{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## theory"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Let's get the result in this easy case."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Suppose $z = xW = [W_{s1} ... W_{sD}]$ where $x$ is one-hot encoded and $x_s = 1$. Then we have $\\frac{\\partial J}{\\partial W_{ij}} = \\sum_k{\\frac{\\partial J}{\\partial z_k} \\frac{\\partial z_k}{\\partial W_{ij}}}$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "This is equal to $0$ if $s \\neq i$. Also $\\frac{\\partial z_k}{\\partial W_{sj}} = 0$ if $k \\neq j$ and $1$ otherwise. So $\\frac{\\partial J}{\\partial W_{sj}} = \\frac{\\partial J}{\\partial z_j}$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "In other words: $\\frac{\\partial J}{\\partial W} = \\begin{bmatrix}0 & ... & 0\\\\ \\frac{\\partial J}{\\partial z_1} & ... & \\frac{\\partial J}{\\partial z_D} \\\\ 0 & ... & 0 \\end{bmatrix}$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "So we just need to fill in $s^{th}$ row of our gradient with the upstream gradient. "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## implementation"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "np.random.seed(42)\n",
    "V, D = 10, 2\n",
    "x = [3]\n",
    "W = np.random.randn(V, D)\n",
    "dW = np.zeros_like(W)\n",
    "dout = np.random.randn(1, D)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[ 1.46564877, -0.2257763 ]])"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "dout"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [],
   "source": [
    "np.add.at(dW, x, dout)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[ 0.        ,  0.        ],\n",
       "       [ 0.        ,  0.        ],\n",
       "       [ 0.        ,  0.        ],\n",
       "       [ 1.46564877, -0.2257763 ],\n",
       "       [ 0.        ,  0.        ],\n",
       "       [ 0.        ,  0.        ],\n",
       "       [ 0.        ,  0.        ],\n",
       "       [ 0.        ,  0.        ],\n",
       "       [ 0.        ,  0.        ],\n",
       "       [ 0.        ,  0.        ]])"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "dW"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "This concludes our short analysis in this simple case."
   ]
  }
 ],
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