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6

$begingroup$

There are two grid boards with the dimension of 9x9 and 14x14, consist of all white squares. You are supposed to color some squares with a red color on the conditions below:

• In a board, if two square has a common edge, they are neighbor squares.


• Every square needs to have at least 2 colored neighbor squares (excluding itself)

so

What is the least number of red squares you can have with the condition above?

logical-deduction strategy optimization

share|improve this question

asked 8 hours ago

OrayOray

17k44 gold badges3939 silver badges171171 bronze badges

$endgroup$

add a comment | 

6

$begingroup$

There are two grid boards with the dimension of 9x9 and 14x14, consist of all white squares. You are supposed to color some squares with a red color on the conditions below:

• In a board, if two square has a common edge, they are neighbor squares.


• Every square needs to have at least 2 colored neighbor squares (excluding itself)

so

What is the least number of red squares you can have with the condition above?

logical-deduction strategy optimization

share|improve this question

asked 8 hours ago

OrayOray

17k44 gold badges3939 silver badges171171 bronze badges

$endgroup$

add a comment | 

$begingroup$

There are two grid boards with the dimension of 9x9 and 14x14, consist of all white squares. You are supposed to color some squares with a red color on the conditions below:

• In a board, if two square has a common edge, they are neighbor squares.


• Every square needs to have at least 2 colored neighbor squares (excluding itself)

so

What is the least number of red squares you can have with the condition above?

logical-deduction strategy optimization

share|improve this question

asked 8 hours ago

OrayOray

17k44 gold badges3939 silver badges171171 bronze badges

$endgroup$

share|improve this question

asked 8 hours ago

OrayOray

17k44 gold badges3939 silver badges171171 bronze badges

share|improve this question

asked 8 hours ago

OrayOray

17k44 gold badges3939 silver badges171171 bronze badges

asked 8 hours ago

OrayOray

17k44 gold badges3939 silver badges171171 bronze badges

asked 8 hours ago

OrayOray

17k44 gold badges3939 silver badges171171 bronze badges

4 Answers
4

active

oldest

votes

5

$begingroup$

I think the answer for $14 times 14$ is

$112$

Achieved as follows

While the best I've achieved for $9 times 9$ is

$50$

Achieved as follows

share|improve this answer

edited 8 hours ago

answered 8 hours ago

hexominohexomino

55.1k55 gold badges161161 silver badges253253 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  good start but not optimal ones
  $endgroup$
  – Oray
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  At least you can blank out your boards' corners.
  $endgroup$
  – msh210
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @msh10 If we blank out the board's corners, the ones adjacent two those now only have one red square.
  $endgroup$
  – hexomino
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hexomino 9x9 seems right :)
  $endgroup$
  – Oray
  7 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

I assume

the boards are independent

Then one solution would be

171 red squares.

As shown below

for 51 squares

And

for 120 squares

No idea if this is minimal though

share|improve this answer

answered 8 hours ago

SteveVSteveV

7,29822 gold badges66 silver badges3636 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that for example cell x=9,y=14 doesn't have to be red
  $endgroup$
  – Matti
  8 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

9*9 solution:

45 reds

Looks like:

14*14 solution:

110 reds

Looks like:

share|improve this answer

answered 5 hours ago

kanookanoo

2,08033 silver badges2929 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In your 9x9, there are eight red squares only adjacent to one red square and the one in the centre is adjacent to none. In your 14 x 14, the two red squares in the centre are adjacent to no red squares,
  $endgroup$
  – hexomino
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh crap, apologies, I thought that the rule was that all white squares needed to be adjacent to two red squares
  $endgroup$
  – kanoo
  5 hours ago
  

  
  
  
  

add a comment | 

-2

$begingroup$

The answer for 14X14 is 98. Think about a chess board
As for 9X9 would be 40

share|improve this answer

edited 6 hours ago

answered 6 hours ago

ValAstaValAsta

5666 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Can you show how to fill in the grid to demonstrate this?
  $endgroup$
  – hexomino
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Make a square grid on ms word. color alternate blocks. For 9X9 color the side with 4 blocks. If you do it correctly it will end at 4 blocks as well. I dont know how to post an image. OR i would have
  $endgroup$
  – ValAsta
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There's an "insert image" button at the top of the answer box. It's the sixth one along.
  $endgroup$
  – F1Krazy
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Now that I can see what you're referring to, I'm afraid this doesn't fit the requirements of the question. Every square needs to be adjacent to two coloured squares, which includes the coloured squares themselves. The coloured squares in your solution are adjacent to zero other coloured squares. (I'm not the downvoter, btw)
  $endgroup$
  – F1Krazy
  6 hours ago
  
  

  
  
  
  

add a comment | 

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5

$begingroup$

I think the answer for $14 times 14$ is

$112$

Achieved as follows

While the best I've achieved for $9 times 9$ is

$50$

Achieved as follows

share|improve this answer

edited 8 hours ago

answered 8 hours ago

hexominohexomino

55.1k55 gold badges161161 silver badges253253 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  good start but not optimal ones
  $endgroup$
  – Oray
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  At least you can blank out your boards' corners.
  $endgroup$
  – msh210
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @msh10 If we blank out the board's corners, the ones adjacent two those now only have one red square.
  $endgroup$
  – hexomino
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hexomino 9x9 seems right :)
  $endgroup$
  – Oray
  7 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

I think the answer for $14 times 14$ is

$112$

Achieved as follows

While the best I've achieved for $9 times 9$ is

$50$

Achieved as follows

share|improve this answer

edited 8 hours ago

answered 8 hours ago

hexominohexomino

55.1k55 gold badges161161 silver badges253253 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  good start but not optimal ones
  $endgroup$
  – Oray
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  At least you can blank out your boards' corners.
  $endgroup$
  – msh210
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @msh10 If we blank out the board's corners, the ones adjacent two those now only have one red square.
  $endgroup$
  – hexomino
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hexomino 9x9 seems right :)
  $endgroup$
  – Oray
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I think the answer for $14 times 14$ is

$112$

Achieved as follows

While the best I've achieved for $9 times 9$ is

$50$

Achieved as follows

share|improve this answer

edited 8 hours ago

answered 8 hours ago

hexominohexomino

55.1k55 gold badges161161 silver badges253253 bronze badges

$endgroup$

share|improve this answer

edited 8 hours ago

answered 8 hours ago

hexominohexomino

55.1k55 gold badges161161 silver badges253253 bronze badges

answered 8 hours ago

hexominohexomino

55.1k55 gold badges161161 silver badges253253 bronze badges

answered 8 hours ago

hexominohexomino

55.1k55 gold badges161161 silver badges253253 bronze badges

3

$begingroup$

I assume

the boards are independent

Then one solution would be

171 red squares.

As shown below

for 51 squares

And

for 120 squares

No idea if this is minimal though

share|improve this answer

answered 8 hours ago

SteveVSteveV

7,29822 gold badges66 silver badges3636 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that for example cell x=9,y=14 doesn't have to be red
  $endgroup$
  – Matti
  8 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

I assume

the boards are independent

Then one solution would be

171 red squares.

As shown below

for 51 squares

And

for 120 squares

No idea if this is minimal though

share|improve this answer

answered 8 hours ago

SteveVSteveV

7,29822 gold badges66 silver badges3636 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that for example cell x=9,y=14 doesn't have to be red
  $endgroup$
  – Matti
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I assume

the boards are independent

Then one solution would be

171 red squares.

As shown below

for 51 squares

And

for 120 squares

No idea if this is minimal though

share|improve this answer

answered 8 hours ago

SteveVSteveV

7,29822 gold badges66 silver badges3636 bronze badges

$endgroup$

share|improve this answer

answered 8 hours ago

SteveVSteveV

7,29822 gold badges66 silver badges3636 bronze badges

answered 8 hours ago

SteveVSteveV

7,29822 gold badges66 silver badges3636 bronze badges

answered 8 hours ago

SteveVSteveV

7,29822 gold badges66 silver badges3636 bronze badges

0

$begingroup$

9*9 solution:

45 reds

Looks like:

14*14 solution:

110 reds

Looks like:

share|improve this answer

answered 5 hours ago

kanookanoo

2,08033 silver badges2929 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In your 9x9, there are eight red squares only adjacent to one red square and the one in the centre is adjacent to none. In your 14 x 14, the two red squares in the centre are adjacent to no red squares,
  $endgroup$
  – hexomino
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh crap, apologies, I thought that the rule was that all white squares needed to be adjacent to two red squares
  $endgroup$
  – kanoo
  5 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

9*9 solution:

45 reds

Looks like:

14*14 solution:

110 reds

Looks like:

share|improve this answer

answered 5 hours ago

kanookanoo

2,08033 silver badges2929 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In your 9x9, there are eight red squares only adjacent to one red square and the one in the centre is adjacent to none. In your 14 x 14, the two red squares in the centre are adjacent to no red squares,
  $endgroup$
  – hexomino
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh crap, apologies, I thought that the rule was that all white squares needed to be adjacent to two red squares
  $endgroup$
  – kanoo
  5 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

9*9 solution:

45 reds

Looks like:

14*14 solution:

110 reds

Looks like:

share|improve this answer

answered 5 hours ago

kanookanoo

2,08033 silver badges2929 bronze badges

$endgroup$

share|improve this answer

answered 5 hours ago

kanookanoo

2,08033 silver badges2929 bronze badges

answered 5 hours ago

kanookanoo

2,08033 silver badges2929 bronze badges

answered 5 hours ago

kanookanoo

2,08033 silver badges2929 bronze badges

-2

$begingroup$

The answer for 14X14 is 98. Think about a chess board
As for 9X9 would be 40

share|improve this answer

edited 6 hours ago

answered 6 hours ago

ValAstaValAsta

5666 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Can you show how to fill in the grid to demonstrate this?
  $endgroup$
  – hexomino
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Make a square grid on ms word. color alternate blocks. For 9X9 color the side with 4 blocks. If you do it correctly it will end at 4 blocks as well. I dont know how to post an image. OR i would have
  $endgroup$
  – ValAsta
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There's an "insert image" button at the top of the answer box. It's the sixth one along.
  $endgroup$
  – F1Krazy
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Now that I can see what you're referring to, I'm afraid this doesn't fit the requirements of the question. Every square needs to be adjacent to two coloured squares, which includes the coloured squares themselves. The coloured squares in your solution are adjacent to zero other coloured squares. (I'm not the downvoter, btw)
  $endgroup$
  – F1Krazy
  6 hours ago
  
  

  
  
  
  

add a comment | 

-2

$begingroup$

The answer for 14X14 is 98. Think about a chess board
As for 9X9 would be 40

share|improve this answer

edited 6 hours ago

answered 6 hours ago

ValAstaValAsta

5666 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Can you show how to fill in the grid to demonstrate this?
  $endgroup$
  – hexomino
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Make a square grid on ms word. color alternate blocks. For 9X9 color the side with 4 blocks. If you do it correctly it will end at 4 blocks as well. I dont know how to post an image. OR i would have
  $endgroup$
  – ValAsta
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There's an "insert image" button at the top of the answer box. It's the sixth one along.
  $endgroup$
  – F1Krazy
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Now that I can see what you're referring to, I'm afraid this doesn't fit the requirements of the question. Every square needs to be adjacent to two coloured squares, which includes the coloured squares themselves. The coloured squares in your solution are adjacent to zero other coloured squares. (I'm not the downvoter, btw)
  $endgroup$
  – F1Krazy
  6 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

The answer for 14X14 is 98. Think about a chess board
As for 9X9 would be 40

share|improve this answer

edited 6 hours ago

answered 6 hours ago

ValAstaValAsta

5666 bronze badges

$endgroup$

share|improve this answer

edited 6 hours ago

answered 6 hours ago

ValAstaValAsta

5666 bronze badges

answered 6 hours ago

ValAstaValAsta

5666 bronze badges

answered 6 hours ago

ValAstaValAsta

5666 bronze badges