Le FCSC sans se fatiguer

2022-05-08T00:00:00+00:00

C'est le printemps, la saison du France Cybersecurity Challenge</a> et donc de mon billet de blog annuel, apparemment. Organisé par l'ANSSI</a>, l'agence de cybersécurité française, le FCSC est l'occasion de sélectionner les membres de l'équipe de France pour l'ECSC, le challenge européen. Le challenge est généreusement aussi ouvert aux vieux comme moi qui ne sont pas éligibles pour raisons d'âge 1</a></sup>.</p>

La diversité et l'originalité des épreuves en font facilement l'un de mes concours préférés. Je m'étais beaucoup amusé</a> avec les épreuves hardware l'année dernière, et cette année j'ai eu à nouveau beaucoup de plaisir dans de multiples catégories d'épreuves, mais surtout en crypto.</p>

Comme c'est de tradition, je prolonge le plaisir en me prêtant au jeu des writeups</em> pour quelques unes des épreuves que j'ai le plus appréciées.</p>

Hardware</h2> Mommy Morse (77 résolutions)</h3>
Il est demandé de transmettre un signal en morse, où les points et les traits sont représentés par une « porteuse pure à une fréquence de 5 kHz », et les espacements par une fréquence de 1 kHz. Un signal d'exemple est fourni, et comme j'aime bien visualiser, j'ai fait un graphe:</p>

point trait point point ESPACE point etc..</em></figcaption> </figure>
Comme d'habitude, je trouve que tracer l'argument complexe du signal aide bien, parce que dans ce cas il évolue de manière linéaire (modulo le modulo $2\pi$), avec une pente qui dépend de la fréquence de la porteuse. C'est exactement la manière dont je m'y suis pris pour créer le signal dont j'avais besoin : j'ai créé l'argument qu'il fallait, que j'ai mis dans une exponentielle complexe et c'est passé.</p>
Crypto</h2>
Hash-ish (71 résolutions) et Khal Hash (10 résolutions)</h3>
Sûrement mes deux épreuves préférées :)</p>
Hash-ish donne un entier `H</code> et demande de trouver deux entiers a</code> et b</code> tels que la valeur de hash((a, b))</code> soit H</code> (hash</code> étant la fonction de hachage par défaut de Python).</p>`
`Mais comment ça fonctionne, hash</code>, d'abord ? Pour les tuples,` `son code est là</a> ou bien simplifié2</a></sup> ici en Python (PH1</code>, PH2</code> et PH5</code> sont des constantes définies dans le code de CPython, et MOD</code> vaut $2^{64}$) :</p>`
def </span>myhash</span>(</span>integer_tuple</span>: Tuple[int, </span>...</span>]): </span> acc = </span>PH5 </span> </span>for </span>element </span>in </span>integer_tuple: </span> acc = (acc + element * </span>PH2</span>) % </span>MOD </span> acc = ((acc << </span>31</span>) | (acc >> </span>33</span>)) % </span>MOD </span> acc = (acc * </span>PH1</span>) % </span>MOD </span> acc = (acc + </span>len</span>(t) ^ (</span>PH5 </span>^ </span>3527539</span>)) % </span>MOD </span> </span>return </span>acc </span></code></pre> C'est donc cette fonction qu'il faut inverser (on sait ce qu'elle doit retourner, et on veut savoir quel argument lui donner pour qu'elle fasse ça).</p>
Entre Z3</a>, un « prouveur de théorèmes » développé par Microsoft Research</em>. Plus précisément, c'est un solveur SMT</a>. Sans rentrer dans les détails, un solveur SMT est une généralisation d'un solveur SAT</a>. Alors qu'un solveur SAT détermine la satisfiabilité de formules logiques (avec des portes logiques et des booléens), un solveur SMT détermine la satisfiabilité de formules impliquant des objets plus complexes, comme des nombres ou des structures de données. Ici, c'est exactement ce qu'il faut : on a une formule mathématique à résoudre sur les entiers modulo $2^{64}$.</p>
Z3 possède des variables de type entier, mais vu qu'ici on a des opérations sur des bits, j'ai préféré utiliser les `BitVec</code>s, qui sont comme leur nom l'indique des tableaux de bits de taille fixée, et sur lesequels on peut quand même faire des opérations arithmétiques facilement.</p>`
def </span>solve</span>(</span>h</span>): </span> N = </span>64 </span> s = z3.</span>Solver</span>() </span> </span># a and b are the integers that we are looking for </span> a = z3.</span>BitVec</span>('</span>a</span>', N) </span> b = z3.</span>BitVec</span>('</span>b</span>', N) </span> </span># they need to be less than 260, or they don't hash to themselves </span> </span># ULT = Unsigned Less Than: `a < 260` doesn't work unless we also add `0 <= a` </span> s.</span>add</span>(z3.</span>ULT</span>(a, </span>2</span></span>60</span>)) </span> s.</span>add</span>(z3.</span>ULT</span>(b, </span>2</span></span>60</span>)) </span> </span># hashing the first integer </span> acc0 = z3.</span>BitVec</span>('</span>acc0</span>', N) </span> s.</span>add</span>(acc0 == </span>PH5 </span>+ a * </span>PH2</span>) </span> acc1 = z3.</span>BitVec</span>('</span>acc1</span>', N) </span> </span># LShR stands for Logical Shift Right, >> being the arithmetic one </span> s.</span>add</span>(acc1 == (acc0 << </span>31</span>) | z3.</span>LShR</span>(acc0, </span>33</span>)) </span> acc2 = z3.</span>BitVec</span>('</span>acc2</span>', N) </span> s.</span>add</span>(acc2 == acc1 * </span>PH1</span>) </span> </span># hashing the second integer </span> acc3 = z3.</span>BitVec</span>('</span>acc3</span>', N) </span> s.</span>add</span>(acc3 == acc2 + b * </span>PH2</span>) </span> acc4 = z3.</span>BitVec</span>('</span>acc4</span>', N) </span> s.</span>add</span>(acc4 == (acc3 << </span>31</span>) | z3.</span>LShR</span>(acc3, </span>33</span>)) </span> acc5 = z3.</span>BitVec</span>('</span>acc5</span>', N) </span> s.</span>add</span>(acc5 == acc4 * </span>PH1</span>) </span> </span># the final hash value should be the one specified </span> s.</span>add</span>(h == acc5 + (</span>2 </span>^ (</span>PH5 </span>^ </span>3527539</span>))) </span> </span># calling the solver </span> sol = s.</span>check</span>() </span> </span>if </span>str</span>(sol) == '</span>sat</span>': </span> </span># a solution was found: return it </span> model = s.</span>model</span>() </span> </span>return </span>(model.</span>evaluate</span>(a).</span>as_long</span>(), model.</span>evaluate</span>(b).</span>as_long</span>()) </span></code></pre> Appelée avec h = 42</code> (au hasard), la fonction retourne le tuple suivant: (471024597674449489, 129304832457527322)</code>. On vérifie :</p>
>>> </span>hash</span>((</span>471024597674449489</span>, </span>129304832457527322</span>)) </span>42 </span></code></pre> Ça suffisait pour passer Hash-ish. En revanche, le grand frère Khal Hash est plus coriace. En effet, il demande toujours de trouver un tuple d'entiers dont le hash est une valeur donnée, mais avec la contrainte que les entiers doivent être des caractères ascii (compris entre 0 et 127). Or, deux entiers ne suffisent plus : en effet, le hash étant une valeur de 64 bits, il faut environ le même nombre de paramètres pour avoir une chance, soit environ 9 ou 10 entiers (puisque chacun a 7 bits utilisables et que $64 / 7 \approx 9$). Z3 est malheureusement dépassé par ces contraintes (le problème est beaucoup plus profond</em> : non seulement il y a plus de variables intermédiaires, mais aussi le chemin est plus long entre le hash final et l'entrée à trouver).</p> J'ai donc écrit une version optimisée de la fonction hash</code> en Rust, qui tourne en une poignée de nanosecondes pour une entrée de 10 entiers. C'est rapide ! Il ne reste plus qu'à essayer des tuples au hasard jusqu'à tomber sur la valeur qu'on voulait. Pas vrai ? Pas vrai ?</em></p>
On peut essayer d'estimer combien de temps ça peut prendre : la probabilité de tomber sur la valeur qu'on veut en hachant un tuple est environ $2^{-64}$. Donc la probabilité de ne pas tomber dessus est de $1-2^{-64}$. La probabilité de ne jamais être tombé dessus après avoir haché $n$ tuples est donc $(1-2^{-64})^n$. Quelle valeur doit avoir $n$ pour que cette probabilité soit assez basse, disons à $1/2$ ? La réponse est $n \approx 2^{64}$, à peu de choses près. Même avec la fonction optimisée, ça prendrait plus de 2000 ans de hacher ce nombre de tuples`3</a></sup>. Le FCSC ne durant que 10 jours, il faut trouver une autre technique.</p>`
`La clé ici était de se rendre compte que les opérations de la fonction de hash sont toutes inversibles. En effet, reprenons la fonction ligne par ligne:</p>`
acc = (acc + element * </span>PH2</span>) % </span>MOD </span></code></pre> Pour inverser cette addition, il suffit de retirer element * PH2</code> modulo MOD</code> à acc</code>.</p> acc = ((acc << </span>31</span>) | (acc >> </span>33</span>)) % </span>MOD </span></code></pre> Cette ligne réalise en fait une rotation des bits de acc</code>, de 33 bits vers la droite (ou 31 vers la gauche). Pour l'inverser, il suffit donc de faire la même rotation à l'envers.</p> acc = (acc * </span>PH1</span>) % </span>MOD </span></code></pre> Pour inverser cette multiplication, on multiplie acc</code> par l'inverse modulaire de PH1</code> modulo MOD</code>. On peut calculer facilement cet inverse en python avec la fonction pow</code> du langage: pow(PH1, -1, MOD)</code>.</p> acc = (acc + </span>len</span>(t) ^ (</span>PH5 </span>^ </span>3527539</span>)) % </span>MOD </span></code></pre> La dernière ligne est également facile à inverser puisqu'il s'agit encore juste d'une addition.</p> Tout cela signifie qu'en partant de la valeur finale connue du hash, on peut prendre les éléments du tuple à l'envers et revenir à la valeur initiale de l'accumulateur acc</code> (PH5</code>). Mais chercher dans un sens ou dans l'autre ne change rien en termes de temps de calcul. L'astuce est de chercher par les deux bouts et de faire se retrouver la recherche au milieu.</p> </figure> Imaginons qu'on commence par le faisceau de droite, et qu'on enregistre tous ses points d'arrivée. Maintenant, le faisceau de gauche peut se raccorder non plus juste au point final, mais à chacun des $n$ points du faisceau de droite. On arrive ainsi à $n \approx 2^{32}$ au lieu des $2^{64}$ de la première méthode, pour la même probabilité de succès. C'est un sacré progrès, et ça se calcule à présent en quelques secondes ou minutes au lieu de miliers d'années4</a></sup>.</p> En pratique, j'ai calculé tous les tuples de taille 4 en partant d'un côté, et cherché les tuples de taille 6 en partant de l'autre côté, et je suis tombé sur une solution en une ou deux minutes.</p> T-Rex (67 résolutions)</h3> Je passe des détails, mais en gros on nous donnait le résultat de $F^{31337}(k)$ où $F$ est l'application $x \mapsto (2x+1)x$ modulo $2^{128}$, et on veut remonter à $k$ qui est la clé de chiffrement du message. Après m'être rendu compte que $F$ était bijective, j'ai passé un bon moment à essayer de l'inverser à la main sans succès, avant de réaliser que je pouvais utiliser là aussi Z3 (31337 fois de suite).</p> Gaston La Paffe (29 résolutions)</h3> On avait ici un algorithme de signature à base de polynomes, qui fonctionnait de la manière suivante (là aussi, j'omets des détails) :</p> s1</code> et s2</code> des polynomes choisis au hasard, constituent la clé privée ;</li> On nous donne a</code> et t</code>, polynomes tels que t = as1 + s2</code> ;</li> Pour signer un message m</code>, y1</code> et y2</code> sont d'abord générés aléatoirement. Puis on calcule c = H(ay1 + y2, m)</code> où H</code> est une fonction à sens unique. Enfin, on retourne z1 = s1c + y1</code> et z2 = s2c + y2</code>, ainsi que y1</code> « par erreur ».</li> </ul> Quelques calculs permettent de se rendre compte que connaître y1</code> permet de trouver aussi y2</code>:</p> y2 = a(z1 - y1) + z2 - tc </span></code></pre> et qu'à partir de là, on a les équations</p> s1 * c = z1 - y1 </span>s2 * c = z2 - y2 </span></code></pre> dans lesquelles on connaît toutes les variables à part les secrets s1</code> et s2</code>.</p> Pour trouver les coefficients des secrets, j'ai choisi d'utiliser ortools</code>, une suite de solveurs développée par Google. Z3 aurait peut-être pu aussi fonctionner, mais il s'agissait ici en fait d'un problème de programmation linéaire sur les entiers, ce sur quoi ortools</code> est plus efficace.</p> Side Channels and Fault Attacks</h2> Never Skip Class Nor Multiplication (85 résolutions)</h3> On avait accès à un oracle RSA chiffrant des messages choisis, à l'aide d'une fonction d'exponentiation modulaire « buguée » de manière à ce qu'on puisse choisir quelle multiplication elle allait sauter. L'objectif était de déterminer la clé privée $d$ (l'exposant de l'opération $c = m^d \mod n$).</p> La question à se poser est la suivante : Si je saute la multiplication du bit $k$ de la clé $d$, quel est l'exposant $d'$ qui a alors été utilisé ?</p> Exposant d initial Exposant d' réel </span>N-1 K 0 skip K N-1 K 0 </span>[ dH ] 0 [ dL ] -----> [ dH ] 0 [ dL ] </span> </span>N-1 K 0 skip K N-1 K 0 </span>[ dH ] 1 [ dL ] -----> [ dH ] 0 [ dL ] </span></code></pre> On remarque facilement que sauter un bit nul ne modifie pas l'exposant, alors que sauter un bit non-nul le change. Il suffit alors de demander à chiffrer le même message en sautant tous les bits un par un, et de comparer ces résultats à la situation où on ne saute aucun bit. On a alors une correspondance directe entre les résultats et la clé.</p> Never Skip Class Nor Squaring (70 résolutions)</h3> Même principe, mais la fonction est buguée de manière à sauter l'étape de mise au carré. La question est la même : si on saute le bit $K$ de l'exposant $d$, quel exposant $d'$ a-t-on alors vraiment utilisé ?</p> Exposant d initial Exposant d' réel </span>N-1 K 0 skip K N-2 K 0 </span>[ dH ] 0 [ dL ] -----> [ dH ] [ dL ] </span> </span>N-1 K 0 skip K N-2 K 0 </span>[ dH ] 1 [ dL ] -----> [ dH + 1 ] [ dL ] </span></code></pre> Grâce à quelques dessins dans ce genre, on se rend compte que peu importe la valeur du bit $K$, on a toujours $d - d' = d_H \cdot 2^K$. Cette observation permet de déterminer $d$ bit par bit. En effet, si on appelle $c$ la vraie valeur de $m^d \mod n$, en sautant le bit $N-2$ on obtient pour $m^{d'}$ deux valeurs possibles selon que le bit $N-1$ est nul ou non : soit $c$ (s'il est nul), soit $c\cdot m^{-2^{N-2}}$ (s'il ne l'est pas). Armé de cette information, on saute alors le bit $N-3$ pour déterminer la valeur du bit $N-2$, etc.</p> Misc</h2> Guess Me Too (68 résolutions)</h3> Un nombre de 128 bits est tiré au hasard et il faut le deviner. On ne peut poser que 136 questions de la forme « Parmi les bits numéros $x_0$, $x_1$, $x_2$, etc. du nombre secret, quelle est la parité du nombre de bits non nuls ? » (en choisissant les $x_i$). De plus, on sait qu'une des réponses sera fausse mais on ne sait pas laquelle.</p> Il semble logique de commencer par demander les 128 bits un à un. On a déjà presque 6% de chance d'avoir deviné le nombre (si la réponse incorrecte tombe dans les 8 autres questions), donc on pourrait déjà s'arrêter là et faire plusieurs tentatives jusqu'à ce que ça marche (17 fois en moyenne).</p> Pour deviner le nombre à coup sûr, il faut faire de la correction d'erreur.</p> On commence par demander la parité de la somme de tous les bits, et on compare avec le résultat des 128 premières questions. Si les résultats concordent, on sait qu'on n'est pas encore tombé sur la réponse incorrecte. Sinon, c'est qu'il y a une erreur soit dans les questions individuelles, soit dans la question de contrôle. Pour la déterminer, on demande alors la parité de la somme de tous les bits impairs, que l'on compare à nouveau au résultat des questions individuelles. On sait que le résultat de cette question sera correct, puisqu'on a déjà détecté une erreur et qu'il ne peut y en avoir qu'une seule. Si les résultats concordent, on sait que l'erreur est dans les bits pairs, sinon c'est qu'elle est dans les bits impairs. On peut alors continuer à préciser la position de l'erreur en demandant la parité de la somme des bits impairs des bits suspects d'être erronés, jusqu'à ce qu'il n'y ait plus qu'une seule possibilité.</p> En pratique, toutes les questions doivent être posées en même temps au début. Qu'à cela ne tienne, les huit questions supplémentaires à poser qui fonctionnent dans tous les cas sont les suivantes :</p> 11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111 </span>10101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010 </span>11001100110011001100110011001100110011001100110011001100110011001100110011001100110011001100110011001100110011001100110011001100 </span>11110000111100001111000011110000111100001111000011110000111100001111000011110000111100001111000011110000111100001111000011110000 </span>11111111000000001111111100000000111111110000000011111111000000001111111100000000111111110000000011111111000000001111111100000000 </span>11111111111111110000000000000000111111111111111100000000000000001111111111111111000000000000000011111111111111110000000000000000 </span>11111111111111111111111111111111000000000000000000000000000000001111111111111111111111111111111100000000000000000000000000000000 </span>11111111111111111111111111111111111111111111111111111111111111110000000000000000000000000000000000000000000000000000000000000000 </span></code></pre> Si on regarde par exemple la ligne 11001100...</code>, elle contient bien tous les bits impairs des bits impairs, ainsi que tous les bits impairs des bits pairs, et ainsi de suite.</p> Coda</h2> Z3 et ortools</code> se sont avérés être un arsenal efficace cette année, me permettant de résoudre 5 challenges (dont 4 de crypto) sans trop d'effort. Ils rendent aussi service en dehors du FCSC : ce sont des outils à connaître.</p> ^{1</sup> et qui ne l'ont d'ailleurs jamais été :'(</p> </div> ^{2</sup> notamment en profitant du fait que hash(n) == n</code> pour tout n</code> entier inférieur à $2^{60}$</p> </div> ^{3</sup> Et au bout du compte, on aurait encore une chance sur deux de ne pas avoir trouvé la bonne valeur…</p> </div> ^{4</sup> C'est un exemple d'application du paradoxe des anniversaires</a>, un grand classique de la cryptanalyse, qui, en échange de la mémoire utilisée pour stocker des hashes, divise l'exposant du nombre de calculs à faire par deux.</p> </div>}}}}

Guessing lottery game

2019-05-15T00:00:00+00:00

I found a fun problem on reddit</a> some time ago:</p>

A number is randomly selected between two bounds. Each player can make one guess, the closest to the selected number wins.</p> </blockquote>
The author of this comment notes that they "always use a random number generator for [their] pick rather than guessing to avoid accidentally clustering with others". This relies on the idea illustrated by this reddit thread, that people tend to favour the same numbers when asked to make a random choice, thus generating non-uniform distributions that can create lumps. Picking a number from a uniform distribution might then increase the chances to pick it in a less preferred region, and thus, the chances to be the closest from the target.</p>
But could we do better than that? What is the optimal strategy?</p>
Against unbiased opponents</h1>
We could start assuming that the opponents are unbiased and all pick numbers from the uniform distribution on the allowed range. Knowing that, where should we play?</p>
First of all, we should decide on a couple of details on the lottery that we will be modelling: discrete (we can pick any integer from $1</script> to <script type="math/tex">N</script> included), or continuous (we can pick any real number between <script type="math/tex">0</script> and <script type="math/tex">1</script> ), and what should we do in case of tie. These rules lead to slightly different behaviours, but I suggest we study both the discrete and continuous cases. Ties will not happen in the continuous case, but in the discrete case, we could either</p> <ul> <li>give a whole prize to everyone tied first;</li> <li>make another lottery among only the people tied first, and repeatedly until only one person remains;</li> <li>split the prize equally between the people tied first;</li> <li>decide that everyone lost.</li> </ul> <p>The second case is a bit more delicate to study because it would be reasonable to allow a change of strategy between the rounds, depending on the number of opponents, so I propose that we restrain ourselves to the simple cases where ties cause the prize to be either multiplied, split, or destroyed.</p> <h2 id="discrete-case">Discrete case</h2> <p>We are searching for the winning expectancy <script type="math/tex">\mathbf{E}(N,K,C)</script> against <script type="math/tex">K</script> opponents, if we play at <script type="math/tex">C</script> (between <script type="math/tex">1</script> and <script type="math/tex">N</script> ).</p> <p>Assume the selected number is <script type="math/tex">S</script> . Then we can count the number <script type="math/tex">N_=</script> of numbers <script type="math/tex">X</script> at the same distance from the target as we are: such that <script type="math/tex">|X-S| = |C-S|</script> . This will be <script type="math/tex">1</script> or <script type="math/tex">2</script> . We can also count the number <script type="math/tex">N_\gt</script> of numbers <script type="math/tex">X</script> farther from the target than we are: such that <script type="math/tex">|X-S| \gt |C-S|</script> . If <script type="math/tex">E</script> of our opponents are at the same distance from the target as we are, and all the others are farther, then in case we split the prize, we win <script type="math/tex">\frac{1}{E+1}</script> of the prize. This happens with probability <script type="math/tex">\binom{K}{E} \left(\frac{N_=}{N}\right)^E \left(\frac{N_\gt}{N}\right)^{K-E}</script> for given <script type="math/tex">C</script> and <script type="math/tex">S</script> . The total winning expectancy is then</p> <script type="math/tex;mode=display">\mathbf{E}(N,K,C) = \frac{1}{N}\sum_{S=1}^{N}{ \sum_{E=0}^{K}{ \frac{\binom{K}{E} \left(\frac{N_=(C,S)}{N}\right)^E \left(\frac{N_>(C,S)}{N}\right)^{K-E}}{E+1} } }</script> <figure> <figcaption style="text-align: center"><em>Expectancy to win in the uniform discrete case if the prize is split between the exaequos, for N=10.</em></figcaption> </figure> <p>From this graph, we indeed win every time when we have no opponent, we win less often as the number of opponents increases, and picking the middle region seems in general to be a good choice.</p> <p>But as the number of opponents goes up, the winning expectancy goes down and it becomes more difficult to distinguish the strategy. To make it appear more clearly, I will normalize each curve such that their integrals are equal (more precisely I divide them by the total winning expectancy for this number of opponents).</p> <figure> <figcaption style="text-align: center"><em>Strategies for the uniform discrete case if the prize is split between the exaequos, for N=10.</em></figcaption> </figure> <p>What jumps to the eye is that for enough opponents, the best strategy is not to play in the middle any more. We can also note that the strategies against 1 or 2 opponents are identical. Finally, when the number of opponents becomes very large, the best strategy becomes to chose uniformly: indeed in this case all the numbers will probably have been chosen by several people, so that we only win if we pick exactly the selected number.</p> <p>Similar reasoning allows to obtain formulas for the other rules:</p> <figure> <figcaption style="text-align: center"><em>Strategies for the uniform discrete case, for N=10.</em></figcaption> </figure> <h2 id="continuous-case">Continuous case</h2> <p>A closed-form formula is easier to find for the continuous case: following a similar reasoning, the winning expectancy if we play at <script type="math/tex">x\in[0,1]</script> against <script type="math/tex">K</script> opponents is</p> <script type="math/tex;mode=display">\mathbf{E}(K, x) = \frac{2+x^K(1+K-x\,(2+K))+(1-x)^K(x\,(2+K)-1)}{2\,(1+K)}</script> <figure> <figcaption style="text-align: center"><em>Expectancy to win in the continuous case.</em></figcaption> </figure> <figure> <figcaption style="text-align: center"><em>Strategies in the continuous case.</em></figcaption> </figure> <p>A small advantage seems to subsist in the continuous case, since we can make a guess as close to the border as we want. The positions of the two peaks of the strategy <script type="math/tex">p(K)</script> and <script type="math/tex">1-p(K)</script> are such that <script type="math/tex">p(K)\sim\frac{2}{2+K}</script> when <script type="math/tex">K</script> tends to infinity, and <script type="math/tex">(1+K)\\,\mathbf{E}(K,p(K)) = (1+K)\\,\mathbf{E}(K,1-p(K)) \to 1+\frac{1}{2\\,\mathrm{e}^2} \approx 1.068</script> , the demonstration of these properties is left as an exercise to the reader.</p> <h1 id="back-to-the-real-world">Back to the real world</h1> <p>Ok but the point of the reddit post was that humans don't pick numbers uniformly. What does everything become if the selected number is still uniformly random, but the opponents pick at random according to the distribution given in the post (favouring 7 and choosing infrequently 10 or 1)?</p> <figure> <figcaption style="text-align: center"><em>Strategies for the real world, for N=10.</em></figcaption> </figure> <p>Depending on the rules of the lottery, it is either beneficial to pick the 7, or to avoid the 7. Until at some point, the less preferred numbers 10 and 1 become the best options, at least in the versions that encourage winning alone (all ties lose, and to a lesser extent all ties split).</p> <p>But as nice as all this seems, I wouldn't encourage betting your house with this strategy. Indeed, it assumes random opponents and is not protected against smart opponents which might observe you play a couple of times and adapt their strategy accordingly. To handle this kind of adversaries, we can't play the same number every time: a constant strategy would be too easy to exploit. We then need a mixed strategy, which is a probability distribution over our possible actions. This will be for another post.</p> </article> <article> <h1>Stockholm visit guide</h1> <p>2019-05-06T00:00:00+00:00</p> <p>I spent two weeks in Stockholm, one of them dedicated to visiting the city. Here are some notes that could be useful for travelers. It is not at all exhaustive though, but just the result of my tastes and peregrinations. I would appreciate to know if you think I missed interesting places!</p> <figure> <figcaption style="text-align: center"><em>The Gustav Vasa church in Vasastan.</em></figcaption> </figure> <h1 id="if-you-are-in-a-hurry-my-top-3">If you are in a hurry, my top 3</h1> <ul> <li><em>Östermalms saluhall</em>;</li> <li><em>Nordiska museet</em> after 13 o'clock;</li> <li><em>Historiska museet</em>.</li> </ul> <h1 id="general-tips">General tips</h1> <ul> <li>1 EUR <script type="math/tex">\approx</script> 10 SEK, local currency can be withdrawn at the airport (and at least with my bank, it is cheaper to withdraw directly in SEK rather than accept the offered conversion in EUR, which includes a significant change tax);</li> <li>"Hello" = "Hej" or "Hejhej", pronounced not too far from the american "Hey"; "Thank you" = "Tack", but almost everyone is willing and able to help in English anyway;</li> <li>Museums have a tendency to open late (10-11 AM) and close early (4-5 PM);</li> <li>Single metro tickets bought at machines in metro stations are very expensive (45 SEK). If you buy one anyway, you obtain a paper ticket that you are supposed to show to the clerk to enter the metro. It is cheaper to buy a rechargable card for 20 SEK (lasts for life) at the counter and charge it with trips.</li> <li>Some metro stations are really pretty. But in general I prefer walking, which is especially suitable in Stockholm since the city is not too large. Walking is also the only way to stumble on random places that one wouldn't have discovered otherwise. Be aware though that Stockholm is a relatively hilly city and some of the paved streets in the old city are in a rather poor condition, so good shoes and a good physical condition are appropriate.</li> <li>Stockholm is spread over a handful of islands of different sizes and ambiances, whose main ones are: <ul> <li>the continent, <ul> <li>Kungsholmen ("The King's islet"), where my hostel was, but not the island that I found the most interesting,</li> <li>Gamla Stan (the old city) with the King's palace, a lot of old paved streets, tiny passages and many tourists. Consequently, it is not the place where the food is the cheapest. <ul> <li>Riddarholmen ("The Knights' islet"), a small island with mostly administrative buildings and courthouses, but it is prettier than it sounds,</li> </ul> </li> <li>Skeppsholmen ("The islet of the ships"),</li> <li>Djurgården ("The animal park"), also nicknamed "museum island",</li> <li>Södermalm.</li> </ul> </li> </ul> </li> </ul> <h1 id="points-of-interest">Points of interest</h1> <h2 id="continent">Continent</h2> <figure> <figcaption style="text-align: center"><em>Stockholm's 18th century observatory.</em></figcaption> </figure> <ul> <li><a href="https://www.openstreetmap.org/#map=18/59.33458/18.09006">Historiska museet</a> Museum of the history of Sweden, free and interesting;</li> <li><a href="https://www.openstreetmap.org/#map=19/59.33594/18.07908">Östermalms saluhall</a> Covered market, closed on Sundays. It is a <em>very</em> nice place to buy all kinds of fancy food (fish, meat, bread), and also to eat traditional Swedish meals that can be heated up and served on the spot. It is in a temporary location until winter 2020, while the historical building is being renovated. Without contest one of my very favourite places in Stockholm;</li> <li><a href="https://www.openstreetmap.org/#map=17/59.34182/18.05457">The observatory hill</a> Nice point of view to see the sunrise over Stockholm;</li> <li><a href="https://www.openstreetmap.org/#map=19/59.34792/18.07239">KTH library</a> seems open to everyone, is a pretty nice building and is a nice place to study, or prepare the rest of the day;</li> <li><a href="https://www.openstreetmap.org/#map=19/59.35008/18.06806">KTH R1</a> The dismantled (and safe) remains of the first nuclear reactor in Sweden. It is now used as a culture center and exhibitions and parties are regularly organized there.</li> </ul> <h2 id="kungsholmen">Kungsholmen</h2> <figure> <figcaption style="text-align: center"><em>Stockholm's city hall.</em></figcaption> </figure> <ul> <li><a href="https://www.openstreetmap.org/#map=18/59.32742/18.05477">The city hall</a> is open to visits during the summer, but the view from under the archs is very pretty all year long. This is where the Nobel gala dinner is served on December 10th. Fun fact: the restaurant is open all year long and can serve the menu from any given year's gala, on the porcelain actually used for the banquet.</li> </ul> <h2 id="gamla-stan">Gamla Stan</h2> <figure> <figcaption style="text-align: center"><em>The Swedish parliament, not exactly on Gamla Stan but close enough.</em></figcaption> </figure> <p>Not my favourite part of the city: too many tourists.</p> <ul> <li><a href="https://www.openstreetmap.org/#map=19/59.32677/18.07168">The royal palace</a>, the official residence of the royal family (but they don't usually live here) can be visited. As for the city hall, it is a massive building but finely decorated inside. Watching the changing of the guard in the palace's courtyard is a fun attraction;</li> <li><a href="https://www.openstreetmap.org/#map=19/59.32527/18.07066">The Nobel museum</a> was a little bit underwhelming but I probably had too high expectations. It is rather small, presents two showcases about Alfred Nobel, a temporary exhibition and a permanent collection of some of the personal items that the Nobel laureates are invited to donate to the museum. Among them, the scarf that Malala Yousafzai wore during her speech at UN, Geim and Novoselov's original scotch tape dispenser, or the sample jar Barry Marshall drank his <em>Helicobacter pylori</em> broth from, but also a trumpet, a calculator, an abacus belonging to the laureates… Most notably, some of the prize diplomas were also exposed. Each of them is unique, and are expectedly some of the finest works of art there are, drawn and colored by hand by Swedish artists. They also have a video room with short introductions to the work of the laureates playing in random order: I could have spent hours in this room…</li> </ul> <h2 id="riddarholmen">Riddarholmen</h2> <ul> <li><a href="https://www.openstreetmap.org/#map=19/59.32466/18.06461">Riddarholmskyrkan</a> The visit costs a couple dozens of crowns, but I recommend it. This church is the sepulchre of the royal family. It is better to see it after the visit of <em>Historiska museet</em> to have some of the genealogy in mind (it doesn't help that everyone is named Gustav).</li> </ul> <h2 id="skeppsholmen">Skeppsholmen</h2> <ul> <li><a href="https://www.openstreetmap.org/#map=19/59.32738/18.08182">The museum of east-asian art</a>: free and open late (20 o'clock). I have seen a nice exhibition on traditional paper making in asian countries, and the Japan section of the museum appealed to me. The museum also hosts an extensive collection of ancient chinese books (not that I could read them);</li> <li><a href="https://www.openstreetmap.org/#map=19/59.32631/18.08400">Moderna museet</a> that I didn't have time to visit.</li> </ul> <h2 id="djurgarden">Djurgården</h2> <figure> <figcaption style="text-align: center"><em>Nordiska museet, the museum of Nordic life.</em></figcaption> </figure> <ul> <li><a href="https://www.openstreetmap.org/#map=18/59.32912/18.09367">Nordiska museet</a>: go in the afternoon, it is free after 13 o'clock. This is the museum of nordic life and exposes the daily life of Swedes during the last 6 centuries. The building itself looks like a fairy tale castle. Really worth the visit;</li> <li><a href="https://www.openstreetmap.org/#map=18/59.32808/18.09126">Vasamuseet</a> exposes the local Titanic: a battle ship that sunk in 1628 in her maiden trip and was resurfaced in 1961;</li> </ul> <figure> <figcaption style="text-align: center"><em>A traditional sami storehouse, elevated to protect the food from animals and snow.</em></figcaption> </figure> <ul> <li><a href="https://www.openstreetmap.org/#map=16/59.3265/18.1043">Skansen</a> is an old and large open air museum / ecomuseum / zoo. The public is mostly Swedish: families, or school classes. I understood that Stockholmers generally buy a ticket allowing entry all year long, and come back regularly for the animations proposed all along the year around the traditional holidays. I visited over Easter and have seen for example a Swedish grandma prepare traditional Easter cakes. Furthermore, the shop is nice, and there is a couple of restaurants and hot dog stands inside: I would recommend to visit Skansen in the morning, eat there and wait for 13 o'clock to go visit <em>Nordiska museet</em> just in front;</li> <li>The botanical garden was unfortunately closed for restauration.</li> </ul> <h2 id="sodermalm">Södermalm</h2> <figure> <figcaption style="text-align: center"><em>View of Kungsholmen from Södermalm.</em></figcaption> </figure> <ul> <li><a href="https://www.openstreetmap.org/#map=17/59.31999/18.04984">This part of the island</a>, besides being a nice picnic place, has very old traditional wooden houses once occupied by the working class. The view from this bank to the north side of the city is alone worth the trip;</li> <li><a href="https://www.openstreetmap.org/#map=19/59.31961/18.07116">Stockholm's city museum</a>, closed for restauration until a couple of days after my return date…</li> </ul> <h2 id="further">Further</h2> <figure> <figcaption style="text-align: center"><em>Lindholmen from Vinterviken.</em></figcaption> </figure> <ul> <li>The furthest I went by foot is <a href="https://www.openstreetmap.org/#map=16/59.3113/17.9859">Vinterviken</a>, a very bucolic place once owned by Nobel where he installed his dynamite factory. The beauty of the landscape contrasts with the rocks torn by the experiments. The place seems to be appreciated by Stockholmers since I saw many of them jogging in the park, or going out with a baby stroller. The old acid factory was converted into an exhibition center, and a really nice restaurant. You can take a tray, order one of the three meals of the day, grab bread and water and go eat it outside at one of the picnic tables, as a reward for the walk to arrive here from Stockholm. I warmly recommend it.</li> </ul> <h2 id="even-further">Even further</h2> <ul> <li>I wish I had the time to go pick a rock at <a href="https://www.openstreetmap.org/#map=18/59.42643/18.35363">Ytterby</a>'s ytterbium quarry but that will be for another time.</li> </ul> <figure> <figcaption style="text-align: center"><em>One of the cherry trees of Kungsträdgården, a square that I discovered completely randomly on my last day.</em></figcaption> </figure> </article> <article> <h1>The perfect speed bump</h1> <p>2018-10-20T00:00:00+00:00</p> <p>As he regularly does, Jason Cole recently delighted the readers of his blog with <a href="https://jasmcole.com/2018/10/07/the-perfect-speed-bump/">an interesting problem</a> that I will quickly sum up.</p> <h1 id="the-problem">The problem</h1> <blockquote> <p>You have been assigned the task to use a given amount of concrete <script type="math/tex">A</script> to build a speed bump over a section of a road of length <script type="math/tex">2\\,\ell</script> . But being a regular user of the road, you are trying to do your job while minimizing the jolt felt by a driver passing above it.</p> </blockquote> <p>We will consider the road as a 2D curve and the car as a point-like particle on this curve. We are then searching for a function <script type="math/tex">x \mapsto y(x)</script> , with domain and values <script type="math/tex">[-\ell,\ell] \to \mathbb{R}</script> . We will assume that the road outside of this interval lies at <script type="math/tex">y=0</script> ; the continuity conditions are then <script type="math/tex">y(-\ell) = y(\ell) = 0</script> , and we will also impose the nullity of the first derivatives of <script type="math/tex">y</script> at these two points, which we will note with a parenthesized subscript <script type="math/tex">y_{(x)}(-\ell) = y_{(x)}(\ell) = 0</script> .</p> <p>The author chose to minimize the integral of the square of the vertical acceleration felt by the car passing over the bump, <script type="math/tex">\int_0^T{a_y(t)^2\mathrm{d}t}</script> .</p> <h1 id="analytical-results">Analytical results</h1> <p>Going quickly over the details because they already are in the original post, conservation of energy and geometrical considerations give</p> <script type="math/tex;mode=display">\vec{v}(x) = \sqrt{\frac{{v_0}^2-2\,g\,y}{1+{y_{(x)}}^2}} \left(\vec{x}+y_{(x)}\,\vec{y}\right)</script> <p>which, derived with respect to time, yields</p> <script type="math/tex;mode=display">\vec{a}(x) = \frac{{v_0}^2-2\,g\,y}{1+{y_{(x)}}^2}\left[ -\left(\frac{g\,y_{(x)}}{{v_0}^2-2\,g\,y}+\frac{y_{(x)}\,y_{(x,x)}}{1+{y_{(x)}}^2}\right)\vec{x} +\left(\frac{y_{(x,x)}}{1+{y_{(x)}}^2}-\frac{g\,{y_{(x)}}^2}{{v_0}^2-2\,g\,y}\right)\vec{y} \right]</script> <p>The quantity that we want to minimize is then</p> <script type="math/tex;mode=display">\begin{aligned} \int_0^T{{a_y}^2\mathrm{d}t} &= \int_{-\ell}^{\ell}{\frac{{a_y}^2}{v_x}\mathrm{d}x}\\ &= \int_{-\ell}^{\ell}{\left(\frac{{v_0}^2-2\,g\,y}{1+{y_{(x)}}^2}\right)^{3/2} \left(\frac{y_{(x,x)}}{1+{y_{(x)}}^2}-\frac{g\,{y_{(x)}}^2}{{v_0}^2-2\,g\,y}\right)^2\mathrm{d}x} \end{aligned}</script> <p>The author then proceeds to two approximations. The first one is the limit of "infinite velocity", which seems more appropriate for some drivers than some others, but is actually justified for everyone if we compute the quantity <script type="math/tex">\frac{{v_0}^2}{2\\,g\\,y} \approx 35 \gg 1</script> . This approximation allows to set <script type="math/tex">g</script> to <script type="math/tex">0</script> :</p> <script type="math/tex;mode=display">\begin{aligned} \int_0^T{{a_y}^2\mathrm{d}t} &\approx \int_{-\ell}^{\ell}{\left(\frac{{v_0}^2}{1+{y_{(x)}}^2}\right)^{3/2} \left(\frac{y_{(x,x)}}{1+{y_{(x)}}^2}\right)^2\mathrm{d}x}\\ &\approx {v_0}^3\int_{-\ell}^{\ell}{\frac{{y_{(x,x)}}^2}{\left(1+{y_{(x)}}^2\right)^{7/2}}\mathrm{d}x} \end{aligned}</script> <p>The second one is that of small angles, consisting of neglecting <script type="math/tex">y_{(x)}</script> in front of <script type="math/tex">1</script> . This ones strips the criterion even further:</p> <script type="math/tex;mode=display">\begin{aligned} \int_0^T{{a_y}^2\mathrm{d}t} &\approx {v_0}^3\int_{-\ell}^{\ell}{{y_{(x,x)}}^2\mathrm{d}x} \end{aligned}</script> <p>such that the problem becomes tractable. Indeed, applying the <a href="https://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation">Euler-Lagrange equation</a> on the Lagrangian <script type="math/tex">\mathcal{L}(y,y_{(x)},y_{(x,x)}) = {v_0}^3{y_{(x,x)}}^2-\lambda y</script> yields the optimal solution of this twice simplified problem:</p> <script type="math/tex;mode=display">\frac{\ell}{A}y(x) = \frac{15}{16}\left(1-\left(\frac{x}{\ell}\right)^2\right)^2</script> <p>He finally programs a stochastic optimizer with the exact equations, and looks how this solution changes. He didn't observe many changes, and concluded that the approximations were not too coarse.</p> <h1 id="numerical-optimization-of-the-exact-problem">Numerical optimization of the exact problem</h1> <p>I decided to challenge this observation, and proceeded to submit the problem to <a href="https://www.bocop.org/">Bocop</a>, an optimal control solver developed by the <a href="https://www.inria.fr/en">Inria</a> team <a href="https://team.inria.fr/commands/">COMMANDS</a>. Bocop takes as input a system of ordinary differential equations involving one or several control variables, a criterion to optimize, and possibly constraints, and determines the temporal profile of the control variables that optimizes the criterion while satisfying the constraints.</p> <p>For example, <a href="https://www.bocop.org/micro-swimmer-in-low-reynolds-number-fluid/">it is able to determine</a> how a model micro-swimmer should twist in order to move fast. Or how a rocket should <a href="https://www.bocop.org/goddard/">adapt its thrust during flight</a> to go the highest.</p> <p>Bocop is designed to models dynamical systems (with time derivatives), and even though the equation of motion of the car involves time derivatives, it seemed more natural here to remove the time and only reason with the independent variable <script type="math/tex">x</script> . Since we go numeric, let's make the problem dimensionless with the change of variables <script type="math/tex">\hat{x} = \frac{x}{\ell}</script> , <script type="math/tex">\hat{y} = y\frac{\ell}{A}</script> , <script type="math/tex">\hat{A} = \frac{A}{\ell^2}</script> , <script type="math/tex">\hat{g} = g\frac{\ell}{{v_0}^2}</script> . The criterion becomes</p> <script type="math/tex;mode=display">\frac{\ell}{\hat{A}^2\,{v_0}^3}\int_0^T{{a_y}^2\mathrm{d}t} = \int_{-1}^1{\left(\frac{1-2\,\hat{A}\,\hat{g}\,\hat{y}}{1+\hat{A}^2{\hat{y}_{(\hat{x})}}^2}\right)^{3/2} \left(\frac{\hat{y}_{(\hat{x},\hat{x})}}{1+\hat{A}^2{\hat{y}_{(\hat{x})}}^2}- \frac{\hat{A}\,\hat{g}\,{\hat{y}_{(\hat{x})}}^2}{1-2\,\hat{A}\,\hat{g}\,\hat{y}}\right)^2\mathrm{d}\hat{x}}</script> <p>and from now on, I will drop all the hats and work only in reduced variables.</p> <p>The system can be modelled with a system of four first-order ordinary differential equations: <script type="math/tex">y(x)</script> the profile of the road, <script type="math/tex">y_{(x)}(x)</script> its first derivative, <script type="math/tex">c(x)</script> the criterion and <script type="math/tex">Y(x) = \int_{-1}^x{y(u)\mathrm{d}u}</script> the quantity of concrete used. The control variable is <script type="math/tex">y_{(x,x)}</script> , the second derivative of the profile. The dynamics is straightforward:</p> <script type="math/tex;mode=display">\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &= y_{(x)}\\ \frac{\mathrm{d}y_{(x)}}{\mathrm{d}x} &= y_{(x,x)}\\ \frac{\mathrm{d}c}{\mathrm{d}x} &= \left(\frac{1-2\,A\,g\,y}{1+A^2{y_{(x)}}^2}\right)^{3/2} \left(\frac{y_{(x,x)}}{1+A^2{y_{(x)}}^2}-\frac{A\,g\,{y_{(x)}}^2}{1-2\,A\,g\,y}\right)^2\\ \frac{\mathrm{d}Y}{\mathrm{d}x} &= y \end{aligned}</script> <p>and is implemented in the <code>dynamics.tpp</code> file</p> <pre data-lang="c++" style="background-color:#2b303b;color:#c0c5ce;" class="language-c++ "><code class="language-c++" data-lang="c++"><span>Tdouble yxx = control[</span><span style="color:#d08770;">0</span><span>]; </span><span>Tdouble y = state[</span><span style="color:#d08770;">0</span><span>]; </span><span>Tdouble yx = state[</span><span style="color:#d08770;">1</span><span>]; </span><span>Tdouble c = state[</span><span style="color:#d08770;">2</span><span>]; </span><span>Tdouble Y = state[</span><span style="color:#d08770;">3</span><span>]; </span><span style="color:#b48ead;">double</span><span> g = constants[</span><span style="color:#d08770;">0</span><span>]; </span><span style="color:#b48ead;">double</span><span> A = constants[</span><span style="color:#d08770;">1</span><span>]; </span><span> </span><span>Tdouble vx = </span><span style="color:#96b5b4;">sqrt</span><span>((</span><span style="color:#d08770;">1</span><span>-</span><span style="color:#d08770;">2</span><span>*A*g*y)/(</span><span style="color:#d08770;">1</span><span>+A*A*yx*yx)); </span><span>Tdouble ay = yxx/(</span><span style="color:#d08770;">1</span><span>+A*A*yx*yx) - A*g*yx*yx/(</span><span style="color:#d08770;">1</span><span>-</span><span style="color:#d08770;">2</span><span>*A*g*y); </span><span> </span><span>state_dynamics[</span><span style="color:#d08770;">0</span><span>] = yx; </span><span style="color:#65737e;">// dy/dx </span><span>state_dynamics[</span><span style="color:#d08770;">1</span><span>] = yxx; </span><span style="color:#65737e;">// dyx/dx </span><span>state_dynamics[</span><span style="color:#d08770;">2</span><span>] = vx*vx*vx*ay*ay; </span><span style="color:#65737e;">// dc/dx </span><span>state_dynamics[</span><span style="color:#d08770;">3</span><span>] = y; </span><span style="color:#65737e;">// dY/dx </span></code></pre> <p>The criterion is the final state of <script type="math/tex">c</script> :</p> <pre data-lang="c++" style="background-color:#2b303b;color:#c0c5ce;" class="language-c++ "><code class="language-c++" data-lang="c++"><span>criterion = final_state[</span><span style="color:#d08770;">2</span><span>]; </span></code></pre> <p>The boundary conditions take care of the continuity conditions, initialize the criterion, and enforce the area constraint on the profile:</p> <pre data-lang="c++" style="background-color:#2b303b;color:#c0c5ce;" class="language-c++ "><code class="language-c++" data-lang="c++"><span style="color:#b48ead;">double</span><span> A = constants[</span><span style="color:#d08770;">1</span><span>]; </span><span>boundary_conditions[</span><span style="color:#d08770;">0</span><span>] = initial_state[</span><span style="color:#d08770;">0</span><span>]; </span><span style="color:#65737e;">// y(-1) = 0 </span><span>boundary_conditions[</span><span style="color:#d08770;">1</span><span>] = final_state[</span><span style="color:#d08770;">0</span><span>]; </span><span style="color:#65737e;">// y(1) = 0 </span><span>boundary_conditions[</span><span style="color:#d08770;">2</span><span>] = initial_state[</span><span style="color:#d08770;">1</span><span>]; </span><span style="color:#65737e;">// yx(-1) = 0 </span><span>boundary_conditions[</span><span style="color:#d08770;">3</span><span>] = final_state[</span><span style="color:#d08770;">1</span><span>]; </span><span style="color:#65737e;">// yx(1) = 0 </span><span>boundary_conditions[</span><span style="color:#d08770;">4</span><span>] = initial_state[</span><span style="color:#d08770;">2</span><span>]; </span><span style="color:#65737e;">// c(-1) = 0 </span><span>boundary_conditions[</span><span style="color:#d08770;">5</span><span>] = initial_state[</span><span style="color:#d08770;">3</span><span>]; </span><span style="color:#65737e;">// Y(-1) = 0 </span><span>boundary_conditions[</span><span style="color:#d08770;">6</span><span>] = final_state[</span><span style="color:#d08770;">3</span><span>] - </span><span style="color:#d08770;">1</span><span>; </span><span style="color:#65737e;">// Y(1) = 1 </span></code></pre> <p>Finally we can run the solver, let's first check if we can reach the analytical solution of the simple problem (<code>Tdouble vx = 1, ay = yxx;</code>):</p> <figure> <figcaption style="text-align: center"><em>Analytical and numerical solutions of the simplified problem.</em></figcaption> </figure> <p>It seems to work, and the objective value <script type="math/tex">22.5</script> is consistent with what we could have expected from the analytical result. Now let's try the exact problem, with different values of <script type="math/tex">A</script> and <script type="math/tex">g</script> :</p> <figure> <figcaption style="text-align: center"><em>Numerical solutions of the exact problem.</em></figcaption> </figure> <p>Note that "realistic" values of <script type="math/tex">A</script> and <script type="math/tex">g</script> could be, respectively, <script type="math/tex">0.1</script> and <script type="math/tex">0.15</script> . In this regime, the two approximations (infinite speed, and small angles) are indeed very reasonable. However, for higher bumps, the exact solution starts to diverge noticeably from the approximation. Could we guess the analytical form of the solution? That seems difficult in the general case, but we could try it in the limit of infinite speed (<script type="math/tex">g=0</script> ). Actually, the objective value for the solution calculated by Bocop for the condition <script type="math/tex">A=1</script> , <script type="math/tex">g=0</script> is <script type="math/tex">6.30</script> , which is suspiciously close to an interesting number, and the profile itself looks like four arcs of circles. Coincidence?</p> <h1 id="hypothesis-for-the-infinite-speed-limit">Hypothesis for the infinite speed limit</h1> <p>Let's do the maths in the infinite speed limit, for the function</p> <script type="math/tex;mode=display">\mathrm{circles}(x) = \begin{cases} \frac{1}{2}-\sqrt{\frac{1}{4}-(x+1)^2}&, -1 \leq x \leq -\frac{1}{2}\\ \frac{1}{2}+\sqrt{\frac{1}{4}-x^2}&, -\frac{1}{2} \leq x \leq \frac{1}{2}\\ \frac{1}{2}-\sqrt{\frac{1}{4}-(x-1)^2}&, \frac{1}{2} \leq x \leq 1 \end{cases}</script> <p>The objective is reduced to <script type="math/tex">\mathrm{Obj}\_{g=0}[y] = \int_{-1}^{1}{\frac{{y_{(x,x)}}^2}{(1+A^2{y_{(x)}}^2)^{7/2}}\mathrm{d}x}</script> .</p> <p>Mathematica helps and indeed <script type="math/tex">\mathrm{Obj}_{A=1,g=0}[\mathrm{circles}] = 2\\,\pi</script> . To know if it is actually the optimal profile, we need to plug it into the Euler-Lagrange equation. Spoilers, it is not.</p> <p>This profile is not optimal, but we can at least compare it with <script type="math/tex">\mathrm{approx}(x) = \frac{15}{16}(1-x^2)^2</script> , the analytical solution of the approximated problem. Using the same criterion, we get <script type="math/tex">\mathrm{Opt}_{A=1,g=0}[\mathrm{approx}] = 10.39</script> , which proves that the four circles are a better profile for the limit of infinite speed, at <script type="math/tex">A=1</script> . We can also compare these profiles as a function of <script type="math/tex">A</script> :</p> <figure> <figcaption style="text-align: center"><em>Comparison of the fitnesses of the two profiles.</em></figcaption> </figure> <p>It was expected that the polynomial would win at low <script type="math/tex">A</script> since low <script type="math/tex">A</script> means small angles. As we have seen, the solution in circles is better at high <script type="math/tex">A</script> . What was more difficult to understand for me was that the circles become infinitely bad at low <script type="math/tex">A</script> (like <script type="math/tex">\frac{16}{15\\,a^4}</script> ). I think that the reason for this behaviour is the two points of infinite slope at <script type="math/tex">x= \pm\frac{1}{2}</script> , which never disappear even when the profile gets flattened by a decreasing <script type="math/tex">A</script> , and create two discontinuity points where most of the acceleration is taken.</p> <p>Can we find a better solution? Is the optimal solution tractable in the limit of infinite speed? What if we wanted to minimize the maximum of the acceleration over the period (the infinite norm) instead of the L2 norm? Perhaps for another time!</p> </article> </main></body></html>$

Sociologie de comptoir</h2>
Bon, soyons clairs, ce genre de graphique ne montre pas grand-chose de plus que ce qu'un observateur attentif de la vie politique française sait déjà. Mais ils ont l'avantage d'être jolis (contrairement aux observateurs attentifs de la vie politique française).</p>

Crypto</h2>

Misc</h2>

Coda</h1>
J'ai eu de la chance que 15 lignes de Python suffisent pour ces épreuves cette année, mais pour en savoir plus je recommande fortement la lecture de writeups plus sérieux (exemple</a>) où on apprend à faire les choses correctement.</p>

Suboptimal

Réélections

Le FCSC sans se fatiguer

Élections

La radio pour les nuls

Graphology

Guessing lottery game

Suboptimal

Réélections

Sociologie de comptoir</h2> Bon, soyons clairs, ce genre de graphique ne montre pas grand-chose de plus que ce qu'un observateur attentif de la vie politique française sait déjà. Mais ils ont l'avantage d'être jolis (contrairement aux observateurs attentifs de la vie politique française).</p>

Le FCSC sans se fatiguer

Crypto</h2>

Gaston La Paffe (29 résolutions)</h3> On avait ici un algorithme de signature à base de polynomes, qui fonctionnait de la manière suivante (là aussi, j'omets des détails) :</p>

Misc</h2>

Élections

La radio pour les nuls

Coda</h1> J'ai eu de la chance que 15 lignes de Python suffisent pour ces épreuves cette année, mais pour en savoir plus je recommande fortement la lecture de writeups plus sérieux (exemple</a>) où on apprend à faire les choses correctement.</p>

Graphology

Multidimensional scaling</h3> To resolve the tensions created by the edges in the high dimensional space, we would like to have the euclidean distances between pairs of nodes inversely proportional to the weight of the edge.space.</p> Let's start simple and assume that:</p>

Guessing lottery game

Sociologie de comptoir</h2>
Bon, soyons clairs, ce genre de graphique ne montre pas grand-chose de plus que ce qu'un observateur attentif de la vie politique française sait déjà. Mais ils ont l'avantage d'être jolis (contrairement aux observateurs attentifs de la vie politique française).</p>

Coda</h1>
J'ai eu de la chance que 15 lignes de Python suffisent pour ces épreuves cette année, mais pour en savoir plus je recommande fortement la lecture de writeups plus sérieux (exemple</a>) où on apprend à faire les choses correctement.</p>