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    <item rdf:about="https://ocw.cs.pub.ro/ppcarte/doku.php?id=aa:lab:sol:1&amp;rev=1729109894&amp;do=diff">
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        <dc:date>2024-10-16T23:18:14+03:00</dc:date>
        <title>aa:lab:sol:1</title>
        <link>https://ocw.cs.pub.ro/ppcarte/doku.php?id=aa:lab:sol:1&amp;rev=1729109894&amp;do=diff</link>
        <description>Solutii Mașina Turing

1. 


 // Scrieți o Mașină Turing care primește un șir binar și verifică dacă începe
 // și se termină cu simboluri distincte (e.g. 100110110).

//-------CONFIGURATION
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expect1, 0
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        <dc:date>2025-10-12T18:49:36+03:00</dc:date>
        <title>aa:lab:sol:2</title>
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        <description>Soluții notații asimptotice și analiză amortizată

Notații asimptotice

1. Dați exemple de câte o funcție din următoarele clase de complexitate:

$ O(n)$




$Ω(log(n))$




$θ(n^2)$



$ɷ(1/n)$




$ o(3^n)$



2. Verificați valoarea de adevăr a următoarelor propoziții:

$ √n∈O(logn)$

=&gt;Fals$ logn∈O(log(log n))$$ n∈O(√n logn)$$ n+logn∈θ(n)$$ log(nlogn)∈θ(logn)$$ √n∈ɷ(log n)$</description>
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        <dc:date>2025-10-19T21:59:40+03:00</dc:date>
        <title>aa:lab:sol:3</title>
        <link>https://ocw.cs.pub.ro/ppcarte/doku.php?id=aa:lab:sol:3&amp;rev=1760900380&amp;do=diff</link>
        <description>Exercitiul 1

Metoda Agregatelor

ArrayList - inserare

Raspuns: ĉi = 3

Explicatie: Incepem cu un array cu capacitate 1 si 0 elemente (vom nota cost insert = ci, copiere n elemente = cpyn.

	*  insert1: c = ci = 1
	*  insert2: c = cpy1 + ci = 2
	*</description>
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        <title>aa:lab:sol:4</title>
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        <description>Soluții Laboratorul 4

Enunțuri

Demonstrați următoarele proprietăți folosind inducție structurală:

	*  1. ∀l∈List.reverse(reverse(l))=l
	*  2. ∀r∈Ring,size(move(r))=size(r)  
	*  3. ∀i∈N,∀r∈Ring,element(i,move(r))=element(i+1,r)\[
\mathsf{reverse}(\mathsf{reverse}(\mathsf{Void})) 
\overset{(\text{REV1})}{=} 
\mathsf{reverse}(\mathsf{Void}) 
\overset{(\text{REV1})}{=} 
\mathsf{Void}
\]\[
\mathsf{reverse}(\mathsf{reverse}(l)) = l
\]\[
\mathsf{reverse}(\mathsf{reverse}(\mathsf{Cons}(x, l))) = \ma…</description>
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        <dc:date>2025-11-01T11:55:08+03:00</dc:date>
        <title>aa:lab:sol:5</title>
        <link>https://ocw.cs.pub.ro/ppcarte/doku.php?id=aa:lab:sol:5&amp;rev=1761990908&amp;do=diff</link>
        <description>Soluții Laborator 5: Analiza probabilistă în Hashing

Această pagină conține explicații și răspunsuri pentru întrebările din laboratorul 5.

----------

Exemplul 1: Paradoxul zilelor de naștere

Întrebare 1:
Pentru ce valoare a lui n probabilitatea ca două persoane să aibă aceeași zi de naștere depășește 0.5?$$
P(\text{distincte}) = \frac{k!}{(k-n)! k^n}
$$$ → P(\text{coliziune}) \approx 0.41$$ → P(\text{coliziune}) \approx 0.507$$ → P(\text{coliziune}) \approx 0.706$$$
E[T] = n H_n = n (1 + \tf…</description>
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        <dc:date>2025-11-11T12:10:50+03:00</dc:date>
        <title>aa:lab:sol:6</title>
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        <description>Soluții Laboratorul 6

1. 
a. binary_search:

După verificările realizate în timp constant, problema este împărțită în două subprobleme de dimensiune înjumătățită, din care doar una este aleasă. (Pentru simplitate, putem considera termenul constant = 1) $ T(n) = T(n/2) + 1, T(1) = 1 $$ T\left(\frac n{2^h}\right) + 1 $$ h $$ T(1) $$ \log n $$$ \left(\sum_{i=0}^{\log n - 1} 1\right) + T(1) = \log n + 1 = \Theta(\log n) $$$ T(n) = \Theta(\log n) $$ \exists(c_1, c_2 &gt; 0)(N \geq 0): \forall(n\geq N):…</description>
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        <dc:date>2025-11-16T23:00:31+03:00</dc:date>
        <title>aa:lab:sol:7</title>
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        <description>+

1.

	*  $ T_a(n) = 2T_a(n-1) + 1$

 nu se poate aplica master 


	*  $ T_b(n) = T_b(n-1) + 1$

 nu se poate aplica master 

Prin metoda arborelui obtinem n niveluri continand cate o constanta si un nod, deci $ T_b(n) = n * \Theta(1) = \Theta(n) $.

	*  $ T_c(n) = 2T_c(n/2) + \log(n)$

 Metoda substitutiei 

Vrem sa demonstram ca $ T_c(n) = \Theta(n), \ presupunem \ T_c(1) = 1$ 

$ \Leftrightarrow \exists c_1, c_2 &gt; 0 $ a.i. $ c_1 n  \le T_c(n) \le c_2 n$

$ Ca \ sa \ demonstram \ ca \ T_c(n) …</description>
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        <dc:date>2025-12-04T12:08:55+03:00</dc:date>
        <title>aa:lab:sol:8</title>
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        <description>Soluții Laboratorul 8

Soluții

1. Definiții și axiome

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

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

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

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

-…</description>
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        <dc:date>2025-12-08T22:35:50+03:00</dc:date>
        <title>aa:lab:sol:9</title>
        <link>https://ocw.cs.pub.ro/ppcarte/doku.php?id=aa:lab:sol:9&amp;rev=1765226150&amp;do=diff</link>
        <description>Exerciții

1.

\(
\mathrm{height}(\mathrm{Nil}) = 0 \\
\mathrm{height}(\mathrm{Node(x, l, r)}) = 1 + \mathrm{max}(\mathrm{height}(l), \mathrm{height}(r))
\)




\(
\mathrm{depth}(y, \mathrm{Nil}) = \mathrm{ABSENT} \\
\mathrm{depth}(y, \mathrm{Node(x, l, r)}) = 
            \begin{cases}
            0, &amp; \text{dacă } x = y \ \\
            1 + \mathrm{depth}(y, l), &amp; \text{dacă } \mathrm{depth}(y, l) \ne \mathrm{ABSENT} \ \\
            1 + \mathrm{depth}(y, r), &amp; \text{dacă } \mathrm{depth}(y, r…</description>
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        <dc:date>2025-12-15T15:04:16+03:00</dc:date>
        <title>aa:lab:sol:10</title>
        <link>https://ocw.cs.pub.ro/ppcarte/doku.php?id=aa:lab:sol:10&amp;rev=1765803856&amp;do=diff</link>
        <description></description>
    </item>
    <item rdf:about="https://ocw.cs.pub.ro/ppcarte/doku.php?id=aa:lab:sol:11&amp;rev=1704750927&amp;do=diff">
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        <dc:date>2024-01-08T23:55:27+03:00</dc:date>
        <title>aa:lab:sol:11</title>
        <link>https://ocw.cs.pub.ro/ppcarte/doku.php?id=aa:lab:sol:11&amp;rev=1704750927&amp;do=diff</link>
        <description>Analiza amortizata

1. Stack

Naiv: La un momnt dat pot fi k elemente in stack, deci cea mai costisitoare operatie va fi mpop cu complexitate temporala O(k). 
Daca avem n operatii in secventa (avem deci  $ n \leq k $ ), vom avea maxim  $ n * O(k) = O(n^2) $ . 
Dar totusi nu putem face atatea  $ mpop(k) $$ pop $$ S $$ mpop(k) $$ S $$ k $$ (k-1) $$ mpop(k-1) $$ pop $$ push $$ 1 $$ mpop $$ k $$ cost(S) = (k-1) * 1 + (k - 1) = 2k - 2 $$ cost(op) = cost(S) / k = (2k - 2) / k \leq 2 \Rightarrow cost(o…</description>
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        <dc:date>2026-01-19T13:01:34+03:00</dc:date>
        <title>aa:lab:sol:12</title>
        <link>https://ocw.cs.pub.ro/ppcarte/doku.php?id=aa:lab:sol:12&amp;rev=1768820494&amp;do=diff</link>
        <description>Abordări practice pentru probleme NP-Complete</description>
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