HDTenEightyP/NetNews huggingface dataset:
Organization: Stanford Linear Accelerator Center
Date: Friday, 18 Dec 1992 10:03:32 PST
From: <SHERE@SLACVM.SLAC.STANFORD.EDU>
Newsgroups: rec.puzzles
Subject: Monty Hall Puzzle RIP

Alright, I've had just about all bickering I can stand on this
Monty Hall drivel and I'm going to put it to bed.  You folks are so
busy outsmarting yourselves that this should be called the Monty
Python Puzzle; however, if you cut through the pseudo-intellectual
double talk about probabilities and how they may, or may not,
change after being shown a door and put together a simple chart
detailing each possible permutation of how the game can progress
you will have the following:

      Permutation:  A B C D E F G H I J K L M N O P Q R S T U V W X

Prize behind door:  1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3
Originally choose:  1 1 1 1 2 2 3 3 1 1 2 2 2 2 3 3 1 1 2 2 3 3 3 3
       Door shown:  2 2 3 3 3 3 2 2 3 3 1 1 3 3 1 1 2 2 1 1 1 1 2 2
   Change or Stay:  S C S C S C S C S C S C S C S C S C S C S C S C
     Final choice:  1 3 1 2 2 1 3 1 1 2 2 3 2 1 3 2 1 3 2 3 3 2 3 1
      Win or Lose:  W L W L L W L W L W W L W L L W L W L W W L W L

You will note that out of the twenty-four permutations of this game
you can win exactly twelve ways, six ways by staying with your
original choice (ACKMUW) and six ways by changing your mind
(FHJPRT). You can also lose exactly twelve ways, six ways by
staying with your original choice (EGIOQS) and six ways by changing
your mind (BDLNVX).  Now everybody, repeat after me; half the time
we stay with our original choice we win, half the time we change
our minds we win, half the time we win, PERIOD!  You have no
problems with the fact that if Monty reveals what is behind one of
the doors and then lets you choose between the two remaining that
your chances are fifty-fifty.  Yet this is exactly what's happening
here!  The only difference is SEMANTICS!  Whether you "change your
mind" or "stay with your original choice" you are STILL CHOOSING
BETWEEN ONE OF TWO DOORS.  FIFTY-FIFTY.

Flames and arguments expected, oh well.

 Lee
 Standard disclaimer applies, my thoughts are my own.
Xref: sparky rec.puzzles:7973 sci.math:17035
From: J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger)
Newsgroups: rec.puzzles,sci.math
Subject: Re: Circular graph problem
Date: 16 Dec 92 17:21:45 GMT
Organization: Dartmouth College, Hanover, NH
X-Posted-From: InterNews1.0b1@newshost.dartmouth.edu

In article <1992Dec16.113500.1157@client24.comlab.ox.ac.uk>
mnewman@comlab.ox.ac.uk (Matthew Newman) writes:

> Consider circular graphs with N nodes, such that there is an integer
> at each node. Consider those integers K such that for some graph G
> each integer in [1..k] is the sum of integers at the nodes of some
> connected sub-graph G. For example with N=4 I can get K=11 with the
> following graph :     (1)--(2)
>                        |    | 
>                       (3)--(5) 
> 
> and I can do better than this and get the maximum possible k=13 like
> this :     (1)--(2)    and like this:   (1)--(3)
>             |    |                       |    |
>            (4)--(6)                     (2)--(7)

I don't understand why these last two count as valid -- in the first,
there's no combination that totals 4, while in the second there's no
combination that totals 7.  Or am I just reading yet another problem
incorrectly?!?


--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: creech@hardy.u.washington.edu (Bill Creech)
Subject: Re: Ask Marilyn
Organization: University of Washington
Date: Mon, 14 Dec 1992 19:29:36 GMT

Organization: University of Washington, Seattle
Keywords: 

In article <92349.120251HERSCH@auvm.american.edu> HERSCH@auvm.american.edu writes:
>Does everyone agree that Marilyn vos Savant's answer to the pancake
>probability problem in yesterday's Parade magazine was wrong? The
>answer is really 50-50. Either you've picked the pancake with 2 brown
>sides, in which case the other side is brown, or you've picked the
>pancake with one brown side, in which case the other side is not
>brown. 50-50.
>
>H.
>
>Herschel Browne
>"The" American University

	I don't.






















>



	I don't think Marilyn is wrong at all.  Her pancake
	paradox is just a variation on the Monty Hall puzzle.
	There are three pancakes:

	A: Both sides are white.
	B: Both sides are brown.
	C: ONe side of each color.

	Suppose you were blindfolded and pulled out a pancake
	and laid it on the table.  What are the chances that both
	sides match?  2/3.  Exactly the chances of selecting A or
	B.

	Now remove the blindfold.  Does seeing a brown side faceup
	change the odds?  No way.  The reason is that pancake B will
	be brown side up every time while pancake C will show brown
	only half of the time.

	This same "paradox" is known to all good bridge players in
	this type of situation:
				North
				K 10 9 8 7

				South
				A 6 5 4

	Suppose you play the A, West follows low and East plays the
	Jack.  Now you play the 4, West follows low again.  There is
	one card remaining in the suit.  If West has it, playing the 
	Ten will win; if East has it, playing the King will win.  What is
	the best chance?  A priori, East is slightly more likely to have
	begun with QJ, but with QJ East will sometimes play the Q but
	with J alone, he must play the Jack.  Just like Marilyn's pan-
	cakes, the odds are about 2/1 to play the 10.  


________________________________________________________________
|                             |                                 |
|      Bill Creech            |                                 |
|      creech@u.washington.edu|                                 |
|_____________________________|_________________________________|  















Newsgroups: rec.puzzles
Subject: Re: Ask Marilyn
Summary: 
Expires: 
Followup-To: 
Distribution: 
Organization: University of Washington, Seattle
Keywords: 

In article <92349.120251HERSCH@auvm.american.edu> HERSCH@auvm.american.edu writes:
>Does everyone agree that Marilyn vos Savant's answer to the pancake
>probability problem in yesterday's Parade magazine was wrong? The
>answer is really 50-50. Either you've picked the pancake with 2 brown
>sides, in which case the other side is brown, or you've picked the
>pancake with one brown side, in which case the other side is not
>brown. 50-50.
>
>H.
>
>Herschel Browne
>"The" American University
From: marcelk@blade.stack.urc.tue.nl (Marcel van Kervinck)
Newsgroups: rec.puzzles
Subject: Re: 8-queen problems (revisited?)
Date: 18 Dec 92 12:59:28 GMT
Organization: MCGV Stack, Eindhoven University of Technology, the Netherlands.

Here's my little program to solve the problem for N queens on
a NxN board. (0 < N <= MAXSIZE) It's a simple backtracker and it
uses three arrays to keep track of the lines that are already
occupied. (vert[] for vertical lines, dia1[] and dia2[] for
the diagonals) For more speed it is worth noting that these
arrays can be held in registers, if you convert it to assembler
by hand. I always use this program to test the pure processor
speed of computer systems.
Hope you like it,

Marcel van Kervinck
marcelk@stack.urc.tue.nl

-----8<------cut-------here-----8<-------cut------here------8<-----

/*------------------------------------*/
/*        Queens on chessboard        */
/*   Written by Marcel van Kervinck   */
/*        C version : May 1992        */
/*------------------------------------*/

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>

typedef int bool;
#define FALSE 0
#define TRUE 1

#define MAXSIZE 16

int size;
bool doshow;
long counter = 0L;
int r[MAXSIZE];
bool vert[MAXSIZE];
bool dia1[2*MAXSIZE-1];
bool dia2[2*MAXSIZE-1];

long starttime, endtime;

void init() 
{
    int i;
    for (i=0; i<size; i++)
        vert[i] =
        dia1[size-1-i] = dia1[size-1+i] =
        dia2[size-1-i] = dia2[size-1+i] = TRUE;
}

void show()
{
    int i,j;
    for (i=0; i<size; i++)
    {
        for (j=0; j<size; j++)
            printf("%2c", (r[i]==j) ? '*' : '-');
        printf("\n");
    }
    printf("%ld\n\n", counter);
}

void try(int y)
{
    int x;
    if (y==size)
    {
        counter++;
        if (doshow) show();
    }
    else
    {
        for (x=0; x<size; x++)
        {
            if (vert[x] && dia1[y+x] && dia2[size-1-y+x])
            {
                vert[x] = dia1[y+x] = dia2[size-1-y+x] = FALSE;
                r[y] = x;
                try(y+1);
                vert[x] = dia1[y+x] = dia2[size-1-y+x] = TRUE;
            }
        }
    }
}

main (int argc, char *argv[])
{
    printf("\nQueens - C version - By Marcel van Kervinck - May 1992\n");
    if (argc==2 || argc==3 && !strcmp(argv[1],"-q"))
    {
        doshow = argc==2;
        size = atoi(argv[argc-1]);
        if (0<size && size<=MAXSIZE)
        {
            init();
            time(&starttime);
            try(0);
            time(&endtime);
            printf("Total number of solutions: %ld "
                   "(%ld seconds)\n\n", counter, endtime-starttime);
            return 0;
        }
    }
    printf("Usage:  %s [-q] <size>\n\n", argv[0]);
    return 0;
}
From: Dennis.Matney@f3.n8120.z86.toadnet.org (Dennis Matney)
Newsgroups: rec.puzzles
Subject: Re: "concentration camp"
Date: Fri, 11 Dec 92 06:55:00 cdt
Organization: FidoNet node 86:8120/3 - Sometimes BBS, Portage IN


 To: ST102315@brownvm.brown.edu (Jake)

 Ja> What do you get when you multiply six by nine?

Why 42, natch...



... 6 x 9 = 42 ... (in base 13)
___ Blue Wave/QWK v2.11


--- TosScan 1.00
 * Origin: The Sometimes BBS (219-763-2031) (86:8120/3)
--  
Dennis Matney - via Toadnet/UUCP Gateway 86:86/200
UUCP: ....!wupost!8120!3!Dennis.Matney
INTERNET: Dennis.Matney@f3.n8120.z86.toadnet.org
Newsgroups: rec.puzzles
From: costley@solo.eng.hou.compaq.com (Brett Costley)
Subject: Re: Ask Marilyn
Organization: Compaq Computer Corp.
Distribution: na
Date: Mon, 14 Dec 1992 19:33:32 GMT

In article <92349.120251HERSCH@auvm.american.edu> <HERSCH@auvm.american.edu> writes:
>Does everyone agree that Marilyn vos Savant's answer to the pancake
>probability problem in yesterday's Parade magazine was wrong? The
>answer is really 50-50. Either you've picked the pancake with 2 brown
>sides, in which case the other side is brown, or you've picked the
>pancake with one brown side, in which case the other side is not
>brown. 50-50.
>
>H.
>
>Herschel Browne
>"The" American University

But you are more likely to have picked the pancake with two brown sides 
than with just one brown side giving a better than 50/50 chance that
the other side is also brown.


Newsgroups: rec.puzzles,sci.math
From: chappell@symcom.math.uiuc.edu (Glenn Chappell)
Subject: Re: Circular graph problem
Organization: Math Dept., University of Illinois at Urbana/Champaign
Date: Wed, 16 Dec 1992 23:42:09 GMT

In article <1992Dec16.113500.1157@client24.comlab.ox.ac.uk> mnewman@comlab.ox.ac.uk (Matthew Newman) writes:
>Consider circular graphs with N nodes, such that there is an integer
>at each node. Consider those integers K such that for some graph G
>each integer in [1..k] is the sum of integers at the nodes of some
>connected sub-graph G.
.
.
>and for N=8 I can get K=53 with 1-4-22-7-3-6-2-12-

I believe that should be "K=57".

>The question is, what is the largest K I can get for a given N ?

I have seen a similar problem (probably in M. Gardner's "Mathematical
Games" in Scientific American) involving markings on a ruler. Given an
unmarked stick K units long, how many marks do you need to put on the
stick to be able to measure any integral number of units from 1 to K?

Here, K is your K, the number of marks is your N-1, but the graph isn't
circular, it's "linear", e.g.

        (1)--(3)--(2)

where the numbers correspond to distances between marks, and edges of
the graph correspond to marks on the ruler.

In fact, we can generalize your problem to any simple graph with a
finite number of vertices: Given such a graph G, what is the largest
positive integer K so that we can assign positive integers to the
vertices of K, s.t. each integer in [1..K] is the sum of integers at
the vertices of some connected subgraph of G? (Are we using "subgraph"
correctly here?? My first real course in graph theory doesn't start
for another month :-).

So, for example, if G is the complete graph on N vertices, then K is
2^n-1 (assign 1,2,...,2^(N-1) to the vertices). It is easy enough to see
that if G, H have the same vertex set, while the edge set of G is a
proper subset of the edge set of H, then K (for G) <= K (for H) [maybe
"K" is a not a good letter to use, it being so common is graph theory
already....] Clearly, then, if G has N vertices, then K < 2^N.

				Glenn Chappell  <><
				chappell@math.uiuc.edu
Newsgroups: rec.puzzles
From: wvanweer@wsuaix.csc.wsu.edu (Wayne VanWeerthuizen)
Subject: Paper Folding
Organization: Washington State University
Date: Tue, 15 Dec 92 07:17:49 GMT

      /========/   This is a (unit) cube with striped sides. On the 
     /========//   unseen sides, the lines run parallel the those on
    /========///   the opposite (seen) side.  (Eg. on Top of bottom
    ! ! ! ! !///   stripes run left-right; on front and back, stripes 
    ! ! ! ! !///   run up-down; on left and right, stripes run
    ! ! ! ! !//    front-back.)
    ! ! ! ! !/  
 
   Now everyone should know a unit cube can be made by folding a flat 
  paper of 6 unit-squares.
  
  But, if the paper has parallel stripes on it (ie. stripes run the
  same way in all unit squares.) how can the above cube be made?
  The paper must (before folding) be one continuous flat sheet.
  What is the minimum area required? 
 
-- 
 ---------------------------------------------------------------------------
        Wayne VanWeerthuizen    205 Scott Hall     Pullman, WA  99163
           wvanweer@wsuaix.csc.wsu.edu      (509) 335-1835 (voice)   
 ---------- Only YHVH is God.  Jesus is the Messiah, Son of YHVH. ----------
From: chris@questrel.com (Chris Cole)
Newsgroups: rec.puzzles
Subject: Re: 8-queen problems (revisited?)
Date: 15 Dec 92 18:11:55 GMT
Organization: Questrel, Inc.

In article <Bz9x2H.KDv@umassd.edu> rolls@cis.umassd.edu (Frank Pinto) writes:
>Hello all
>
>If this is old, forgive-me.
>
>Has anyone heard of the 8-queens problem?
>What is the algorithm?

This question is in the rec.puzzles Archive.

To request a copy of the index to the Archive, send a letter to
archive-request@questrel.com
containing the line:
send index

The index will be mailed via return email to the address in your
request's "From:" line.  If you are unsure of this address, and cannot
edit this line, then include in your message BEFORE the first "send" line
the line:

return_address <your_return_email_address>

The Archive has been posted to news.answers.  News.answers is archived in
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Postings are located in the anonymous ftp directory
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		Lyon, Grenoble)

Note that the periodic posting archives on pit-manager.mit.edu are
also accessible via Prospero and WAIS (the database name is "usenet"
on port 210).

From the Archive, the answer is:
********
games/chess/queens.s
********
92.  The following program uses a backtracking algorithm to count positions:

#include <stdio.h>

static int count = 0;

void try(int row, int left, int right) {
   int poss, place;
   if (row == 0xFF) ++count;
   else {
      poss = ~(row|left|right) & 0xFF;
      while (poss != 0) {
         place = poss & -poss;
         try(row|place, (left|place)<<1, (right|place)>>1);
         poss &= ~place;
         }
      }
   }

void main() {   
   try(0,0,0);
   printf("There are %d solutions.\n", count);
   }
--
Tony Lezard IS tony@mantis.co.uk OR tony%mantis.co.uk@uknet.ac.uk
OR EVEN arl10@phx.cam.ac.uk if all else fails.
Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: Monte Hall II <SPOILER>
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Fri, 18 Dec 1992 17:27:17 GMT

In article <1992Dec18.153808.4204@nntp.nta.no> stein@hal.nta.no (Stein Kulseth) writes:
>|> 1.  If doors A, B, and C (door A leftmost, door C rightmost) have the
>|> probabilities a, b, c with a+b+c=1 of having the good prize behind
>|> them, what is your best strategy?

<SPOILER>

[...]
>|> 2.  If when Monte is faced with two empty doors instead of choosing
>|> uniformly he chooses the leftmost door with probability p, the
>|> rightmost with probability q and p+q=1, what is your best strategy
>|> now?
[...]
>|> 3.  What if both conditions hold?
>
>choose door with least probability(again, say a), and switch.
>
>And again the actual winning probabilities when switching may vary,
>they are c/(c+ap) or b/(b+aq), but the average probability will
>equal b+c.

Not only that but any other strategy is worse:  If your strategy tells
you always to stay with door A the best you can do is just probability a;
whereas if you stay or switch depending on which door Monty opens,
you get  pa + b  or  qa + c  (depending on which of two cases you use).
In no case can you order the probabilities a,b,c to make any of these
results greater than the sum of the two largest of a,b,c.

(This completes the analysis from my last message.)

-- David Karr (karr@cs.cornell.edu)
From: clong@romulus.rutgers.edu (Chris Long)
Newsgroups: rec.puzzles
Subject: Monte Hall II
Date: 18 Dec 92 11:50:20 GMT
Organization: Rutgers Univ., New Brunswick, N.J.


There are three doors, behind one of which is a good prize and behind
the other two are booby prizes (each door equally likely to have the
good prize).  You pick a door, and then of the two remaining doors
Monte picks one containing a booby prize and shows it to you, and
offers you the opportunity to switch.  We all know by now that if Monte
choose uniformly at random if there is more than one door with a booby
prize to choose from you should always switch and your probability of
winning is 2/3.

1.  If doors A, B, and C (door A leftmost, door C rightmost) have the
probabilities a, b, c with a+b+c=1 of having the good prize behind
them, what is your best strategy?

2.  If when Monte is faced with two empty doors instead of choosing
uniformly he chooses the leftmost door with probability p, the
rightmost with probability q and p+q=1, what is your best strategy
now?

3.  What if both conditions hold?
-- 
Chris Long, 265 Old York Rd., Bridgewater, NJ  08807-2618

"S.B., with an I.Q. of 161, failed to complete his course of study,
running away instead with his professor's wife."
H. J. Eysenck, _Know Your Own I.Q._
Newsgroups: rec.puzzles
From: hallyb@globbo.enet.dec.com (John Hallyburton)
Subject: Re: Monty Hall Puzzle RIP
Summary: Read this.  Act rationally.  Now what?
Organization: Twin Peaks Municipal Software Works
Date: Fri, 18 Dec 1992 19:45:11 GMT


<SHERE@SLACVM.SLAC.STANFORD.EDU> writes...
>Alright, I've had just about all bickering I can stand on this
>Monty Hall drivel and I'm going to put it to bed.
 [...sophisticated but erroneous analysis deleted...]
>Flames and arguments expected, oh well.

Suppose instead of 3 doors I have arranged (laid out) 52 cards, face down.
You select one, keeping it face down.  I turn over 50 cards to show that
none of them is the Jack of Diamonds.  You now have the choice of keeping 
your selected card or switching to the one other unknown card.

Same basic problem, but more extreme - right?  Think you have a 50-50 shot
at having originally chosen the Jack of Diamonds?  Want to put money on it?
--
						"The Smart Money was on Goliath"
						John Hallyburton 43 55'N 71 45'W
From: huhr@formentera.uni-paderborn.de (Holger Uhr)
Newsgroups: rec.puzzles
Subject: Re: Christmas riddle
Date: 15 Dec 92 16:22:57 GMT
Organization: Universitaet Paderborn, Germany
NNTP-Posting-Host: formentera.uni-paderborn.de
X-Newsreader: TIN [version 1.1 PL7]

record@force.ssd.lmsc.lockheed.com wrote:
: In article <3305@nlsun1.oracle.nl>, asymes@nl.oracle.com (Andy Symes) writes:
: >In article <1992Dec9.083159@gruc19.nor.chevron.com>, hjiwa@gruc19.nor.chevron.com (Santa Claus Ho Ho Ho) writes:
: >|> What's the perfect Christmas gift for the man who has everything ???
: >|> What's the perfect Christmas gift for the woman who has everything ???
: >|> 
: >|> HO, HO, HO
: >
: A box to put it in!  Oh, maybe not.  I guess the man already has that.
: 
Neither of them can have a box to put everything in,
because you can't put a box into itself.

-- 
|   Holger Uhr  ***   Engernweg 32  ***  W-4790 Paderborn   ***   Germany   |
|E-Mail: huhr@uni-paderborn.de                            Phone: 05251-58950|
|How to become immortal: Read this signature tomorrow and follow its advice.|
Date: Thursday, 17 Dec 1992 04:50:47 EST
From: <RVESTERM@vma.cc.nd.edu>
Newsgroups: rec.puzzles
Subject: Re: Big Numbers
Distribution: world
 <92351.123112RVESTERM@vma.cc.nd.edu>,<92351.132601RVESTERM@vma.cc.nd.edu>
 <1gp8ttINNik@gap.caltech.edu>

yes, yes.  obviously, throw the word "uniquely" where appropriate.
"the positive integers" doesn't uniquely describe a number.

bob vesterman.
From: bcash@crchh410.BNR.CA (Brian Cash)
Subject: Re: Serial representation of arbitrary data
Reply-To: bcash@crchh410.BNR.CA (Brian Cash)
Organization: Bell-Northern Research Ltd.
Date: Tue, 15 Dec 1992 03:30:40 GMT

From: warwick@cs.uq.oz.au (Warwick Allison)
The problem is this:

	Represent a sequence of arbitrary length binary values.

Do it in as short a number of bits as possible.

For example, one solution is to represent

	10010010
	1001
        1010010
	11
	110111
==================================================
How about 4 bit binary coded length with an extra "and" bit.

Like this:

1000 0 10010010 0100 0 1001 0111 0 1010010 0010 0 11 0110 0 110111
  L  +    V       L  +  V     L   +   V      L  +  V   L  +   V

the "and" bit (the lone zeros above) would concat. the 4-bit segments
for lengths longer than 16 bits (code 0000 to equal 16 in a single 
4-bit length but keep it as zero for longer lengths).

ex.  10110101011010010101101110010101
would be 

0001 1 0000 0 10110101011010010101101110010101
  L  +   L  +       V


and 1010001011010101 
would be

0000 0 1010001011010101 
  L  +      V


How's that?

Brian /-|-\ (I didn't recheck bit lengths or binary values, so no flames
on those
             please!)
Newsgroups: rec.puzzles