From 2f28ab4909d770fd75cdc214f49fddfe4b04de07 Mon Sep 17 00:00:00 2001
From: Crowdin Bot <support+bot@crowdin.com>
Date: Tue, 28 May 2024 04:34:57 +0000
Subject: [PATCH] New Crowdin translations by GitHub Action

---
 i18n/zh-Hant/basics/passwords-overview.md | 6 +++---
 1 file changed, 3 insertions(+), 3 deletions(-)

diff --git a/i18n/zh-Hant/basics/passwords-overview.md b/i18n/zh-Hant/basics/passwords-overview.md
index 69e4cb0a..e6896db5 100644
--- a/i18n/zh-Hant/basics/passwords-overview.md
+++ b/i18n/zh-Hant/basics/passwords-overview.md
@@ -82,11 +82,11 @@ Diceware 是一種創建密碼短語的方法，這些密短口令易於記憶
 
 為了證明 diceware 密語的強度，我們將使用前面提到的七個單詞密語（`viewable fastness reluctant squishy seventeen shown pencil`）和 [EFF 的大型單詞列表](https://eff.org/files/2016/07/18/eff_large_wordlist.txt)作例子。
 
-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>
+判斷 diceware 口令密語強度的衡量標準是確定它有多少熵。 Diceware 密碼短語中每個單字的熵計算如下 <math> <mrow> <msub> <mtext>記錄(log)</mtext> <mn>2</mn> </msub> <mo form="prefix" stretchy="false">(</mo> <mtext>WordsInList</mtext> <mo form="postfix" stretchy="false">)</mo> </mrow> </math> 密碼短語的整體熵計算如下： <math> <mrow> <msub> <mtext>記錄(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>
 
-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>).
+因此，上述列表中的每個單字都會產生約 12.9 位元的熵（<math> <mrow> <msub> <mtext>記錄(log)</mtext> <mn>2</mn> </msub> <mo form="prefix" stretchy="false">(</mo> <mn>7776</mn> <mo form="postfix" stretchy="false">)</mo> </mrow> </math>），從它衍生出的七字密碼有約 90.47 位元的熵（<math> <mrow> <msub> <mtext>記錄(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>).
 
-[EFF 的大型單字清單](https://eff.org/files/2016/07/18/eff_large_wordlist.txt)包含 7776 個獨特單字。 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>.
+[EFF 的大型單字清單](https://eff.org/files/2016/07/18/eff_large_wordlist.txt)包含 7776 個獨特單字。 要計算可能的密碼短語的數量，要做的就是 <math> <msup> <mtext>WordsInList</mtext> <mtext>WordsInPhrase</mtext> </msup> </math>，或者在我們的例子中， <math><msup><mn>7776</mn><mn>7</mn></msup></math>.
 
 讓我們從這個角度來看：使用 \[EFF 的大型單詞列表\](https://eff.org/files/2016/07/18/eff_large_wordlist.txt) 的七個單詞的口令密短大約有1,719,070,799,748,422,500,000,000 種組合。