Pentominoes Numbers
Can you better?
Mail a 'number' to
o.d.m@fulladsl.be
Maybe we send you a chocolate 'number'
George Sicherman: "5 of your digits lack unique solutions.
Permit me to remedy this."
Large
numbers in Sudoku and pentominos
We asked Helmut if he could think of something fun
with "70" (for the anniversary of Aad van de Wetering November 16, 2014) and
we got a wonderful sudoku connected with a lot pentominoproblems.
The following Sudoku has the additional property that the values in all
coloured cells sum up to 70.
Correct solutions earn eternal fame.
Send your solution to:
o.d.m@fulladsl.be
Name  Country 
Helmut Postl  Austria 
Aad Thoen  The Netherlands 
Aad van de Wetering  The Netherlands 
Odette De Meulemeester  Belgium 
Sander Waalboer  The Netherlands 
George Sicherman  USA 
Nico Looije  The Netherlands 
Bob Henderson  USA 
Matthijs Coster  The Netherlands 
Peter Jeuken  The Netherlands 
George Sicherman:"I solved them by computer. Doing it by hand would take
too much trial and error, and I already have enough trial and error in my
life!"
Bob Henderson: "I found the following results by modifying my Sudoku
solver to run until the marked cells had the correct sums. I did not search
for any more solutions: I have no doubt that Helmut designed them to have
only one!"
Furthermore, there are 60 white cells, so they can be
covered by 12 pentominoes. Unfortunately, it is not possible to use each
pentomino exactly once.
But there are some interesting possibilities:
1)Choose 4 pentominoes and use each one three times. The choice is unique,
and there are
two trivially connected solutions.
2) The biggest number of different pentominoes used is 9.
There are several possibilities. One among them is unique: it is without the
N pentomino.
Here is the solution.
3)The smallest number of pentominoes used is 3. Each of
them can be used in an arbitrary number of copies.
The solution is unique.
4) The biggest number of copies of a single pentomino is 8. The remaining 4
pieces need not be different. There are two trivially connected solutions.
Another idea is to glue the two horizontal edges together so to make a
cylinder. Unfortunately, there are still no solutions. The same holds when
the two vertical edges are glued together. But combining both (so to make a
torus) yields 105 solutions. There are two unique ones among them: one with
the fewest (4) and one with the most (9) pentominoes wrapping around.


Helmut send a
text file with all the solutions.
Let W = number of wrapped pentominoes, S = number of solutions for this
case. Then we have the following possibilities:
(W,S) = (4,5), (5,14), (6,33), (7,42), (8,10), (9,1).
You can find the solutions as above: Search for “4A” resp. “9A” to get the
solutions with the fewest resp. most wrapping pentominoes.
Helmut made a similiar
for 64. (October 10, 2014)
The following Sudoku has the additional property that the values in all
coloured cells sum up to 64.
Correct solutions earn eternal fame.
Send your solution to:
o.d.m@fulladsl.be
Name  Country 
Helmut Postl  Austria 
Aad Thoen  The Netherlands 
Aad van de Wetering  The Netherlands 
Odette De Meulemeester  Belgium 
Sander Waalboer  The Netherlands 
George Sicherman  USA 
Nico Looije  The Netherlands 
Bob Henderson  USA 
Martin Friedeman  The Netherlands 
Aad van de Wetering: "Voor de
64 (ook een idee) had ik geen software, met Xudoku alle oplossingen gemaakt
en Excel laten uitrekenen welke som 64 had voor de gevraagde vakjes. Dat was
er één, alle andere hadden een grotere som."
Sander Waalboer zal daar wel software voor hebben want op 10 min stuurde
hij zowel een oplossing voor "64" als voor "70" terug
Nico Looije: "Dit is een leuk idee. Ik moest een minimale programmatische
aanpassing doen om ze op te lossen"
Furthermore, there are 60 white cells (ignoring the isolated cell in the
digit 6), so they can be covered by 12 pentominoes. Unfortunately, it is not
possible to use each pentomino exactly once.
But there are some interesting possibilities:
1)Choose 6 pentominoes and use each one twice. There are many
possibilities. Try a set without the L or the P pentomino.
Without the Lpentomino: 2
combinations.
INPTWY: 3
solutions
INPUVY: 1
solution
Without the Ppentomino : 2 combinations.
ILNTVY: 2 solutions
ILNTVW: 1 solution
2) Choose 4 pentominoes and use each one three times. The choice is unique,
and there are
three trivially connected solutions .
3)The biggest number of different pentominoes used is 10. There are many
possibilities. Three among them are unique:
one without the L pentomino
one without the N
one with the X
4) The smallest number of pentominoes used is 4. There are many
possibilities.:
One among them is unique: it uses the Z pentomino.
5)The biggest number of copies of a single pentomino is 8. The remaining 4
pieces need not be different. There are three essentially different
possibilities, but all with the same 8 copy pentomino.
There are
12 solutions.
Another idea is to glue the two horizontal and/or vertical edges together so
to make a cylinder or a torus.
W = number of wrapped pentominoes, S =
number of solutions for the corresponding case.
1) Connect the vertical edges (vertical cylinder): There are 22 solutions.
(W,S) = (2,3), (3,10), (4,7), (5,2).
2) Connect the horizontal edges (horizontal cylinder): There are 12
solutions.
(W,S) = (3,3), (4,9).
3) Connect the horizontal and vertical edges (torus): There are
6577 solutions (including the 34 cylinders).
(W,S) = (2,16), (3,179), (4,852), (5,2029), (6,1926), (7,1087), (8,434),
(9,51), (10,3).


Helmut added the wrap information to
the first line of each solution. Here is an example: "2768 HVH.H.V.BB.. 4V
5H 2B 7A"
2768 is the solution number.
The next string consists of 12 characters, one for each pentomino in
alphabetical order (FILNPTUVWXYZ).
‘V’ = only vertical wrap, ‘H’ = only horizontal wrap, ‘B’ = both wraps, ‘.’
= no wrap.
The following four numbers are the numbers of pentominoes with the indicated
wrap kind: ‘V’ = vertical, ‘H’ = horizontal, ‘B’ = both, ‘A’ = any. V
includes also B (since a pentomino with both wraps does also wrap
vertically), H includes also B (for the same reason), and A can be computed
as A = V + H – B.
The given example means that I and U wrap only vertically, F, L and P wrap
only horizontally, and W and X wrap in both ways. There are 4 pentominoes
that wrap vertically (IUWX), 5 horizontally (FLPWX), 2 in both ways (WX) and
7 anyway (FILPUWX).
Large numbers in Sudoku
The following sudoku puzzles are
not with pentominos but they are funny. The Sudoku can be used at one or
other occasion.
Correct solutions earn eternal fame.
Send your solution to:
o.d.m@fulladsl.be









Ontvangen van Aad van de Wetering op 10 oktober 2017 
In the splendid book "Exotische
sudoku's" by Aad Thoen and Aad van de Wetering is more: 19, 24, 32, 37,44, 50, 52, 64, 66, 69, 83
and 88.
You can order this book e.g. at Standaard Boekhandel.
Here are two
sudoku puzzles written by Aad van de Wetering to put someone in the
spotlight .


For "65"
Aad made this pento sudoku with 65
cells.
Three 5x5 concatenated sudoku's a1e5, e1i5, i1m5
Each pentomino must have the numbers 1, 2,3,4 and 5
Solvers: Aad van de Wetering  The Netherlands
Odette De Meulemeester  Belgium
Martin Friedeman  The Netherlands
Nico Looije  The Netherlands
Bob Henderson  USA
George Sicherman  USA
Aad Thoen  The Netherlands