In the example below the mines are to the left of the ones. We figure this out by a process of elimination. If the square to the left of the 2 was a mine then the 1s would both have their maximum number of mines leaving no free square for the second mine adjacent to the 2 (since they would also be adjacent to the 1s). Having eliminated the middle square we are left with two squares we require for the 2 so they are both mined.
Azpiko adibidean, minak ‘bat’en ezkerretara daude. Ondorio horretara eliminazio-prozesu baten ondoren iritsi gara. 2aren ezkerreko laukia mina bat balitz, 1ek mina kopuru gorena izango lukete eta ez lukete lauki librerik utziko 2aren ondoko bigarren minarentzat (1en alboan egongo bailitzake). Erdiko laukia ezabatu dugunean, bi lauki gelditzen zaizkigu 2rako; beraz, biak minak dira.