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# September 10

## Friends meeting at the park

Three people come and meet at the park each day. Each comes 7 out of 10 days. What is the chance of all three coming? Only two? Only one? None? Thanks. I'm stumped. Please try to answer in a way that a complete idiot (me) will understand. Actually, the reason isn't so imporant. It is the actual % chance that I'm after. Many many thanks. :) Anna Frodesiak (talk) 03:32, 10 September 2014 (UTC)

We'd have to start with the assumption that each person showing up is an independent event. This probably isn't correct, as they may all avoid rainy days, or all show up when they planned to meet. But, if we assume each has an independent chance of showing up 70% of the time, then the chances of all or none showing up are:
Zero: 0.33 = 0.027
Three: 0.73 = 0.343

Now the chances of 1 or 2 showing up are complicated by the fact that a different one or two might show up, so we have to account for all the ways that can happen. In this case there are 3 ways 1 person can show up (A, B or C) and 3 ways 2 people can show up (AB, AC, or BC):
One: 3(0.320.71) = 0.189
Two: 3(0.310.72) = 0.441

To check our work, add them all up and you should get 1.0, or, if we multiply all the numbers by 100, we get percentages: 2.7% + 34.3% + 18.9% + 44.1% = 100%. StuRat (talk) 04:42, 10 September 2014 (UTC)

Wow! You are a super-genius. I am very impressed. I sort of figured out the zero and three part, but got stuck on the one and two. A thousand thanks for your help. :) :) :) Yay StuRat! And yay refdesk. The best kept secret on the Internet. :) Anna Frodesiak (talk) 07:37, 10 September 2014 (UTC)
You're quite welcome. Here it is, presented in the tree diagram format mentioned below (or as close as I can get using ASCII text):
                            I N D E P E N D E N T   E V E N T S
+-----------------------------------------------------------------------+
Person A: |     P R E S E N T   ( 0 . 7 )     |      A B S E N T   ( 0 . 3 )      |
+-----------------+-----------------+-----------------+-----------------+
Person B: |  Present (0.7)  |   Absent (0.3)  |  Present (0.7)  |   Absent (0.3)  |
+--------+--------+--------+--------+--------+--------+--------+--------+
Person C: | P (0.7)| A (0.3)| P (0.7)| A (0.3)| P (0.7)| A (0.3)| P (0.7)| A (0.3)|
+--------+--------+--------+--------+--------+--------+--------+--------+
/ |.7×.7×.7|.7×.7×.3|.7×.3×.7|.7×.3×.3|.3×.7×.7|.3×.7×.3|.3×.3×.7|.3×.3×.3|
Probability   +--------+--------+--------+--------+--------+--------+--------+--------+
\ | 0.343  | 0.147  | 0.147  | 0.063  | 0.147  | 0.063  | 0.063  | 0.027  |
+--------+--------+--------+--------+--------+--------+--------+--------+
# Present: |    3   |    2   |    2   |    1   |    2   |    1   |    1   |    0   |
+--------+--------+--------+--------+--------+--------+--------+--------+

3 present = 0.343                         = 34.3%
2 present = 0.147 + 0.147 + 0.147 = 0.441 = 44.1%
1 present = 0.063 + 0.063 + 0.063 = 0.189 = 18.9%
0 present = 0.027                         =  2.7%
------
100.0%

Note that tree diagrams are only practical for a small number of events, with a small number of possible outcomes for each event. Here we have 3 events, with two outcomes each (3 people who can be present or absent), making for 23 or 8 possible outcomes. If we had 10 events with 2 outcomes each, that would give us 1024 possible outcomes, or if we had 3 events with 10 possible outcomes each, that would give us 1000 possible outcomes. Either would be way too big to draw as a tree. But, if you can draw a tree, it can help to visualize dependencies on events, as well as if all events are independent. For example, let's say person A and B are a couple, and always are present (0.7) or absent (0.3) at the same time. The presence of person C (0.7) remains an independent event:
                                D E P E N D E N T   E V E N T S   ( A = B )
+-----------------------------------------------------------------------+
Person A: |     P R E S E N T   ( 0 . 7 )     |      A B S E N T   ( 0 . 3 )      |
+-----------------+-----------------+-----------------+-----------------+
Person B: |  Present (1.0)  |   Absent (0.0)  |  Present (0.0)  |   Absent (1.0)  |
+--------+--------+--------+--------+--------+--------+--------+--------+
Person C: | P (0.7)| A (0.3)| P (0.7)| A (0.3)| P (0.7)| A (0.3)| P (0.7)| A (0.3)|
+--------+--------+--------+--------+--------+--------+--------+--------+
/ |.7×1×.7 |.7×1×.3 |.7×0×.7 |.7×0×.3 |.3×0×.7 |.3×0×.3 |.3×1×.7 |.3×1×.3 |
Probability   +--------+--------+--------+--------+--------+--------+--------+--------+
\ |  0.49  |  0.21  |    0   |    0   |    0   |    0   |  0.21  |  0.09  |
+--------+--------+--------+--------+--------+--------+--------+--------+
# Present: |    3   |    2   |    2   |    1   |    2   |    1   |    1   |    0   |
+--------+--------+--------+--------+--------+--------+--------+--------+

3 present = 0.49             = 49%
2 present = 0.21 + 0 + 0     = 21%
1 present = 0    + 0 + 0.21  = 21%
0 present = 0.09             =  9%
----
100%

StuRat (talk) 13:58, 10 September 2014 (UTC)
You should also take a look at Binomial distribution. -- Meni Rosenfeld (talk) 09:05, 10 September 2014 (UTC)
You mean me? That page could be upside down and scrambled and would make as much sense to me. Thank you, though. :) Anna Frodesiak (talk) 09:16, 10 September 2014 (UTC)
Yeah, I've noticed that page isn't very newbie-friendly. But it describes the general way to solve problems like the one you've presented. If there are n different things which can either happen or not, each with probability p, and they are independent, then the probability that exactly k of them will happen is $\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}$, where n! is the factorial. In your case, the things that happen are each person showing up, $n=3$ and $p=0.7$. -- Meni Rosenfeld (talk) 12:49, 10 September 2014 (UTC)
A good way to understand this sort of thing "visually" is with a tree diagram. That article is just a stub, but the linked BBC page has some good examples. If you take their 3-coin-tosses example, and replace the 3 tosses with the arrival or non-arrival of each of the 3 people (and change the 0.5 probabilities of heads and tails to 0.7 and 0.3) then you should end up with the same answers as above. AndrewWTaylor (talk) 13:22, 10 September 2014 (UTC)
Holy moly. I'm actually understanding this. I sort of lost it half way through, but was getting it. I will read it again tomorrow after a big coffee. This is very nice. I never understand stuff like this. I have the IQ of lichen. Anna Frodesiak (talk) 14:47, 10 September 2014 (UTC)
Awesome, glad we could help. StuRat (talk) 16:57, 10 September 2014 (UTC)

# September 11

## is any of Spinoza's Ethics mathematically rigorous? (to the standard of published proofs.)

Hi,

Spinoza's Ethics takes the form of extreme mathematical rigor. I was wondering if any of it were rigorous enough to be published as a mathematical proof, or, on the contrary, does it just take this form with all the real convincing 'power' couched in the terms and language themselves, which are left undefined?

I hope you see my question. If indeed it contains real (rigorous) proofs, is there any chance that you can quote (or produce) me a small 'lemma'-like (or even smaller!) rigorous argument from that work, to show how we can translate it into mathematics and treat it as such?

What I mean is that clearly it takes the form of proofs with premises, logical steps, and conclusions - but are these vacuous? Could we translate any of htis into a proof in a computer language, for example?

I've only just glanced at it but for me it seems that there is no logic or rigor used at all, and in fact the form is highly misleading, as it makes it seem as though there are definitions that are being applied, whereas there are no such definitions and instead we are left with undefined terms like perfection and God that are useless in a formal context. however this is just my impression!!! I've had the same, mistaken, impression, of highly rigorous works that could be understood well and translated directly into code.

Therefore I would like your opinion about whether Spinoza's Ethics is of this kind of work, and, if so, I wonder if you could produce for me either a quotation or your own synthesis of a very short (perhaps trivial) but rigorous "proof" from it.

Thank you kindly! --212.96.61.236 (talk) 01:42, 11 September 2014 (UTC)

One can look at the logical structure and try to formalize it; see for instance [1]. But as a review of that paper states [2], there is more to Spinoza's Ethics than logical inference. --Mark viking (talk) 02:43, 11 September 2014 (UTC)
Well, of course there is more as it is a religious text, not a proof. But it takes the form of a proof. The question is: is that form rigorous? Is there anything interesting or good about what it proves rigorously? For example, I could write a whole treatise on color coordination in interior decorating, what colors go together and what don't, what needs to match with what, I can define premises and conclusions, such as the maximum number of colors that can be in touch with each other without dissonance, etc. It will all be absolutely meaningless gibberish! Color simply isn't the kind of thing that is amenable to reasoning about rigorously. Period. I could also write the same thing about physics. But in this case it would be highly meaningful. Physics is the kind of thing you can reason about. Now what about Spinoza's Ethics (the work) - is it like an axiomatic treatment of interior decorating (i.e. meaningless gibberish) or like an axiomatic treatment of quantum mechanics (fully meaningful and formalizable)? Given that in a sense you could say Spinoza's Ethics (the work) is a work of physics, if we treat it as such does it (in parts) attain modern standards of rigor? Could you produce kind of an extract of one, or a synthesis of one? Thanks. 212.96.61.236 (talk) 02:50, 11 September 2014 (UTC)
An axiomatic treatment of interior decorating can be absolutely rigorous, as far as the math, in which case, those who accept that the axioms truthfully correspond to decorating will also be compelled, rationally, to accept that any theorems apply equally as much. Axiomatic systems of physics work pretty much the same (though, physics isn't axiomatized, or, at least, it is not done from axioms). I don't much about the specific case of Spinoza, but you are essentially asking if he makes any logical errors, if taken at face value. It sounds, more so, like you are asking if his theories are correct, or have correspondence, in the way that physics does - that question has nothing to do with axiomatics and logical structure, however, and, really, is just asking if his premises are sound, which is not a mathematics question at all; and, as for physics, it does not correspond to reality because its structure is mathematical (nor is that a defining aspect of what physics is, to be honest).Phoenixia1177 (talk) 03:40, 11 September 2014 (UTC)
Right, but my point is there is not a single axiom in interior decorating that anyone would accept, or even so much as entertain, for even a second. Any axiom can be false under certain conditions - i.e. leads "logically" to a conclusion that you would, however, not accept. So, obviously a system of axioms which is the null set does not make for a very interesting axiomatic system. But is theology the same way? Or, in the case of Spinoza, does he use axioms that are in some sense interesting and perhaps have correspondence, and then does he reason from them in a logically sound way? Or, is it just pseudomathematical/logical? (Following the form, but without any chance of correspondence.) 212.96.61.236 (talk) 04:01, 11 September 2014 (UTC)
You're not asking about mathematics, though, you're either asking if Spinoza had true axioms or if those axioms were interesting, neither has anything to do with mathematics and logic, it has everything to do with philosophy and ethics and the world. Even if nobody believed some set of axioms for "interior design" those axioms would be every bit as legit as the axioms of ZFC, or any other system. A set of axioms is a system, that is it, the content of those axioms and what they mean is not the purview of mathematics, as a subject. This is especially so when you are talking about axioms for something philosophical. I don't know, personally, if his reasoning is logically sound, but you might have better luck trying at the humanities desk as this is really a philosophy question - or just try googling criticisms of Spinoza's work, you're sure to find something far more salient.Phoenixia1177 (talk) 06:47, 11 September 2014 (UTC)
You've hit at the crux of my question by stating, "A set of axioms is a system...the content of those axioms and what they mean is not the purview of mathematics". That might be true for a real set of axioms, but not everything that is labelled that way is an axiomatic system. Here is an example. I've prepared a farcical "axiomatic system" and some proofs, below. They're just the first few sentences of our Color article. What do you think?

What do you think of the above?

What you SHOULD think is that it's total nonesense, it doesn't even try to look like an axiomatic system. It's just borrowing the form, like gibberish or gobbledegoock or Lorem ipsem. It only might look like an axiomatic system at a brief glance.

It's obviously NOT actually an axiomatic system!

So, my problem/question is that to me, Spinoza's Ethics seems the same way (at a first impresssion). I was wondering if it actually was that way - or if, on the contrary, it really is an axiomatic system and some proofs within it.

So, which is it? Is it like my farcical example? Or is it more? Does it meet logical/mathematical rigor, is it nonesense (from a mathematical point of view), much as my sentences above are. Note that my sentences aren't actual total nonesense - they're quoted from the Wikipedia color article after all. 213.246.165.17 (talk) 09:18, 11 September 2014 (UTC)

If you rearranged the "axioms" above so that they were all sentences, they would be axioms, just pointless ones as far as that goes. As for the propositions, if the terms all traced back (that's not even that necessary, really), that would work too. The only issue is the logic of your proofs wouldn't be standard logic (I'm sure you could concoct some goofy deduction rules too, if you really wanted, why not?). That's kind of the problem with your whole question, any set of sentences can be "axioms", as long as they are some form of declarative essentially. So, when you ask if Spinoza's are nonsense, do you, literally, mean to ask if they satisfy being declarative sentences relating terms and if he was capable of following basic logic? Most philosophical work is going to be logically valid (as a general rule), the debate is over soundness. I'm not trying to be a jerk, but as far as mathematical requirements go, the answer should be immediately obvious upon reading it, the terms can be complete nonsense, the relationships all bullshit, as long as axioms aren't of the form "Is it good to steal, ever?" or "Stop!", I'm sure someone can whip something up in symbols. For example:
1. All lines love at least 3 distinct circles.
2. Some circle loves a line.
3. Every circle is loved by some line.
4. If puppy A loves puppy B loves puppy C, puppy A loves puppy C.
5. Every circle is a puppy.
6. Every line is a puppy.
Is a system of axioms. And we can deduce that there is some line that loves some other line, and that there are at least 3 circles if there is a line. That's all perfectly valid and fine, mathematically - of course, it's all meaningless gibberish as far as humans go (I imagine, but who knows? Maybe it has a neat model - I doubt it).Phoenixia1177 (talk) 09:47, 11 September 2014 (UTC)
One thing to keep in mind here is that the standards of rigor have changed significantly since Spinoza's time. There was a time when you wouldn't be considered an educated person unless you could recite the 47th proposition of Euclid upon demand, and The Elements where hugely influential a result. Spinoza modeled Ethics on it in an attempt to bring the same sense of certainty to his conclusions as was perceived in be in the propositions of Euclid. But much has happened since then and The Elements would not stand up to modern standards of rigor. For on thing Euclid's geometry was trying describe the universe as it actually exists, at least in some idealized Platonic sense. The general theory of relativity says the universe doesn't behave as described. Also, modern analysis has found many hidden assumptions in the Elements and a truly rigorous axiomatization (see Hilbert's axioms e.g.) requires many more axioms than Euclid gave. Finally, the whole viewpoint of mathematics changed from a description of the world to a purely logical construct. People of Spinoza's time would not be familiar with the concept of symbolic logic, but today's standards of rigor require that mathematical reasoning can, at least in theory, be stated in symbolic form. Euclid and Spinoza basically start by saying "here is a bunch of things we can all agree about what they are, and here is a list of things we can all agree are true about them, now let's see what conclusions we can draw." The definitions used are not definitions in a rigorous sense but descriptions that enable the reader and author to agree on what things are being talked about. An example of this type of definition might be "A cat is a small fuzzy creature that sometimes lives in people's houses." But mathematically this definition is nonsense, just as the definitions of point and line given in Euclid are more or less nonsense by modern standards. A mathematician would say "Small relative to what? What does it mean for a thing to be fuzzy? What is a house? etc." At some point people realized that in order to avoid circularity a mathematical theory would have to include undefined concepts, from which other terms could be defined. But Spinoza doesn't take that approach, instead starting out with definitions involving things like "essence", "nature", "conceivable" which the reader is supposed to already understand. So, the short answer to the original question is no, at least by modern standards, it's not mathematically rigorous, but then it's hard to see how someone from the 17th century could produce something of the kind that would be. Whether it was rigorous by 17th century standards is something you'd have to get from contemporaries. Leibniz was a philosopher in his spare time so perhaps he had something relevant to say about it. --RDBury (talk) 11:49, 11 September 2014 (UTC)

Thank you for the responses! There is a lot to read there. Let me ask some basic background questions. 1) As a mathematician, do you find Spinoza's Ethics any more convincing than the "some circles love a line" axiomatic system and proofs? Or is it equally gobbledegook. 2) Although it's not formally rigorous by today's standards, is Euclid convincing to modern mathematicians, i.e. can they follow and rely on those proofs, within the system that Euclid set up? (Despite its being insufficiently formal). 3) A clarification on your analogy. Is this still an equally valid axiomatic system:

1. All lines love at least 3 distinct circles.
2. [+] There is at least 1 line.
3. [+] No lines love at least 3 distinct circles.
4. Some circle loves a line.
5. [+] No circle loves a line.
6. Every circle is loved by some line.
7. [+] No circle is loved loved by some line.
8. [+] There is at least 1 circle
9. If puppy A loves puppy B loves puppy C, puppy A loves puppy C.
10. Every circle is a puppy.
11. [+] Every circle is NOT a puppy.
12. [+] There is at least 1 circle
13. Every line is a puppy.

And for good measure:

1. [+] No line exists
2. [+] There is no circle

Are we still good? Even though I've now added literal contradictions, just the same sentences with a "not" in them, as axioms?

Does at least this make the axiomatic system nonsensical? 213.246.165.17 (talk) 10:14, 12 September 2014 (UTC)

If you're using classical inference rules, a formal system that has both P and ¬P as axioms for some proposition P is uninteresting since you can trivially prove anything in it (ex falso quodlibet), but it is still a formal system. Its proofs are valid, though boring. There are paraconsistent logics with weaker inference rules that allow for nontrivial, potentially interesting formal systems in which classically contradictory propositions are provable, but I'm not sure that's relevant here.
Spinoza's Ethics is clearly not formally rigorous (i.e., mechanically checkable), but very few published mathematical proofs are either. They can only be checked by specialists who understand the concepts well enough to fill in the gaps. I have an intuition for plane geometry that enables me to follow Euclid's proofs, though they aren't always entirely convincing since it's not always obvious that the constructions work in the general case (see e.g. Euclid's Elements#Criticism). Spinoza's "proofs" look like gobbledegook to me, but it may be that someone else understands them. The very first reply to your question linked to a paper ("The Logical Structure of Spinoza's Ethics, Part I" by Charles Jarrett) that actually attempts to encode Spinoza's proofs in a formal system, so apparently Jarrett understands them, or thinks he does. -- BenRG (talk) 00:27, 13 September 2014 (UTC)
(Neither) Euclid, nor Spinoza, is perfectly laid out in a modern sense, but I think it would certainly be possible to do so, if one were inclined. For Spinoza it might take a little more effort since the intended content is not mathematical, but it would, really, just be a matter of organizing terms to reflect how they are being used by him. The better question is if it would be worth it; personally, all it would do is codify and make explicit the reasoning, but the real question is if the results actually mean anything or have any worthwhile philosophical content - and I don't know if they do, but if they do, formalizing them isn't going to make that any more apparent, probably less. As for your add on axioms: yes, that is a formal system, having a contradiction does not mean that the system is not axiomatic, or somehow not allowed (just as being meaningless doesn't make it not a formal system). Indeed, the axioms of set theory may contradict each other, and we have no way of proving that they do not, so if contradictions are not allowed, then for many common systems we would have to say that we aren't sure if they actually are formal systems; obviously, this wouldn't be a good position to take. When a system has a contradiction, it is called inconsistent.Phoenixia1177 (talk) 04:31, 13 September 2014 (UTC)

## Statistical statement

Message from a "stop smoking" campaign: "Stop smoking for 28 days and you're 5 times more likely to stop for good."

It seems to me that this statement is nonsensical. Am I missing anything? 86.129.18.104 (talk) 03:15, 11 September 2014 (UTC)

I think it means that people who stop for 28 days are 5 times more likely to quit for good, compared to the ones that don't make it 28 days. Bubba73 You talkin' to me? 03:35, 11 September 2014 (UTC)
There's probably a better, more formal way to say it. Suppose that, as seems likely, the probability of a relapse strictly decreases with length of abstinence. Say $P(n)$ is the probability of never smoking again after not smoking for n days. Clearly $P(0)$ is near zero, and $P(28)$ is greater than that, but five times what? What is q such that $P(28) = 5 P(q)$ ? —Tamfang (talk) 03:59, 11 September 2014 (UTC)
I take the meaning to be $q = 0$. That is, it's comparing to the probability before you even try. Another issue is whether "5 times more" implies a factor of 5, as Tamfang assumed, or a factor of 6. --65.94.51.64 (talk) 04:19, 11 September 2014 (UTC)
I agree it's nonsensical. It's like a "Live long!" campaign that says, "Live to the age of 80 and you're 30 times more likely to live to the age of 85." Well, great. That helps someone live to 85 how? The hard part of smoking is probably quitting smoking for 28 days. In fact, it sounds like it's 80% of the hard part." 213.246.165.17 (talk) 08:11, 11 September 2014 (UTC)
• Hmmm, it seems that no one else saw quite the same fundamental illogicality as me, so let me explain the way I see it. Let's say, for the sake of argument, that if you stop smoking for 28 days you have a 50% chance of stopping for good. The statement then implies that if you DON'T stop smoking for 28 days you have a 10% chance of stopping for good. To me, this obviously cannot be correct. If you don't stop smoking for 28 days then you have NO chance of stopping for good. In order to give up for good, you HAVE to stop for 28 days. Any further thoughts? 86.160.86.83 (talk) 11:00, 11 September 2014 (UTC)
I think it's trying to convey a statement about conditional probability-- $\mathbb{P}(\mathrm{stop for good (i.e. day X+1 \to \infty) }|\geq 28 \mathrm{days \; smoke \; free \; on\; day \;X)} \approx 5 \cdot \mathbb{P}(\mathrm{stop for good}| < 28 \mathrm{days \; smoke \; free \; on\; day \;X})$. While the English phrasing might be awkward, I sincerely doubt that the stat isn't drawn from some fairly legit study. They're just struggling to get it into a snappy ad campaign. SemanticMantis (talk) 14:03, 11 September 2014 (UTC)
Wouldn't that calculation depend on X, though? Let's take a concrete example. Say ten people try to stop smoking, and the number of days they last is {1, 2, 5, 10, 25, 40, 60, permanently, permanently, permanently}, then how would you do that calculation? Where the original statement has "5 times", what factor would this data yield? 86.160.86.83 (talk) 19:20, 11 September 2014 (UTC)
No, I was just specifying it for clarity, thinking of X as a calendar date when they did the survey. Mark viking seems to (mostly) share my interpretation below. The difference is, I'm lumping everything less than 28 days together, rather than comparing to zero-days-stopped. This would be an easy way to "cherry pick" the data to find a nice statistic, just keep dividing the pool into two groups based on Y days stopped, until you get the multiplier that you want. SemanticMantis (talk) 23:09, 11 September 2014 (UTC)
I interpret this as the conditional probability of quitting forever given that you stopped for 28 days is five times the probability of quitting given no such 28 day stoppage, i.e, the probability of quitting after 0 days of stoppage, right at the start. It makes sense to me, as after 28 days, most of the the physical withdrawal effects are probably gone and the psychological habit may be broken, too. It looks to be part of the Stoptober campaign, but I could not find a source for the statistic. --Mark viking (talk) 19:54, 11 September 2014 (UTC)
Perhaps the missing or unclear information, then, is "5 times more likely than what?"? I read it as "5 times more likely than if you don't stop for 28 days". In your interpretation, I suppose it would be "5 times more likely than when you start out". Is that right? 86.160.86.83 (talk) 20:38, 11 September 2014 (UTC)
Presumably it means that if you picked two persons at random from a group of recently quit smokers, one of whom had just quit that day and the other had quit 28 days ago, the person who had quit 28 days ago would be five times more likely to quit for good than the person who had just quit. Doctors are not known for any Bayesian subtlety when they make statements like this. Sławomir Biały (talk) 20:46, 11 September 2014 (UTC)
Why compare to the 0 day quitter? pooling all days-quit<28 as I did above makes more sense to me... SemanticMantis (talk) 23:09, 11 September 2014 (UTC)
Because the statement will certainly be false if you do that. Why should the 28th day be so much more important than the 27th? Dbfirs 09:57, 14 September 2014 (UTC)

## Confusion Over Open Problem's Meaning

On the article linked in the subject line, it asks "Is there a logic satisfying the interpolation theorem which is compact?", I'm assuming the reference is to Craig interpolation, but FO is compact and satisfies it, so I'm not sure what it is asking. I don't have access to the source, unfortunately.Phoenixia1177 (talk) 10:27, 11 September 2014 (UTC)

The statement in the WP article is not correct. From the chapter referenced, the open problem is Is there a logic L which satisfies both the Beth property and Δ-interpolation, is compact but does not satisfy the interpolation property? The interpolation here looks like Craig interpolation, but I have little knowledge of this field, so don't trust me on that. --Mark viking (talk) 21:57, 11 September 2014 (UTC)
Is this a case for WP:SOFIXIT then? SemanticMantis (talk) 23:10, 11 September 2014 (UTC)
Phoenixia1177 just did. --Mark viking (talk) 23:34, 11 September 2014 (UTC)
Thank you for the response:-)Phoenixia1177 (talk) 03:15, 12 September 2014 (UTC)

## puzzle from game question...

In this game there is a puzzle as follows. There are 16 items, four each of four colors and four each of different types of shells. Call the colors A-D and the shell types 1-4. They are placed in a 4x4grid at random and the game is won if the top row is A1-A4 in order, second row B1-B4 in order and so on. Legal moves are as follows: Two items may be switched if and only if the cells border each other vertically, horizontally or diagonally *and* the two items share a characteristic (color or type of shell). Can it be won from any starting position? (If moves are only allowed Horizontally and Vertically, then the Order 4 Graeco-Latin square would be a losing starting position) If the puzzle can be won from any position, is there anything like a strategy to win it in as few moves as possible?Naraht (talk) 12:59, 11 September 2014 (UTC)

I had a go at solving it starting from a random position and thought it made a nice puzzle. It's similar to but slightly harder than the 15 puzzle, much easier than Rubik's cube. I think this starting position
A1 B2 C1 D2
C3 D4 A3 B4
A2 B1 C2 D1
C4 D3 A4 B3

leaves you with no moves allowed, which would mean it's not always possible to win from any starting position, but it looks like such configurations are very rare. Anyway, I'm sensing commercial possiblities, maybe a cell-phone app? --RDBury (talk) 11:23, 12 September 2014 (UTC)
I agree that that starting position has no moves. (and rotating the columns/rows and/or rotating the numbers/letters also would give a no move position, so a few more than just that one). It is a (small) part of a game that my wife downloaded a few days ago, I'll try to find the name. Any ideas for solution strategy?11:55, 12 September 2014 (UTC)
Somehow I thought this was a game you made, my mistake. I don't know about as few moves as possible, but the way I solved it was a lot like the 15-puzzle. First get all the 1's in the top row. Getting 3 out of 4 isn't too hard to do ad hoc so the tricky part is the 4th. To make things a bit easier, do the corners first so you have something B1, C1, *, A1 in the first row. You need to get D1 where the * is, but let's say you have B4 there instead. Work D4 into one of the three squares below and diagonal from the * (again using whatever ad hoc moves will work, then swap it in to get B1, C1, D4, A1. Now work D1 into a position where you can swap it with D4 and do so to get B1, C1, D1, A1. Once all the 1's are in the top row you can swap them around to get them in the right order. With the top row done do the same with the first column using more or less the same method. Continue with the second row and second column and you're down a 2 by 2 square in the lower right. But 2 by 2 is easy so you're done. The trick is that when you have three out of four in the first row, it does no good at all to try to move the fourth one into position before getting a square in that you can swap with it. Not a very specific method I know but it would be a lot of work to write out and verify a detailed algorithm. I'm guessing that for finding the minimum number of moves it would be hard to improve upon brute force enumeration of possible moves, but 16! is very large so it may not be possible. You could probably get some good upper bounds with a clever algorithm though. Just for a ballpark value I did another randomly generated on and it took about 55 moves, but that's probably not optimal. --RDBury (talk) 18:04, 12 September 2014 (UTC)

# September 13

## 3rd and 4th degree Bézier curve

Hi,
the computer's fonts use Bézier curve to represent glyphs.
The creator of the font divide the complex curve to sub-curve, which he then describe with Bézier curve technique.
My question is whether there is a technique that can take 4-th degree collection of Bézier curve, and create 3-rd degree collection of Bézier curve that looks the same.Exx8 (talk) 01:10, 13 September 2014 (UTC)

You can't in general turn a 4th degree Bézier into exact 3rd degree ones. So your 'looks the same' needs to be some sort of approximation. They do lend themselves to mathematical reasoning - for instance you can always subdivide a Bézier curve exactly into smaller ones of the same degree and your rule may lend itself to checking easily if the sub curve can be approximated closely enough. A computing solution might be to simply generate lots of points close to each other and then approximate them with cubic Bézier curves with smoothing just like one would for freehand drawing. There could easily be a computer package already written to do this. Dmcq (talk) 06:27, 13 September 2014 (UTC)
TrueType fonts just use quadratic Bézier curves and OpenType fonts uses cubic Bézier curves. You should be able to get a good enough approximation to your curves for use with fonts. I would suspect most font designers would do this by eye adjusting control points to get the look they want rather than use an deterministic algorithm. A good discussion of Bezier curves is A Primer on Bézier Curves sections 29 and 30 near the end might help.--Salix alba (talk): 08:18, 13 September 2014 (UTC)
My naïve suggestion: Cut up your original curve so that there is a control point wherever the tangent angle is a multiple of π/4, and wherever the curvature reaches an extreme or crosses zero; these are the points most important to the appearance of the glyph. On each segment of the arc, make the derivatives of the cubic match (in angle and magnitude) those of the original curve at the control points. (I have a sporadic project involving even more arcane curves, and this is how I plan to convert them to cubic fonts.) —Tamfang (talk) 02:05, 14 September 2014 (UTC)

## Verifying that the cts dual space is subspace of the algebraic dual space

Hi,

It's mentioned on the page for dual spaces that if a vector space is topological, the continuous dual space is a linear subspace of the algebraic dual space.

I tried to verify this directly from the subspace axioms:

Let $V$ be a topological v.s over $F$. Assume $f\in Hom(V,F)$ is cts, and let $k\in F$. Then

$(kf)(x)=k(f(x))=(\phi\circ f)(x)$

(where $\phi$ is multiplication by $k$), by the definition of scalar multiplication on $Hom(V,F)$

Since scalar multiplication on a topological field is cts, as is the composition of cts functions, it follows that $kf$ is continuous.

But I can't seem to show closure under addition. Have I overlooked something, or is there a better approach?

Neuroxic (talk) 06:28, 13 September 2014 (UTC)

Perhaps I'm missing something, this isn't really my area, but by definition of a topological field, addition is continuous, so if f and g are V->k, then f + g is continuous. Clearly, f + g is a linear functional, so the continuous dual is closed under addition. That the continuous dual is a subset of the algebraic dual is immediate, so it is a linear subspace. Again, I apologize if there is something that I am glossing over, I haven't touched any of this stuff in a year, or two.Phoenixia1177 (talk) 06:36, 13 September 2014 (UTC)
"by definition of a topological field, addition is continuous, so if f and g are V->k, then f + g is continuous" this is the bit I'm having trouble with. Would you be able to expand on this more precisely?
Neuroxic (talk) 10:15, 13 September 2014 (UTC)
The definition of a topological field k includes that the basic field operations are continuous. For any topological space X and continuous f,g:X -> k, f + g is continuous because it is composition of functions; explicitly: +(f, g). Continuous linear functionals are linear maps V -> k that are continuous. Thus, pointwise addition will yield, another, continuous map. It is obvious that the sum of any two linear functionals is a linear functional, thus, the sum of two continuous linear functionals is another one. Hence, closure under addition. --Essentially, functions add using pointwise addition, since addition is continuous, addition of functions outputs continuous functions.Phoenixia1177 (talk) 10:28, 13 September 2014 (UTC)
"For any topological space X and continuous f,g:X -> k, f + g is continuous because it is composition of functions; explicitly: +(f, g)."
This part doesn't make sense to me, and was one of the reasons I was stuck. I tried to use the idea of composing cts functions to preserve continuity (that worked to show closure under scalar multiplication) but wasn't able to do it. I thought I couldn't write $(f+g)(x)=(\tilde f \circ g)(x)$ for some $\tilde f$ because $\tilde f$ depends on $g(x)$, not on $x$. How do you write $f(x)+g(x)$ as a composition of two cts functions?
Neuroxic (talk) 12:24, 13 September 2014 (UTC)
The function +:k x k -> k is continuous, so f + g is +(f, g), it is a continuous function of two variables composed with f and g. Since all the operations are continuous, the end result is. As a more basic example, consider continuous f,g from a topological space into the reals, their sum is continuous for the same reason - it's the continuity of addition that gives this result. The same principle applies here, just X replaced with the dual space and the reals with a topofield.Phoenixia1177 (talk) 12:32, 13 September 2014 (UTC)
Epic fail. I forgot that a function $(f,g)$ is cts if $f,g$ are cts. I know the proof now. Thanks.
Neuroxic (talk) 12:37, 13 September 2014 (UTC)

# September 14

## Star angle

Imagine pointed stars made to match the notation used at Schläfli symbol#Regular polygons .28plane.29.

For example the pentagram is {5,2}.

How do I determine the angle within the points from this pair of numbers? How do I determine the exterior angle where lines from two points meet?

Thanks -- SGBailey (talk) 19:53, 14 September 2014 (UTC)

I think the notation for the pentagram is actually (5/2). The angle for the star polygon with symbol (n/k) is π(1-2k/n) radians. So for example the angle for the pentagram is π(1-4/5)=π/5=36°. --RDBury (talk) 00:53, 15 September 2014 (UTC)
Both the pentagram and the Schlafli article use {} not (). -- SGBailey (talk) 06:40, 15 September 2014 (UTC)
Yes, but it contains one rational number $\frac{5}{2}$, not two integers separated by a comma. —Tamfang (talk) 06:53, 15 September 2014 (UTC)
And the sum of internal angles for a star polygon with symbol {n/k} is 2 π (n - 1). From this and the convex internal angles it is easy to see that the concave internal angles are π (1 + 2 (k-1)/n), or equivalently, the corresponding external angles are π (1 - 2 (k-1)/n). Icek (talk) 01:09, 15 September 2014 (UTC)
So for {5,2) in degrees we have 180 * ( 1 + 2 * (2-1) / 5) = 180 * 1.4. That has to be wrong. I'm currently thinking 180 * ( n - 2k ) / n -- SGBailey (talk) 06:40, 15 September 2014 (UTC)
No, it isn't wrong for the internal concave angles of the pentagram. Measure it. The regular pentagon has internal angles of 108 degrees. Think of the pentagram as a pentagon with 5 triangles attached; then from RDBury's formula for the outer angle of such a triangle (36 degrees), you can calculate that the sum of the other 2 angles of such a triangle must be 144 degrees, therefore one such angle is 72 degrees. The internal concave angle of the pentagram can be thought of as the sum of the internal angle of the pentagon and 2 angles of 72 degrees each from the adjacent triangles. That sums to 252 degrees, the same answer as from my formula. Icek (talk) 11:20, 15 September 2014 (UTC)
Your equation is wrong. The answer for the {5,2} pentagram is 36° (or equivalent in radians). Your equation doesn't give 36°. It may be the correct formula for some other aspect of the shape, but not for the angle of the points of the star. -- SGBailey (talk) 15:26, 15 September 2014 (UTC)
RDBury's formula is perfectly adequate for that angle. I only added the formula for the concave angles, and its complement, and the latter would be your "exterior angle where lines from two points meet" (but only if the two points are adjacent): π (1 - 2 (k-1)/n). Icek (talk) 01:30, 16 September 2014 (UTC)
I don't know where Icek's formula comes from. The sum of internal angles of a convex polygon is (n-2)π, as you can see by decomposing it into triangles, not 2(n-1)π.
The vertex of a regular {x} polygon, whether x is integer or not (convex or star polygon respectively), represents a turn of 2π/x, and the internal angle is π minus this, whence RDBury's formula π(1-2/x).
Another way to look at it: consider the triangle formed by the center and one edge of the star. The angle at the center is 2π/x, so the sum of the other two angles – each of which is half of the interior angle at a vertex – is π-2π/x. —Tamfang (talk) 07:05, 15 September 2014 (UTC)
The sum of the internal angles of a star-shaped polygon with n corners is (n-2)π as well: You can consider the triangles formed by the center and one edge of the star, and there are n of them. The sum of angles of these triangles is n π. Subtract the angles at the center (which sum to 2π), and you are left with the sum of the internal angles of the star-shaped polygon.
But the star-shaped polygon with symbol {n/k} should have 2n corners, as long as k > 1 and the fraction n/k is irreducible, hence my formula valid only for these special cases: 2π (n-1).
The rest follows from subtracting the n pointed angles and then dividing the rest by n, because there are n concave angles. Icek (talk) 11:20, 15 September 2014 (UTC)

# September 15

## closed group of operations

Hello. I am interested in the six functions $f(z) = z,~z/(z-1),~1-z,~1/z,~1/(1-z),~1-1/z$. These seem to be closed under composition and form a group which must be isomorphic to S3 because it is not commutative. It is a subgroup of the Mobius group of transformations. But does it have a special name? Is it the only subgroup of order six? I am looking for a couple of sentences to describe unambiguously its relationship to the Mobius group of transformations. Thanks, Robinh (talk) 08:13, 15 September 2014 (UTC)

That's only five functions ... (I've added some spaces after your commas to ease counting.) -- SGBailey (talk)
(OP) corrected, now there are six. It looks better now it's got spaces, but I didn't think math typesetting was sensitive to whitespace. Robinh (talk) 08:47, 15 September 2014 (UTC)
(I think for WP math mode, as in LaTeX, any whitespace >1 renders the same as whitespace=1, but whitespace=0 is different. I took the liberty of testing/confirming this in your original question, as you can see if you look at the source now :) SemanticMantis (talk) 14:18, 15 September 2014 (UTC)
In the context, Robinh's suspicion seems to have been correct: whitespace only has significance in LaTeX if it changes the parsing. It does not change the display spacing; for that you need an explicit spacing (e.g. ~, which I've taken the liberty of including) —Quondum 15:24, 15 September 2014 (UTC)
It is not a normal subgroup, so cannot be the only subgroup of order six. However, within $PSL(2,\mathbb Z)$ this is the group of automorphisms of the quadratic form $x^2+y^2+xy$ whose zeros (projectively) are the cube roots of unity. This group is called the anharmonic group. Sławomir Biały (talk) 12:18, 15 September 2014 (UTC)
As for uniqueness, you can generate an infinite family of 6-function sets with the same group structure by taking the one that you found and using the mapping from the Mobius group to 2-dimensional matrix groups and changing basis. For example, here's one that (I hope) also gives you S3:
$f(z) = z, -z, \frac{3+z}{1-z}, \frac{3+z}{z-1}, \frac{3-z}{1+z}, \frac{z-3}{1+z};$
For subgroups of order 6 which are unrelated to the S3 that you found, you can quite easily find representations of Z6 (e.g. the group generated by $f(z)=e^{i\pi/3}z$). 129.234.186.11 (talk) 16:50, 15 September 2014 (UTC)

(OP) thanks everyone. So it's not normal (Sławomir, how did you discover this? Did you compose a general element of the Mobius group with one of the transformations above and observe that it made a function not in my set of six? Or is it obvious to you in a more direct way?). And therefore not unique because I can consider $g^{-1}Hg$ for some $g\in M$. It says in Cross-ratio that "The anharmonic group is the group of order 6 generated by λ ↦ 1/λ and λ ↦ 1 − λ. It is abstractly isomorphic to S3 . . .". This is close to what I'm seeking. But what does "abstractly isomorphic to S3" mean? Does this statement have connotations not implied by plain old "isomorphic"? Robinh (talk) 20:13, 15 September 2014 (UTC)

I take "abstractly isomorphic" to mean that there is no particular implied realization of the group as a group of permutations of three objects. However, it is easy to see that they do, in fact, act as permutations on the points {0,1,∞}, so this gives an explicit isomorphism of the group with a symmetric group and it might be helpful if the article cleared that up. As for the non-normality, I just did it by conjugating by some random thing in SL(2,Z) (any element of infinite order will work). But if you believe what I said about the quadratic form, it is also "obvious" from that point of view. Sławomir Biały (talk) 21:57, 15 September 2014 (UTC)
OK thanks Sławomir. I see what abstractly isomorphic means now. I am having difficulty understanding the relationship between the Mobius group and your quadratic form $x^2+y^2+xy$. Also, would it help to consider the group of six functions and their group action on the set {0,1,∞}? Robinh (talk) 22:17, 15 September 2014 (UTC)

## Graph theory:

Is there a special name, or a recognized short expression/description, for denoting the following type of directed graphs?

1. Let G be a directed graph, and let v be a vertex - from which there is an edge in G - and to which there is an edge in G.
2. For every u there is w, such that if G has an edge from u to v, then G has both an edge from v to w and an edge from u to w.
3. For every w there is u, such that if G has an edge from v to w, then G has both an edge from u to v and an edge from u to w.

HOOTmag (talk) 14:25, 15 September 2014 (UTC)

## Relation theory:

Is there a special name, or a recognized short expression/description, for denoting the following type of relations?

1. Let R be a binary relation, and let v be an object for which there are both u satisfying R(u,v) and w satisfying R(v,w).
2. For every u there is w, such that if R(u,v) then both R(v,w) and R(u,w).
3. For every w there is u, such that if R(v,w) then both R(u,v) and R(u,w).

HOOTmag (talk) 14:25, 15 September 2014 (UTC)

## determine interior angle of triangle give the coordinates of the vertices.

So I was asked to calculate the interior angles of a triangle given the x coordinates P(2,0) Q(0,3) and R(3,4)

Though this is a basic geometry question, this was in the context of a chapter where we are learning the dot product of vectors.

Anyway, I feel like I understand how to solve the problem up to a point, but I must have made mistakes somewhere.

First thing I did was calculate the length and slope of the sides PQ, QR, and RP.
PQ length = sqrt (13) with slope of -2/3

QR length = sqrt (10) with slope of 1/3

RP length = sqrt (17) with slope of 4

Next I used the slopes of each sides to convert the sides to vectors. I called

vector based on PQ vector A <3, -2>

vector based on QR vector B <3, 1>

vector based on RP vector C <1, 4>

Then I used the identity that says that the dot product of vectors A and B = |A||B| cos theta, where theta is the angle formed by the two vectors

The dot product of A and B is 7 (associated with vertex Q of the triangle)

The dot product of B and C is 7 (associated with vertex R of the triangle)

The dot product of C and A is -5 (associated with vertex P of the triangle)

Therefore, I thought, that

the cosine of theta for angle Q is 7/sqrt (130) - > theta = approx 52 degrees

the cosine of theta for angle R is 7/sqrt (170) - > theta = approx 57 degrees

the cosine of theta for angle P is -5/sqrt (221) - > theta = approx 109.7 degrees

These are not the correct answers! they don't even add up to 180

I looked in the back of my book and apparently my value for angle R = approx 57 degrees is correct, but the other two are wrong. So maybe I did something wrong with my vector arithmetic or coordinate geometry?--Jerk of Thrones (talk) 21:36, 15 September 2014 (UTC)

Your first vector had the i and j components switched. I also don't think you should use slope to determine vectors. After all, slope can be infinite, and vectors representing the sides of a triangle can't. Instead, find vectors as the difference between each pair of coords:
P = (2,0)
Q = (0,3)
R = (3,4)

P-Q vector = (2-0,0-3) = (2,-3)
R-Q vector = (3-0,4-3) = (3,1)
R-P vector = (3-2,4-0) = (1,4)

You could also use the Q-P, R-Q, and P-R vectors, which would just point in the opposite directions. Now, which directions the vectors point are rather critical when using the dot products. If one vector is pointing the wrong way at a point, you will get the supplementary angle at that point. If your angles still don't add up to 180, then that's probably the issue. I don't think this particular triangle will have any angles over 90, so if you get any, try subtracting it from 180 to get the correct angle.
(BTW, the "R-P vector" would normally just be called the "RP vector", but I wanted to make it a bit clearer by adding the minus sign.) StuRat (talk) 00:55, 16 September 2014 (UTC)