Knapsack 0-1 Python binary & rosettacode & WE

[DANILIN]

Knapsack 0-1 Python binary & rosettacode & WE

Classic Knapsack problem is solved in many ways

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

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:

   # 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