78 lines
2.4 KiB
MiniZinc
78 lines
2.4 KiB
MiniZinc
%-----------------------------------------------------------------------------%
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% Radiation problem, MiniZinc 2.0.4 version
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%
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%-----------------------------------------------------------------------------%
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%-----------------------------------------------------------------------------%
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% Parameters
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%-----------------------------------------------------------------------------%
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int: m; % Rows
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int: n; % Columns
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set of int: Rows = 1..m;
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set of int: Columns = 1..n;
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% Intensity matrix
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array[Rows, Columns] of int: Intensity;
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set of int: BTimes = 1..Bt_max;
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int: Bt_max = max(i in Rows, j in Columns) (Intensity[i,j]);
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int: Ints_sum = sum(i in Rows, j in Columns) (Intensity[i,j]);
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%-----------------------------------------------------------------------------%
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% Variables
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%-----------------------------------------------------------------------------%
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% Total beam-on time
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var 0..Ints_sum: Beamtime;
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% Number of shape matrices
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var 0..m*n: K;
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% N[b] is the number of shape matrices with associated beam-on time b
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array[BTimes] of var 0..m*n: N;
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% Q[i,j,b] is the number of shape matrices with associated beam-on time
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% b that expose cell (i,j)
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array[Rows, Columns, BTimes] of var 0..m*n: Q;
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%-----------------------------------------------------------------------------%
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% Constraints
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%-----------------------------------------------------------------------------%
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% For FD/LP hybrid solving, all these should go the LP solver
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% (with a suitable linearisation of the 'max' expressions).
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constraint
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Beamtime = sum(b in BTimes) (b * N[b])
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/\
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K = sum(b in BTimes) (N[b])
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/\
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forall(i in Rows, j in Columns)
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( Intensity[i,j] = sum([b * Q[i,j,b] | b in BTimes]) )
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/\
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forall(i in Rows, b in BTimes)
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( upper_bound_on_increments(N[b], [Q[i,j,b] | j in Columns]) );
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predicate upper_bound_on_increments(var int: N_b, array[int] of var int: L) =
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N_b >= L[1] + sum([ max(L[j] - L[j-1], 0) | j in 2..n ]);
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%
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% Good linear version:
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% let { array[min(index_set(L))..max(index_set(L))] of var int: S } in
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% N_b >= L[1] + sum([ S[j] | j in 2..n ]) /\
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% forall([ S[j] >= L[j] - L[j-1] /\ S[j] >= 0 | j in 2..n ]);
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%-----------------------------------------------------------------------------%
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output [
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"Beamtime = \(Beamtime);\n",
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"K = \(K);\n",
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"N = \(N);\n",
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"Q = array3d(\(Rows), \(Columns), \(BTimes), \(Q));\n"
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];
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%-----------------------------------------------------------------------------%
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