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Hello, I need the key point of your algorithm is not really clear
> • We start with the root such that all edges are possibly
> here or not. The upper bound is the sum of the $n \choose 2$
> heaviest edges, while the associated solution is the empty
> subgraph, the lower bound is thus 0.
>
> • Then at each iteration we create two children for the node
> with maximum lower bound (i.e. density of the associated
> solution). Suppose the node is at depth $i$ in the tree, we
> keep the decisions made on the first $i-1$ edges and create
> two children, one where the $i^\text{th}$ edge is included and one
> where it is not.
I need to understand this part in clear way if that is possible please

> I need to understand this part in clear way if that is possible please
I see. I agree it is not perfectly clear, sorry about that.
Can you try to understand it with the branch and bound [wikipedia page](https://en.wikipedia.org/wiki/Branch_and_bound), [the slides](https://drive.google.com/file/d/0B6cGK503Ibt0Qlg3bUVKRnFBTG8/view) and [the code](https://github.com/maxdan94/HkS)?
If it is still not clear after that, please come back and I'll try to phrase a better explanation ASAP.

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##To take away:##
- This paper is about a slight improvement of the $k$-clique Algorithm of Chiba and Nishizeki
- The performance in practice on sparse graphs is impressive
- The parallelization is non-trivial and the speedup is nearly optimal up to 40 threads
- Authors generate a stream of k-cliques to compute "compact" subgraphs
- A parallel C code is available here: https://github.com/maxdan94/kClist
##Suggestions to extend this work:##
- Can we find a node ordering better than the core ordering?
- Generate a stream of $k$-cliques to compute other quantities?
- Generalize the algorithm to $k$-motifs?
- Parallelization on higher order $k$-cliques if more threads are available?

Slides of the talk: https://drive.google.com/file/d/15MVJ2TzkdsHcyF6tE4VeYQqH8bU0kzDE/view

> Another extension: can we guarantee a given order on the output stream? that it is uniformly random, for instance?
I think that this is a very interesting and open question! I have tried to generate a stream of k-cliques such that the order is random by modifying the kClist algorithm. But I was not able to do so.
I wanted to do that in order to optimize a function depending on the k-cliques using stochastic gradient descent. I have found that using a random ordering lead to a faster convergence than using the order the k-cliques are outputed by the kClist algorithm.
Here is what I've tried:
- If you have enough RAM, then you can of course store all k-cliques and do a [random permutation](https://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle). But, since you mention "steam", I do not think that this is the case for you.
- You can use another node-ordering (different from the core-ordering) to form the DAG. You can use, for instance, a random node ordering. You may lose the theoretical upperbound on the running time, but you will see that, in practice, the algorithm is still very fast (say twice slower than with the core ordering (but this depends on the input graph and k, you may also find some settings where it is actually faster than with the core ordering)). The order the k-cliques are stream will then change, but it will not be uniform at random.
- Once you have formed the DAG using the node ordering (core ordering or any other ordering), you do not need to process the nodes in that same order. You can use another random ordering for that. It will add some randomness in the stream, but the order will still not be uniform at random.
Please let me know if you have any better ideas.

> Another extension: can we guarantee a given order on the output stream? that it is uniformly random, for instance?
One possible way to do is using a buffer, however, it remains non uniform. A buffer of size n/100 can be filled at first using the first n/100 output. Afterwards, one K-clique is randomly selected to be « outputted » and replaced with a new k-clique. The larger the buffer, the closer the ouput will be to a uniformly random output.

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Apparently, a single 'mov' assembler instruction is already Turing complete.
Implementation of mov-only compilator is [available](https://github.com/xoreaxeaxeax/movfuscator/blob/master/README.md).
> The M/o/Vfuscator (short 'o', sounds like "mobfuscator") compiles programs into "mov" instructions, and only "mov" instructions. Arithmetic, comparisons, jumps, function calls, and everything else a program needs are all performed through mov operations; there is no self-modifying code, no transport-triggered calculation, and no other form of non-mov cheating.

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