Mixing
Minimum number of moves required to obtain chess like coloring.
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I've been preparing for a competition and there is this problem that I cannot solve. Can you please help me and also tell me how to do similar problems if they appear in the future?
Problem:
geometry arithmetic
edited Jan 3 at 3:02
David G. Stork
10.4k21332
asked Jan 3 at 1:15
The ThunderkingThe Thunderking
61
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add a comment |
$begingroup$
I've been preparing for a competition and there is this problem that I cannot solve. Can you please help me and also tell me how to do similar problems if they appear in the future?
Problem:
geometry arithmetic
edited Jan 3 at 3:02
David G. Stork
10.4k21332
asked Jan 3 at 1:15
The ThunderkingThe Thunderking
61
$endgroup$
add a comment |
$begingroup$
I've been preparing for a competition and there is this problem that I cannot solve. Can you please help me and also tell me how to do similar problems if they appear in the future?
Problem:
geometry arithmetic
edited Jan 3 at 3:02
David G. Stork
10.4k21332
asked Jan 3 at 1:15
The ThunderkingThe Thunderking
61
$endgroup$
I've been preparing for a competition and there is this problem that I cannot solve. Can you please help me and also tell me how to do similar problems if they appear in the future?
Problem:
geometry arithmetic
geometry arithmetic
edited Jan 3 at 3:02
David G. Stork
10.4k21332
asked Jan 3 at 1:15
The ThunderkingThe Thunderking
61
edited Jan 3 at 3:02
David G. Stork
10.4k21332
asked Jan 3 at 1:15
The ThunderkingThe Thunderking
61
edited Jan 3 at 3:02
David G. Stork
10.4k21332
edited Jan 3 at 3:02
David G. Stork
10.4k21332
edited Jan 3 at 3:02
David G. Stork
10.4k21332
10.4k21332
asked Jan 3 at 1:15
The ThunderkingThe Thunderking
61
asked Jan 3 at 1:15
The ThunderkingThe Thunderking
61
asked Jan 3 at 1:15
The ThunderkingThe Thunderking
61
61
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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I am unsure what the "general procedure" should be for this kind of problem, but I can help for this specific one.
Note that for the $5times 5$ square shown, there are $12$ black squares. Since none of the black squares are neighbors, we know that the minimum number of moves to take the starting configuration to the final configuration is at least $12$. Thus, if we can demonstrate a sequence of moves of length $12$ which accomplishes our goal, we will have proven that the minimum is exactly $12$.
SPOILERS AHEAD!
So what sequence of $12$ moves accomplishes our goal? Well, we can "click" each corner twice to get $8$ black squares while leaving the corners white (since two "clicks" will negate each other on the same square), and we can get the remaining $4$ black squares in the center by "clicking" the very middle square $4$ times.
answered Jan 3 at 1:28
ItsJustASeriesBroItsJustASeriesBro
1563
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add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
I am unsure what the "general procedure" should be for this kind of problem, but I can help for this specific one.
Note that for the $5times 5$ square shown, there are $12$ black squares. Since none of the black squares are neighbors, we know that the minimum number of moves to take the starting configuration to the final configuration is at least $12$. Thus, if we can demonstrate a sequence of moves of length $12$ which accomplishes our goal, we will have proven that the minimum is exactly $12$.
SPOILERS AHEAD!
So what sequence of $12$ moves accomplishes our goal? Well, we can "click" each corner twice to get $8$ black squares while leaving the corners white (since two "clicks" will negate each other on the same square), and we can get the remaining $4$ black squares in the center by "clicking" the very middle square $4$ times.
answered Jan 3 at 1:28
ItsJustASeriesBroItsJustASeriesBro
1563
$endgroup$
add a comment |
$begingroup$
I am unsure what the "general procedure" should be for this kind of problem, but I can help for this specific one.
Note that for the $5times 5$ square shown, there are $12$ black squares. Since none of the black squares are neighbors, we know that the minimum number of moves to take the starting configuration to the final configuration is at least $12$. Thus, if we can demonstrate a sequence of moves of length $12$ which accomplishes our goal, we will have proven that the minimum is exactly $12$.
SPOILERS AHEAD!
So what sequence of $12$ moves accomplishes our goal? Well, we can "click" each corner twice to get $8$ black squares while leaving the corners white (since two "clicks" will negate each other on the same square), and we can get the remaining $4$ black squares in the center by "clicking" the very middle square $4$ times.
answered Jan 3 at 1:28
ItsJustASeriesBroItsJustASeriesBro
1563
$endgroup$
add a comment |
$begingroup$
I am unsure what the "general procedure" should be for this kind of problem, but I can help for this specific one.
Note that for the $5times 5$ square shown, there are $12$ black squares. Since none of the black squares are neighbors, we know that the minimum number of moves to take the starting configuration to the final configuration is at least $12$. Thus, if we can demonstrate a sequence of moves of length $12$ which accomplishes our goal, we will have proven that the minimum is exactly $12$.
SPOILERS AHEAD!
So what sequence of $12$ moves accomplishes our goal? Well, we can "click" each corner twice to get $8$ black squares while leaving the corners white (since two "clicks" will negate each other on the same square), and we can get the remaining $4$ black squares in the center by "clicking" the very middle square $4$ times.
answered Jan 3 at 1:28
ItsJustASeriesBroItsJustASeriesBro
1563
$endgroup$
I am unsure what the "general procedure" should be for this kind of problem, but I can help for this specific one.
Note that for the $5times 5$ square shown, there are $12$ black squares. Since none of the black squares are neighbors, we know that the minimum number of moves to take the starting configuration to the final configuration is at least $12$. Thus, if we can demonstrate a sequence of moves of length $12$ which accomplishes our goal, we will have proven that the minimum is exactly $12$.
SPOILERS AHEAD!
So what sequence of $12$ moves accomplishes our goal? Well, we can "click" each corner twice to get $8$ black squares while leaving the corners white (since two "clicks" will negate each other on the same square), and we can get the remaining $4$ black squares in the center by "clicking" the very middle square $4$ times.
answered Jan 3 at 1:28
ItsJustASeriesBroItsJustASeriesBro
1563
edited Jan 3 at 1:35
answered Jan 3 at 1:28
ItsJustASeriesBroItsJustASeriesBro
1563
answered Jan 3 at 1:28
ItsJustASeriesBroItsJustASeriesBro
1563
answered Jan 3 at 1:28
ItsJustASeriesBroItsJustASeriesBro
1563
1563
add a comment |
add a comment |
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