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CodeExplorer

CodeExplorer

https://0xec.blogspot.com/2016/08/solving-weasel-keygenme-by-kao.html

On 6/4/2018 at 6:01 PM, GautamGreat said:

I already read that, that is not a proper solution. 

 

  for (int b = 0; b < 49; b++)
   for (int c = b + 1; c < 50; c++)
    if (huge[y++])
     if (calc[b] & calc[c]) bit = !bit;

the expression will be negated only if calc=1 and calc[c]=1 and huge[y++]
this is easier in reverse order from the end to the begin:
last time checks last two bits: calc[49]&calc[48]

Still trying to find solution.
 

CodeExplorer

CodeExplorer

https://0xec.blogspot.com/2016/08/solving-weasel-keygenme-by-kao.html

On 6/4/2018 at 6:01 PM, GautamGreat said:

I already read that, that is not a proper solution. 

 

  for (int b = 0; b < 49; b++)
   for (int c = b + 1; c < 50; c++)
    if (huge[y++])
     if (calc[b] & calc[c]) bit = !bit;

the expression will be negated only if calc=1 and calc[c]=1 and huge[y++]
this is easier in reverse order from the end to the begin:
last time checks last two bits: calc[49]&calc[48]

Still trying to find solution.
 

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