Knapsack 0-1 Python binary & rosettacode & WE Classic Knapsack problem is solved in many ways Contents: rosettacode.org/wiki/Knapsack_problem Long read: rosettacode.org/wiki/Knapsack_problem/0-1 My newest program synthesizes all ciphers from 0 & 1 adding an extra register and 0 remain on left in cipher Number of comparisons decreases from N! to 2^N for example N=10 & N!=3628800 >> 2^N=1024 Random values origin are automatically assigned quantity and quality and integral of value is obtained and in general: integral of quantity and quality and it is possible to divide only anyone will not understand Code: n=5; N=n+1; G=5; a=2**N # KNAPSACK 0-1 DANILIN L=[];C=[];e=[];j=[];q=[];s=[] # rextester.com/BCKP19591 d=[];L=[1]*n;C=[1]*n;e=[1]*a j=[1]*n;q=[0]*a;s=[0]*a;d=[0]*a from random import randint for i in range(0,n): L[i]=randint(1,3) C[i]=10+randint(1,9) print(i+1,L[i],C[i]) print() for h in range(a-1,(a-1)//2,-1): b=str(bin(h)) e[h]=b[3:len(b)] for k in range (n): j[k]=int(e[h][k]) q[h]=q[h]+L[k]*j[k]*C[k] d[h]=d[h]+L[k]*j[k] if d[h]<= G: print(e[h], G, d[h], q[h]) print() max=0; m=1 for i in range(a): if d[i]<=G and q[i]>max: max=q[i]; m=i print (d[m], q[m], e[m]) Main thing is very brief and clear to even all Results is reduced manually: Code: # Mass Cost 1 2 12 2 3 17 3 1 14 4 3 17 5 1 13 Chifer Mass Cost 11000 5 5 75 01001 5 4 64 00111 5 5 78 !!! 00110 5 4 65 00101 5 2 27 Mass MAX Chifer 5 78 00111
