## A knapsack riddle [#2]?

**S**till about this allocation riddle of the past week, and still with my confusion about the phrasing of the puzzle, when looking at a probabilistic interpretation of the game, rather than for a given adversary’s y, the problem turns out to search for the maximum of

where the Y’s are Binomial B(100,p). Given those p’s, this function of x is available in closed form and can thus maximised by a simulated annealing procedure, coded as

utility=function(x,p){ ute=2*pbinom(x[1]-1,100,prob=p[1])+ dbinom(x[1],100,p[1]) for (i in 2:10) ute=ute+2*i*pbinom(x[i]-1,100,prob=p[i])+ i*dbinom(x[i],100,p[i]) return(ute)} #basic term in utility baz=function(i,x,p){ return(i*dbinom(x[i],100,p[i])+ i*dbinom(x[i]-1,100,p[i]))} #relies on a given or estimated p x=rmultinom(n=1,siz=100,prob=p) maxloz=loss=0 newloss=losref=utility(x,p) #random search T=1e3 Te=1e2 baza=rep(0,10) t=1 while ((t<T)||(newloss>loss)){ loss=newloss i=sample(1:10,1,prob=(10:1)*(x>0)) #moving all other counters by one xp=x+1;xp[i]=x[i] #corresponding utility change for (j in 1:10) baza[j]=baz(j,xp,p) proz=exp(log(t)*(baza-baza[i])/Te) #soft annealing move j=sample(1:10,1,prob=proz) if (i!=j){ x[i]=x[i]-1;x[j]=x[j]+1} newloss=loss+baza[j]-baza[i] if (newloss>maxloz){ maxloz=newloss;argz=x} t=t+1 if ((t>T-10)&(newloss<losref)){ t=1;loss=0 x=rmultinom(n=1,siz=100,prob=p) newloss=losref=utility(x,p)}}

which seems to work, albeit not always returning the same utility. For instance,

> p=cy/sum(cy) > utility(argz,p) [1] 78.02 > utility(cy,p) [1] 57.89

or

> p=sy/sum(sy) > utility(argz,p) [1] 82.04 > utility(sy,p) [1] 57.78

Of course, this does not answer the question as intended and reworking the code to that purpose is not worth the time!

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