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1. The dictionary holds bad, dad and mad. Which pattern does NOT match any stored word?

..dma.p.db..

2. Why must the search try every child at a dot, instead of following just one child?

A dot is stored in the trie as its own child, so the search follows that childA dot can stand for any letter, so the search does not know which child is right and must try each oneEvery child of a node holds the same letter, so the first one is always enoughTrying every child first makes the search faster than following one child

3. Using the WordDictionary class from this lesson, this code runs: ``

cpp
WordDictionary words;
for (std::string_view w : {"moon", "mood", "book"}) words.addWord(w);
std::cout << std::boolalpha << words.search("moo") << '\n';

`` What does it print?

true0false1

4. Using the same class, this code runs: ``

cpp
WordDictionary fox;
for (std::string_view w : {"fox", "fig", "fan"}) fox.addWord(w);
std::cout << std::boolalpha << fox.search("f.g") << '\n';

`` What does it print?

truefalseThe program fails with an assertionThe output is empty

5. Which statement describes the bug in this dot branch? ``

cpp
static bool searchFrom(const Node* node, std::string_view pattern, std::size_t pos) {
    if (pos == pattern.size()) return node->isEnd;
    if (pattern[pos] == '.') {
        for (const auto& child : node->children) {
            if (child) return searchFrom(child.get(), pattern, pos + 1);
        }
        return false;
    }
    // plain-letter case omitted
}

``

It checks isEnd before the pattern is used up, so it ends words too earlyIt turns the dot into a child index, so it reads memory outside the arrayIt has no base case, so it recurses forever on an empty patternIt returns the result of the first non-empty child, so later children are never tried

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