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| /* | ||
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| Diet problem in Picat. | ||
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| From the GLPK model diet.mod | ||
| """ | ||
| STIGLER'S NUTRITION MODEL | ||
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| This model determines a least cost diet which meets the daily | ||
| allowances of nutrients for a moderately active man weighing 154 lbs. | ||
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| References: | ||
| Dantzig G B, "Linear Programming and Extensions." | ||
| Princeton University Press, Princeton, New Jersey, 1963, | ||
| Chapter 27-1. | ||
| """ | ||
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| This Picat model was created by Hakan Kjellerstrand, hakank@gmail.com | ||
| See also my Picat page: http://www.hakank.org/picat/ | ||
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| */ | ||
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| import mip. | ||
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| main => go. | ||
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| go => | ||
| a(A), | ||
| b(B), | ||
| NumNutr = A[1].length, | ||
| NumFood = A.length, | ||
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| X = new_list(NumFood), | ||
| X :: 0.0..1.0, | ||
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| Cost #= sum(X), | ||
| Cost #>= 0, | ||
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| % nutrient balance (units) | ||
| foreach(N in 1..NumNutr) | ||
| sum([A[F,N] * X[F] : F in 1..NumFood]) #= B[N] | ||
| end, | ||
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| solve($[min(Cost)], X), | ||
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| printf("\nTotal cost: $%0.4f\n", Cost), | ||
| % println(x=X), | ||
| println("Dollars of food f to be purchased daily:"), | ||
| Food = ["Wheat", "Cornmeal", "Cannedmilk", "Margarine", "Cheese", "Peanut-B", "Lard", | ||
| "Liver", "Porkroast", "Salmon", "Greenbeans", "Cabbage", "Onions", "Potatoes", "Spinach", | ||
| "Sweet-Pot", "Peaches", "Prunes", "Limabeans", "Navybeans"], | ||
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| foreach({F,C} in zip(Food, X)) | ||
| printf("%-10w: $%0.4f\n", F, C) | ||
| end, | ||
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| nl. | ||
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| % Calorie 3 % thousands | ||
| % Protein 70 % grams | ||
| % Calcium 0.8 % grams | ||
| % Iron 12 % milligrams | ||
| % Vitamin-A 5 % thousands IUs | ||
| % Vitamin-B1 1.8 % milligrams | ||
| % Vitamin-B2 2.7 % milligrams | ||
| % Niacin 18 % milligrams | ||
| % Vitamin-C 75 % milligrams ; | ||
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| b(B) => B = [3.0, 70.0, 0.8, 12.0, 5.0, 1.8, 2.7, 18.0, 75.0]. | ||
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| a(A) => | ||
| A = [ | ||
| % Calorie Protein Calcium Iron Vitamin-A Vitamin-B1 Vitamin-B2 Niacin Vitamin-C | ||
| % (1000) (g) (g) (mg) (1000IU) (mg) mg) (mg) (mg) | ||
| [44.7, 1411.0, 2.0, 365.0, 0.0, 55.4, 33.3, 441.0, 0.0], | ||
| [36.0, 897.0, 1.7, 99.0, 30.9, 17.4, 7.9, 106.0, 0.0], | ||
| [ 8.4, 422.0, 15.1, 9.0, 26.0, 3.0, 23.5, 11.0, 60.0], | ||
| [20.6, 17.0, 0.6, 6.0, 55.8, 0.2, 0.0, 0.0, 0.0], | ||
| [ 7.4, 448.0, 16.4, 19.0, 28.1, 0.8, 10.3, 4.0, 0.0], | ||
| [15.7, 661.0, 1.0, 48.0, 0.0, 9.6, 8.1, 471.0, 0.0], | ||
| [41.7, 0.0, 0.0, 0.0, 0.2, 0.0, 0.5, 5.0, 0.0], | ||
| [ 2.2, 333.0, 0.2, 139.0, 169.2, 6.4, 50.8, 316.0, 525.0], | ||
| [ 4.4, 249.0, 0.3, 37.0, 0.0, 18.2, 3.6, 79.0, 0.0], | ||
| [ 5.8, 705.0, 6.8, 45.0, 3.5, 1.0, 4.9, 209.0, 0.0], | ||
| [ 2.4, 138.0, 3.7, 80.0, 69.0, 4.3, 5.8, 37.0, 862.0], | ||
| [ 2.6, 125.0, 4.0, 36.0, 7.2, 9.0, 4.5, 26.0, 5369.0], | ||
| [ 5.8, 166.0, 3.8, 59.0, 16.6, 4.7, 5.9, 21.0, 1184.0], | ||
| [14.3, 336.0, 1.8, 118.0, 6.7, 29.4, 7.1, 198.0, 2522.0], | ||
| [ 1.1, 106.0, 0.0, 138.0, 918.4, 5.7, 13.8, 33.0, 2755.0], | ||
| [ 9.6, 138.0, 2.7, 54.0, 290.7, 8.4, 5.4, 83.0, 1912.0], | ||
| [ 8.5, 87.0, 1.7, 173.0, 86.8, 1.2, 4.3, 55.0, 57.0], | ||
| [12.8, 99.0, 2.5, 154.0, 85.7, 3.9, 4.3, 65.0, 257.0], | ||
| [17.4, 1055.0, 3.7, 459.0, 5.1, 26.9, 38.2, 93.0, 0.0], | ||
| [26.9, 1691.0, 11.4, 792.0, 0.0, 38.4, 24.6, 217.0, 0.0]]. | ||
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| /* | ||
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| Triangle problem in Picat. | ||
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| From the lecture notes Modelling with Constraints | ||
| http://www.cse.unsw.edu.au/~cs4418/2008/Lectures/Modelling2.ppt, page 45ff | ||
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| This Picat model was created by Hakan Kjellerstrand, hakank@gmail.com | ||
| See also my Picat page: http://www.hakank.org/picat/ | ||
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| */ | ||
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| import mip. | ||
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| main => go. | ||
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| go => | ||
| temp(Temp), | ||
| N = Temp.length, | ||
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| % Note .vars() here. | ||
| Temp.vars() :: 0.0..100.0, | ||
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| foreach(I in 2..N-1,J in 2..N-1) | ||
| % Temp[I,J] #= 1.0/4.0*(Temp[I-1,J] + Temp[I+1,J] + Temp[I,J-1] + Temp[I,J+1]) | ||
| Temp[I,J]*4.0 #= (Temp[I-1,J] + Temp[I+1,J] + Temp[I,J-1] + Temp[I,J+1]) | ||
| end, | ||
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| solve(Temp), | ||
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| foreach(Row in Temp) | ||
| foreach(T in Row) | ||
| printf("%10.3f ", T) | ||
| end, | ||
| nl | ||
| end, | ||
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| HalfN = N div 2, | ||
| printf("Tempature at Temp[%d,%d]: %0.3f\n", HalfN, HalfN, Temp[HalfN,HalfN]), | ||
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| nl. | ||
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| temp(Temp) => | ||
| Temp = | ||
| [[100.0,100.0,100.0,100.0,100.0,100.0,100.0], | ||
| [ 0.0, _, _, _, _, _, 0.0], | ||
| [ 0.0, _, _, _, _, _, 0.0], | ||
| [ 0.0, _, _, _, _, _, 0.0], | ||
| [ 0.0, _, _, _, _, _, 0.0], | ||
| [ 0.0, _, _, _, _, _, 0.0], | ||
| [ 0.0, _, _, _, _, _, 0.0]]. |
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| Original file line number | Diff line number | Diff line change |
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| /* | ||
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| Knapsack problem (Rosetta Code) in Picat. | ||
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| From | ||
| http://rosettacode.org/wiki/Knapsack_problem/Unbounded | ||
| """ | ||
| A traveller gets diverted and has to make an unscheduled stop in what turns out to | ||
| be Shangri La. Opting to leave, he is allowed to take as much as he likes of the | ||
| following items, so long as it will fit in his knapsack, and he can carry it. He knows | ||
| that he can carry no more than 25 'weights' in total; and that the capacity of his | ||
| knapsack is 0.25 'cubic lengths'. | ||
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| Looking just above the bar codes on the items he finds their weights and volumes. | ||
| He digs out his recent copy of a financial paper and gets the value of each item. | ||
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| Item Explanation Value (each) weight Volume (each) | ||
| ------------------------------------------------------------ | ||
| panacea (vials of) Incredible healing properties 3000 0.3 0.025 | ||
| ichor (ampules of) Vampires blood 1800 0.2 0.015 | ||
| gold (bars) Shiney shiney 2500 2.0 0.002 | ||
| ---------------------------------------------------------------------------- | ||
| Knapsack For the carrying of - <=25 <=0.25 | ||
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| He can only take whole units of any item,, but there is much more of any item than | ||
| he could ever carry | ||
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| How many of each item does he take to maximise the value of items he is carrying | ||
| away with him? | ||
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| Note: | ||
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| 1. There are four solutions that maximise the value taken. Only one need be given. | ||
| """ | ||
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| This Picat model was created by Hakan Kjellerstrand, hakank@gmail.com | ||
| See also my Picat page: http://www.hakank.org/picat/ | ||
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| */ | ||
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| import mip. | ||
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| main => go. | ||
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| go => | ||
| Value = [3000.0, 1800.0, 2500.0 ], | ||
| Weight = [ 0.3, 0.2, 2.0 ], | ||
| Volume = [ 0.025, 0.015, 0.002], | ||
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| N = Value.length, | ||
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| X = new_list(N), | ||
| X :: 0..1000, | ||
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| Z #= sum([X[I]*Value[I] : I in 1..N]), | ||
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| foreach(I in 1..N) | ||
| X[I] #>= 0 | ||
| end, | ||
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| knapsack(Volume, X, 0.25), | ||
| knapsack(Weight, X, 25.0), | ||
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| solve($[max(Z)], X), | ||
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| println(z=Z), | ||
| println(x=X), | ||
| Items = ["panacea","ichor","gold"], | ||
| foreach({Item,Num} in zip(Items,X), Num > 0) | ||
| printf("Take %d of %w\n", Num,Item) | ||
| end, | ||
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| println("\nTotal volume:"), | ||
| println(sum([X[I]*Volume[I] : I in 1..N])), | ||
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| println("Total weight:"), | ||
| println(sum([X[I]*Weight[I] : I in 1..N])), | ||
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| println("Total cost:"), | ||
| println(sum([X[I]*Value[I] : I in 1..N])), | ||
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| nl. | ||
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| knapsack(W, Take, WTMax) => | ||
| sum([W[I]*Take[I] : I in 1..W.length]) #<= WTMax. |
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| Original file line number | Diff line number | Diff line change |
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| @@ -0,0 +1,47 @@ | ||
| /* | ||
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| Least square in Picat. | ||
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| Least square problem from the Swedish book | ||
| Optimeringslara, page 286f | ||
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| This Picat model was created by Hakan Kjellerstrand, hakank@gmail.com | ||
| See also my Picat page: http://www.hakank.org/picat/ | ||
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| */ | ||
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| import mip. | ||
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| main => go. | ||
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| go => | ||
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| % temperature | ||
| T = [20.0, 30.0, 80.0,125.0,175.0,225.0,275.0,325.0,360.0,420.0,495.0,540.0,630.0,700.0], | ||
| % percentage gas | ||
| F = [0.0,5.8 ,14.7,31.6,43.2,58.3,78.4,89.4,96.4,99.1,99.5,99.9,100.0,100.0], | ||
| N = T.length, | ||
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| M = 4, | ||
| A = new_list(M+1), % equation (index 0..M) | ||
| A :: -100.0..100.0, | ||
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| Z :: -1000.0..1000.0, | ||
| Z #= sum([F[I] - (A[0+1] + A[1+1]*T[I] + A[2+1]*T[I]*T[I] + A[3+1]*T[I]*T[I]*T[I] + A[4+1]*T[I]*T[I]*T[I]*T[I]) : I in 1..N]), | ||
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| A[0+1] + 20.0*A[1+1] + 20.0*20.0*A[2+1] + 20.0*20.0*20.0*A[3+1] + 20.0*20.0*20.0*20.0*A[4+1] #= 0.0, | ||
| A[0+1] + 700.0*A[1+1] + 700.0*700.0*A[2+1] + 700.0*700.0*700.0*A[3+1] + 700.0*700.0*700.0*700.0*A[4+1] #= 100.0, | ||
| foreach(I in 1..N) | ||
| (A[1+1] + 2.0*A[2+1]*T[I] + 3.0*A[3+1]*T[I]*T[I] + 4.0*A[4+1]*T[I]*T[I]*T[I]) #>= 0.0 | ||
| end, | ||
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| solve($[min(Z)],A), | ||
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| println(z=Z), | ||
| println(a=A), | ||
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| nl. | ||
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