Soluții Laboratorul 8

\( \mathrm{insert} : \mathbb{E} \times \mathrm{BTree} \to \mathrm{BTree} \)

\( \mathrm{insert}(x, \mathrm{Nil}) \overset{(\text{INS1})}{=} \mathrm{Node}(x, \mathrm{{Nil}}, \mathrm{Nil}) \)

\( \mathrm{insert}(x, \mathrm{Node}(y,l,r)) \overset{(\text{INS2})}{=} \mathrm{Node}(y, \mathrm{insert}(x,l), r), x \le y \ \)

\( \mathrm{insert}(x, \mathrm{Node}(y,l,r)) \overset{(\text{INS3})}{=} \mathrm{Node}(y, l, \mathrm{insert}(x,r)), x > y \)

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\( \mathrm{min} : \mathrm{BTree} \setminus \{\mathrm{Nil}\} \to \mathbb{E} \)

\( \mathrm{min}(\mathrm{Node}(x, \mathrm{Nil}, r)) \overset{(\text{MIN1})}{=} x \)

\( \mathrm{min}(\mathrm{Node}(y, \mathrm{Node}(x,l,r), r')) \overset{(\text{MIN2})}{=} \mathrm{min}(\mathrm{Node}(x,l,r)) \)

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\( \mathrm{max} : \mathrm{BTree} \setminus \{\mathrm{Nil}\} \to \mathbb{E} \)

\( \mathrm{max}(\mathrm{Node}(x, l, \mathrm{Nil})) \overset{(\text{MAX1})}{=} x \)

\( \mathrm{max}(\mathrm{Node}(y, l', \mathrm{Node}(x,l,r))) \overset{(\text{MAX2})}{=} \mathrm{max}(\mathrm{Node}(x,l,r)) \\ \)

Vrem să demonstrăm \( \mathrm{isBST}(t) \implies \mathrm{isBST}(\mathrm{insert}(x,t)), \forall t \in \mathrm{BTree}, \forall x \in \mathbb{E} \)

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Caz de bază: \(t = \mathrm{Nil}\)

\( \mathrm{insert}(x, \mathrm{Nil}) \overset{(\text{INS1})}{=} \mathrm{Node}(x, \mathrm{Nil}, \mathrm{Nil}), \forall x \in \mathbb{E} \)

\( \begin{aligned} \mathrm{isBST}(\mathrm{Nil}) &\overset{(\text{BST})}{=} \mathrm{isBSTBetween}(\mathrm{Nil}, -\infty, \infty) \\ &\overset{(\text{BST1})}{=} \text{true}, \forall x \in \mathbb{E} \end{aligned} \)

\( \begin{aligned} \mathrm{isBST}(\mathrm{Node}(x, \mathrm{Nil}, \mathrm{Nil})) &\overset{(\text{BST})}{=} \mathrm{isBSTBetween}(\mathrm{Node}(x, \mathrm{Nil}, \mathrm{Nil}), -\infty, \infty) \\ &\overset{(\text{BST2})}{=} (-\infty \le x \le \infty) \;\land\; \mathrm{isBSTBetween}(\mathrm{Nil}, -\infty, x) \;\land\; \mathrm{isBSTBetween}(\mathrm{Nil}, x, \infty) \\ &\overset{(\overline{\mathbb{E}})}{=} \text{true} \;\land\; \mathrm{isBSTBetween}(\mathrm{Nil}, -\infty, x) \;\land\; \mathrm{isBSTBetween}(\mathrm{Nil}, x, \infty) \\ &\overset{(\text{BST1})}{=} \text{true}, \forall x \in \mathbb{E} \end{aligned} \)

Deci cazul de bază este adevărat \(\forall x \in \mathbb{E}\).

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Pas de inducție: \(t = \mathrm{Node}(y, l, r)\)

Presupunem \(\mathrm{isBST}(t)\) \( \overset{(\text{BST})}{\Leftrightarrow} \mathrm{isBSTBetween}(t, -\infty, \infty) \overset{(\text{BST2})}{\Leftrightarrow} (lo \le y \le hi) \;\land\; \mathrm{isBSTBetween}(l, lo, y) \;\land\; \mathrm{isBSTBetween}(r, y, hi) \ (\text{II}) \)

\( \begin{aligned} \forall lo, y \in \mathbb{E}: &\quad \mathrm{isBSTBetween}(l, lo, y) \overset{(\text{II1})}{\implies} \mathrm{isBSTBetween}(\mathrm{insert}(x,l), lo, y) \\ \forall y, hi \in \mathbb{E}: &\quad \mathrm{isBSTBetween}(r, y, hi) \overset{(\text{II2})}{\implies} \mathrm{isBSTBetween}(\mathrm{insert}(x,r), y, hi) \end{aligned} \)

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Caz 1: \(x \le y\)

\( \mathrm{insert}(x, \mathrm{Node}(y,l,r)) \overset{(\text{INS2})}{=} \mathrm{Node}(y, \mathrm{insert}(x,l), r) \)

\( \begin{aligned} \mathrm{isBSTBetween}(\mathrm{Node}(y, \mathrm{insert}(x,l), r), lo, hi) &\overset{(\text{BST2})}{=} (lo \le y \le hi) \;\land\; \mathrm{isBSTBetween}(\mathrm{insert}(x,l), lo, y) \;\land\; \mathrm{isBSTBetween}(r, y, hi) \\ &\overset{(\text{II1})}{=} (lo \le y \le hi) \;\land\; \mathrm{isBSTBetween}(l, lo, y) \;\land\; \mathrm{isBSTBetween}(r, y, hi) \\ &\overset{(\text{II})}{=} \text{true} \end{aligned} \)

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Caz 2: \(x > y\)

\( \mathrm{insert}(x, \mathrm{Node}(y,l,r)) \overset{(\text{INS3})}{=} \mathrm{Node}(y, l, \mathrm{insert}(x,r)) \)

\( \begin{aligned} \mathrm{isBSTBetween}(\mathrm{Node}(y, l, \mathrm{insert}(x,r)), lo, hi) &\overset{(\text{BST2})}{=} (lo \le y \le hi) \;\land\; \mathrm{isBSTBetween}(l, lo, y) \;\land\; \mathrm{isBSTBetween}(\mathrm{insert}(x,r), y, hi) \\ &\overset{(\text{II2})}{=} (lo \le y \le hi) \;\land\; \mathrm{isBSTBetween}(l, lo, y) \;\land\; \mathrm{isBSTBetween}(r, y, hi) \\ &\overset{(\text{II})}{=} \text{true} \end{aligned} \)

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Deci \(\mathrm{isBST}(t) \implies \mathrm{isBST}(\mathrm{insert}(x,t)), \forall x \in \mathbb{E}, \forall t \in \mathrm{BTree}\)