/// <summary>
/// Returns the optimal squared freedom histogram.
/// </summary>
public static Histogram OptimalSquaredFreedom(int histSize, ICollection distribution)
{
if(distribution.Count < Math.Max(histSize, 2))
{
throw new ArgumentException(Resources.InvalidOperationHistogramNotEnoughPoints);
}
// "values" contains the sorted distribution.
double[] values = new double[distribution.Count];
distribution.CopyTo(values, 0);
Array.Sort(values);
// 'optimalCost[i,k]' contains the optimal costs for an
// histogram with the 'i+1' first values and 'k' buckets.
double[,] optimalCost = new double[values.Length, histSize];
// 'lastBucketIndex[i,k]' contains the index of the first
// value of the last bucket for optimal histogram comprising
// the 'i+1' first values and 'k' buckets.
int[,] lastBucketIndex = new int[values.Length, histSize];
// "One bucket" histograms initialization
for(int i = 0; i < values.Length; i++)
{
optimalCost[i, 0] =
(values[i] - values[0]) * (values[i] - values[0]) * (i + 1);
}
// "One value per bucket" histograms initialization
for(int k = 0; k < histSize; k++)
{
// optimalCost[k, k] = 0;
lastBucketIndex[k, k] = k;
}
// ----- Dynamic programming part -----
// Loop on the number of buckets
// (note that there are 'k+1' buckets)
for(int k = 1; k < histSize; k++)
{
// Loop on the number of considered values
// (note that there are 'i+1' considered values)
for(int i = k; i < values.Length; i++)
{
optimalCost[i, k] = double.PositiveInfinity;
// Loop for finding the optimal boundary of the last bucket
// ('j+1' is the index of the first value in the last bucket)
for(int j = (k - 1), avg = (k - 1); j < i; j++)
{
double currentCost = optimalCost[j, k - 1] +
(values[i] - values[j + 1]) * (values[i] - values[j + 1]) * (i - j);
if(currentCost < optimalCost[i, k])
{
optimalCost[i, k] = currentCost;
lastBucketIndex[i, k] = j + 1;
}
}
}
}
// ----- Reconstitution of the histogram -----
Histogram histogram = new Histogram();
int index = values.Length - 1;
for(int k = (histSize - 1); k >= 0; k--)
{
histogram.Add(new Bucket(
values[lastBucketIndex[index, k]],
values[index],
index - lastBucketIndex[index, k] + 1
));
index = lastBucketIndex[index, k] - 1;
}
return histogram;
}
/// <summary>
/// Returns the optimal dispersion histogram.
/// </summary>
public static Histogram OptimalDispersion(int bucketCount, ICollection distribution)
{
if(distribution.Count < Math.Max(bucketCount, 2))
{
throw new ArgumentException(Resources.InvalidOperationHistogramNotEnoughPoints);
}
// "values" contains the sorted distribution.
double[] values = new double[distribution.Count];
distribution.CopyTo(values, 0);
Array.Sort(values);
// 'optimalCost[i,k]' contains the optimal costs for an
// histogram with the 'i+1' first values and 'k' buckets.
double[,] optimalCost = new double[values.Length, bucketCount];
// 'lastBucketIndex[i,k]' contains the index of the first
// value of the last bucket for optimal histogram comprising
// the 'i+1' first values and 'k' buckets.
int[,] lastBucketIndex = new int[values.Length, bucketCount];
// 'prefixSum[i]' contains the sum of the 'i-1' first values.
double[] prefixSum = new double[values.Length + 1];
// Initialization of the prefix sums
for(int i = 0; i < values.Length; i++)
{
prefixSum[i + 1] = prefixSum[i] + values[i];
}
// "One bucket" histograms initialization
for(int i = 0, avg = 0; i < values.Length; i++)
{
while((avg + 1) < values.Length &&
values[avg + 1] < prefixSum[i + 1] / (i + 1))
{
avg++;
}
optimalCost[i, 0] =
prefixSum[i + 1] - 2 * prefixSum[avg + 1]
+ (2 * avg - i + 1) * (prefixSum[i + 1] / (i + 1));
}
// "One value per bucket" histograms initialization
for(int k = 0; k < bucketCount; k++)
{
// optimalCost[k, k] = 0;
lastBucketIndex[k, k] = k;
}
// ----- Dynamic programming part -----
// Loop on the number of buckets
// (note that there are 'k+1' buckets)
for(int k = 1; k < bucketCount; k++)
{
// Loop on the number of considered values
// (note that there are 'i+1' considered values)
for(int i = k; i < values.Length; i++)
{
optimalCost[i, k] = double.PositiveInfinity;
// Loop for finding the optimal boundary of the last bucket
// ('j+1' is the index of the first value in the last bucket)
for(int j = (k - 1), avg = (k - 1); j < i; j++)
{
while((avg + 1) < values.Length &&
values[avg + 1] < (prefixSum[i + 1] - prefixSum[j + 1]) / (i - j))
{
avg++;
}
double currentCost = optimalCost[j, k - 1] +
prefixSum[i + 1] + prefixSum[j + 1] - 2 * prefixSum[avg + 1]
+ (2 * avg - i - j) * (prefixSum[i + 1] - prefixSum[j + 1]) / (i - j);
if(currentCost < optimalCost[i, k])
{
optimalCost[i, k] = currentCost;
lastBucketIndex[i, k] = j + 1;
}
}
}
}
// ----- Reconstitution of the histogram -----
Histogram histogram = new Histogram();
int index = values.Length - 1;
for(int k = (bucketCount - 1); k >= 0; k--)
{
histogram.Add(new Bucket(
values[lastBucketIndex[index, k]],
values[index],
index - lastBucketIndex[index, k] + 1
));
index = lastBucketIndex[index, k] - 1;
}
return histogram;
}
public static Histogram OptimalSquaredFreedom(int histSize, ICollection distribution)
{
if (distribution.Count < Math.Max(histSize, 2))
{
throw new ArgumentException("Not enough points in the distribution.");
}
// "values" contains the sorted distribution.
double[] values = new double[distribution.Count];
distribution.CopyTo(values, 0);
Array.Sort(values);
// 'optimalCost[i,k]' contains the optimal costs for an
// histogram with the 'i+1' first values and 'k' buckets.
double[,] optimalCost = new double[values.Length, histSize];
// 'lastBucketIndex[i,k]' contains the index of the first
// value of the last bucket for optimal histogram comprising
// the 'i+1' first values and 'k' buckets.
int[,] lastBucketIndex = new int[values.Length, histSize];
// "One bucket" histograms initialization
for (int i = 0; i < values.Length; i++)
{
optimalCost[i, 0] =
(values[i] - values[0]) * (values[i] - values[0]) * (i + 1);
}
// "One value per bucket" histograms initialization
for (int k = 0; k < histSize; k++)
{
// optimalCost[k, k] = 0;
lastBucketIndex[k, k] = k;
}
// ----- Dynamic programming part -----
// Loop on the number of buckets
// (note that there are 'k+1' buckets)
for (int k = 1; k < histSize; k++)
{
// Loop on the number of considered values
// (note that there are 'i+1' considered values)
for (int i = k; i < values.Length; i++)
{
optimalCost[i, k] = double.PositiveInfinity;
// Loop for finding the optimal boundary of the last bucket
// ('j+1' is the index of the first value in the last bucket)
for (int j = (k - 1), avg = (k - 1); j < i; j++)
{
double currentCost = optimalCost[j, k - 1] +
(values[i] - values[j + 1]) * (values[i] - values[j + 1]) * (i - j);
if (currentCost < optimalCost[i, k])
{
optimalCost[i, k] = currentCost;
lastBucketIndex[i, k] = j + 1;
}
}
}
}
// ----- Reconstitution of the histogram -----
Histogram histogram = new Histogram();
int index = values.Length - 1;
for (int k = (histSize - 1); k >= 0; k--)
{
histogram.Add(new Bucket(values[lastBucketIndex[index, k]],
values[index], index - lastBucketIndex[index, k] + 1));
index = lastBucketIndex[index, k] - 1;
}
//histogram.JoinBuckets();
return(histogram);
}
/// <summary>Returns the optimal variance histogram.</summary>
/// <param name="histSize">The number of buckets in the histogram.</param>
/// <param name="distribution"><c>double</c> elements expected.</param>
/// <remarks>Requires a computations time quadratic to
/// <c>distribution.Length</c>.</remarks>
public static Histogram OptimalVariance(int bucketCount, ICollection distribution)
{
if (distribution.Count < bucketCount)
{
throw new ArgumentException("Not enough points in the distribution.");
}
// "values" contains the sorted distribution.
double[] values = new double[distribution.Count];
distribution.CopyTo(values, 0);
Array.Sort(values);
// 'optimalCost[i,k]' contains the optimal costs for an
// histogram with the 'i+1' first values and 'k' buckets.
double[,] optimalCost = new double[values.Length, bucketCount];
// 'lastBucketIndex[i,k]' contains the index of the first
// value of the last bucket for optimal histogram comprising
// the 'i+1' first values and 'k' buckets.
int[,] lastBucketIndex = new int[values.Length, bucketCount];
// 'prefixSum[i]' contains the sum of the 'i-1' first values.
double[] prefixSum = new double[values.Length + 1],
// 'sqPrefixSum' contains the sum of the 'i-1' first squared values.
sqPrefixSum = new double[values.Length + 1];
// Initialization of the prefix sums
for (int i = 0; i < values.Length; i++)
{
prefixSum[i + 1] = prefixSum[i] + values[i];
sqPrefixSum[i + 1] = sqPrefixSum[i] + values[i] * values[i];
}
// "One bucket" histograms initialization
for (int i = 0; i < values.Length; i++)
{
optimalCost[i, 0] = sqPrefixSum[i + 1] -
prefixSum[i + 1] * prefixSum[i + 1] / (i + 1);
}
// "One value per bucket" histograms initialization
for (int k = 0; k < bucketCount; k++)
{
// optimalCost[k, k] = 0;
lastBucketIndex[k, k] = k;
}
// ----- Dynamic programming part -----
// Loop on the number of buckets
// (note that there are 'k+1' buckets)
for (int k = 1; k < bucketCount; k++)
{
// Loop on the number of considered values
// (note that there are 'i+1' considered values)
for (int i = k; i < values.Length; i++)
{
optimalCost[i, k] = double.PositiveInfinity;
// Loop for finding the optimal boundary of the last bucket
// ('j+1' is the index of the first value in the last bucket)
for (int j = (k - 1); j < i; j++)
{
double currentCost = optimalCost[j, k - 1] + sqPrefixSum[i + 1] - sqPrefixSum[j + 1]
- (prefixSum[i + 1] - prefixSum[j + 1]) * (prefixSum[i + 1] - prefixSum[j + 1]) / (i - j);
if (currentCost < optimalCost[i, k])
{
optimalCost[i, k] = currentCost;
lastBucketIndex[i, k] = j + 1;
}
}
}
}
// ----- Reconstitution of the histogram -----
Histogram histogram = new Histogram();
int index = values.Length - 1;
for (int k = (bucketCount - 1); k >= 0; k--)
{
histogram.Add(new Bucket(values[lastBucketIndex[index, k]],
values[index], index - lastBucketIndex[index, k] + 1));
index = lastBucketIndex[index, k] - 1;
}
//histogram.JoinBuckets();
return(histogram);
}
/// <summary>
/// Returns the optimal dispersion histogram.
/// </summary>
public static Histogram OptimalDispersion(int bucketCount, ICollection distribution)
{
if (distribution.Count < Math.Max(bucketCount, 2))
{
throw new ArgumentException(Resources.InvalidOperationHistogramNotEnoughPoints);
}
// "values" contains the sorted distribution.
double[] values = new double[distribution.Count];
distribution.CopyTo(values, 0);
Array.Sort(values);
// 'optimalCost[i,k]' contains the optimal costs for an
// histogram with the 'i+1' first values and 'k' buckets.
double[,] optimalCost = new double[values.Length, bucketCount];
// 'lastBucketIndex[i,k]' contains the index of the first
// value of the last bucket for optimal histogram comprising
// the 'i+1' first values and 'k' buckets.
int[,] lastBucketIndex = new int[values.Length, bucketCount];
// 'prefixSum[i]' contains the sum of the 'i-1' first values.
double[] prefixSum = new double[values.Length + 1];
// Initialization of the prefix sums
for (int i = 0; i < values.Length; i++)
{
prefixSum[i + 1] = prefixSum[i] + values[i];
}
// "One bucket" histograms initialization
for (int i = 0, avg = 0; i < values.Length; i++)
{
while ((avg + 1) < values.Length &&
values[avg + 1] < prefixSum[i + 1] / (i + 1))
{
avg++;
}
optimalCost[i, 0] =
prefixSum[i + 1] - 2 * prefixSum[avg + 1]
+ (2 * avg - i + 1) * (prefixSum[i + 1] / (i + 1));
}
// "One value per bucket" histograms initialization
for (int k = 0; k < bucketCount; k++)
{
// optimalCost[k, k] = 0;
lastBucketIndex[k, k] = k;
}
// ----- Dynamic programming part -----
// Loop on the number of buckets
// (note that there are 'k+1' buckets)
for (int k = 1; k < bucketCount; k++)
{
// Loop on the number of considered values
// (note that there are 'i+1' considered values)
for (int i = k; i < values.Length; i++)
{
optimalCost[i, k] = double.PositiveInfinity;
// Loop for finding the optimal boundary of the last bucket
// ('j+1' is the index of the first value in the last bucket)
for (int j = (k - 1), avg = (k - 1); j < i; j++)
{
while ((avg + 1) < values.Length &&
values[avg + 1] < (prefixSum[i + 1] - prefixSum[j + 1]) / (i - j))
{
avg++;
}
double currentCost = optimalCost[j, k - 1] +
prefixSum[i + 1] + prefixSum[j + 1] - 2 * prefixSum[avg + 1]
+ (2 * avg - i - j) * (prefixSum[i + 1] - prefixSum[j + 1]) / (i - j);
if (currentCost < optimalCost[i, k])
{
optimalCost[i, k] = currentCost;
lastBucketIndex[i, k] = j + 1;
}
}
}
}
// ----- Reconstitution of the histogram -----
Histogram histogram = new Histogram();
int index = values.Length - 1;
for (int k = (bucketCount - 1); k >= 0; k--)
{
histogram.Add(new Bucket(
values[lastBucketIndex[index, k]],
values[index],
index - lastBucketIndex[index, k] + 1
));
index = lastBucketIndex[index, k] - 1;
}
return(histogram);
}
/// <summary>Returns the optimal variance histogram.</summary>
/// <param name="histSize">The number of buckets in the histogram.</param>
/// <param name="distribution"><c>double</c> elements expected.</param>
/// <remarks>Requires a computations time quadratic to
/// <c>distribution.Length</c>.</remarks>
public static Histogram OptimalVariance(int bucketCount, ICollection distribution)
{
if (distribution.Count < bucketCount)
throw new ArgumentException("Not enough points in the distribution.");
// "values" contains the sorted distribution.
double[] values = new double[distribution.Count];
distribution.CopyTo(values, 0);
Array.Sort(values);
// 'optimalCost[i,k]' contains the optimal costs for an
// histogram with the 'i+1' first values and 'k' buckets.
double[,] optimalCost = new double[values.Length, bucketCount];
// 'lastBucketIndex[i,k]' contains the index of the first
// value of the last bucket for optimal histogram comprising
// the 'i+1' first values and 'k' buckets.
int[,] lastBucketIndex = new int[values.Length, bucketCount];
// 'prefixSum[i]' contains the sum of the 'i-1' first values.
double[] prefixSum = new double[values.Length + 1],
// 'sqPrefixSum' contains the sum of the 'i-1' first squared values.
sqPrefixSum = new double[values.Length + 1];
// Initialization of the prefix sums
for (int i = 0; i < values.Length; i++)
{
prefixSum[i + 1] = prefixSum[i] + values[i];
sqPrefixSum[i + 1] = sqPrefixSum[i] + values[i] * values[i];
}
// "One bucket" histograms initialization
for (int i = 0; i < values.Length; i++)
optimalCost[i, 0] = sqPrefixSum[i + 1] -
prefixSum[i + 1] * prefixSum[i + 1] / (i + 1);
// "One value per bucket" histograms initialization
for (int k = 0; k < bucketCount; k++)
{
// optimalCost[k, k] = 0;
lastBucketIndex[k, k] = k;
}
// ----- Dynamic programming part -----
// Loop on the number of buckets
// (note that there are 'k+1' buckets)
for (int k = 1; k < bucketCount; k++)
// Loop on the number of considered values
// (note that there are 'i+1' considered values)
for (int i = k; i < values.Length; i++)
{
optimalCost[i, k] = double.PositiveInfinity;
// Loop for finding the optimal boundary of the last bucket
// ('j+1' is the index of the first value in the last bucket)
for (int j = (k - 1); j < i; j++)
{
double currentCost = optimalCost[j, k - 1] + sqPrefixSum[i + 1] - sqPrefixSum[j + 1]
- (prefixSum[i + 1] - prefixSum[j + 1]) * (prefixSum[i + 1] - prefixSum[j + 1]) / (i - j);
if (currentCost < optimalCost[i, k])
{
optimalCost[i, k] = currentCost;
lastBucketIndex[i, k] = j + 1;
}
}
}
// ----- Reconstitution of the histogram -----
Histogram histogram = new Histogram();
int index = values.Length - 1;
for (int k = (bucketCount - 1); k >= 0; k--)
{
histogram.Add(new Bucket(values[lastBucketIndex[index, k]],
values[index], index - lastBucketIndex[index, k] + 1));
index = lastBucketIndex[index, k] - 1;
}
//histogram.JoinBuckets();
return histogram;
}