HDTenEightyP/NetNews huggingface dataset:
From: j1w@icf.hrb.com (Doug ('Oscar') Works)
Newsgroups: rec.puzzles
Subject: Need anagram maker in Fortran or Turbo
Date: 8 Jan 93 13:46:30 GMT
Organization: HRB Systems, Inc.

I need an anagram maker to run in either VAX Fortran or else PC based Turbo C
or Turbo Pascal.  Anyone have info to help?  Thanks.

                                                      - Doug

................................................................................
................................................................................
....    |         "He is no fool who gives what he cannot keep,             ....
....  --+--              to gain what he cannot lose."                      ....
....    |                                           - Jim Elliot            ....
....    |                                                                   ....
.......................  Internet: j1w@icf.hrb.com .............................
................................................................................
    J. Douglas Works * HRB Systems, Inc. * State College, PA  16804  USA
Newsgroups: rec.puzzles
From: jerryd@PROBLEM_WITH_INEWS_GATEWAY_FILE (Jerry Donovan)
Subject: Re: Manhole covers - why circular?
Date: Mon, 11 Jan 1993 22:23:23 GMT
Organization: Hewlett-Packard / Boise, Idaho
X-Newsreader: TIN [version 1.1.4 PL7]

Reidar J. Husmo (radar@grumpy) wrote:
: Robert Weinberg (robertw@informix.com) wrote:
: : Manhole covers are manufactured in circular shape, rather than oval or square, to  
: : satisfy one paramount condition.
: : 
: : What is the condition, and why does a circular shape satisfy it?
: : 
: : What other shapes would also satisfy the condition?
: : 
: Ok, lots of people have already given the correct answer(s) to the first
: question, so I won't bother with that.

Really?  I don't think so.  All the answers I've seen that appear to considered
correct assume that the shape is constant diameter.  Any shape will work iff the
lip or ledge of the hole is smaller than the smallest cross section of the lid.
ie. I could make a manhole with a wide enough lip to make sure any shape manhole
cover would not fall through.  Of course the usabiltiy of such a hole and cover
would become suspect.

Therfore doesn't the real answer have more to do with minimizing material
while maximizing the real hole area such that the cover won't fall through?

Jerry Donovan
From: 95csk_2@williams.edu (Jumping Bean)
Newsgroups: rec.puzzles
Subject: Re: 12 bolt puzzle HELP!
Date: 13 Jan 1993 01:30:09 GMT
Organization: Williams College, Williamstown, MA
Distribution: world
NNTP-Posting-Host: bigbird.cc.williams.edu

Capt Scott (retired) writes:
>I was told a puzzle recently and it is driving me spare.
>I have a feeling it is probably  standard puzzle so if anyone
>can help me ....
>
>This is it:-
>	You have 12 bolts, totally identical except 1 is either 
>lighter or heavier. Using a balance-scale 3 times find the 
>Dud.
>
>Any help appreciated
>Scott
>
>-- 
>-----------------------------------------------------------------
> Scott Gilhooly             |The only thing more dangerous than |
> Glasgow University         |a programmer with a screwdriver is |
> gilhoosg@dcs.gla.ac.uk     |a user with an idea                |

Wow. We just discussed this today in my Winter Study course at
Williams College - puzzle solving.  I think it may be a FAQ, but
here's our solution:
 
Let the numbers 1 - 12 represent the 12 bolts:

1 2 3 4 5 6 7 8 9 10 11 12

Divide into three groups of four:

(1 2 3 4)    (5 6 7 8)    (9 10 11 12)

Weigh (1 2 3 4) vs (5 6 7 8).  If they weigh the same, then bolts 1-8
weigh the same, so one of 9-12 weighs more or less than the others.
Weigh (9 10 11) vs (1 2 3) (normal weight).  If it weighs
the same, then 12 is the odd bolt, and you can easily find out if it
is heavy or light using the last weighing.  Otherwise, 9, 10, and 11
is either heavier or lighter than 1, 2, and 3, and the odd bolt will
be heavier or lighter correspondingly.  Weigh 9 vs 10.  If they are
equal, then bolt eleven is odd (and you know if it is heavy or light)
otherwise, you know from weighing 9 vs 10 which is not normal, since
you know if the odd one is heavy or light.  

Now, assume that (1 2 3 4) does not balance (5 6 7 8).  We may assume WLOG
that (1 2 3 4) is heavier.  Thus, either one of (1 2 3 4) is too heavy, or
one of (5 6 7 8) is too light.  Weigh (1 2 5) vs (3 4 6).  (Note that
(1 2 5) is essentially the same as (3 4 6) - 1, 2, 3, 4 *might* be too
heavy, 5, 6 *might* be too light) If they balance, then either 7 or 8
is too light.  In this case, weigh 7 vs 8, and the lighter of the
two is the odd one.  Otherwise, assume WLOG that (1 2 5) is heavier
than (3 4 6).  Then either 1 or 2 is too heavy, or 6 is too light.
Weigh 1 vs 2.  If one of them is heavier, it is the odd bolt, and is
too heavy, otherwise 6 is too light.
 

Sorry if this is too complicated, but it is complete (I'm convinced,
at least) and it even tells if the odd bolt is too heavy or too light
(Ooh, Aah :-)  I hope this helps.

-Chris


-- 
- Chris Kollett
  Christopher.S.Kollett@williams.edu
  WUFO - 1992 Division II Champs!
Newsgroups: rec.puzzles
From: deghare@daisy.uwaterloo.ca (Dave Hare)
Subject: Re: I'm bored with manhole covers
Organization: University of Waterloo
Date: Tue, 12 Jan 1993 01:43:27 GMT

In article <1993Jan11.103323.2169@EE.Surrey.Ac.UK> @ee.surrey.ac.uk (null) writes:
>As I was going up Guildford High Street on Saturday, I walked over a manhole cover.
>It was square.
>
>Can someone explain what is going wrong?

I can think of several possible explanations:

 1) You are in England.  (Mad dogs etc.)

 2) It wasn't really an access cover at all.  For example, was there any
    obvious way for it to be removed?  Storm drain covers are typically
    rectangular, but are not meant for frequent removal.

 3) It was part of a government make-work project: Change all access 
    covers (and, no doubt, all pipes as well) to square, then back to 
    round, then back to square...  Possibly the city works department
    found themselves with surplus funds near the end of the fiscal year.

 4) You are in England.

Dave Hare
From: torbenm@diku.dk (Torben AEgidius Mogensen)
Newsgroups: rec.puzzles
Subject: Re: 19 (possible spoiler)
Date: 7 Jan 93 13:26:09 GMT
Organization: Department of Computer Science, U of Copenhagen

ct@dde.dk (Claus Tondering) writes:

>What computer operating system could be spelled thus:

>		19

AIX. A=1 (position in alphabet) and IX=9 (roman numerals).

	Torben Mogensen (torbenm@diku.dk)
Newsgroups: rec.puzzles
From: rolls@cis.umassd.edu (Frank Pinto)
Subject: Re: closed form for a sum
Organization: University of Massachusetts Dartmouth
Date: Wed, 13 Jan 1993 04:38:31 GMT

If you mean by evaluation "the limit"
then "it is " 0

Since sin can onlybe from -1, 1
the denominator will grow much "faster"
which bring the ratio 
to a very small number, namely = 0

Frank Pinto
Newsgroups: rec.puzzles
From: acheng@ncsa.uiuc.edu (Albert Cheng)
Subject: Re: Alternate cover shape [SPOILER]
Originator: acheng@shalom.ncsa.uiuc.edu
Organization: Nat'l Ctr for Supercomp App (NCSA) @ University of Illinois
Date: Thu, 7 Jan 1993 19:47:00 GMT


In article <106207@bu.edu>, tvaughan@buphyk.bu.edu (Timothy Vaughan) writes:
>I just thought of a non-circular shape that won't fall through.
>I don't know what it is called, but it is the shape of the rotor
>in a Wankel engine.  It has constant diameter, although not
>constant radius.  It could be described nonrigorously as a "rounded
>triangle."  I will try to derive the equation for it.

As someone has pointed out, the height of a triangle is smaller than
the base length, it is possible for it to fall through close to the
base.  But make it concave in like star-shape would work.  All
odd-pointed star shapes would not fall through.  (Pretty hard to climb
through.)
Newsgroups: rec.puzzles
From: hiraga@Csli.Stanford.EDU (Yuzuru Hiraga)
Subject: Re: An(other) *easy* puzzle. (COMMENT AND ADDITION)
Organization: Stanford University CSLI
Date: Mon, 11 Jan 1993 08:44:49 GMT

In article <1993Jan9.030824.13624@Csli.Stanford.EDU> I wrote
>A plane is totally colored with two colors.
>* Show that there exist three points of same color that
>  (when connected) form an equilateral triangle.

The answers posted so far, although different in appearance,
can all be related to the fact that:
Assuming there can't be such a triangle, then with a regular hexagon
ABCDEF and its center O:
      A B
     F O C
      E D
       x
A,B, E,D must be the same color, and F,O,C must be the other color
(thus either FCx or EDx will be the same color).

Are there solutions that are radically different from this?

>* How about one with a 1cm (no, not inch :-) edge?

Add the following problem:

* Does there exist an isosceles triangle ABC with AB=AC=1cm?

# This, actually is a give-away to the previous problem:
# well, there IS a two-coloring of a plane where no equilateral triangle
# of a given size can exist.  Show one (this, actually, is easy).

-Yuzuru Hiraga
From: landman@hal.COM (Howard Landman)
Newsgroups: rec.puzzles
Subject: Re: New Year's problem
Date: 10 Jan 93 06:31:58 GMT
Organization: HaL Computer Systems, Inc.
NNTP-Posting-Host: bang.hal.com

In article <1htrk2INNrql@mirror.digex.com> kfl@access.digex.com (Keith F. Lynch) writes:
>What property did 1961 have, that no year after it will have until 6009?

Too easy.  I figured that out in 1961 when I was 8 or 9 years old.
The previous year with that property was 1881.  Of course, one might
argue that '69  and '88 and '96 have some of the same property ...

>What property did 1987 have, that no year after it will have until 2013?

I'll take a wild guess without even checking.  It's prime?

	Howard A. Landman
Newsgroups: rec.puzzles
From: ghazel@eskimo.com (Geoff Hazel)
Subject: Re: Manhole covers - why circular?
Organization: >>> Eskimo North (206)-FOR-EVER <<<
Date: Wed, 6 Jan 1993 17:30:05 GMT

In article <C0EHKG.AL5@watdragon.uwaterloo.ca> deghare@daisy.uwaterloo.ca (Dave Hare) writes:
>In article <1993Jan5.173440.10482@informix.com> robertw@informix.com (Robert Weinberg) writes:
>>Manhole covers are manufactured in circular shape, rather than oval or square, to  
>>satisfy one paramount condition.
>>
>>What is the condition, and why does a circular shape satisfy it?
>
>Actually, access covers are circular for many reasons, and I often use this
>as a game with kids to get them to stretch their brains.
>
>I don't know which reason you have in mind, but here are some:
>
>    - It is orientation free: it will fit in any position
             ^^^^^^^^^^^^^^^^^

In fact, I read that they were round in order to minimize the number
of decisions the sewer workers had to make.







-- 
Geoff Hazel                  ghazel@eskimo.com
Edmark Corporation           206-556-8445
Box 3218, Redmond, WA        "the syntax is painful"
From: mikem@ll.mit.edu
Newsgroups: rec.puzzles
Subject: Re: Help me with MH puzzle, also, a new puzzle.
Date: 6 Jan 93 21:33:51 GMT
Organization: MIT Lincoln Laboratory

In article <1i7cb6INNe0@mirror.digex.com> adric@access.digex.com (William Johnson) writes:
wj> When playing Russian Roulette, each member has a revolver with 6 chambers
wj> and 1 bullet (may have been 2 bullets, I don't think it changes the answer,
wj> just the likelihood of death).  Each member then fires the gun twice into
wj> his/her own head.  Members may spin the revolver before each shot.  Which
wj> option gives the best chance of survival:
wj> 	A) Spin shoot spin shoot
wj> 	B) Spin shoot shoot
wj> 	C) Shoot spin shoot
wj> 	D) Shoot shoot.
wj> 

In article <1i8at3INNjbl@iskut.ucs.ubc.ca>  leon@unixg.ubc.ca (Leon ter Beek) writes:
ltb> A=C if you didn't spin before you start the next round. 

Wait a minute.  What's this about starting another round?  The problem states,
"Each member then fires the gun twice into his/her own head."  Twice.  Two shots.
That's it.  Nothing more.  Furthermore, each member has his/her *own* *gun*, so
the result of any other player's move is irrelevant too, since you use your own
gun, not the gun just fired by another player.

ltb> Spinning makes the events independent, ie. next shot same probability as
ltb> first shot if you survive the first one.

I'll go along with that.

ltb> A: P(survive) = 5/6*5/6= 25/36
ltb> B: P(survive) = 5/6*4/6= 20/36 Second shot only 4 empty chamber left.
ltb> C: P(survive) = 3/6*3/6= 9/36  Don't spin after preceding round.
ltb> D: P(survive) = 3/6*2/6= 6/36

Whoa!  Hold on!  Why is first probability in C and D only 3/6?  What assumption
are you making about the state of the gun prior to the first shot?  What is the
basis for your making this assumption?

wj> Ok, it seems obvious to me that the two options where you shoot twice
wj> without a spin in the middle are worst.  Because assuming you survive
wj> the first shot (because if you didn't, it's over), you have now reduced
wj> your chances of surviving the next chamber.

Makes perfect sense to me.

wj> But I don't see the difference between A and C, unless in case C you know
wj> in advance that where you put the bullet either was or was not in the
wj> next chamber to be fired.  This may have been the answer, but I don't think
wj> so, because reading their answer sheet, the probabilities weren't as clear
wj> cut as that.  Any suggestions?

The missing piece of information in this puzzle is whether you know anything
about the state of the gun before you choose an option.  But the puzzle does
not fill in this missing information.  It's therefore not complete, and you
can't answer it based on the information provided.

Bad puzzle!  Bad, bad puzzle!  No biscuit!

-------------------------------------------------------------------------------
Michael Maciolek - mikem@ll.mit.edu --- VOICE (617)981-3174 - FAX (617)981-0189
Network Engineer - MIT Lincoln Laboratory - Computer Telecommunications Systems
-------------------------------------------------------------------------------
	    "It is not an optical illusion, it just looks like one."
From: robertw@informix.com (Robert Weinberg)
Newsgroups: rec.puzzles
Subject: Manhole covers - why circular?
Date: 5 Jan 93 17:34:40 GMT
Organization: Informix Software, Inc.

Manhole covers are manufactured in circular shape, rather than oval or square, to  
satisfy one paramount condition.

What is the condition, and why does a circular shape satisfy it?

What other shapes would also satisfy the condition?

+================================+================================+
|         Rob Weinberg           |  Workstation publications tech |
|   Email : robertw@informix.com |        Graphic designer        |
|   Voice : 415-926-6754         |     <imagine your ad here>     |
+================================+================================+
From: tkacik@hobbes.hobbes.cs.gmr.com (Tom Tkacik)
Newsgroups: rec.puzzles
Subject: Re: Manhole covers - why circular?
Date: 5 Jan 93 20:51:30 GMT
Reply-To: tkacik@hobbes.hobbes.cs.gmr.com
Organization: GM Research Labs

In article 93Jan5134917@kariba.bbn.com, nichael@bbn.com (Nichael Cramer writes:
> robertw@informix.com (Robert Weinberg) writes:
> 
>    Manhole covers are manufactured in circular shape, rather than oval or square, to  
>    satisfy one paramount condition.
>    What is the condition ...
>
> The cover can't fall down the hole.
> 
>    ... and why does a circular shape satisfy it?

>    What other shapes would also satisfy the condition?
> 
> Given the above, for any fully convex shape, I don't _think_ there are
> any others.

A circle is not the only constant diameter closed shape.
An equalateral triangle with arcs on each side (whose center is the opposite
vertex, also has a constant diameter.
To see what this looks like, check out a rotor in a wankel engine.
It has this shape.

Tom Tkacik
GM Research Labs
tkacik@hobbes.cs.gmr.com
tkacik@kyzyl.mi.org
From: stein@hal.nta.no (Stein Kulseth)
Newsgroups: rec.puzzles
Subject: Re: Bus Stop Puzzle
Date: 7 Jan 93 14:18:39 GMT
Organization: Norwegian Telecom Research

In article <C0HD3p.1u4@dcs.ed.ac.uk>, tk@dcs.ed.ac.uk (Tommy Kelly) writes:
|> Every morning, I take a train to the city, then have to take a bus to
|> Uni.  There are two bus stops within easy walking distance of the train
|> station.
|> 
|> Bus stop 1 is 4mins 20 seconds away, and is the earlier on the route (the bus
|> visits it before bus stop 2).
|> 
|> Bus stop 2 is 3mins 20 seconds away.
|> 
|> The bus takes about 30 seconds to travel from stop 1 to stop 2.
|> 
|> Bus interarrival time is 15 minutes, and can effectively be at any point
|> in their run when I begin my walk from the train station.
|> 
|> a. On average, which bus stop should I walk to minimize my time in travelling
|>    to the University?

Bus stop 2
 
|> b. In general, for times :
|> 	Tn - time to walk to stop n
|> 	dn - time for bus to travel from stop n to stop n+1
|> 	l  - bus interarrival time
|> 
|> 	which stop should I choose?

The one that maximises Sum(d1...dn) - Tn

-- 
stein.kulseth@nta.no (Norwegian Telecom Research)
   'When murders are committed by mathematics, they can be solved by
   mathematics. Most of them aren't, and this one wasn't'
   - Nick Charles (Dashiell Hammett's "The Thin Man")
From: dik@cwi.nl (Dik T. Winter)
Newsgroups: rec.puzzles
Subject: Re: I'm bored with manhole covers
Date: 11 Jan 93 21:10:48 GMT
Organization: CWI, Amsterdam

In article <1993Jan11.103323.2169@EE.Surrey.Ac.UK> @ee.surrey.ac.uk (null) writes:
 > As I was going up Guildford High Street on Saturday, I walked over a manhole cover.
 > It was square.
 > 
 > Can someone explain what is going wrong?
 > 
As may be clear you do not exist.  You are null.  So square manhole covers do
not exist.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl
From: rmurphy@aludra.usc.edu (Bob Murphy)
Newsgroups: rec.puzzles
Subject: Re: An(other) *easy* puzzle. <<SPOILER>>
Date: 10 Jan 1993 21:00:53 -0800
Organization: University of Southern California, Los Angeles, CA
NNTP-Posting-Host: aludra.usc.edu

In article <1993Jan9.030824.13624@Csli.Stanford.EDU> hiraga@Csli.Stanford.EDU (Yuzuru Hiraga) writes:
>
>A plane is totally colored with two colors.
>* Show that there exist three points of same color that
>  (when connected) form an equilateral triangle.
>
>-YH

My proof is probably easier to show in person but here goes:

Suppose there is no equilateral triangle within the plane such that
its three vertices are all the same color.

Pick two points, A and B, of the same color.  WLOG, suppose that
A and B are both red.  Now refer to the following diagram:

              C     D     E
              *     *     *

                 A     B     F
                 *     *     *

                    G
                    *

Since A and B are both red, and ABD is an equilateral triangle, we
must have that D is blue.

Since A and B are both red, and ABG is an equilateral triangle, we 
must have that G is blue.

Since G is blue, and CGE is an equilateral triangle, we have that
C and E cannot both be blue.  WLOG, suppose that E is red.

Since B and E are both red, and BEF is an equilateral triangle, we
must have that F is blue.

But this now gives us the equilateral triangle DGF whose vertices
are all blue.

This is a contradiction.

Thus, there must be an equilateral triangle in the plane such that
its vertices are all the same color.

                           Bob Murphy (rmurphy@aludra.usc.edu)
From: eahu099@orion.oac.uci.edu (Asako Mariather Kurosaka-Black)
Subject: Inverter problem
Newsgroups: rec.puzzles
Organization: University of California, Irvine
Date: 7 Jan 93 21:26:11 GMT

I was reading the FAQ and found the inverter problem, but it never explains
how to invert three independent inputs with two inverters and unlimited
ANDs and ORs.  Can someone enlighten me?
From: grabiner@math.harvard.edu (David Grabiner)
Newsgroups: rec.puzzles
Subject: Re: Bus Stop Puzzle
Date: 8 Jan 93 20:49:54 GMT
Organization: /etc/organization
In-reply-to: tk@dcs.ed.ac.uk's message of 7 Jan 93 10:57:24 GMT


Here's another fairly easy puzzle involving bus schedules.

While walking east along a major bus route, I am passed by an eastbound
bus every 24 minutes, and by a westbound bus every 8 minutes.

1. How often do the buses run?  (It isn't rush hour, so the buses run
with the same frequency in both directions.)

2. Are they local or express buses?

--
David Grabiner, grabiner@zariski.harvard.edu
"We are sorry, but the number you have dialed is imaginary."
"Please rotate your phone 90 degrees and try again."
Disclaimer: I speak for no one and no one speaks for me.
Newsgroups: rec.puzzles
From: djoyce@black.clarku.edu (Dave Joyce)
Subject: Re: Manhole covers - why circular?
Organization: Clark University (Worcester, MA)
Date:  5 Jan 93 18:20:48 GMT

In <1993Jan5.173440.10482@informix.com> robertw@informix.com
(Robert Weinberg) writes:

>Manhole covers are manufactured in circular shape, rather than oval 
>or square, to satisfy one paramount condition.
>
>What is the condition, and why does a circular shape satisfy it?
>
>What other shapes would also satisfy the condition?

There must be another condition besides what you're looking for.  After all,
a sphere would satisfy that condition.  Even a big cork would do.   So, a 
secondary condition, not the one you're looking for, must be that the shape
be nearly flat, with just enough of a rim to fit over the hole.
-- 
David E. Joyce				Dept. Math. & Comp. Sci.
Internet:  djoyce@black.clarku.edu	Clark University
BITnet:    djoyce@clarku		Worcester, MA 01610-1477
From: johnsbra@tmlh (Havard Johnsbraten)
Newsgroups: rec.puzzles
Subject: Solution wanted on a nice Christmas problem
Date: 7 Jan 93 09:21:22 GMT
Organization: tmlh


A couple of weeks before Christmas the following nice problem was posed on 
this net:

  Does it exist a countable set of disjoint line segments of length 1 such 
  that no line segment of length 1 can be added without touching one of the 
  line segments in the set?
  The problem is to be solved for
      i) open line segments
     ii) half open line segments
    iii) closed line segments
 
Now I should like to see the complete solution of this exiting problem.

For open intervals the answer i clearly "yes"; there are several simple 
patterns here.  For half open and closed intervals I think the answers must 
be "no", but the solution is not so trivial as proposed in an earlier post 
message.

If the problem is solved, please let us have the solution.  If the problem 
is still open, it deserves to be given to the sci.math conference.

 o            o
Havard Johnsbraten, Telemark College of Education, Notodden, Norway
E-mail  haavard.johnsbraaten@tmlh.no  or
G=haavard;S=johnsbraaten;O=tmlh;P=uninett;A= ;C=no
From: ct@dde.dk (Claus Tondering)
Newsgroups: rec.puzzles
Subject: Re: 19 (SPOILER)
Date: 8 Jan 93 09:24:20 GMT
Organization: Dansk Data Elektronik A/S

ct@dde.dk (Claus Tondering) writes:

>What computer operating system could be spelled thus:

>		19

The answer I had in mind was UNIX. (UN=1 in French, IX=9 in Roman
numerals.)

In fact you might also spell UNIX thus: -9.

-9 is the opposite of 9. Opposites are in English indicated by the
prefix 'un-'.

-- 
Claus Tondering
Dansk Data Elektronik A/S, Herlev, Denmark
E-mail: ct@dde.dk
Newsgroups: rec.puzzles