How many squares are in this image? Is there a method to check?

Question

In this image I have counted 14 but others say 18.

Is there a method to check exactly?

Answer 1

[I will give two methods. The SECOND one is the better one.]

Method? Well, sort of, you can look at each "atom" piece and count how many squares that is the upper right hand corner of.

A)Top right square -> top right square; engulf the the rectangles for a 3 by 3, the whole 4x4 = 3.

B) top middle rectangle -> 2x2 square; 3x3 square = 2.

C) top left square -> top left square = 1.

D) middle side rectangle -> 2x2 square; 3x3 = 2.

E) next square in the (2,2) spot -> 1x1;2x2;3x3 = 3

F) (3,2) spot ->1x1; 2x2 = 2

G) (4,2) spot -> 1x1 = 1

H) (3,2) spot -> 1x1; 2x2 = 2

I) (3,3) spot -> 1x1 = 1

J) (4,1) spot -> 1x1 = 1

K) low middle rectangle = 0

L) (4,4) spot -> 1x1 = 1

So total: 18

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This i a simplification of simply figuring out the squares in a complete grid and subtracting the one that rectangles make impossible.

A 4x4 grid will have: 16 1x1 squares; 9 2x2 squares (as there are 3 squares in each of the top 3 rows that can be an upper right hand corner of a 3x3 square), 4 3x3 squares, and 1 4x4 square.

So an n x n grid will have $\sum k^2$ total squares.

In this case 16 + 9 + 4 + 1 = 30.

The first top rectangle eliminates 2 1x1 squares and 1 2x2 square. So only 27 possible.

The left side rectangle eliminates 2 1x1 squares and 1 2x2 square. So only 24 possible.

The right side triangle eliminates 2 1x1 squares (1 2x2 that was already eliminated) and a 3x3. So only 21 possible.

The bottom rectangle eliminates 2 1x1 squares, 2 2x2 squares. So only 18 possible squares.

Answer 2

Thm: Total squares s= 13#-8

Where # is the number of hashes. A hash being division of a square by a set of orthogonal equally spaced (including overlapping ; as is the initial case ) parallel lines such that the corner square is exactly one fourth the area of the centre square.(if it exists) Each hash thus dividing the centre square of the outer hash.

Here #=2 thus s=18

Inn case we extend the outer square as well we are looking at 31 squares.

A154609 @OEIS gives the sequence with # starting from 0

Also given by recurrence relationship s(#) = 2s(#-1)-s(#-2) where s(1)=5 and s(2)=18.

Answer 3

19 are yall skipping the big one on the outside? Can't be an even number as only one outside square. I counted 19

Answer 4

Well the answer is 18. How I do these problems is first i count the 1x1 squares. There is 8 of them. Then I count the 2x2 squares. There is 5 of them. Then 3x3 squares. There is four of them. And then the 4x4 square. Total 18. In each square size I go from left to right top to bottom.

Answer 5

In each open space, I counted the number of squares that shared the upper left corner. The total is $18$.

Another way. (I see that fleablood did it this way first)

If all of the lines were drawn, there would be $1+4+9+16 = 30\;\;$ squares.

So how many squares could not be drawn because of missing lines.

size    count  description (upper left corners)
1x1     8      B C E H I L N O 
2x2     4      A C I K
3X3     0
4X4     0
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total  12 

So $30 - 12 = 18\;$ squares can be drawn.