Question

我正在尝试为Diffie-Hellman密钥交换计算一些值。我理解这些概念，但我正在使用的数据让我质疑这个方法。

我已经获得了5个值来处理p和g：

    // p and g
    BigInteger p = new BigInteger(DH_PRIME, 16);
    BigInteger g = new BigInteger(DH_GEN, 16);

    // My Private Key
    BigInteger a = new BigInteger(
            "5E8B02F2B2E9C96E0C359ECD14EB1B29EBDD61E70A61E4" +
            "2F0836A5974963E96D91F1462B699C222BC92BC068E9DC" +
            "E5C78E4349D28DDCB6D0ED2C41F7CD8AF2418C8AE27B69" +
            "09484DED7F0C5B4C286D9C36518FA5953974741B3A6F75" +
            "7B59A41A5CA0B74EFD919BB7ED8CCEC9CB3BC4B4F8D15D" +
            "16DC4642E54691904B2F35B969", 16);

    // My Public Key
    BigInteger A = new BigInteger(
            "85F04DD00345642AD12B65BD1A7C38728BFF0B8E281DDB" +
            "6AC4F2739E82A02145DAABF23D173C933913B1F8440597" +
            "10E9125591569DE427EAE1D269ACCBFA3305069DEB7622" +
            "D1DA3AD9820D11BD24FDCCE5381D2DF99BDA314394738D" +
            "FCBE210EAE247B1303E79297FF746CD919E189F6A5776E" +
            "6ECC24C8900DE0F38F159072DE", 16);

    // Their Private Key
    BigInteger b = new BigInteger(
            "42111D3A7ECAA6A83E503825F38629AD9754D93370D681" +
            "AEFEE152329D8DAE6C20732C5A7B6393DEDDB62753CEEF" +
            "AE0A5E1BD037A5A32364CE1375442E58997C2918563EE5" +
            "D7452373847AABAD5A5D02DF289B3A0B9096A375AE509F" +
            "16363B4573A5CCCDFFF2B60459D52C0E5280853000CE62" +
            "68560A95111723AF5916CC8376", 16);

    // Their Public Key
    BigInteger B = new BigInteger(
            "71257BA7758CDE21480706CA55861F5FE6122E5B879420" +
            "80F3E384890284FD62341B90A1B60FB44ADD61031D6AAC" +
            "3D5B267F1435B0765AC289040B63B242EED82863FD18BB" +
            "637757EDF44BA4589E0CE99D192E902C16EF1A89E7E7C1" +
            "C2EB5A6A8AB3E3E4F6B8A9CACCA4B8F6C4E20D12626797" +
            "5406CF9151D57BEEAE32C33CD8", 16);

    // The Shared Secret
    BigInteger secret = new BigInteger(
            "834A9D0434D817735589F22A4633FB6DD3E530DBA1EA2B" +
            "BB9E1ACCB438084513087F5DA00EA86ED53164D8893B81" +
            "A9C8DF65BC189CF6830D271E1A3E504CBCB25714164B51" +
            "9C75F6E0ADF41BB07E7F8C4FB7B9960D813E6577A73252" +
            "EEA9C139CDDA606D51122170E71636E7849149618C8238" +
            "A226128821F0B668490BDCA82F", 16);

使用这篇文章作为指南： https://security.stackexchange.com/questions/45963/diffie-hellman-key-exchange-in-plain-english

1）我的假设是否正确，'a'是我身边的密码，'b'是他们身边的密码？

2）鉴于示例数据，我是否正确将十六进制表示转换为十进制值？以下是我的结果：

 a = 66390362007035946406218401401207761013475548285294105112105139639190607639308478292266684826306137190981682856073453566963077570668348041252633322979431955395580457467551843233313476557017872071709128383858840166213588753166690401421078658676432871207970713843501321433541674906101053126631545366776741476713
 A = 94054944803568781365809132293092437685411466113799940774655767034465070801688227674438324625217997304495304798217968020105674208988107083532555808216365925310845077730469696561179874023824545031639895643922574243862372138887930362438243722132412662233732339999845565810626073575967836399703588412025249755870
 b = 46393721373616931425713842738733091490641584087681350012011014830147740618840619997608221051635856275110730538390991643443828273505863608898989702862901548035408503167446323013096534700981785904886213090032981937144110592283933751272337196723467356342510104910060731712678136878986285012929934142772165247862
 B = 79454116791030243835993774846060329772022864579356590515844871048029546733381452221063689783318067017491406528317462838099472851672248798338610594948647929971365872809557181659701202351270701590108182485073251154126367917793952098022309258299793944660722596621675214128052755659890352643244396810687568100568
 secret = 92195997420654412005403859326763427963568159942225029128672319592580542441945382083896651021886036463236361034857638584808142466991971457292885469363097913358292550515467126779618590503912963769256823553075620480922531451229861911226149279841977690852350735509236250226638964983746622786925283953012622796847

3）在计算结果时，这些值是否有意义？我很难让这个函数与如此大的数字一起工作：

A = g ^ a mod p

鉴于'a'长约308位，看起来太大了，没有任何意义。我无法在BigInteger上设置如此大的数字作为指数，因为它只允许我传递一个int值，其中我的'a'绝对超出范围。

Answer 1

你的前两个问题是正确的。

对于第三个，请使用modPow(BigInteger exponent, BigInteger modulo)，所以

BigInteger A = g.modPow(a,p);

具有非常大数字的Diffie-Hellman计算

1 个答案: