L0=stuf(L0,13,1,chr(asc(subs(L0,13,1))-4))
    endi
    =fwri(bb,L0)
    L0=frea(aa,2)
    =fwri(bb,L0+frea(aa,doubval(L0)))
    L0=frea(aa,4)
    =fwri(bb,L0+frea(aa,doubval(L0)+256*doubval(subs(L0,3))+4))

    dime skey(4,2) && 48 06 B3 00
    skey(1,1)=4
    skey(1,2)=8
    skey(2,1)=0
    skey(2,2)=6
    skey(3,1)=11
    skey(3,2)=3
    skey(4,1)=0
    skey(4,2)=0

    L0=0
    do whil !feof(aa)
    L0=iif(L0=4,1,L0+1)
    L1=asc(frea(aa,1))
    =fwri(bb,chr(16*mod(int(L1/16)-skey(L0,1),16)+mod(mod(L1,16)-skey(L0,2),16)))
    endd

    othe
    ?[yet unknown (to me gpi-type]
    _o=-14
    endc

    =fclo(bb),fclo(aa)
    retu _o

    func doubval
    para _
    retu asc(_)+256*asc(subs(_,2))[/hide]
    Last edited by sbog; 30th June 2013 at 21:59. Reason: Code changed in order to fully correspond both Cyclops and Fuel types POIs :)

  10. #29
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    Quote Originally Posted by sbog View Post
    By some reasons i shall not distribute full standalone executable, - only small one for VFP.9 (but user must install FoxPro libraries from elsewhere). If it is ok - i can do this.
    OK! If WinHex or sbog have some reason do not want to make a full standalone executable - Microsoft Excel with copy-paste do it in any case
    Love your wife? Buy yourself a GPS for answering to her favorite question: "Dear, where are you?".

  11. #30
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    I have posted two new gpi decryption and encryption tools as below.

    [Only registered and activated users can see links. ]

 

 
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