HDTenEightyP/NetNews huggingface dataset:
From: radcliff@csd4.csd.uwm.edu (David G Radcliffe)
Newsgroups: rec.puzzles
Subject: Re: simple number puzzle
Date: 22 Dec 1992 16:48:27 GMT
Organization: UW-Milwaukee Math Department
NNTP-Posting-Host: 129.89.7.4

Given a positive integer n, in how many ways can it be expressed
as the sum of (one or more) consecutive positive integers?  The
surprising answer is that the number of such representations is
equal to the number of odd divisors of n. 

First we find the number of ways of representing n as the sum of
consecutive integers.  To write n as the sum of an odd number of
consecutive integers, we need to solve (a-b) + . . . + (a+b) = n.
That is, a(2b+1)=n.  The number of solutions of this equation is
equal to the number of odd divisors of n.  Similarly, to write n
as the sum of an even number of consecutive integers, we need to
solve (a-b+1) + . . . + (a+b) = n; that is, (2a+1)b=n.  Again,
the number of solutions of this equation is equal to the number of
odd divisors of n.  The total number of representations is thus
twice the number of odd divisors of n.

If n=a+...+b then n=(1-a)+...+b.  Define an equivalence relation
on the set of all solutions by declaring that {a,...,b} ~ {1-a,...,b}.
Each equivalence class has two elements, and exactly one of these
elements contains only positive integers.  Therefore, half of
the solutions to n=a+...+b are positive.  So, the number of solutions
to n=a+...+b in positive integers is equal to the number of odd
divisors of n.

--
David Radcliffe                              radcliff@csd4.csd.uwm.edu
From: egs@cs.Princeton.EDU.Princeton.EDU (Emin Gun Sirer)
Newsgroups: rec.puzzles
Subject: Re: Guessing at random numbers.
Date: 24 Dec 92 10:23:42 GMT
Reply-To: egs@cs.Princeton.EDU.Princeton.EDU (Emin Gun Sirer)
Organization: Department of Computer Science, Princeton University
Originator: egs@volkl

Matt Crawford writes:
>I assume that we play the game for several iterations.
>
>As long as you don't alter the distribution that your generator picks
>from between iterations....
>
>I guess "high" or "low" according to whether the number you give me is
>higher or lower than the median of the numbers given so far.

	A good guess given my incomplete original description of the problem.
Do not assume that we play the game for more than one iteration (though I
would like to hear your ideas on multiple-iteration games). The answer
I have can be "proved" to be better than average even for single games.

	Also, I can change my distribution for each iteration. If the two
numbers I pick are equal, I discard one and pick a new one. Degenerate
distributions (e.g. rand(){return 1}) are not allowed. This is not an adversary
problem, however. I will not look at your guess history and modify my
behaviour (and doing so shouldn't help you in nay way I can foresee) to 
thwart your scheme. In fact, as I said before, we may have no history at
all and still play with better than average odds (for you).

	I hope this covers everything, sorry about the lack of rigor in the
first description.

	Gun.
From: kfl@access.digex.com (Keith F. Lynch)
Newsgroups: rec.puzzles
Subject: "Vesterman's paradox"
Date: 28 Dec 1992 19:07:12 GMT
Organization: Express Access Public Access UNIX, Greenbelt, Maryland USA
Distribution: world
NNTP-Posting-Host: access.digex.com

Bob Vesterman, <RVESTERM@vma.cc.nd.edu> claims that it's a paradox that:

1) "The smallest positive integer too small to be specified in 150 ASCII
   characters" doesn't exist, since if it did, it has just been specified
   in fewer than 150 characters.

2) There are a finite number of ASCII character strings with length 150
   or less.

3) There are an infinite number of positive integers.

The resolution lies in the fact that there's no unique mapping of
character strings onto integers.  "The smallest..." only makes sense
in terms of some mapping.  One mapping would be to use nothing but the
digits 0 through 9, in the usual way.  In that case, 10^151 is the
smallest positive integer which cannot be specified in 150 characters.
If we use base 256 notation, using the full eight-bit ASCII set, then
256^150 is the answer.  These two mappings have the virtue of being able
to represent any positive integer less than this limit.

What if our mapping includes the digits 0 through 9 and the symbols
+ - * / ^ to be used in the usual way?  Then the highest number that
can be represented is 9^9^9^9^9...9^9^99 (note that the number one less
than this cannot be represented).

Since there are no restrictions on possible mappings, there is no highest
integer that can be represented in 150 characters.  Or in 1 character.
I could take hundreds of megabytes describing what number I mean by "N",
and then giving the string "N", representing whatever mind-bogglingly
large number I defined it as.

Of course you can devise meta-mappings, i.e. languages for describing
mappings, but that doesn't change the essense of this argument.

Keith Lynch, kfl@access.digex.com
Newsgroups: rec.puzzles
From: ags@seaman.cc.purdue.edu (Dave Seaman)
Subject: Re: sequences
Organization: Purdue University
Date: Wed, 23 Dec 1992 18:48:09 GMT

In article <DBH.92Dec23172137@wombat.doc.ic.ac.uk> dbh@doc.ic.ac.uk (Denis  
Howe) writes:
> No, but:
> 
>     ..., nne, nbne
> 
> Is easy.  Maybe yours is Dutch?  Can anyone else solve either of
> these?

I don't know what language yours is in, but in English, after NNE comes  
either NbE or NEbN (depending on which direction you are turning). There  
is no such thing as NbNE, but there is a NbNW. That's a movie by  
Hitchcock.

--
Dave Seaman
ags@seaman.cc.purdue.edu
From: carl@SOL1.GPS.CALTECH.EDU (Carl J Lydick)
Newsgroups: rec.puzzles
Subject: Re: Direction of the sun
Date: 27 Dec 1992 10:06:47 GMT
Organization: HST Wide Field/Planetary Camera
Distribution: world
Reply-To: carl@SOL1.GPS.CALTECH.EDU
NNTP-Posting-Host: sol1.gps.caltech.edu

In article <1992Dec24.012636.17115@Csli.Stanford.EDU>, hiraga@Csli.Stanford.EDU (Yuzuru Hiraga) writes:
>
># Heard from a Japanese colleague -- Sorry again if this is in the FAQL
>
>Two boys, A and B, are watching the sunset.
>A points to the sun and says:
>"This is the direction where the sun is right now."
>B replies:
>"But don't you know that the rays take over 8 minutes to reach the Earth?
>The sun is actually already below the horizon."
>Which is correct?  Why? (Make it understandable to a ten year-old.)

The one who says that the sun is below the horizon.  Just to help us in this
exercise, a fighter plane flies overhead just as the conversation is happening. 
It's flying at Mach 10.  A few seconds later the boys hear the sonic boom, and
boy B says, "See, that fighter plane way over there caused the sonic boom."
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CARL@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXen and VMS.  That's what I get paid for.  My
understanding of astronomy is purely at the amateur level (or below).  So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it.  If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.
From: J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger)
Newsgroups: rec.puzzles
Subject: Re: simple number puzzle
Date: 22 Dec 92 01:06:21 GMT
Organization: Dartmouth College, Hanover, NH
X-Posted-From: InterNews1.0b1@newshost.dartmouth.edu

Yuzuru Hiraga writes:

A simple number puzzle for Christmas...
# sorry if this is in the FAQ: ours just expired.

What positive integer cannot be expressed as a sum of 2 or more
consecutive integers?

I believe the answer is 2^n power (where n is an integer) can't be
expressed as the sum of consecutive integers, but all other numbers
can.

**End of quoted material.

Proof:

1. All odd integers can be expressed as the sum of two consecutive
integers.

2. 4 consecutive integers will give all numbers with exactly one factor
of 2.

3. x number of consecutive integers, where x is prime (and not equal to
2), will give you all multiples of x (starting at x^2+x/2).

Numbers lower than (x^2+x)/2 fall into two cases:
a) Odd numbers, covered above.
b) Even numbers.  As these have only one multiple of 2, they will
obviously be a number that has a remainder of 2 when divided by 4 (ie.
case #2 above).

This solves all numbers except for those whose only factors are 2 (ie.
powers of 2).

As I am leaving to go home for Christmas tomorrow morning, anyone who
has the rest of the proof should email me directly with it.

Thanks!

--Ted Schuerzinger
email: .zed@Dartmouth.EDU
"I should have known it would be bad vodka when all the label said was
'Russian Vodka'."
Newsgroups: rec.puzzles
From: waknispv@necs.Vanderbilt.EDU (Prashant V. Waknis)
Subject: A number sequence matching puzzle
Organization: School of Engineering, Vanderbilt University
Date: Mon, 28 Dec 1992 17:57:46 GMT



Consider any two pairs of numbers. (n1, n2) and (n3, n4). 
Let (n1 * n2)/ (n1 +n2) = (n3 *n4) / (n3 + n4) = C.

For example, n1 = 12, n2  = 60, n3 = 15, n4 = 30.

Now, look at the following chart.

12 | 0    12    24    36   48      60 ....
60 | 0                             60 ....
-------------------------------------------
15 | 0       15    30     45       60  ....
30 | 0             30              60  ....

Can you see how the chart is made? Basically, starting with 0, each 
the sequences are constructed.
Now try to match the numbers  (on the right of the |s) that are above 
the horizontal line to the once that are  below  the horizontal line. 
The rule is the following.
Assume that the numbers start appearing on the scene  from left to right (in
increasing order)

So, in the above example,
all zeros appear first, get matched to other zeros above/below the line.
12 comes, waits till 15 comes (for 3 time units), they match up. 
24 comes, waits till 30 comes (for 6 time units), and matches to one 
of the 30s.
Remaining 30 waits for 6 units, and matches to 36.
45 waits for 48 (3 units) and they match up.
All 60s match. 
The cycle repeats.

My hypotheis is, in this matching, the max. waiting period (in this case 6)
can't exceed C ( =  (n1 * n2)/ (n1 +n2), in this case 10).

Is this hypothesis right? Can somebody prove/disprove it?


-- Prashant
(waknispv@vuse.vanderbilt.edu)
From: leon@unixg.ubc.ca (Leon ter Beek)
Newsgroups: rec.puzzles
Subject: Revised Monty Hall problem
Date: 1 Jan 1993 00:02:48 GMT
Organization: Chemistry Dept., University of British Columbia, Vancouver, B.C., Canada
NNTP-Posting-Host: unixg.ubc.ca


I've made up another problem, similar to the Monty Hall problem. The
similarity is very obvious, but it does give it a different twist,
though.

Disclaimer: First, I have been reading this group for just two days, so
excuse me if a similar extension has already been posted.
Second, I don't claim this to be a smart problem (what do you expect ;-)).

There are five doors, behind two of them a car, and behind each of the
other three doors a goat. You pick a door and the quizmaster (who knows
what's behind the doors) opens a door with a goat behind it.
1) Should you change doors?
2) Why is the sum of the probability of changing and not changing not
equal to one?

Happy Fireworks!

Leon ter Beek
----------------------------------------------------------------------
"De wereld is mijn vaderland          Email: leon@nematic.chem.ubc.ca
 en wetenschap mijn religie"          Fax:   +1-604-82-22847
   -Christiaan Huygens-               Phone: +1-604-82-22603
Newsgroups: rec.puzzles
From: costley@solo.eng.hou.compaq.com (Brett Costley)
Subject: Re: simple number puzzle
Organization: Compaq Computer Corp.
Date: Tue, 22 Dec 1992 17:23:38 GMT

In article <1992Dec22.010621.19817@dartvax.dartmouth.edu> J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger) writes:
>Yuzuru Hiraga writes:
>
>A simple number puzzle for Christmas...
># sorry if this is in the FAQ: ours just expired.
>
>What positive integer cannot be expressed as a sum of 2 or more
>consecutive integers?
>
>I believe the answer is 2^n power (where n is an integer) can't be
>expressed as the sum of consecutive integers, but all other numbers
>can.
>
"proof" deleted

2^0 = 1 = 0+1
2^1 = 2 = -1+0+1+2
2^2 = 4 = (-3)+(-2)+(-1)+0+1+2+3+4
2^n = [-(n-1)]+...+0+...+n
From: kfl@access.digex.com (Keith F. Lynch)
Newsgroups: rec.puzzles
Subject: Re: Crossword Puzzle Generation
Date: 28 Dec 1992 19:30:05 GMT
Organization: Express Access Public Access UNIX, Greenbelt, Maryland USA
NNTP-Posting-Host: access.digex.com
Keywords: crossword

In article <1992Dec17.151405.8820@hubcap.clemson.edu> jayc@hubcap.clemson.edu (Jay Crawford) writes:
> Where can I find an algorithm to use in generating crossword puzzles?
> Any software available to do this?

Robert T. Morris has an excellent program for this.

He is much better known for the infamous 1988 Internet worm.

I don't know if he's on the net.  I think he lives in Cambridge, Mass.

Keith Lynch, kfl@access.digex.com
Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: "Vesterman's paradox"
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Tue, 29 Dec 1992 01:47:41 GMT

Regarding "Vesterman's paradox":

In article <1hnj90INN4f5@mirror.digex.com> kfl@access.digex.com (Keith F. Lynch) writes:
>The resolution lies in the fact that there's no unique mapping of
>character strings onto integers.  "The smallest..." only makes sense
>in terms of some mapping.
>[...]
>Since there are no restrictions on possible mappings, there is no highest
>integer that can be represented in 150 characters.

What if the description is rephrased as follows:

    the smallest positive integer that cannot be specified uniquely
    in plain English using no more than 150 ASCII characters

Now the only mapping of interest is the one that maps strings of
characters to unambiguously specified integers via the semantics of
English.  (If the string does not describe an integer, or if two
educated speakers of English might disagree on the integer specified,
then the string is not in the function's domain.)  

*Now* how is the paradox resolved?  (Hint: quite easily, in my
opinion.)

-- David Karr (karr@cs.cornell.edu)



Newsgroups: rec.puzzles
From: msb@sq.sq.com (Mark Brader)
Subject: Re: FAQ list
Organization: SoftQuad Inc., Toronto, Canada
Date: Thu, 31 Dec 92 20:33:31 GMT


I guess Chris Cole, who maintains it, must be on holiday, so I'll
take it upon myself to answer the recent questions about the FAQ list.

The rec.puzzles FAQ list is now called the rec.puzzles archive, because
it includes answers to a lot of puzzles that are not really all that
frequently asked.  Because of this, it's also too big to be regularly
reposted.  It was posted once in its entirety to news.answers and
rec.puzzles a few months ago, which means that if you have access to
anonymous FTP, you can get it from a site that archives news.answers.
One of these is pit-manager.mit.edu (18.172.1.27), where the directory
to look in is pub/usenet/news.answers/puzzles-faq.

The archive is also available through an email server.  Send a message
containing two lines of text:

return_address YOUR-ADDRESS-GOES-HERE
send index

to the address archive-request@questrel.com, and it will send you a
message of about 60K bytes containing an index of all the items, and a
list of FTP sites.  If you omit the return_address line, it will use
your From: address for the mail.

There has been talk of making a true FAQ list, with only the most
frequently occurring puzzles on it, but nobody has done this.
-- 
Mark Brader		    "Howeb45 9 qad no5 und8ly diturvrd v7 7jis dince
SoftQuad Inc., Toronto	     9 qas 8mtillihemt mot ikkfavpur4d 5esoyrdeful
utzoo!sq!msb, msb@sq.com     abd fill if condif3nce on myd3lf."      -- Cica

This article is in the public domain.
Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: Direction of the sun
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Mon, 28 Dec 1992 05:11:38 GMT

In article <1hjv7nINN34t@gap.caltech.edu> carl@SOL1.GPS.CALTECH.EDU writes:
>In article <1992Dec24.012636.17115@Csli.Stanford.EDU>, hiraga@Csli.Stanford.EDU (Yuzuru Hiraga) writes:
>>Two boys, A and B, are watching the sunset.
>>A points to the sun and says:
>>"This is the direction where the sun is right now."
>>B replies:
>>"But don't you know that the rays take over 8 minutes to reach the Earth?
>>The sun is actually already below the horizon."
>>Which is correct?  Why? (Make it understandable to a ten year-old.)
>
>The one who says that the sun is below the horizon.  Just to help us in this
>exercise,
the great ship that the boys saw sailing off into the sunset suddenly falls
off the edge of the world and disappears from sight.

Well, if you're going to go with the Ptolemaic system, why not go whole-hog
and postulate a flat Earth as well?

-- David Karr (karr@cs.cornell.edu)


Newsgroups: rec.puzzles
From: egs@cs.Princeton.EDU (Emin Gun Sirer)
Subject: Guessing at random numbers.
Originator: egs@volkl
Reply-To: egs@cs.Princeton.EDU (Emin Gun Sirer)
Organization: Department of Computer Science, Princeton University
Date: Thu, 24 Dec 1992 01:56:25 GMT

Here's a paradoxical (at least to me) puzzle that I could not locate
in the FAQ. Enjoy.

I pick two numbers, randomly, and tell you one of them. You are supposed
to guess whether this is the lower or higher one of the two numbers I
picked. Can you come up with a method of guessing that does better than
picking the response "low" or "high" randomly (i.e. probability to guess
right > .5) ?

Nothing is known about my random number generation algorithm. I give you
the higher or lower number with equal probability, so trying to find patterns
in my behaviour is useless (unless you believe that past coin tosses affect
the outcome of future tosses :-).

Thanks to Steve Gortler for this mind-twister.

Gun.
From: elnitsky@math.lsa.umich.edu (Serge Elnitsky)
Newsgroups: rec.puzzles
Subject: rec.puzzles FAQ
Date: 29 Dec 1992 13:05:03 GMT
Organization: University of Michigan, Mathematics Department, Ann Arbor
NNTP-Posting-Host: sirius.math.lsa.umich.edu

In article <19921228.140236.87@almaden.ibm.com> IACHETTA@BCRVMPC2.VNET.IBM.COM writes:
>Would some kind soul please send me the FAQ for rec.puzzles.
     
   Ditto.
From: kenneths@tropicana.ai.mit.edu (Kenneth J. Schneider)
Newsgroups: rec.puzzles
Subject: Re: Christmas Songs
Date: 22 Dec 1992 22:09:51 GMT
Organization: MIT Artificial Intelligence Lab
Distribution: world
NNTP-Posting-Host: tropicana.ai.mit.edu

>>13. A miniature Nazarene village.

>O Little Town of Bethelehem

i don't think bethelehem (sp?) is a Nazarene village. Nazareth was another city,
where Jesus was raised.


Newsgroups: rec.puzzles
From: rabino@eng.umd.edu (Michael Rabinovitz)
Subject: Re: List of words
Date: Sun, 27 Dec 92 01:26:28 GMT
Organization: College of Engineering, University of Maryland, College Park

In article <1hiebjINN5ld@controversy.math.lsa.umich.edu> elnitsky@math.lsa.umich.edu (Serge Elnitsky) writes:
>  I don't think I've posted this here before... If I have, who
>cares. It's original, at any rate.
>
>THE HIT LIST
>
>  All of the 208 words on the list below possess a certain 
>very specific property that is shared by fewer than 300 English 
>words altogether. Find it. (Once you have it, you'll know 
>without a doubt that it's correct.)
> 
	The property all the words have is that they all form new words by
prefixing each one with the letter "s."

-- 
         	                        Mike Rabinovitz 
					rabino@eng.umd.edu
From: lordSnooty@cup.portal.com (Andrew - Palfreyman)
Newsgroups: rec.puzzles
Subject: santa's puzzle
Date: Thu, 24 Dec 92 19:08:40 PST
Organization: The Portal System (TM)
Distribution: world

	Santa's New Automated North Pole Operation 
	------------------------------------------

The workload finally got too much (even for the diligent dwarves), and Santa
decides to build a conveyor and assembly-line system to automate the
wrapping of presents. Wrapping paper goes down to the wrapping area, and
finished, wrapped presents come back.

Twin hoppers in the production area have capacities of 10 kilos for paper and
100 kilos for finished presents, respectively. The dwarves can handle the
paper at a rate of 100 gm/second, and can deliver presents at the rate of
one kilo per second. The hoppers must neither empty nor overflow.

By a clever reversal circuit, Santa optimises his resources by putting the
conveyor to double duty. The conveyor is installed so that four seconds are
required to switch from paper to presents transport, and eight seconds from 
presents to paper. Although somewhat ancient (reindeer-driven), the technology
appears to work just fine. In the interests of efficiency, Santa needs to know:

Q1. What is the necessary capacity of the conveyor?
Q2. How often must the reindeer switch over, in either direction?

...One year later....
The recession bites even at the North Pole; Santa must cut back the conveyor's
capacity to 1200 grams per second. This costs less (he hopes) than an increase
in the hopper size, although he would like to minimise any such increase.

Q3. Which hopper should be increased?
Q4. By how much?
Q5. How often now must the reindeer switch over, in either direction?
 
	merry christmas to all!
							andrew palfreyman @cup.portal.com
From: "eliot gelwan" <eliot.gelwan@channel1.com>
Newsgroups: rec.puzzles
Subject: new year's problem
Date: 3 Jan 93 23:50:47 EST
Reply-To: "eliot gelwan" <eliot.gelwan@channel1.com>
Distribution: rec
Organization: Channel 1 Communications

Mfor 1987, it seems you're looking for a year in which it's
Mpossible to expressthe date and time using the digits 0-9
Mexactly once - this first happened in 198on February 3 at 4:56
MAM.  2/3/1987 04:56

MSo it seems this isn't *really* a valid answer to the 1987/2013
Mpuzzle.  Anyoneelse care to offer a "better" solution?

I think what he's looking for is simply a year where all the
digits are unique. You're complicating things too much.

                   |
...eliot          -*-              Postlink: ->CHANNEL1 [R/O okay]
                   |               Internet: eliot.gelwan@channel1.com

--
Channel 1 (R)   Cambridge, MA

Newsgroups: rec.puzzles
From: arromdee@jyusenkyou.cs.jhu.edu (Ken Arromdee)
Subject: Re: Direction of the sun
Organization: Johns Hopkins University CS Dept.
Date: Thu, 24 Dec 1992 04:26:04 GMT

In article <1992Dec24.012636.17115@Csli.Stanford.EDU> hiraga@Csli.Stanford.EDU (Yuzuru Hiraga) writes:
>Two boys, A and B, are watching the sunset.
>A points to the sun and says:
>"This is the direction where the sun is right now."
>B replies:
>"But don't you know that the rays take over 8 minutes to reach the Earth?
>The sun is actually already below the horizon."
>Which is correct?  Why? (Make it understandable to a ten year-old.)

A is correct.  The sun is always in the same place, and the Earth moves.
So the horizon moves too, and "the point so-and-so distance below the
horizon" is also a moving point.

Therefore, it's meaningless to say that something is in a direction relative
to the horizon unless you specify which horizon, since the horizon moves.
The sun is indeed below the 8-minutes-from-now horizon, but you can't
conclude this means it's below the current one.

># note: ignore subsidiary points like refraction by the atmosphere.
>BONUS: What would the answer be under the Ptolemaic system?

The sun would really be some distance from where it appears to be.
--
"the bogosity in a field equals the bogosity imported from related areas, plus
the bogosity generated internally, minus the bogosity expelled or otherwise
disposed of."  -- K. Eric Drexler

Ken Arromdee (arromdee@jyusenkyou.cs.jhu.edu, arromdee@jhunix.hcf.jhu.edu)
Newsgroups: rec.puzzles