4 ms·

Thanks for clarifying. If you've the time, could you tell me how I should read: non_conflicting = (g) -> ![0...9].some (i) -> g in [s[x][i], s[i][y], box i]

• pathikrit 14y ago
  [0...9].some (i) -> g in [s[x][i], s[i][y], box i] returns true iff it finds some i in [0,9) such that given a guess g, g equals either s[x][i] or s[i][y] or box i or in other words it finds a conflict in either a row or column or box for g assuming it g is at s[x][y] The ! at the start negates it - so the whole method non_conflicting returns true iff g does not conflict with any i. If Arrays in JavaScript had an "all" reducer it could be written like this: non_conflicting = (g) -> [0...9].all (i) -> g not in [s[x][i], s[i][y], box i]