Math Fun

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Re: Math Fun

Postby Silhouette » Wed May 22, 2013 9:01 pm

James S Saint wrote:You guys keep trying to reduce the "perfect logician" to being no more than a calculator wherein someone has to push his buttons. In the case of the ultra android, when someone pushes his 2+2 button sequence, it might well reach out slap him as well as add the numbers.

Nobody is taking perfect logicians as mere calculators. They can think/do anything they like, it's just that nothing else is relevant to whether they leave or not, except the procedure which I have outlined. There are relevant consequences of there either being 100 blues or 99 blues (to each blue), but nothing else. Everything else, slapping included, is perfectly fine, but irrelevant to the exhaustive procedure of the correct solution.

This is enough to prove why no other solutions are valid in general.
Adequate specific disprovals have also been offered as a bonus, such as there being zero logical reason to choose to start counting from any particular number outside of the procedure of the correct solution. Further to this, there is no logical reason for all islanders to know which of the many combinations of e.g. 97 blue islanders are being referred to if they were to start counting from this or any other number.
No amount of knowing of knowing etc. can replace what the Guru says because none of this knowledge can be transferred, out of context, to situations where it no longer applies - such as the thought experiment of what 1 blue would do if there was only 1 blue (who would obviously only know what 1 blue would know and nothing more). It's possible for each of 100 blue-eyed perfect logicians to attribute their knowledge to this sole blue, but inconsistent - and that is the problem. There is going to be a problem with any attempt at a solution that starts from irrelevance in order to "better" solve the problem (relevance as defined by the exhaustive analysis of the ONLY consequences that are to have any affect - as completely covered in the correct solution).

This problem has sufficiently been brought to a close.
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Re: Math Fun

Postby Flannel Jesus » Sat Jun 22, 2013 7:33 pm

Tralix wrote:Topology question sequence:

AE FHIK LMN

What letters come next?

It's loosely related to maths, but then isn't everything.

I hadn't read the beginning of this thread before, I was just going through it and reading your exchange with Abstract. I got up to the point of reading your hint that T is the next letter when I finally figured it out.
The letters have no curves.
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Re: Math Fun

Postby Flannel Jesus » Sat Jun 22, 2013 7:36 pm

Abstract wrote:here is the hint to my problem:
http://primes.utm.edu/lists/small/10000.txt ignore 2 3 5 7

Just figured this one out as well, after looking at your first hint
sum of digits of two-or-more-digit primes
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Re: Math Fun

Postby Flannel Jesus » Sat Jun 22, 2013 7:46 pm

New Problem. Some of you may have heard this in a different form. Avoid spoilers, and avoid looking up the solution.

I'm walking with James to an apartment where 3 girls we know live. Let's call them Rachel, Monica and Phoebe. As we're approaching, we see that their livingroom light is on. We know (you can assume this as an undeniable axiom for the case of this problem) that the light would only be on if at least one of them is home. The other thing we both know is that Monica always leaves the house with Phoebe if she leaves at all -- either she's in the house, or she's out with Phoebe.

James tells me that, based on this information, Rachel by logical necessity MUST be in. His explanation is like this: Suppose that Rachel is out. If Rachel is out, then if Monica is also out Phoebe would have to be in—since someone must be in for the light to be on. However, we know that whenever Monica goes out she takes Phoebe with her, and thus we know as a general rule that if Monica is out, Phoebe is out. So if Rachel is out then the statements "if Monica is out then Phoebe is in" and "if Monica is out then Phoebe is out" would both be true at the same time. As those can't be true at the same time, Rachel must be in.

Seems like a pretty compelling argument, no? The puzzle is to explain specifically what's wrong with it.
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Re: Math Fun

Postby Silhouette » Sat Jun 22, 2013 9:09 pm

Flannel Jesus wrote:The puzzle is to explain specifically what's wrong with it.

Declaration: R, M, P (Rachel, Monica, Phoebe)
State ("in" or "out") abbreviations: Rin/Rout, Min/Mout, Pin/Pout
Condition 1: if Mout, Pout
Condition 2: at least 1 must be "in"

8 permutations:
(1) Rout, Mout, Pout = wrong (at least 1 must be in)
(2) Rout, Mout, Pin = wrong (if Mout, Pout)
(3) Rout, Min, Pout = possible
(4) Rout, Min, Pin = possible
(5) Rin, Mout, Pout = possible (*only case where Mout possible)
(6) Rin, Mout, Pin = wrong (if Mout, Pout)
(7) Rin, Min, Pout = possible
(8) Rin, Min, Pin = possible

James's assertation: if Rout:
if Mout, Pin (condition 2)
BUT if Mout, Pout (condition 1)
(Mout, Pin AND if Mout, Pout) =/= true, therefore Rin.
Problem:
This does not factor in the possibility that (given Rachel is out) Monica is "in": (3) and (4).

Also, anything James says is always wrong.
You got your topology and prime puzzle answers correct btw. I got the first one but not the second, grats.
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Re: Math Fun

Postby Flannel Jesus » Sat Jun 22, 2013 9:27 pm

You got it Sil. Here's a bit more to chew on maybe:
As you said, James Hadn't considered the possibility of Rout Min. What James actually proved is that, if Rout, then Min necessarily. Because Rout AND Mout produces Pout AND Pin, as he showed.

The misunderstanding James had was to think that A -> B and A -> ¬B is a contradictory scenario. B and ¬B is contradictory, but A -> B and A -> ¬B is not contradictory at all (not unless you're also saying A). You can reduce logical implication thusly:
A -> B is the same as saying B v ¬A
So, (A -> B and A -> ¬B) is reducible to (B v ¬A) ^ (¬B v ¬A). The way to make that conjunction true, is of course to make ¬A true. No contradiction at all, it just requires A not to be true.

So, A -> B and A -> ¬B isn't a contradiction, it's just a disproof of A.
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Re: Math Fun

Postby Flannel Jesus » Sat Jun 22, 2013 9:36 pm

Some more that I was just thinking:
James' thought process was obviously that A -> B and A -> ¬B is a contradiction.
To be a true contradiction, it would have to be (A -> B) and ¬(A -> B), and so he may have thought that ¬(A -> B) = (A -> ¬B)
Let's look at what ¬(A -> B) really equals:

As we said, A -> B = B v ¬A
¬(A -> B) = ¬(B v ¬A) = ¬B ^ A

A -> ¬B = ¬B v A

¬B v A =/= ¬B ^ A (although they do look similar)
So ¬(A -> B) =/= (A -> ¬B)
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Re: Math Fun

Postby Flannel Jesus » Sun Aug 11, 2013 8:20 am

Sleeping Beauty volunteers to undergo the following experiment. On Sunday she is given a drug that sends her to sleep. A fair coin is then tossed just once in the course of the experiment to determine which experimental procedure is undertaken. If the coin comes up heads, Beauty is awakened and interviewed on Monday, and then the experiment ends. If the coin comes up tails, she is awakened and interviewed on Monday, given a second dose of the sleeping drug, and awakened and interviewed again on Tuesday. The experiment then ends on Tuesday, without flipping the coin again. The sleeping drug induces a mild amnesia, so that she cannot remember any previous awakenings during the course of the experiment (if any). During the experiment, she has no access to anything that would give a clue as to the day of the week. However, she knows all the details of the experiment.

Each interview consists of one question, “What is your credence now for the proposition that our coin landed heads?”

[hint: A viable answer is to remove the abstract term 'credence' and turn the situation into something more...tangible. If you so wish.]
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Re: Math Fun

Postby Sauwelios » Mon Aug 12, 2013 12:44 pm

When she's interviewed, there's a .333[bar] chance that it's Monday and the coin has landed heads, a .333[bar] chance that it's Monday and the coin has landed tails, and a .333[bar] chance that it's Tuesday and the coin has landed tails. This means that her credence for the proposition that the coin landed heads is logically one third full each interview.
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Re: Math Fun

Postby Silhouette » Wed Sep 25, 2013 8:52 pm

For the following:

0 0 0 = 6
1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6

Insert any mathematical operators (except ones that involve writing a number) into the left hand side of each of the above equations, such that they become correct.
No manipulation of any number shapes or outside-of-the-box thinking is required to do this. Use of brackets/parentheses is encouraged for clarity.
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Re: Math Fun

Postby Carleas » Fri Sep 27, 2013 4:27 am

Still missing:
8 8 8=6
9 9 9=6

The rest:
(0!+0!+0!)!=6
(1+1+1)!=6
2+2+2=6
(3x3)-3=6
(4-(4/4))!=6
5+(5/5)=6
6+6-6=6
7-(7/7)=6


I think I need to incorporate another operation...
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Re: Math Fun

Postby Silhouette » Fri Sep 27, 2013 8:00 pm

Good effort! :)

8s are the hardest, btw.

Good job thinking of using factorials - that was the least obvious one to me.

You're right about needing to incorporate another operation to finish things off.

Don't open the next tab if you want to see if you can do it all without any help....
Here's a clue, squirt. Another operation would neaten up the 4s a little while you're at it.
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Re: Math Fun

Postby Carleas » Sun Sep 29, 2013 11:53 pm

I figured out a cheat-y way that solves all of them:
( ( d/dx(8) )! + ( d/dx(8) )! + ( d/dx(8) )! )!

That works no matter what number you use, because the derivative of any constant is 0, and the factorial of 0 is 1. It's a bit of a cheat because d/dx is arguably adding another number (dx).

It also makes the problem boring.

Still no legit solution to 8 and 9.
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Re: Math Fun

Postby Silhouette » Mon Sep 30, 2013 12:01 am

Carleas wrote:I figured out a cheat-y way that solves all of them:
( ( d/dx(8) )! + ( d/dx(8) )! + ( d/dx(8) )! )!

That works no matter what number you use, because the derivative of any constant is 0, and the factorial of 0 is 1. It's a bit of a cheat because d/dx is arguably adding another number (dx).

It also makes the problem boring.

Still no legit solution to 8 and 9.

Haha!

Wow, it's amazing the creativity that difficult problems can drive us to.
The function you're looking for really is fairly elementary.
Another hint, should you need it:
Start with the 9s. Think about what you can do to a number like that, especially in light of a number I've already mentioned in my previous hint.
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Re: Math Fun

Postby Carleas » Mon Sep 30, 2013 1:51 am

Hmmm...
I think you're suggesting taking the square root, which I wrote off because I saw it as incorporating another number: 2.

If that's what you're suggesting, 9 is easy: ( 9/sqrt(9) ) +sqrt(9) = 6
and 8 - sqrt( sqrt( 8+8 ) = 6
But why not cube root + cube root + cube root?
And if we're doing that, why not x root of c + c + c, where x is whatever irrational number root results in 6?
Or log base y where y is a number that when raised to the 6th power equals c + c + c?

I don't know why I treated the square root as adding a number, but not the factorial. As I think about it, I find the line between adding another number and just adding an operator to be fuzzy.
After all, you can always define a new operator. So we could define the unary operator B such that Bc=(c x 0) + 6.
B(c + c + c) = 6 for all c.
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Re: Math Fun

Postby Flannel Jesus » Mon Sep 30, 2013 7:41 pm

I consider square root to have a hidden number as well, Carleas. It's raising to the power of 1/2. Just because we happen to have a symbol that doens't look like a number that means 'raise to the power of 1/2' doesn't mean it's not using a number. I mean, I could make a symbol that means Raise to the power of 0 and multiply by 2...right? I could just invent that symbol. Maybe it could be the dollar sign. 8£ + 8£ + 8£ = 6. BOOM! Realistically, that's just as valid as using square root.
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Re: Math Fun

Postby James S Saint » Mon Sep 30, 2013 7:57 pm

The point to the puzzle was to stick to conventional mathematics as it currently is with its current symbols. Adding new creative "operators" isn't acceptable.
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Re: Math Fun

Postby Silhouette » Mon Sep 30, 2013 8:32 pm

Carleas wrote:Hmmm...
I think you're suggesting taking the square root, which I wrote off because I saw it as incorporating another number: 2.

If that's what you're suggesting, 9 is easy: ( 9/sqrt(9) ) +sqrt(9) = 6
and 8 - sqrt( sqrt( 8+8 ) = 6
But why not cube root + cube root + cube root?
And if we're doing that, why not x root of c + c + c, where x is whatever irrational number root results in 6?
Or log base y where y is a number that when raised to the 6th power equals c + c + c?

I don't know why I treated the square root as adding a number, but not the factorial. As I think about it, I find the line between adding another number and just adding an operator to be fuzzy.
After all, you can always define a new operator. So we could define the unary operator B such that Bc=(c x 0) + 6.
B(c + c + c) = 6 for all c.

Yes, I was suggesting that you use the square root, since it can be denoted by √ which does not require using a number. Contrast this with other roots such as the cubed root that requires using the number "3" however you present it. The square root may also be represented by raising a quantity to the power "1/2", which involves using a number, but using the √ symbol remains within the designated rules. The ! symbol to denote "factorial" is fine for the same reason, even if its expansion would involve adding other numbers.

Perhaps I didn't explain that rule sufficiently.

Also, only the integers provided at the beginning of the problem are the only ones allowed - so no "x root of c + c + c, where x is whatever irrational number root results in 6" or "log base y where y is a number that when raised to the 6th power equals c + c + c". And no use-defined operators either! Lol. Only conventional mathematical ones - I should have been more explicit on that one too :P Nice idea though. I've enjoyed all of your contributions to this problem, even though some of them did not fall inside the rules I was trying to communicate.

( 9/sqrt(9) ) +sqrt(9) = 6 and 8 - sqrt( sqrt( 8+8 ) = 6 are both lovely.

I would even say they're better than the solutions I was expecting, which were:
(√9 x √9) - √9 = 6 and (√(8/8 + 8 ))! = 6

Though I do prefer √4 + √4 + √4 = 6 for the 4s.

So Carleas has solved them all =D>

A shame nobody else contributed.

EDIT:
I got distracted writing this reply, as usual, so by the time I previewed it there were more contributions.
JSS, that's correct.
FJ:
By that argument, all conventional mathematical operators would involve extra numbers if fully expanded. My bad for not explicitly stating exclusively conventional mathematical operators in the rules. Carleas has already experimented with user-defined operators in much the same way as you suggest, though maybe you resisted opening the tabs which show this.

Not to say both of your arguments are invalid - they're good points. I just didn't mean to imply that you could use them when I described the problem. Both square roots and factorials were supposed to be fine, along with addition, subtraction, multiplication and division. No numbers *have* to be used when using certain conventional mathematical operators for each of these.
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Re: Math Fun

Postby Helandhighwater » Tue Oct 29, 2013 2:23 am

I came to this problem late but I will say I did like the way Carleas called differentials a cheat, I don't think they were since they don't use numbers per se but assume general solutions to the differences between one value and another with a related dependant value and rule to evaluate it. In fact every single one of the values could be solved by differentials, substitution rules etc, addition of one number is irrelevant if you use the right maths, in fact you can basically solve an value of summation in a differential without an operator other than d/dx, it's kinda too easy though if you know A' level plus maths, and you can see why it would not be allowed. Kudos to the solutions that weren't "cheats". :)
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Re: Math Fun

Postby Flannel Jesus » Thu May 01, 2014 8:46 pm

I've got a 'puzzle' so to speak, but it's pretty open ended.

I've got a data set: I've tested drug A and drug B on both women and men. For the sake of simplicity, we'll just say that each subject either died or healed (just a binary measurement of the effect of the drug on each individual).

The percentage of men who were healed by drug A was higher than the percentage of men who were healed by drug B.
The percentage of women who were healed by drug A was higher than the percentage of women who were healed by drug B.

But, the percentage of people overall who were healed by drug B was higher than the percentage of people healed by drug A.

How is this possible? What does a data set have to look like to produce this result?

And no, it's not a trick question -- the answer is not anything like "Many of the participants were hermaphrodites or transvestites." Just keep it simple: all participants are men or women.

Bonus points if you know the name of this mathematical anomaly.
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Re: Math Fun

Postby Carleas » Fri May 02, 2014 2:18 am

I spent a while on this before I gave up and googled it, and learned that I had added some assumptions that were getting in my way.

This first entabment will only discuss assumptions, but I tab them because talking about them can sometimes suggest the answer; purists may want to avoid:
I had assumed that the treatment group sizes were the same, so that the number of individuals given each drug were the same, and thus that the number of men given drug A was the same as the number of men given drug B, and ditto for women. The trial sizes and compositions can be different.


This next entabment will discuss the answer:
I'd never heard of Simpson's paradox before this, so thanks for pointing it out. They have some interesting real-world examples in that article. Particularly interesting was the example of gender-biased admissions.

I'm having difficulty succinctly expressing what must be true of the data set for this to occur.. I know that the compositions need to be reversed, so for example, more women participate in the drug A trial and more men participate in the drug B trial. And I intuitively grasp the "vector interpretation" given in the wikipedia article, especially the image:
Image

But I need to think more on this to distil what's going on. Every time I try to express the criteria that are necessary, it sounds like a tautological restatement of the question itself. I know the two trial groups must be of different sizes and ratios of men to women, and I think it is expressible as some relationship between the success group of one drug for one gender to the failure group of the other drug for the other gender.
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Re: Math Fun

Postby Flannel Jesus » Fri May 02, 2014 8:17 am

Carleas: That's the one

I would explain it like this:
It's about weighting (think weighted averages).

Women have a higher healing rate than men. A has a higher healing rate than B.
If the healing rate of Women taking B is higher than the healing rate of Men taking A, and A is disproportionately taken by men while B is disproportionately taken by women, this (with some more mathematical constraints) can allow for the Simpson's Paradox to occur.

I've taken a screenshot of a journal article about it that you might find enlightening:
simspon.jpg
simspon.jpg (149.87 KiB) Viewed 1744 times
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Re: Math Fun

Postby Carleas » Tue Apr 14, 2015 4:05 pm

A new one circulating the net, similar to other problems that have been discussed here.

I edited lightly, but the NY Daily News wrote:Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates:

May 15
May 16
May 19
June 17
June 18
July 14
July 16
August 14
August 15
August 17

Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively. [The following conversation ensues:]

Albert: I do not know when Cheryl’s birthday is, but I know that Bernard does not know too.
Bernard: At first I don’t know when Cheryl’s birthday is, but I know now.
Albert: Then I also know when Cheryl’s birthday is.

So when is Cheryl’s birthday?
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Re: Math Fun

Postby phoneutria » Tue Apr 14, 2015 4:20 pm

August 14

Edit; lol no, that can't be it... has to be either June 18 or July 16

Edit2: July 16
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Re: Math Fun

Postby Carleas » Tue Apr 14, 2015 4:35 pm

Yep.

I'm curious why you said it has to be June 18 or July 16. I eliminated May and June first and went from there, I'd be interested to know your reasoning.
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