6

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 9-16 字节长的数据。不少于。不再。

是否有一些指令（RFC4880或其他地方）指定如何处理任何长度的纯文本数据？也许我错过了某种填充？PKCS1?

我很确定我将数据格式化为正确散列，因为 RFC 4880 sec 5.2.4 中的说明针对文本文档，只需将所有内容替换\n为\r\n并添加预告片。由于我的测试值是单行数据，因此无需替换任何内容

除非另有说明，否则所有这些值都以 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

在职的：

-----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，即使相关值已经发布）

-----BEGIN PGP PUBLIC KEY BLOCK-----
Version: BCPG v1.39
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=sStS
-----END PGP PUBLIC KEY BLOCK-----
4

1 回答 1

4

没有足够的时间查看详细信息，但我猜您正在正确应用（或不应用）填充。这将导致对某些输入长度产生正确的结果，但对其他输入长度则不然。

我想我会对此进行更多研究，但我想在赏金线下找到一些东西:)

编辑：好的，发现一个错误。不知道你为什么会得到它，但如果它是固定的，那么正确的答案就会出现。在您不工作的示例中，您将w(s^-1 mod q) 计算为

w =  702955297882281869313155599553522395227576660460 // s^-1 mod q

但我明白了

w =  702955297882281869313155599553522395227576660458

减2！真的，真的很接近的价值观。可以证明我的是对的：

s * your_w mod q = 308227306159276200906356361486529830038073078504
s * my_w mod q = 1

如果你插入这个 w 值，你会得到

u1 = 536931432138658080437983667536052790245747416035
u2 = 591698847955233800072578903940910445457030802333
v = (g^u1 * y^u2) % p % q = 666804200764671083282351405489424949903645052927
r == v

希望有帮助。

于 2012-07-09T19:59:52.677 回答