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--- aa:lab:sol:8 [2025/12/01 03:22]
mihnea.gheorghe Draft soluții BST
+++ aa:lab:sol:8 [2025/12/04 12:08] (current)
mihnea.gheorghe fix notation consistency
@@ Line 63: / Line 63: @@
 \)
-=== 2. a) \(\forall x \in \mathbb{E}, \forall t \in \mathrm{BTree} : \mathrm{isBST}(t) \implies \mathrm{isBST}(\mathrm{insert}(x,t))\) ===
+=== 2. Soluție Python ===
+<file python bst_path.py>
+import random
+class Node:
+    def __init__(self, x):
+        self.x = x
+        self.left = None
+        self.right = None
+def insert(root, x):
+    if root is None:
+        return Node(x)
+    if x <= root.x:
+        root.left = insert(root.left, x)
+    else:
+        root.right = insert(root.right, x)
+    return root
+def search_path_length(root, x):
+    steps = 0
+    current = root
+    while current is not None:
+        steps += 1
+        if x == current.x:
+            return steps
+        elif x < current.x:
+            current = current.left
+        else:
+            current = current.right
+    return steps
+def average_search_path(n):
+    """Construiește un BST dintr-o permutare aleatoare și măsoară media."""
+    perm = list(range(1, n + 1))
+    random.shuffle(perm)
+    root = None
+    for x in perm:
+        root = insert(root, x)
+    lengths = [search_path_length(root, x) for x in perm]
+    return sum(lengths) / n
+def experiment(trials=2000, n=100):
+    total = 0
+    for _ in range(trials):
+        total += average_search_path(n)
+    return total / trials
+for n in [10, 20, 50, 100, 200, 500, 1000]:
+    avg = experiment(trials=1000, n=n)
+    print(f"n = {n} → lungime medie ≈ {avg:.3f}")
+</file>
+=== 3. a) \(\forall x \in \mathbb{E}, \forall t \in \mathrm{BTree} : \mathrm{isBST}(t) \implies \mathrm{isBST}(\mathrm{insert}(x,t))\) ===
 Vrem să demonstrăm
@@ Line 81: / Line 138: @@
 \begin{aligned}
 \mathrm{isBST}(\mathrm{Nil})
-&\overset{(\text{BST})}{=} \mathrm{isBSTBetween}(\mathrm{Nil}, -\infty, \infty) \\
+&\overset{(\text{BST})}{=} \mathrm{isBSTBetween}(\mathrm{Nil}, -\infty, +\infty) \\
 &\overset{(\text{BST1})}{=} \text{true}, \forall x \in \mathbb{E}
 \end{aligned}
@@ Line 89: / Line 146: @@
 \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{BST})}{=} \mathrm{isBSTBetween}(\mathrm{Node}(x, \mathrm{Nil}, \mathrm{Nil}), -\infty, +\infty) \\
-&\overset{(\text{BST2})}{=} (-\infty \le x \le \infty)
+&\overset{(\text{BST2})}{=} (-\infty \le x \le +\infty)
 \;\land\;
 \mathrm{isBSTBetween}(\mathrm{Nil}, -\infty, x)
 \;\land\;
-\mathrm{isBSTBetween}(\mathrm{Nil}, x, \infty) \\
+\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) \\
+\mathrm{isBSTBetween}(\mathrm{Nil}, x, +\infty) \\
 &\overset{(\text{BST1})}{=} \text{true}, \forall x \in \mathbb{E}
 \end{aligned}
@@ Line 112: / Line 169: @@
 \(
 \overset{(\text{BST})}{\Leftrightarrow}
-\mathrm{isBSTBetween}(t, -\infty, \infty)
+\mathrm{isBSTBetween}(t, -\infty, +\infty)
 \overset{(\text{BST2})}{\Leftrightarrow}
-(lo \le y \le hi) \;\land\;
+(-\infty \le y \le +\infty) \;\land\;
-\mathrm{isBSTBetween}(l, lo, y) \;\land\;
+\mathrm{isBSTBetween}(l, -\infty, y) \;\land\;
-\mathrm{isBSTBetween}(r, y, hi) \
+\mathrm{isBSTBetween}(r, y, +\infty) \
 (\text{II})
 \)