osu/osu.Game/Beatmaps/IBeatmapDifficultyInfo.cs

115 lines
5.2 KiB
C#

// Copyright (c) ppy Pty Ltd <contact@ppy.sh>. Licensed under the MIT Licence.
// See the LICENCE file in the repository root for full licence text.
using System;
namespace osu.Game.Beatmaps
{
/// <summary>
/// A representation of all top-level difficulty settings for a beatmap.
/// </summary>
public interface IBeatmapDifficultyInfo
{
/// <summary>
/// The default value used for all difficulty settings except <see cref="SliderMultiplier"/> and <see cref="SliderTickRate"/>.
/// </summary>
const float DEFAULT_DIFFICULTY = 5;
/// <summary>
/// The drain rate of the associated beatmap.
/// </summary>
float DrainRate { get; }
/// <summary>
/// The circle size of the associated beatmap.
/// </summary>
float CircleSize { get; }
/// <summary>
/// The overall difficulty of the associated beatmap.
/// </summary>
float OverallDifficulty { get; }
/// <summary>
/// The approach rate of the associated beatmap.
/// </summary>
float ApproachRate { get; }
/// <summary>
/// The base slider velocity of the associated beatmap.
/// This was known as "SliderMultiplier" in the .osu format and stable editor.
/// </summary>
double SliderMultiplier { get; }
/// <summary>
/// The slider tick rate of the associated beatmap.
/// </summary>
double SliderTickRate { get; }
/// <summary>
/// Maps a difficulty value [0, 10] to a two-piece linear range of values.
/// </summary>
/// <param name="difficulty">The difficulty value to be mapped.</param>
/// <param name="min">Minimum of the resulting range which will be achieved by a difficulty value of 0.</param>
/// <param name="mid">Midpoint of the resulting range which will be achieved by a difficulty value of 5.</param>
/// <param name="max">Maximum of the resulting range which will be achieved by a difficulty value of 10.</param>
/// <returns>Value to which the difficulty value maps in the specified range.</returns>
static double DifficultyRange(double difficulty, double min, double mid, double max)
{
if (difficulty > 5)
return mid + (max - mid) * DifficultyRange(difficulty);
if (difficulty < 5)
return mid + (mid - min) * DifficultyRange(difficulty);
return mid;
}
/// <summary>
/// Maps a difficulty value [0, 10] to a linear range of [-1, 1].
/// </summary>
/// <param name="difficulty">The difficulty value to be mapped.</param>
/// <returns>Value to which the difficulty value maps in the specified range.</returns>
static double DifficultyRange(double difficulty) => (difficulty - 5) / 5;
/// <summary>
/// Maps a difficulty value [0, 10] to a two-piece linear range of values.
/// </summary>
/// <param name="difficulty">The difficulty value to be mapped.</param>
/// <param name="range">The values that define the two linear ranges.
/// <list type="table">
/// <item>
/// <term>od0</term>
/// <description>Minimum of the resulting range which will be achieved by a difficulty value of 0.</description>
/// </item>
/// <item>
/// <term>od5</term>
/// <description>Midpoint of the resulting range which will be achieved by a difficulty value of 5.</description>
/// </item>
/// <item>
/// <term>od10</term>
/// <description>Maximum of the resulting range which will be achieved by a difficulty value of 10.</description>
/// </item>
/// </list>
/// </param>
/// <returns>Value to which the difficulty value maps in the specified range.</returns>
static double DifficultyRange(double difficulty, (double od0, double od5, double od10) range)
=> DifficultyRange(difficulty, range.od0, range.od5, range.od10);
/// <summary>
/// Inverse function to <see cref="DifficultyRange(double,double,double,double)"/>.
/// Maps a value returned by the function above back to the difficulty that produced it.
/// </summary>
/// <param name="difficultyValue">The difficulty-dependent value to be unmapped.</param>
/// <param name="diff0">Minimum of the resulting range which will be achieved by a difficulty value of 0.</param>
/// <param name="diff5">Midpoint of the resulting range which will be achieved by a difficulty value of 5.</param>
/// <param name="diff10">Maximum of the resulting range which will be achieved by a difficulty value of 10.</param>
/// <returns>Value to which the difficulty value maps in the specified range.</returns>
static double InverseDifficultyRange(double difficultyValue, double diff0, double diff5, double diff10)
{
return Math.Sign(difficultyValue - diff5) == Math.Sign(diff10 - diff5)
? (difficultyValue - diff5) / (diff10 - diff5) * 5 + 5
: (difficultyValue - diff5) / (diff5 - diff0) * 5 + 5;
}
}
}