Imagine you are a computer scientist who enjoys attending conferences or workshops within the year. Sadly, your travel budget is limited, so you must select a subset of events you can travel to. When you are aware of all possible events and their costs at the beginning of the year, you can select the subset of the possible events that maximizes your happiness and is within your budget. On the other hand, if you are blind about the options, you will likely have a hard time when trying to decide if you want to register somewhere or not, and will likely regret decisions you made in the future. These scenarios can be modeled by knapsack variants, either by an offline or an online problem. However, both scenarios are somewhat unrealistic: Usually, you will not know the exact costs of each workshop at the beginning of the year. The online version, however, is too pessimistic, as you might already know which options there are and how much they cost roughly. At some point, you have to decide whether to register for some workshop, but then you are aware of the conference fee and the flight and hotel prices. We model this problem within the setting of online knapsack problems with estimates: in the beginning, you receive a list of potential items with their estimated size as well as the accuracy of the estimates. Then, the items are revealed one by one in an online fashion with their actual size, and you need to decide whether to take one or not. In this article, we show a best-possible algorithm for each estimate accuracy δ (i.e., when each actual item size can deviate by ± δ from the announced size) for both the simple knapsack (also known as subset sum problem) and the simple knapsack with removability.

Abstract We analyze the competitive ratio and the advice complexity of the online unbounded knapsack problem. An instance is given as a sequence of n items with a size and a value each, and an algorithm has to decide whether or not and how often to pack each item into a knapsack of bounded capacity. The items are given online and the total size of the packed items must not exceed the knapsack’s capacity, while the objective is to maximize the total value of the packed items. While each item can only be packed once in the classical knapsack problem (also called the 0-1 knapsack problem), the unbounded version allows for items to be packed multiple times. We show that the simple unbounded knapsack problem, where the size of each item is equal to its value, allows for a competitive ratio of 2. We also analyze randomized algorithms and show that, in contrast to the 0-1 knapsack problem, one uniformly random bit cannot improve an algorithm’s performance. More randomness lowers the competitive ratio to less than 1 . 736 , but it can never be below 1 . 693 . In the advice complexity setting, we measure how many bits of information (so-called advice bits) the algorithm has to know to achieve some desired solution quality. For the simple unbounded knapsack problem, one advice bit lowers the competitive ratio to $$\varvec{3/2}$$ 3 / 2 . While this cannot be improved with fewer than $$\varvec{\log }_{\varvec{2}} \varvec{n} $$ log 2 n advice bits for instances of length n , a competitive ratio of $$\varvec{1}\varvec{+}\varvec{\varepsilon }$$ 1 + ε can be achieved with $$\varvec{O}\varvec{(}\varvec{\varepsilon }^{\varvec{-1}} \varvec{\cdot }\varvec{\log }\varvec{(}\varvec{n}\varvec{\varepsilon }^{\varvec{-1}}\varvec{))}$$ O ( ε - 1 · log ( n ε - 1 ) ) advice bits for any $$\varvec{\varepsilon }\varvec{>}\varvec{0}$$ ε > 0 . We further show that no amount of advice bounded by a function $$\varvec{f(n)}$$ f ( n ) allows an algorithm to be optimal. We also study the online general unbounded knapsack problem and show that it does not allow for any bounded competitive ratio for both deterministic and randomized algorithms, as well as for algorithms using fewer than $$\varvec{\log }_{\varvec{2}} \varvec{n}$$ log 2 n advice bits. We also provide a surprisingly simple algorithm that uses $$\varvec{O}\varvec{(}\varvec{\varepsilon }^{\varvec{-1}} \varvec{\cdot }\varvec{\log }\varvec{(}\varvec{n}\varvec{\varepsilon }^{\varvec{-1}}\varvec{))}$$

In the Online Simple Knapsack Problem we are given a knapsack of unit size 1. Items of size smaller or equal to 1 are presented in an iterative fashion and an algorithm has to decide whether to permanently reject or include each item into the knapsack without any knowledge about the rest of the instance. The goal is then to pack the knapsack as full as possible. In this work, we introduce a third option additional to those of packing and rejecting an item, namely that of reserving an item for the cost of a fixed fraction α of its size. An algorithm may pay this fraction in order to postpone its decision on whether to include or reject the item until after the last item of the instance was presented. While the classical Online Simple Knapsack Problem does not admit any constantly bounded competitive ratio in the deterministic setting, we find that adding the possibility of reservation makes the problem constantly competitive, with varying competitive ratios depending on the value of α. We give upper and lower bounds for the whole range of reservation costs, with tight bounds for costs up to 1/6 - an area that is strictly 2-competitive - , for costs between √2-1 and 1 - an area that is strictly (2+α)-competitive up to ϕ -1, and strictly 1/(1-α)-competitive above ϕ-1, where ϕ is the golden ratio. With our analysis, we find a counterintuitive characteristic of the problem: Intuitively, one would expect that the possibility of rejecting items becomes more and more helpful for an online algorithm with growing reservation costs. However, for higher reservation costs above √2-1, an algorithm that is unable to reject any items tightly matches the lower bound and is thus the best possible. On the other hand, for any positive reservation cost smaller than 1/6, any algorithm that is unable to reject any items performs considerably worse than one that is able to reject.

Abstract In online edge- and node-deletion problems the input arrives node by node and an algorithm has to delete nodes or edges in order to keep the input graph in a given graph class $$\Pi $$ Π at all times. We consider only hereditary properties $$\Pi $$ Π , for which optimal online algorithms exist and which can be characterized by a set of forbidden subgraphs $${{\mathcal{F}}}$$ F and analyze the advice complexity of getting an optimal solution. We give almost tight bounds on the Delayed Connected $${{\mathcal{F}}}$$ F -Node-Deletion Problem, where all graphs of the family $${\mathcal{F}}$$ F have to be connected and almost tight lower and upper bounds for the Delayed $$H$$ H -Node-Deletion Problem, where there is one forbidden induced subgraph H that may be connected or not. For the Delayed $$H$$ H -Node-Deletion Problem the advice complexity is basically an easy function of the size of the biggest component in H. Additionally, we give tight bounds on the Delayed Connected $${\mathcal{F}}$$ F -Edge-Deletion Problem, where we have an arbitrary number of forbidden connected graphs. For the latter result we present an algorithm that computes the advice complexity directly from $${\mathcal{F}}$$ F . We give a separate analysis for the Delayed Connected $$H$$ H -Edge-Deletion Problem, which is less general but admits a bound that is easier to compute.

Many graph properties are expressible in first order logic. Whether a graph contains a clique or a dominating set of size k are two examples. For the solution size as its parameter the first one is W[1]-complete and the second one W[2]-complete meaning that both of them are hard problems in the worst-case. If we look at both problem from the aspect of average-case complexity, the picture changes. Clique can be solved in expected FPT time on uniformly distributed graphs of size n, while this is not clear for Dominating Set. We show that it is indeed unlikely that Dominating Set can be solved efficiently on random graphs: If yes, then every first-order expressible graph property can be solved in expected FPT time, too. Furthermore, this remains true when we consider random graphs with an arbitrary constant edge probability. We identify a very simple problem on random matrices that is equally hard to solve on average: Given a square boolean matrix, are there k rows whose logical AND is the zero vector? The related Even Set problem on the other hand turns out to be efficiently solvable on random instances, while it is known to be hard in the worst-case.