Cover Chessboard with Domino Tiles
I have a question about a mathematical riddle which I already solved but still looking for a shorter/simpler solution:
Following problem: We consider a standard $8 times 8$ chessboard and we cover it (completely!) with dominos of size $2 times 1$ (therefore every domino tile cover exactly $2$ fields).
The question is if we can find a cover such that there doesn't exist a $2 times 2$ subsquare which is covered exactly by two domino tiles or in other words in the cover there don't occure two "direct" neighbour domino tiles from following shape:
I have it already solved in following way: I claim that such covering isn't possible.
Argue via contradiction: Assume that it's possible. Consider the $2 times 2$ squares of the chess board and consider the partial cover of directly neighboured $2 times 2$ squares. If a cover as above really exist then up to symmetry on the common edge of the two neighboured squares there could only occure two following cases (here only the vertical pairs; horizontally: analogous):
The two neighboured squares share a common domino tile (the orange one)
they don't share any domino tile on the common edge
Now there are exactly $24$ such pairings between neighboured $2 times 2$ squares (note we don't consider the diagonal neighbour pairs).
Now we count all domino tiles in following way:
-each pair of neighbour squares which share a unique common domino tile contribute a $1$ (the orange one)
-each pair of neighbour squares don't share a common domino tile contribute a $1$ with the unique tile beeing fully contained in only one square and intersecting the common edge (the grey one).
That's all. But then we get only $24$ tiles althought there are $32$. Contradiction.
I guess this argument works but I think that it's too cumbersome. Does anybody have an easier / not too circumstaneous way to show it?
recreational-mathematics puzzle
locked by quid♦ Dec 26 at 21:45
This question is locked in view of our policy about contest questions. Questions originating from active contests are locked for the duration of the contest, with answers hidden from view by soft-deletion. Please see the comments below for references to the originating contest.
comments disabled on deleted / locked posts / reviews |
I have a question about a mathematical riddle which I already solved but still looking for a shorter/simpler solution:
Following problem: We consider a standard $8 times 8$ chessboard and we cover it (completely!) with dominos of size $2 times 1$ (therefore every domino tile cover exactly $2$ fields).
The question is if we can find a cover such that there doesn't exist a $2 times 2$ subsquare which is covered exactly by two domino tiles or in other words in the cover there don't occure two "direct" neighbour domino tiles from following shape:
I have it already solved in following way: I claim that such covering isn't possible.
Argue via contradiction: Assume that it's possible. Consider the $2 times 2$ squares of the chess board and consider the partial cover of directly neighboured $2 times 2$ squares. If a cover as above really exist then up to symmetry on the common edge of the two neighboured squares there could only occure two following cases (here only the vertical pairs; horizontally: analogous):
The two neighboured squares share a common domino tile (the orange one)
they don't share any domino tile on the common edge
Now there are exactly $24$ such pairings between neighboured $2 times 2$ squares (note we don't consider the diagonal neighbour pairs).
Now we count all domino tiles in following way:
-each pair of neighbour squares which share a unique common domino tile contribute a $1$ (the orange one)
-each pair of neighbour squares don't share a common domino tile contribute a $1$ with the unique tile beeing fully contained in only one square and intersecting the common edge (the grey one).
That's all. But then we get only $24$ tiles althought there are $32$. Contradiction.
I guess this argument works but I think that it's too cumbersome. Does anybody have an easier / not too circumstaneous way to show it?
recreational-mathematics puzzle
locked by quid♦ Dec 26 at 21:45
This question is locked in view of our policy about contest questions. Questions originating from active contests are locked for the duration of the contest, with answers hidden from view by soft-deletion. Please see the comments below for references to the originating contest.
This is part of an ongoing contest! (Bundeswettbewerb Mathematik)
– Hagen von Eitzen
Dec 26 at 19:44
@HagenvonEitzen: Hi, sorry, I wasn't conscious about it. I just accidentally encountered encountered this problem made me curious. Generally, I would suggest to delete the question but I think that it would not be ok towards the author or the excellent answer below.
– KarlPeter
Dec 26 at 21:32
Feel free to flag for restoring after the contest is over (march 4th 2019, I think).
– quid♦
Dec 26 at 21:46
comments disabled on deleted / locked posts / reviews |
I have a question about a mathematical riddle which I already solved but still looking for a shorter/simpler solution:
Following problem: We consider a standard $8 times 8$ chessboard and we cover it (completely!) with dominos of size $2 times 1$ (therefore every domino tile cover exactly $2$ fields).
The question is if we can find a cover such that there doesn't exist a $2 times 2$ subsquare which is covered exactly by two domino tiles or in other words in the cover there don't occure two "direct" neighbour domino tiles from following shape:
I have it already solved in following way: I claim that such covering isn't possible.
Argue via contradiction: Assume that it's possible. Consider the $2 times 2$ squares of the chess board and consider the partial cover of directly neighboured $2 times 2$ squares. If a cover as above really exist then up to symmetry on the common edge of the two neighboured squares there could only occure two following cases (here only the vertical pairs; horizontally: analogous):
The two neighboured squares share a common domino tile (the orange one)
they don't share any domino tile on the common edge
Now there are exactly $24$ such pairings between neighboured $2 times 2$ squares (note we don't consider the diagonal neighbour pairs).
Now we count all domino tiles in following way:
-each pair of neighbour squares which share a unique common domino tile contribute a $1$ (the orange one)
-each pair of neighbour squares don't share a common domino tile contribute a $1$ with the unique tile beeing fully contained in only one square and intersecting the common edge (the grey one).
That's all. But then we get only $24$ tiles althought there are $32$. Contradiction.
I guess this argument works but I think that it's too cumbersome. Does anybody have an easier / not too circumstaneous way to show it?
recreational-mathematics puzzle
I have a question about a mathematical riddle which I already solved but still looking for a shorter/simpler solution:
Following problem: We consider a standard $8 times 8$ chessboard and we cover it (completely!) with dominos of size $2 times 1$ (therefore every domino tile cover exactly $2$ fields).
The question is if we can find a cover such that there doesn't exist a $2 times 2$ subsquare which is covered exactly by two domino tiles or in other words in the cover there don't occure two "direct" neighbour domino tiles from following shape:
I have it already solved in following way: I claim that such covering isn't possible.
Argue via contradiction: Assume that it's possible. Consider the $2 times 2$ squares of the chess board and consider the partial cover of directly neighboured $2 times 2$ squares. If a cover as above really exist then up to symmetry on the common edge of the two neighboured squares there could only occure two following cases (here only the vertical pairs; horizontally: analogous):
The two neighboured squares share a common domino tile (the orange one)
they don't share any domino tile on the common edge
Now there are exactly $24$ such pairings between neighboured $2 times 2$ squares (note we don't consider the diagonal neighbour pairs).
Now we count all domino tiles in following way:
-each pair of neighbour squares which share a unique common domino tile contribute a $1$ (the orange one)
-each pair of neighbour squares don't share a common domino tile contribute a $1$ with the unique tile beeing fully contained in only one square and intersecting the common edge (the grey one).
That's all. But then we get only $24$ tiles althought there are $32$. Contradiction.
I guess this argument works but I think that it's too cumbersome. Does anybody have an easier / not too circumstaneous way to show it?
recreational-mathematics puzzle
recreational-mathematics puzzle
edited Dec 25 at 19:30
asked Dec 25 at 18:59
KarlPeter
5951314
5951314
locked by quid♦ Dec 26 at 21:45
This question is locked in view of our policy about contest questions. Questions originating from active contests are locked for the duration of the contest, with answers hidden from view by soft-deletion. Please see the comments below for references to the originating contest.
locked by quid♦ Dec 26 at 21:45
This question is locked in view of our policy about contest questions. Questions originating from active contests are locked for the duration of the contest, with answers hidden from view by soft-deletion. Please see the comments below for references to the originating contest.
This is part of an ongoing contest! (Bundeswettbewerb Mathematik)
– Hagen von Eitzen
Dec 26 at 19:44
@HagenvonEitzen: Hi, sorry, I wasn't conscious about it. I just accidentally encountered encountered this problem made me curious. Generally, I would suggest to delete the question but I think that it would not be ok towards the author or the excellent answer below.
– KarlPeter
Dec 26 at 21:32
Feel free to flag for restoring after the contest is over (march 4th 2019, I think).
– quid♦
Dec 26 at 21:46
comments disabled on deleted / locked posts / reviews |
This is part of an ongoing contest! (Bundeswettbewerb Mathematik)
– Hagen von Eitzen
Dec 26 at 19:44
@HagenvonEitzen: Hi, sorry, I wasn't conscious about it. I just accidentally encountered encountered this problem made me curious. Generally, I would suggest to delete the question but I think that it would not be ok towards the author or the excellent answer below.
– KarlPeter
Dec 26 at 21:32
Feel free to flag for restoring after the contest is over (march 4th 2019, I think).
– quid♦
Dec 26 at 21:46
This is part of an ongoing contest! (Bundeswettbewerb Mathematik)
– Hagen von Eitzen
Dec 26 at 19:44
This is part of an ongoing contest! (Bundeswettbewerb Mathematik)
– Hagen von Eitzen
Dec 26 at 19:44
@HagenvonEitzen: Hi, sorry, I wasn't conscious about it. I just accidentally encountered encountered this problem made me curious. Generally, I would suggest to delete the question but I think that it would not be ok towards the author or the excellent answer below.
– KarlPeter
Dec 26 at 21:32
@HagenvonEitzen: Hi, sorry, I wasn't conscious about it. I just accidentally encountered encountered this problem made me curious. Generally, I would suggest to delete the question but I think that it would not be ok towards the author or the excellent answer below.
– KarlPeter
Dec 26 at 21:32
Feel free to flag for restoring after the contest is over (march 4th 2019, I think).
– quid♦
Dec 26 at 21:46
Feel free to flag for restoring after the contest is over (march 4th 2019, I think).
– quid♦
Dec 26 at 21:46
comments disabled on deleted / locked posts / reviews |
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This is part of an ongoing contest! (Bundeswettbewerb Mathematik)
– Hagen von Eitzen
Dec 26 at 19:44
@HagenvonEitzen: Hi, sorry, I wasn't conscious about it. I just accidentally encountered encountered this problem made me curious. Generally, I would suggest to delete the question but I think that it would not be ok towards the author or the excellent answer below.
– KarlPeter
Dec 26 at 21:32
Feel free to flag for restoring after the contest is over (march 4th 2019, I think).
– quid♦
Dec 26 at 21:46