LightBoard free response 3

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.

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

2019 AP CS Exam Free Response Answers

3 thoughts on “LightBoard free response

  1. Richard May 21,2019 5:37 pm

    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 May 21,2019 6:51 pm

      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 May 28,2019 10:45 am

    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

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