And what's your secret for achieving the best score ?
Concours de rentrée 2020 - défi Python de Xuanwu
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Re: Concours de rentrée 2020 - défi Python de Xuanwu
by commandblockguy » 20 Oct 2020, 21:10
I'll take the Numworks calc (Sagittarius).
As for my strategy, I basically found a path through the maze by hand that had as few points as possible, without worrying about my score too much. I made sure to take a very shallow angle when exiting so that the path would be long enough to get the maximum score.
I then tested every single combination of angles and distances that would put me within some distance (if I remember correctly, about 1.5 tiles, though I changed it several times) of my original points, while still allowing me to get the maximum score. This was feasible because you get points off for each digit after the first decimal place, or before the units place, which means that there are only 100 angles and 100 distances to try for each point. The exception for this is the last point, which has to be more than 10 units away in order to meet the length requirement. This took a few hours to run in its entirety, and I was left with a few thousand paths. Prior to the scoring change, all of these paths would have left me with a score of 72, but after the scoring change it was now possible to get a fractional score.
The fractional part of the score was determined based on the sine of the sum of the angles and distances, which means that the input to sin() could be no more precise than to the tenths place. For the range of numbers that this sum could possibly be without adding a full point to my score, the tenth whose sine is closest to -1 is 11.0, and sin(11) is = -0.999990207.
Unfortunately, none of the few thousand paths that I had found added up to 11.0. However, I realized that I was only calling a_droite() in my path, and that if I replaced a_droite(x) with a_gauche(-x), x would be subtracted from the sum rather than added to it, without actually changing the validity of the path. I tried out every combination of a_droite and a_gauche calls for each of my few thousand paths, and found one that added up to 11.0, giving me a score of 71.0000097934493.
Here's the code that I used to find this path.
You'll notice that this is still very slightly off from my actual score - I realized that I could slightly decrease sin(sum) by slightly decreasing sum, and that I could slightly decrease sum by subtracting a tiny amount from each distance and angle. Normally, this would cause my score to increase by several full points because of the decimal places after the tenths place, but because the score is rounded to 5 decimal places before the length of the number is calculated, if the difference is less than 0.000005, you aren't penalized. So, I subtracted 0.0000049999... from each distance, to get a sum of 10.99991, a sine of -0.99999060081, and a final score of 71.00000939918644. I believe that this is within a floating-point error of the ideal solution.
Here's the solution that I submitted, for reference.
I actually tested all of this on a computer, as I didn't have my EP with me. I ported polycalc to use cv2 for drawing stuff.
I also tried to submit a few, uh, let's call them
After that, I tried subclassing the float class so that it would return an empty string. Technically, this wasn't modifying the internal state of laby.py
. I could have also returned a custom string class with a modified __len__ method, but in CPython __len__ must return a positive integer, which also gets converted into a regular integer, not one that I could subclass. Interestingly, MicroPython allows you to return a negative length for __len__, so I could have probably gotten a score of negative infinity if the admins didn't shut that idea down .
Finally, I also found a solution that's far better than the others that I decided not to submit since I was already in first. Here's the link if you want to give it a try for yourself.
As for my strategy, I basically found a path through the maze by hand that had as few points as possible, without worrying about my score too much. I made sure to take a very shallow angle when exiting so that the path would be long enough to get the maximum score.
I then tested every single combination of angles and distances that would put me within some distance (if I remember correctly, about 1.5 tiles, though I changed it several times) of my original points, while still allowing me to get the maximum score. This was feasible because you get points off for each digit after the first decimal place, or before the units place, which means that there are only 100 angles and 100 distances to try for each point. The exception for this is the last point, which has to be more than 10 units away in order to meet the length requirement. This took a few hours to run in its entirety, and I was left with a few thousand paths. Prior to the scoring change, all of these paths would have left me with a score of 72, but after the scoring change it was now possible to get a fractional score.
A "heatmap" of all the paths I found. The color of each point represents the last "target" point it was able to get to, with purple being the final point, outside of the maze.
The fractional part of the score was determined based on the sine of the sum of the angles and distances, which means that the input to sin() could be no more precise than to the tenths place. For the range of numbers that this sum could possibly be without adding a full point to my score, the tenth whose sine is closest to -1 is 11.0, and sin(11) is = -0.999990207.
Unfortunately, none of the few thousand paths that I had found added up to 11.0. However, I realized that I was only calling a_droite() in my path, and that if I replaced a_droite(x) with a_gauche(-x), x would be subtracted from the sum rather than added to it, without actually changing the validity of the path. I tried out every combination of a_droite and a_gauche calls for each of my few thousand paths, and found one that added up to 11.0, giving me a score of 71.0000097934493.
Here's the code that I used to find this path.
You'll notice that this is still very slightly off from my actual score - I realized that I could slightly decrease sin(sum) by slightly decreasing sum, and that I could slightly decrease sum by subtracting a tiny amount from each distance and angle. Normally, this would cause my score to increase by several full points because of the decimal places after the tenths place, but because the score is rounded to 5 decimal places before the length of the number is calculated, if the difference is less than 0.000005, you aren't penalized. So, I subtracted 0.0000049999... from each distance, to get a sum of 10.99991, a sine of -0.99999060081, and a final score of 71.00000939918644. I believe that this is within a floating-point error of the ideal solution.
Here's the solution that I submitted, for reference.
I actually tested all of this on a computer, as I didn't have my EP with me. I ported polycalc to use cv2 for drawing stuff.
I also tried to submit a few, uh, let's call them
gimmicky
, solutions. My first instinct was to try an integer overflow, but to my disappointment, Python doesn't actually have those. The numbers, they just keep going
.After that, I tried subclassing the float class so that it would return an empty string. Technically, this wasn't modifying the internal state of laby.py
. I could have also returned a custom string class with a modified __len__ method, but in CPython __len__ must return a positive integer, which also gets converted into a regular integer, not one that I could subclass. Interestingly, MicroPython allows you to return a negative length for __len__, so I could have probably gotten a score of negative infinity if the admins didn't shut that idea down .Finally, I also found a solution that's far better than the others that I decided not to submit since I was already in first. Here's the link if you want to give it a try for yourself.
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Re: Concours de rentrée 2020 - défi Python de Xuanwu
by critor » 20 Oct 2020, 23:44
Thanks for your reply and comprehensive details @commandblockguy.
Which NumWorks bag ?
Which NumWorks poster ?
And which Xcas goodies ?
T-shirt ? Hat ? Mousepad ? Sticker ?
Which NumWorks bag ?
Which NumWorks poster ?
And which Xcas goodies ?
T-shirt ? Hat ? Mousepad ? Sticker ?
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critorAdmin
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Re: Concours de rentrée 2020 - défi Python de Xuanwu
by commandblockguy » 21 Oct 2020, 00:01
I'll take the third bag and the fourth poster, and a large shirt.
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Re: Concours de rentrée 2020 - défi Python de Xuanwu
by TIny_Hacker » 21 Oct 2020, 00:48
Very nice @commandblockguy, I remember you bragging about it and that (non-submitted) score is INSANELY lower than the others! Now I feel so dumb not going that path...
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Re: Concours de rentrée 2020 - défi Python de Xuanwu
by critor » 21 Oct 2020, 07:51
Thanks @commandblockguy.
Also I forgot, which TI-Planet sticker ?
Also I forgot, which TI-Planet sticker ?
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Re: Concours de rentrée 2020 - défi Python de Xuanwu
by critor » 21 Oct 2020, 07:58
Also to everybody, please make a post your details/explanations/secrets.
Aussi à tous, svp faites un post avec vos détails / explications / secrets.
Aussi à tous, svp faites un post avec vos détails / explications / secrets.
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Re: Concours de rentrée 2020 - défi Python de Xuanwu
by Pavel » 21 Oct 2020, 10:38
critor wrote:Also to everybody, please make a post your details/explanations/secrets.
Aussi à tous, svp faites un post avec vos détails / explications / secrets.
Voici une explication de mon approche pour trouver une solution à ce défi.
J'ai commencé par essayer de comprendre comment la consommation est calculée. Voici les parties du code Python montrant tous les composants des calculs de consommation:
- Code: Select all
def fix_angle(a):
return a * 2 * asin(1) / pi
def a_gauche(a):
global state
state[7] += a
state[5] += 5 + cout(a)
def avancer(l):
global state
state[7] += l
state[5] += 8 + cout(l)
while(l > 0):
l -= .25
state[6] += (state[6] < 200)
def aller_selon(f):
global state
state = [0, .5, 0, 0, .5, 0, 0, 0]
f()
state[5] += sin(fix_angle(state[7])) - state[6] // 2
print('Consommation : ' + str(state[5]))
Apparemment, la consommation se compose des trois éléments suivants:
- Code: Select all
state[5] = consommation de base
state[6] = distance totale parcourue multipliée par quatre
state[7] = angle total + distance totale
Ensuite, j'ai testé la fonction 'cout()' qui est utilisée dans les calculs de consommation:
- Code: Select all
>>> from laby import *
>>> cout(5)
3
>>> cout(0.5)
3
>>> cout(0.05)
4
>>> cout(5e-3)
5
>>> cout(5e-4)
6
>>> cout(5e-5)
7
>>> cout(5e-6)
3
Il semble donc que la consommation puisse être minimisée en combinant les éléments suivants:
- minimiser la quantité de manoeuvres
- minimiser 'sin(state[7])'
- maximiser la distance totale parcourue
- ne pas utiliser plus d'un chiffre décimal après la virgule décimale
- essayer d'exploiter le faible coût du 'cout(5e-6)'
- Code: Select all
from math import pi, sin
l = []
for i in range(-10, 11):
a = round(-pi/2 + 2 * pi * i, 1)
l.append((a, sin(a)))
l.sort(key=lambda e: e[1])
for e in l:
print('a = %5.1f -> sin(a) = %f' % e)
Voici le résultat:
- Code: Select all
a = -64.4 -> sin(a) = -0.999996
a = -26.7 -> sin(a) = -0.999994
a = 11.0 -> sin(a) = -0.999990
a = 48.7 -> sin(a) = -0.999986
a = 42.4 -> sin(a) = -0.999934
De petites valeurs de 'state[7]' sont donc nécessaires pour une faible consommation. Vu que 'state[7]' est la somme des angles et des distances et que les distances ne peuvent pas être négatives, alors les valeurs des angles passés à la fonction 'a_gauche()' doivent être négatives.
Une fois que tout ça est devenu plus ou moins clair, j'ai modifié le code laby.py pour en faire une version qui peut être contrôlée comme un jeu vidéo avec les touches haut/bas/gauche/droite sur le clavier:
https://gitea.planet-casio.com/Pavel/labygui
En utilisant cette approche, j'ai obtenu la solution suivante:
- Code: Select all
from laby import *
def chemin():
for e in [(-6.6, 2.3), (-7.3, 6.3), (-5.2, 4.6), (-4.8, 4.1), (-7.3, 6.3), (-8.3, 3.8), (-5.1, 3.8), (-2.3, 4.3), (9.1, 13.3)]:
a_gauche(e[0])
avancer(e[1])
aller_selon(chemin)
La consommation correspondant à cette solution est de 71.0000097934493.
À ce stade, je n'ai toujours pas exploité le faible coût du 'cout(5e-6)'. J'ai donc essayé d'ajouter ou de soustraire 5e-6 des distances et des angles. La solution qui m'a donné le score le plus bas (71.00000939918644) était la suivante:
- Code: Select all
from laby import *
def chemin():
for e in [(-6.6, 2.3), (-7.3, 6.3), (-5.2, 4.6), (-4.8, 4.1), (-7.3, 6.3), (-8.3, 3.8), (-5.1, 3.8), (-2.3, 4.3), (9.1, 13.3)]:
a_gauche(e[0] - 5e-6)
avancer(e[1] - 5e-6)
aller_selon(chemin)
Au moment où j'ai trouvé cette solution, elle était déjà soumise par commandblockguy et j'ai dû trouver la deuxième meilleure solution (71.00000939918645) en ajoutant de très petites valeurs aux angles et aux distances:
- Code: Select all
from laby import *
def chemin():
for e in [(-6.6, 2.3), (-7.3, 6.3), (-5.2, 4.6), (-4.8, 4.1), (-7.3, 6.3), (-8.3, 3.8), (-5.1, 3.8), (-2.3, 4.3), (9.1, 13.3)]:
a_gauche(e[0] - 5e-6 + 1e-13)
avancer(e[1] - 5e-6 + 1e-13)
aller_selon(chemin)
Last edited by Pavel on 21 Oct 2020, 10:57, edited 1 time in total.
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Re: Concours de rentrée 2020 - défi Python de Xuanwu
by cent20 » 21 Oct 2020, 10:50
Bravo pour ces explications d'une clarté lumineuse.
Je viens de comprendre que du coup, j'aurais pu fabriquer le score 71,00000939918646, oui enfin il fallait déjà arriver à faire l'intégralité du raisonnement pour cela.
Et merci pour le simulateur PC, une aide inestimable.
Je viens de comprendre que du coup, j'aurais pu fabriquer le score 71,00000939918646, oui enfin il fallait déjà arriver à faire l'intégralité du raisonnement pour cela.
Et merci pour le simulateur PC, une aide inestimable.
NumWork, Python : -
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Re: Concours de rentrée 2020 - défi Python de Xuanwu
by critor » 21 Oct 2020, 15:43
critor wrote:Thanks @commandblockguy.
Also I forgot, which TI-Planet sticker ?
Ok, 1er autoccollant, merci.
@Pavel, merci pour ton explication et félicitations. Que nous prends-tu ?
- 12 lots:Capricorne ♑calculatrice+ 1Casio Graph 90+Epack de goodies+ 1Casiogoodieau choix + 1Xcaspack de goodiesTI-Planet&Planète Casio
- 12 lots:Bélier ♈solution d'émulationau choix + 1CasioouCDde vidéosCasio fx-CG20catalogue de produitsau choix + 1Casiopack de goodies+ 1Casiogoodieau choix + 1Xcaspack de goodiesTI-Planet&Planète CasioCDde54vidéos pour plus de8hpar , pourfx-CG20mais compatibleGraph 90+E, pourWindows/Mac.
Détail dessolutions d'émulationau choix :Casio- clé USB 8 God'émulation permanente au choix, à jour avec 3 émulateurs pourWindows:fx-92+ Spéciale Collège+Graph 35+E II 3.30+Graph 90+E 3.40
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- clé monolithique EMTEC 2020(7,7 Gode capacité réelle)
- clé à capuchon
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- fx-CP400 Managerpour /
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- clé
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Détail descalculatricesau choix :TI-83 Premium CE Edition Python- TI-83 Premium CE Edition Pythonsous blister version E
- TI-83 Premium CE Edition Pythonsous blister version E avec autocollant masquant sceau
[info]Détail des packs de goodies communs accompagnants les lots :
- 1 styloau choixCasio
- 1 clé USBau choix :Casio
- clé USB Casio fx-CP400+E-4,019 Go
- clé USB Casio-4,018 Go
- clé USB
- 1 batterieUSBCasio
- 1 histoire intégrale imprimée du manga- épisode 1, 2 ou 3 au choix - 5 pages ci-dessous + page de fin secrète :Casio Academy - ClassWiz Edition - Function Hero
- Episode 0 : Kaito(pour référence)
- Episode 1 : Takuma
- Episode 2 : Emi
- Episode 3 : Azusa
- Episode 0 :
- 1 manuelau choixNumWorks(N0100ouN0110)
- 1 cahier d'activitésNumWorks SNT 2nde
- 1 sacau choixNumWorks(N0100versions1.0-1.5,N0100versions1.6+, ouN0110)
- 1 cahierNumWorks
- 1 posterau choix :NumWorks
- format A0:(118,9×84,1 cm²)NumWorks N0100- roulé
- format A2:(42×59,4 cm²)
- NumWorks N0100:Eduscol / Ministère de l'Education Nationale- roulé - brillant
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- NumWorks N0100:@Pims / @qabosse / @antalpilipiliet ses collègues d'EPS- roulé
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- format
- 1 styloNumWorks
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- 1 porte-documentsTI
- 1 posterau choix :TIplié
- format ANSI-D:(55,9×86,4 cm²)TI-73 Explorer
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- format 55,75×83,5 cm²:TI-Nspire CX,TI-Nspire CX CAS
- format
- 1 cléau choix :USBTI
- clé USB T3 Francebleue -2 Gode capacité nominale
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- clé USB
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Aperçus de quelques
cahiers d'activités
au choix:TI-83 Premium CE Python
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1
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Xcas
- 1 autocollantau choixTI-Planet
- 1 autocollantPlanète Casio
- 1 compte premiumTI-Planet
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