<p>From the engeneering point of view, when not allowing any irrationalities like the golden ratio, the practical thershold should be 4,109890672858455 the exact one as long you don't consider any theoretical 1 million variables formulas or such.</p>

CC BY-SA 4.0
2021-01-01T08:09:24.937Z
2021-01-01T08:09:24.937Z
Eugen
<p>In light of the Ding--Sly--Sun verification of the 1-step Replica Symmetry Breaking picture for kSAT (when k is large enough) I think experts would now be pretty surprised if the MPZ/MMZ-conjectured formula for the 3SAT satisfiability threshold (approximate value: 4.2667) is incorrect.</p>

CC BY-SA 3.0
2017-06-11T02:49:47.143Z
I think this is the [paper](https://arxiv.org/abs/1411.0650) mentioned in this answer, by Jian Ding, Allan Sly, and Nike Sun (118 pages!).
Moot
CC BY-SA 3.0
2016-11-15T14:14:21.053Z
2016-11-15T14:14:21.053Z
Ryan O'Donnell
<p>The critical ratio of clauses to variables for random 3-SAT is more than 3 and less than 6, and seems to be commonly described as "around 4.2" or "around 4.25".  <a href="http://www.sciencemag.org/content/297/5582/812.abstract">Mezard, Parisi, and Zecchina prove</a> (in the physics sense) that the critical ratio is 4.256, whereas <a href="http://arxiv.org/pdf/cond-mat/0207194v3.pdf">the first and third authors prove</a> that it is 4.267.</p>

<pre><code>What is the range of values that the critical ratio could possibly take?
</code></pre>

<p>The motivation for me asking this question is that if the ratio could be $2+\sqrt{5} \approx 4.236$, then the standard reduction of 3-SAT to NAE-3-SAT (transforming $m$ clauses and $n$ variables into $2m$ clauses and $m+n+1$ variables) gives a ratio of $\phi$, which seems unlikely but would be pretty cool.</p>

CC BY-SA 3.0
2013-12-04T22:33:09.937Z
You need to define what it means for a ratio to be critcal.
Tyson Williams
CC BY-SA 3.0
2013-12-05T03:30:05.307Z
I think this is standard terminology: the critical ratio is the real number $\alpha$ such that random 3-SAT formulas with $< \alpha n$ clauses are almost surely satisfiable, and random 3-SAT formulas with $> \alpha n$ clauses are almost surely not satisfiable. almost surely here means the probability goes to $1$ as $n \to \infty$
Sasho Nikolov
CC BY-SA 3.0
2013-12-05T16:11:22.357Z
possible duplicate of [What are the current best known upper and lower bounds on the (un)satisfiability threshold for random k-sat and/or 3-sat?](http://cstheory.stackexchange.com/questions/1816/what-are-the-current-best-known-upper-and-lower-bounds-on-the-unsatisfiability)
András Salamon
CC BY-SA 3.0
2014-01-08T22:11:38.673Z
I contend that the question is distinct.  I'm looking for some estimate of the set of values that, say, would not shock any expert in the community.  I don't think anything below 4, for example, qualifies.  (I realize that this is, to an extent, subject to individual mileage.)
Andrew D. King
<sat><phase-transition><random-k-sat>
Random 3-SAT: What is the consensus experimental range of the threshold?
Andrew D. King
# Random 3-SAT: What is the consensus experimental range of the threshold?

(Asked by: username_0 on 2013-12-04)

The critical ratio of clauses to variables for random 3-SAT is more than 3 and less than 6, and seems to be commonly described as "around 4.2" or "around 4.25".  Mezard, Parisi, and Zecchina prove (in the physics sense) that the critical ratio is 4.256, whereas the first and third authors prove that it is 4.267.
What is the range of values that the critical ratio could possibly take?

The motivation for me asking this question is that if the ratio could be $2+\sqrt{5} \approx 4.236$, then the standard reduction of 3-SAT to NAE-3-SAT (transforming $m$ clauses and $n$ variables into $2m$ clauses and $m+n+1$ variables) gives a ratio of $\phi$, which seems unlikely but would be pretty cool.

**Tags:** sat, phase-transition, random-k-sat
> **Comments:**
> **username_1**: You need to define what it means for a ratio to be critcal.  
> **username_2**: I think this is standard terminology: the critical ratio is the real number $\alpha$ such that random 3-SAT formulas with $< \alpha n$ clauses are almost surely satisfiable, and random 3-SAT formulas with $> \alpha n$ clauses are almost surely not satisfiable. almost surely here means the probability goes to $1$ as $n \to \infty$  
> **username_3**: possible duplicate of [What are the current best known upper and lower bounds on the (un)satisfiability threshold for random k-sat and/or 3-sat?](http://cstheory.stackexchange.com/questions/1816/what-are-the-current-best-known-upper-and-lower-bounds-on-the-unsatisfiability)  
> **username_0**: I contend that the question is distinct.  I'm looking for some estimate of the set of values that, say, would not shock any expert in the community.  I don't think anything below 4, for example, qualifies.  (I realize that this is, to an extent, subject to individual mileage.)  

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## 2 answers

### Answer by username_5 on 2016-11-15. Score: 10 | Accepted answer

In light of the Ding--Sly--Sun verification of the 1-step Replica Symmetry Breaking picture for kSAT (when k is large enough) I think experts would now be pretty surprised if the MPZ/MMZ-conjectured formula for the 3SAT satisfiability threshold (approximate value: 4.2667) is incorrect.

> **Comments:**
> **username_6**: I think this is the [paper](https://arxiv.org/abs/1411.0650) mentioned in this answer, by Jian Ding, Allan Sly, and Nike Sun (118 pages!).  

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### Answer by username_4 on 2021-01-01. Score: -5

From the engeneering point of view, when not allowing any irrationalities like the golden ratio, the practical thershold should be 4,109890672858455 the exact one as long you don't consider any theoretical 1 million variables formulas or such.


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