Knapsack 0-1 & Python & qbasic qb64 & WE

[DANILIN]

Knapsack 0-1 & python & qbasic qb64 & WE

Classic Knapsack problem is solved in many ways

Contents: http://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=5 N!=120 >> 2^N=32

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

Program qb64 write results to directory

Main thing is very brief and clear to even all

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n=5; N=n+1; G=5; a=2**N         # N=7: L=5: a = 2^(N+1): 'knapsack.bas DANILIN L=[];C=[];e=[];j=[];q=[];s=[]       # Dim L(N), C(N), j(N), q(a), q$(a), d(a) d=[];L=[1]*n;C=[1]*n;e=[1]*a    j=[1]*n;q=[0]*a;s=[0]*a;d=[0]*a   from random import randint  # Randomize Timer for i in range(0,n):        # For i=1 To N:     L[i]=randint(1,3)       # L(i)=Int(Rnd*3+1)     C[i]=10+randint(1,9)    # C(i)=10+Int(Rnd*9)     print(i+1,L[i],C[i])    # Print i, L(i), C(i) print()   for h in range(a-1,(a-1)//2,-1):    # For h=a-1 To 0 Step -1  #     b=str(bin(h))     e[h]=b[3:len(b)]            # binary without origin               for k in range (n):         # For k=1 To N:         j[k]=int(e[h][k])       # j(k)=Val(Mid$(q$(h), k, 1)) ' from chifer                      q[h]=q[h]+L[k]*j[k]*C[k]    # q(h)=q(h)+L(k)*j(k)*C(k) ' 0 or 1         d[h]=d[h]+L[k]*j[k]     # d(h)=d(h)+L(k)*j(k)   Next         #print(e[h],L[k],j[k],C[k],d[h], q[h])               if d[h]<= G:            # If d(h) <= L Then         print(e[h], G, d[h], q[h])      # Print #1, d(h), q(h), q$(h)   Next print()     max=0; m=1              # max=0: m=1: for i in range(a):          # For i=1 To a     if d[i]<=G and q[i]>max:        # If d(i)<=L Then If q(i) > max         max=q[i]; m=i           # Then max=q(i): m=i    Next print (d[m], q[m], e[m])        # Print #1,: Print #1, d(m), q(m), q$(m): End

Output:
# Mass Cost
1  2  12
2  3  17
3  1  14
4  3  17
5  1  13

Chifer Mass Cost           
11000  5  5  75
10101  5  4  51
01001  5  4  64
00111  5  5  78 !!!
00110  5  4  65
00101  5  2  27
00000  5  0  0

Mass MAX Chifer
5     78     00111

Only Knapsack 0-1 binary Python

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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])

[Larz60+]

moved to code sharing

[DANILIN]

which python compilers are best online?

I know 3 compilers online

of these 1 is embedded on websites and forums


kakie kompilyatory piton onlayn luchschie?
znayu 3 kompilyatora onlayn
iz nih 1 vstraivaetsya na sayty i forumy
какие компиляторы питон онлайн лучшие?
знаю 3 компилятора онлайн
из них 1 встраивается на сайты и форумы


I know 3 compilers online

of these 1 is embedded on websites and forums

which python compilers are best online?