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

2. 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.
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3. 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
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4. Banknotes (one currency)
Solution: take the biggest (till A).
A = 200
Opt = 3
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5. 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?
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6. 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.
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7. 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.
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8. 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.
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9. 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.
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10. 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).
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11. 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.
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12. 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.
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13. 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).
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14. 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.
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15. 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.
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16. Single swap – equilibrium point
For given b, assume #A<#B, (before swap).
After swap: #B<=#A
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17. 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).
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18. 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!
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19. 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”.
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20. 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.
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21. 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)).
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22. Degenerate case Breaking “symmetric” inputs.
What if 3 boxes (lines) intersect at the same point?
Reduction to a combinatorial problem.
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23. Reduction:
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24. 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.
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25. The 2D NP hardness
Suggestions ??
Tricky reduction.
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26. 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.
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27. 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 …
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28. 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)
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29. 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.
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30. 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.
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31. 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.
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32. 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).
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33. Degenerate case, for m=3
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34. 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.
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35. 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.
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36. 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?
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37. 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…) ?
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38. FIN benmoshe@cs.bgu.ac.il www.cs.bgu.ac.il/~benmoshe/PHD/safeBox
2/23/2019