#### 2013-04-09: Branch and Bound in Racket

The source for this post is online at 2013-04-09-bb-in-racket.rkt.

Categories: Racket Algorithms

Based on the fun I had with the last post on tree search, I thought this week I would do a generic branch and bound algorithm in Racket. We’ll make it generic by using Racket’s unit system.

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Branch and bound is an optimization algorithm that is based on being able to have accurate estimates of the bounds on the utility of different sets of choices. The key is to be able to provide the following functions:

 (define-signature bb-client^ (option? split upper-bound lower-bound inject extract))

where split takes an option set and returns some number of subsets, while the upper-bound and lower-bound functions return the respective bounds on the subset. For convenience with option set representations that are different than the "natural" representation of the problem, inject converts the latter to the former and extract does the reverse.

##### 1Trival problem instance

A trivial instance of this unit is based on binary search trees, where the option set data structure explicitly records the current bounds based on the binary search invariant. (This example is especially trivial because the branch-and-bound will just go all the way to the left, because it will find the minimal value in the search space, i.e. the tree.)

 (struct node (key left right)) (define-unit binary-bb@ (import) (export bb-client^) (struct option (lower node upper)) (define (inject n) (option -inf.0 n +inf.0)) (define extract option-node) (define (split os) (match os [(option lower #f upper) empty] [(option lower (node key left right) upper) (list (option lower left key) (option key (node key key key) key) (option key right upper))])) (define upper-bound option-upper) (define lower-bound option-lower))

The branch and bound algorithm itself is parameterized over this interface and is a single function that takes a search space and returns the optimal solution.

Here’s an example of how it should work:

 (define (binary-insert k bt) (match bt [#f (node k #f #f)] [(node nk left right) (if (< k nk) (node nk (binary-insert k left) right) (node nk left (binary-insert k right)))])) (define the-tree (foldr binary-insert #f (shuffle (build-list 1000 add1)))) (check-equal? (node-key (find binary-bb@ the-tree)) 1)

##### 2Branch and bound

The shell of the branch and bound function is:

<bb> ::=
 (define (find client@ start) (define count 0) (define-values/invoke-unit client@ (import) (export bb-client^)) (define minimum-upper-bound +inf.0) (define best #f) (printf "visited ~a candidates\n" count) (extract best))

We initialize the client and maintain the best upper bound found so far and the candidate which had that bound. We store these in the mutable variables minimum-upper-bound and best. At the end of the algorithm, we extract the underlying solution from this best option. (For curiousity’s sake, we’ll keep a count of how many options we looked at.)

We update these values whenever we discover something better:

 (when (<= candidate-upper minimum-upper-bound) (set! minimum-upper-bound candidate-upper) (set! best candidate))

We keep track of the possible candidates, in order of the best lower bound, using a priority queue.

 (define queue (make-heap (λ (x y) (<= (lower-bound x) (lower-bound y))))) (heap-add! queue (inject start)) (for ([candidate (in-heap/consume! queue)]) (set! count (add1 count)) )

The body of the loop simply determines if the candidate should be pruned because its lower bound is worse than the minimum upper bound. If the candidate is worth visiting, then we determine if it is a leaf, i.e. when its upper and lower bounds match. Otherwise, we add all its children to the queue.

 (define candidate-lower (lower-bound candidate)) (when (<= candidate-lower minimum-upper-bound) (define candidate-upper (upper-bound candidate)) (cond [(= candidate-lower candidate-upper) ] [else (heap-add-all! queue (split candidate))]))

When we run our binary tree test, indeed it works. I find that it averages about 25 candidates, which represents the average depth.

##### 3A more interesting problem

A far more interesting use of branch and bound is to solve something like the knapsack problem. This problem is when you have a set of items, each with a value and a weight, and you must select the optimal set of items, given a weight constraint. In defining this client, we will parameterize the unit with the weight constraint.

 (struct item (value weight) #:transparent) (define (knapsack-bb@ W) (unit (import) (export bb-client^) ))

The key here is to represent options as a set of fixed items, that all children of the option must use, and a set of free items, that children may or may not use. Initially, all items are free and at the end, we only take the fixed items.

 (struct option (fixed free) #:transparent) (define (inject l) (option empty l)) (define (extract o) (option-fixed o))

Since our branch and bound finds the minimum, we have to swap the sign when we compute values, because we want the most value. The lower bound is the sum of the values of the fixed and free, while the upper bound is just the fixed, because we have to choose them.

 (define (upper-bound o) (match-define (option fixed free) o) (* -1 (items-value fixed))) (define (lower-bound o) (match-define (option fixed free) o) (* -1 (+ (items-value fixed) (items-value free))))

We can enforce the constraint on the weight inside of the splitting function by not producing children that violate the constraint.

 (define (split o) (match-define (option fixed free) o) (match free [(list) empty] [(list-rest fst more) (define next-fixed (cons fst fixed)) (list* (option fixed more) (if (> (items-weight next-fixed) W) empty (list (option next-fixed more))))]))

We can test this use of branch and bound like this, using a list of 30 items with values from 1 to 10 and weights of either 10 or 5:

 (define items (append (build-list 10 (λ (i) (item (add1 i) 10))) (build-list 10 (λ (i) (item (add1 i) 5))) (build-list 10 (λ (i) (item (add1 i) 5))))) (check-equal? (find (knapsack-bb@ 10) items) (list (item 10 5) (item 10 5)))

Our branch and bound finds an optimum by only looking at 1,709 candidates out of the 1,073,741,824 (2^30) options.

Exercise: This knapsack instance has a problem: the bound and split operations do not work in constant time, because they sum the values and weights of different lists. Change the code to cache these computations in the option structure.

Exercise: This knapsack instance considers the items in the order that they are given. Change it so it tries the greedy solution first by sorting the items.

If you’d like to use this code at home, you should put it in this order:

<*> ::=
 (require rackunit racket/unit racket/match racket/list data/heap) (define (items-value is) (for/sum ([i (in-list is)]) (item-value i))) (define (items-weight is) (for/sum ([i (in-list is)]) (item-weight i)))