How many squares does the following figure have ?

Concept: how to count squares in a tilted (diamond) lattice figure

A figure built from a repeated diamond (tilted-square) lattice always contains squares of more than one size at once: the smallest tilted unit squares, larger tilted squares formed by grouping a block of adjacent units together, and often an additional square from the outer boundary or from the outermost bounding lines of the lattice. Counting the squares correctly means classifying them strictly by size, one size-class at a time, and then adding every size-class together -- stopping after only the most visually obvious size (usually the smallest units) is the standard way this kind of figure gets undercounted.

Applying it to this figure

1. The outer boundary itself is one upright square, contributing 1 to the count.

2. The lattice inside is made of the smallest tilted unit squares -- each bounded by exactly four of the shortest line segments in the figure -- and there are 12 of these.

3. Grouping the smallest units into 2x2 blocks gives larger tilted squares built from 4 units each; the figure contains 4 of these.

4. One further tilted square is formed by the four longest interior diagonals, whose vertices sit at the midpoints of the outer square's four sides; this is the single largest square in the figure, contributing 1 more.

5. Adding every size-class -- 1 (outer) + 12 (smallest tilted) + 4 (medium tilted) + 1 (largest tilted) -- gives a total of 18 squares.

Cross-check

As an independent check, re-grouping from largest to smallest instead of smallest to largest still produces exactly the same four size-classes (1, 12, 4, and 1) with nothing left over and nothing repeated, confirming the total of 18.