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index.html
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<!doctype html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0, maximum-scale=1.0, user-scalable=no">
<title>Secret sharing</title>
<link rel="stylesheet" href="reveal/reset.css">
<link rel="stylesheet" href="reveal/reveal.css">
<link id="theme" rel="stylesheet" href="reveal/theme/white.css">
<link rel="stylesheet" href="fonts/linlibertine/linlibertine.css">
<link rel="stylesheet" href="fonts/noto-sans/noto-sans.css">
<link rel="stylesheet" href="fonts/roboto/roboto.css">
<style>
:root {
--r-main-font: "Libertinus Sans";
--r-heading-font: "Libertinus Serif Display";
}
</style>
<style>
cite {
display: none;
}
.reveal.show-notes > .slides {
border: thin solid var(--r-main-color);
}
.r-stretch > div {
height: 100%;
}
.r-stretch > div > svg {
min-height: 100%;
min-width: 100%;
}
.reveal span.r-stack {
display: inline-grid;
}
.side-elem {
z-index: 100;
position: fixed;
left: 12px;
font-size: small;
color: var(--r-main-color);
background-color: var(--r-background-color);
opacity: 0.70;
}
.reveal .pause-overlay {
background-color: var(--r-background-color);
}
.reveal .slides section .fragment.semi-highlight {
opacity: 0.5;
visibility: inherit;
}
.reveal .slides section .fragment.semi-highlight.current-fragment {
opacity: 1;
}
#setup {
top: 12px;
display: none;
}
.reveal.show-notes > #setup {
display: unset;
}
#cite {
bottom: 12px;
}
#cite a:hover, #cite a:visited, #cite a:active, #cite a:link {
color: inherit;
text-decoration: none;
}
.MJXc-display, .MathJax_Display {
margin: 0 !important;
}
.reveal p {
text-align: left;
}
.reveal ul {
display: block;
}
.reveal ol {
display: block;
}
.reveal blockquote {
width: 85%;
}
.reveal ::selection {
color: reset;
background: #ff9966;
text-shadow: none;
}
.reveal section.img_container {
width: 100%;
height: 100%;
margin: 0;
padding: 0;
}
.reveal section img {
display: block;
margin: auto;
height: auto;
max-height: 100%;
width: auto;
max-width: 100%;
}
@media screen and (orientation: portrait) {
.reveal.show-notes .speaker-notes {
top: 100%;
left: 0;
height: 50vh;
width: 100%;
}
.reveal.show-notes {
max-width: none;
max-height: 50vh;
}
}
.columns {
display: flex;
}
.columns > .col {
flex: 1;
}
.columns > p:first-child {
margin-right: 1em;
}
.columns > p:last-child {
margin-left: 1em;
}
.columns > .col > ol {
margin: var(--r-block-margin) 0 var(--r-block-margin) 1em;
}
svg {
height: 100%;
width: 100%;
}
.scrollable-slide {
overflow-y: auto !important;
}
/*.reveal-viewport {
overflow-y: auto;
}*/
::-webkit-scrollbar {
width: 6px;
}
::-webkit-scrollbar-track {
-webkit-box-shadow: inset 0 0 6px rgba(0,0,0,0.3);
}
::-webkit-scrollbar-thumb {
background-color: #333;
}
::-webkit-scrollbar-corner {
background-color: #333;
}
</style>
</head>
<body>
<div class="reveal">
<div id="setup" class="side-elem">
<select onchange="document.documentElement.style.setProperty('--r-main-font', this.value)">
<option value="sans-serif" disabled selected>Шрифт</option>
<option value="Libertinus Sans">Без засечек</option>
<option value="Libertinus Serif">С засечками</option>
<option value="Noto Sans">Noto Sans</option>
<option value="Roboto">Roboto</option>
</select>
<select onchange="document.getElementById('theme').setAttribute('href', this.value)">
<option value="reveal/theme/white.css" disabled selected>Тема</option>
<option value="reveal/theme/white.css">Светло</option>
<option value="reveal/theme/black.css">Темно</option>
</select>
<span>
Презентацию листать при помощи пробела, с шифтом чтобы назад. За справкой нажми <kbd>?</kbd>
</span>
</div>
<div id="cite" class="side-elem"></div>
<div class="slides">
<section data-markdown
data-separator="^\n\*\*\*\n"
data-separator-vertical="^\n---\n"
data-separator-notes="^Notes:">
<script type="text/template">
# Разделение <br/> секрета <!-- .element: class="r-fit-text" -->
Notes:
Доклад будет о разделении секрета.
---
# План
1. Введение
2. $(n,n)$-схема
3. Схема Блэкли
4. Многочлен Лагранжа (★)
5. Схема Шамира
6. Атаки (★)
7. Реализация простых структур доступа
8. Связь с теорией матроидов
9. Реализация произвольных структур доступа (★)
Notes:
Рассказать нужно много.
***
# Зачем это надо?
<cite>G. R. BLAKLEY, "Safeguarding cryptographic keys," 1979</cite>
<div class="r-stack r-stretch">
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Notes:
1. Есть важный секрет. _(клик)_ Например, приватный ключ.
Некоторые компании весь бизнес строят на том, что есть ключ и репутация.
2. Конечно, его ни в коем случае нельзя случайно потерять.
Если на главу компании упадёт кирпич, то это не должно быть катастрофой.
Поэтому давайте его разделим между разными людьми _(клик)_
3. _(клик)_ С другой стороны, конкуренты тоже не против получить ключ и что-то сделать.
Что, если один из сотрудников отдаст ключ?
Ну или хакеры там украдут?
4. _(клик)_ Поэтому хочется сделать так, чтобы мы могли восстановить секрет, а враги — нет.
---
# Схема разделения секрета
<div class="columns">
<div class="col">
Схема разделения секрета (СРС) разделяет секрет $s$ между $n$ участниками.
1. Дилер берёт секрет и делит его на доли
2. Они раздаются участникам — $n$ штук
3. Затем какое-то подмножество участников собираются вместе
4. И они вместе восстанавливают секрет обратно
Две важные фазы: разделение и восстановление
</div>
<div class="col">
<div style="height: 750px">
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style="stroke-width:0.264583" x="157.7173" y="101.40268">n</tspan></text> </g> <g id="g28943" transform="translate(2.5958687)"> <circle style="fill:var(--r-main-color);fill-rule:evenodd;stroke:var(--r-main-color);stroke-width:1.165;stop-color:var(--r-main-color)" id="path28597" cx="91.973381" cy="98.845581" r="1" /> <circle style="fill:var(--r-main-color);fill-rule:evenodd;stroke:var(--r-main-color);stroke-width:1.165;stop-color:var(--r-main-color)" id="path28597-5" cx="99.595566" cy="98.845581" r="1" /> <circle style="fill:var(--r-main-color);fill-rule:evenodd;stroke:var(--r-main-color);stroke-width:1.165;stop-color:var(--r-main-color)" id="path28597-6" cx="107.21775" cy="98.845581" r="1" /> </g> </g> <g id="g28962-0" transform="translate(-1.0183312,39.136322)" /> <g id="g32813" transform="translate(1.0183369)"> <g id="g28957-2" transform="translate(-1.0183312,39.136322)"> <circle style="fill:none;fill-rule:evenodd;stroke:var(--r-main-color);stroke-width:1.165;stop-color:var(--r-main-color)" id="path1016-15-9" cx="73.413055" cy="99.624496" r="10" /> <text xml:space="preserve" style="fill:var(--r-main-color);font-size:8px;line-height:1.25;font-family:sans-serif;-inkscape-font-specification:'sans-serif, Normal';stroke-width:0.264583" x="71.132576" y="102.03675" id="text20067-3"><tspan sodipodi:role="line" id="tspan20065-9" style="stroke-width:0.264583" x="71.132576" y="102.03675">i₁</tspan></text> </g> <circle style="fill:none;fill-rule:evenodd;stroke:var(--r-main-color);stroke-width:1.165;stop-color:var(--r-main-color)" id="path1016-7-8" cx="130.9698" cy="99.624496" r="10" transform="translate(-1.0183312,39.136323)" /> <text xml:space="preserve" style="fill:var(--r-main-color);font-size:8px;line-height:1.25;font-family:sans-serif;-inkscape-font-specification:'sans-serif, Normal';stroke-width:0.264583" x="128.63661" y="102.03675" id="text22139-8" transform="translate(-1.0183312,39.136323)"><tspan sodipodi:role="line" id="tspan22137-5" style="stroke-width:0.264583" x="128.63661" y="102.03675">iₖ</tspan></text> <g id="g28943-8" transform="translate(1.5775375,39.136322)"> <circle style="fill:var(--r-main-color);fill-rule:evenodd;stroke:var(--r-main-color);stroke-width:1.165;stop-color:var(--r-main-color)" id="path28597-56" cx="91.973381" cy="98.845581" r="1" /> <circle style="fill:var(--r-main-color);fill-rule:evenodd;stroke:var(--r-main-color);stroke-width:1.165;stop-color:var(--r-main-color)" id="path28597-5-1" cx="99.595566" cy="98.845581" r="1" /> <circle style="fill:var(--r-main-color);fill-rule:evenodd;stroke:var(--r-main-color);stroke-width:1.165;stop-color:var(--r-main-color)" id="path28597-6-1" cx="107.21775" cy="98.845581" r="1" /> </g> </g> <text xml:space="preserve" style="fill:var(--r-main-color);font-size:8px;line-height:1.25;font-family:sans-serif;-inkscape-font-specification:'sans-serif, Normal';stroke-width:0.264583" x="29.332674" y="53.073647" id="text34489"><tspan sodipodi:role="line" id="tspan34487" style="stroke-width:0.264583" x="29.332674" y="53.073647">Разделение:</tspan></text> <text xml:space="preserve" style="fill:var(--r-main-color);font-size:8px;line-height:1.25;font-family:sans-serif;-inkscape-font-specification:'sans-serif, Normal';stroke-width:0.264583" x="14.133604" y="175.23087" id="text36341"><tspan sodipodi:role="line" id="tspan36339" style="stroke-width:0.264583" x="14.133604" y="175.23087">Восстановление:</tspan></text> <rect style="fill:none;fill-rule:evenodd;stroke:var(--r-main-color);stroke-width:1.165;stop-color:var(--r-main-color)" id="rect36907" width="46.511631" height="19.834822" x="78.935623" y="163.67149" /> <g id="path37222"> <path style="color:var(--r-main-color);fill:var(--r-main-color);fill-rule:evenodd;-inkscape-stroke:none" d="M 77.570312,60.099609 48.535156,90.310547 48.943359,90.701172 77.978516,60.490234 Z" id="path40507" /> <g 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Notes:
Итак, схемы разделения секрета описывают, как нужно разделять и восстанавливать секрет.
В общем, секрет всегда передаётся какому-то дилеру, который этот секрет затем разделяет.
Дилером может выступать либо человек, который придумал или уже знает секрет, либо какой-то компьютер, в котором мы уверены. Дилеру мы всегда полностью доверяем.
Так вот, он секрет разделяет на несколько долей, которые затем раздаются участникам. На этом фаза разделения завершается.
Потом происходит восстановление секрета. Для этого некоторое подмножество участников собираются вместе — может и вообще все — которые вместе восстанавливают секрет из своих долей. Конечно же, восстановленный должен совпадать с исходным.
---
# Кратко об определениях
1. Разрешенное множество — те участники, которые могут восстановить секрет.
2. Структура доступа — совокупность всех разрешенных множеств. Должна быть монотонной.
3. Секрет разделяется на **доли**.
4. Схема называется **совершенной**, если недостаточное число долей (не входящие ни в одно разрешенное множество) не дают **никакой** информации о секрете.
5. Схема называется **идеальной**, если каждая доля содержит не больше информации, чем содержится в секрете.
Notes:
Давайте сразу кратко пробегусь по определениям. Более подробно будет позднее, когда появятся хорошие примеры.
1. Если группа участников могут восстановить секрет, то эта группа называется разрешенным подмножеством.
2. Структура доступа — совокупность всех разрешенных множеств. Монотонное всегда подразумевается, и означает что добавление участника в группу не может лишать их возможности восстановить секрет. Об этом будем говорить уже ближе к концу доклада.
3. Долями называем то, что получают участники
4. Совершенность. Мы хотим, чтобы, скажем, конкуренты, вообще ничего не знали о секрете по тем долям, которые у них есть. Т.е. если группа участников не образует разрешенное множество, то они совсем ничего не знают о секрете. Нам интересны именно такие схемы.
5. Идеальность. Если каждая доля содержит не больше информации, чем секрет, то схема будет идеальной.
Лучше стараться обращать на эти свойства внимание.
---
# Пороговая $(n, k)$-схема
Это такая схема разделения доступа, что любые $k$ участников из $n$ могут восстановить секрет.
Другими словами, чтобы получить секрет, нужно хотя бы $k$ участников.
Множество разрешенное, если в нём не меньше $k$ участников.
Notes:
Таким схемам посвящена практически половина доклада. Очень простые — если имеется больше $k$ участников, то секрет восстановить можно. Более сложные будут сильно позже.
---
# Пример
Разделим между **тремя** участниками, чтобы только вместе они могли восстановить
<div class="columns">
<div class="col img_container">
<img src="images/projections.gif">
</div>
<div class="col">
- **Секрет** ∈ { Шар, Куб, Цилиндр }
- **Доля** ∈ { Круг, Квадрат }
Любых двух проекций недостаточно, чтобы восстановить секрет
Это $(3,3)$-схема. $n=k=3$
</div></div>
Notes:
Доклад начинается с простых $(n,n)$-схем, где только все участники вместе могут восстановить секрет.
Вот тут очень простой геометрический пример. Мы прячем секрет в коробку и даём трём участникам три проекции секрета на грани коробки. Здесь так получается, что если только два участника попытаются восстановить секрет, то они никогда не смогут сделать это с полной уверенностью. Всегда будет два подходящих варианта. Но когда соберутся трое — они с полной уверенностью смогут восстановить секрет. Разделили ровно так, как и нужно.
***
<div class="r-stack">
# $(n, n)$-схема
</div>
$k = n$. То есть $n$ участников только <b>все вместе</b> могут получить секрет.
- Идея 1: нарезать секрет на доли.
- Недостаток: конкуренты легко взломают перебором
- Никогда такую схему использовать не будем
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</div>
Notes:
Начнём с построения простейшей схемы. Это не $(n, k)$-схема, а попроще.
Первая идея очень простая. Есть какой-то длинный секрет, мы его можем просто разрезать на доли и раздать участникам. Тогда если они объединятся, то смогут с легкостью восстановить его. Казалось бы, всё хорошо и просто.
Однако здесь есть большой недостаток: конкуренты, если обладают $k-1$ долями — на одну меньше нужного — запросто могут перебрать все значения оставшейся неизвестной. Их не так уж и много, особенно с ростом числа долей.
Мы этого **совсем** не хотим, поэтому давайте строго определим, чего именно мы не хотим.
---
# Совершенность
<cite>"On secret sharing systems", E. Karnin, J. Greene and M. Hellman, 1983</cite>
Хотим, чтобы конкуренты ничего не знали о секрете. Даже если знают $k-1$ долю.
По-другому: знание $k-1$ долей ничего не даёт <!-- .element: class="fragment" -->
- $s$ — секрет <!-- .element: class="fragment" -->
- $v_i$ — доля секрета ($i = 0…n$) <!-- .element: class="fragment" -->
- $H(s \mid v_{i_1}, v_{i_2}, …, v_{i_m})$ — сколько энтропии в $s$, если знаем эти доли <!-- .element: class="fragment" -->
- $H(s)$ — сколько энтропии в секрете, если мы совсем ничего про него не знаем <!-- .element: class="fragment" -->
<div class="fragment">
Хотим:
$$
\begin{array}{l l}
H(s \mid v_{i_1}, v_{i_2}, …, v_{i_m}) = H(s), &\text{ при } m < k \\\\
H(s \mid v_{i_1}, v_{i_2}, …, v_{i_m}) = 0, &\text{ при } m ≥ k
\end{array}
$$
</div>
Notes:
Итак, "хорошие" схемы, в которых конкуренты совершенно ничего не знают о секрете, называются совершенными.
Так и запишем: _(клик)_ знание $k-1$ долей ничего не даёт
Теперь давайте попробуем это записать более строго. Для этого надо ввести пару обозначений _(2 клика)_.
Переходим к сложной часть. _(клик)_.
Здесь упоминается энтропия, причём условная. Это мера неопределённости — как много мы ещё не знаем о значении при таких-то условиях.
Грубо говоря, вот мы знаем столько-то долей секрета. Как много надо угадывать, чтобы узнать его весь?
Вспоминая старую плохую схему, если мы знаем много долей, то нам остается угадать только один маленький.
Очевидно, это гораздо быстрее чем угадать весь секрет.
_(клик)_. Следующее — количество энтропии в секрете, когда мы вообще ничего не знаем. Грубо, как сложно угадать весь секрет целиком.
Вообще, запросто могу переписать при помощи условных вероятностей, если кому-то не очень понятно.
_(клик)_
И теперь можем записать что именно мы хотим.
- Если мы знаем меньше $k$ долей, то мы знаем о секрете ровно столько же, как если бы не знали совсем ничего
- А если мы знаем достаточно долей, то знаем и весь секрет полностью. Никаких сомнений в его значении нету.
---
# Настоящая $(n,n)$-схема
Идея:
$ \displaystyle
s = v_1 + v_2 + … + v_n
$
<div class="fragment columns">
<p>Алгоритм: </p>
<div class="col">
0. Пусть $s ∈ 𝔽$
1. Генерируем совершенно случайные $v_2, …, v_n$ тоже из $𝔽$. <!-- .element: class="fragment" -->
2. Находим $v_1 = s - v_2 - … - v_n$ <!-- .element: class="fragment" -->
3. Раздаём эти доли участникам <!-- .element: class="fragment" -->
</div></div>
**Почему она совершенная?** <!-- .element: class="fragment" -->
Допустим: знаем $v_1, …, v_{n-1}$ — все кроме (без потери общности) последнего. <!-- .element: class="fragment" -->
<div class="fragment columns">
<p>Тогда: </p>
<div class="col">
1. Есть $|𝔽|$ вариантов для последней доли.
2. Каждый вариант даёт свой уникальный $s$. <!-- .element: class="fragment" -->
3. Угадать значение доли из $|𝔽|$ также сложно, как угадать секрет (тоже из $𝔽$). <!-- .element: class="fragment" -->
</div></div>
Notes:
Теперь реализуем подходящую схему. Хотим секрет разделить между $n$ участниками.
Идея очень проста: пусть все доли в сумме дают этот самый секрет.
_(клик)_ Во-первых, договоримся что секрет взялся из какого-то поля $𝔽$. Совершенно не важно из какого.
_(клик)_ Для реализации достаточно сгенерировать все числа кроме одного. Они должны быть из того же поля.
_(клик)_ Затем остаётся единственным образом определить одно оставшееся.
_(клик)_ И всё, доли готовы.
_(клик)_ Следующий шаг: покажем что она действительно совершенна.
_(клик)_ Допустим самое худшее — знаем все доли кроме одного.
_(клик)_ Тогда мы можем попытаться найти последний подбором.
_(клик)_ Но их ровно столько, сколько и возможных секретов.
_(клик)_ Получается, что ровно с тем же успехом мы могли бы пытаться угадать секрет.
Видно, что знания долей никак не помогают. А значит схема совершенна.
---
Теперь и секрет, и доли из одного поля, т.е. они содержат одинаково информации (если они выбраны случайно). Это интересно.
<div class="fragment">
# Теорема
<cite>E. Karnin, J. Greene and M. Hellman, "On secret sharing systems", 1983</cite>
Пусть схема совершенная, а $v_i∈V$ — доля секрета $s∈S$. <br/>Тогда $H(v_i) ≥ H(s)$.
</div><div class="fragment">
**Следствие:** Если секрет и доли выбираются случайно, то $|V| ≥ |S|$.
</div><div class="fragment">
**Доказательство следствия**
1. Известно, что знание $k-1$ доли не даёт никакой информации о секрете
2. Также известно, что зная $k$ долей можно единственным образом восстановить секрет.
3. Если $|V| < |S|$, то зная $k-1$ долей, мы можем получить лишь $|V|$ значений секрета, по одному на каждое возможное значение доли.
4. Но ведь секрет — случайно выбран из $|S|$ возможных вариантов! Пришли к противоречию: такая схема не совершенна.
</div>
Notes:
Заметили, что размер доли вырос? В плохой и неправильной схеме он был в $n$ раз меньше секрета, а теперь совпадает с ним?
Так вот, существует подходящая теорема _(клик)_.
Она говорит, что у совершенной схемы каждая доля не может содержать меньше информации, чем целый секрет.
У неё есть довольно интересное следствие _(клик)_.
А именно, мощность множества с долью обычно не может быть меньше, чем множество с секретом.
На практике это означает, что если у вас секрет занимает столько-то бит, то и каждая доля будет не меньше.
Теперь давайте это докажем. Будем доказывать только следствие, потому что доказательство самой теоремы слишком сложно для этого доклада, да и не нужно. Будем считать, что числа у нас всегда случайные, это обычно правда.
Если в двух словах, то когда мы знаем $k-1$ долей, то, чтобы восстановить секрет, достаточно перебрать все значения оставшейся доли. И если этих значений меньше, чем значений секрета, то это противоречит тому, что знание $k-1$ долей ничего не даёт.
---
## Разделение длинных секретов
Есть строка: `I like secret sharing systems`. Она длинная (29 символов).
Можем её рассмотреть, как набор байтов
```
49 20 6c 69 6b 65 20 73 65 63 72 65 74 20 73 68 61 72 69 6e 67 20 73 79 73 74 65 6d 73
```
И каждый байт разделять отдельно, используя группу $ℤ_{256}$:
- $49_{16} = 73 = 10 + 63$
- $20_{16} = 32 = 50 + 238$
- $6c_{16} = 108 = … $
Такое разделение всё равно останется совершенным.
***
# Схема Блэкли
<cite>G. R. BLAKLEY, "Safeguarding cryptographic keys" 1979</cite>
<div class="columns">
<div class="col">
Это $(n, k)$-схема. Из $n$ участников достаточно только $k$.
Описана Джорджем Блэкли в начале июня 1979 года.
</div>
<div class="col">
<img src="images/george-blakley.jpg">
</div></div>
Notes:
Мы реализовали простейшую $(n, n)$-схему, но она не даёт права потерять никакую долю. Обязательно все должны присутствовать.
На самом деле, с её помощью можно делать удивительные вещи, но это не очень эффективно и будет в самом конце доклада. А пока что поговорим про схему Блэкли.
---
## Схема Блэкли
**Идея:** $k$ гиперплоскостей пересекаются в $k$-мерном пространстве в точке.
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</div>
<div class="fragment">
**Алгоритм**:
0. Секрет $s∈𝔽$. <span class="fragment">Возьмём пространство $𝔽^k$</span>
1. Выберем случайные и независимые $x_2, x_3, …, x_k$ <!-- .element: class="fragment" -->
2. Секретная точка: $(s, x_2, x_3, …, x_k)$ <!-- .element: class="fragment" -->
3. Провести через неё $n$ случайных гиперплоскостей и раздать их участникам. <!-- .element: class="fragment" -->
</div>
Notes:
Идея этой схемы заключается в том, что гиперплоскости пересекаются только в одной точке.
На примере трёхмерного пространства как раз показано, что две плоскости пересекаются по прямой, а три — уже в точке.
_(клик)_ Теперь пройдёмся по алгоритму. Как и раньше, возьмём секрет из поля $𝔽$. _(клик)_ К нему сразу добавляем $k$-мерное пространство, в котором и будет всё происходить.
_(клик)_ Дальше снова генерируем случайные $k-1$ точку. Удивительно, но это продолжает совпадать с простейшей схемой.
_(клик)_ И вот теперь начались отличия. Как мы помним, плоскости пересекаются в точке. Значит какая-то точка пространства должна кодировать секрет. Ну и вот мы её выбрали. Одна её координата содержит секрет, а остальные выбраны совершенно случайно.
_(клик)_ Дальше мы должны провести через эту точку $n$ произвольных плоскостей. Вот и всё.
---
## Пример
- Секрет из поля $ℤ_{23}$ будет $s = 18$
- Действуем в пространстве $ℤ_{23}^3$ — три участника могут восстановить
- Выберем случайную точку $(\mathbf{18}, 7, 20)$ — секрет в первой координате
- Проведём плоскости через эту точку:
* $21x_1 + x_2 + 15x_3 = 18$
* $11x_1 + 9x_2 + x_3 = 5$
* $3x_1 + 6x_2 + 8x_3 = 3$
Именно этой информацией обладают участники — каждый знает ровно одну плоскость.
<div class="fragment">
- Пересечём их:
$$ \begin{pmatrix}
21 & 1 & 15 \\\\ 11 & 9 & 1 \\\\ 3 & 6 & 8
\end{pmatrix} \vec{x} = \begin{pmatrix}
18 \\\\ 5 \\\\ 3
\end{pmatrix} $$
- Решение СЛАУ: $\vec{x} = \begin{pmatrix}\mathbf{18} \\\\ 7 \\\\ 20\end{pmatrix}$; отсюда секрет $s=18$
</div>
Notes:
Берём какой-то секрет — здесь он из поля $ℤ_{23}$ — и хотим его разделить. Восстановить могут любые три участника, поэтому пространство трёхмерное.
Для этого помещаем его в какую-то случайную точки и проводим через неё три случайных плоскости — по одной на каждого участника. Эти плоскости и будем раздавать участникам.
На этом с разделением всё, теперь давайте восстановим обратно _(клик)_.
У нас есть ьтри плоскости, надо найти их общую точку пересечения. Для этого достаточно будет составить вот такую СЛАУ на точку и решить её любым удобным способом. Поскольку мы действуем в поле, то какое-нибудь решение найдется.
И вот решение – как раз та точка, которые мы сгенерировали раньше. Берём первую её координату и получаем секрет обратно.
---
## Схема Блэкли: совершенность
- Дана $k-1$ плоскость, пересекаются по прямой ($\cong 𝔽$)
- Но нас интересует только одна координата <!-- .element: class="fragment" data-fragment-index="0" -->
<div class="r-stack">
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</div>
- На прямой столько же точек, сколько и значений секрета! <!-- .element: class="fragment" data-fragment-index="1" -->
Notes:
Давайте покажем, почему эта схема совершенна.
То есть что даже зная $k-1$ плоскость, угадывать секрет не становится легче.
С одной стороны, вполне известно, что секретная точка лежит на прямой.
Казалось бы, это облегчает задачу. И это было бы так, если бы были важны все координаты точки.
Поскольку сам секрет находится в первой координате точки, то можно рассмотреть только эту координату у прямой.
И тогда будет легко видеть, что прямая содержит столько же точек, сколько и ось координат.
То есть столько же точек, сколько и значений секрета.
---
## Схема Блэкли
**Секрет**: $s ∈ 𝔽$.
**Доля**: $k$-мерная плоскость
<div class="fragment">
$$
A_1x_1 + A_2x_2 + … + A_kx_k + A_0 = 0
$$
</div>
<div class="fragment">
**Плоскость**: набор $(A_1, A_2, …, A_k) ∈ 𝔽^{k+1}$
Но секрет-то только из $𝔽$! Доля в $k+1$ раз больше.
</div>
Notes:
Теперь посмотрим что же у нас получилось.
А получились плоскости. Каждую из них можно записать таким образом: _(клик)_.
Таким образом, каждая плоскость записывается при помощи ровно $k+1$ коэффициента, каждый из нашего поля: _(клик)_.
В этой схеме так получилось, что размер доли в $k+1$ раз больше, чем размер самого секрета.
***
# Идеальность
Секрет $s ∈ S$, доля $v_i ∈ V$.
- Из теоремы: $H(v_i) ≥ H(s)$
- Если $H(v_i) = H(s)$, то схема идеальна
Схема идеальна тогда, когда доля содержит ровно столько же информации, сколько и секрет.
Notes:
Теперь давайте определим новое свойство.
Идеальность — когда размер доли ровно такой же, как и размер секрета.
Если схема совершенна, то меньше он быть и не может, мы это доказали раньше.
Поскольку в схеме Блэкли размер доли гораздо больше, то она не идеальна. Совершенна, но не идеальна.
---
## <u>Неправильная</u> схема
<cite>Почему-то описана на некоторых сайтах</cite>
Хотим сделать схему Блэкли её идеальной.
- Секрет хранится только в одной координате. <!-- .element: class="fragment" -->
* $s ∈ 𝔽$
- А доля гораздо больше <!-- .element: class="fragment" -->
* $v_i ∈ 𝔽^k$
- Идея: распределить секрет по всем координатам: $s ∈ 𝔽^k$ <!-- .element: class="fragment" -->
* Нельзя: на прямой всего лишь $|𝔽|$ точек, хотя вариантов секрета $|𝔽|^k$ <!-- .element: class="fragment" -->
<li class="fragment">Такая схема <span style="color: #ff2c2d;">не будет совершенной!</span></li>
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</div>
Notes:
Теперь давайте рассмотрим вариацию схемы, которая является идеальной.
_(клик)_ Во-первых, секрет хранится только в одной координате.
_(клик)_ Но вот размер доли гораздо больше, и нам это не нравится.
_(клик)_ Что будет, если распределить секрет по всем координатам точки? Ведь тогда как раз размеры сравняются.
Как думаете, так можно сделать?
_(клик)_ Но тогда $k-1$ плоскость пересекается по прямой. И на этой прямой уже гораздо меньше точек.
А значит секрет становится гораздо проще найти, даже если не знать все доли.
_(клик)_ Такая схема будет несовершенной.
Именно такой вариант схемы Блэкли почему-то описан на русской википедии и какой-то криптовики.
Эта схема аналогична самой первой, где мы просто нарезали секрет на доли. Конечно, это нам не подходит.
Ни в коем случае так не делайте. Я это показываю во многом из-за того, что почему-то на некоторых русскоязычных сайтах приводится именно такая реализация схемы. Она неправильная.
***
# Многочлен Лагранжа
<cite><a href="https://ru.wikipedia.org/wiki/Интерполяционный_многочлен_Лагранжа">ru.wikipedia.org/wiki/Интерполяционный_многочлен_Лагранжа</a></cite>
<div class="fragment">
**Хотим**: провести многочлен не более $n-1$ степени через $n$ точек
Например, **прямая** — многочлен первой степени — легко строится по **двум** точкам.
**Дано**: точки $(x_1, y_1), (x_2, y_2), …, (x_n, y_n)$.
</div><div class="fragment">
**Очень простая идея:**
Можно составить СЛАУ на $c_i$, подставив точки в $c_0 + c_1 x + c_2 x^2 + … + c_{n-1}x^{n-1}$.
Здесь $n$ неизвестных, $n$ уравнений, а следовательно решение единственно.
</div><div class="fragment">
Но есть явная формула: $\displaystyle f({\color{red} x}) = \sum_{j=1}^n y_j \prod_{\substack{i=0\\\\i≠j}} \frac{{\color{red} x} - x_i}{x_j - x_i}$ — многочлен Лагранжа
</div>
Notes:
Теперь немного поговорим про многочлен Лагранжа. Он нам потом пригодится. _(клик)_
Вообще, довольно известный факт, что через $n$ точек можно построить многочлен $n-1$ степени.
_(клик)_ Для этого достаточно составить СЛАУ и найти этот самый многочлен.
_(клик)_ Но это не так эффективно — иногда интересно узнать значение только в одной точке — так что давайте всё-таки рассмотрим многочлен Лагранжа. Он тоже очень простой.
---
## Очевидное решение
<cite><a href="https://ru.wikipedia.org/wiki/Интерполяционный_многочлен_Лагранжа">ru.wikipedia.org/wiki/Интерполяционный_многочлен_Лагранжа</a></cite>
Просто скажем следующее:
$$
f(x) = \begin{cases}
y_1, &x = x_1 \\\\
y_2, &x = x_2 \\\\
… \\\\
y_n, &x = x_n \\\\
\text{что-нибудь},& x \notin \\{x_1, x_2, …, x_n\\}
\end{cases}
$$
<div class="fragment">
И если бы у нас была такая функция:
$$
\ell_j(x) = \begin{cases}
1,& x = x_i, i=j \\\\
0,& x = x_i, i≠j \\\\
\text{что-нибудь}& x \notin \\{x_1, x_2, …, x_n\\}
\end{cases}
$$
</div>
<div class="fragment">
Тогда бы мы разбили на сумму:
$$
f(x) = y_1 \ell_1(x) + y_2\ell_2(x) + … + y_n \ell_n(x)
$$
</div>
Notes:
Итак, во-первых, мы хотим как-то построить вот такую функцию.
Важно лишь то, чтобы она проходила через некоторые точки, а всё остальное — как получится.
_(клик)_ Но строить такую большую функцию довольно сложно, поэтому давайте упростим себе жизнь и сделаем функцию $\ell$, она поменьше.
В одной из нужных точек она равна единице, в других нужных — равна нулю, а во всех остальных нас значение не волнует.
_(клик)_ И если мы вдруг сможем построить такую функцию $\ell$, то первую большую можно будет легко разбить на вот такую сумму.
Здесь довольно очевидно, что она будет проходить через все нужные точки.
Для каждой интересующей нас точки только одна из $\ell$ будет равна единице, а все остальные равны нулю.
И эту единственную единицу мы умножаем на требуемое значение функции в нужной нам точке. Вот и всё.
---
## Такая функция есть!
<cite><a href="https://ru.wikipedia.org/wiki/Интерполяционный_многочлен_Лагранжа">ru.wikipedia.org/wiki/Интерполяционный_многочлен_Лагранжа</a></cite>
Напомню:
<span class="r-stack">
<span class="fragment fade-out" data-fragment-index="1">$
\ell_j(x) = \begin{cases}
1,& x = x_j \\\\
0,& x = x_i, i≠j \\\\
\text{что-нибудь},& x \notin \\{x_1, x_2, …, x_n\\}
\end{cases}
$</span>
<span class="fragment fade-in-then-out" data-fragment-index="1">$
\ell_j(x) = \begin{cases}
\color{orange}{1},& \color{orange}{x = x_j} \\\\
0,& x = x_i, i≠j \\\\
\text{что-нибудь},& x \notin \\{x_1, x_2, …, x_n\\}
\end{cases}
$</span>
<span class="fragment fade-in-then-out" data-fragment-index="2">$
\ell_j(x) = \begin{cases}
1,& x = x_j \\\\
\color{orange}{0},& \color{orange}{x = x_i, i≠j} \\\\
\text{что-нибудь},& x \notin \\{x_1, x_2, …, x_n\\}
\end{cases}
$</span>
<span class="fragment fade-in-then-out" data-fragment-index="3">$
\ell_j(x) = \begin{cases}
1,& x = x_i, i=j \\\\
0,& x = x_i, i≠j \\\\
\color{orange}{\text{что-нибудь}},& \color{orange}{x \notin \\{x_1, x_2, …, x_n\\}}
\end{cases}
$</span>
</span>
<div class="fragment" data-fragment-index="0">
Выглядит она как произведение дробей:
$$
\ell_j({\color{red} x}) = \prod_{\substack{i=0\\\\i≠j}} \frac{{\color{red} x} - x_i}{x_j - x_i}
$$
</div>
- Если $x = x_j$, то каждая дробь равна $1$, и вся функция тоже. <!-- .element: class="fragment fade-in-then-semi-out" data-fragment-index="1" -->
- Если $x = x_i, i≠j$, то найдётся $i$, такой что $x_i = x$. А значит один из числителей будет таким: $x_i - x_i = 0$. И всё произведение обнулится. <!-- .element: class="fragment fade-in-then-semi-out " data-fragment-index="2" -->
- Все точки различны и $j≠i$, а значит мы не делим на ноль. <!-- .element: class="fragment fade-in-then-semi-out " data-fragment-index="3" -->
Notes:
К счастью, такая маленькая функция $\ell$ существует, и сейчас покажу как она работает.
_(клик)_ Выглядит она как вот такое сложное произведение дробей. Здесь $i$ из произведения пробегает по всем, кроме $j$.
_(клик)_ Смотрим на первое требование. Функция при $x=x_j$ должна быть равна единице. Здесь это вполне выполняется.
Каждая дробь будет иметь один и тот же числитель и знаменатель, поэтому все они равны единице.
_(клик)_ Двигаемся дальше. Надо чтобы во всех остальных точках функция была равна нулю.
Здесь это обеспечивается за счёт того, что мы перебираем все эти остальные точки и вычитаем $x_i$ из аргумента.
Тогда в этих точках хотя бы один числитель обнулится, а вместе с ним обнулится и всё произведение.
_(клик)_ Ну и наконец, функция везде определена. Поскольку предполагается, что все иксы различны, то мы на ноль никогда не делим.
---
## Итог
<cite><a href="https://ru.wikipedia.org/wiki/Интерполяционный_многочлен_Лагранжа">ru.wikipedia.org/wiki/Интерполяционный_многочлен_Лагранжа</a></cite>
Интерполяционный многочлен Лагранжа:
$$
f({\color{red} x}) = \sum_{j=1}^n y_j \prod_{\substack{i=0\\\\i≠j}} \frac{{\color{red} x} - x_i}{x_j - x_i}
$$
- Проходит через точки $(x_1, y_1), …, (x_n, y_n)$
- Степень не больше $n$
Notes:
Ну и вот итог. Интерполяционный многочлен Лагранжа выглядит вот таким вот образом. Как мы показали, он проходит через все нужные точки (если они различны).
Поскольку в каждом произведении не больше $n-1$ линейного множителя, то и весь многочлен тоже не больше $n-1$ степени.
Остается только разобраться с единственностью.
---
## Единственность
<cite><a href="https://ru.wikipedia.org/wiki/Интерполяционный_многочлен_Лагранжа">ru.wikipedia.org/wiki/Интерполяционный_многочлен_Лагранжа</a></cite>
> Через $n$ точек проходит единственный многочлен степени не больше $n-1$.
**Доказательство:**
- Пусть $f(x)$ и $q(x)$ — два многочлена, оба проходят через одинаковые $n$ точек
- Тогда $f(x) - q(x)$ имеет не меньше $n$ нулей — в точках через которые они проходят.
- А ещё $f(x) - q(x)$ тоже степени не больше $n-1$.
- Но $f(x) - q(x)$ не может иметь $n$ нулей! Это же больше его степени.
- А значит $f(x) - q(x) = 0$ ∎
Notes:
Доказательство очень простое.
Действуем от противного. Пусть есть два многочлена, которые оба проходят через одни и те же $n$ точек.
Тогда рассмотрим их разность. Она будет тоже многочленом степени не больше $n-1$, как и исходные два.
Но по основной теореме алгебры такой многочлен не может иметь более чем $n-1$ нуль.
А здесь у нас их $n$, поскольку $f$ и $q$ точно совпадают в $n$ точках.
Получаем что $f - q$ равно нулю, то есть $f = q$, что и требовалось доказать.
***
# Схема Шамира
<cite>Shamir, A. (1979). How to share a secret.</cite>
<div class="columns">
<div class="col">
Опубликована в ноябре 1979 года, всего через полгода после схемы Блэкли.
**Идея:** через $k$ точек можно провести единственный многочлен. Пусть $s = f(0)$.
Доли — точки на многочлене. Секрет — его значение в нуле.
</div><div class="col">
<img src="images/Adi_Shamir.jpg" style="height:700px">
</div>
</div>
Notes:
Итак, переходим к практически полезной схеме. Схема Шамира была опубликована в ноябре 1979 года криптографом Ади Шамиром, через примерно полгода после схемы Блэкли. Та была представлена в начале июня. Шамир тот же самый, который участвовал в создании RSA, и в схеме нулевого разглашения.
Идея схемы Шамира очень проста.
По $k$ долей мы всегда можем построить единственный многочлен степени $k-1$.
И секрет тогда будет равен его значению в нуле.
А вот по $k-1$ долей уже нельзя ничего говорить о том, какой может быть многочлен и где находится секрет.
---
## Реализация
<cite>Shamir, A. (1979). How to share a secret.</cite>
1. Выбрать достаточно большое поле (секрет должен поместится).
2. Сгенерировать случайный многочлен степени $k-1$
$$f(x) = c_1x^{k-1} + c_2x^{k-2} + … + c_{k-2}x^2 + c_{k-1} x + s$$
Свободный член равен $s$, благодаря чему $f(0) = s$
3. Посчитать значения в точках $1,2,3,…,n$ и раздать их. То есть $v_i = f(i)$. Координата $x$ точек — публичная информация.
4. По $k$ любых из этих $n$ точек можно построить $f$ обратно. Тогда $s = f(0)$.
<iframe data-preload class="fragment invert-if-dark" style="width: 100%; height: 720px;" data-src="https://www.desmos.com/calculator/8qgjticahg?embed"></iframe>
Notes:
Подробная схема такова.
1. Сначала надо выбрать какое-то поле. Чаще всего используется поле вычетов по большому простому модулю. Но вообще-то говоря, сгодится любое. Просто чем меньше поле, тем меньше значений для секрета и тем легче взломать перебором. Хотя обычно на размер секрета влиять нельзя. Также в маленьком поле будет мало точек, которые можно раздать. Например, в $ℤ_7$ никак нельзя выбрать 10 разных точек на многочлене.
2. Затем генерируем случайные коэффициенты для многочлена. Здесь они обозначены как $c_i$.
Свободный же член делаем равным $s$, за счёт чего можно очень легко сделать так, чтобы $f(0) = s$.
Вообще-то говоря, не обязательно использовать именно $x=0$ для секрета, просто иначе генерация многочлена становится немного менее тривиальной. Можно, например, сгенерировать $k-1$ точку и интерполировать многочлен через них и точку с секретом. Но мы так делать не будем, у нас всегда секрет в нуле.
3. Дальше надо выбрать любые $n$ точек на многочлене и посчитать их.
Обычно выбирают просто по порядку $x = 1, 2, 3$ и т.д., но вообще-то говоря это совершенно не важно, ведь здесь эти координаты — публичная информация.
4. Восстановление секрета происходит очень просто. Нужно всего лишь выбрать любые $k$ точек и построить многочлен обратно.
Ну а дальше секрет легко находится. Вообще-то говоря, лучше не искать многочлен отдельно, а просто подставить $x=0$ d многочлен Лагранжа и посчитать значение в этой точке.
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