# Tag Info

7

Here's the relevant paragraph from the MiniSAT paper: The decision phase will continue until either all variables have been assigned, in which case we have a model, or a conflict has occurred. On conflicts, the learning procedure will be invoked and a conflict clause producted. The trail will be used to undo decision, one level at a time, until ...

6

Backtracking is a general algorithm "that incrementally builds candidates to the solutions, and abandons each partial candidate ("backtracks") as soon as it determines that the candidate cannot possibly be completed to a valid solution." (Wikipedia). So, basically, what you do is build incrementally all permutations. As soon as as you build a single ...

6

You described the simplest case of the neural network, where the center neuron has only one output $a$, which is connected to the final loss function. In general, there can be several outputs and the total error signal coming to this node equals the sum over all output connections. Plus the error messages across these connections are different. And the ...

6

(Throughout this answer I use $\phi = \frac{1+\sqrt{5}}{2}$, which is the Golden Ratio.) Solution: If $s\leq n$, the attackers can win in one day. If $s+dpw<\phi^2n$ and $s<2n$, the attackers can win in $2+\max(k,0)$ days, where $$k=\left\lceil\frac{\log_\phi{\left(\frac{n+(dpw+s-2n)\phi^{-1}}{n-(dpw+s-2n)\phi}\right)}}{4}\right\rceil.$$ If $dpw<n$...

6

Overview of the problem If you takes teenagers as vertices of a graph, and have an edge whenever the two teenagers are compatible. This gives you an undirected graph, and what you need is a Hamiltonian path in this graph (a path that contains every node exactly once). Maybe searching the web on this abstract version of the problem will yield more ...

5

The idea of the backtracking algorithm is simple, though somewhat cumbersome to express. Perhaps it's easiest to explain it working through the example in the question. We start by putting $T_1$ on chair 1. We then put $T_2$ on chair 2. Then we put $T_3$ on chair 3, and we discover a conflict. So we backtrack, replacing $T_3$ with the next available student. ...

5

This answer begins with a first section explaining the problem in more general terms. The direct answer to the question for the example given by the OP is then given in a second section that illustrate the discussion in the first. You may well want to skip the first section and go direcly to the second one, depending on whether you like to start with basic ...

5

One of the best solution is likely based on a linear programming relaxation or direct integer programming. For the latter, the branching and backtracking will be implicit, and you won't have to manage it yourself. I have seen it solved in two ways using this technique. We can slightly improve your bounding algorithm as well. The textbook method Using ...

4

Introduce a new variable $Q$, whose domain is $\{0000,0001,0010,\dots,1111\}$. It represents the value of $C1,C2,C3,P$. For instance, if $Q=0001$, that means that $C1=0$, $C2=0$, $C3=1$, $P=0$. Then, you add a constraint that the first bit of $Q$ is equal to $C1$ (that's a binary constraint), a constraint that the second bit of $Q$ is equal to $C2$ (...

4

Your optimal alignment seems to be an optimal local alignment, the best substring match. Needleman-Wunsch is for global alignment. With the simple program by Eddy that you can find on the internet I have determined a global score of 0. This is better than the gobal score you get: -5 (for 10 gaps and 5 matches). Sequence X: GCATGCU Sequence Y: GATTACA ...

4

Yes, these two heuristics does sound like inconsistent. Most Constrained Variable (MCV) (also called MRV for Minimum Remaining Values) tries to reduce the size of the next branch to search while Least Constraining Values tries to enlarge the size of the next branch to branch. However, if you take a close a look, they both serve the same goal, which is, ...

3

This is a classic Dynamic Programming problem. Basically, dynamic programming is a way of turning a recursive algorithm into an iterative one with better runtime, by saving previous results that we might need again. In this case, we can write the problem as two recursive functions: maxSumSubtractFirst(startIndex): if startIndex is past the end of the ...

3

The idea here is to implement the following recursive definition of the set $A_n$ of all binary strings of length $n$: $$A_0 = \{\epsilon\}, \qquad A_n = \{x0 : x \in A_{n-1}\} \cup \{x1 : x \in A_{n-1}\}.$$ Here $\epsilon$ is the empty string. You can also start at $A_1 = \{0,1\}$, but your program actually has $0$ as a base case. Starting with this ...

3

Here's an idea for generating instances with a non-unique solution that will be difficult to solve by backtracking but actually are quite easy to solve: Let $M$ be $h$ tall and $2w$ wide. Take the integers ${1, 2, ... w}$ and fill $M$ with them randomly. For $i = 1$ to $w$, replace column $w + i$ to be the column with only the value $i$. We know that $M$ ...

3

Consider that there might be something wrong with your experimental setup. The theoretical running time of an algorithm is based on an ideal model of a computer, but in practice there are lots of non-ideal behaviours. The running time of an experiment can vary based on lots of factors, and will be different from run-to-run. You should repeat the experiment ...

3

The problem is strongly $NP$-complete by reduction from 3-Partition. Given a set of $3n$ integers with total sum $3M$, we want to determine whether they can be partitioned into $n$ triples, each having sum $M$. Note that we can assume that no four integers have sum $\leq M$. Suppose a day is $3M+n-1$ units of time long. We want to visit $n-1$ museums $m_1,\... 2 The obvious answer is: take a snapshot (checkpoint) of your state before applying the arc consistency inferences; recursively explore that option; and then if it is a failure, restore the state back to your snapshot. Whether this is efficient depends upon the size of the state and the amount of work done during the recursive call. For this particular ... 2 for example 1 and 2 relative to 3 are placed correctly, but 1 relative to 2 is placed incorrectly? You have already tested that 1 and 2 are placed correctly, otherwise you wouldn't call backtrack(3). Recall that sol[i] = j means that the queen in column$i$is put at row$j$. By calling backtrack(k+1) you have already verified that every queen in$1 \dots ...

2

If it were me, I would encode this as an instance of ILP (integer linear programming) and let an off-the-shelf ILP solver deal with the backtracking and pruning. Introduce $n$ variables $x_1,\dots,x_n$, where $x_i=1$ means that you include the $i$th circle in the subset and $x_i=0$ means you don't include it. Add the following constraints: $0 \le x_i \le ... 2 The intervening assignments between the current assignment and the backjump point might have been arrived at via considerable backtracking and backjumping themselves and it would be wasteful to repeat all that work. But in the context of a modern solver that work may not need to be repeated, because: Any conflicts that occurred during the intervening ... 2 Yes, it is common for a SAT solver to combine several techniques, e.g., random restarts with smart backtracking or other tricks. More generally, nothing stops you from throwing everything and anything at a problem you want to solve. (The reason a book might talk very specifically about say "probabilistic algorithms" instead of "here's how you solve problem X ... 2 The problem is NP-hard, by a straightforward reduction from the partition problem. Therefore, you should not expect any efficient algorithm. However, there is a pseudo-polynomial-time algorithm using dynamic programming (see pseudo-polynomial-time algorithm for the partition problem for inspiration). 2 Why did it fail? Look at the code cache = new int[size+1][volume+1];. Suppose size is 1 and volume is$10^9$. One int uses 4 bytes. So this cache will use$4 \times 2 \times 10^9$bytes, which is about 8G, which is way much bigger than the limit, 32768K. So the online judge will show you out-of-memory error. Could a DP solution without backtracking be ... 2 Integer linear programming can potentially be more efficient. ILP solvers normally combine backtracking with many other methods, so for many problems ILP might be faster than backtracking alone. There are no hard-and-fast rules about which will do better on every problem, though. It's not just implementation in C++. 2 The exponential running time of backtracking for the$N$-queens problem is an upper bound, which means that the algorithm may do better sometimes. If the algorithm is 'lucky' and tries good positions for the queens first, it can find a solution with doing less backtracking compared when trying bad positions first. Whether the algorithm is lucky or not ... 2 In general Warnsdorff's rule is just a heuristic that guides the search. It is still possible that the search hits a dead-end and we are forced to backtrack. So let us consider the$n \times n$chessboard now. Warnsdorff's rule (nor any other method) won't find a solution for$n < 5$as a solution exists precisely when$n \geq 5$. Given that$n \geq 5$, ... 2 When a conflict is found during a recursive constraint satisfaction search there may be assignments and inferences in the call stack that have no connection to the conflict. Instead of backtracking through these ineffectual assignments at exponential cost it is better to jump backward over them in the call stack directly to the most recent assignment that ... 1 Sorry, but there is no solution. (Assuming that all games are 2 vs 2.) How many matches does each player play? They play with each other player exactly once, one such player per match, for a total of$n-1$matches. They play against each other player exactly once, two such players per match, for a total of$\frac{n-1}{2}$matches. This implies$n-1=\frac{...

1

Think about the chess problem or more simple 8-Queen problem are Tic Tac Toe game. The states are not binary trees. As seen from Quoara ; A state space tree is a tree constructed from all of the possible states of the problem as nodes, connected via state transitions from some initial state as root to some terminal state as a leaf.

1

There are only $9^9$ possible ways to fill each cell with one of the 9 tiles. That's a bit under 400 million possibilities. So, I suggest a simple approach: enumerate all of them, and for each, check whether or not it meets your conditions. That should run pretty fast: you should be able to very quickly test a candidate configuration to see if it ...

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