From 03a74138930cb27ef9d435fc04026e35de97e609 Mon Sep 17 00:00:00 2001 From: fortierq Date: Sun, 5 Nov 2023 20:10:55 +0000 Subject: [PATCH] deploy: 89771d284647f9838ada3b6cf6a1da5c1847c44b --- .../dl/algo/prog_dyn/tp/tp1/tp_prog_dyn.ipynb | 340 ------------------ dl/algo/prog_dyn/tp/tp1/tp_prog_dyn.html | 172 --------- searchindex.js | 2 +- 3 files changed, 1 insertion(+), 513 deletions(-) diff --git a/_sources/dl/algo/prog_dyn/tp/tp1/tp_prog_dyn.ipynb b/_sources/dl/algo/prog_dyn/tp/tp1/tp_prog_dyn.ipynb index 19fffd5c..960af842 100644 --- a/_sources/dl/algo/prog_dyn/tp/tp1/tp_prog_dyn.ipynb +++ b/_sources/dl/algo/prog_dyn/tp/tp1/tp_prog_dyn.ipynb @@ -27,21 +27,6 @@ "````" ] }, - { - "cell_type": "markdown", - "metadata": { - "tags": [ - "cor" - ] - }, - "source": [ - "````{admonition} Solution\n", - ":class: tip, dropdown\n", - "$$\\binom{0}{k} = 0$$\n", - "$$\\binom{n}{0} = 1, \\text{si } n \\neq 0$$\n", - "````" - ] - }, { "cell_type": "markdown", "metadata": {}, @@ -51,28 +36,6 @@ "````" ] }, - { - "cell_type": "markdown", - "execution_count": 1, - "metadata": { - "tags": [ - "cor" - ] - }, - "source": [ - "````{admonition} Solution\n", - ":class: tip, dropdown\n", - "``` python\n", - "def binom_rec(n, k): # voir cours\n", - " if k == 0:\n", - " return 1\n", - " if n == 0:\n", - " return 0\n", - " return binom_rec(n - 1, k - 1) + binom_rec(n - 1, k)\n", - "```\n", - "````" - ] - }, { "cell_type": "code", "execution_count": 2, @@ -119,31 +82,6 @@ " return ..." ] }, - { - "cell_type": "markdown", - "execution_count": 4, - "metadata": { - "tags": [ - "cor" - ] - }, - "source": [ - "````{admonition} Solution\n", - ":class: tip, dropdown\n", - "``` python\n", - "def binom_dp(n, k):\n", - " M = [[0]*(k + 1) for _ in range(n + 1)]\n", - " for i in range(0, n + 1):\n", - " M[i][0] = 1 # cas de base\n", - "\n", - " for i in range(1, n + 1):\n", - " for j in range(1, k + 1):\n", - " M[i][j] = M[i - 1][j - 1] + M[i - 1][j]\n", - " return M[n][k]\n", - "```\n", - "````" - ] - }, { "cell_type": "code", "execution_count": 5, @@ -187,31 +125,6 @@ " return aux(n, k)" ] }, - { - "cell_type": "markdown", - "execution_count": null, - "metadata": { - "tags": [ - "cor" - ] - }, - "source": [ - "````{admonition} Solution\n", - ":class: tip, dropdown\n", - "``` python\n", - "def binom(n, k):\n", - " d = {}\n", - " def aux(i, j):\n", - " if j == 0: return 1\n", - " if i == 0: return 0\n", - " if (i, j) not in d:\n", - " d[(i, j)] = aux(i - 1, j - 1) + aux(i - 1, j)\n", - " return d[(i, j)]\n", - " return aux(n, k)\n", - "```\n", - "````" - ] - }, { "cell_type": "markdown", "metadata": {}, @@ -241,27 +154,6 @@ "````" ] }, - { - "cell_type": "markdown", - "metadata": { - "tags": [ - "cor" - ] - }, - "source": [ - "````{admonition} Solution\n", - ":class: tip, dropdown\n", - "Si $a_k$ est utilisée : il faut encore rendre $n - a_k$ euros avec les pièces $a_1$, ..., $a_k$ (on a le droit d'utiliser plusieurs fois $a_k$), d'où $r(n, k) = r(n - a_k, k) + 1$.\n", - "\n", - "Dans le cas général, on considère les deux possibilités et on conserve le minimum : \n", - "$$\n", - " r(n, k) = min(r(n, k - 1), r(n - a_k, k) + 1)\n", - "$$\n", - "\n", - "Remarque : on ne peut utiliser $a_k$ pour rendre $n$ euros que si $n \\geq a_k$. Si $n < a_k$, on a donc $r(n, k) = r(n, k - 1)$.\n", - "````" - ] - }, { "cell_type": "markdown", "metadata": {}, @@ -272,33 +164,6 @@ "````" ] }, - { - "cell_type": "markdown", - "execution_count": 1, - "metadata": { - "tags": [ - "cor" - ] - }, - "source": [ - "````{admonition} Solution\n", - ":class: tip, dropdown\n", - "``` python\n", - "def rendu(L, n):\n", - " k = len(L) # nombre de pièces\n", - " M = [[0]*(k + 1) for _ in range(n + 1)]\n", - " for i in range(1, n + 1):\n", - " M[i][0] = float(\"inf\")\n", - " for j in range(1, k + 1):\n", - " if i - L[j - 1] >= 0:\n", - " M[i][j] = min(M[i][j - 1], 1 + M[i - L[j - 1]][j])\n", - " else:\n", - " M[i][j] = M[i][j - 1]\n", - " return M[-1][-1]\n", - "```\n", - "````" - ] - }, { "cell_type": "code", "execution_count": 8, @@ -328,43 +193,6 @@ "````" ] }, - { - "cell_type": "markdown", - "execution_count": 9, - "metadata": { - "tags": [ - "cor" - ] - }, - "source": [ - "````{admonition} Solution\n", - ":class: tip, dropdown\n", - "``` python\n", - "def rendu_memo(L, n):\n", - " k = len(L)\n", - " d = {}\n", - " def aux(i, j):\n", - " if (i, j) in d:\n", - " return d[(i, j)]\n", - " if i == 0:\n", - " return 0\n", - " if j == 0:\n", - " return float(\"inf\")\n", - " if i - L[j - 1] >= 0:\n", - " d[(i, j)] = min(aux(i, j - 1), 1 + aux(i - L[j - 1], j))\n", - " else:\n", - " d[(i, j)] = aux(i, j - 1)\n", - " return d[(i, j)]\n", - " return aux(n, k)\n", - "rendu_memo([1, 2, 5], 7)\n", - "```\n", - "``` \n", - "2\n", - "\n", - "```\n", - "````" - ] - }, { "cell_type": "markdown", "metadata": {}, @@ -395,23 +223,6 @@ "````" ] }, - { - "cell_type": "markdown", - "execution_count": 5, - "metadata": { - "tags": [ - "cor" - ] - }, - "source": [ - "````{admonition} Solution\n", - ":class: tip, dropdown\n", - "``` python\n", - "M = [[1, 0, 0, 0], [0, 0, 1, 1], [0, 1, 1, 1], [0, 1, 0, 1]]\n", - "```\n", - "````" - ] - }, { "cell_type": "markdown", "metadata": {}, @@ -428,28 +239,6 @@ "````" ] }, - { - "cell_type": "markdown", - "execution_count": 6, - "metadata": { - "tags": [ - "cor" - ] - }, - "source": [ - "````{admonition} Solution\n", - ":class: tip, dropdown\n", - "``` python\n", - "def est_carre(M, x, y, k):\n", - " for i in range(x, x + k):\n", - " for j in range(y, y + k):\n", - " if M[i][j] != 1:\n", - " return False\n", - " return True\n", - "```\n", - "````" - ] - }, { "cell_type": "code", "execution_count": 11, @@ -468,29 +257,6 @@ "````" ] }, - { - "cell_type": "markdown", - "execution_count": 9, - "metadata": { - "tags": [ - "cor" - ] - }, - "source": [ - "````{admonition} Solution\n", - ":class: tip, dropdown\n", - "``` python\n", - "def contient_carre(M, k):\n", - " n = len(M)\n", - " for i in range(n - k + 1):\n", - " for j in range(n - k + 1):\n", - " if est_carre(M, i, j, k):\n", - " return True\n", - " return False\n", - "```\n", - "````" - ] - }, { "cell_type": "code", "execution_count": 12, @@ -509,28 +275,6 @@ "````" ] }, - { - "cell_type": "markdown", - "execution_count": 13, - "metadata": { - "tags": [ - "cor" - ] - }, - "source": [ - "````{admonition} Solution\n", - ":class: tip, dropdown\n", - "``` python\n", - "def max_carre1(M):\n", - " n = len(M)\n", - " for k in range(n, 0, -1):\n", - " if contient_carre(M, k):\n", - " return k\n", - " return 0\n", - "```\n", - "````" - ] - }, { "cell_type": "code", "execution_count": 14, @@ -560,22 +304,6 @@ "````" ] }, - { - "cell_type": "markdown", - "metadata": { - "tags": [ - "cor" - ] - }, - "source": [ - "````{admonition} Solution\n", - ":class: tip, dropdown\n", - "- `est_carre(M, x, y, k)` est en $O(k^2)$. \n", - "- `contient_carre(M, k)` appelle O($n$) fois `est_carre`, donc est en $O(n^2 k^2)$. \n", - "- `max_carre1(M)` appelle `contient_carre` pour $k = 1, 2, ..., n$, donc est de complexité $\\sum_{k=1}^n O(n^2 k^2) = O(n^3 \\sum_{k=1}^n k^2)$. Comme $\\sum_{k=1}^n k^2 = \\frac{n(n+1)(2n+1)}{6} = O(n^3)$, la complexité totale est $\\boxed{O(n^6)}$.`\n", - "````" - ] - }, { "cell_type": "markdown", "metadata": {}, @@ -593,22 +321,6 @@ "````" ] }, - { - "cell_type": "markdown", - "metadata": { - "tags": [ - "cor" - ] - }, - "source": [ - "````{admonition} Solution\n", - ":class: tip, dropdown\n", - "`c[0][y] = 0` si `m[0][y] = 0` et `c[0][y] = 1` sinon. \n", - "De même pour `c[x][0]`. \n", - "Remarque : `c[0][y]` et `c[x][0]` sont donc les mêmes valeurs que `m[0][y]` et `m[x][0]`, on peut donc initialiser `c` comme une copie de `m`.\n", - "````" - ] - }, { "cell_type": "markdown", "metadata": {}, @@ -618,20 +330,6 @@ "````" ] }, - { - "cell_type": "markdown", - "metadata": { - "tags": [ - "cor" - ] - }, - "source": [ - "````{admonition} Solution\n", - ":class: tip, dropdown\n", - "`c[x][y] = 0`.\n", - "````" - ] - }, { "cell_type": "markdown", "metadata": {}, @@ -650,29 +348,6 @@ "````" ] }, - { - "cell_type": "markdown", - "execution_count": 15, - "metadata": { - "tags": [ - "cor" - ] - }, - "source": [ - "````{admonition} Solution\n", - ":class: tip, dropdown\n", - "``` python\n", - "def max_carre2(m):\n", - " c = m.copy()\n", - " for i in range(len(m)):\n", - " for j in range(len(m[0])):\n", - " if m[i][j] == 1:\n", - " c[i][j] = 1 + min(c[i - 1][j], c[i][j - 1], c[i - 1][j - 1])\n", - " return max(max(l) for l in c)\n", - "```\n", - "````" - ] - }, { "cell_type": "code", "execution_count": 16, @@ -702,21 +377,6 @@ "````" ] }, - { - "cell_type": "markdown", - "metadata": { - "tags": [ - "cor" - ] - }, - "source": [ - "````{admonition} Solution\n", - ":class: tip, dropdown\n", - "`max_carre2(m)` est en $\\boxed{O(n^2)}$ à cause des deux boucles `for` imbriquées. \n", - "C'est donc beaucoup mieux que `max_carre1(m)` qui est en $O(n^6)$.\n", - "````" - ] - }, { "cell_type": "markdown", "metadata": {}, diff --git a/dl/algo/prog_dyn/tp/tp1/tp_prog_dyn.html b/dl/algo/prog_dyn/tp/tp1/tp_prog_dyn.html index 2f34c260..ce2c629b 100644 --- a/dl/algo/prog_dyn/tp/tp1/tp_prog_dyn.html +++ b/dl/algo/prog_dyn/tp/tp1/tp_prog_dyn.html @@ -602,28 +602,10 @@

Coefficient binomialQuestion

Que peut-on prendre comme cas de base ?

-
-

Solution

-
-\[\binom{0}{k} = 0\]
-
-\[\binom{n}{0} = 1, \text{si } n \neq 0\]
-

Question

Écrire une fonction récursive binom_rec(n, k) renvoyant \(\binom{n}{k}\) à partir de la formule ci-dessus. Expliquer pourquoi la complexité de cette fonction est très mauvaise.

-
-

Solution

-
def binom_rec(n, k): # voir cours
-    if k == 0:
-        return 1
-    if n == 0:
-        return 0
-    return binom_rec(n - 1, k - 1) + binom_rec(n - 1, k)
-
-
-
binom_rec(20, 4)
@@ -655,20 +637,6 @@ 
Coefficient binomial
-
-

Solution

-
def binom_dp(n, k):
-    M = [[0]*(k + 1) for _ in range(n + 1)]
-    for i in range(0, n + 1):
-        M[i][0] = 1 # cas de base
-
-    for i in range(1, n + 1):
-        for j in range(1, k + 1):
-            M[i][j] = M[i - 1][j - 1] + M[i - 1][j]
-    return M[n][k]
-
-
-
binom_dp(20, 4)
@@ -697,20 +665,6 @@ 
Coefficient binomial
-
-

Solution

-
def binom(n, k):
-    d = {}
-    def aux(i, j):
-        if j == 0: return 1
-        if i == 0: return 0
-        if (i, j) not in d:
-            d[(i, j)] = aux(i - 1, j - 1) + aux(i - 1, j)
-        return d[(i, j)]
-    return aux(n, k)
-
-
-

Rendu de monnaie#

@@ -731,36 +685,11 @@

Rendu de monnaie\(a_k\) est utilisée ou non, on a dans le cas général : \(r(n, k) = \min(..., ...)\).

-
-

Solution

-

Si \(a_k\) est utilisée : il faut encore rendre \(n - a_k\) euros avec les pièces \(a_1\), …, \(a_k\) (on a le droit d’utiliser plusieurs fois \(a_k\)), d’où \(r(n, k) = r(n - a_k, k) + 1\).

-

Dans le cas général, on considère les deux possibilités et on conserve le minimum : -$\( - r(n, k) = min(r(n, k - 1), r(n - a_k, k) + 1) -\)$

-

Remarque : on ne peut utiliser \(a_k\) pour rendre \(n\) euros que si \(n \geq a_k\). Si \(n < a_k\), on a donc \(r(n, k) = r(n, k - 1)\).

-

Question

En déduire une fonction rendu(L, n) par programmation dynamique renvoyant le nombre minimum de pièces requises pour rendre n euros, où L est la liste des pièces.
On remplira une matrice M pour que M[i][j] contienne le nombre minimum de pièces pour rendre i euros en utilisant les j premières pièces de L.

-
-

Solution

-
def rendu(L, n):
-    k = len(L) # nombre de pièces
-    M = [[0]*(k + 1) for _ in range(n + 1)]
-    for i in range(1, n + 1):
-        M[i][0] = float("inf")
-        for j in range(1, k + 1):
-            if i - L[j - 1] >= 0:
-                M[i][j] = min(M[i][j - 1], 1 + M[i - L[j - 1]][j])
-            else:
-                M[i][j] = M[i][j - 1]
-    return M[-1][-1]
-
-
-
rendu([1, 2, 5], 7)
@@ -777,32 +706,6 @@ 
Rendu de monnaieQuestion

 
Réécrire la fonction précédente par mémoïsation plutôt que par programmation dynamique.
-
-

Solution

-
def rendu_memo(L, n):
-    k = len(L)
-    d = {}
-    def aux(i, j):
-        if (i, j) in d:
-            return d[(i, j)]
-        if i == 0:
-            return 0
-        if j == 0:
-            return float("inf")
-        if i - L[j - 1] >= 0:
-            d[(i, j)] = min(aux(i, j - 1), 1 + aux(i - L[j - 1], j))
-        else:
-            d[(i, j)] = aux(i, j - 1)
-        return d[(i, j)]
-    return aux(n, k)
-rendu_memo([1, 2, 5], 7)
-
-
-
2
-
-
-
-

Plus grand carré dans une matrice#

@@ -814,12 +717,6 @@

Plus grand carré dans une matriceQuestion

Définir M en Python.

-
-

Solution

-
M = [[1, 0, 0, 0], [0, 0, 1, 1], [0, 1, 1, 1], [0, 1, 0, 1]]
-
-
-

Méthode naïve#

@@ -827,17 +724,6 @@

Méthode naïveQuestion

Écrire une fonction est_carre telle que est_carre(m, x, y, k) détermine si la sous-matrice de m de taille \(k \times k\) et dont la case en haut à gauche a pour coordonnées (x, y) ne possède que des 1.

-
-

Solution

-
def est_carre(M, x, y, k):
-    for i in range(x, x + k):
-        for j in range(y, y + k):
-            if M[i][j] != 1:
-                return False
-    return True
-
-
-
assert(est_carre(M, 1, 2, 2) and not est_carre(M, 1, 1, 2))
@@ -849,18 +735,6 @@ 
Méthode naïveQuestion

 
Écrire une fonction contient_carre telle que contient_carre(m, k) renvoie true si m contient un carré de 1 de taille \(k\), false sinon.
-
-

Solution

-
def contient_carre(M, k):
-    n = len(M)
-    for i in range(n - k + 1):
-        for j in range(n - k + 1):
-            if est_carre(M, i, j, k):
-                return True
-    return False
-
-
-
assert(contient_carre(M, 2) and not contient_carre(M, 3))
@@ -872,17 +746,6 @@ 
Méthode naïveQuestion

 
Écrire une fonction max_carre1 telle que max_carre1(m) renvoie la taille maximum d’un carré de 1 contenu dans m.
-
-

Solution

-
def max_carre1(M):
-    n = len(M)
-    for k in range(n, 0, -1):
-        if contient_carre(M, k):
-            return k
-    return 0
-
-
-
max_carre1(M)
@@ -899,34 +762,16 @@ 
Méthode naïveQuestion

 
Quelle est la complexité de max_carre1(m) dans le pire cas ?
-
-

Solution

-
    -

  • est_carre(M, x, y, k) est en \(O(k^2)\).

    • -

  • contient_carre(M, k) appelle O(\(n\)) fois est_carre, donc est en \(O(n^2 k^2)\).

    • -

  • max_carre1(M) appelle contient_carre pour \(k = 1, 2, ..., n\), donc est de complexité \(\sum_{k=1}^n O(n^2 k^2) = O(n^3 \sum_{k=1}^n k^2)\). Comme \(\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} = O(n^3)\), la complexité totale est \(\boxed{O(n^6)}\).`

    • -
-

On va construire une matrice c telle que c[x][y] est la taille maximum d’un carré de 1 dans m dont la case en bas à droite est m[x][y] (c’est à dire un carré de 1 qui contient m[x][y] mais aucun m[i][j] si \(i > x\) ou \(j > y\)).
Par exemple, c[1][2] = 1 et c[2][3] = 2 pour la matrice \(M\) ci-dessus.

Question

Que vaut c[0][y] et c[x][0] ?

-
-

Solution

-

c[0][y] = 0 si m[0][y] = 0 et c[0][y] = 1 sinon.
-De même pour c[x][0].
-Remarque : c[0][y] et c[x][0] sont donc les mêmes valeurs que m[0][y] et m[x][0], on peut donc initialiser c comme une copie de m.

-

Question

Que vaut c[x][y] si m[x][y] = 0 ?

-
-

Solution

-

c[x][y] = 0.

-

Question

Montrer que, si m[x][y] = 1, c[x][y] = 1 + min(c[x-1][y], c[x][y-1], c[x-1][y-1]).

@@ -935,18 +780,6 @@

Méthode naïveQuestion

En déduire une fonction max_carre2 telle que max_carre2(m) renvoie la taille maximum d’un carré de 1 contenu dans m, ainsi que les coordonnées de la case en haut à gauche d’un tel carré.

-
-

Solution

-
def max_carre2(m):
-    c = m.copy()
-    for i in range(len(m)):
-        for j in range(len(m[0])):
-            if m[i][j] == 1:
-                c[i][j] = 1 + min(c[i - 1][j], c[i][j - 1], c[i - 1][j - 1])
-    return max(max(l) for l in c)
-
-
-
max_carre2(M)
@@ -963,11 +796,6 @@ 
Méthode naïveQuestion

 
Quelle est la complexité de max_carre2(m), en fonction des dimensions de m? Comparer avec max_carre1(m).
-
-

Solution

-

max_carre2(m) est en \(\boxed{O(n^2)}\) à cause des deux boucles for imbriquées.
-C’est donc beaucoup mieux que max_carre1(m) qui est en \(O(n^6)\).

-

Pour ceux qui ont fini#

diff --git a/searchindex.js b/searchindex.js index e842ada8..2b695d85 100644 --- a/searchindex.js +++ b/searchindex.js @@ -1 +1 @@ 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