1+ {
2+ "cells" : [
3+ {
4+ "cell_type" : " markdown"
5+ "metadata" : {},
6+ "source" : [
7+ " # 1A - Enoncé 5 novembre 2025 - comptabilité schtroumph\n "
8+ " \n "
9+ " Correction de l'examen du 5 novembre 2025.\n "
10+ " \n "
11+ " Toutes les questions valent 2 points."
12+ ]
13+ },
14+ {
15+ "cell_type" : " markdown"
16+ "metadata" : {},
17+ "source" : [
18+ " ## Q1"
19+ ]
20+ },
21+ {
22+ "cell_type" : " code"
23+ "execution_count" : 3 ,
24+ "metadata" : {},
25+ "outputs" : [
26+ {
27+ "data" : {
28+ "text/plain" : [
29+ " np.int64(45)"
30+ ]
31+ },
32+ "execution_count" : 3 ,
33+ "metadata" : {},
34+ "output_type" : " execute_result"
35+ }
36+ ],
37+ "source" : [
38+ " import numpy as np\n "
39+ " \n "
40+ " \n "
41+ " def distance(table1, table2):\n "
42+ " return np.abs(table1 - table2).sum()\n "
43+ " \n "
44+ " \n "
45+ " table1 = np.array([[8, 7], [18, 18], [6, 8]])\n "
46+ " table2 = np.array([[18, 18], [7, 9], [8, 6]])\n "
47+ " distance(table1, table2)"
48+ ]
49+ },
50+ {
51+ "cell_type" : " markdown"
52+ "metadata" : {},
53+ "source" : [
54+ " ## Q2"
55+ ]
56+ },
57+ {
58+ "cell_type" : " code"
59+ "execution_count" : 6 ,
60+ "metadata" : {},
61+ "outputs" : [
62+ {
63+ "data" : {
64+ "text/plain" : [
65+ " [(0, 1, 2, 3),\n "
66+ " (0, 1, 3, 2),\n "
67+ " (0, 2, 1, 3),\n "
68+ " (0, 2, 3, 1),\n "
69+ " (0, 3, 1, 2),\n "
70+ " (0, 3, 2, 1),\n "
71+ " (1, 0, 2, 3),\n "
72+ " (1, 0, 3, 2),\n "
73+ " (1, 2, 0, 3),\n "
74+ " (1, 2, 3, 0),\n "
75+ " (1, 3, 0, 2),\n "
76+ " (1, 3, 2, 0),\n "
77+ " (2, 0, 1, 3),\n "
78+ " (2, 0, 3, 1),\n "
79+ " (2, 1, 0, 3),\n "
80+ " (2, 1, 3, 0),\n "
81+ " (2, 3, 0, 1),\n "
82+ " (2, 3, 1, 0),\n "
83+ " (3, 0, 1, 2),\n "
84+ " (3, 0, 2, 1),\n "
85+ " (3, 1, 0, 2),\n "
86+ " (3, 1, 2, 0),\n "
87+ " (3, 2, 0, 1),\n "
88+ " (3, 2, 1, 0)]"
89+ ]
90+ },
91+ "execution_count" : 6 ,
92+ "metadata" : {},
93+ "output_type" : " execute_result"
94+ }
95+ ],
96+ "source" : [
97+ " import itertools\n "
98+ " \n "
99+ " \n "
100+ " def permutations(n):\n "
101+ " return list(itertools.permutations(list(range(n))))\n "
102+ " \n "
103+ " \n "
104+ " permutations(4)"
105+ ]
106+ },
107+ {
108+ "cell_type" : " markdown"
109+ "metadata" : {},
110+ "source" : [
111+ " ## Q3, Q4"
112+ ]
113+ },
114+ {
115+ "cell_type" : " code"
116+ "execution_count" : 23 ,
117+ "metadata" : {},
118+ "outputs" : [
119+ {
120+ "data" : {
121+ "text/plain" : [
122+ " array([[ 6, 8],\n "
123+ " [18, 18],\n "
124+ " [ 8, 7]])"
125+ ]
126+ },
127+ "execution_count" : 23 ,
128+ "metadata" : {},
129+ "output_type" : " execute_result"
130+ }
131+ ],
132+ "source" : [
133+ " def permute_ligne_ou_colonne(table, permutation, axis):\n "
134+ " if axis == 0:\n "
135+ " if len(table.shape) == 2:\n "
136+ " return table[permutation, :]\n "
137+ " return table[list(permutation)]\n "
138+ " return table[:, permutation]\n "
139+ " \n "
140+ " \n "
141+ " permute_ligne_ou_colonne(table1, [2, 1, 0], axis=0)"
142+ ]
143+ },
144+ {
145+ "cell_type" : " markdown"
146+ "metadata" : {},
147+ "source" : [
148+ " ## Q5"
149+ ]
150+ },
151+ {
152+ "cell_type" : " code"
153+ "execution_count" : 24 ,
154+ "metadata" : {},
155+ "outputs" : [
156+ {
157+ "data" : {
158+ "text/plain" : [
159+ " ((1, 0, 2), (1, 0))"
160+ ]
161+ },
162+ "execution_count" : 24 ,
163+ "metadata" : {},
164+ "output_type" : " execute_result"
165+ }
166+ ],
167+ "source" : [
168+ " def optimise(table1, table2):\n "
169+ " best = None\n "
170+ " for p1 in permutations(table1.shape[0]):\n "
171+ " for p2 in permutations(table1.shape[1]):\n "
172+ " t = permute_ligne_ou_colonne(permute_ligne_ou_colonne(table1, p1, 0), p2, 1)\n "
173+ " d = distance(t, table2)\n "
174+ " if best is None or d < best:\n "
175+ " best = d\n "
176+ " perms = p1, p2\n "
177+ " return perms\n "
178+ " \n "
179+ " \n "
180+ " optimise(table1, table2)"
181+ ]
182+ },
183+ {
184+ "cell_type" : " markdown"
185+ "metadata" : {},
186+ "source" : [
187+ " ## Q6\n "
188+ " \n "
189+ " Si $i$ et $j$ sont les dimensions de deux tables, c'est $O((i!)(j!))$."
190+ ]
191+ },
192+ {
193+ "cell_type" : " markdown"
194+ "metadata" : {},
195+ "source" : [
196+ " ## Q7"
197+ ]
198+ },
199+ {
200+ "cell_type" : " code"
201+ "execution_count" : 25 ,
202+ "metadata" : {},
203+ "outputs" : [
204+ {
205+ "data" : {
206+ "text/plain" : [
207+ " ((0, 1), (1, 0, 2))"
208+ ]
209+ },
210+ "execution_count" : 25 ,
211+ "metadata" : {},
212+ "output_type" : " execute_result"
213+ }
214+ ],
215+ "source" : [
216+ " def optimise_vecteur(vec1, vec2):\n "
217+ " best = None\n "
218+ " for p1 in permutations(vec1.shape[0]):\n "
219+ " t = permute_ligne_ou_colonne(vec1, p1, 0)\n "
220+ " d = distance(t, vec2)\n "
221+ " if best is None or d < best:\n "
222+ " best = d\n "
223+ " perm = p1\n "
224+ " return perm\n "
225+ " \n "
226+ " \n "
227+ " def optimise_fast(table1, table2):\n "
228+ " return (\n "
229+ " optimise_vecteur(table1.sum(axis=0), table2.sum(axis=0)),\n "
230+ " optimise_vecteur(table1.sum(axis=1), table2.sum(axis=1)),\n "
231+ " )\n "
232+ " \n "
233+ " \n "
234+ " optimise_fast(table1, table2)"
235+ ]
236+ },
237+ {
238+ "cell_type" : " markdown"
239+ "metadata" : {},
240+ "source" : [
241+ " Le coût est en $O(i!) + O(j!)$. Pas nécessairement optimal mais beaucoup plus rapide."
242+ ]
243+ },
244+ {
245+ "cell_type" : " markdown"
246+ "metadata" : {},
247+ "source" : []
248+ },
249+ {
250+ "cell_type" : " code"
251+ "execution_count" : null ,
252+ "metadata" : {},
253+ "outputs" : [],
254+ "source" : [
255+ " import numpy as np\n "
256+ " \n "
257+ " \n "
258+ " coef = np.random.rand(5).reshape((-1, 1))\n "
259+ " coef[:, 0] = 1\n "
260+ " print(coef)\n "
261+ " t1 = np.ones((5, 1)) @ np.array([[5, 1, 2, 3, 4]], dtype=np.float32)\n "
262+ " t2 = np.ones((5, 1)) @ np.array([[2, 3, 1, 4, 5]], dtype=np.float32)\n "
263+ " M = np.array(\n "
264+ " [\n "
265+ " [0, 0, 1, 0, 0],\n "
266+ " [0, 0, 0, 1, 0],\n "
267+ " [0, 1, 0, 0, 0],\n "
268+ " [0, 0, 0, 0, 1],\n "
269+ " [1, 0, 0, 0, 0],\n "
270+ " ]\n "
271+ " ).T\n "
272+ " \n "
273+ " # Il faut diminuer le nombre de solutions pour n'en garder qu'une.\n "
274+ " for i in range(5):\n "
275+ " t1[i, t1[i, :] == i] = i + 0.01\n "
276+ " t2[i, t2[i, :] == i] = i + 0.01\n "
277+ " \n "
278+ " assert (t1 @ M - t2).max() < 1e-6\n "
279+ " \n "
280+ " m = np.linalg.lstsq(t1, t2)\n "
281+ " print(t1)\n "
282+ " print(t2)\n "
283+ " print(M)\n "
284+ " print((m[0] * 100).astype(int) / 100)"
285+ ]
286+ }
287+ ],
288+ "metadata" : {
289+ "kernelspec" : {
290+ "display_name" : " this312"
291+ "language" : " python"
292+ "name" : " python3"
293+ },
294+ "language_info" : {
295+ "codemirror_mode" : {
296+ "name" : " ipython"
297+ "version" : 3
298+ },
299+ "file_extension" : " .py"
300+ "mimetype" : " text/x-python"
301+ "name" : " python"
302+ "nbconvert_exporter" : " python"
303+ "pygments_lexer" : " ipython3"
304+ "version" : " 3.12.3"
305+ }
306+ },
307+ "nbformat" : 4 ,
308+ "nbformat_minor" : 2
309+ }
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