pclub-graphics-hire-test.tex

gistfile1.txt

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\lhead{\small \textbf{Graphics Recruitment Test}}
\rhead{\small Total Marks: 50}
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\begin{document}

\begin{center}
    {\LARGE \textbf{Graphics Spring Camp}} \\[0.2cm]
    {\large Selection Test \hfill \textbf{Max Marks: 50}}
\end{center}

\noindent
\fbox{
    \begin{minipage}{\dimexpr\textwidth-2\fboxsep-2\fboxrule\relax}
        \vspace{0.2cm}
        \noindent
        % Adjusted columns to fix wrapping and margins:
        % Col 1: 0.17\textwidth (Wider for "Priority")
        % Col 2: 0.30\textwidth (Slightly narrower)
        % Col 3: 0.15\textwidth (Standard)
        % Col 4: 0.30\textwidth (Slightly narrower)
        \begin{tabular}{@{}p{0.17\textwidth} p{0.30\textwidth} p{0.15\textwidth} p{0.30\textwidth}@{}}
            \textbf{Name:} & \hrulefill & \textbf{Roll No:} & \hrulefill\hspace{0.5cm} \\[1.5em]
            \textbf{Priority (1-3):} & \hrulefill & \textbf{Start/End:} & \underline{\hspace{1.8cm}} / \underline{\hspace{1.8cm}} \\
        \end{tabular}
        \vspace{0.1cm}
    \end{minipage}
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\vspace{0.5cm}

% --- CONTENT START ---

\section*{Passage 1: Ray-Sphere Intersection Geometry}
We define a Ray as $P(t) = O + tD$, where $O$ is the origin and $D$ is the direction.
We define a Sphere with Center $C$ and Radius $r$ as the set of points where $\|P - C\|^2 = r^2$.
To find the intersection, we substitute the Ray equation into the Sphere equation. This gives us a quadratic equation for $t$.

\begin{enumerate}
    \item \textbf{[2 Marks] Geometric Miss Condition:}
    Without relying on calculating the discriminant ($b^2 - 4ac$), describe the condition for "No Intersection" purely in terms of the geometry of vectors $O, D, C$ and scalar $r$. (Conceptually, when does the ray miss?)
    \answerbox{2cm}

    \item \textbf{[5 Marks] Solving for $t$:}
    The standard quadratic formula is $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
    In the live session, we implemented the code to find the value of $t$.
    \textbf{Derive the formula} or show the logical steps to find the intersection point $t$ below. \textit{Use this box for rough work; you will be graded on your derivation.}
    \answerbox{6cm}

    \item \textbf{[2 Marks] The Choice of Root:}
    The quadratic formula produces two solutions: $t_1$ (using $-\sqrt{\dots}$) and $t_2$ (using $+\sqrt{\dots}$). In our code, we specifically looked for the root with the \textbf{minus} sign first. Why do we prefer $t = -b - \sqrt{\dots}$?
    \answerbox{1.5cm}
\end{enumerate}

\section*{Passage 2: Constructive Solid Geometry \& SDFs}
Modern rendering often uses \textbf{Signed Distance Fields (SDFs)}. An SDF is a function \texttt{float map(vec3 p)} that returns the distance from point $p$ to the closest surface.
\begin{itemize}
    \item If $map(p) > 0$: You are outside the object.
    \item If $map(p) < 0$: You are \textbf{inside} the object.
\end{itemize}

\begin{lstlisting}[language=C]
float sdSphere(vec3 p, float r) { return length(p) - r; }
\end{lstlisting}

\begin{enumerate}[resume]
    \item \textbf{[3 Marks] Intersection Logic:}
    To combine two shapes (Shape A with distance $d_1$, Shape B with distance $d_2$), we use min/max functions. 
    \texttt{min(d1, d2)} creates the \textbf{Union}.
    
    Which function creates the \textbf{Intersection}? \textbf{Explain why} that specific logic results in the intersection of the two shapes.
    \answerbox{3cm}

    \item \textbf{[2 Marks] Subtraction Logic:}
    If you wanted to cut Shape B out of Shape A (Subtraction), which formula would you use?
    \begin{itemize}
        \item[a)] $d_1 - d_2$
        \item[b)] $\max(d_1, -d_2)$
        \item[c)] $\min(d_1, -d_2)$
    \end{itemize}

    \item \textbf{[2 Marks] Inflation (Isosurfaces):}
    If we calculate a new distance field $d_{new} = d_{cube} - 0.1$, describe exactly how the visual shape of the cube changes, particularly at the sharp corners.
    \answerbox{1.5cm}
\end{enumerate}

\section*{Passage 3: The Rendering Equation}
The fundamental integral that governs how light behaves is:

$$ L_o(p, \omega_o) = L_e(p, \omega_o) + \int_{\Omega} f_r(p, \omega_i, \omega_o) \, L_i(p, \omega_i) \, (\omega_i \cdot n) \, d\omega_i $$

\begin{enumerate}[resume]
    \item \textbf{[2 Marks] The Cosine Term:}
    What physical phenomenon does the term $(\omega_i \cdot n)$ account for?
    \answerbox{1.5cm}

    \item \textbf{[2 Marks] The F Term:}
    What does the term $f_r$ (the BRDF) represent in this equation?
    \answerbox{1.5cm}

    \item \textbf{[2 Marks] The L Term:}
    What does $L_i(p, \omega_i)$ represent?
    \answerbox{1.5cm}

    \item \textbf{[2 Marks] Color Mixing:}
    A ray hits a pure \textbf{Red} wall ($1.0, 0.0, 0.0$) and then bounces to hit a pure \textbf{Blue} wall ($0.0, 0.0, 1.0$). What is the resulting color of that ray?
    \begin{itemize}
        \item[a)] Purple ($0.5, 0.0, 0.5$)
        \item[b)] Black ($0.0, 0.0, 0.0$)
        \item[c)] White ($1.0, 1.0, 1.0$)
    \end{itemize}
\end{enumerate}

\section*{Passage 4: Implementation Details}

\begin{enumerate}[resume]
    \item \textbf{[2 Marks] Fuzz Factor:}
    We implemented metal reflection as: \texttt{ray\_dir = reflected + fuzz * random\_vector}.
    If we want to simulate \textbf{Chalk} (matte), what is the ideal value for \texttt{fuzz}?
    \answerbox{1cm}

    \item \textbf{[3 Marks] Shadows:}
    We did not write any specific code like \texttt{if (is\_blocked) return black}. How did shadows appear in our final render?
    \answerbox{2cm}

    \item \textbf{[3 Marks] Recursion Limit:}
    We set \texttt{\#define MAX\_DEPTH 10}. If we changed this to \texttt{1}, we would essentially see the Sky Gradient where there is nothing. But what would we see where the \textbf{Glass/Mirror} spheres are?
    \answerbox{2cm}

    \item \textbf{[2 Marks] Missing Keyword:}
    In GLSL, we defined \texttt{bool hit\_sphere(Ray r, out Hit rec)}. Why does the image break if we remove the \texttt{out} keyword?
    \answerbox{1.5cm}

    \item \textbf{[2 Marks] Aspect Ratio:}
    If we strictly mapped coordinates from 0 to 1 without correcting for Aspect Ratio, what happens to the spheres on a wide monitor?
    \answerbox{1.5cm}

    \item \textbf{[2 Marks] The Epsilon:}
    In \texttt{world\_hit}, we ignored hits closer than $0.001$. Why not just use $0.0$?
    \answerbox{1.5cm}

    \item \textbf{[3 Marks] Reflection Math:}
    Write the vector formula to calculate the reflection vector $R$, given incoming vector $I$ and normal $N$.
    \answerbox{2cm}

    \item \textbf{[2 Marks] Gamma Correction:}
    We applied \texttt{col = sqrt(col);} at the end. Why is this necessary for monitors?
    \answerbox{1.5cm}
\end{enumerate}

\section*{Passage 5: Rust Comprehension}

\begin{enumerate}[resume]
    \item \textbf{[4 Marks] Iterator Invalidation:}
    The rule in Rust is: \textit{You cannot modify a collection while iterating over it.}
    Explain why the following code is memory unsafe and why the compiler prevents it:
\begin{lstlisting}[language=Rust]
let mut list = vec![1, 2, 3, 4];
for num in &list {
    if *num == 2 {
        list.push(5); // Compiler Error
    }
}
\end{lstlisting}
    \answerbox{2.5cm}

    \item \textbf{[3 Marks] Move Semantics vs Copy:}
    In Rust, complex types (like Strings) "Move" ownership, meaning the old variable becomes invalid. However, primitive types (like integers) "Copy".
    
    Given this code:
\begin{lstlisting}[language=Rust]
fn main() {
    let mut a = 2;
    let mut T = [1, a, 3]; // We put 'a' into an array
    println!("{}", a);     // We try to print 'a'
}
\end{lstlisting}
    
    Does this code compile? If so, what does it print? If not, why?
    \answerbox{2cm}
\end{enumerate}

\end{document}