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Author Topic: Question Hash Sha256 Decrypt  (Read 322 times)
GaboBegue
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September 18, 2017, 04:36:30 PM
 #1

Hello:

I have a question about sha256 or sha512 decrypt:

Suppose I want to decrypt this hash sha 256: "6A09E667BB67AE853C6EF372A54FF53A510E527F9B05688C1F83D9AB5BE0CD19"

Step 1:

A= (h0) 6A09E667 + ? = 6A09E667 -------> ? = 6A09E667 - 6A09E667 = 00000000
B= (h1) BB67AE85 + ? = BB67AE85 -------> ? = BB67AE85 - BB67AE85 = 00000000
C= (h2) 3C6EF372 + ? = 3C6EF372 -------> ? = 3C6EF372 - 3C6EF372 = 00000000
D= (h3) A54FF53A + ? = A54FF53A -------> ? = A54FF53A - A54FF53A = 00000000
E= (h4) 510E527F + ? = 510E527F -------> ? = 510E527F - 510E527F = 00000000
F= (h5) 9B05688C + ? = 9B05688C -------> ? = 9B05688C - 9B05688C = 00000000
G= (h6) 1F83D9AB + ? = 1F83D9AB -------> ? = 1F83D9AB - 1F83D9AB = 00000000
H= (h7) 5BE0CD19 + ? = 5BE0CD19 -------> ? = 5BE0CD19 - 5BE0CD19 = 00000000

Round 64 =

A = 0000000
B = 0000000
C = 0000000
D = 0000000
E = 0000000
F = 0000000
G = 0000000
H = 0000000

Round 63=

A = 0000000
B = 0000000
C = 0000000
D = ?
E = 0000000
F = 0000000
G = 0000000
H = ?

Being Round 64 E = 00000000 and E = D + [Ch(E,F,G) + Ma(A,B,C) + Σ0(A) + Σ1(E) + Wt + Kt]

Is it possible that one of these is a negative hexadecimal number?

Otherwise, E could not be worth "00000000".

Or am I in procedural error?  Huh

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September 18, 2017, 05:37:40 PM
 #2

Hello:

I have a question about sha256 or sha512 decrypt:

Suppose I want to decrypt this hash sha 256: "6A09E667BB67AE853C6EF372A54FF53A510E527F9B05688C1F83D9AB5BE0CD19"
Hashes are not encryption; they cannot be decrypted or reversed. If they could, then the hash function would be broken.

Is it possible that one of these is a negative hexadecimal number?

Otherwise, E could not be worth "00000000".

Or am I in procedural error?  Huh
Each of the 4 byte numbers are 32-bit unsigned integers, so they cannot be negative. The addition of these are done modulo 2^32 so the numbers you sum will need to be some multiple of 2^32 in order for E to be 0x00000000.

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September 18, 2017, 05:55:38 PM
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Thank you for your help!

I now understand  Wink
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