• a cow (maybe a robot-cow?) (unregistered)

    if(cow==frist) ++cnt;

  • trainbrain27 (unregistered)

    Your password only has to be strong enough that the adversary uses a 0-day or backdoor. Nothing is completely secure, and at some point you wind up trading off usability. An air gap is much safer, but this is my browsing machine, not my <redacted> machine. Hopefully we're wise enough to make the right choice.

  • Jay (unregistered)

    So what was happening is that "cnt" was being incremented every time it ran into a character that was used somewhere else in the string. It was checking the character at each position and every time split returned more than one string, it would increment. If the character "." showed up twice in the password, "cnt" would be incremented by 2 .

    The length of the example password is 63 characters. Half of that length would be 31.5. The string "6Y9^}Ky.SK50ZR84.p,5u$380(G;m;bI%NZG%zHd?lOStqRzS}Z?t;8qSg;[gy@" has over 40 character positions containing characters that occur elsewhere in the string (at least 19 characters are not unique and several occur more than twice).

  • Officer Johnny Holzkopf (unregistered)

    The password complexity is true - refering to "if (passwordComplexity == true && len > 8)"... yes.

  • Rocky (unregistered)

    Repeated characters can be bad, but forcing you not to repeat characters also means the password has less entropy.

  • tdwtf (unregistered)

    One part of cracking the enigma code was the observation that no character was mapped to itself.

  • eric bloedow (unregistered)
    Comment held for moderation.
  • LZ79LRU (unregistered) in reply to trainbrain27
    Comment held for moderation.

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