4 ms·

or perhaps he can break RSA in polynomial time or something like that

by sswam 17y ago

deleted 17y ago
[deleted]
noamsml 17y ago
Pretty sure that's the same thing as proving P=NP.
- meami 17y ago
  HMAC/SHA-512 Tests HMAC/SHA-512 Short ASCII Text: HMAC/SHA-512 Short Hex Key:value="0123456789:;<=>?@ABC0123456789:;<=>?@ABC" HMAC/SHA-512 Short Hex Result: value="74ed1131b8e37c0e18829b7ae7d99925664cfe055c2d01fa00d0f152ac321a50f3ef1ee91a36fd6248de60ede4196a4f9e5acca4981a09a91a0745d674ca11d3" HMAC/SHA-512 Med ASCII Text: HMAC/SHA-512 Med Hex Key: value="000102030405060708090a0b0c0d0e0f101112131415161718191a1b1c1d1e1f202122232425262728292a2b2c2d2e2f303132333435363738393a3b3c3d3e3f000102030405060708090a0b0c0d0e0f101112131415161718191a1b1c1d1e1f202122232425262728292a2b2c2d2e2f303132333435363738393a3b3c3d3e3f" HMAC/SHA-512 Med Hex Result: HMAC/SHA-512 Med Correct:value="e3c46b747b407ae103d75d94cc2247d6fef607a019e9dca6a1183966a163eb19317e7642da27b6e6ed22fea8224c493d5f20a458f83a555b94264253ddceb242" HMAC/SHA-512 Large Hex Text: value="53616D706C65202333" HMAC/SHA-512 Large Hex Key:value="505152535455565758595a5b5c5d5e5f606162636465666768696a6b6c6d6e6f707172737475767778797a7b7c7d7e7f808182838485868788898a8b8c8d8e8f909192939495969798999a9b9c9d9e9fa0a1a2a3a4a5a6a7a8a9aaabacadaeafb0b1b2b3505152535455565758595a5b5c5d5e5f606162636465666768696a6b6c6d6e6f707172737475767778797a7b7c7d7e7f808182838485868788898a8b8c8d8e8f909192939495969798999a9b9c9d9e9fa0a1a2a3a4a5a6a7a8a9aaabacadaeafb0b1b2b3" HMAC/SHA-512 Large Hex Result: HMAC/SHA-512 Large Correct: value="a2a8627f166e8c52d95fbf5c55271d466d43b73696e94977d08262f021d9a916f9d10f09f40db64e038c2ed3b16fbb9e61082c9173acfa86445612ccd4d8836c"
- Locke1689 17y ago
  No it's not. The factoring problem/RSA problem have never been proven to be NP-complete. Shor's algorithm shows that it's NP, but not NP-complete (or hard). I think it's also in co-NP.
  - deleted 17y ago
    [deleted]
    - PG-13 17y ago
      Yeah.... but like he said, RSA being in P does not imply that P=NP because RSA has not be shown (and isn't believed) to be NP-complete. If P=NP, then RSA would of course be in P.

or perhaps he can break RSA in polynomial time or something like that

by sswam 17y ago

or perhaps he can break RSA in polynomial time or something like that

deleted 17y ago

[deleted]

noamsml 17y ago

Pretty sure that's the same thing as proving P=NP.

meami 17y ago
HMAC/SHA-512 Tests HMAC/SHA-512 Short ASCII Text: HMAC/SHA-512 Short Hex Key:value="0123456789:;<=>?@ABC0123456789:;<=>?@ABC" HMAC/SHA-512 Short Hex Result: value="74ed1131b8e37c0e18829b7ae7d99925664cfe055c2d01fa00d0f152ac321a50f3ef1ee91a36fd6248de60ede4196a4f9e5acca4981a09a91a0745d674ca11d3" HMAC/SHA-512 Med ASCII Text: HMAC/SHA-512 Med Hex Key: value="000102030405060708090a0b0c0d0e0f101112131415161718191a1b1c1d1e1f202122232425262728292a2b2c2d2e2f303132333435363738393a3b3c3d3e3f000102030405060708090a0b0c0d0e0f101112131415161718191a1b1c1d1e1f202122232425262728292a2b2c2d2e2f303132333435363738393a3b3c3d3e3f" HMAC/SHA-512 Med Hex Result: HMAC/SHA-512 Med Correct:value="e3c46b747b407ae103d75d94cc2247d6fef607a019e9dca6a1183966a163eb19317e7642da27b6e6ed22fea8224c493d5f20a458f83a555b94264253ddceb242" HMAC/SHA-512 Large Hex Text: value="53616D706C65202333" HMAC/SHA-512 Large Hex Key:value="505152535455565758595a5b5c5d5e5f606162636465666768696a6b6c6d6e6f707172737475767778797a7b7c7d7e7f808182838485868788898a8b8c8d8e8f909192939495969798999a9b9c9d9e9fa0a1a2a3a4a5a6a7a8a9aaabacadaeafb0b1b2b3505152535455565758595a5b5c5d5e5f606162636465666768696a6b6c6d6e6f707172737475767778797a7b7c7d7e7f808182838485868788898a8b8c8d8e8f909192939495969798999a9b9c9d9e9fa0a1a2a3a4a5a6a7a8a9aaabacadaeafb0b1b2b3" HMAC/SHA-512 Large Hex Result: HMAC/SHA-512 Large Correct: value="a2a8627f166e8c52d95fbf5c55271d466d43b73696e94977d08262f021d9a916f9d10f09f40db64e038c2ed3b16fbb9e61082c9173acfa86445612ccd4d8836c"
Locke1689 17y ago
No it's not. The factoring problem/RSA problem have never been proven to be NP-complete. Shor's algorithm shows that it's NP, but not NP-complete (or hard). I think it's also in co-NP.
- deleted 17y ago
  [deleted]
  - PG-13 17y ago
    Yeah.... but like he said, RSA being in P does not imply that P=NP because RSA has not be shown (and isn't believed) to be NP-complete. If P=NP, then RSA would of course be in P.