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שאלה 1

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השיחה עם chatGPT:

<iframe width="800" height="600" src="https://chat.openai.com/share/2f5a54f1-054a-4f23-bfca-501c3288d16e"></iframe>

הקוד שקיבלתי מchatGPT:

public class TimeComplexityExercise {
    public static void main(String[] args) {
        for (int i = 0; i < N; i += 2) {
            System.out.println("Hello world");
        }

        for (int i = 0; i < Math.sqrt(N); i++) {
            for (int j = 0; j < N; j += 3) {
                System.out.println("Hello world");
            }
        }

        for (int i = 0; i < Math.log(N); i++) {
            for (int j = 0; j < Math.pow(N, 0.25); j++) {
                System.out.println("Hello world");
            }
        }

        for (int i = 0; i < N; i *= 3) {
            System.out.println("Hello world");
        }

        for (int i = 0; i < Math.sqrt(N); i++) {
            for (int j = 0; j < Math.sqrt(N); j++) {
                System.out.println("Hello world");
            }  
        }
    }
}

שלב ב׳

חישוב פונקציית זמן ריצה

loop inner loop line
N/2 + 1 for (int i = 0; i < N; i += 2) {
N/2 System.out.println("Hello world");
sqrt(N) + 1 for (int i = 0; i < Math.sqrt(N); i++) {
sqrt(N) * (N/3 + 1) N/3+1 for (int j = 0; j < N; j += 3) {
sqrt(N) * N/3 N/3 System.out.println("Hello world");
log(N) + 1 for (int i = 0; i < Math.log(N); i++) {
log(N) * (pow(N, 0.25) + 1) pow(N, 0.25) + 1 for (int j = 0; j < Math.pow(N, 0.25); j++) {
log(N) * pow(N, 0.25) pow(N, 0.25) System.out.println("Hello world");
log_3(N) + 1 for (int i = 1; i < N; i *= 3) { changed i = 1 for valid loop condition
log_3(N) System.out.println("Hello world");
sqrt(N) + 1 for (int i = 0; i < Math.sqrt(N); i++) {
sqrt(N) * (sqrt(N) + 1) sqrt(N) + 1 for (int j = 0; j < Math.sqrt(N); j++) {
sqrt(N) * sqrt(N) sqrt(N) System.out.println("Hello world");
<iframe width="800" height="600" src="https://chat.openai.com/share/d8234381-d901-4595-81b4-3ddc9125e582"></iframe>

$$\text{T(N)} = (\frac{N}{2} + 1 + \frac{N}{2}) + ((\sqrt{N} + 1) + (\sqrt{N} \cdot (\frac{N}{3} + 1)) + (\sqrt{N} \cdot \frac{N}{3})) + ((\log{N} + 1) + (\log{N} \cdot (\sqrt[4]{N} + 1)) + (\log{N} \cdot \sqrt[4]{N})) + ((\log_3{N} + 1) + \log_3{N}) + ((\sqrt{N} + 1) + (\sqrt{N} \cdot (\sqrt{N} + 1)) + (\sqrt{N} \cdot \sqrt{N})) $$ $$ \text{T(N)} = (N+1)+\left( \left( 2+\frac{2\cdot N}{3} \right)\cdot \sqrt{ N }+1 \right)+(2\cdot\log N\cdot(1+\sqrt[4]{N})+1)+ (2\cdot \log_3{N}+1)+(2\cdot\sqrt{ N }+2\cdot N+1) $$ $$ \text{T(N)} = \sqrt{ N } \cdot\left( 2+\frac{2\cdot N}{3} \right)+2\cdot \log N\cdot(1+\sqrt[4]{ N })+2\cdot \log_3{N}+2\cdot \sqrt{ N }+3N+5 $$

שלב ג׳

חישוב חסם עליון של פונקציית זמן ריצה:

<iframe width="800" height="600" src="https://chat.openai.com/share/9282e3c8-0746-4946-9219-f65d91937870"></iframe> $$ \text{T(N)} = \sqrt{ N } \cdot\left( \cancel{ 2 }+\frac{2\cdot N}{3} \right)+2\cdot \log N\cdot(\cancel{ 1 }+\sqrt[4]{ N })+2\cdot \log_3{N}+2\cdot \sqrt{ N }+3N+5 $$ $$ \text{T(N)}=\cancel{ \frac{2}{3} }\cdot N^{1.5}+\cancel{ 2\cdot \log N\cdot \sqrt[4]{ N } }+\cancel{ 2\cdot \log_3{N} }+\cancel{ 2\cdot \sqrt{ N } }+\cancel{ 3N }+\cancel{ 5 } $$ $$ O(N^{1.5}) $$

שאלה 2

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<iframe width="800" height="600" src="https://chat.openai.com/share/53132c7e-1b75-4c82-9864-05b07cd73b8a"></iframe>

$$ f(x)=x^3+ 2⋅e^x+ log(x) $$ $$ f(x)=3x^2−\frac{e ^{2x }}{x}​+log2​(x!) $$ $$ f(x)=\sqrt{ x }+e^{2x}\cdot \log_{10}{(x+1)} $$ $$ f(x)=\frac{e^x}{x^2+1}+\log_3{(x!)}-x^3 $$ $$ f(x)=\frac{x^4}{4!}+3\cdot \ln (x+2)+e^{0.5x} $$ $$ f(x)=\frac{e^x-e^{-x}}{2x}+\ln(2x^2+1) $$$$ f(x)=2x^3+\frac{e^{0.5x}}{\sqrt{ x }}-\log_2{(x!)} $$ $$ f(x)=\ln(x^2+1)+\frac{e^{2x}}{2x}-\frac{x^3}{e!} $$ $$ f(x)=4x^4 -e^{0.5x}\cdot \ln(x+3)-\frac{1}{x!} $$ $$ f(x)=\frac{e^x}{x^2}+\frac{x^3}{3!}-\log_{e^2}{(x+1)} $$

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<iframe width="800" height="600" src="https://chat.openai.com/share/ec22937e-1c1c-411f-9e8e-5490796ec3c2"></iframe>

$$ f(x)=\cancel{ x^3 }+C2⋅e^x+\cancel{ log(x) } $$ $$O(e^x)$$

$$ f(x)=\cancel{ 3x^2 }−\frac{e ^{2x }}{x}​+\cancel{ log2​(x!) } $$ $$O(e^{2x})$$

$$ f(x)=\cancel{ \sqrt{ x } }+e^{2x}\cdot \log_{10}{(x+1)} $$ $$O(e^{2x})$$

$$ f(x)=\frac{e^x}{x^2+1}+\cancel{ \log_3{(x!)} }-\cancel{ x^3 } $$ $$O(e^x)$$

$$ f(x)=\cancel{ \frac{x^4}{4!} }+\cancel{ 3\cdot \ln (x+2) }+e^{0.5x} $$ $$O(e^{0.5x})$$

$$ f(x)=\frac{e^x-e^{-x}}{2x}+\cancel{ \ln(2x^2+1) } $$ $$O(e^x)$$

$$ f(x)=\cancel{ 2x^3 }+\frac{e^{0.5x}}{\sqrt{ x }}-\cancel{ \log_2{(x!)} } $$ $$O(e^{0.5x})$$

$$ f(x)=\cancel{ \ln(x^2+1) }+\frac{e^{2x}}{2x}-\cancel{ \frac{x^3}{e!} } $$ $$O(e^{2x})$$

$$ f(x)=\cancel{ 4x^4 }-e^{0.5x}\cdot \ln(x+3)-\cancel{ \frac{1}{x!} } $$ $$O(e^{0.5x})$$

$$

f(x)=\frac{e^x}{x^2}+\cancel{ \frac{x^3}{3!} }-\cancel{ \log_{e^2}{(x+1)} } $$ $$O(e^x)$$

שלב ג׳

<iframe width="800" height="600" src="https://chat.openai.com/share/669089ce-e0d7-4062-ac3a-250566dc6ef2"></iframe>
  1. $$O(e^{0.5x})$$
  2. $$O(e^x)$$
  3. $$O(e^{2x})$$

שאלה 3

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<iframe width="800" height="600" src="https://chat.openai.com/share/fc15cafc-8b15-4383-a316-09a441924fb3"></iframe>

סיכום: כאשר אלגוריתם עובר כל המערך המקרה הטוב והרע יהיו בדרך כלל שווים.

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<iframe width="800" height="600" src="https://chat.openai.com/share/d85f1094-3690-4f57-a045-9909a6dbaca0"></iframe>

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