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 ?