Mothai59 and David21490, please allow me to answer in English since my French is limited. This puzzle gets complicated in the bottom, since the upper part is doable by single line logic. In the bottom you will find the following: 1. The row 25 (r25) can be partially done using the "two-line logic". 2. When solving row 24 you can blacken column 4 and 5 (c4-c5) by single logic. Then you cannot blacken c6 becasue it forces to blacken c5-r23, which blocks the line of 3 on r23. Hence you blacken c3r24. 3. Later on when you look at c5, and faced to blacken r23 or r25 you have to go with r23. R25 is not valid because it forces to blacken c21r23, c20r23 and c19r23 and c18r22, which then would leave no room to play the missing spot on c20 because you need at least two spaces on the row side.
Hope this helps, and my apologies ofr the language.
Très sympa et 100% logique, merci et bienvenue Jmestradaabud. Pour ma part, j?ai surtout travaillé les deux dernières lignes et, une fois les deux tiers supérieurs placés, j?ai utilisé une hypothèse spg par symétrie sur le 15 pour finir ;)
Very nice and 100% logical, thanks and welcome there Jmestradaabud. For my part, I worked with the last two rows and, when I placed the top two thirds, I assumed without loss of generality that R25C6 was black thanks to symmetry ;)