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Nov 28th, 2014
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  1. solve;
  2. printf "The knapsack contains:\n";
  3. printf {(i,s,p) in I: x[i] == 1} " %i", i;
  4.  
  5. printf("\n");
  6.  
  7. printf(sum{ i in 1..N}(A[i]*x[i]));
  8.  
  9. printf(4*PI * sum{ i in 1..N} (A[i]*x[i]) /
  10. (
  11. 2* sum{i in 1..N-1, j in i+1..N}
  12. (P[i]*P[j]* y[i,j]) # i < j
  13. + sum{i in 1..N}
  14. (P[i]*P[i]* x[i]) # i = j
  15.  
  16. -4 * sum{i in 1..N-2, j in i+1..N-1, k in j+1..N}
  17. (P[i]*S[j,k]* w[i,j,k]) # j< k --> i < j < k
  18. -4 * sum{i in 1..N-1, j in i+1..N}
  19. (P[i]*S[i,j]* y[i,j]) # j< k --> i = j < k
  20. -4 * sum{i in 1..N-2, j in i+1..N-1, k in j+1..N}
  21. (P[j]*S[i,k]* w[i,j,k]) # j< k --> j < i --> j < i < k
  22. -4 * sum{i in 1..N-1, j in i+1..N}
  23. (P[j]*S[i,j]* y[i,j]) # j< k --> j < i --> j < i = k
  24. -4 * sum{i in 1..N-2, j in i+1..N-1, k in j+1..N}
  25. (P[k]*S[i,j]* w[i,j,k]) # j< k --> j < i --> j < k < i
  26.  
  27. +8 * sum{i in 1..N-3, j in i+1..N-2, k in j+1..N-1, l in k+1..N}
  28. (S[i,k]*S[j,l]*z[i,j,k,l]) # i < k < j < l
  29.  
  30. +8 * sum{i in 1..N-2, j in i+1..N-1, k in j+1..N}
  31. (S[i,k]*S[j,k]*w[i,j,k]) # i < k < j = l
  32.  
  33. +8 * sum{i in 1..N-3, j in i+1..N-2, k in j+1..N-1, l in k+1..N}
  34. (S[i,l]*S[j,k]*z[i,j,k,l]) # i < k < l < j
  35.  
  36. +8 * sum{i in 1..N-2, j in i+1..N-1, k in j+1..N}
  37. (S[i,j]*S[j,k]*w[i,j,k]) # i < k = j < l
  38.  
  39. +8 * sum{i in 1..N-3, j in i+1..N-2, k in j+1..N-1, l in k+1..N}
  40. (S[i,j]*S[k,l]*z[i,j,k,l]) # i < j < k < l
  41.  
  42. +8 * sum{i in 1..N-2, j in i+1..N-1, k in j+1..N}
  43. (S[i,j]*S[i,k]*w[i,j,k]) # i = k < j < l
  44.  
  45. +4 * sum{i in 1..N-1, j in i+1..N}
  46. (S[i,j]*S[i,j]*y[i,j]) # i = k < j = l
  47. );)
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