public void GetProbabilityValuesForHistogramTest()
{
// We will pass in a sorted sequence of 10 pairs of values, each pair has an iterator
// numbered 0-9, and a corresponding value between 0-4, which repeats after each increment
// of five by the iterator, so each pair has one duplicate value in the sequence
// (e.g., (0,0),(1,1),(2,2),...(5,0),(6,1),(7,2),...(9,4) ).
// The GetProbabilityValues method calculates the number of packet counts (the pair values),
// then calculates the probability for each pair value, and returns a sorted sequence of
// pairs with the probability for each pair value.
// So we should get back a sequence of five pairs with the probabilities for the original
// pair values. For example, since the pair values total 20, the probability for a value
// of 2 is 2/20 = 0.1. So the returned sequence should be (0,0),(1,0.1),(2,0.2),(3,0.3),(4,0.4).
// Arrange
Dictionary<int, int> histogram = new Dictionary<int, int>();
for (int i = 0; i < 10; i++)
{
histogram.Add(i,i % 5);
}
CalculateProbability cp = new CalculateProbability(histogram);
// Act
Dictionary<int, decimal> sortedProbabilities = new Dictionary<int, decimal>();
sortedProbabilities = cp.GetProbabilityValues();
bool results = true;
if(sortedProbabilities.Count != 5)
{
results = false;
}
decimal expectedProbability = 0M;
foreach (var pair in sortedProbabilities)
{
if(pair.Value != expectedProbability)
{
results = false;
}
expectedProbability = expectedProbability + 0.1M;
}
// Assert
Assert.IsTrue(results);
}
