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I wrote an online single player strategy game called Trek and released it on GH pages a few days ago. Link is here: https://vivegi.github.io/Trek https://vivegi

by vivegi 5y ago

I wrote an online single player strategy game called Trek and released it on GH pages a few days ago. Link is here: https://vivegi.github.io/Trek https://vivegi.github.io/Trek It is free and online. No user registrations. No trackers. No 3rd party scripts or dependencies and the entire web app is self-contained. Just HTML/CSS/JS + media assets. It is based on interactive 3-SAT solving although I deliberately made a point of not mentioning it on the game's home page to keep it appealing to the widest possible audience, not just CompSci/SAT folks. Would love to hear what you and your 4/yo think of it. Of course, open to collabs.

fladrif 5y ago
Hey, I'm looking through it now, but the instructions aren't super clear to me. I'm not sure what you're trying to say here. > The selected pieces must not contain both a and of the same color. The following example shows a couple X's selected with the same color in different rows. Are you trying to say O's and X's cannot share a color?
- vivegi 5y ago
  Yes. You need to select one piece from each row. Selections in different rows cannot have an O and X of the same color. If you think of this in terms of 3-SAT, the O's are positive literals (like a, b, c etc.,) and X's are negative literals (like a', b', c' etc.,) and the color is used to denote the underlying variable (ie., a, b etc.,). Each row is a disjunctive clause (eg: a + b' + c, where + denotes Logical OR or disjunction). Consecutive rows represent a conjunction (eg: (a + b' + c).(a + b + c'). The condition ensures that we don't permit a selection such as b.b' or c.c' since they would evaluate to False in Boolean algebra. The goal is to get a selection that doesn't contain a literal and its complement. Due to the distributive law in Boolean algebra, solving SAT is equivalent to finding a term that doesn't vanish to False when the formula is fully expanded out. Thanks for trying it out.

by vivegi 5y ago

fladrif 5y ago

Hey, I'm looking through it now, but the instructions aren't super clear to me. I'm not sure what you're trying to say here. > The selected pieces must not contain both a and of the same color. The following example shows a couple X's selected with the same color in different rows. Are you trying to say O's and X's cannot share a color?

vivegi 5y ago
Yes. You need to select one piece from each row. Selections in different rows cannot have an O and X of the same color. If you think of this in terms of 3-SAT, the O's are positive literals (like a, b, c etc.,) and X's are negative literals (like a', b', c' etc.,) and the color is used to denote the underlying variable (ie., a, b etc.,). Each row is a disjunctive clause (eg: a + b' + c, where + denotes Logical OR or disjunction). Consecutive rows represent a conjunction (eg: (a + b' + c).(a + b + c'). The condition ensures that we don't permit a selection such as b.b' or c.c' since they would evaluate to False in Boolean algebra. The goal is to get a selection that doesn't contain a literal and its complement. Due to the distributive law in Boolean algebra, solving SAT is equivalent to finding a term that doesn't vanish to False when the formula is fully expanded out. Thanks for trying it out.