How many squares are drawn if foo(0,0,5,2) is called? The provided text describes a function in an educational context regarding computer programming and graphics.---**Explanation of the Function:**The text defines a recursive function `foo` that uses an auxiliary function called `drawSquare`. The `drawSquare(double x, double y, double halfLen)` function is responsible for drawing a square centered at coordinates `(x, y)` with a side length that is double the given `halfLen`. For instance, `drawSquare(0,0,5)` would draw a square centered at the origin `(0,0)` with a side length of `10`.**Detailed Breakdown of the Code:**```javapublic static void foo(double x, double y, double len, int n) { if (n < 1) return; drawSquare(x, y, len); foo(x - len/2, y - len/2, len/2, n - 1); foo(x - len/2, y + len/2, len/2, n - 1); foo(x + len/2, y + len/2, len/2, n - 1); foo(x + len/2, y - len/2, len/2, n - 1);}```**Function Behavior:**- **Base Case:** The recursion stops when `n < 1`. In this case, the function simply returns without doing anything. - **Recursive Case:** - First, it draws a square centered at `(x, y)` with side length `len`. - Then, it makes four recursive calls to itself, each time subdividing the current square into four smaller squares and recentering. - The square's center for the recursive calls is adjusted by half of the current length (`len/2`) to cover all four quadrants: - Bottom-left - Top-left - Top-right - Bottom-rightThis method is often used to create fractal patterns or divide graphical space efficiently.--- This explanation provides a detailed view of the recursive structure and graphical operations performed by the `foo` function, essential for understanding recursion and graphical transformations in programming.

Given, drawSquare (double x, double y, double halfLen) draws a square with each side of length 2×halfLen with centre at (x,y). foo(double x, double y, double len, int n) { if (n<1) return; drawSquare (x, y, len); foo (x-len/2, y-len/2, len/2, n-1); foo (x-len/2, y+len/2, len/2, n-1); foo (x+len/2, y+len/2, len/2, n-1); foo (x+len/2, y-len/2, len/2, n-1); }foo (0, 0, 5, 2) Here, x=0, y=0, len 5, n=2 n>1 so the if condition is invalid. drawSquare (0, 0, 5); This function draws a square with each side of length 2×5=10 and centre at (0,0). foo (0-5÷2, 0-5/2, 5÷2, 2-1); ⇒foo (-2.5, -2.5, 2.5, 1); foo (-2.5, 2.5, 2.5, 1); foo (2.5, 2.5, 2.5, 1); foo (2.5, -2.5, 2.5, 1);foo (-2.5, -2.5, 2.5, 1) Here, x=-2.5, y=-2.5, len=2.5, n=1 Since n=1, the if condition is not valid. drawSquare (-2.5,-2.5, 2.5); This function draws a square with each side of length 2×2.5=5 and centre at (-2.5,-2.5). foo (-3.75, -3.75, 1.25, 0); foo (-3.75, -1.25, 1.25, 0); foo (-1.25, -1.25, 1.25, 0); foo (-1.25, -3.75, 1.25, 0); Since n<0, the above 4 foo functions will not draw any square. Similarly, foo (-2.5, 2.5, 2.5, 1), foo (2.5, 2.5, 2.5, 1) and foo (2.5, -2.5, 2.5, 1) will also draw 1 square each. So, the total number of squares drawn if foo(0,0,5,2) is called = 1+1+1+1+1=5. Answer. 5.