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I would like to ask a question related to chess solving (and ofc. related to humanoid robots with passion for chess).
Let's imagine I have a robot friend named Borislav that has solved the chess and knows all the perfect moves. Let's say I bet Borislav for $20 that he can't mate me in X moves (the X stands for variable). Even though I'm terrible at chess I am able to prepare one specific tree of responses to my opponent's moves. Now what number should X stand for if I want to be 100% sure that I won't lose my precious $20? What is the longest known line that doesn't allow my opponent to checkmate me even if he plays the perfect game?Thanks for your replies!
There's no "forced" opening sequence that you can use to survive. Your opponent, even if playing perfect chess, has numerous equally viable options to choose from. So you'd have to have perfect memory recall of hundreds of thousands of opening lines in order to safely assume you'll reach X number of moves if X is say greater than 12. Now with experience and skill, you can apply fundamental concepts to last quite some time even against a powerful engine like Houdini before being checkmated.
I will say that you can safely avoid losing in the first couple moves if you studied the Fool's Mate though. ;)
Well, the Caro-Kann classical variation is a pretty long line with few options to deviate. But then again, your opponent isn't forced to play 3. Nc3 entering that line and could just play e5, putting you in completely different terrain.
The question presumes that chess has been "solved" which hasn't happened yet. Therefore I believe the question has no logically correct answer. What you could ask is "which opening as white/black has best record against a certain opening as black/white?".
Not to mention the advance variation is lately considered to be white's best try at theoretically refuting the Caro-Kann (though even here black can draw in the right lines).
Also, this assumes white will play e4. Since there are plenty of stagnant variations like the Petroff, e4 has no more claim to being better than d4, c4, g3, etc. So you still have to have an entire repertoire to be prepared for everything. Hence, there's no specific answer on how long a beginner could ensure they would last against perfect play. It's a gamble that has odds increasing with each higher number for X.
There's also the chance that he could play a bad move on purpose just to throw you off.
Thank you for your replies
How about aproaching this issue through computers? I know that computers solved chess for 6 and less pieces. Is there any info on the internet for how many moves was chess solved? Or is this a waste of time because of so many possibilities?
I know Borislav well. He's not that sneaky. Although he once licked my f-pawn with his robot tongue, so I wouldn't play king's gambit.
It's a waste of time. Checkers is solved but the solutions are so numerous that no one cares to know them.
Any game is a ''known line'' and any sequence may have holes, even in the opening. It's not unreasonable to assume for now with best play the game should be a draw, so I wonder if this is even a relevant question. You didn't mention who was even playing white.
Some opening lines can certainly go 30+ moves, but you don't know if they are bullet proof or if you will even get that on the board. So one of the problems is you don't really know what will happen in advance.
+1 on Firebrand's posts and Dutchday. They explain why the question according to todays knowledge only has speculative answers.
I'm sorry to inform you that I've just lost my $20 to my humanoid-solved-chess-robot-friend. Thanks for your thoughts anyway.
I would bet I could last at least 40 moves; if my objective is to last the longest possible, no to reasonably try to win. But, I am looking forward meeting Borislav.
This is a great question from mathematical point of view! I will think over it and will also discuss with other mathematicians!!
You don't understand enough mathematics!
I imagine it wouldn't be too difficult to reach a position against a computer by using anti-computer tactics such as entirely closing the position in which you can both just shuffle back and forth indefinitely. Either that, or reach a forced draw in which either side has a perpetual.
Let's assume that Borislav has generated the entire search-space (every possible position from starting), and evaluated the value of every possible leaf node (i.e., the very bottom of the tree) as a value of infinity (white won) or -infinity (black won) or 0 (draw). Let's also assume that he calculates the value of non-leaf nodes using the standard minimax algorithm.
If Borislav had truly solved chess, then Borislav would have been able to propagate the value of the non-leaf nodes all the way back to the root of the tree (the starting position)
The answer really depends on the value of this node. Since all leaf nodes are either 'infinity', '-infinity' or 0 (draw), then the value of the root node has to be one of those.
If the value of the root node is 0 (i.e., white cannot force a mate, and neither can black), then X is unlimited.
If the value of the root node is infinity (i.e., white can always force a mate), then the answer is: the number of steps along the shortest line between root node and a leaf node where all values are infinity (divided by two, because the tree assumes that each of black's and white's are 2 separate moves, whereas chess assumes that a "move" is a move by each player). Given that some openings have been worked out to over 30 (chess) moves, I would assume that the value of X is at least 40. The value of X is directly dependent on us solving chess and traversing that all-infinity path. Until we do, we can only slowly increment our current value of X as we play more and more games.
I can't imagine the value of the root node being -infinity.
That being said, the answer here implies that both Borislav and yourself are playing perfectly, not just Borislav. If you're "terrible at chess" (i.e., you're help-mating Borislav), then the value of X is 2 (if Borislav is Black) or 3 (if Borislav is white), using the Fool's mate.
But Borislav is not a computer, he's a genius. Follow the thread, please.
Okay, I'll bite. You're an idiot. Your move.
But THE Borislav is most certainly a computer.
2/10/2016 - Gregoriev, 1925
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