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Ingy döt Net 2023-07-01 11:58:00 -04:00
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Task/Knapsack-problem-Unbounded/11l/knapsack-problem-unbounded.11l Normal file
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T Bounty
Int value
Float weight, volume
F (value, weight, volume)
(.value, .weight, .volume) = (value, weight, volume)
V panacea = Bounty(3000, 0.3, 0.025)
V ichor = Bounty(1800, 0.2, 0.015)
V gold = Bounty(2500, 2.0, 0.002)
V sack = Bounty( 0, 25.0, 0.25)
V best = Bounty( 0, 0, 0)
V current = Bounty( 0, 0, 0)
V best_amounts = (0, 0, 0)
V max_panacea = Int(min(sack.weight I/ panacea.weight, sack.volume I/ panacea.volume))
V max_ichor = Int(min(sack.weight I/ ichor.weight, sack.volume I/ ichor.volume))
V max_gold = Int(min(sack.weight I/ gold.weight, sack.volume I/ gold.volume))
L(npanacea) 0 .< max_panacea
L(nichor) 0 .< max_ichor
L(ngold) 0 .< max_gold
current.value = npanacea * panacea.value + nichor * ichor.value + ngold * gold.value
current.weight = npanacea * panacea.weight + nichor * ichor.weight + ngold * gold.weight
current.volume = npanacea * panacea.volume + nichor * ichor.volume + ngold * gold.volume
I current.value > best.value & current.weight <= sack.weight & current.volume <= sack.volume
best = current
best_amounts = (npanacea, nichor, ngold)
print(‘Maximum value achievable is ’best.value)
print(‘This is achieved by carrying (one solution) #. panacea, #. ichor and #. gold’.format(best_amounts[0], best_amounts[1], best_amounts[2]))
print(‘The weight to carry is #2.1 and the volume used is #.3’.format(best.weight, best.volume))