{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# 用python告诉你为什么十二平均律有12个音" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "作为一名学音乐的艺术生，为什么平均律现行的是12个音而不是其他的选择，我们需要一个合理的解释。后现代解构主义喜欢把所有的锅都甩给“文化上下文”跟“历史的偶然性”，这是很不负责任的表现。\n", "\n", "![](res/人文三连.svg)\n", "\n", "本文的目的是通过简单的数学和python可视化工具来解释为什么平均律是12个音。以下是正文。\n", "\n", "任何数学工具都要从人的需求开始，因此我们先从音乐的角度说人话。\n", "\n", "从听感的角度，我们将音程分为两种：\n", "\n", "- 和谐音程\n", "- 不和谐音程\n", "\n", "这里我们有了第一个需求：**和谐音程比不和谐音程有更高的优先级，也就是说我们优先考虑让和谐音程尽量和谐**。\n", "\n", "因为我们的生理构造以及各种神秘的原因，和谐音程对应的都是这种形式的比例：\n", "\n", "$$\\frac{k+1}{k}, k = \\{1,2,3,...\\}$$\n", "\n", "例如：\n", "\n", "- $\\frac21$ = 纯八度\n", "- $\\frac32$ = 纯五度\n", "- $\\frac43$ = 纯四度\n", "- $\\frac54$ = 大三度\n", "- $\\frac65$ = 小三度" ] }, { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [], "source": [ "from enum import Enum\n", "class INTERVALS(Enum):\n", " P8 = 2/1\n", " P5 = 3/2\n", " P4 = 4/3\n", " M3 = 5/4\n", " m3 = 6/5" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "因此我们想要一个n-平均律，使得这一平均律最好地照顾到这些和谐音程。\n", "\n", "本文的目的就是解释为什么当$n=12$的时候是最优解。\n", "\n", "## 定义损失函数\n", "\n", "怎么样算“照顾到这些和谐音程”？我们需要找到一种衡量n-平均律跟这些和谐音程之间的“距离”，来判定n-平均律的优劣。\n", "\n", "当我们定义了一个n-平均律，我们将八度分为了n等份，同时也生成了n个音程：\n", "\n", "- 第1个音程：$2^{\\frac1n}$\n", "- 第2个音程：$2^{\\frac2n}$\n", "- ...\n", "- 第i个音程：$2^{\\frac{i}{n}}$\n", "\n", "我们想要的就是：**对于任何一个和谐音程，存在一个n平均律中的音程，使得这两个音程之间的距离尽量小**。\n", "\n", "例如，我们熟悉的12平均律（$n=12$），对于纯五度，我们可以找到12平均律中的第7个音程：\n", "\n", "$$2^{\\frac{7}{12}} = 1.4983070768766815$$\n", "\n", "而这跟理想中的纯五度\n", "\n", "$$\\frac32 = 1.5$$\n", "\n", "非常接近。" ] }, { "cell_type": "code", "execution_count": 2, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "0.0016929231233184794" ] }, "execution_count": 2, "metadata": {}, "output_type": "execute_result" } ], "source": [ "import numpy as np\n", "np.abs(2**(7/12) - 1.5)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "因此定义损失函数非常简单：遍历上文枚举出的和谐音程，对于任何一个和谐音程，找出在n-平均律中离这一和谐音程“距离”最近的一个音程然后计算“距离”。\n", "\n", "既然上文提到了“距离”，我们可以很直观地直接用欧氏距离（毕竟从几何到统计学的MSE都在用这个）：\n", "\n", "$$d(\\vec x,\\vec y) = \\sqrt{(x_1 - y_1)^2 + ... + (x_k - y_k)^2}$$\n", "\n", "也就是说\n", "\n", "$$\\text{损失函数}(\\text{n-平均律}) = \\sqrt{(\\text{纯八度跟最接近的音程的距离})^2 + (\\text{纯五度跟最接近的音程的距离})^2 + ...}$$" ] }, { "cell_type": "code", "execution_count": 3, "metadata": {}, "outputs": [], "source": [ "def loss(n):\n", " loss = 0\n", " for interval in INTERVALS:\n", " x = 2**(np.arange(0,n + 1) / n) - interval.value\n", " x = np.abs(x)\n", " x = np.min(x)\n", " x = np.square(x)\n", " loss = loss + x\n", " return np.sqrt(loss)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "我们现在来看一下效果。" ] }, { "cell_type": "code", "execution_count": 4, "metadata": {}, "outputs": [], "source": [ "import matplotlib.pyplot as plt\n", "import pandas as pd" ] }, { "cell_type": "code", "execution_count": 5, "metadata": {}, "outputs": [ { "data": { "image/png": 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" ] }, "metadata": { "needs_background": "light" }, "output_type": "display_data" } ], "source": [ "x = np.arange(2,25)\n", "y = np.array(list(map(loss,x)))\n", "data = pd.DataFrame({'n':x,'loss':y})\n", "\n", "fig, ax = plt.subplots()\n", "ax.set_xticks(data['n'].head(20))\n", "ax.bar(data['n'].head(20),data['loss'].head(20))\n", "ax.set(xlabel='n', ylabel='loss',\n", " title='n-equal temperament')\n", "ax.grid()\n", "plt.show()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 进阶：两个可以考虑的问题\n", "\n", "从上面已经可以看出来12对应着一个极小值，这也解释了为什么我们会选择12。\n", "\n", "然而美中不足的是：19一样也很小，也就是说，19这一竞争者的存在让我们看不到12独特的光辉。\n", "\n", "接下来我们可以用一个邪恶的招数：**通过修改“比赛规则”让12成为内定的冠军**。\n", "\n", "### 不同的音程并非生而平等\n", "\n", "我们可以提出第二个（伪）需求：**不同的和谐音程重要程度也不一样**。例如，或许我们觉得顾及纯五度比顾及小三度更加重要，也就是说我们更需要P5精准而不是m3。\n", "\n", "从数学的角度来说，我们需要的是加入权重$\\alpha_1,...,\\alpha_k$。\n", "\n", "$$\\text{损失函数}(\\text{n-平均律}) = \\sqrt{\\alpha_1(\\text{纯八度跟最接近的音程的距离})^2 + \\alpha_2(\\text{纯五度跟最接近的音程的距离})^2 + ...}$$\n", "\n", "权重该如何取值？一个很heuristic也是最为自然的办法就是直接用音程自身的值，因为我们发现**和谐音程的大小跟重要性呈正比**。\n", "\n", "例如纯五度比小三度重要，恰好纯五度对应的$\\frac32$比小三度对应的$\\frac65$要大。\n", "\n", "因此公式就成了\n", "\n", "$$\\text{损失函数}(\\text{n-平均律}) = \\sqrt{\\frac21(\\text{纯八度跟最接近的音程的距离})^2 + \\frac32(\\text{纯五度跟最接近的音程的距离})^2 + ...}$$" ] }, { "cell_type": "code", "execution_count": 6, "metadata": {}, "outputs": [], "source": [ "def loss(n):\n", " loss = 0\n", " for interval in INTERVALS:\n", " x = 2**(np.arange(0,n + 1) / n) - interval.value\n", " x = np.abs(x)\n", " x = np.min(x)\n", " x = interval.value * np.square(x)\n", " loss = loss + x\n", " return np.sqrt(loss)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### n的取值\n", "\n", "另外一个直觉则是，n-平均律，n越大越不现实，毕竟作曲家和听众都会懵逼：比如如果一个八度有100万个音，那么我们大家都会集体懵逼。因此从regularization的角度来讲，我们可以加入对n的penalty：我们想要的是**小的损失函数，但是n也最好不要大**。" ] }, { "cell_type": "code", "execution_count": 7, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
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nloss
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" ], "text/plain": [ " n loss\n", "17 19 0.009908\n", "20 22 0.012017\n", "22 24 0.014834\n", "10 12 0.014834\n", "13 15 0.023361\n", "19 21 0.028531\n", "18 20 0.030169\n", "21 23 0.031397\n", "15 17 0.033355\n", "14 16 0.034023\n", "16 18 0.039706\n", "12 14 0.040861\n", "5 7 0.040936\n", "8 10 0.041996\n", "11 13 0.051487\n", "7 9 0.053746\n", "9 11 0.066498\n", "6 8 0.073648\n", "3 5 0.088890\n", "4 6 0.128209\n", "1 3 0.129295\n", "2 4 0.133091\n", "0 2 0.284371" ] }, "execution_count": 7, "metadata": {}, "output_type": "execute_result" } ], "source": [ "data.sort_values(by='loss') # 从这里可以看出来，另一个要考虑的feasibility问题就是n的大小：n越大越不现实，因此也可以加上一个penalty" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "所以一个最简单的方式就是把这个对n的regularization penalty放到损失函数中，**n越大损失越大**。\n", "\n", "一个最简单且自然的选择就是选择n本身当做系数相乘\n", "\n", "$$\\text{损失函数}(\\text{n-平均律}) = n \\sqrt{\\frac21(\\text{纯八度跟最接近的音程的距离})^2 + \\frac32(\\text{纯五度跟最接近的音程的距离})^2 + ...}$$" ] }, { "cell_type": "code", "execution_count": 8, "metadata": {}, "outputs": [], "source": [ "def loss(n):\n", " loss = 0\n", " for interval in INTERVALS:\n", " x = 2**(np.arange(0,n + 1) / n) - interval.value\n", " x = np.abs(x)\n", " x = np.min(x)\n", " x = interval.value * np.square(x)\n", " loss = loss + x\n", " return n * np.sqrt(loss)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "看一下效果：" ] }, { "cell_type": "code", "execution_count": 9, "metadata": {}, "outputs": [ { "data": { "image/png": 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\n", 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" ] }, "metadata": { "needs_background": "light" }, "output_type": "display_data" } ], "source": [ "x = np.arange(2,25)\n", "y = np.array(list(map(loss,x)))\n", "data = pd.DataFrame({'n':x,'loss':y})\n", "\n", "fig, ax = plt.subplots()\n", "ax.set_xticks(data['n'].head(20))\n", "ax.bar(data['n'].head(20),data['loss'].head(20))\n", "ax.set(xlabel='n', ylabel='loss',\n", " title='n-equal temperament')\n", "ax.grid()\n", "plt.show()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "如上，12这个数字成了唯一的最优解。" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "草稿\n", "\n", "## 进阶：量化trade-off\n", "\n", "平均律本质上就是trade-off：牺牲纯五纯四的纯度换取transposition-invariance。\n", "\n", "我们接下来要找的就是一个最小的irreducible量：也就是说任何temperament都最少会损失“这么多”的“这个量”。" ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.6.8" } }, "nbformat": 4, "nbformat_minor": 4 }