“At random” means that all possible clauses are equally likely to be chosen
This lovely graph is due to Bart Selman of Cornell University. Here, k=3. On the x-axis, we have the clause-to-variable ratio. The y-axis stands for the fraction of all formulas that are satisfiable, and the relative running time of the usual industrial strength SAT solver for solving the instance. It certainly looks like the truly hard instances of Sat lie here, around the phase transition point. (This point is empirically 4.26, but it has not been rigorously proven.) Phase transition: analogous to the physical transition from liquid to gas: it occurs at a certain critical “temperature”
Cluster: Just means that the satisfying assignments are all very tightly “close” together. Hamming distance: the distance between two assignments is the number of bits in which they differ
So think of the space of all satisfying assignments as points in n-dimensional space, and we connect two points with a line if they are within distance 1 of each other. Then here is a “cartoon” of what a typical solution space looks like in the RSB phase.... We have many clusters which are all “far” from each other in space. The RSB satisfiable phase has been rigorously shown to exist for k >= 9 Deolalikar’s proof focuses on formulas arising from this RSB satisfiable phase. These are the ones he considers to have a solution spaces with “complex” structure. And indeed it is this RSB satisfiable phase that is considered to contain “hard to satisfy” formulas, since empirically, the known SAT algorithms tend to get tripped up on these formulas (there is some debate among the statistical physicists about this, though!)
People tried many different tricks to try to get ahold of what’s going on in the proof. One thing they stumbled on was the following. Make sure your proof doesn’t prove too much! (In particular, make sure it does not prove false statements)
Now what does the solution space to F’ look like? Well, a little thought shows that it is nothing more than a translation of the solution space of F! Furthermore, the all-zero assignment satisfies F’, so we have turned a “hard” formula into an “easy” one, without changing the solution space!
We were excited because the proof strategy is new. The mere fact that no one tried it before gave it a chance of working.
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