cal*_*pto 6 hash implementation cryptography pgp dsa
我终于设法验证了一些简单的PGP签名消息块.但是,我发现由于某种原因,我的实现限制了我验证长度为9-16个字节的数据.不会少.不再.
是否有某些指令(RFC4880或其他地方)指定如何处理任何长度的明文数据？也许我错过了某种填充？PKCS1？
我敢肯定我被格式化的数据正确地散列,因为在RFC 4880秒5.2.4说明书说的文本文档,只需更换所有\n与\r\n添加的拖车.由于我的测试值是单行数据,因此不需要替换任何内容
除非另有说明,否则所有这些值均以10为基数:
// DSA public key values
p = 175466718616740411615640156350265486163809613514213656685227237159351776260193236923030228927905671867677337184318134702903960237546408302010360724274436019639502405323187799029742776686067449287558904042137172927936686590837020160292525250748155580652384740664931255981772117478967314777932252547256795892071
q = 809260232002608708872165272150356204306578772713
g = 127751900783328740354741342100721884490035793278553520238434722215554870393020469115393573782393994875216405838455564598493958342322790638050051759023658096740912555025710033120777570527002197424160086000659457154926758682221072408093235236853997248304424303705425567765059722098677806247252106481642577996274
y = 172935968966072909036304664996424500241381878537444332146572958203083745609400290814117451480512268901233962890933482206538294509037615827035398352528065134903071886710296983781453184598843331365336270501467458073523376152406987560592548479865116940266729198119357206749848310472131186772143408998928864559411
不工作:
-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
abcd
-----BEGIN PGP SIGNATURE-----
Version: BCPG v1.39
iFsEARECABsFAk/tB28UHGFiYyA8bWFrY21AYWFhLmNvbT4ACgkQMFIlRc933Ya2
RwCfdMyI08Iz0rDXVHOPlGA3s5Y9j/8An2He7+hHjWfGJNoOJT7gAxqJaoLo
=I2rT
-----END PGP SIGNATURE-----
data hashed (in hex): 6162636404011102001b05024fed076f141c616263203c6d616b636d406161612e636f6d3e04ff00000021
r = 666804200764671083282351405489424949903645052927
s = 558743769080942454889260816818443017172325925608
w = 702955297882281869313155599553522395227576660460 // s^-1 mod q
u1 = 190417717173929082607343542521304347388874234334
u2 = 306786785479358548892951170619047936651163362761
v = g^u1 * y^u2 % p % q = 737052148656331043521702886300418501784667890334
v != r
工作:
-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
0123456789abcdef
-----BEGIN PGP SIGNATURE-----
Version: BCPG v1.39
iFsEARECABsFAk/tCE0UHGFiYyA8bWFrY21AYWFhLmNvbT4ACgkQMFIlRc933YYG
IQCfercgPsXFnah6otgQdEMbv9OeCgIAnRIyOLirbqSlBugBT6Ex/Adz4+7L
=bzab
-----END PGP SIGNATURE-----
data hashed (in hex): 3031323334353637383961626364656604011102001b05024fed084d141c616263203c6d616b636d406161612e636f6d3e04ff00000021
r = 700580719365380086754774917458461236187098909186
s = 103881812262595813943381509986903840453887782603
w = 178510125628083028184051840492924307896586330444 // s^-1 mod q
u1 = 78831508775508876446567239486098677466912246622
u2 = 572875590470993668032596348682349224460207395691
v = g^u1 * y^u2 % p % q = 700580719365380086754774917458461236187098909186
v == r
我没有在哈希中包含哪些数据/我做错了什么？
编辑:这是所要求的公钥,即使相关的值已经发布了)
-----BEGIN PGP PUBLIC KEY BLOCK-----
Version: BCPG v1.39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=sStS
-----END PGP PUBLIC KEY BLOCK-----
没有足够的时间来查找详细信息，但我猜您正在正确应用（或未应用）填充。这将导致某些输入长度出现正确的结果，但对于其他输入长度则不然。
我想我会更多地研究这个，但我想在赏金线下得到一些东西:)
编辑：好的，发现一个错误。不知道为什么你会得到它，但如果它被修复，那么正确的答案就会出现。在您的不起作用的示例中，您将w(s^-1 mod q) 计算为
w = 702955297882281869313155599553522395227576660460 // s^-1 mod q
但我得到
w = 702955297882281869313155599553522395227576660458
减2！不过，价值观确实非常接近。并且可以证明我的想法是正确的：
s * your_w mod q = 308227306159276200906356361486529830038073078504
s * my_w mod q = 1
如果你代入这个 w 值，你就会得到
u1 = 536931432138658080437983667536052790245747416035
u2 = 591698847955233800072578903940910445457030802333
v = (g^u1 * y^u2) % p % q = 666804200764671083282351405489424949903645052927
r == v
希望有帮助。