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  • Appoxo@lemmy.dbzer0.comEnglish
    1·
    1 year ago

    I would assume with arrays they will use a different way to calculate parity or have higher redundancy to compensate the risk.

    • Frezik@lemmy.blahaj.zoneEnglish
      2·
      1 year ago

      If there’s higher redundancy, then they are already giving up on density.

      We’ve pretty much covered the likely ways to calculate parity.