{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# [Newton's method](https://en.wikipedia.org/wiki/Newton%27s_method)\n",
    "Let's use the Newton-Raphson method for finding the root of an equation to write a function that can compute the square root of a number.\n",
    "\n",
    "Cf. [\"Why Functional Programming Matters\" by John Hughes](https://www.cs.kent.ac.uk/people/staff/dat/miranda/whyfp90.pdf)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "from notebook_preamble import J, V, define"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## A Generator for Approximations\n",
    "\n",
    "To make a generator that generates successive approximations let’s start by assuming an initial approximation and then derive the function that computes the next approximation:\n",
    "\n",
    "       a F\n",
    "    ---------\n",
    "        a'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### A Function to Compute the Next Approximation\n",
    "\n",
    "This is the equation for computing the next approximate value of the square root:\n",
    "\n",
    "$a_{i+1} = \\frac{(a_i+\\frac{n}{a_i})}{2}$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "    a n over / + 2 /\n",
    "    a n a    / + 2 /\n",
    "    a n/a      + 2 /\n",
    "    a+n/a        2 /\n",
    "    (a+n/a)/2\n",
    "\n",
    "The function we want has the argument `n` in it:\n",
    "\n",
    "    F == n over / + 2 /"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Make it into a Generator\n",
    "\n",
    "Our generator would be created by:\n",
    "\n",
    "    a [dup F] make_generator\n",
    "\n",
    "With n as part of the function F, but n is the input to the sqrt function we’re writing. If we let 1 be the initial approximation:\n",
    "\n",
    "    1 n 1 / + 2 /\n",
    "    1 n/1   + 2 /\n",
    "    1 n     + 2 /\n",
    "    n+1       2 /\n",
    "    (n+1)/2\n",
    "\n",
    "The generator can be written as:\n",
    "\n",
    "    23 1 swap  [over / + 2 /] cons [dup] swoncat make_generator\n",
    "    1 23       [over / + 2 /] cons [dup] swoncat make_generator\n",
    "    1       [23 over / + 2 /]      [dup] swoncat make_generator\n",
    "    1   [dup 23 over / + 2 /]                    make_generator"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "scrolled": true
   },
   "outputs": [],
   "source": [
    "define('gsra == 1 swap [over / + 2 /] cons [dup] swoncat make_generator')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[1 [dup 23 over / + 2 /] codireco]\n"
     ]
    }
   ],
   "source": [
    "J('23 gsra')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Let's drive the generator a few time (with the `x` combinator) and square the approximation to see how well it works..."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "23.0000000001585\n"
     ]
    }
   ],
   "source": [
    "J('23 gsra 6 [x popd] times first sqr')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Finding Consecutive Approximations within a Tolerance\n",
    "\n",
    "From [\"Why Functional Programming Matters\" by John Hughes](https://www.cs.kent.ac.uk/people/staff/dat/miranda/whyfp90.pdf):\n",
    "\n",
    "\n",
    "> The remainder of a square root finder is a function _within_, which takes a tolerance and a list of approximations and looks down the list for two successive approximations that differ by no more than the given tolerance.\n",
    "\n",
    "(And note that by “list” he means a lazily-evaluated list.)\n",
    "\n",
    "Using the _output_ `[a G]` of the above generator for square root approximations, and further assuming that the first term a has been generated already and epsilon ε is handy on the stack...\n",
    "\n",
    "       a [b G] ε within\n",
    "    ---------------------- a b - abs ε <=\n",
    "          b\n",
    "\n",
    "\n",
    "       a [b G] ε within\n",
    "    ---------------------- a b - abs ε >\n",
    "       b [c G] ε within\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Predicate\n",
    "\n",
    "    a [b G]             ε [first - abs] dip <=\n",
    "    a [b G] first - abs ε                   <=\n",
    "    a b           - abs ε                   <=\n",
    "    a-b             abs ε                   <=\n",
    "    abs(a-b)            ε                   <=\n",
    "    (abs(a-b)<=ε)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [],
   "source": [
    "define('_within_P == [first - abs] dip <=')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Base-Case\n",
    "\n",
    "    a [b G] ε roll< popop first\n",
    "      [b G] ε a     popop first\n",
    "      [b G]               first\n",
    "       b"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [],
   "source": [
    "define('_within_B == roll< popop first')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Recur\n",
    "\n",
    "    a [b G] ε R0 [within] R1\n",
    "\n",
    "1. Discard a.\n",
    "2. Use `x` combinator to generate next term from `G`.\n",
    "3. Run `within` with `i` (it is a `primrec` function.)\n",
    "\n",
    "Pretty straightforward:\n",
    "\n",
    "    a [b G]        ε R0           [within] R1\n",
    "    a [b G]        ε [popd x] dip [within] i\n",
    "    a [b G] popd x ε              [within] i\n",
    "      [b G]      x ε              [within] i\n",
    "    b [c G]        ε              [within] i\n",
    "    b [c G]        ε               within\n",
    "\n",
    "    b [c G] ε within"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [],
   "source": [
    "define('_within_R == [popd x] dip')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Setting up\n",
    "\n",
    "The recursive function we have defined so far needs a slight preamble: `x` to prime the generator and the epsilon value to use:\n",
    "\n",
    "    [a G] x ε ...\n",
    "    a [b G] ε ..."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [],
   "source": [
    "define('within == x 0.000000001 [_within_P] [_within_B] [_within_R] primrec')\n",
    "define('sqrt == gsra within')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Try it out..."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "6.0\n"
     ]
    }
   ],
   "source": [
    "J('36 sqrt')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "4.795831523312719\n"
     ]
    }
   ],
   "source": [
    "J('23 sqrt')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Check it."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "22.999999999999996"
      ]
     },
     "execution_count": 11,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "4.795831523312719**2"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "4.795831523312719"
      ]
     },
     "execution_count": 12,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "from math import sqrt\n",
    "\n",
    "sqrt(23)"
   ]
  }
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