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user2284570
user2284570
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I have the following data (represented as hex from binary below) where the lower bytes is controlled by attacker in the second case :

0x00: 000000000000000000000000cf269986 ????????????????
0x10: da781407b0eeeac3ea79ac1c9d857d38 ?x???????y?????8
0x20: 00000000000000000000000000000000 ????????????????
0x30: 00000000000000000000000000000001 ????????????????
0x40: 00000000000000000000000000000000 ????????????????

which does 0x1350c327906a2da929ed2a625e133cdfc3aaa945d7540e93b1266293dd2c1d11.

0x00: 000000000000000000000000cf269986 ????????????????
0x10: da781407b0eeeac3ea79ac1c9d857d38 ?x???????y?????8
0x20: 00000000000000000000000000000000 ????????????????
0x30: 00000000000000000000000000000002 ????????????????
0x40: 00000000000000000000000000000000 ????????????????

which does 0xf11930e4cc6cdaaaa9da21243f512da3fba796ffa7c74870669af9cee3a06c0c. Because the use of 0x2 from 0x1 left shift a single bit by one (but both inputs keep the same length).

There’sEven if the change is about a single byte, there’s no way to modify the first 20 bytes of the second case so the hash is equal to 0x1350c327906a2da929ed2a625e133cdfc3aaa945d7540e93b1266293dd2c1d11(the hash of first case) without requiring a computational power not available on the short term, right ?

I have the following data (represented as hex from binary below) where the lower bytes is controlled by attacker in the second case :

0x00: 000000000000000000000000cf269986 ????????????????
0x10: da781407b0eeeac3ea79ac1c9d857d38 ?x???????y?????8
0x20: 00000000000000000000000000000000 ????????????????
0x30: 00000000000000000000000000000001 ????????????????
0x40: 00000000000000000000000000000000 ????????????????

which does 0x1350c327906a2da929ed2a625e133cdfc3aaa945d7540e93b1266293dd2c1d11.

0x00: 000000000000000000000000cf269986 ????????????????
0x10: da781407b0eeeac3ea79ac1c9d857d38 ?x???????y?????8
0x20: 00000000000000000000000000000000 ????????????????
0x30: 00000000000000000000000000000002 ????????????????
0x40: 00000000000000000000000000000000 ????????????????

which does 0xf11930e4cc6cdaaaa9da21243f512da3fba796ffa7c74870669af9cee3a06c0c. Because the use of 0x2 from 0x1 left shift a single bit by one (but both inputs keep the same length).

There’s no way to modify the first 20 bytes of the second case so the hash is equal to 0x1350c327906a2da929ed2a625e133cdfc3aaa945d7540e93b1266293dd2c1d11(the hash of first case) without requiring a computational power not available on the short term, right ?

I have the following data (represented as hex from binary below) where the lower bytes is controlled by attacker in the second case :

0x00: 000000000000000000000000cf269986 ????????????????
0x10: da781407b0eeeac3ea79ac1c9d857d38 ?x???????y?????8
0x20: 00000000000000000000000000000000 ????????????????
0x30: 00000000000000000000000000000001 ????????????????
0x40: 00000000000000000000000000000000 ????????????????

which does 0x1350c327906a2da929ed2a625e133cdfc3aaa945d7540e93b1266293dd2c1d11.

0x00: 000000000000000000000000cf269986 ????????????????
0x10: da781407b0eeeac3ea79ac1c9d857d38 ?x???????y?????8
0x20: 00000000000000000000000000000000 ????????????????
0x30: 00000000000000000000000000000002 ????????????????
0x40: 00000000000000000000000000000000 ????????????????

which does 0xf11930e4cc6cdaaaa9da21243f512da3fba796ffa7c74870669af9cee3a06c0c. Because the use of 0x2 from 0x1 left shift a single bit by one (but both inputs keep the same length).

Even if the change is about a single byte, there’s no way to modify the first 20 bytes of the second case so the hash is equal to 0x1350c327906a2da929ed2a625e133cdfc3aaa945d7540e93b1266293dd2c1d11(the hash of first case) without requiring a computational power not available on the short term, right ?

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kodlu
kodlu
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Swaping Swapping a single bit inside a 40 bytes inputs fed to keccak256. Is it safe to assume no change in the first 20 bytes can result in the same hash?

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user2284570
user2284570
  • 333
  • 2
  • 16

Swaping a single bit inside a 40 bytes inputs fed to keccak256. Is it safe to assume no change in the first 20 bytes can result in the same hash?

I have the following data (represented as hex from binary below) where the lower bytes is controlled by attacker in the second case :

0x00: 000000000000000000000000cf269986 ????????????????
0x10: da781407b0eeeac3ea79ac1c9d857d38 ?x???????y?????8
0x20: 00000000000000000000000000000000 ????????????????
0x30: 00000000000000000000000000000001 ????????????????
0x40: 00000000000000000000000000000000 ????????????????

which does 0x1350c327906a2da929ed2a625e133cdfc3aaa945d7540e93b1266293dd2c1d11.

0x00: 000000000000000000000000cf269986 ????????????????
0x10: da781407b0eeeac3ea79ac1c9d857d38 ?x???????y?????8
0x20: 00000000000000000000000000000000 ????????????????
0x30: 00000000000000000000000000000002 ????????????????
0x40: 00000000000000000000000000000000 ????????????????

which does 0xf11930e4cc6cdaaaa9da21243f512da3fba796ffa7c74870669af9cee3a06c0c. Because the use of 0x2 from 0x1 left shift a single bit by one (but both inputs keep the same length).

There’s no way to modify the first 20 bytes of the second case so the hash is equal to 0x1350c327906a2da929ed2a625e133cdfc3aaa945d7540e93b1266293dd2c1d11(the hash of first case) without requiring a computational power not available on the short term, right ?