Fast Additive Constant Approximation Algorithms for Safe Deposit Boxes problem in 2D and 3D. Boaz Ben-Moshe Yefim Dinitz 2/23/2019

Safe Deposit Boxes problem Given: a set S of n safe boxes, where the i-th box contains ai Dollars, bi Francs, etc. and numbers (restrictions - constrains): A, B, etc. We wish to find the minimal subset of S, such that the total Dollar value is at least A, Franc value at least B, etc. A variant of the knapsack problem, with a fixed cost (of 1) of taking any item. 2/23/2019

Banknotes (one currency) There are banknotes, seen on the table. We would like to take the smallest amount of banknotes, which sums up to at least A. A = 200 2/23/2019

Banknotes (one currency) Solution: take the biggest (till A). A = 200 Opt = 3 2/23/2019

Two currencies: Motivation Shortest link path: Given a set (V) of 2D vectors, find the smallest subset of V, that performs a path from “start” into the half open rectangle “end”. What may be a strategy? 2/23/2019

Two currencies: Motivation Killing processes problems: Given a set S of processes find the smallest subset of S, such that: Killing these processes, by OS, will free at least A cpu and B space. 2/23/2019

The state and results Safe Deposit Box problem (SDB) is NP-hard, beginning from the case of two currencies [DK]. There is a polynomial approximation solution with a constant additive error [DK]. Our new algorithms lead to significant improvements of running time. 2/23/2019

Theorem [DK] Consider m currencies instance of SDB. Let (a : b ...) be the exchange rate for each currency. The value of the i-th box, w.r.t this rate, is aai + bbi + … (in, say, CAD). For any non-degenerate instance of SDB, there exist a certain exchange rate such that the opt+(m-1) most valuable boxes forms a feasible solution. 2/23/2019

Algorithmic proof (sketch) SDB is an ILP problem. Relax it to LP, i.e. let the variables xi be fractional, between 0 and 1. Consider an optimal vertex of the feasible polyhedron. The value is a lower bound for ILP. At most m variables are fractional. Round them up. The resulting solution is feasible and integer, and the error is at most m-1. The optimum of the dual LP problem is the exchange rate as in Theorem. 2/23/2019

Running time For any (constant) m, the running time of the algorithm of [DK] is O(nm+1). The algorithm is of the type primal-dual. There is a specific algorithm of [DK] for the case m=2; it runs in O(n2). Our result for m=2: O(n2)  O(n log2 n). Our result for m=3: O(n4)  O(n2 log2 n). 2/23/2019

Our tools (shown for m=2) Let us normalize any rate to the form (1:b) Given an exchange rate (1:b) we can sort the boxes according to there values. Let #A(b) be the number of best boxes (according to the b-order) needed to reach A. Similarly for #B(b). Obviously, max{#A,#B} is an upper bound, for the optimum (it is a feasible solution). Main Lemma: For any rate (1:b), min{#A,#B} is a lower bound. 2/23/2019

Lower bound proof Obviously, the total (1:b) value of any feasible solution is at least A+bB. Note that min{#A,#B}-1 most valuable boxes, w.r.t. (1:b), do not reach this total value bound A+bB. So, no min{#A,#B}-1 other boxes do; that is, they cannot form a feasible solution, as required. 2/23/2019

Monotonicity of #A and #B Let us sweep b from 0 to ∞, and trace changes in the box order. When two box values become equal, value(i)=value(j), so that before that, box i was better, and after that, box j is better, then holds ai>aj and bi<bj. Assume General Position (for now). 2/23/2019

Geometric representation: A straight line for each box: value(i) = ai+ bbi. When box j becomes better than box i (a swap point): ai>aj & bi<bj. 2/23/2019

2D Binary Search Observation: #A and #B are monotone. Consider the equilibrium point, where a single swap turns #A=<#B into #A>=#B. There, the lower and upper bounds for opt differ by at most one! Thus we obtain an (opt+1) solution. 2/23/2019

Single swap – equilibrium point For given b, assume #A<#B, (before swap). After swap: #B<=#A 2/23/2019

Algorithm of [DK]: There are n*(n-1)/2 different orders (between b swap points). Compute all b values. Sweep all the b values from left to right. O(n^2 log(n)), algorithm. With some more CG mechanism: O(n^2). 2/23/2019

If only we could… Avoid computing all O(n2) b values. Use the convex (monotone) property to find the equilibrium point. Perform binary search over b values (sorted by the x coordinate). Introducing Slope Selection! 2/23/2019

Slope selection Parametric search: known (CG) technique: [Megiddo 83], [Cole 89], [Katz & Sharir 93]. Given a set of n straight lines, the k’th intersection (sorted by x), can be found in O(n*log(n)) run time. We use this algorithm as a “black box”. 2/23/2019

2D algorithm (improved) 1. Compute the next pivot: b‘ (using Slope Selection). 2. Compute #A, #B according to b’. 3. If #A - #B >1 goto 1 - search up 4. If #A - #B < -1 goto 1 - search down 5. else done. 2/23/2019

Results (general position) Running time: Finding the next b: O(n*log(n)). Computing #A, #B for given b: O(n*log(n)). Binary search steps: O(log(n)). Total time: O(n*log^2(n)). 2/23/2019

Degenerate case Breaking “symmetric” inputs. What if 3 boxes (lines) intersect at the same point? Reduction to a combinatorial problem. 2/23/2019

Reduction: 2/23/2019

Solving the reduced problem: All boxes have the same value. Sorted by increasing A (decreasing B) Move a sliding window of size min(#A’, #B’), step by step. Till an equilibrium point is found. 2/23/2019

The 2D NP hardness Suggestions ?? Tricky reduction. 2/23/2019

The 2D NP hardness Reduction to perfect sum: given a set (S) on numbers and a value T is there a subset of S which sums up to A. Spit to n (|S|) different problems: 1<k<n  k’th instance: Dollars: values: S, constraint: A. Francs: values: {smax-si}, constraint: B = k*smax–A. 2/23/2019

The 2D NP hardness If exist a k for which: the safe box optimal* solution is of size k  there is a perfect sum for A (the set of Dollars). If there is a perfect sum for A. Otherwise, the Observe that any safe box optimal* solution must contains at least k boxes (k>=(A+B)/smax) Therefore the safe box optimal* solution will find it … 2/23/2019

The 3D case - overview Each box is a 3D plane. Two exchange rates(b,g). O(n^3) different orders. Main lemma holds. Goal: to find the equilibrium point for which: min(#A,#B,#C)+2 <= max(#A,#B,#C) 2/23/2019

The 3D case - overview 3D algorithm overview: Two level Binary Search, over (b, g). Compute g (BS). Find the matching equilibrium b (2D). If not the g equilibrium point, goto 1. 2/23/2019

Three currencies case Generalization to 3D is not that obvious: For each fixed g, we balance not A and B currencies, but A+gC (an artificial currency) and B - in order to maintain monotonicity. The critical lines in the (b : g) plane are the projections of the intersection lines of n planes ai+bbi+gci. searching for the equilibrium value of g. Binary search over O(n^4) g values. 2/23/2019

Three currencies case (continued) Generalization to 3D is not that obvious: Eventually, we find two neighboring values of g, where #A and #C flips (from the late to early satisfaction). However, this is not all, yet, since the A+gC constraint is only a tool . Hence, a certain fine tuning is needed, between the above two values of g. 2/23/2019

Running time, for three currencies Since there are n2 straight lines, a request to the Slope Selection method costs O(n2 log n). Processing of each line (g is fixed) finding the 2D equilibrium costs O(n log^2 n). Thus, the total time is O(n2 log2 n). 2/23/2019

Degenerate case, for m=3 2/23/2019

Degenerate case, for m=3 A nice combinatorial problem arises, at the g-equilibrium: All values ai+bbi+gci=V (a constant); k=(A+B+C)/V is a lower bound; We look for a feasible set of k+2 boxes. Observation: for any k boxes at least one restriction must be satisfied. 2/23/2019

A combinatorial way for finding a solution of at most k+2 boxes Compute all two constrained solutions: P'AB, P'AC, P'BC Compute TAB , TBC We step swaps from TAB to Tbc maintaining k+1<=#B<=k+2. Hence, either A, or C must be satisfied, at any step. At the equilibrium, #A,#C<=k+1, #B<=k+2. 2/23/2019

Possible directions of research Parametric search ??: for the specific problem  find b (and not k): Open the black box: Reducing the runtime to O(n log(n)). Lower bound 2D: W(n) , W(n log(n)) ?? Lower bound 3D: ? 3sum hard? 2/23/2019

Possible directions of research Simply generalization of the 3D method – leads to inefficient runtime (too many exchange rates) KD: LP approach: could the algorithm ne reduces to O(nm log…) ? 2/23/2019

FIN benmoshe@cs.bgu.ac.il www.cs.bgu.ac.il/~benmoshe/PHD/safeBox 2/23/2019