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Author Topic: SHA-1 Hash Collision awarded 2.48 BTC  (Read 258 times)
TheArchaeologist (OP)
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June 29, 2020, 10:02:45 AM
Merited by pooya87 (1), ABCbits (1)
 #1

Hi all,

I like a good puzzle now and then and I stumbled upon this script rewarding anyone who could find a SHA-1 collision:
Code:
$ btc decodescript 6e879169a77ca787
{
    "asm" : "OP_2DUP OP_EQUAL OP_NOT OP_VERIFY OP_SHA1 OP_SWAP OP_SHA1 OP_EQUAL",
    "type" : "nonstandard",
    "p2sh" : "37k7toV1Nv4DfmQbmZ8KuZDQCYK9x5KpzP"
}
In simpler terms: Provide two values which give the same result when hashed with the SHA-1 algorithm.

On February 23, 2017 all the unspent outputs which match the decodescript were claimed:
Code:
txid: 8d31992805518fd62daa3bdd2a5c4fd2cd3054c9b3dca1d78055e9528cff6adc
total claimed: 2.48000000 BTC
So obviously a collision was found. Fun-fact: when you dive into the SHA-1 collision history you find that a collision was found on that exact date for the first time. So someone was quick on applying it to this challenge. By the way: Google said it took about 6500 years of CPU-capacity or about 110 year of single GPU capacity to find the collision.

The SigScript has the following value:
Code:
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
6e879169a77ca787

I tried to look at the two values which hash to the same SHA-1 result but was unable to find a match. So my question is: which values are in there which I should compare to successfully find a collision?



Sooner or later you're going to realize, just as I did, that there's a difference between knowing the path and walking the path
domob
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June 29, 2020, 10:57:31 AM
 #2

See here: https://bitcointalk.org/index.php?topic=293382.0

Use your Namecoin identity as OpenID: https://nameid.org/
Donations: 1domobKsPZ5cWk2kXssD8p8ES1qffGUCm | NMC: NCdomobcmcmVdxC5yxMitojQ4tvAtv99pY
BM-GtQnWM3vcdorfqpKXsmfHQ4rVYPG5pKS | GPG 0xA7330737
TheArchaeologist (OP)
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July 03, 2020, 11:58:41 AM
 #3

Sorry for the late response, I was on a trip for a few days.

Can you point me to where the actual values I was asking for are in that topic? I read all the replies and the linked articles from there but I could'nt find what I was looking for (or maybe I missed it).

So to be clear: I'm looking for the following

Input 1 - > SHA 1 -> Hash #1
Input 2 - > SHA 1 -> Hash #2

Where Hash #1 == Hash #2 (otherwise it wouldn't be a collision).  In that case, what are the values for Input 1 and Input 2?


Sooner or later you're going to realize, just as I did, that there's a difference between knowing the path and walking the path
pooya87
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July 03, 2020, 12:17:07 PM
Merited by TheArchaeologist (2), ABCbits (1)
 #4

Input 1 - > SHA 1 -> Hash #1
Input 2 - > SHA 1 -> Hash #2

you have already posted both values yourself inside SigScript in your first post.
the first line is the Input 1 and second line is Input 2 and third line is the redeem script (each are similar looking PDF files).
Code:
input1: 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
input2: 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
redeem: 6e879169a77ca787

SHA-1= f92d74e3874587aaf443d1db961d4e26dde13e9c

you can paste them here and see the hash result for each value: https://emn178.github.io/online-tools/sha1.html
remember to change Input type to hex.

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TheArchaeologist (OP)
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July 03, 2020, 12:30:28 PM
 #5

@pooya87: Thanks, I had the exact same thought before so I pasted the values in an online SHA-1 generator but I guess it was treated as text instead of hex so I got two different values. Even tried it backward (Little Endian to Big Endian) but again no collision.

Sometimes it's just too simple Wink Thanks for helping me realize I'm not crazy after all..

Sooner or later you're going to realize, just as I did, that there's a difference between knowing the path and walking the path
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