Fibonacci Numbers and the Raw Materials Karl Lieberherr

Use two relations only R1(x) = x R2(x,y)= !x or !y consider symmetric formula (gives the worst raw material) n variables, x = multiplicity of constraints using R1, y = multiplicity of constraints using R2.

Playing with the weights x: a x: b x: c x: d y: !a !b y: !a !c y: !b !c y: !a !d y: !b !d y: !c !d x =1, y=1 best assignment a=1, b=0, c=0: (1+6)/10=7/10 x = 2, y=1 a=1, b=0, c=0: (2+6)/14=8/14=0.57 best assignment a=1, b=1, c=0: (4+5)/14=9/14=0.64

Insight By choosing the weights appropriately, we can produce harder raw materials.

Consider only R1 and R2 How to choose the raw materials? n variables fraction k*x+(bin(n,2)-bin(k,2))*y/(n*x+bin(n,2)*y) f(n,a,k)=k*a+(bin(n,2)-bin(k,2))/(n*a+bin(n,2)) use bin(n,2)=n*n/2

Derivation of Fibonacci sequence for raw material f1(n,a,k)= 2*k*a+n^2-k^2/(2*n*a+n^2) max if k=a f2(n,a)=(a^2+n^2)/(2*n*a+n^2) min if a/n=h, h^2+h-1=0, h=0.618… a[n]=n*F[n]/F[n+1] F[n] is the n-th Fibonacci number