https://bugs.kde.org/show_bug.cgi?id=478808
--- Comment #1 from Jazeix Johnny <jaz...@gmail.com> --- Thank you for the report! I would go with the opposite fix (but let's discuss it if you think it should work as you propose). To be sure I understand well the issue, let's assume: G = Green, B = Blue, Y = Yellow and the solution of the next examples is GBG. Actual: > the user inputs GYY, there will be 1 black peg and 1 white pegs (as the last > correct one is G and it's already placed at the beginning). For me, this behaviour is correct (as we expect the G to be at 2 positions). > the user inputs YGY, there will be only 1 misplaced token. For me, here we should display two white pegs (else the child does not know it should be placed twice). If we go the opposite way and only place one black peg if one of the two positions is found, the children won't have a clue they should reuse the same one (as they could expect to have been warn with a white peg). -- You are receiving this mail because: You are watching all bug changes.
