50. Anniversary
ContestTo our call for a nice match statement we received many responses.
Many thanks for all the great problems we have got.
Therefore, we thought we let the solver choose.
We hope that we have not forgotten anyone. Else send an email.
This is probably our last competition.
We start with the problem of Kate Jones because we got the following mail "Attached is a "50" puzzle I designed some time ago, one copy of which we will be happy to give as a prize for your winner. It is a $45 value and contains all the polyominoes 1 through 5. It has many solutions and is based on our Poly-5 puzzle, www.gamepuzzles.com/polycub2.htm#P5"
We thank Kate and Kadon Enterprises for the prize.
Problem of
Kate Jones.
Fit all the polyominoes from 1 through 5 into the grid within the
number 50, so that no two of the tetrominoes, trominoes, the domino and
monomino touch each other, not even at the corners. It's not difficult
but very pretty!
Kate:"I would like to know whether it was solved by a computer or by
a human!"
To help you with solving we made
an
excel file
and also
a geogebra applet.
o.d.m@fulladsl.be
Flat Poly calculated that there are many different solutions.
You can download FlatPoly at
http://home.wxs.nl/~avdw3b/aad.html
The idea of our 34th competition is also by Kate Jones.
Problem of Bob Harris
(A) Choose a pentomino and make 13 copies of it, colored yellow, and choose a tetromino, and make 4 copies of it, colored black.
(B) Place the pieces so that a 7x7 square is covered to a "depth" of 1. The yellow pieces are "positives" and count as +1; the black pieces are "negatives" and count as -1. Pieces may be placed on top of each other and wherever this occurs a black cell will cancel a yellow one. For example, a cell covered by a single yellow piece has a depth of 1, as will have a cell covered by two yellows and one black piece.
Pieces needn't stay within the 7x7 square, but any cell outside the square must have a depth of zero (which usually means that it is covered by a single yellow and a single black piece).
Bob's motivation for this problem: "I've been investigating polycube puzzles made from several copies of two (or three) pieces. One such puzzle has been on my website for years:
http://www.bumblebeagle.org/polycubes/7andF.solution.gif
Now, if I want to have a heptacube and a pentacube fill a 4x4x4, how many copies of each would I need? This is solved by 7a + 5b = 64, looking for positive integer solutions. Integer solutions are a=7-5k b=3+7k (for some integer k), and restricting this to positive integers the only solutions are a,b=7,3 or a,b=2,10.
Looking at the same sort of thing in 2D, suppose I want to pack a 7x7 with pentominoes and tetrominoes. Then I have 5a+4b = 49 with integer solutions (a,b)=(9-4k,1+5k). Solutions in positive integers are (9,1), (5,6), (1,11). But there are also solutions with negative integers. In this example is (a,b) = (13,-4), or 13 pentominoes and negative 4 tetrominoes.
We choose the L-pentomino and the L-tetromino.
Use the mouse pointer to see the example
Can
you do it with
the
other
pentominoes?"
Problem of
Martin Watson.
The idea of our
21th
competition is also of Martin Watson.
Place the 12 pentominoes in a 12x12 square so
that no two pentominoes
touch each other, not even corner to corner.
Count how many edges of the square (12x4=48 in total) are touched by the
pentominoes. What are the maximum and minimum values? In the example
below this value is 32 (1+1+4+3 = 9 units at the top edge, and so on.)
Notice that a pentomino cell which covers a corner of the square counts
for 2 edges.
Problem of Peter Jeuken
Peter created this pentomino puzzle especially for us.
12 sets of 12 pentominoes and 12 sets of 5 tetrominoes together form a
32x32 square where the four corners (4x4 squares) are missing.
If you prefer to see it larger, then click on the picture.
Cut the figure into 12 pieces so that each piece contains exactly 12
different pentominoes and 5 different tetrominoes.
Only cuts along the grid lines are allowed. Pentominoes and tetrominoes
must not be cut in two.
If you find more than one solution, then submit the solution with the
smallest total perimeter of the 12 pieces.
Problem of Aad Thoen : Pentomino-
Place all pentominoes as islands in a sea such that the area of this sea should be as small as possible. The pentominoes must not touch each other, not even at the corners.
In our example the sea area is 122 units.
A variant of this problem is us potpourriproblem 11 (Dutch) where the sea has to be a rectangle.
Another problem is that we allow the pentominoes to be peninsulas.
Problem of Aad van de Wetering
We would like to thank Aad van de Wetering very much for doing a great job in establishing our pentomino site.
1 + 1 = 3
This problem was created by our friend Peter Torbijn to whom we dedicate the whole competition.
Enclose a set of pentominoes with two other sets. Make the area of the circumscribed rectangle as small as possible.
In our example the rectangle is 16 x 14.
You can find another solution of Michael Keller on http://www.solitairelaboratory.com/pentmino.html.
This solution is a rectangle of 18 x 13. One can do much better.
Pento-Product-Xudoku
On request, because we know we have many visitors who love sudoku, Aad van de Wetering made for us this great hint free xudoku.
Put the numbers from 1 to 8 in each row, in each column and in the two diagonals.
The products of the numbers in each pentomino are as follows:
L:420, X:96, U:1080, V:11760, Z:576, Y:336, F:1440, W:84, P:720, I:1120, T:384 , N:1680
Would you like a hint? Mail to o.d.m@fulladsl.be
'50. Fifty-Fifty' Competition by Edo Timmermans
We asked Edo to create a competition for our 50th Jubilee because we consider him to be one of our most creative problem creators. (see the Pentomino postcard competition and the Isle of count Penteviticole)
His competition consists of 2 parts. Take a red 10x10-square with a diagonal green square inside having 50% of the surface of the 10x10-square. The pentominoes are to be placed within the big square in such a way that they are all partially red and partially green.
In the first 50% of the competition, you are invited to search for solutions where 10 pentiminoes are 50% red and 50% green. Also, all pentominoes together must be 50% red and 50% green.
The examples show that this is not very difficult when the remaining 2 pentominoes are allowed to be 10% red or 10 % green. That is why these 2 pentominoes must minimally be 20% red as well as minimally 20% green, yet the closer to 50% the better, of course.
In the second 50% of the competition you are invited to look for a symmetrical solution. The demand that 10 pentominoes must be half red and half green is dropped, but each pentomino must still be partially red and partially green.
The example is not a valid solution, only because the W-pentomino is entirely green.
Computer programmers are invited to look for symmetrical solutions where at least 50% of the pentominoes are 50% red and 50% green.
Pentorigami-Problem of Bob Henderson
Cut 12 pentominos from stiff paper of different colors.
For your convenience, we put 6 copies of each pentomino on a sheet.
So you get different sets of pentominoes
F I L N P T U V W X Y Z
Cut 1 of the 4 grid lines that meet at the center of P.
There are eight pentominoes which can be folded into an open cube (a cube without top face). They are F,L,N,T,W,X,Y and Z.
Take 6 of these 8 open cubes and join them at the edges of their openings. They will look like 6 cubes attached to the faces of a seventh (hidden) cube.
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Now take the I,P,U and V pentominoes together with the two remaining
pentominoes. Fold and join these 6 pieces to build the same shape as you
did before with the other 6 pentominoes.
To build the second shape, all six pentominoes may be folded arbitrarily
along the grid lines, they need not form open cubes.
3D-Problem of Michael Dowle
Cover a cube of edge length √10 with a pentomino set.
During several
years
students of KSO
Glorieux Ronse (Belgium) cover
a
cube
for
their
work
of
Pythagoras.
http://www.pentomino.wirisonline.net/versnijdenopkubus.html
http://home.planet.nl/~avdw3b/kubuspoly.html
C. J. Bouwkamp wrote in 1998 a book (which we have received from him) "Tiling the surface of the cube by 12 identical pentominoes"
Also on the following sites you can find coverings of a cube.
http://www.ericharshbarger.org/pentominoes/article_06.html
http://www.tzingaro.com/artelectric/pentominoes06.html
http://home.educities.edu.tw/proteon/note131.htm
http://www.iread.it/lz/pag3_eng.html
Cover two congruent triangular dipyramids of shorter edge length √10 and longer edge length√20 with a pentomino set.
This we found nowhere else.
And now Michael’s beautiful contest:
We should now be able to visualise easily the folding of pentominoes around the surfaces of cubes and triangular dipyramids. This problem requires us to find a pentomino arrangement which covers the surface of a cube (edge length √10) and which may be divided into halves each of which covers the surface of a triangular dipyramid. (shorter edge length √10/longer edge length√20). No repositioning of the pentominoes is allowed. The separation of the cube into two and formation of two triangular dipyramids is illustrated here.
Pentominoes with a rope
Our inspiration comes from the great book "Denkwaar" of Jaap Klauwen (in Dutch).
Thanks to Aad Thoen for sending us this book.
The problem: Place all 12 pentominoes on a grid such that the vertices of the pentominoes coincide with grid points, each pentomino touches another one along at least one side unit, and the 12 pentominoes form a single connected shape. Put a rope around the shape and pull it tight.
Place the pentominoes such that the circumference (the length of the rope) is a maximum.
For the calculations of the lengths of the rope parts use the Pythagorean theorem.
In the above example, the circumference of the rope (green line) is 71.98 (rounded to two decimals) .
This value shall be maximized.
Problem of Alexandre Owen Muniz
Alexandre sent us an email with a nice problem but we did not understand it well. We ask for advice to Helmut Postl and we got of him a very nice explanation.
The transformation of a polyomino means that you choose an arbitrary single square of the polyomino and move it to some other place to make a new polyomino. For example, transform an L-tetromino into a T-tetromino:
Now you can transform the T-tetromino again to some other tetromino, and so on. In this way you can build a tetromino chain. If this chain has the special property that you use every tetromino exactly once and end up with the starting tetromino, you have built a cycle. In this case it does not matter which polyomino is used as the starting piece since a cycle has no beginning and no end.
In the following example, the squares of the tetrominoes all have individual colours so to better recognize where a single square has been moved:
This cycle consists of all five tetrominoes in the sequence (L, N, Q, T, I), and you can transform the tetrominoes in this way forever and you will stay in this cycle.
Remark: The final L has a different colouring than the starting L, but this is irrelevant. The colouring was just to visualize the movements of the squares, and the placement of the single squares is not important, just the polyomino as a whole is concerned.
In a cycle, the final polyomino must have the same orientation as the starting polyomino, i.e. it must not be rotated or reflected. But it may be translated! In the example above, there is no translation. So running the transformations through this cycle again and again will keep the tetrominoes at the same place.
In the following example, the final piece is translated by the vector (2,1):
So running through this cycle again and again will move the polyominoes further and further away.
The speed of the movement is measured by the vector by which a polyomino is translated within one cycle. In the example above this is (2,1). The speed now is the sum of the absolute values of the two coordinates, this is 3. This measure is more plausible than the usual Euclidean distance (which is sqrt(5) in this example) because it is more natural to regard only horizontal and vertical shifts instead of additionally diagonal ones. Alexandre calls it the taxicab distance, because it describes the distance which a taxicab has to go along the streets (grid lines). I like this name, it shows immediately what is meant.
The speed of the cycle in the first example is of course 0.
The problem of Alexandre
a) Take the 12 pentominoes and build a cycle where the speed is as big as possible.
b) Since this task may be too easy, but on the other hand, the 35 hexominoes may be too difficult, take the 18 one-sided pentominoes to build a cycle (again with maximal speed). Unlike as before, the pentominoes here must not be flipped over.
Remark: Do you know Conway’s Game of Life? The rules are rather different, and they are deterministic for a given pattern. But these polyomino cycles resemble the movement of a so called glider, this is a Life-pattern consisting of 5 cells which after 4 generations reappears shifted by the vector (1,1).
For
this
anniversary-contest
we have many prices.
With each contest you can win a 'PENGUINS
on ice'.
More
information about
this
fun
game
on
the site of Raf Peeters,
the
maker
of
this
game.
The
game is of
Smart.
Kate Jones and
Kadon Enterprises
give
a
50-puzzle
for the
winner
of her problem.
This time
Texas
Instruments gives us two TI-Nspire.
With many thanks to:Texas Instruments who sponsored each time a TI-Nspire.
Eureka bvba gives 16 Pentadivio puzzle.
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Vierkant voor wiskunde gives three abacus.
Kangaroo
gives 25 'vouw4vlakken'
Publisher 'Plantyn' gives 10 sticks of 1GB
Each contestant earns eternal fame.
Whomever send in a solution is free to request a
pentominoset.
If you do wish to receive one, please let us know.
The draw of the winners will take place on 31
December
2011
Send your solution to:
o.d.m@fulladsl.be