如何使用loco进行基本优化

Question

我正在尝试使用loco进行优化的基本示例.

我有一个成本向量,其索引对应于多个时隙的整数值,并希望最小化时隙的不同子集的成本之和.

请参阅下面的我的尝试,但由于所选插槽与成本之间没有"链接",因此无效.

(def costs [10 10 20 20 30 30 40 40 10 10])

(let [slot-vars (for [i (range 5)] ($in [:slot i] 1 10))
      cost-vars (for [i (range 10)] ($in [:cost i] 10 40))]
  (solution
   (concat
    slot-vars
    cost-vars
    [($distinct (for [i (range 5)] [:slot i]))]
    (for [i (range 5)]
      ($= [:cost i] (get costs i))))
   :minimize (apply $+ (for [i (range 5)] [:slot i]))))

Answer 1

首先，我不完全确定我理解这个问题的要点，因为显然解决方案是只取列表中的 5 个最小值。但如果你想让机车做到这一点，我同意背包约束是一个方便的工具。与迈克在评论中所说的相反，我认为使用背包进行最小化没有任何障碍。令权重全部为 1，并强制权重之和为 5，以便从 10 个插槽中选择 5 个。我使用该变量[:include i]来指示插槽 i 是否应包含在子集中（1 表示 true，0 表示 false）。我们希望最小化 :include 变量向量和成本向量的点积。

(def costs [10 10 20 20 30 30 40 40 10 10])
(def weights (repeat 10 1))

(def include-vars (for [i (range 10)] [:include i]))
(def include-constraints (for [i (range 10)] ($in [:include i] 0 1)))

(def model
  (concat
   include-constraints
   [($knapsack weights costs include-vars 5 :total)
    ($in :total 0 (apply + costs))]))

(solution model :minimize :total)

结果是：

{[:include 4] 0, [:include 6] 0, [:include 9] 1, [:include 1] 1, [:include 3] 0, [:include 8] 1, :total 60, [:include 0] 1, [:include 7] 0, [:include 2] 1, [:include 5] 0}