# LightBoard free response3

LightBoard free response problem from the 2019 AP Computer Science A Exam.

LightBoard is #4 from the from the 2019 AP Computer Science A Free Response problems.

https://apcentral.collegeboard.org/pdf/ap19-frq-computer-science-a.pdf?course=ap-computer-science-a

## Part (a) – LightBoard constructor

```public LightBoard(int numRows, int numCols)
{
lights = new boolean[numRows][numCols];

for(int r = 0; r < lights.length; r++)
for(int c = 0; c < lights[0].length; c++)
if(Math.random() <= 0.4)
lights[r][c] = true;
}
```

## Part (b) – evaluateLight method

```public boolean evaluateLight(int row, int col)
{
int onInColumn = 0;

for(int r = 0; r < lights.length; r++)
if(lights[r][col])
onInColumn++;

if(lights[row][col])
{
if(onInColumn % 2 == 0)
return false;
}
else
{
if(onInColumn % 3 == 0)
return true;
}

return lights[row][col];
}
```

## 3 thoughts on “LightBoard free response”

1. Richard

Hey Mr. Horn,
I’m curious as to why in part (a) you would have to write “less than or equal to .4” because I thought that the random number generator would generate a random double with a value of 0.0 up to but not including 1.0. If you included .4, wouldn’t you be giving the light a 41% chance of being turned on instead of a 40% chance or does the random generator include the value of 1.0?

• Brandon Horn

Hi Richard,

I’m not convinced that < vs <= matters. Less than would catch 0.399… and less than or equal to would also catch 0.4. I think the difference is insignificant.

2. Angel Perez

Hello Mr. Horn,

I actually have the same answer as yours (including equality) and experienced a variety of answers from my students (as well as from other teachers). I look forward to the AP reading this coming week to hear more about this.

Best regards,

Mr. Perez