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i18n/it/basics/passwords-overview.md
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@ -82,11 +82,11 @@ We recommend using [EFF's large wordlist](https://eff.org/files/2016/07/18/eff_l
To demonstrate how strong diceware passphrases are, we'll use the aforementioned seven word passphrase (`viewable fastness reluctant squishy seventeen shown pencil`) and [EFF's large wordlist](https://eff.org/files/2016/07/18/eff_large_wordlist.txt) as an example.
Un parametro per determinare la forza di una passphrase diceware è la sua entropia. L'entropia per parola in una frase segreta Diceware è calcolata come $\text{log}_2(\text{WordsInList})$ e l'entropia complessiva della frase segreta è calcolata come $\text{log}_2(\text{WordsInList}^\text{WordsInPhrase})$.
One metric to determine the strength of a diceware passphrase is how much entropy it has. The entropy per word in a diceware passphrase is calculated as <math> <mrow> <msub> <mtext>log</mtext> <mn>2</mn> </msub> <mo form="prefix" stretchy="false">(</mo> <mtext>WordsInList</mtext> <mo form="postfix" stretchy="false">)</mo> </mrow> </math> and the overall entropy of the passphrase is calculated as: <math> <mrow> <msub> <mtext>log</mtext> <mn>2</mn> </msub> <mo form="prefix" stretchy="false">(</mo> <msup> <mtext>WordsInList</mtext> <mtext>WordsInPhrase</mtext> </msup> <mo form="postfix" stretchy="false">)</mo> </mrow> </math>
Dunque, ogni parola nell'elenco suddetto risulta in circa 12,9 bit di entropia ($\text{log}_2(7776)$), e una frase segreta di sette parole da esso derivaata contiene circa 90,47 bit di entropia ($\text{log}_2(7776^7)$).
Therefore, each word in the aforementioned list results in ~12.9 bits of entropy (<math> <mrow> <msub> <mtext>log</mtext> <mn>2</mn> </msub> <mo form="prefix" stretchy="false">(</mo> <mn>7776</mn> <mo form="postfix" stretchy="false">)</mo> </mrow> </math>), and a seven word passphrase derived from it has ~90.47 bits of entropy (<math> <mrow> <msub> <mtext>log</mtext> <mn>2</mn> </msub> <mo form="prefix" stretchy="false">(</mo> <msup> <mn>7776</mn> <mn>7</mn> </msup> <mo form="postfix" stretchy="false">)</mo> </mrow> </math>).
The [EFF's large wordlist](https://eff.org/files/2016/07/18/eff_large_wordlist.txt) contains 7776 unique words. Per calcolare la quantità di frasi segrete possibili, tutto ciò che dobbiamo fare è $\text{WordsInList}^\text{WordsInPhrase}$ o, nel nostro caso, $ 7776^7 $.
The [EFF's large wordlist](https://eff.org/files/2016/07/18/eff_large_wordlist.txt) contains 7776 unique words. To calculate the amount of possible passphrases, all we have to do is <math> <msup> <mtext>WordsInList</mtext> <mtext>WordsInPhrase</mtext> </msup> </math>, or in our case, <math><msup><mn>7776</mn><mn>7</mn></msup></math>.
Let's put all of this in perspective: A seven word passphrase using [EFF's large wordlist](https://eff.org/files/2016/07/18/eff_large_wordlist.txt) is one of ~1,719,070,799,748,422,500,000,000,000 possible passphrases.