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