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Entropy strength for cards assumes no replacement

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This commit is contained in:
Ian Coleman
2016-11-16 11:19:49 +11:00
parent b6dbc2a1ae
commit 02f05d3e46
2 changed files with 31 additions and 1 deletions
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31
src/js/index.js
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@@ -818,7 +818,7 @@
strength = strength + " - " + z.feedback.warning;
};
}
var bitsStr = entropy.binaryStr.length;
var bitsStr = getNumberOfEntropyBits(entropy);
var wordCount = Math.floor(entropy.binaryStr.length / 32) * 3;
DOM.entropyFiltered.html(entropy.cleanHtml);
DOM.entropyType.text(entropy.base.str);
@@ -830,6 +830,35 @@
DOM.entropyBitsPerEvent.text(Math.log2(entropy.base.asInt).toFixed(2));
}
function getNumberOfEntropyBits(entropy) {
var bitsStr = entropy.binaryStr.length.toString();
// If using cards, assume they are not reused, thus additional entropy
// decreases as more cards are used. This means entropy is measured
// using n!, not base^n.
// eg the second last card can be only one of two, not one of fifty two
// so the added entropy for that card is only one bit at most
if (entropy.base.asInt == 52) {
var totalCombos = factorial(52);
var remainingCards = 52 - entropy.base.parts.length;
var remainingCombos = factorial(remainingCards);
var currentCombos = totalCombos.divide(remainingCombos);
bitsStr = currentCombos.toString(2).length.toString();
}
return bitsStr
}
// Depends on BigInteger
function factorial(n) {
if (n == 0) {
return 1;
}
f = BigInteger.ONE;
for (var i=1; i<=n; i++) {
f = f.multiply(new BigInteger(i));
}
return f;
}
var networks = [
{
name: "Bitcoin",