Material
added 30 September 2003
Robert
Abbott: "I've been working on a new series of mazes, called "Eyeball Mazes."
They are at this address: http://www.logicmazes.com/eyeball.html."
Of course, just like anything Robert Abbott does, these are very cool,
and are highly interesting to study. So, take a gander at Eyeball Mazes as the
Puzzle of the Week. (It's one of the year's most popular puzzles.
Solvers.)
Patrick Hamlyn and Andrew Clarke are the world's
top polyformists, but they were perplexed by a seemingly simple (to
them) task -- making a rectangle with the 227 balanced, non-holey
octominoes. Andrew Clarke found the below non-solution.
Patrick found over 77 million different ways to place 226 of the
pieces. Was there a way to prove a solution is impossible? Andrew
Clarke: "Classify the balanced pieces as even if they can be divided up
into dominoes with an even number in each direction and odd if they are
1-3 in direction. For those which cannot be divided into dominoes add a
domino so that the new shape can be divided into dominoes. In this
division remove one domino that is in the same orientation as the added
domino and then classify as before. If my count is correct there is an
odd number of odd pieces and, since any rectangle will be even, no
rectangle can be made." Beautiful.
Material
added 26 September 2003
Eric Solomon: The Telescope maze offers a
new puzzle of some interest. It is rather similar in some ways to
Andrea's tilt mazes, and the first 15 levels are fairly easy. The
mazes are based on 'telescopes' which push and pull a ball across an
orthogonal grid with the object of getting it into a hole.
I think there must be an efficient algorithm based on the minimum
and maximum extension points of the telescopes for solving any maze of
this type.
Karl Scherer has made a grand page about WireWorld.
Geoff Morley sent me good data for Mrs. Perkins Quilts.
Material
added 15 September 2003
Square-free numbers
have no repeated factors. One thing I learned from Steven Finch's
new book, Mathematical
Constants, is that the probability that a number is squarefree is
6/Pi2 ~ .607927.
Months ago, I pointed out my Neglected Gaussian
page to Steven. That had a nice result on the GCD of two random
numbers. After looking through his book, he wrote back to me, and
asked if I saw the last page ... an extra page after the index that
talks about the Collins-Johnson result. He liked the result so
much that added it the day before he sent the book off to the printer.
It's a great reference book, and is loaded with lots of the wonders in
mathematics.
ListPlot[FoldList[Plus, 0, Table[If[SquareFreeQ[n], 1,
0], {n, 1, 100000}]]/Range[100001], PlotStyle -> PointSize[.001],
AspectRatio -> 1/7]
The above is the percentage of square-free
numbers for regular digits. The Collins-Johnson result implies
that the probability that a random Gaussian integer is squarefree is
6/Pi2/Catalan
~ .663701. Here's a plot for the 100000 smallest (by Abs)
Gaussian Integers.
ComplexCircle = Take[Sort[Flatten[Table[a + b I, {a,
-200, 200}, {b, -200, 200}]], Abs[#1] <; Abs[#2] &], 100000];
ListPlot[FoldList[Plus, 0, Table[If[Max[Map[Last,
FactorInteger[ComplexCircle[[n]]]]] == 1, 1, 0], {n, 1,
100000}]]/Range[100001], PlotStyle -> PointSize[.001], AspectRatio
-> 1/7]
The publishers of The DaVinci Code have started up a themed puzzle set.
Codefun.com
has some beautiful combinations of genetic code (with 20 amino acids)
and the icosahedron.
If A+B = REACQUAINTS, and B+C =
RESUSCITATING, what are A, B, and C, if A&C are related? Solved by Bryce Herdt. A similar
piece of
wordplay: {flea, Dr. Seuss, cabin, duster} and {self-assured, business
card, disturbance, defaulters}.
The preorder for The Fool's Errand sequel
is somewhere on Cliff Johnson's
site.
I've been looking at Turing Machines again
recently, and I've decided it's time to settle Sigma(5). The
last time it was seriously attacked was in 1989, on a 33MHz
machine. Another solvable problem might be to resolve the 11
unresolved cases for Paterson's Worms.
If you would like to assist with either of these, let me know.
On 27 September, I'll be attending a small puzzle party in the San Francisco area. I'm allowed to invite a few people, so write to me if you'd be interested in attending.
Robert Wainwright sent me some solutions
he found in 1993 for the Mrs. Perkin's Quilt problem.
{29,24,24,16,13,13,7,6,6,5,5,4,4,3,3,1},
{52,39,39,28,24,24,15,9,8,7,7,6,6,5,4,4,1,1},
{91,63,63,52,39,39,24,24,15,9,8,7,7,6,6,5,4,4,1,1}. Here is the
solution to the side-154 square, in 20 squares. Can it be beaten
in a non-trivial way?
Robert Wainwright's solution for 20 squares.
Material
added 11 September 2003
I did some experiments with 51-star
flags. You can see my favorite effort here.
You also might like a tighter packing.
Andrea Gilbert has made a number of
additions to clickmazes.com,
including 4D mazes, parallel-universe mazes, and counter-step mazes.
Material
added 5 September 2003
Erich Friedman's Math
Magic this month discusses one of my favorite unsolved problems -- pairwise
touching polyforms. I'd love for someone to give a definitive
answer for the 4x4x4 box, or the 2x5x5. If you can solve or extend
anything there, please send Erich your findings. What is the smallest
box for 9?
Gordon Bower -- "You mentioned liking
close approximations on this week's Mathpuzzle page update (and have
posted a variety of them in the past.) This, and the diophantine
inequality puzzle of the week before, reminded me of a fun little
exercise I indulged in in the spring of 2000. The question of the week
for the now-defunct student math club, of which I was an advisor, was
"is there a power of 2 that begins with ten sevens in a row?" (Yes,
there are infinitely many of them beginning with any finite string of
digits you want.) But no-one bothered to try to actually FIND such a
number. The next week I had to give a short presentation explaining how
I proved that 2^40193336864 = 7.777777777996 x 10^12099400021. (No harm
in me revealing the answer, since the fun is in discovering the method
of finding the solution for any sequence of digits you want.) It makes
a fun party game (if you are a sufficiently warped-minded
mathematician) to find an n such that 2^n starts with your girlfriend's
birthdate or suchlike and surprise her with it. Must warn you that mine
was not overly impressed by it, though." [Can you figure out the
technique? Answers and Solvers. Bodo
Zinser sent a very nice DOS program]
A long unsolved problem, due to Ramanujan,
involves the parity
of partitions, in the partition
function. Here's a picture of it, using the Mathematica
code ( MultipleListPlot[{FoldList[Plus, 0,
Table[2Mod[PartitionsP[n], 2] - 1, {n, 1, 10000}]], FoldList[Plus, 0,
Table[Mod[PartitionsP[n], 3] - 1, {n, 1, 10000}]], FoldList[Plus, 0,
Table[Mod[PartitionsP[n], 4] - 1.5, {n, 1, 10000}]]}, PlotJoined
-> True, SymbolShape -> {None, None,, None}, PlotStyle -> {Hue[.01], Hue[.35], Hue[.65]}, AspectRatio ->
1/5]; ). For even/odd parity, in red, the plot shifts up one for
odd, and down one for even. The unsolved part -- is there a
simple formula for Mod[PartitionP[n], 2]? At the moment, the value of
Mod[PartitionP[10^100], 2] seems unknowable.
Running totals of PartitionP parity, in Mathematica 5,
then processed
through Irfanview.
Jorge Luis
Mireles and Micheal
Reid invite people to submit solutions and improvements to their
polyform
equivalency charts.
A photorealistic painting of a puzzle can
be seen at the Steve
Mills gallery.
Puzzle hunts have a number of good sites. You might enjoy the Golden Tickets at T-Hunts.com.
Material
added 25 August 2003
I picked up the Futurama
Season 2 DVD (I'm
a fan of the show), and I happened upon the following dialog between
Bender and a robot named Flexo.
Bender: Hey,
brobot, what's your serial number? Flexo: 3370318
Bender: No
way! Mine's 2716057!
(Everyone laughs)
Fry: I don't get
it. Bender and Flexo:
We're both expressible as the sum of two cubes! Wahoo!
David X Cohen (audio
commentary): I invite the zealous viewer to check that
claim. It is true! There's a trick to it, the numbers are
all integers, but it's a little tricky.
That sounds like a great little challenge. As a warning,
if you search for the numbers, you'll find the answer. If you'd
like to try trickier numbers, here a a few more numbers that are the
sum of two cubes: 6669, 7222, 119041, 2716343, 2741256, 3370087,
6017193, 6742008,
9016488, 16776487, 24375176.
Can you figure out all the
tricks? Answer and Solvers.
On a similar but unrelated name, beautiful
graphical images can be made with Fractorama,
which gives code samples for making fractal variations of many types.
Jerry Slocum, with the help of an
international team, has tracked down much origin and history for The Tangram Book. A
precursor was the Sei Shonagon's Wisdom Plates, shown below (the pieces
make a square in two different ways.) The Tangram was invented between
1796 and 1802 in China by Yang-cho-chu-shih. He published the
book Ch'i ch'iao t'u
(Pictures using
seven clever pieces). The first European publication of Tangrams
was in 1817.
The word Tangram itself was coined by Dr. Thomas Hill in 1848 for his
book Geometrical Puzzles for the
Young. He became the president of Harvard in 1862, and
also invented the game Halma. Jerry tracked down the sets owned
by Poe, Napoleon, and others. Martin Gardner: "This will surely
be the classic reference on
the topic for many decades to come."
A new large prime, 1176694131072+1,
has been discovered by Daniel Heuer. This Generalized Fermat prime has
795695 digits, and is currently number 5 on
the list.
The continually updated list is maintained by Chris Caldwell at The Largest
Known Primes page.
Andreas Gammel has long been working on
the Teabag
Problem -- "What is the maximum volume that a teabag can
hold?" You can look at Andreas' page,
which describes the Inflatulator.
Michael W Ecker has recently updated his
page for his magazine, Recreational &
Educational Computing. He's continually published it since
1986, and it has long been associated with the Journal of Recreational
Mathematics.
Karl Scherer, the master creator for Zillions of Games,
has put together a WireWorld
explorer.
Larry Brash's Anagrammy page is well
worth a look if you haven't seen it. I particularly liked an
extremely apt anagram found in June by David Bourke ... ?? ~
RECONSIDERABLE. (Answer) Another
beauty, by Joe Fathallah, is "Rats and
mice ~ in cat's dream."
Andrew Glassner
has a very nice page, one part of it is the series of Graphics Gems,
which has been made available for free.
I like close approximations. One I
found recently is 95^(1/12) ~ 19/13. Comments.
Previous Puzzles of the week are here.
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