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TrustPilotChallenge/WhiteRabbit/VectorsProcessor.cs

199 lines
8.4 KiB

namespace WhiteRabbit
{
using System;
using System.Collections.Generic;
using System.Collections.Immutable;
using System.Linq;
using System.Numerics;
internal sealed class VectorsProcessor
{
private const byte MaxComponentValue = 8;
private const int LeastCommonMultiple = 840;
// Ensure that permutations are precomputed prior to main run, so that processing times will be correct
static VectorsProcessor()
{
PrecomputedPermutationsGenerator.HamiltonianPermutations(0);
}
public VectorsProcessor(Vector<byte> target, int maxVectorsCount, Vector<byte>[] dictionary)
{
if (Enumerable.Range(0, Vector<byte>.Count).Any(i => target[i] > MaxComponentValue))
{
throw new ArgumentException($"Every value should be at most {MaxComponentValue} (at most {MaxComponentValue} same characters allowed in the source string)", nameof(target));
}
this.Target = target;
this.MaxVectorsCount = maxVectorsCount;
this.Dictionary = ImmutableArray.Create(FilterVectors(dictionary, target).ToArray());
}
private Vector<byte> Target { get; }
private int MaxVectorsCount { get; }
private ImmutableArray<VectorInfo> Dictionary { get; }
// Produces all sequences of vectors with the target sum
#if SINGLE_THREADED
public IEnumerable<int[]> GenerateSequences()
#else
public ParallelQuery<int[]> GenerateSequences()
#endif
{
return GenerateUnorderedSequences(this.Target, GetVectorNorm(this.Target, this.Target), this.MaxVectorsCount, this.Dictionary, 0)
#if !SINGLE_THREADED
.AsParallel()
#endif
.Select(Enumerable.ToArray)
.SelectMany(GeneratePermutations);
}
// We want words with more letters (and among these, words with more "rare" letters) to appear first, to reduce the searching time somewhat.
// Applying such a sort, we reduce the total number of triplets to check for anagrams from ~62M to ~29M.
// Total number of quadruplets is reduced from 1468M to mere 311M.
// And total number of quintuplets becomes reasonable 1412M.
// Also, it produces the intended results faster (as these are more likely to contain longer words - e.g. "poultry outwits ants" is more likely than "p o u l t r y o u t w i t s a n t s").
// This method basically gives us the 1-norm of the vector in the space rescaled so that the target is [1, 1, ..., 1].
private static int GetVectorNorm(Vector<byte> vector, Vector<byte> target)
{
var norm = 0;
for (var i = 0; target[i] != 0; i++)
{
norm += (LeastCommonMultiple * vector[i]) / target[i];
}
return norm;
}
private static VectorInfo[] FilterVectors(Vector<byte>[] vectors, Vector<byte> target)
{
return Enumerable.Range(0, vectors.Length)
.Where(i => Vector.GreaterThanOrEqualAll(target, vectors[i]))
.Select(i => new VectorInfo(vectors[i], GetVectorNorm(vectors[i], target), i))
.OrderByDescending(vectorInfo => vectorInfo.Norm)
.ToArray();
}
// This method takes most of the time, so everything related to it must be optimized.
// In every sequence, next vector always goes after the previous one from dictionary.
// E.g. if dictionary is [x, y, z], then only [x, y] sequence could be generated, and [y, x] will never be generated.
// That way, the complexity of search goes down by a factor of MaxVectorsCount! (as if [x, y] does not add up to a required target, there is no point in checking [y, x])
private static IEnumerable<ImmutableStack<int>> GenerateUnorderedSequences(Vector<byte> remainder, int remainderNorm, int allowedRemainingWords, ImmutableArray<VectorInfo> dictionary, int currentDictionaryPosition)
{
if (allowedRemainingWords > 1)
{
var newAllowedRemainingWords = allowedRemainingWords - 1;
// E.g. if remainder norm is 7, 8 or 9, and allowedRemainingWords is 3,
// we need the largest remaining word to have a norm of at least 3
var requiredRemainderPerWord = (remainderNorm + allowedRemainingWords - 1) / allowedRemainingWords;
for (var i = FindFirstWithNormLessOrEqual(remainderNorm, dictionary, currentDictionaryPosition); i < dictionary.Length; i++)
{
var currentVectorInfo = dictionary[i];
if (currentVectorInfo.Vector == remainder)
{
yield return ImmutableStack.Create(currentVectorInfo.Index);
}
else if (currentVectorInfo.Norm < requiredRemainderPerWord)
{
break;
}
else if (Vector.LessThanOrEqualAll(currentVectorInfo.Vector, remainder))
{
var newRemainder = remainder - currentVectorInfo.Vector;
var newRemainderNorm = remainderNorm - currentVectorInfo.Norm;
foreach (var result in GenerateUnorderedSequences(newRemainder, newRemainderNorm, newAllowedRemainingWords, dictionary, i))
{
yield return result.Push(currentVectorInfo.Index);
}
}
}
}
else
{
for (var i = FindFirstWithNormLessOrEqual(remainderNorm, dictionary, currentDictionaryPosition); i < dictionary.Length; i++)
{
var currentVectorInfo = dictionary[i];
if (currentVectorInfo.Vector == remainder)
{
yield return ImmutableStack.Create(currentVectorInfo.Index);
}
else if (currentVectorInfo.Norm < remainderNorm)
{
break;
}
}
}
}
// BCL BinarySearch would find any vector with required norm, not the first one; or would find nothing if there is no such vector
private static int FindFirstWithNormLessOrEqual(int expectedNorm, ImmutableArray<VectorInfo> dictionary, int offset)
{
var start = offset;
var end = dictionary.Length - 1;
if (dictionary[start].Norm <= expectedNorm)
{
return start;
}
if (dictionary[end].Norm > expectedNorm)
{
return dictionary.Length;
}
// Norm for start is always greater than expected norm, or start is the required position; norm for end is always less than or equal to expected norm
// The loop always ends, because the difference always decreases; if start + 1 = end, then middle will be equal to start, and either end := middle = start or start := middle + 1 = end.
while (start < end)
{
var middle = (start + end) / 2;
var newNorm = dictionary[middle].Norm;
if (dictionary[middle].Norm <= expectedNorm)
{
end = middle;
}
else
{
start = middle + 1;
}
}
return start;
}
private static IEnumerable<T[]> GeneratePermutations<T>(T[] original)
{
var length = original.Length;
foreach (var permutation in PrecomputedPermutationsGenerator.HamiltonianPermutations(length))
{
var result = new T[length];
for (var i = 0; i < length; i++)
{
result[i] = original[permutation[i]];
}
yield return result;
}
}
private struct VectorInfo
{
public VectorInfo(Vector<byte> vector, int norm, int index)
{
this.Vector = vector;
this.Norm = norm;
this.Index = index;
}
public Vector<byte> Vector { get; }
public int Norm { get; }
public int Index { get; }
}
}
}