After all that work on the title (I assume trying to increase chance of "hotness"), I'm surprised you didn't have a ready-to-post reference answer. Fast answers are a big boost to the arbitrary hotness score :P
@MartinBüttner I've actually been working on creating a web version of a variation of Waving Hands I made a few months ago. I just uploaded it to my website (see the list of spells with Space, scroll with up and down arrow keys). Just thought you might be interested :P
@Doorknob冰 I was considering writing one myself ("when I've got some time...") seeing how unmaintained and outdated all the existing implementations look
@PeterTaylor I've invited you to another game, I want to try out something else, if you don't mind.
actually, never mind, feel free to ignore it
I wanted to test if a two-hand gesture in a previous round can be used for two spells in the current round. but there are no spells that have a two-hand gesture before the current turn, for which casting them twice is distinguishable from casting them once.
(although I guess one could try to cast a fire and an ice elemental at the same time)
"Given a list of gestures for each hand, determine which spells could be cast in the current turn"
I get it I guess.
The list is a list of gestures that have already been completed, except for the last gesture which is the gesture that is being made this turn.
@feersum Ok, it definitely does not say "what could be cast with the last gesture in the list". It says "what could be cast for the current turn", which could very well be after the last gesture in the list.
Score a Mahjong Hand
So you've built a solver for which tiles you need to complete a Mahjong hand, but you've heard that certain hands are worth more than others and sometimes a hand that scores too low can't even win.
So, you decide to build a program that will score hands for you.
This time,...
hm yeah... I was thinking you could just work out how much of a peperoni you need in each slice, then start with some tangent, and work out where to cut this (or the next) peperoni and skip across peperoni-free sections. but the potential of peperonis overlapping the origin makes this a lot harder.
get perpendicular distance from line to centre, if it's closer than the radius, work out the intersections as perpendicular offsets from the radius with simple trigonometry
yes, although you don't need to min/max mu1 and mu2, because you know which one is greater
{{\[Mu] ->
a Cos[t] + b Sin[t] -
Sqrt[-a^2 - b^2 + 2 r^4 + (a - b) (a + b) Cos[2 t] +
2 a b Sin[2 t]]/Sqrt[2]},
{\[Mu] ->
a Cos[t] + b Sin[t] +
Sqrt[-a^2 - b^2 + 2 r^4 + (a - b) (a + b) Cos[2 t] +
2 a b Sin[2 t]]/Sqrt[2]}}
it's probably what you've got, just with some double-angle formulae applied
also, when the question gave 1 complex example of 1 pep... and 10 people. we have to consider that we cannot simply make 10 cuts of θ angle where 10θ is the angle spanned by that pepperoni
I don't think I can get a functional from from NIntegrate (which I need, in order to invert it, so I can find the angles, where the covered area hits the next person's share)
and if I just integrate around the entire thing I just get 0 for some reason, so it doesn't seem to work anyway
Mathematica, 215 bytes
f=(A=Pi.01Length@#2/#;l=m/.Solve[Norm[{a,b}-m{Cos@t,Sin@t}]==.1,m];B=0;k=(l/.{a->#,b->#2})&@@@#2;Print@0;((B+=If[Im@#<0,0,d(Max[#2,0]^2-Max[#,0]^2)/2])&@@@(k/.{t->#});If[B>A,Print@#;B-=A])&/@Range[0,2Pi,d=1.*^-5];)&
Ungolfed:
f = (
A = Pi .01 Length@#2/#;
l = m /....
Python + matplotlib, 688
from pylab import*
C=cross
P,M=eval("map(array,input()),"*2)
P,N=[[P[0]]+L+[P[-1]]for L in P,M]
W=[.5]*len(P)
def T(a,c,b):
I=[(H[0]**2,id(n),n)for n in N for H in[(C(n-a,b-a),C(n-b,c-b),C(n-c,a-c))]if(min(H)*max(H)>=0)*H[1]*H[2]]
if I:d=max(I)[2];A=T(a,c,d);B=T(d,c,b)...
C++, 275,000,000+
We'll refer to pairs whose magnitude is accurately representable, such as (x, 0), as honest pairs and to all other pairs as dishonest pairs of magnitude m, where m is the pair's wrongly-reported magnitude.
The first program in the previous post used a set of tightly related cou...
if the last turn has the same gesture on both hands, check which spells can be cast with that two-hand gesture and cast them individually. (unless it's P and P, then only output Surrender)
and in any case, look at which spells can be cast on each individual hand, and output the cartesian product
currently, no two different spell combination can lead up to a situation where both the spell are ending in current turn and both require double hand gestures
Score a Mahjong Hand
So you've built a solver for which tiles you need to complete a Mahjong hand, but you've heard that certain hands are worth more than others and sometimes a hand that scores too low can't even win.
So, you decide to build a program that will score hands for you.
This time,...
Mathematica, 215 bytes
f=(A=Pi.01Length@#2/#;l=m/.Solve[Norm[{a,b}-m{Cos@t,Sin@t}]==.1,m];B=0;k=(l/.{a->#,b->#2})&@@@#2;Print@0;((B+=If[Im@#<0,0,d(Max[#2,0]^2-Max[#,0]^2)/2])&@@@(k/.{t->#});If[B>A,Print@#;B-=A])&/@Range[0,2Pi,d=1.*^-5];)&
Ungolfed:
f = (
A = Pi .01 Length@#2/#;
l = m /....
