HDTenEightyP/NetNews huggingface dataset:
From: bmesmer@occs.cs.oberlin.edu (Carl Muckenhoupt)
Subject: Re: Dollar sign (was: FAQed again)
Organization: Oberlin College Computer Science
Date: Thu, 9 Apr 1992 22:45:22 GMT

In article <15760003@hpspdla.spd.HP.COM> bruceh@hpspdla.spd.HP.COM (Bruce Hayek) writes:
   I was under the (I suppose false) impression that the dollar sign
   was the result of superimposing the United States' initials.      

   Enventually the loop fell of the bottom of the "U" leaving an "S"
   with two vertical lines through it, then one vertical line.  Don't
   ask where I got that from, but it seems plausible since the dollar
   is more often drawn with two vertical lines.            

   (And actually, Objectivist author Ayn Rand in the book "Atlas
   Shrugged" makes a reference to this origin.)

I missed the beginning of this thread. I suppose that you are replying to
someone's statement that it comes from the insignia imprinted on Spanish
coins when America was being colonized, picturing two pillars with a ribbon
wrapped around them? If not, consider this a fresh theory; certainly the
symbol predates the United States.
I just had to reply because of the Ayn Rand reference. I read "Atlas
Shrugged" in a spirit of belligerence, and was quite pleased when I caught
her in a clear factual error.

ObPuzzle: You are on an island where everyone either says only true
statements or false statements. You can ask one person one question.
Imagine it. Regardless of whether that person knows the answer or not,
she/he is determined by mysterious forces either to give the correct
answer or an incorrect one. What question do you ask to best take
advantage of this oracular power, taking into account that you might be
talking to one of the liars?
--

Carl Muckenhoupt        bmesmer@occs.cs.oberlin.edu
"As a lyricist you become sensitive to certain things. You become aware
that there aren't very many words that rhyme with 'love'. There are a lot
more words that rhyme with 'hate' than with 'love'. You've got 'above'
and 'dove' and 'glove' and 'shove' and that's about it."
                        - Peter Shickele
From: rose@fsu1.cc.fsu.edu (Kermit Rose)
Newsgroups: rec.puzzles
Subject: sum of series 1 - 1/2 + 1/3 - 1/4 + ...
Date: 3 Apr 92 15:49:55 GMT
Article-I.D.: mailer.1992Apr3.104955.21943
Reply-To: rose@fsu1.cc.fsu.edu
Organization: Florida State University
News-Software: VAX/VMS VNEWS 1.3-4

This series 1 - 1/2 + 1/3 - 1/4 + ... is the integral from 0 to x of 
1/(1+x) .
1/(1+x) = 1 - x + x^2 - x^3 + x^4 - ...
the integral = x - (x^2)/2 + (x^3)/3 - (x^4)/4 +...
The series in question is this integral evaluated at x=1.
That is, the series is the integral of 1/(1+x) from 0 to 1.
The integral of 1/(1+x) is ln(1+x) for x>0.
the series converges to ln(2)

Kermit@FSU or rose@fsu1.cc.fsu.edu.bitnet
If you want to be sure I see your response, please e-mail me.
-----------------------------------------------------------------------
This is the signature file of Kermit Rose.

Believing in dichotomies leads to hardening of your logic.

What determines the existence of Free Will?
I strive to believe only what is true. I have faith that whatever is 
true will also be logically consistent and knowing truth will increase my 
capacity to be kind.
Peace is the way to peace.  Hostility is the way to hostility.
Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: Dying young
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Wed, 8 Apr 1992 21:21:33 GMT

In article <1992Apr8.073536.9063@descartes.waterloo.edu> phsieh@lambert.waterloo.edu (Paul Hsieh) writes:
>In article <farrell.702613794@coral.cs.jcu.edu.au> farrell@coral.cs.jcu.edu.au (John Farrell) writes:
>>In <7452@uqcspe.cs.uq.oz.au> warwick@cs.uq.oz.au (Warwick Allison) writes:
>>  Well, are we talking about optimal for you or optimal for everyone? ...
>>Before you
>>can talk about an optimal strategy you have to define what optimal means.
>
>I cannot believe you jumped on this posting with little else to add but 
[various ad hominem remarks omitted]
>If you take "Friendless"'s [...] approach
>what's to say you don't have 99 other putzs [sic] doing the same thing?  The
>probability of living in such a scenario?  0.

We each interpreted optimal in our own way.  Friendless interpreted it to
mean "maximize your own chances of survival."  The analysis of the two
strategies then is simple:

    Strategy    Chance of personal survival
      push                0
   don't push             x

I don't know what the value of x is and neither do you, but it is a
probability and so is at least 0.  So given a choice of 0 or x,
clearly we maximize the output by choosing x.  This is true no matter
how many times the experiment is repeated, who is in the other rooms,
what you know about them, what they know about you, or any other spin
you could place on this problem.  (These things can affect the value
of x, sure, but they can NEVER make it less than 0, hence x is always
the maximum of the set {0,x}, even if x = 0.)

--------

Now even if you interpret "optimal" as "most lives saved, regardless
of whether mine is one of them" (which was my chosen interpretation,
by the way), as to the various "rational solutions" that are
equivalent to staking your fate on a roll of dice, all seem to depend
on the assumption that whatever rationale you come up with, the other
100 people reasoned exactly the same way.  The debate on this question
seems to me adequate evidence to conclude that this assumption is
FALSE, i.e. whatever you decide, most of the other people will reason
differently than you do (though ultimately they can take only one of
two actions).

-- David Karr
From: fisher@qut.edu.au
Newsgroups: rec.puzzles
Subject: Re: Brick Problem (crude ASCII)
Date: 8 Apr 92 12:41:29 EST
Organization: Queensland University of Technology

In article <orange.702580580@cunews>, orange@alfred.carleton.ca (Paul Vallee) writes:
> 
> The object is to draw a curvilinear line through every wall of every
> brick, without going through any given wall more than *once*.
> 
As has been said, and it is obvious once it has been pointed out, any
brick with an odd number of walls must contain one end of the line.
Hence, counting the outside space as another "brick", such puzzles can
only be solved without a trick if the number of bricks with odd numbers
of walls is 2 or 0. (In the latter case both ends of the line must be in
one brick.) 

Depending on the exact wording of the puzzle the trick answer is to say 
that you can go through a wall from end to end.

Hence a trick answer to the second puzzle:

>       a       b
>   |-------W-------|
> c |   d   |   e   | f
>   |----|--X--|----|
> g |  h |  i  |  j | k
>   |----|--Y--|----|
> l |   m   |   n   | o
>   |-------Z-------|
>       p       q
> 
is:   a d c g h d i h m l p m i YZ q n o k j n i j e f b e i XW

which passes through every wall just once. (There are many other similar 
answers.)

Bill
Newsgroups: rec.puzzles
From: farrell@coral.cs.jcu.edu.au (John Farrell)
Subject: Re: The puzzle generator
Organization: James Cook University
Date:  7 Apr 92 06:38:46 GMT

In <1992Apr6.202833.730@hellgate.utah.edu> tolman%asylum.utah.edu@cs.utah.edu (Kenneth Tolman) writes:
>Of course, most of these will be nonsense.  For example, all the puzzles
>where no one says anything about anyone will be worthless.  Also some
>of these will have multiple solutions.  I have no idea how many of each
>size is "valid" and "one solutioned".  

  As it turns out, the simple versions of these problems, where you can only
make statements of the form "X tells the truth", and "Y lies" are boring.
In the solution, there are two groups of people: liars and truthtellers.
A statement by X that Y tells the truth means that X and Y are in the same
group. A statement by X that Y is a liar means that X and Y are in different
groups. Hence if there are n people, you only need (n-1) statements to divide
the people into a unique partitioning. Any other statements will be redundant
or inconsistent.
  Now here's the annoying bit. Just from the testimony of the people, you
can't tell which are the liars and which are the truthtellers! You can find
the two groups, but you need additional information to decide which to trust.

>   What would be a good next addition beyond each player just saying other
>players are true or false?
>    (possibly using "or", some sort of ordering, alternating true false, etc)

  A simple statement type which could be added is "X is the same as me", or
"Y is different from me". If I say "X is different from me", it means that X
is a liar, no matter what I am. This allows a unique solution. If there are n
people, n-1 statements of the form (tells the truth/lies) and 1 statement of
this form can provide a unique solution.

>   Would anyone have a use for such a program?  Maybe in a MUD?

  Well I've just turned it into a great student assignment :-).

>   In the puzzle maker above, how can the computer tell if the solution is
>obvious?

  Now that I know how, they're all obvious :-(.


Friendless
From: dwr2560@zeus.tamu.edu (RING, DAVID WAYNE)
Newsgroups: rec.puzzles
Subject: Re: Brick Problem (crude ASCII)
Date: 12 Apr 92 04:36:00 GMT
Organization: Texas A&M University, Academic Computing Services
News-Software: VAX/VMS VNEWS 1.41

>>Hence, counting the outside space as another "brick", such puzzles can
>>only be solved without a trick if the number of bricks with odd numbers
>>of walls is 2 or 0. (In the latter case both ends of the line must be in
>>one brick.) 

jerryd@hpgrla.gr.hp.com (Jerry Donovan) writes...
>1 will also work.

Hmmm... the total number of brick-sides is odd... now that's a good trick,
considering that each wall has two sides. ;)

Dave Ring
dwr2560@zeus.tamu.edu
Xref: sparky rec.puzzles:3579 soc.culture.indian:13787
From: amorgan@Xenon.Stanford.EDU (Crunchy Frog)
Newsgroups: rec.puzzles,soc.culture.indian
Subject: Re: A simple number puzzle
Date: 9 Apr 92 17:12:43 GMT
Article-I.D.: CSD-News.1992Apr9.171243.6338
Organization: Computer Science Department, Stanford University.

In article <1992Apr09.150049.15977@eng.umd.edu> 
  russotto@eng.umd.edu (Matthew T. Russotto) writes:
>In article <1992Apr8.042047.20639@wuecl.wustl.edu> 
>  bs@wucs1.wustl.edu (Bala SWAMINATHAN) writes:
>>
>>Did the original article imply any where that LIVE humans can have only
>>integral ages? OR did the very use of the word *age* imply an integer?
>>
>>If the age can be counted in months (say 26 years and 5 months etc.)
>>then there are more solutions.
>
>Everyone I know, when asked their age, will either give a
>non-numerical answer ("What business is it of yours?" or some variant
>:-) ) or will state their age in years.

Guess you don't hang around kids younger than 10 too often.  Ask a 6
year old how old they are and you will get something like "6 and 3/4".
 
>Matthew T. Russotto	russotto@eng.umd.edu	russotto@wam.umd.edu

C "22 1/2" Frog
Newsgroups: rec.puzzles
From: surya@watson.ibm.com (Surya Mohanty)
Subject: Re: More Math Problems
Date: Thu, 09 Apr 1992 15:30:27 GMT
Reply-To: surya@rothko.watson.ibm.com (Surya Mohanty)
Disclaimer: This posting represents the poster's views, not necessarily those of IBM
Organization: IBM T.J. Watson Research Center

Note that the sequence
101 , 10101, 1010101, ....
can be viewed as 
100**1 +1, 100**2 + 100**1 + 1, 100**3 + 100**2 + 100**1 +1 ....
that is, 
the k-th term in the sequence is 
100**k + 100**(k-1) + 100**(k-2) + ...+ 100**(1) + 1
= (100)**(k+1) - 1
  ----------------
    11 * 9
= (10)**(2k+2) - 1
  ----------------
    11 * 9
= ((10)**(k+1) - 1)*((10)**(k+1) +1)
   ---------------------------------
       11*9
thus either 11 and 9 divide the numerator. Either they both divide the same 
factor in the numerator or different factors in the numerator. In any case, 
after dividing, they leave the numerators as a product of two integers.
Only in the case of k = 1, one of the integers is 1. Thus there is exactly one 
prime in the above sequence that is 101.

varification:
10101 is divisible by 3
1010101 = 101*10001
101010101 = 11111*9091
etc....
Surya Mohanty.
From: dseal@armltd.co.uk (David Seal)
Newsgroups: rec.puzzles
Subject: Re: Math Problem
Date: 10 Apr 92 03:01:24 GMT
Distribution: rec
Organization: A.R.M. Ltd, Swaffham Bulbeck, Cambs, UK

In article <9204071337.AA25632@ucbvax.Berkeley.EDU>
BANDA@AUSVM1.VNET.IBM.COM ("Venu Prabhakar Banda") writes:

>Problem: Show that (4444 ** 4444) is divisible by 9

Since 4444 is not divisible by the prime 3, no integer power of it is
divisible by 3 either. So (4444 ** 4444) is not divisible by 3, and so
certainly can't be divisible by 9.

If your problem really *is* to show that (4444 ** 4444) is divisible by 9,
you have got a difficult problem indeed :-)

David Seal
dseal@armltd.co.uk

All opinions are mine only...
Newsgroups: rec.puzzles
From: Dave Dodson <dodson@convex.COM>
Subject: Re: The Foreign Counterfeiter
Originator: dodson@bach.convex.com
Reply-To: dodson@convex.COM (Dave Dodson)
Organization: Engineering, CONVEX Computer Corp., Richardson, Tx., USA
Date: Wed, 15 Apr 1992 16:12:26 GMT
X-Disclaimer: This message was written by a user at CONVEX Computer
              Corp. The opinions expressed are those of the user and
              not necessarily those of CONVEX.

In article <YdtPlVK00iV1M2Gc8M@andrew.cmu.edu>, ss8c+@andrew.cmu.edu (Simon Szykman) writes:
> 
> 1.    If and only if m and n have a certain property, there is
> a number X, such that Fred can buy ANYTHING that costs more than
> X bux.  What is that property? 
> 
> 2.    What is X?

If m and n are relatively prime (i.e., gcd(m,n) = 1) then Fred can buy
anything costing not less than (m-1)*(n-1) with his m and n bux.  If m
and n are not relatively prime, then there are infinitely many amounts
that cannot be achieved with m and n bux; e.g. if m = 2 and n = 4, Fred
cannot buy anything costing an odd number of bux.

----------------------------------------------------------------------

Dave Dodson		                             dodson@convex.COM
Convex Computer Corporation      Richardson, Texas      (214) 497-4234
Newsgroups: rec.puzzles,soc.culture.indian
From: uds@parvathi.cs.wayne.edu (Seetamraju)
Subject: Re: A simple number puzzle
Organization: Wayne State University
Date: Wed, 8 Apr 1992 05:59:02 GMT

In article <1992Apr7.202400.26466@Cadence.COM> hemmady@Cadence.COM (Shankar Hemmady) writes:
>
>"My child's age raised to itself gives my age.
>If my wife is younger than me by my child's age
>and if we are LIVE humans, then what are our ages?"

1.	Question : Did you forget your child's date of birth?
2.	Answer : If you and your wife were fed, since childhood, american
		 meat, that everyone knows to contain overdosage of growth
		hormones --- then,since you and your wife sexually matured
		much much earlier than the rest of the us chooths,  x=4.


Seetamraju, Uday Bhaskar Sarma
(email : seetam @ ss0 . eng . wayne . edu)
Newsgroups: rec.puzzles
From: macster@IASTATE.EDU (Michael C Mccarty)
Subject: NOTE: Bar Trick Clarification!!!!
Reply-To: macster@IASTATE.EDU (Michael C Mccarty)
Organization: Iowa State University
Date: Wed, 8 Apr 1992 08:42:28 GMT

	In an earlier post, I presented a puzzle as follows:

>INTERESTING-YET-SIMPLE BAR TRICK: Take two straws, melt them together at the
>ends at approximately 90 degree angle.  Take a third straw and prop the fused
>straws up to form a pyramid (Note: the third straw is not fused to the other
>straws, it merely supports them).  Now, taking a pencil, pick up all three
>straws SIMULTANEOUSLY, USING ONLY THE PENCIL, NOTHING ELSE!

	Please note that I did not say that the pencil has been sharpened, so please do
not assume that it is so. In other words, the fact that it is a pencil does not
enter into the solution.

	However, if you did use that in your solution, you are getting close.

				Mike
From: dant@logos.wr.tek.com (Dan Tilque)
Newsgroups: rec.puzzles
Subject: Re: order of numbers?
Date: 7 Apr 92 02:44:39 GMT
Distribution: rec
Organization: Upper Left Coast Liberation Army

pfalstad@phoenix.Princeton.EDU (Paul Falstad) writes:
>dant@logos.wr.tek.com (Dan Tilque) wrote on 6 Apr 92 17:01:18 GMT:
>>> > 	2, 9, 8, 4, 5, 7, 6, 3, 1
>
>Spoilers...

>Latin.

Right the first time.

>
>The Romance languages are similar:
>5, 2, 8, 9, 4, 7, 6, 3, 1
>and
>5, 2, 9, 8, 4, 7, 6, 3, 1

There's 5 main Romance languages plus some minor ones, so you have to
be a bit more specific.  I'm sure that neither sequence is French and
probably not Rumanian, but any of Italian, Spanish or Portuguese.

>The zero really wasn't invented when Latin was spoken.  My Latin dictionary
>says that you should use "nihil", meaning "nothing", when translating
>the word "zero".

That's right.  I remember now.

>>	8, 3, 1, 5, 9, 7, 6, 4, 2
>I'll guess German, based on where the "2" is.

Right again.

>>All four of these lists, 7 comes in the sixth position.  I wonder if this
>>is only a coincidence.
>
>Aren't all these languages derived from a common ancestor?  Indo-European,
>or something.

Finnish is not IE; the others are.  But even among related languages,
the exact letters that represent related words can be significantly
different.  For example: consider "8" in four languages:

Enlgish   German   French   Latin
 eight     acht     huit     octo

Except for the 't' at or near the end of each, the spellings are quite
diverse.  The pronunciation of the Latin and German are fairly close,
even though Latin is closer to French and German is closer to English.
So a single word or small set of words from related languages are not
necessarily close in spelling or pronunciation.

>Two more:
>2, 1, 9, 6, 7, 8, 5, 4, 3
>1, 4, 5, 7, 3, 8, 6, 2, 9

Sorry, I've run out of languages.

---
Dan Tilque	--	dant@logos.WR.TEK.COM
Newsgroups: rec.puzzles
From: farrell@coral.cs.jcu.edu.au (John Farrell)
Subject: Re: Dying young
Organization: James Cook University
Date:  7 Apr 92 01:10:11 GMT

In <1992Apr6.175229.23741@hellgate.utah.edu> tolman%asylum.utah.edu@cs.utah.edu (Kenneth Tolman) writes:
>You are put in a room by evil Nazi's.  There is a button in front of you.  If
>you press the button, you will die.  However, there are 100 other people in
>similar rooms.  If no one pushes their button before 12:00, everyone dies.  You
>have no way to know if the other people have pushed theirs or not. 
>It is now 11:50. What do you do?

  Well I'm damned if I'm gonna press the button! I'll sit down and pray. I'm
not usually a religious man, but this may be the time to start.


Friendless
Newsgroups: rec.puzzles
From: macster@IASTATE.EDU (Michael C Mccarty)
Subject: Re: SPOILERS: Isaac Asimov's mysteries puzzle
Reply-To: macster@IASTATE.EDU (Michael C Mccarty)
Organization: Iowa State University
Date: Mon, 13 Apr 1992 08:07:24 GMT

In article <1992Apr11.014450.25843@bmerh2.bnr.ca>, coop1w13@bnr.ca (Mike
Hammond) writes:
> Spoilers for a tough puzzle. (The only reason I know this is because I
> happened to read that particular issue of Isaac Asimov's Science Fiction
> Magazine. By the way, isn't this a puzzle of Martin Gardener?)
> 
> In article <3847@shum.huji.ac.il>, yehuda@shuldig.huji.ac.il (Jonathan
Berlinger) writes:
> |> Donn't know if anyone heard this one yet, so:
> |> 
> |> Why does>	Christmas = Halloween
> |>
              [...deleted...] 
> 
> Well, December 25 is 12/25, while October 31 is 10/31. Note that 25 in base
> 12 is equal to 31 in base 10. Thus Christmas=Hallowe'en.
> 
            [...deleted...]

	I hate to sound blatantly ignorant, but, how does:
                  25 (base 12) = 31 (base 10)?
        Wouldn't that be 29 = 31, or am I just being stupid?

				Mike
From: carabalo@phoenix.Princeton.EDU (David G. Caraballo)
Newsgroups: rec.puzzles
Subject: Primes in sequences...
Date: 13 Apr 92 16:33:22 GMT
Organization: Princeton University
Originator: news@ernie.Princeton.EDU

I canceled my post about the sequence 1, 11, 111, 1111, ..., since it contained
a typo that invalidated the proof.  In fact, I have been unable to construct
a proof that there are only a finite number of primes in that sequence, and I
am not entirely sure this is the case.  If a_n is the term with n 1's, then
a_n is clearly composite if n is composite:  If r>1 divides n, then a_r divides
a_n; thus, it suffices to check a_n for prime values of n.

Q1:  What is the least odd prime p for which a_p is prime?  

Someone asked about primes in arithmetic sequences (which the above is not).
The pertinent result is Dirichlet's theorem:  If gcd(a,b)=1, where a and b
are positive integers, then the sequence a, a+b, a+2b,..., a+nb, ...
contains infinitely many primes.

For those who want a similar problem that is actually fairly easy, here goes:

Q2:  How many primes are there in the sequence 1, 101, 10101, 1010101,
101010101, ... , where the first and last digits (the same, in the case of 1)
are 1, and the digits alternate between 0 and 1?

David
From: killer@psykoman.cs.hut.fi (Kalle Kivimaa)
Newsgroups: rec.puzzles
Subject: Re: The negative Harmonic series
Date: 14 Apr 92 07:04:04 GMT
Organization: Helsinki University of Technology, CS-lab

In <1992Apr13.152928.25582@cs.cornell.edu> karr@cs.cornell.edu (David Karr) writes:
>In article <1992Apr11.020521.16082@Cadence.COM> hemmady@Cadence.COM (Shankar Hemmady) writes:
>>	1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 ....
>[...intermediate steps omitted...]
>>   = 	(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...)
>>     	- (1 + 1/2 + 1/3 + 1/4 + ...)
>>   =    ZERO!!
>How about the fact that 1 + 1/2 + 1/3 + ... sums to infinity?  Then you
>are subtracting infinity from infinity, which is a well-known fallacy.

You can group the terms in a way that it really sums to zero, like this:
(1-1) + (1/2-1/2) + (1/3-1/3) + (1/4-1/4) +...

Again, there is no fallacy, it is a property of the alternating and 
conditionaly convergent series.
--
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* The night sky over the planet Krikkit is the least interesting sight *
* in the entire Universe.  -- Douglas Adams, HHGTG                     *
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Newsgroups: rec.puzzles