2015-11-03 48 views
2

我想为Diffie-Hellman密钥交换计算一些值。我理解这些概念,但我正在处理的数据让我质疑这种方法。Diffie-Hellman计算与非常大的数字

他们给了我5个值与P和G的工作:

// 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导: https://security.stackexchange.com/questions/45963/diffie-hellman-key-exchange-in-plain-english

1)我的假设正确的,“一”是秘密号码在我身边,'B'是他们身边的秘密号码?

2)给出示例数据,我是否正确将十六进制表示转换为十进制值?这里是我的结果:

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

3)当计算结果时,这些值是否有意义?我有一个很难得到这个功能以如此大量的工作:

A = G ^一个模p

鉴于“A”是一些308位长,似乎过大,使任何意义。我不能在BigInteger上设置如此大的数字,因为它只会允许我传递一个int值,其中'a'绝对超出范围。

回答

2

对于您的前两个问题,您是正确的。

对于第三之一,使用modPow(BigInteger exponent, BigInteger modulo),所以

BigInteger A = g.modPow(a,p); 
+0

惊人。谢谢! –