examples/preffix_tree.ipynb

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  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Преффиксное дерево, Ахо-Корасик и их применение в языковым моделях\n",
    "Существует две наиболее универсальные структуры данных для хранения произвольных объектов -- это хэш-таблицы и бинарные деревья поиска. Для хранения строк с целью какого-либо последующего поиска интересны графовые структуры, вот например"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "from interactive_visualization.graph_utils import Node, Arc, Graph\n",
    "from interactive_visualization.animation_utils import animate_list, animate_dict"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [],
   "source": [
    "def simple_tree(words):\n",
    "    arcs = []\n",
    "    root = 0\n",
    "    state = 1\n",
    "    for word in words:\n",
    "        prev = root\n",
    "        for letter in word:\n",
    "            arcs.append(Arc(prev, state, letter))\n",
    "            prev = state\n",
    "            state += 1\n",
    "    return Graph(arcs)        "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "words = [\n",
    "    'мама',\n",
    "    'мыла',\n",
    "    'раму',\n",
    "    'мылом',\n",
    "]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
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     "execution_count": 4,
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     "output_type": "execute_result"
    }
   ],
   "source": [
    "simple_tree(words).Visualize()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Каждому слову в списке соответствует путь в этом графе из $0$ до некоторого листа и наоборот, конкатенируя символы, написанные на переходах мы получаем исходные слова. Само по себе это дерево не имеет ценности, зато имеет его более компактный вариант -- преффиксное дерево. Ключевое соображение заключается в следующем: вершины $1, 5, 13$ можно объединить, так как переходы $0\\rightarrow 1, 0\\rightarrow 5, 0\\rightarrow 13$ помечены одним и тем же символом, после такого объединения набор строк на путях от корня к листьям не поменяется. Если попробовать выжать как можно больше из этого соображения, то мы получим <i><b>преффиксное дерево</b> (trie, prefix tree)</i>."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [],
   "source": [
    "class TrieNode:\n",
    "    def __init__(self, parent, word=None):\n",
    "        # Хранение этого поля очень расточительно, используется исключительно\n",
    "        # для демонстрации ленивого подсчета суффиксных ссылок\n",
    "        self.parent = parent\n",
    "        # Здесь аналогично, это поле чаще всего избыточно\n",
    "        self.word = word\n",
    "        self.children = {}\n",
    "        self.suff_link = None"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [],
   "source": [
    "def init_trie():\n",
    "    trie = [TrieNode(-1)]\n",
    "    return trie"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [],
   "source": [
    "def add_word(trie, word):\n",
    "    _num = 0\n",
    "    for ch in word:\n",
    "        if not ch in trie[_num].children:\n",
    "            _n = len(trie)\n",
    "            trie[_num].children[ch] = _n\n",
    "            trie.append(TrieNode((_num, ch)))\n",
    "        _num = trie[_num].children[ch]\n",
    "    trie[_num].word = word"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [],
   "source": [
    "def make_trie(words):\n",
    "    trie = init_trie()\n",
    "    for word in words:\n",
    "        add_word(trie, word)\n",
    "    return trie"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [],
   "source": [
    "def to_graph(trie):\n",
    "    arcs = []\n",
    "    for i, node in enumerate(trie):\n",
    "        for c, nextstate in node.children.items():\n",
    "            arcs.append(Arc(i, nextstate, c))\n",
    "        if node.suff_link is not None and node.suff_link != 0:\n",
    "            arcs.append(Arc(i, \n",
    "                            node.suff_link, \n",
    "                            attributes={\"constraint\" : \"False\", \"style\" : \"dashed\"}))\n",
    "            \n",
    "    return Graph(arcs)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
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   "source": [
    "trie = make_trie(words)\n",
    "to_graph(trie).Visualize()"
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  },
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   "metadata": {},
   "source": [
    "В терминах конечных автоматов процедура получения этого дерева из простого, изображенного ранее -- это детерминизация. В случае набора слов эта процедура довольно тривиально и её проще поддерживать по мере добавления слова, в процедуре add_word мы сначала пройдем некоторую часть слова, которая уже есть в дереве, а потом создадим ответвление специально под добавляемое слово. В общем случае (регулярные языки) ипользуется специальный алгоритм  построение [супермножества](https://en.wikipedia.org/wiki/Powerset_construction). Так или иначе, преффиксное дерево обладает ключевым свойством: из одного состояния не может быть 2 и более переходов, помеченных одинаковым символом. Это свойство позволяет делать эффективный поиск по дереву"
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   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Для визуализации\n",
    "def init(graph):\n",
    "    graph.nodes[0].SetColor('blue')\n",
    "\n",
    "def step_forward(graph, node, letter):\n",
    "    node.SetColor('green')    \n",
    "    for arc in node.arcs:\n",
    "        if arc.weight == letter or (arc.weight is None and (letter is None or letter == \"\")):\n",
    "            arc.SetColor('green')\n",
    "            graph.nodes[arc.end].SetColor('blue')\n",
    "            return\n",
    "    # Хак, связанный с визуализацией суффиксных ссылок\n",
    "    graph.nodes[0].SetColor('blue')\n",
    "\n",
    "def fail(graph):\n",
    "    for node in graph.nodes.values():\n",
    "        if node.attributes['color'] != 'black':\n",
    "            node.SetColor('red')\n",
    "        for arc in node.arcs:\n",
    "            if arc.attributes['color'] != 'black':\n",
    "                arc.SetColor('red')\n",
    "                \n",
    "def finish_find(node):\n",
    "    node.SetColor('green')\n",
    "                \n",
    "def reset(graph):\n",
    "    for node in graph.nodes.values():\n",
    "        node.SetColor('black')\n",
    "        for arc in node.arcs:\n",
    "            arc.SetColor('black')\n",
    "    \n",
    "######################################\n",
    "\n",
    "def find(trie, word):\n",
    "    graph = to_graph(trie)\n",
    "    init(graph)\n",
    "    slices = [graph.Visualize()]\n",
    "    \n",
    "    state = 0\n",
    "    for letter in word:\n",
    "        if letter in trie[state].children:\n",
    "            step_forward(graph, graph.nodes[state], letter)\n",
    "            slices.append(graph.Visualize())\n",
    "            ###\n",
    "            state = trie[state].children[letter]            \n",
    "        else:\n",
    "            fail(graph)\n",
    "            slices.append(graph.Visualize())\n",
    "            reset(graph)\n",
    "            return (False, slices)\n",
    "        \n",
    "    if trie[state].word != word:\n",
    "        fail(graph)\n",
    "        slices.append(graph.Visualize())\n",
    "        reset(graph)\n",
    "        return(False, slices)\n",
    "    finish_find(graph.nodes[state])\n",
    "    slices.append(graph.Visualize())\n",
    "    reset(graph)\n",
    "    return (True, slices)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "False\n"
     ]
    },
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "877da61d33c54dde96460e75c5740f43",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "HBox(children=(Button(description='Prev', style=ButtonStyle()), Button(description='Next', style=ButtonStyle()…"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "7fe3971fb4604b35ba762c3ac7b7ae32",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "interactive(children=(IntSlider(value=0, description='step', max=4), Output()), _dom_classes=('widget-interact…"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "result, slices = find(trie, 'рама')\n",
    "print(result)\n",
    "animate_list(slices);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Здесь стоит отметить, что при поиске следующей вершины нужно каким-то образом искать ребро по соответствующей букве. Это возвращает нас к хэш таблицам и бинарным деревьям поиска. Стоит отметить, однако, что набор символом заранее известен и ограничен, что довольно сильно облегчает задачу поиска. В итоге поиск строки разбивается на несколько последовательных поисков символов в небольших таблицах, кроме того в случае отстутствия искомого слова в дереве мы просматриваем строку не целиком а только ту часть, которая в дереве содержится. В случае с хеш таблицей мы делаем подсчет хеша -- обычно это линейный проход по строке, -- один запрос поиска по хешу и финальное честное сравнение строк в случае совпадения хешей."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Алгоритм Ахо-Корасик\n",
    "Основная задача, которую решает этот алгоритм: \"дан набор слов и строка. Нужно в строке найти все вхождения слов из словаря\". Важно то, что строка -- это не текст, который разбивается на слова, нужно искать именно как подстроки. Самый простой способ решить эту задачу -- по очереди пытаться найти каждое из слов, пытаясь сопоставить слово всем подстрокам нужной длины"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [],
   "source": [
    "search_words = ['ма', 'мы', 'ам', 'рама']\n",
    "sentence = 'мама мыла раму'"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [],
   "source": [
    "def find_substrings_simple(words, sentence):\n",
    "    result = dict()\n",
    "    comparison_count = 0\n",
    "    n = len(sentence)\n",
    "    for word in words:\n",
    "        # Сюда будем складывать интервалы, соответствующие подстрокам равным word\n",
    "        intervals = []\n",
    "        for i in range(n - len(word) + 1):\n",
    "            match = True\n",
    "            for j, letter in enumerate(word):\n",
    "                comparison_count += 1\n",
    "                if letter != sentence[i + j]:\n",
    "                    match = False\n",
    "                    break\n",
    "            if match:\n",
    "                intervals.append((i, i + len(word)))\n",
    "        result[word] = intervals\n",
    "    return (result, comparison_count)                "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "мама мыла раму\n",
      "Number of character comparisons: 65\n",
      "ма:(0, 2), (2, 4)\n",
      "мы:(5, 7)\n",
      "ам:(1, 3), (11, 13)\n",
      "рама:\n"
     ]
    }
   ],
   "source": [
    "result, comparison_count = find_substrings_simple(search_words, sentence)\n",
    "print(sentence)\n",
    "print('Number of character comparisons:', comparison_count)\n",
    "for word, intervals in result.items():\n",
    "    print(word, end=':')\n",
    "    print(', '.join(map(str, intervals)))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Представления множества в виде преффиксного дерева уже сильно помогает ускорить этот базовый подход. Если немного перегруппировать проверки в базовом подходе (поменять местами два внешних цикла), то получается, что мы по очереди для каждой позиции пытаемся проверить, не соответствует ли подстроки, начинающиеся в этой позиции искомым словам? Эту группу проверок можно сделать гораздо быстее с помощью преффиксного дерева, достаточно использовать процедуру на подобие той, что мы делали для поиска"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [],
   "source": [
    "def match_at_position(trie, sentence, position):\n",
    "    slices = []\n",
    "    graph = to_graph(trie)\n",
    "    init(graph)\n",
    "    slices.append(graph.Visualize())\n",
    "    state = 0\n",
    "    matched_words = []\n",
    "    comparisons = 0\n",
    "    for i in range(position, len(sentence)):\n",
    "        comparisons += 1\n",
    "        if sentence[i] in trie[state].children:\n",
    "            step_forward(graph, graph.nodes[state], sentence[i])\n",
    "            slices.append(graph.Visualize())\n",
    "            \n",
    "            state = trie[state].children[sentence[i]]\n",
    "        else:\n",
    "            fail(graph)\n",
    "            slices.append(graph.Visualize())\n",
    "            break\n",
    "        \n",
    "        if trie[state].word is not None:\n",
    "            matched_words.append(trie[state].word)\n",
    "            \n",
    "    return (matched_words, slices, comparisons)\n",
    "\n",
    "def match_suffix(trie, sentence):\n",
    "    return match_at_position(trie, sentence, position=0)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {},
   "outputs": [],
   "source": [
    "def find_substrings_trie(words, sentence):\n",
    "    search_trie = make_trie(words)\n",
    "    suffixes = [sentence[i:] for i in range(len(sentence))]\n",
    "    \n",
    "    visualization_result = {}\n",
    "    result = {}\n",
    "    \n",
    "    comparison_total = 0\n",
    "    for i, suffix in enumerate(suffixes):\n",
    "        matched_words, slices, comparisons_pos = match_suffix(search_trie, suffix)\n",
    "        comparison_total += comparisons_pos\n",
    "        for word in matched_words:\n",
    "            if word not in result:\n",
    "                result[word] = []\n",
    "            result[word].append((i, i + len(word)))\n",
    "        visualization_result[suffix] = slices\n",
    "        \n",
    "    return (result, visualization_result, comparison_total)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Total number of letters in searched words: 10\n",
      "Number of character dictionary queries: 30\n",
      "ма:(0, 2), (2, 4)\n",
      "ам:(1, 3), (11, 13)\n",
      "мы:(5, 7)\n"
     ]
    }
   ],
   "source": [
    "result, visualization_result, comparison_total = find_substrings_trie(search_words, sentence)\n",
    "    \n",
    "# Сложность построения преффикс-дерева линейно по суммарной длине всех входящих в него слов\n",
    "print('Total number of letters in searched words:', sum(map(len, search_words)))\n",
    "print('Number of character dictionary queries:', comparison_total)\n",
    "for word, intervals in result.items():\n",
    "    print(word, end=':')\n",
    "    print(', '.join(map(str, intervals)))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "afe86cbc4e73417bb93b36c85ac19ed4",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "interactive(children=(Select(description='Суффикс', options=('мама мыла раму', 'ама мыла раму', 'ма мыла раму'…"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "animate_dict(visualization_result, 'Суффикс');"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Что более важно, в этом подходе возможно использовать сравнения, которые были сделаны на предыдущих суффиксах, чтобы аккуратно к этому прейти стоит сначала разобрать случай, когда мы ищем только одно слово"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Преффикс функция\n",
    "Одной из ключевых концепций, ведущих к эффективным алгоритмам поиска подстрок, является <i><b>преффикс-функция</b></i>. Для строки $S=[S_0, S_1, \\ldots, S_{n-1}]$ преффкс-функция $P$ определяется как\n",
    "$$\n",
    "P[i]=\\max\\{~j~|~j\\leq i,~S[0:j]=S[i-j+1:i+1]~\\}\n",
    "$$\n",
    "Иначе говоря $P[i]$ -- это максимальная длина нетривиального суффикса, который совпадает с преффиксом. Например для строки `abacaba` преффикс функция выглядит следующим образом\n",
    "$$\n",
    "abacaba\n",
    "$$\n",
    "$$\n",
    "0010123\n",
    "$$\n",
    "Если мы каким-то образом смогли вычислить преффикс-функцию, то с её помощью мы можем довольно легко без дополнительных сравнений получить все суффиксы, совпадающие с преффиксами, причем в убывающем порядке. По определенияю $P[i]$\n",
    "$$\n",
    "S[0:P[i]]=S[i-P[i]+1:i+1]\n",
    "$$\n",
    "Учитывая посмотрим на преффикс-функцию в позиции $P[i]-1$:\n",
    "$$\n",
    "S[0:P[P[i]-1]]=S[P[i]-P[P[i]-1]:P[i]]=S[i-P[P[i]-1]+1:i+1]\n",
    "$$\n",
    "Первое равенство выполняется в силу $P[P[i]-1]$, второе -- в силу $P[i]$. Вот пример на строке `abacaba`: $P[6]=3$, т.е. $S[0:3]=S[4:7]$. При этом $P[2]=1$, что указывает на то, что $S[0:1]=S[6:7]$. Ключевой момент здесь заключается в том, что $P[P[i]-1]$ дает нам наибольший префикс, совпадающиий с суффиксом для $P[i]-1$, а значит также указывает на следующий по возрастанию. Таким образом длины всех возможных совпадающих преффиксов-суффиксов в позиции $i$ -- это $P[i]\\rightarrow P[P[i]-1] \\rightarrow P[P[P[i]-1]-1]\\rightarrow \\ldots$.\n",
    "\n",
    "У преффикс-функции есть еще одно очень простое свойство\n",
    "$$\n",
    "P[i]\\leq P[i-1]+1.\n",
    "$$\n",
    "Следует это из очень простого факта\n",
    "$$\n",
    "S[0:P[i]]=S[i-P[i]+1:i+1]\\Rightarrow S[0:P[i]-1]=S[i-P[i]+1:i]\n",
    "$$\n",
    "Этих двух свойств достаточно, чтобы эффективно вычислить преффикс функцию слева направо наподобие динамическому программированию: если мы посчитали $P$ для $0\\ldots i$, то для подсчета $P[i+1]$ нужно взять максимальный совпадающий с суффиксом преффикс для $i$ такой, что следующий за ним символ совпадает с $S[i+1]$. При этом мы умеем перебирать совпадающие преффиксы-суффиксы в убывающем по длине порядке, поэтому первый, который нам попадется с совпадением следующего символа -- это нужный нам преффикс."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {},
   "outputs": [],
   "source": [
    "def preffix_function(s):\n",
    "    p = [0]\n",
    "    k = 0\n",
    "    for letter in s[1:]:\n",
    "        while k > 0 and s[k] != letter:\n",
    "            k = p[k - 1]\n",
    "        if s[k] == letter:\n",
    "            k += 1\n",
    "        p.append(k)\n",
    "    return p"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "abacaba\n",
      "0010123\n"
     ]
    }
   ],
   "source": [
    "s = 'abacaba'\n",
    "print(s)\n",
    "print(''.join(map(str, preffix_function(s))))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Анализ скорости работы\n",
    "Несмотря на вложенный цикл оказывается, что эта процедура линейна по количеству символов в `s`. Дело в том, что `k` может увеличиваться на каждой итерации внешнего цикла не более, чем на единицу в строке (в условном операторе), а уменьшаться только при `k = p[k - 1]`. Из этого следует, что количество итераций внутреннего цикла суммарно не превосходит длину строки."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Поиск подстроки в строке с помощью преффикс-функции\n",
    "Самый простой способ использовать преффикс-функцию для поиска вхождений `t` в качестве подстроки `s` -- это посчитать преффикс-функцию для строки `t#s`, где `#` -- это какой-то символ, который гарантированно не встречается ни в `s`, ни в `t`. Тогда очевидно, что значения преффикс-функции не могут превышать длину `t` и при этом позиции, где значения достигают длины `t` соответствтуют вхождениям `t` в `s`. Тем не менее можно поступить более разумно, не строить дополнительную строку, а построить преффикс-функцию только для `t`, после чего пройтись с её помощью по `s`. Это довольно известный подход, его обычно называют алгоритмом <i><b>Кнута-Морриса-Пратта</b></i>."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {},
   "outputs": [],
   "source": [
    "def find_substring_kmp(s, t):\n",
    "    result = []\n",
    "    m = len(t)\n",
    "    \n",
    "    p = preffix_function(t)\n",
    "    k = 0\n",
    "    \n",
    "    for i, letter in enumerate(s):\n",
    "        while k > 0 and t[k] != letter:\n",
    "            k = p[k - 1]\n",
    "        if s[k] == letter:\n",
    "            k += 1\n",
    "        if k == m:\n",
    "            result.append((i - k + 1, i + 1))\n",
    "            k = p[k - 1]\n",
    "    return result"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "ма\n",
      "мама мыла раму\n",
      "[(0, 2), (2, 4)]\n"
     ]
    }
   ],
   "source": [
    "print(search_words[0])\n",
    "print(sentence)\n",
    "print(find_substring_kmp(sentence, search_words[0]))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Обобщение преффикс-функции на случай нескольких строк\n",
    "Основная идея алгоритма Кнута-Морриса-Пратта заключается в том, чтобы для каждой позиции `i` строки `s` считать максимальную длину суффикса, совпадающего с преффиксом `t`, в этом нам помогает предподсчет преффикс-функции для `t`. Преффикс-функция имеет естественное обобщение на преффиксное дерево: каждой вершине дерева соответствует некоторая строка, давайте для каждой вершины посчитаем другую, которая соответствует наибольшему нетривиальному суффиксу этой строки, это и будет наш аналог преффикс-функции для дерева. Полученные переходы обычно называют суффиксными ссылками, они обладают все теми же двумя свойствами, что и преффикс-функция, по считать её можно так же по аналогии начиная с наиболее близких вершин постепенно переходя к более дальним (в порядке обхода в ширину): допустим мы пытаемся посчитать суффиксную ссылку $v$, к которой ведет переход $u\\rightarrow v$, помеченный символом `c`. Достаточно переходить по уже посчитанным суффиксным ссылкам начиная с $u$, пока не упремся в корень или не найдем вершину, из которой есть переход по `c`, итоговая вершина и будет соответствовать суффиксной ссылке $v$. Альтернативным эффективным способом является рекурсия с запоминанием: в первый раз когда мы делаем запрос на подсчет суффиксной ссылки -- вычисляем её и запоминаем, чтобы в следующий раз выдать уже посчитанный результат, попутно при вычислениях используем эту логику для тех суффиксных ссылок, который нам нужны для подсчета. У этого алгоритма есть недостаток: необходимо помнить родителя вершины."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {},
   "outputs": [],
   "source": [
    "from queue import Queue\n",
    "# Подсчет всех ссылок обходом в ширину\n",
    "def calc_suff_links_all(trie):\n",
    "    q = Queue()\n",
    "    q.put(0)\n",
    "    trie[0].suff_link = 0\n",
    "    while not q.empty():\n",
    "        cur = q.get()\n",
    "        for letter, child in trie[cur].children.items():\n",
    "            if cur == 0:\n",
    "                trie[child].suff_link = 0\n",
    "            else:\n",
    "                _tmp = trie[cur].suff_link\n",
    "                while _tmp != 0:\n",
    "                    if letter in trie[_tmp].children:\n",
    "                        break\n",
    "                    _tmp = trie[_tmp].suff_link\n",
    "                if letter in trie[_tmp].children:\n",
    "                    _tmp = trie[_tmp].children[letter]\n",
    "                trie[child].suff_link = _tmp\n",
    "            q.put(child)\n",
    "        \n",
    "\n",
    "# Рекурсивная реализация подсчета одной суффиксной ссылки, использует поле parent\n",
    "def calc_suff_link(trie, i):\n",
    "    if i == 0:\n",
    "        return 0\n",
    "    if not trie[i].suff_link is None:\n",
    "        return trie[i].suff_link\n",
    "    _p = trie[i].parent[0]\n",
    "    _ch = trie[i].parent[1]\n",
    "    if _p == 0:\n",
    "        trie[i].suff_link = 0\n",
    "        return 0\n",
    "    _tmp = calc_suff_link(trie, _p)\n",
    "    while _tmp != 0:\n",
    "        if _ch in trie[_tmp].children:\n",
    "            break\n",
    "        _tmp = calc_suff_link(trie, _tmp)\n",
    "    if _ch in trie[_tmp].children:\n",
    "        _tmp = trie[_tmp].children[_ch]\n",
    "    trie[i].suff_link = _tmp\n",
    "    return _tmp"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "metadata": {},
   "outputs": [],
   "source": [
    "trie = make_trie(words)\n",
    "calc_suff_links_all(trie)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 26,
   "metadata": {},
   "outputs": [
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       "<text text-anchor=\"middle\" x=\"348.49\" y=\"-46.84\" font-family=\"Times New Roman,serif\" font-size=\"14.00\">11</text>\r\n",
       "</g>\r\n",
       "<!-- 10&#45;&gt;11 -->\r\n",
       "<g id=\"edge15\" class=\"edge\">\r\n",
       "<title>10:&#45;&gt;11:</title>\r\n",
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       "<text text-anchor=\"middle\" x=\"307\" y=\"-54.34\" font-family=\"Times New Roman,serif\" font-size=\"14.00\">у</text>\r\n",
       "</g>\r\n",
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       "<title>13:&#45;&gt;1:</title>\r\n",
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       "</g>\r\n",
       "</g>\r\n",
       "</svg>\r\n"
      ],
      "text/plain": [
       "<graphviz.dot.Digraph at 0x27b7da211c0>"
      ]
     },
     "execution_count": 26,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "to_graph(trie).Visualize()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Суффиксные ссылки обозначены пунктирными линиями, для того, чтобы рисунок был чище, суффиксные ссылки, ведущие в корень, не отображены. Наконец осталось проделать аналогичную процедуру на строке, в которой мы что-то ищем"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 27,
   "metadata": {},
   "outputs": [],
   "source": [
    "def find_substrings_aho_corasick(words, sentence):\n",
    "    search_trie = make_trie(words)\n",
    "    calc_suff_links_all(search_trie)\n",
    "    graph = to_graph(search_trie)\n",
    "    init(graph)\n",
    "        \n",
    "    result = dict()\n",
    "    slices = [graph.Visualize()]\n",
    "    comparison_count = 0\n",
    "    \n",
    "    state = 0\n",
    "    for i, letter in enumerate(sentence):\n",
    "        reset(graph)\n",
    "        graph.nodes[state].SetColor(\"blue\")\n",
    "        # Здесь код структурно немного отличается от реализации Кнута-Морриса-Пратта выше, чтобы\n",
    "        # избежать лишнего сравнения, так как здесь сравнения мы используем в качестве меры\n",
    "        # качества алгоритма\n",
    "        while state != 0:\n",
    "            comparison_count += 1\n",
    "            if letter not in search_trie[state].children:\n",
    "                step_forward(graph, graph.nodes[state], \"\")\n",
    "            \n",
    "                ###\n",
    "                state = search_trie[state].suff_link\n",
    "            else:\n",
    "                need_check = False\n",
    "                break\n",
    "        \n",
    "        if state == 0:\n",
    "            comparison_count += 1\n",
    "            if letter in search_trie[state].children:\n",
    "                step_forward(graph, graph.nodes[state], letter)\n",
    "                state = search_trie[state].children[letter]\n",
    "        else:\n",
    "            step_forward(graph, graph.nodes[state], letter) \n",
    "            state = search_trie[state].children[letter]\n",
    "\n",
    "\n",
    "        slices.append(graph.Visualize())\n",
    "        \n",
    "        # Здесь важно пройтись по всем суффиксным ссылкам, потому что в одной позиции\n",
    "        # могут заканчиваться несколько искомых подстрок. Все они как раз генерируются \n",
    "        # проходом по суффиксным ссылкам\n",
    "        tmp = state\n",
    "        while tmp != 0:\n",
    "            if search_trie[tmp].word is not None:\n",
    "                if search_trie[tmp].word not in result:\n",
    "                    result[search_trie[tmp].word] = []\n",
    "                result[search_trie[tmp].word].append((i - len(search_trie[tmp].word) + 1, i + 1))\n",
    "            tmp = search_trie[tmp].suff_link\n",
    "    return (result, slices, comparison_count)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Total number of letters in searched words: 10\n",
      "Number of character dictionary queries: 23\n",
      "ма:(0, 2), (2, 4)\n",
      "ам:(1, 3), (11, 13)\n",
      "мы:(5, 7)\n",
      "мама мыла раму\n"
     ]
    },
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       "HBox(children=(Button(description='Prev', style=ButtonStyle()), Button(description='Next', style=ButtonStyle()…"
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       "interactive(children=(IntSlider(value=0, description='step', max=14), Output()), _dom_classes=('widget-interac…"
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     },
     "metadata": {},
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    }
   ],
   "source": [
    "result, slices, comparison_count = find_substrings_aho_corasick(search_words, sentence)\n",
    "\n",
    "print('Total number of letters in searched words:', sum(map(len, search_words)))\n",
    "print('Number of character dictionary queries:', comparison_count)\n",
    "for word, intervals in result.items():\n",
    "    print(word, end=':')\n",
    "    print(', '.join(map(str, intervals)))\n",
    "print(sentence)\n",
    "animate_list(slices);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Анализ\n",
    "Для первого внутреннего цикла работает тот же аргумент, что и для поиска по преффикс функции, поэтому суммарное число итераций в нем не превосходит длины строки. Для второго ситуация сложнее. На самом деле для \"красивой\" оценки второй внутренний цикл стоит немного изменить при помощи предподсчета. Дело в том, что этот цикл нужно ровно для случаев, когда несколько искомых слов заканчиваются в одной позиции, например если бы мы искали `ма` и `мама` в `мама мыла раму`. Все эти слова можно легко найти, проитерировався по суффиксным ссылкам, но при этом некоторые суффиксы будут соответствовать искомым словам, а некоторые нет. Предподсчитать последовательность переходов по суффиксным ссылкам, ведущим к следующему слову, довольно легко и в этом случае сложность алгоритма будет линейна по суммарной длине всех слов и размеру результата."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Примеры\n",
    "## Исправление ошибок при наборе текста на клавиатуре\n",
    "Предположим, что мы печаем какой-то текст, иногда совершаем ошибки и хотим их по возможности исправлять автоматически. Зная правильно написанные слова мы можем сделать простую модель: каждый раз набирая очередной символ мы с некоторой вероятностью делаем ошибку, с некоторой вероятностью набираем правильный символ. При этом, если мы знаем, какие правильное написание слов, то можем испкать наиболее вероятную последовательность слов с помощью динамического программирования. Далее я приведу пример, как это можно сделать, используя эвристику \"лучевого поиска\" на большом словаре."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 29,
   "metadata": {},
   "outputs": [],
   "source": [
    "import requests, zipfile, io\n",
    "import os\n",
    "# Раскоментируйте следующий код, чтобы скачать и распаковать словарь\n",
    "# url='http://www.ruscorpora.ru/new/ngrams/1grams-3.zip'\n",
    "# r = requests.get(url)\n",
    "# z = zipfile.ZipFile(io.BytesIO(r.content))\n",
    "# z.extractall('data/dict_ru')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 30,
   "metadata": {},
   "outputs": [],
   "source": [
    "min_freq = 30\n",
    "trie_dict_ru = init_trie()\n",
    "with open('data/dict_ru/1grams-3.txt', encoding='utf-8') as f:\n",
    "    for line in f:\n",
    "        freq, word = line.split()\n",
    "        if int(freq) < min_freq:\n",
    "            break\n",
    "        add_word(trie_dict_ru, word)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 31,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "-3.4011973816621555\n"
     ]
    }
   ],
   "source": [
    "prob_mistake = 0.1\n",
    "space_insert_likelihood = prob_mistake / 30\n",
    "probs = {'S' : prob_mistake / 3, 'D' : prob_mistake / 3, 'I': prob_mistake / 3, 'C' : 1.0 - prob_mistake}\n",
    "from numpy import log\n",
    "print(log(probs['S']))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 32,
   "metadata": {},
   "outputs": [],
   "source": [
    "from weakref import WeakValueDictionary\n",
    "\n",
    "# Класс для хранения словных гипотез\n",
    "class HistoryItem:\n",
    "    def __init__(self, word, prev=None, lm_state=None):\n",
    "        self.word = word\n",
    "        self.prev = prev\n",
    "        self.state = lm_state\n",
    "        # Чтобы не мучаться с удалением гипотез, которые умерли\n",
    "        self.next = WeakValueDictionary()\n",
    "        self.hanged = []\n",
    "        \n",
    "    def prolong(self, word, next_lm_state = None):\n",
    "        if word not in self.next:\n",
    "            new_item = HistoryItem(word, self, next_lm_state)\n",
    "            self.next[word] = new_item\n",
    "            return new_item\n",
    "        else:\n",
    "            return self.next[word]\n",
    "        \n",
    "    def hang(self, other):\n",
    "        if self == other or self.prev is None:\n",
    "            return\n",
    "        other.hanged.append(self.prev)\n",
    "        self.prev.next[self.word] = other\n",
    "        \n",
    "    def backtrace(self):\n",
    "        result = [self.word]\n",
    "        _tmp = self.prev\n",
    "        while _tmp is not None:\n",
    "            if _tmp.word is not None:\n",
    "                result.append(_tmp.word)\n",
    "            _tmp = _tmp.prev\n",
    "        return reversed(result)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 33,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Для визуализации процесса поиска гипотез\n",
    "def hypothesis_tree(tokens):\n",
    "    arcs = []\n",
    "    tmp = tokens[0][2]\n",
    "    while tmp is not None and tmp.prev is not None:\n",
    "        tmp = tmp.prev\n",
    "    \n",
    "    q = Queue()\n",
    "    q.put((tmp, 0))\n",
    "    state_num = dict()\n",
    "    states = 1\n",
    "    while not q.empty():\n",
    "        history_item, num = q.get()\n",
    "        for word, next_item in history_item.next.items():\n",
    "            if next_item in state_num:\n",
    "                state = state_num[next_item]\n",
    "            else:\n",
    "                state = states\n",
    "                state_num[next_item] = state\n",
    "                states += 1\n",
    "                q.put((next_item, state))\n",
    "            arcs.append(Arc(num, state, word))\n",
    "            #print(f'Added arc: {num}, {state}, \"\"{word}\"\"')\n",
    "                   \n",
    "    return Graph(arcs).Visualize()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Далее идет основная процедура, в случае обычного динамического программирования у нас отсутствовали условия вида `neg_log_prob < best_neg_log_prob + beam`, иначе говоря в некоторые моменты времени мы отсекаем неправдоподобные гипотезы."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 34,
   "metadata": {},
   "outputs": [],
   "source": [
    "def correct_text_with_beam_search(trie, text, lm=None, lm_scale=1.0, beam=10.0, lattice=False):\n",
    "    hyp_slices = []\n",
    "    tokens = Queue()\n",
    "    root = HistoryItem(None, None, None if lm is None else lm.start)\n",
    "    tokens.put((0, 0.0, root))\n",
    "    prev_best_neg_log_prob = 0.0\n",
    "    for letter in text:\n",
    "        new_tokens = dict()\n",
    "        best_neg_log_prob = float('inf')\n",
    "        \n",
    "        new_tokens[0] = (float('inf'), None)\n",
    "        \n",
    "        # Для обработки \"вставки\" лишнего символа нужен какой-то механизм изменения\n",
    "        # tokens, альтернатно можно использовать dict и обрабатывать состояния в возрастающем\n",
    "        # (топологическом) порядке\n",
    "        while not tokens.empty():\n",
    "            item = tokens.get()\n",
    "            state, neg_log_prob, history_item = item\n",
    "            \n",
    "            if state not in new_tokens:\n",
    "                new_tokens[state] = (float('inf'), None)\n",
    "            if new_tokens[state][0] > neg_log_prob - log(probs['D']):\n",
    "                new_tokens[state] = (neg_log_prob - log(probs['D']), history_item)\n",
    "                best_neg_log_prob = min(best_neg_log_prob, neg_log_prob - log(probs['D']))\n",
    "            if letter == ' ' and trie[state].word is not None:\n",
    "                new_cost = neg_log_prob\n",
    "                if lm is not None:\n",
    "                    next_lm_state, lm_costs = lm.consume_word(history_item.state, trie[state].word)\n",
    "                    lm_cost = sum(lm_costs)\n",
    "                    new_cost += lm_cost * lm_scale\n",
    "                if new_tokens[0][0] > new_cost:\n",
    "                    if lm is not None:\n",
    "                        new_tokens[0] = (new_cost, history_item.prolong(trie[state].word, next_lm_state))\n",
    "                    else:\n",
    "                        new_tokens[0] = (new_cost, history_item.prolong(trie[state].word))\n",
    "                elif lattice and new_tokens[0][0] + beam / 2.5 > new_cost:\n",
    "                    new_item = history_item.prolong(trie[state].word, next_lm_state if lm else None)\n",
    "                    new_item.hang(new_tokens[0][1])\n",
    "            for char, nextstate in trie[state].children.items():\n",
    "                if nextstate not in new_tokens:\n",
    "                    new_tokens[nextstate] = (float('inf'), None)\n",
    "                if letter != char and new_tokens[nextstate][0] > neg_log_prob - log(probs['S']) \\\n",
    "                        and neg_log_prob - log(probs['S']) < best_neg_log_prob + beam:\n",
    "                    new_tokens[nextstate] = (neg_log_prob - log(probs['S']), history_item)\n",
    "                    best_neg_log_prob = min(best_neg_log_prob, neg_log_prob - log(probs['S']))\n",
    "                if letter == char and new_tokens[nextstate][0] > neg_log_prob - log(probs['C']) \\\n",
    "                        and neg_log_prob - log(probs['C']) < best_neg_log_prob + beam:\n",
    "                    new_tokens[nextstate] = (neg_log_prob - log(probs['C']), history_item)\n",
    "                    best_neg_log_prob = min(best_neg_log_prob, neg_log_prob - log(probs['C']))\n",
    "                if neg_log_prob - log(probs['I']) < prev_best_neg_log_prob + beam:\n",
    "                    tokens.put((nextstate, neg_log_prob - log(probs['I']), history_item))\n",
    "\n",
    "        \n",
    "        # Отдельно обрабатываем возможную вставку пробела, потому что при пробеле должно заканчиваться слово\n",
    "        for state, item in new_tokens.items():\n",
    "            neg_log_prob, history_item = item\n",
    "            if history_item is None:\n",
    "                continue\n",
    "            new_cost = neg_log_prob - log(space_insert_likelihood)\n",
    "            if lm is not None:\n",
    "                next_lm_state, lm_costs = lm.consume_word(history_item.state, trie[state].word)\n",
    "                lm_cost = sum(lm_costs)\n",
    "                new_cost += lm_cost * lm_scale\n",
    "                \n",
    "            if trie[state].word is not None and new_tokens[0][0] > new_cost:\n",
    "                if lm is not None:\n",
    "                    new_tokens[0] = (new_cost, \n",
    "                                     history_item.prolong(trie[state].word, next_lm_state))\n",
    "                else:\n",
    "                    new_tokens[0] = (new_cost, \n",
    "                                     history_item.prolong(trie[state].word))\n",
    "                \n",
    "        tokens = Queue()\n",
    "        tokens_list = []\n",
    "        best_token = None\n",
    "        for state, item in new_tokens.items():\n",
    "            if item[0] <= best_neg_log_prob + beam:\n",
    "                tokens.put((state, *item))\n",
    "                tokens_list.append((state, *item))                \n",
    "            if best_token is None or item[0] < best_token[1]:\n",
    "                best_token = (state, *item)\n",
    "        prev_best_neg_log_prob = best_neg_log_prob\n",
    "        \n",
    "        hyp_slices.append(hypothesis_tree(tokens_list))\n",
    "    \n",
    "    result = []\n",
    "    while not tokens.empty():\n",
    "        result.append(tokens.get())\n",
    "               \n",
    "    return result, hyp_slices"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 35,
   "metadata": {},
   "outputs": [],
   "source": [
    "def recover_best_hyp(tokens):\n",
    "    best_token = min([token for token in tokens if token[0] == 0], key=lambda x: x[1])\n",
    "    return best_token[2].backtrace()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 36,
   "metadata": {},
   "outputs": [],
   "source": [
    "tokens, hyp_slices = correct_text_with_beam_search(trie_dict_ru, 'я напсалэто предложеtние с незначительнми ошибкми', beam=7.0)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 37,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "['я', 'напал', 'это', 'предложение', 'с', 'незначительными', 'ошибки']\n"
     ]
    }
   ],
   "source": [
    "print(list(recover_best_hyp(tokens)))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Как видно слова теперь написаны правильно и довольно близко к исходному тексту, но все еще не то, что подразумевалось исходным текстом, в таких случаях помогает языковые модели."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 38,
   "metadata": {},
   "outputs": [
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       "interactive(children=(IntSlider(value=0, description='step', max=48), Output()), _dom_classes=('widget-interac…"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "animate_list(hyp_slices, play=True);"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 39,
   "metadata": {},
   "outputs": [],
   "source": [
    "# hyp_slices[11].render(filename='hypothesis_tree', format='pdf')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# $N$-грамные языковые модели\n",
    "В общем смысле языковая модель (ЯМ) -- это вероятнастная модель, оценивающая правдоподобие тех или иных последовательностей символов над некоторым алфавитом, обычно вероятность последовательности декомпозируется в произведение вероятностей отдельных символов следующим образом\n",
    "$$\n",
    "P(w_1, \\ldots, w_n)=\\prod_{i=1}^nP(w_i~|~w_1, \\ldots, w_{i-1}),\n",
    "$$\n",
    "иначе говоря вероятность последовательностей символов -- это произведение вероятностьей всех символов при условии их левых контекстов. $N$-грамные ЯМ относятся к марковским моделям и ограничивают возможный контекст до $N-1$, гдe $N$ -- порядок ЯМ\n",
    "$$\n",
    "P(w_1, \\ldots, w_n)=\\prod_{i=1}^nP(w_i~|~w_{i-N+1}, \\ldots, w_{i-1}).\n",
    "$$\n",
    "Базовые процесс обучения $N$-грамной ЯМ заключается в том, чтобы посчитать на некотором текстовом корпусе количества встренных последовательностей слов длины вплоть до $N$ и обучить ЯМ методом максимального правдоподобия: если $C(w_1, \\ldots, w_k)$ -- количество раз, сколько последовательность слов $w_1, \\ldots, w_k$ встретилась в тексте, то формулы максимального правдоподобия выглядят следующим образом\n",
    "$$\n",
    "P(w_i~|~w_{i-k+1}, \\ldots, w_{i-1})=\\frac{C(w_{i-k+1}, \\ldots, w_{i-1}, w_{i})}{\\sum_w C(w_{i-k+1}, \\ldots, w_{i-1}, w)}.\n",
    "$$\n",
    "Проблема этого базового подхода заключается в довольно серьезном переобучении: вероятность любой $N$-грамы, которой не было в обучающем тексте равняется нулю. Чтобы как-то решить эту проблему обычно используют следующую модификации для формулы вероятности\n",
    "$$\n",
    "P(w_i~|~w_{i-k+1}, \\ldots, w_{i-1})=\n",
    "\\begin{cases}\n",
    "\\frac{C(w_{i-k+1}, \\ldots, w_{i-1}, w_{i})}{\\sum_w C(w_{i-k+1}, \\ldots, w_{i-1}, w)}, & C(w_{i-k+1}, \\ldots, w_{i-1}, w_{i}) > 0 \\\\\n",
    "P(w_i~|~w_{i-k+2}, \\ldots, w_{i-1}) * b(w_{i-k+2}, \\ldots, w_{i-1}), & C(w_{i-k+1}, \\ldots, w_{i-1}, w_{i}) = 0\n",
    "\\end{cases}\n",
    "$$\n",
    "т.е. в случае наличия данной последовательности в исходном тексте вероятность последовательности оценивается как обычно, в противном случае мы пытаемся оценить вероятность последовательно усекая контекст. Здесь $b(\\cdot)$ -- так называемый <b>backoff weight</b> или просто <b>backoff</b>, величина $b(w_{j}, \\ldots, w_{i})$ может быть интерпретирована как вероятность того (с некоторой нормализацией), что после $(j-i+1)$-грамы $w_{j}, \\ldots, w_{i}$ будет идти слово $w$ такое, что $(j-i)$-грама $w_{j}, \\ldots, w_{i}, w$ отсуствует в обучающем тексте.\n",
    "\n",
    "Здесь я не буду описывать, как происходит подсчет бэкофф весов, давайте подумаем, как такую структуру можно реализовать. Во-первых, $N$-грамы являются посоедовательностью слов, а значит их можно точно представить в виде префиксного дерева: каждому узлу дерева соответствует некоторая $n$-грама. В такой трактовке, бэкофф -- это в точности суффиксная ссылка."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 40,
   "metadata": {},
   "outputs": [
    {
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     },
     "execution_count": 40,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "ngram_simple = make_trie([\n",
    "    ['мама', 'мыла', 'раму'],\n",
    "    ['раму', 'мыла', 'мама'],\n",
    "    ['мыла', 'маму', 'рама']\n",
    "])\n",
    "calc_suff_links_all(ngram_simple)\n",
    "#to_graph(ngram_simple).Visualize().render(filename='graph_images/ngram', format='pdf')\n",
    "to_graph(ngram_simple).Visualize()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 41,
   "metadata": {},
   "outputs": [],
   "source": [
    "class NGramNode:\n",
    "    def __init__(self, final_weight=float('inf')):\n",
    "        self.arcs = dict()\n",
    "        self.backoff = None\n",
    "        self.final_weight = final_weight\n",
    "    \n",
    "    def set_backoff(self, backoff_state, backoff_weight):\n",
    "        self.backoff = (backoff_state, backoff_weight)\n",
    "        \n",
    "    def add_arc(self, nextstate, word_id, neglogprob):\n",
    "        self.arcs[word_id] = (nextstate, neglogprob)\n",
    "        \n",
    "    def set_final(self, final_weight):\n",
    "        self.final_weight = final_weight\n",
    "\n",
    "class NGramModel:\n",
    "    def __init__(self, fst_path, unk_weight=10.0):\n",
    "        nodes = dict()\n",
    "        self.start = None\n",
    "        self.unk_weight = unk_weight\n",
    "        with open(fst_path) as f:\n",
    "            for line in f:\n",
    "                splitted = line.split()\n",
    "                if len(splitted) <= 2:\n",
    "                    state = int(splitted[0])\n",
    "                    if self.start is None:\n",
    "                        self.start = state\n",
    "                    \n",
    "                    if len(splitted) == 2:\n",
    "                        weight = float(splitted[1])\n",
    "                    else:\n",
    "                        weight = 0.0\n",
    "                        \n",
    "                    if state not in nodes:\n",
    "                        nodes[state] = NGramNode(weight)\n",
    "                    else:\n",
    "                        nodes[state].set_final(weight)\n",
    "                else:\n",
    "                    if len(splitted) == 5:\n",
    "                        weight = float(splitted[4])\n",
    "                    else:\n",
    "                        weight = 0.0\n",
    "                    from_state, to_state = map(int, splitted[:2])\n",
    "                    word_id = int(splitted[3])\n",
    "                    if self.start is None:\n",
    "                        self.start = from_state\n",
    "                    \n",
    "                    if from_state not in nodes:\n",
    "                        nodes[from_state] = NGramNode()\n",
    "                    if to_state not in nodes:\n",
    "                        nodes[to_state] = NGramNode()\n",
    "\n",
    "                    if word_id == 0:\n",
    "                        nodes[from_state].set_backoff(to_state, weight)\n",
    "                    else:\n",
    "                        nodes[from_state].add_arc(to_state, word_id, weight)\n",
    "        self.nodes = [None for i in range(max(nodes) + 1)]\n",
    "        for state, node in nodes.items():\n",
    "            self.nodes[state] = node\n",
    "            \n",
    "    def set_symbol_table(self, symbol_table):\n",
    "        self.symbol_table = symbol_table\n",
    "  \n",
    "    def context_length(self, state):\n",
    "        tmp = state\n",
    "        n = 0\n",
    "        while self.nodes[tmp].backoff is not None:\n",
    "            n += 1\n",
    "            tmp = self.nodes[tmp].backoff[0]\n",
    "        return n\n",
    "    \n",
    "    def consume_word(self, state, word):\n",
    "        result = []\n",
    "        word_id = word if self.symbol_table is None else (self.symbol_table[word] if word in self.symbol_table else -1)\n",
    "        current_state = state\n",
    "        \n",
    "        while word_id not in self.nodes[current_state].arcs and self.nodes[current_state].backoff is not None:\n",
    "            result.append(self.nodes[current_state].backoff[1])\n",
    "            current_state = self.nodes[current_state].backoff[0]\n",
    "            \n",
    "        if word_id in self.nodes[current_state].arcs:\n",
    "            arc = self.nodes[current_state].arcs[word_id]\n",
    "            result.append(arc[1])\n",
    "            current_state = arc[0]\n",
    "        else:\n",
    "            result.append(self.unk_weight)\n",
    "        \n",
    "        return current_state, result\n",
    "    \n",
    "def load_symbol_table(table_path):\n",
    "    table = dict()\n",
    "    with open(table_path, encoding='utf-8') as f:\n",
    "        for line in f:\n",
    "            splitted = line.split()\n",
    "            if len(splitted) != 2:\n",
    "                continue\n",
    "            word, word_id = splitted\n",
    "            table[word] = int(word_id)\n",
    "    return table"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 42,
   "metadata": {},
   "outputs": [],
   "source": [
    "# для последующего примера понадобится готовая ЯМ в формате текстовом формате fst\n",
    "# Я воспользовался малой ЯМ из vosk\n",
    "# http://alphacephei.com/kaldi/models/vosk-model-ru-0.10.zip\n",
    "# Раскомментируйте следующий код для загрузки и распаковки архива\n",
    "# API_ENDPOINT = 'https://cloud-api.yandex.net/v1/disk/public/resources/download?public_key={}'\n",
    "# link_r = requests.get(API_ENDPOINT.format('https://disk.yandex.ru/d/14VCQDEWGQJUPQ'))\n",
    "# r = requests.get(link_r.json()['href'])\n",
    "# z = zipfile.ZipFile(io.BytesIO(r.content))\n",
    "# z.extractall('data/vosk_lm_small')\n",
    "vosk_ngram = NGramModel('data/vosk_lm_small/G.txt')\n",
    "words = load_symbol_table('data/vosk_lm_small/words.txt')\n",
    "vosk_ngram.set_symbol_table(words)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 43,
   "metadata": {},
   "outputs": [],
   "source": [
    "from IPython.display import Code\n",
    "def representation(sentence, pos, context, weight, backoff):\n",
    "    pos += 1\n",
    "    bos_sentence = ['<s>', *sentence]\n",
    "    lines = []\n",
    "    lines.append(' '.join(bos_sentence))\n",
    "    context_words = [word if pos - i <= context and pos - i >= 0 else ''.join([' ' for i in word]) for i, word in enumerate(bos_sentence)]\n",
    "    if backoff:\n",
    "        context_words[pos] = ''.join(['-' for i in context_words[pos]])\n",
    "    lines.append(' '.join(context_words))\n",
    "    if weight:\n",
    "        lines.append(' '.join([f'{weight:{len(word)}.2f}' if pos == i else ''.join([' ' for i in word]) for i, word in enumerate(bos_sentence)]))\n",
    "    return Code('\\n'.join(lines))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 44,
   "metadata": {},
   "outputs": [],
   "source": [
    "def sentence_score(sentence, lm):\n",
    "    state = vosk_ngram.start\n",
    "    progress = []\n",
    "    score = 0.0\n",
    "\n",
    "    prev = lm.context_length(state)\n",
    "    for pos, word in enumerate(sentence):\n",
    "        progress.append(representation(sentence, pos - 1, prev - 1, None, False))\n",
    "        state, result = lm.consume_word(state, word)\n",
    "        score += sum(result)\n",
    "        n = lm.context_length(state)\n",
    "        for i, weight in enumerate(result[:-1]):\n",
    "            progress.append(representation(sentence, pos, prev - i, weight, True))\n",
    "        progress.append(representation(sentence, pos, prev - len(result) + 1, result[-1], False))\n",
    "\n",
    "        prev = n\n",
    "    return score, progress"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 45,
   "metadata": {},
   "outputs": [],
   "source": [
    "score, progress = sentence_score(['мама', 'мыла', 'раму', 'мылом', 'и', 'хлоркой'], vosk_ngram)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 46,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "64c431ed969a4126a042c7a63a529125",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "HBox(children=(Button(description='Prev', style=ButtonStyle()), Button(description='Next', style=ButtonStyle()…"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "b9d4ea9c426b44fb9cfeab7efa155030",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "interactive(children=(IntSlider(value=0, description='step', max=15), Output()), _dom_classes=('widget-interac…"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "animate_list(progress);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Попробуем добавить языковую модель к нашему корректору ошибок."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 47,
   "metadata": {},
   "outputs": [],
   "source": [
    "tokens, hyp_slices = correct_text_with_beam_search(trie_dict_ru, 'я напсалэто предложеtние с незначительнми ошибкми', vosk_ngram, lm_scale=0.1, beam=7.0)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 48,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "['я', 'написал', 'это', 'предложение', 'с', 'незначительными', 'ошибки']\n"
     ]
    }
   ],
   "source": [
    "print(list(recover_best_hyp(tokens)))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 49,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "1fce2a50a15b419d9f0402eb1d559106",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "HBox(children=(Play(value=0, interval=200), Button(description='Prev', style=ButtonStyle()), Button(descriptio…"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "3814d78917dc47cabb42053457badb57",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "interactive(children=(IntSlider(value=0, description='step', max=48), Output()), _dom_classes=('widget-interac…"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "animate_list(hyp_slices, play=True);"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 50,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "48.1053915512\n"
     ]
    },
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "47e68739849f402a8cab13fe44901420",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "HBox(children=(Button(description='Prev', style=ButtonStyle()), Button(description='Next', style=ButtonStyle()…"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "0119bd5eda84430690218a9fc5200902",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "interactive(children=(IntSlider(value=0, description='step', max=17), Output()), _dom_classes=('widget-interac…"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "score, progress = sentence_score(['я', 'написал', 'это', 'предложение', 'с', 'незначительными', 'ошибки'], vosk_ngram)\n",
    "print(score)\n",
    "animate_list(progress);"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 51,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "50.7081652512\n"
     ]
    },
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "1595f2a47e1f4f22ba7d8abd12eb8f68",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "HBox(children=(Button(description='Prev', style=ButtonStyle()), Button(description='Next', style=ButtonStyle()…"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "0a4c5cd77d504ba6a120d7c6736035e8",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "interactive(children=(IntSlider(value=0, description='step', max=17), Output()), _dom_classes=('widget-interac…"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "score, progress = sentence_score(['я', 'написал', 'это', 'предложение', 'с', 'незначительными', 'ошибками'], vosk_ngram)\n",
    "print(score)\n",
    "animate_list(progress);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "В данном конкретном случае в ЯМ не оказалось $N$-грамы `незначительными ошибками`, а слово `ошибки` имеет большее правдоподобие, чем `ошибками`."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Lattices\n",
    "Зачастую нам может понадобится не один результат распознавания, а несколько. Одна из самых популярных причин сохранения нескольких хороших гипотез -- последующий рескоринг: ранжирование этих гипотез с помощью более точной языковой модели. Алгоритмы поиска обычно модифифцируются следующим образом для достижения этой цели: иногда вместо того, чтобы отбросить гипотезу мы можем вместо этого подвесить её к какой-то другой. В результате вместо дерева мы получаем граф, в области распознавания речи такой граф принято называть <i>lattice</i>."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 52,
   "metadata": {},
   "outputs": [],
   "source": [
    "tokens, hyp_slices = correct_text_with_beam_search(trie_dict_ru, 'я напсалэто предложеtние с незначительнми ошибкми', vosk_ngram, lm_scale=0.1, beam=7.0, lattice=True)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 53,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "941c400da4014b308fb531d8b151e36c",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "HBox(children=(Play(value=0, interval=200), Button(description='Prev', style=ButtonStyle()), Button(descriptio…"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
      "application/vnd.jupyter.widget-view+json": {
       "model_id": "568a79c662ec4a57a84499b1b14c2863",
       "version_major": 2,
       "version_minor": 0
      },
      "text/plain": [
       "interactive(children=(IntSlider(value=0, description='step', max=48), Output()), _dom_classes=('widget-interac…"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "animate_list(hyp_slices, play=True);"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 54,
   "metadata": {},
   "outputs": [],
   "source": [
    "#hyp_slices[20].render(filename='lattice', format='pdf')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "В системах, основанных на марковских цепях латтисы можно визуилизировать следующим образом"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 55,
   "metadata": {},
   "outputs": [
    {
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     },
     "execution_count": 55,
     "metadata": {},
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   ],
   "source": [
    "graph = Graph([\n",
    "    Arc(0, 1),\n",
    "    Arc(1, 2), \n",
    "    Arc(2, 0),\n",
    "    Arc(1, 1),\n",
    "    Arc(2, 2),\n",
    "    Arc(0, 0)\n",
    "])\n",
    "graph.Visualize()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 56,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "'hmm.pdf'"
      ]
     },
     "execution_count": 56,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "graph.Visualize().render(filename='hmm', format='pdf')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 57,
   "metadata": {},
   "outputs": [
    {
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    "expanded_graph = graph.expand_in_time(4)\n",
    "expanded_graph.Visualize()\n",
    "# expanded_graph.Visualize().render(filename='expanded_hmm', format='pdf')"
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    "Описанный выше поиск можно описать как нахождение самого кратчайшего пути в \"расширенном во времени\" графе. Латтисом же является подграф этого графа."
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          "text/plain": "<s> я написал это предложение с незначительными ошибками\n<s>                                                     "
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          "text/plain": "<s> я написал это предложение с незначительными ошибки\n<s>                                                   "
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