Predavanje 05 - Uvod v numerične metode in sistemi linearnih enačb 1.ipynb

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    "<font size=\"2\"> [github.com/jankoslavic/pypinm](http://github.com/jankoslavic/pypinm)</font>\n",
    "\n",
    "[Prof. dr. Janko Slavič](http://www.ladisk.si/~slavic)\n",
    "\n",
    "<center>\n",
    "<img width=300 src=\"./fig/ladisk.png\">\n",
    "<br>\n",
    "<font size=\"7\" color=\"f00e0e\" face=\"garamond\"><b>Uvod v numerične metode in sistemi linearnih enačb (1)</b></font>\n",
    "<img src=\"\">\n",
    "<font size=\"2\">September 2021</font>\n",
    "</center>"
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    "<h1>Kazalo<span class=\"tocSkip\"></span></h1>\n",
    "<div class=\"toc\"><ul class=\"toc-item\"><li><span><a href=\"#Uvod-v-numerične-metode\" data-toc-modified-id=\"Uvod-v-numerične-metode-1\"><span class=\"toc-item-num\">1&nbsp;&nbsp;</span>Uvod v numerične metode</a></span><ul class=\"toc-item\"><li><span><a href=\"#Zaokrožitvena-napaka\" data-toc-modified-id=\"Zaokrožitvena-napaka-1.1\"><span class=\"toc-item-num\">1.1&nbsp;&nbsp;</span>Zaokrožitvena napaka</a></span></li><li><span><a href=\"#Napaka-metode\" data-toc-modified-id=\"Napaka-metode-1.2\"><span class=\"toc-item-num\">1.2&nbsp;&nbsp;</span>Napaka metode</a></span></li></ul></li><li><span><a href=\"#Uvod-v-sisteme-linearnih-enačb\" data-toc-modified-id=\"Uvod-v-sisteme-linearnih-enačb-2\"><span class=\"toc-item-num\">2&nbsp;&nbsp;</span>Uvod v sisteme linearnih enačb</a></span><ul class=\"toc-item\"><li><span><a href=\"#O-rešitvi-sistema-linearnih-enačb\" data-toc-modified-id=\"O-rešitvi-sistema-linearnih-enačb-2.1\"><span class=\"toc-item-num\">2.1&nbsp;&nbsp;</span>O rešitvi sistema linearnih enačb</a></span></li><li><span><a href=\"#Norma-in-pogojenost-sistemov-enačb\" data-toc-modified-id=\"Norma-in-pogojenost-sistemov-enačb-2.2\"><span class=\"toc-item-num\">2.2&nbsp;&nbsp;</span>Norma in pogojenost sistemov enačb</a></span><ul class=\"toc-item\"><li><span><a href=\"#Primer-slabo-pogojene-matrike\" data-toc-modified-id=\"Primer-slabo-pogojene-matrike-2.2.1\"><span class=\"toc-item-num\">2.2.1&nbsp;&nbsp;</span>Primer slabo pogojene matrike</a></span></li></ul></li><li><span><a href=\"#Numerično-reševanje-sistemov-linearnih-enačb\" data-toc-modified-id=\"Numerično-reševanje-sistemov-linearnih-enačb-2.3\"><span class=\"toc-item-num\">2.3&nbsp;&nbsp;</span>Numerično reševanje sistemov linearnih enačb</a></span></li></ul></li><li><span><a href=\"#Gaussova-eliminacija\" data-toc-modified-id=\"Gaussova-eliminacija-3\"><span class=\"toc-item-num\">3&nbsp;&nbsp;</span>Gaussova eliminacija</a></span><ul class=\"toc-item\"><li><ul class=\"toc-item\"><li><span><a href=\"#Povzetek-Gaussove-eliminacije\" data-toc-modified-id=\"Povzetek-Gaussove-eliminacije-3.0.1\"><span class=\"toc-item-num\">3.0.1&nbsp;&nbsp;</span>Povzetek Gaussove eliminacije</a></span></li></ul></li><li><span><a href=\"#Numerična-zahtevnost\" data-toc-modified-id=\"Numerična-zahtevnost-3.1\"><span class=\"toc-item-num\">3.1&nbsp;&nbsp;</span>Numerična zahtevnost</a></span></li><li><span><a href=\"#Uporaba-knjižnjice-numpy\" data-toc-modified-id=\"Uporaba-knjižnjice-numpy-3.2\"><span class=\"toc-item-num\">3.2&nbsp;&nbsp;</span>Uporaba knjižnjice <code>numpy</code></a></span></li></ul></li><li><span><a href=\"#Nekaj-vprašanj-za-razmislek!\" data-toc-modified-id=\"Nekaj-vprašanj-za-razmislek!-4\"><span class=\"toc-item-num\">4&nbsp;&nbsp;</span>Nekaj vprašanj za razmislek!</a></span></li><li><span><a href=\"#Dodatno\" data-toc-modified-id=\"Dodatno-5\"><span class=\"toc-item-num\">5&nbsp;&nbsp;</span>Dodatno</a></span><ul class=\"toc-item\"><li><span><a href=\"#Primer-simbolnega-reševanja-sistema-linearnih-enačb-v-okviru-sympy\" data-toc-modified-id=\"Primer-simbolnega-reševanja-sistema-linearnih-enačb-v-okviru-sympy-5.1\"><span class=\"toc-item-num\">5.1&nbsp;&nbsp;</span>Primer simbolnega reševanja sistema linearnih enačb v okviru <code>sympy</code></a></span></li></ul></li></ul></div>"
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   "source": [
    "# Uvod v numerične metode"
   ]
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   "source": [
    "Kadar želimo simulirati izbrani fizikalni proces, ponavadi postopamo takole:\n",
    "\n",
    "1. postavimo matematični model*,\n",
    "1. izberemo numerično metodo in njene parametre,\n",
    "1. pripravimo program (pomagamo si z vgrajenimi funkcijami),\n",
    "1. izvedemo izračun, rezultate analiziramo in vrednotimo.\n",
    "\n",
    "\\* Če lahko matematični model rešimo analitično, numerično reševanje ni potrebno."
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    "Matematični model poskušamo rešiti analitično, saj taka rešitev ni obremenjena z napakami. Iz tega razloga se v okviru matematike učimo reševanja sistema enačb, integriranja, odvajanja in podobno. Bistvo **numeričnih metod** je, da matematične modele rešujemo **numerično**, torej na podlagi **diskretnih vrednosti**. Kakor bomo spoznali pozneje, nam numerični pristop v primerjavi z analitičnim omogoča reševanje bistveno obsežnejših in kompleksnejših problemov."
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   "source": [
    "## Zaokrožitvena napaka \n",
    "\n",
    "V nadaljevanju si bomo pogledali nekatere omejitve in izzive numeričnega pristopa. Prva omejitev je, da so v računalniku realne vrednosti vedno zapisane s končno natančnostjo. V Pythonu se števila pogosto zapišejo v dvojni natančnosti s približno 15 signifikantnimi števkami. "
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    "Število z dvojno natančnostjo se v Pythonu imenuje **float64** in je zapisano v spomin v binarni obliki v 64 bitih (11 bitov eksponent in 53 bitov mantisa (1 bit za predznak)). Ker je mantisa definirana na podlagi 52 binarnih števk, se lahko pojavi pri njegovem zapisu *relativna napaka* največ $\\epsilon\\approx2.2\\cdot 10^{-16}$. Ta napaka se imenuje **osnovna zaokrožitvena napaka** in se lahko pojavi pri vsakem vmesnem\n",
    "izračunu!"
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    "Če je korakov veliko, lahko napaka zelo naraste in zato je pomembno, da je njen vpliv na rezultat čim manjši!\n",
    "\n",
    "Spodaj je primer podrobnejših informacij za tip podatkov z dvojno natančnostjo (`float`); pri tem si pomagamo z vgrajenim modulom `sys` za klic parametrov in funkcij python sistema ([dokumentacija](https://docs.python.org/3/library/sys.html)):"
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       "1.0000000000000002e+16"
      ]
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   "source": [
    "import sys\n",
    "sys.float_info.epsilon\n",
    "1e16+2\n",
    "#sys.float_info #preverite tudi širši izpis!"
   ]
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    "Poleg števila z dvojno natančnostjo se uporabljajo drugi tipi podatkov; dober pregled različnih tipov je prikazan v okviru [numpy](https://docs.scipy.org/doc/numpy-1.10.1/user/basics.types.html) in [python](https://docs.python.org/3/library/stdtypes.html) dokumentacije.\n",
    "\n",
    "Tukaj si poglejmo primer tipa `int8`, kar pomeni celo število zapisano z 8 biti (8 bit = 1 byte). Z njim lahko v dvojiškem sistemu zapišemo cela števila od -128 do +127:"
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      "text/plain": [
       "\"Število 1 tipa <class 'numpy.int8'> zapisano v binari obliki:       1\""
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   "source": [
    "import numpy as np\n",
    "število = np.int8(1) # poskušite še števila: -128 in nato 127, 128, 129. Kaj se dogaja?\n",
    "f'Število {število} tipa {type(število)} zapisano v binari obliki:{število:8b}'"
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   "source": [
    "## Napaka metode\n",
    "\n",
    "Poleg zaokrožitvene napake pa se pogosto srečamo tudi z **napako metode** ali **metodično napako**, ki jo naredimo takrat, ko natančen analitični postopek reševanja matematičnega modela zamenjamo s približnim numeričnim."
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   "source": [
    "Pomembna lastnost numeričnih algoritmov je **stabilnost**. To pomeni, da majhna sprememba vhodnih podatkov povzroči majhno spremembo rezultatov. Če se ob majhni spremembi na vhodu rezultati zelo spremenijo, pravimo, da je **algoritem nestabilen**. V praksi torej uporabljamo stabilne algoritme; bomo pa pozneje spoznali, da je stabilnost lahko pogojena tudi z vhodnimi podatki!\n",
    "\n",
    "Poznamo pa tudi nestabilnost matematičnega modela/naloge/enačbe; v tem primeru govorimo o **slabi pogojenosti**."
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   "source": [
    "Med izvajanjem numeričnega izračuna se napake lahko širijo. Posledično je rezultat operacije manj natančen (ima manj zanesljivih števk), kakor pa je zanesljivost podatkov izračuna."
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   "source": [
    "Poglejmo si sedaj splošen pristop k oceni napake. Točno vrednost označimo z $r$, približek z $a_1$; velja $r=a_1+e_1$, kjer je $e_1$ napaka. Če z numeričnim algoritmom  izračunamo bistveno boljši približek $a_2$, velja $r=a_2+e_2$. \n",
    "\n",
    "Ker velja $a_1+e_1=a_2+e_2$, lahko ob predpostavki $\\left|e_1\\right|>>\\left|e_2\\right|$ in $\\left|e_2\\right|\\approx 0$ izpeljemo $a_2-a_1=e_1-e_2\\approx e_1$."
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   "source": [
    "$\\left|a_1-a_2\\right|$ je torej pesimistična ocena absolutne napake,\n",
    "\n",
    "$\\left|\\frac{a_1-a_2}{a_2}\\right|$ pa ocena relativne napake."
   ]
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   "source": [
    "# Uvod v sisteme linearnih enačb\n",
    "\n",
    "Pod zgornjim naslovom razumemo sistem $m$ linearnih enačb ($E_i, i=0, 1,\\dots,m-1$) z $n$ neznankami ($x_j, j=0,1,\\dots,n-1$):\n",
    "$$\n",
    "\\begin{array}{rllllllll}\n",
    "E_0: & A_{0,0}\\,x_0 &+&A_{0,1}\\,x_1&+& \\ldots &+&A_{0,n-1}\\,x_{n-1}&=&b_0\\\\\n",
    "E_1: & A_{1,0}\\,x_0 &+&A_{1,1}\\,x_1&+& \\ldots &+&A_{1,n-1}\\,x_{n-1}&=&b_1\\\\\n",
    "\\vdots && &&& \\vdots\\\\\n",
    "E_{m-1}: & A_{m-1,0}\\,x_0&+&A_{m-1,1}\\,x_1&+& \\ldots &+&A_{m-1,n-1}\\,x_{n-1}&=&b_{m-1}.\\\\\n",
    "\\end{array}\n",
    "$$\n",
    "Koeficienti $A_{i,j}$ in $b_i$ so znana števila. \n",
    "\n",
    "V kolikor je desna stran enaka nič, torej $b_i=0$, imenujemo sistem **homogenem**, sicer je sistem **nehomogen**."
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    "Sistem enačb lahko zapišemo tudi v matrični obliki:\n",
    "\n",
    "$$\\mathbf{A}\\,\\mathbf{x}=\\mathbf{b}$$\n",
    "\n",
    "kjer sta $\\mathbf{A}$ in $\\mathbf{b}$ znana matrika in vektor, vektor $\\mathbf{x}$ pa ni znan. Matriko $\\mathbf{A}$ imenujemo **matrika koeficientov**, vektor $\\mathbf{b}$ **vektor konstant** (tudi: vektor prostih členov ali vektor stolpec desnih strani) in $\\mathbf{x}$ **vektor neznank**. Če matriki $\\mathbf{A}$ dodamo kot stolpec vektor $\\mathbf{b}$, dobimo t. i. **razširjeno matriko** in jo označimo $[\\mathbf{A}|\\mathbf{b}]$."
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   "source": [
    "Opomba glede zapisa:\n",
    "\n",
    "* skalarne spremenljivke pišemo poševno, npr.: $a, A$,\n",
    "* vektorske spremenljivke pišemo z majhno črko poudarjeno, npr.: $\\mathbf{a}$,\n",
    "* matrične  spremenljivke pišemo z veliko črko poudarjeno, npr.: $\\mathbf{A}$."
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   "source": [
    "## O rešitvi sistema linearnih enačb"
   ]
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      "text/html": [
       "\n",
       "        <iframe\n",
       "            width=\"800\"\n",
       "            height=\"300\"\n",
       "            src=\"https://www.youtube.com/embed/7YbyijGUbYw\"\n",
       "            frameborder=\"0\"\n",
       "            allowfullscreen\n",
       "        ></iframe>\n",
       "        "
      ],
      "text/plain": [
       "<IPython.lib.display.YouTubeVideo at 0x1804316c310>"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "from IPython.display import YouTubeVideo\n",
    "YouTubeVideo('7YbyijGUbYw', width=800, height=300)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Če nad sistemom linearnih enačb izvajamo **elementarne operacije**:\n",
    "\n",
    "* množenje poljubne enačbe s konstanto (ki je različna od nič),\n",
    "* spreminjanje vrstnega reda enačb,\n",
    "* prištevanje ene enačbe (pomnožene s konstanto) drugi enačbi, \n",
    "\n",
    "rešitve sistema ne spremenimo in dobimo ekvivalentni sistem enačb."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "S pomočjo elementarnih operacij nad vrsticami matrike $\\mathbf{A}$ jo lahko preoblikujemo v t. i. **vrstično kanonično obliko**:\n",
    "\n",
    "1. če obstajajo ničelne vrstice, so te na dnu matrike,\n",
    "1. prvi neničelni element se nahaja desno od prvih neničelnih elementov predhodnih vrstic,\n",
    "1. prvi neničelni element v *vrstici* imenujemo **pivot** in je enak 1,\n",
    "1. pivot je edini neničelni element v *stolpcu*.\n",
    "\n",
    "\n",
    "**Rang matrike** predstavlja število neničelnih vrstic v vrstični kanonični obliki matrike; število neničelnih vrstic predstavlja **število linearno neodvisnih enačb in je enako številu pivotnih elementov**. **Rang matrike je torej enak številu linearno neodvisnih vrstic matrike**. Transponiranje matrike njenega ranga ne spremeni, zato je rang matrike enak tudi številu linearno neodvisnih *stolpcev* matrike. "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Primer preoblikovanja matrike $\\mathbf{A}$:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[1, 2, 3],\n",
       "       [4, 5, 6],\n",
       "       [7, 8, 9]])"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "import numpy as np #uvozimo numpy\n",
    "A = np.arange(9).reshape((3,3))+1\n",
    "A"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Očitno ima neničelni element `A[0,0]` vrednost 1 in je pivotni element. Prvo vrstico `A[0,:]` pomnožimo z $-4$ in produkt prištejemo drugi vrstici `A[1,:]-4A[0,:]`:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[ 1,  2,  3],\n",
       "       [ 0, -3, -6],\n",
       "       [ 7,  8,  9]])"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "A[1,:] -= A[1,0]*A[0,:]\n",
    "A"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Podobno naredimo za tretjo vrstico:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[  1,   2,   3],\n",
       "       [  0,  -3,  -6],\n",
       "       [  0,  -6, -12]])"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "A[2,:] -= A[2,0]*A[0,:]\n",
    "A"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Drugo vrstico sedaj delimo z `A[1,1]`, da dobimo pivot:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[  1,   2,   3],\n",
       "       [  0,   1,   2],\n",
       "       [  0,  -6, -12]])"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "A[1,:] = A[1,:]/A[1,1]\n",
    "A"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Odštejemo drugo vrstico od ostalih, da dobimo v drugem stolpcu ničle povsod, razen v drugi vrstici vrednost 1:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[  1,   0,  -1],\n",
       "       [  0,   1,   2],\n",
       "       [  0,  -6, -12]])"
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "A[0,:] -= A[0,1]*A[1,:] # odštevanje od prve vrstice\n",
    "A"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[ 1,  0, -1],\n",
       "       [ 0,  1,  2],\n",
       "       [ 0,  0,  0]])"
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "A[2,:] -= A[2,1]*A[1,:] # odštevanje od zadnje vrstice\n",
    "A"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Imamo dve neničelni vrstici; Matrika `A` ima dva pivota in predstavlja dve linearno neodvisni enačbi. Rang matrike je 2."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[ 1,  0, -1],\n",
       "       [ 0,  1,  2],\n",
       "       [ 0,  0,  0]])"
      ]
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "A"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Rang matrike pa lahko določimo tudi s pomočjo `numpy` funkcije `numpy.linalg.matrix_rank` ([dokumentacija](https://docs.scipy.org/doc/numpy/reference/generated/numpy.linalg.matrix_rank.html)):\n",
    "\n",
    "```python\n",
    "matrix_rank(M, tol=None)\n",
    "```\n",
    "\n",
    "kjer je `M` matrika, katere rang iščemo, `tol` opcijski parameter, ki določa mejo, pod katero se vrednosti v algoritmu smatrajo enake nič."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "2"
      ]
     },
     "execution_count": 11,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "np.linalg.matrix_rank(A)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Če velja $r=\\textbf{rang}(\\mathbf{A})=\\textbf{rang}([\\mathbf{A}|\\mathbf{b}])$, potem rešitev **obstaja** (rečemo tudi, da je sistem **konsistenten**).\n",
    "\n",
    "Konsistenten sistem ima:\n",
    "\n",
    "* natanko eno rešitev, ko je število neznank $n$ enako rangu $r$ (rešitev je neodvisna) in \n",
    "* neskončno mnogo rešitev, ko je rang $r$ manjši od števila neznank $n$ (rešitev je odvisna od $n-r$ parametrov)."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Najprej se bomo omejili na sistem $m=n$ linearnih enačb z $n$ neznankami ter velja $n=r$:\n",
    "$$\\mathbf{A}\\,\\mathbf{x}=\\mathbf{b}.$$\n",
    "\n",
    "Pod zgornjimi pogoji je matrika koeficientov $\\mathbf{A}$ nesingularna ($|\\mathbf{A}|\\neq 0$) in sistem ima rešitev:\n",
    "$$\\mathbf{x}=\\mathbf{A^{-1}}\\,\\mathbf{b}.$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Poglejmo si primer sistema, ko so **enačbe linearno odvisne** ($r<n$):"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[1, 2, 1],\n",
       "       [2, 4, 2]])"
      ]
     },
     "execution_count": 12,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "A = np.array([[1 , 2],\n",
    "              [2, 4]])\n",
    "b = np.array([1, 2])\n",
    "Ab = np.column_stack((A,b))\n",
    "Ab"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "S pomočjo `numpy` knjižnice poglejmo sedaj rang matrike koeficientov in razširjene matrike ter determinanto z uporabo `numpy.linalg.det` ([dokumentacija](https://docs.scipy.org/doc/numpy/reference/generated/numpy.linalg.det.html)):\n",
    "\n",
    "```python\n",
    "det(a)\n",
    "```\n",
    "\n",
    "kjer je `a` matrika (ali seznam matrik), katere determinanto iščemo; funkcija `det` vrne determinanto (ali seznam determinant)."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "'rang(A)=1, rang(Ab)=1, število neznank: 2, det(A)=0.0'"
      ]
     },
     "execution_count": 13,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "f'rang(A)={np.linalg.matrix_rank(A)}, rang(Ab)={np.linalg.matrix_rank(Ab)}, \\\n",
    "število neznank: {len(A[:,0])}, det(A)={np.linalg.det(A)}'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Poglejmo še primer, ko **rešitve sploh ni** (nekonsistenten sistem):"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[1, 2, 1],\n",
       "       [2, 4, 1]])"
      ]
     },
     "execution_count": 14,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "A = np.array([[1 , 2],\n",
    "              [2, 4]])\n",
    "b = np.array([1, 1])\n",
    "Ab = np.column_stack((A,b))\n",
    "Ab"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "'rang(A)=1, rang(Ab)=2, število neznank: 2, det(A)=0.0'"
      ]
     },
     "execution_count": 15,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "f'rang(A)={np.linalg.matrix_rank(A)}, rang(Ab)={np.linalg.matrix_rank(Ab)}, \\\n",
    "število neznank: {len(A[:,0])}, det(A)={np.linalg.det(A)}'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "## Norma in pogojenost sistemov enačb"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Numerična naloga je slabo pogojena, če majhna sprememba podatkov povzroči veliko spremembo rezultata. V primeru *majhne spremembe podatkov*, ki povzročijo *majhno spremembo rezultatov*, pa je naloga **dobro pogojena**. \n",
    "\n",
    "Sistem enačb je ponavadi dobro pogojen, če so absolutne vrednosti diagonalnih elementov matrike koeficientov velike v primerjavi z absolutnimi vrednostmi izven diagonalnih elementov."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Za sistem linearnih enačb $\\mathbf{A}\\,\\mathbf{x}=\\mathbf{b}$ lahko računamo **število pogojenosti** (*angl.* condition number):\n",
    "\n",
    "$$\\textrm{cond}(\\textbf{A})=||\\textbf{A}||\\,||\\textbf{A}^{-1}||.$$\n",
    "\n",
    "Z $||\\textbf{A}||$ je označena **norma** matrike."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Obstaja več načinov računanja norme; navedimo dve:\n",
    "\n",
    "* Evklidska norma (tudi Frobeniusova):\n",
    "$$||\\textbf{A}||_e=\\sqrt{\\sum_{i=1}^n\\sum_{j=1}^nA_{ij}^2}$$\n",
    "* Norma vsote vrstic ali tudi neskončna norma:\n",
    "$$||\\textbf{A}||_{\\infty}=\\max_{1\\le i\\le n}\\sum_{j=1}^n |A_{ij}|$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Pogojenost računamo z vgrajeno funkcijo `numpy.linalg.cond` ([dokumentacija](https://docs.scipy.org/doc/numpy/reference/generated/numpy.linalg.cond.html)):\n",
    "\n",
    "```python\n",
    "cond(x, p=None)\n",
    "```\n",
    "\n",
    "ki sprejme dva parametra: matriko `x` in opcijski tip norme `p` (privzeti tip je `None`; v tem primeru se uporabi Evklidska/Frobeniusova norma)."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "-"
    }
   },
   "source": [
    "Če je število pogojenosti majhno, potem je matrika dobro pogojena in obratno - pri slabi pogojenosti se število pogojenosti zelo poveča.\n",
    "\n",
    "Žal je izračun pogojenosti matrike numerično relativno zahteven."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "### Primer slabo pogojene matrike"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Pogledali si bomo slabo pogojen sistem, kjer bomo z malenkostno spremembo na matriki koeficientov povzročili veliko spremembo rešitve."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Matrika koeficientov:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "400002.0000028382"
      ]
     },
     "execution_count": 16,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "A = np.array([[1 , 1],\n",
    "              [1, 1.00001]])\n",
    "np.linalg.cond(A)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Vektor konstant:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[ 1.     ,  1.     ,  3.     ],\n",
       "       [ 1.     ,  1.00001, -3.     ]])"
      ]
     },
     "execution_count": 17,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "b = np.array([3, -3])\n",
    "Ab = np.column_stack((A,b))\n",
    "Ab"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Preverimo rang in determinanto:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "'rang(A)=2, rang(Ab)=2, število neznank: 2, det(A)=1.000000000006551e-05'"
      ]
     },
     "execution_count": 18,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "f'rang(A)={np.linalg.matrix_rank(A)}, rang(Ab)={np.linalg.matrix_rank(Ab)}, \\\n",
    "število neznank: {len(A[:,0])}, det(A)={np.linalg.det(A)}'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Od druge enačbe odštejemo prvo:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[ 1.e+00,  1.e+00,  3.e+00],\n",
       "       [ 0.e+00,  1.e-05, -6.e+00]])"
      ]
     },
     "execution_count": 19,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Ab[1,:] -= Ab[0,:]\n",
    "Ab"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Določimo `x1`:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "-599999.9999960692"
      ]
     },
     "execution_count": 20,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "x1 = Ab[1,2]/Ab[1,1]\n",
    "x1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Preostane še določitev `x0`:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "600002.9999960692"
      ]
     },
     "execution_count": 21,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "x0 = (Ab[0,2] - Ab[0,1]*x1)/Ab[0,0]\n",
    "x0"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Malenkostno spremenimo matriko koeficientov in ponovimo reševanje:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "40002.00007491187"
      ]
     },
     "execution_count": 22,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "A = np.array([[1 , 1],\n",
    "              [1, 1.0001]]) # <= tukaj smo spremenili 1.00001 na 1.0001\n",
    "np.linalg.cond(A)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Ponovimo izračun:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {
    "scrolled": true
   },
   "outputs": [],
   "source": [
    "Ab = np.column_stack((A,b))\n",
    "f'rang(A)={np.linalg.matrix_rank(A)}, rang(Ab)={np.linalg.matrix_rank(Ab)}, \\\n",
    "število neznank: {len(A[:,0])}, det(A)={np.linalg.det(A)}'\n",
    "Ab[1,:] -= Ab[0,:]\n",
    "x1_ = Ab[1,2]/Ab[1,1]\n",
    "x0_ = (Ab[0,2] - Ab[0,1]*x1)/Ab[0,0]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Primerjamo obe rešitvi:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[600002.9999960692, -599999.9999960692]"
      ]
     },
     "execution_count": 24,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "[x0, x1] # prva rešitev"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[600002.9999960692, -60000.00000000661]"
      ]
     },
     "execution_count": 25,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "[x0_, x1_] # druga rešitev"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Ugotovimo, da je malenkostna sprememba enega koeficienta v matriki koeficientov povzročila veliko spremembo v rezultatu. Majhni spremembi podatkov se ne moremo izogniti, zaradi zapisa podatkov v računalniku."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "## Numerično reševanje sistemov linearnih enačb"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "\n",
    "Pogledali si bomo dva, v principu različna pristopa k reševanju sistemov linearnih enačb:\n",
    "\n",
    "A) **Direktni pristop**: nad sistemom enačb izvajamo elementarne operacije, s katerimi predelamo sistem enačb v lažje rešljivega,\n",
    "\n",
    "B) **Iterativni pristop**: izberemo začetni približek, nato pa približek iterativno izboljšujemo."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "slide"
    }
   },
   "source": [
    "# Gaussova eliminacija\n",
    "\n",
    "Predpostavimo, da rešujemo sistem $n$ enačb za $n$ neznank, ki ima rang $n$. Tak sistem je enolično rešljiv. \n",
    "\n",
    "Gaussova eliminacija spada med direktne metode, saj s pomočjo elementarnih vrstičnih operacij sistem enačb prevedemo v zgornje poravnani trikotni sistem (pod glavno diagonalo v razširjeni matriki so vrednosti nič)."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Najprej pripravimo razširjeno matriko koeficientov:\n",
    "$$\\begin{bmatrix}\n",
    "\\mathbf{A}|\\mathbf{b}\n",
    "\\end{bmatrix}=\n",
    "\\left[\\begin{array}{cccc|c}\n",
    "A_{0,0}&A_{0,1}&\\cdots & A_{0,n-1} & b_0\\\\\n",
    "A_{1,0}&A_{1,1}&\\cdots & A_{1,n-1} & b_1\\\\\n",
    "\\vdots&\\vdots&\\ddots & \\vdots & \\vdots\\\\\n",
    "A_{n-1,0}&A_{n-1,1}&\\cdots & A_{n-1,n-1} & b_{n-1}\\\\\n",
    "\\end{array}\\right]\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Gaussovo eliminacijo si bomo pogledali na zgledu:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 26,
   "metadata": {},
   "outputs": [],
   "source": [
    "A = np.array([[ 8., -6, 3],\n",
    "              [-6, 6,-6],\n",
    "              [ 3, -6, 6]])\n",
    "b = np.array([-14, 36, 6])\n",
    "Ab = np.column_stack((A,b))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 27,
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[  8.,  -6.,   3., -14.],\n",
       "       [ -6.,   6.,  -6.,  36.],\n",
       "       [  3.,  -6.,   6.,   6.]])"
      ]
     },
     "execution_count": 27,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Ab"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Korak 0: prvo vrstico pomnožimo z `Ab[1,0]/Ab[0,0]=-6/8` in odštejemo od druge:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[  8.  ,  -6.  ,   3.  , -14.  ],\n",
       "       [  0.  ,   1.5 ,  -3.75,  25.5 ],\n",
       "       [  3.  ,  -6.  ,   6.  ,   6.  ]])"
      ]
     },
     "execution_count": 28,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Ab[1,:] -= Ab[1,0]/Ab[0,0] * Ab[0,:]\n",
    "Ab"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Nato prvo vrstico pomnožimo z `Ab[2,0]/Ab[0,0]=3/8` in odštejemo od tretje:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 29,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[  8.   ,  -6.   ,   3.   , -14.   ],\n",
       "       [  0.   ,   1.5  ,  -3.75 ,  25.5  ],\n",
       "       [  0.   ,  -3.75 ,   4.875,  11.25 ]])"
      ]
     },
     "execution_count": 29,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Ab[2,:] -= Ab[2,0]/Ab[0,0] * Ab[0,:]\n",
    "Ab"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Korak 1: drugo vrstico pomnožimo z `Ab[2,1]/Ab[1,1]=-3.75/1.5` in odštejemo od tretje:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 30,
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[  8.  ,  -6.  ,   3.  , -14.  ],\n",
       "       [  0.  ,   1.5 ,  -3.75,  25.5 ],\n",
       "       [  0.  ,   0.  ,  -4.5 ,  75.  ]])"
      ]
     },
     "execution_count": 30,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Ab[2,:] -= Ab[2,1]/Ab[1,1] * Ab[1,:]\n",
    "Ab"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "S pomočjo pripravljenega modula `moduli/gauss_prikaz.py` lahko postopek tudi animiramo:\n",
    "<img src=\"./fig/gauss_eliminacija.gif\" width=350>"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Dobili smo zgornje trikotno matriko in Gaussova eliminacija je končana. Lahko izračunamo rešitev, torej določimo vektor neznank $x$ z **obratnim vstavljanjem**. \n",
    "\n",
    "Iz zadnje vrstice zgornje trikotne matrike izračunamo $x_2$:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 31,
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[  8.  ,  -6.  ,   3.  , -14.  ],\n",
       "       [  0.  ,   1.5 ,  -3.75,  25.5 ],\n",
       "       [  0.  ,   0.  ,  -4.5 ,  75.  ]])"
      ]
     },
     "execution_count": 31,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Ab"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 32,
   "metadata": {
    "slideshow": {
     "slide_type": "-"
    }
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([  0.        ,   0.        , -16.66666667])"
      ]
     },
     "execution_count": 32,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "x = np.zeros(3) #pripravimo prazen seznam\n",
    "x[2] = Ab[2,-1]/Ab[2,2]\n",
    "x"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "S pomočjo predzadnje vrstice izračunamo $x_1$:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 33,
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[  8.  ,  -6.  ,   3.  , -14.  ],\n",
       "       [  0.  ,   1.5 ,  -3.75,  25.5 ],\n",
       "       [  0.  ,   0.  ,  -4.5 ,  75.  ]])"
      ]
     },
     "execution_count": 33,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Ab"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 34,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([  0.        , -24.66666667, -16.66666667])"
      ]
     },
     "execution_count": 34,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "x[1] = (Ab[1,-1] - Ab[1,2]*x[2]) / Ab[1,1]\n",
    "x"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "S pomočjo prve vrstice nato izračunamo $x_0$:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 35,
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[  8.  ,  -6.  ,   3.  , -14.  ],\n",
       "       [  0.  ,   1.5 ,  -3.75,  25.5 ],\n",
       "       [  0.  ,   0.  ,  -4.5 ,  75.  ]])"
      ]
     },
     "execution_count": 35,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Ab"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 36,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([-14.        , -24.66666667, -16.66666667])"
      ]
     },
     "execution_count": 36,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "x[0] = (Ab[0,3] - Ab[0,1:3]@x[1:]) / Ab[0,0]\n",
    "x"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Preverimo rešitev:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 37,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([0.00000000e+00, 0.00000000e+00, 2.84217094e-14])"
      ]
     },
     "execution_count": 37,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "A @ x - b"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "### Povzetek Gaussove eliminacije"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "V modul ``orodja.py`` shranimo funkciji:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 38,
   "metadata": {},
   "outputs": [],
   "source": [
    "def gaussova_eliminacija(A, b, prikazi_korake = False):\n",
    "    Ab = np.column_stack((A, b))\n",
    "    for p, pivot_vrsta in enumerate(Ab[:-1]):\n",
    "        for vrsta in Ab[p+1:]:\n",
    "            if pivot_vrsta[p]:\n",
    "                vrsta[p:] -= pivot_vrsta[p:]*vrsta[p]/pivot_vrsta[p]\n",
    "            else:\n",
    "                raise Exception('Deljenje z 0.')\n",
    "        if prikazi_korake:\n",
    "            print('Korak: {:g}'.format(p))\n",
    "            print(Ab)\n",
    "    return Ab\n",
    "\n",
    "def gaussova_el_resitev(Ub):\n",
    "    v = len(Ub)\n",
    "    x = np.zeros(v)\n",
    "    for p, pivot_vrsta in enumerate(Ub[::-1]):\n",
    "        x[v-p-1] = (pivot_vrsta[-1] - pivot_vrsta[v-p:-1] @ x[v-p:] ) / (pivot_vrsta[v-p-1])\n",
    "    return x"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "Algoritem, s katerim iz zgornje trikotnega sistema enačb $\\mathbf{U}\\,\\mathbf{x}=\\mathbf{b}$ izračunamo rešitev, imenujemo **obratno vstavljanje** (angl. *back substitution*); $\\mathbf{U}$ je zgornje trikotna matrika.\n",
    "\n",
    "V kolikor bi reševali sistem $\\mathbf{L}\\,\\mathbf{x}=\\mathbf{b}$ in je $\\mathbf{L}$ spodnje trikotna matrika, bi to metodo imenovali **direktno vstavljanje** (angl. *forward substitution*)."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 39,
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[  8.  ,  -6.  ,   3.  , -14.  ],\n",
       "       [  0.  ,   1.5 ,  -3.75,  25.5 ],\n",
       "       [  0.  ,   0.  ,  -4.5 ,  75.  ]])"
      ]
     },
     "execution_count": 39,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "Ub = gaussova_eliminacija(A, b, prikazi_korake=False)\n",
    "Ub"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 40,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([-14.        , -24.66666667, -16.66666667])"
      ]
     },
     "execution_count": 40,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "gaussova_el_resitev(Ub)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "## Numerična zahtevnost"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Numerično zahtevnost ocenjujemo po številu matematičnih operacij, ki so potrebne za izračun. \n",
    "Za rešitev $n$ linearnih enačb tako z Gaussovo eliminacijo potrebujemo približno $n^3/3$ matematičnih operacij. Za določitev neznank $\\mathbf{x}$ potrebujemo še dodatnih približno $n^2$ operacij.\n",
    "\n",
    "Pri Gaussovi eliminaciji smo eliminacijo izvedli samo za člene pod diagonalo; če bi z eliminacijo nadaljevali in jo izvedli tudi za člene nad diagonalo, bi izvedli t. i. *Gauss-Jordanovo* eliminacijo, za katero pa potrebujemo dodatnih približno $n^3/3$ operacij* (kar se šteje kot glavna slabost te metode).\n",
    "\n",
    "\\* Nekaj komentarjev na temo števila numeričnih operacij najdete tukaj: [pinm.ladisk.si](http://pinm.ladisk.si/303/kako-dolo%C4%8Damo-numeri%C4%8Dno-zahtevnost-algoritmov)."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "## Uporaba knjižnjice `numpy`"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Reševanje sistema linearnih enačb z `numpy.linalg.solve` ([dokumentacija](https://docs.scipy.org/doc/numpy/reference/generated/numpy.linalg.solve.html)):\n",
    "\n",
    "```python\n",
    "solve(a, b)\n",
    "```\n",
    "\n",
    "kjer je `a` matrika koeficientov (ali seznam matrik) in je `b` vektor konstant (ali seznam vektorjev). Funkcija vrne vektor (ali seznam vektorjev) rešitev."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 41,
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([-14.        , -24.66666667, -16.66666667])"
      ]
     },
     "execution_count": 41,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "np.linalg.solve(A, b)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "slide"
    }
   },
   "source": [
    "# Nekaj vprašanj za razmislek!"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "1. Za sistem enačb: \n",
    "    $$\n",
    "    \\mathbf{A}=\n",
    "    \\begin{bmatrix}\n",
    "    1 & -4 & 1\\\\\n",
    "    1 & 6 & -1\\\\\n",
    "    2 & -1 & 2\n",
    "    \\end{bmatrix}\n",
    "    \\qquad\n",
    "    \\mathbf{b}=\n",
    "    \\begin{bmatrix}\n",
    "    7\\\\\n",
    "    13\\\\\n",
    "    5\n",
    "    \\end{bmatrix}\n",
    "    $$\n",
    "\n",
    "    najdite rešitev s pomočjo ``np.linalg.solve()``.\n",
    "\n",
    "1. Za zgoraj definirano matriko $\\mathbf{A}$ določite Evklidsko normo (lastni program).\n",
    "1. Za zgoraj definirano matriko $\\mathbf{A}$ določite neskončno normo (lastni program).\n",
    "1. Za zgoraj definirano matriko $\\mathbf{A}$ določite pogojenost (``numpy`` funkcija).\n",
    "1. Definirajte funkcijo ``gauss_elim``, ki za poljubno matriko $\\mathbf{A}$ in vektor $\\mathbf{b}$ izvede Gaussovo eliminacijo (posebej za matriko in posebej za vektor tako, da ne sestavite razširjene matrike $[\\mathbf{A}|\\mathbf{b}]$).\n",
    "1. Definirajte funkcijo ``gauss_elim_x``, ki za razultat funkcije ``gauss_elim`` najde ustrezne vrednosti vektorja $\\textbf{x}$.\n",
    "1. Zgornji funkciji dopolnite s štetjem matematičnih operacij.\n",
    "1. Na sliki je prikazan sistem mas:\n",
    "\n",
    "    <img width=600 src=\"./fig/mase_na_klancu.png\">\n",
    "    \n",
    "    Predpostavite, da se sistem zaradi teže giblje po klancu navzdol z neznanim pospeškom $a$ in da so vrvi napete z neznanimi silami $F_i$. Znane veličine so (sami jih določite): posamično telo ima maso $m_i$, koeficient trenja s podlago $\\mu_i=1$, $g=9,81$m/s$^2$, $\\varphi=55^{\\circ}$.\n",
    "    Določite sistem enačb v primeru dveh teles. Določite matriko koeficientov $\\mathbf{A}$ in vektorja $\\mathbf{b}$ ter $\\mathbf{x}$.\n",
    "1. Za zgoraj definiran sistem mas predpostavite, da je mas 4 (ali več) ter določite matriko koeficientov $\\mathbf{A}$, in vektorja $\\mathbf{x}$ ter $\\mathbf{b}$. Rešite sistem s pomočjo Gaussove elimnacije/LU razcepa ali ``np.linalg.solve``. Preverite pogojenost!\n",
    "1. V sistemu mas dobimo fizikalno nekonsistentno rešitev, če imamo v kateri od vrvi tlačno silo (vrv ne prenese tlačne sile). Preverite, ali je to v vašem primeru res. Ustrezno spremenite koeficient(e) trenja, da se bo to zgodilo."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "# Dodatno"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Poglejte si strani:\n",
    "* [micropython.org](http://www.micropython.org)\n",
    "* [kivy.org](http://www.kivy.org)\n",
    "* [openmodal.com](http://www.openmodal.com)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "source": [
    "## Primer simbolnega reševanja sistema linearnih enačb v okviru `sympy`"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 42,
   "metadata": {},
   "outputs": [],
   "source": [
    "import sympy as sym\n",
    "sym.init_printing()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 43,
   "metadata": {
    "slideshow": {
     "slide_type": "-"
    }
   },
   "outputs": [],
   "source": [
    "A11, A12, A21, A22 = sym.symbols('A11, A12, A21, A22')\n",
    "x1, x2 = sym.symbols('x1, x2')\n",
    "b1, b2 = sym.symbols('b1, b2')\n",
    "A = sym.Matrix([[A11, A12],\n",
    "                [A21, A22]])\n",
    "x = sym.Matrix([[x1],\n",
    "                [x2]])\n",
    "b = sym.Matrix([[b1],\n",
    "                [b2]])"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 44,
   "metadata": {
    "slideshow": {
     "slide_type": "subslide"
    }
   },
   "outputs": [
    {
     "data": {
      "image/png": "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\n",
      "text/latex": [
       "$\\displaystyle \\left[\\begin{matrix}A_{11} x_{1} + A_{12} x_{2}\\\\A_{21} x_{1} + A_{22} x_{2}\\end{matrix}\\right] = \\left[\\begin{matrix}b_{1}\\\\b_{2}\\end{matrix}\\right]$"
      ],
      "text/plain": [
       "⎡A₁₁⋅x₁ + A₁₂⋅x₂⎤   ⎡b₁⎤\n",
       "⎢               ⎥ = ⎢  ⎥\n",
       "⎣A₂₁⋅x₁ + A₂₂⋅x₂⎦   ⎣b₂⎦"
      ]
     },
     "execution_count": 44,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "eq = sym.Eq(A*x,b)\n",
    "eq"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 45,
   "metadata": {
    "slideshow": {
     "slide_type": "-"
    }
   },
   "outputs": [
    {
     "data": {
      "image/png": "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\n",
      "text/latex": [
       "$\\displaystyle \\left\\{ x_{1} : \\frac{- A_{12} b_{2} + A_{22} b_{1}}{A_{11} A_{22} - A_{12} A_{21}}, \\  x_{2} : \\frac{A_{11} b_{2} - A_{21} b_{1}}{A_{11} A_{22} - A_{12} A_{21}}\\right\\}$"
      ],
      "text/plain": [
       "⎧     -A₁₂⋅b₂ + A₂₂⋅b₁       A₁₁⋅b₂ - A₂₁⋅b₁ ⎫\n",
       "⎨x₁: ─────────────────, x₂: ─────────────────⎬\n",
       "⎩    A₁₁⋅A₂₂ - A₁₂⋅A₂₁      A₁₁⋅A₂₂ - A₁₂⋅A₂₁⎭"
      ]
     },
     "execution_count": 45,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "#%%timeit\n",
    "resitev = sym.solve(eq,[x1, x2])\n",
    "resitev"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 46,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": "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\n",
      "text/latex": [
       "$\\displaystyle A_{11} A_{22} - A_{12} A_{21}$"
      ],
      "text/plain": [
       "A₁₁⋅A₂₂ - A₁₂⋅A₂₁"
      ]
     },
     "execution_count": 46,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "A.det()"
   ]
  }
 ],
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