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# In -> V2 -> V3 def Q23(x,rout,rin,p3,F1,F2): return rout/p3*(1-1/(1+x*(1-F1)/rin))*(1-F2) # In -> V3 -> V2 def Q32(x,rout,rin,p3,F1,F2): return rout*(1-1/(1+p3*(1-F1)*(1-F2)*x/rin)) # In -> V2 def Q0(r0,r1,a0In,a1In,a1Out): return a0In*0.997+r0*(1-1/(1+(a1In*0.997-a1Out)/r1)) # The main obstacle swapping on V2 pair is that is not always granted that the maximum theoretical output is # coherent with the "K" check def k(r0,r1,a0In,a1In,a0Out,a1Out): b0=r0+a0In-a0Out b1=r1+a1In-a1Out b0adj = b0*1000-3*a0In b1adj = b1*1000-3*a1In k1 = b0adj*b1adj-1000000*r0*r1>-1 k2 = b0adj*b1adj-1000000*r0*r1 return k2 # I studied on R the behaviour of the K(r0,r1,a0In,a1In,a0Out,a1Out) = (1000*(r0+a0In-a0Out)-3*a0In)*(1000*(r1+a1In-a1Out)-3*a1In)-1000000*r0*r1 # function, which has to be >=0 in order to pass the 'K' revert check in V2 swap() # K(a1In), with all others as parameters, is intermittently constant, with jumps. And non-crescent. # Sometimes its values catch the 0, but other times not: in those cases, when K is negative, the max ouput amount is # found going backwards through the input amount until the previous value for which K is positive. # This is the sense of the ritroso() function # The choice of going backwards exponentially was simply due to power of calculation, the highest amount # would be found going backwards by just one unit at a time def ritroso(x,r0,r1): import math y = x K = k(r0,r1,0,x,Q0(r0,r1,0,x,0),0) ma = float(str(y)[0:4])/10**3 if ma!=0 and y!=0: og = math.trunc(math.log10(y/ma)) else: og = 14 if K<0: es = og while K<0: es-=1 y-=10**(og-es) K = k(r0,r1,0,y,Q0(r0,r1,0,y,0),0) return y else: return y # With the "ritroso" adjustment, the "non-naive" output functions become: # In -> V2 -> V2 def Q22(r0,r1,r2,r3): return Q0(r3,r2,0,ritroso(r3,r2,Q0(r0,r1,0,ritroso(r0,r1,In),0)),0) # where r0/r1 are reserves of token0/token1 of the pair1 and r2/r2 the reserves of token0/token1 of the pair2 # assuming 1 and 3 are the borrowed token # In -> V2 -> V3 def Q23(x,rout,rin,p3,F1,F2): return rout/p3*(1-1/(1+ritroso(x,rout,rin)*(1-F1)/rin))*(1-F2) # In -> V3 -> V2 def Q32(x,rout,rin,p3,F1,F2): return rout*(1-1/(1+p3*(1-F1)*(1-F2)*ritroso(x,rout,rin)/rin)) # The case V3 -> V3 has no "K" issues, the formula is simply def Q33(p1,p2,x,F1,F2): return p1*x*(1-F1)*(1-F2)/p2 # Simply from that formule is evident that, althought counterintuitive, the first pair price must be the min is order # maximize Q33