which would take more moves to win

[bob]

Want to list the BILLIONS of things we can prove. I can prove fine #70 is a forced mate, for example. But if you can't prove something, you can't state that something as a fact either. Which is what has been done here multiple times. A knight handicap is better than no handicap, how much better is, and probably always will be unknown.

One of these "BILLIONS of things we can prove" is the following proof:

I have an EPD file of 5-men TB wins.
Using this file as openings with Komodo, I have the following:
400 Games in 40/1 minute
Komodo No TB self-games: average length of the won game -- 26.0 moves
Komodo 5-men Syzygy self-games: average length of the won game -- 23.6 moves

Therefore the perfect play SHORTENS the length of the game (the path to win) in theoretically won positions, contrary to what you are stating here all the way from the first post in this thread.

Sorry, I have not said otherwise. The point YOU miss here is "how long is the longest non-perfect game?" Yes, perfect play will shorten that. But if neither side plays a blunder? You think it will STILL be shorter? I just happen to believe that the starting position is not won or drawn in 1000 perfect moves by both sides. And I have no idea what that number will be, just that it is doubtful it will be 100. And there are MANY endgames with 500+ moves to the win, and that is just endgames. How long to reach those positions with perfect play first?

[Laskos]

Want to list the BILLIONS of things we can prove. I can prove fine #70 is a forced mate, for example. But if you can't prove something, you can't state that something as a fact either. Which is what has been done here multiple times. A knight handicap is better than no handicap, how much better is, and probably always will be unknown.

One of these "BILLIONS of things we can prove" is the following proof:

I have an EPD file of 5-men TB wins.
Using this file as openings with Komodo, I have the following:
400 Games in 40/1 minute
Komodo No TB self-games: average length of the won game -- 26.0 moves
Komodo 5-men Syzygy self-games: average length of the won game -- 23.6 moves

Therefore the perfect play SHORTENS the length of the game (the path to win) in theoretically won positions, contrary to what you are stating here all the way from the first post in this thread.

Sorry, I have not said otherwise. The point YOU miss here is "how long is the longest non-perfect game?" Yes, perfect play will shorten that. But if neither side plays a blunder? You think it will STILL be shorter? I just happen to believe that the starting position is not won or drawn in 1000 perfect moves by both sides. And I have no idea what that number will be, just that it is doubtful it will be 100. And there are MANY endgames with 500+ moves to the win, and that is just endgames. How long to reach those positions with perfect play first?

Let's recapitulate a bit. It all started with:

a top gm playing komodo with knight handicap

or 2 computers one with a knight handicap but both with access to a 32 piece tablebase

and how many moves is the former ?

Clearly the latter, since they would, by definition, play perfectly.

Are you saying now that a GM would beat Komodo at Knight odds faster than Komodo would beat Komodo? Say, perfect win at Knight odds is 1000 moves. My 5-men result would suggest that Komodo then on average beats Komodo in 1100 moves at Knight odds (it doesn't seem to happen, but for the sake of the argument). Are you suggesting that a GM would beat Komodo not only faster than Komodo beats Komodo (1100 moves), but faster than perfect game (1000 moves)?

Other way around: Komodo beats Komodo at Knight odds in about 65 moves on average. My result with 5-men suggests that perfect play from both sides would result in faster than 65 moves win. Is that Knight handicap such a peculiar "won endgame" that it needs 1000 moves of perfect play instead of less than 65 suggested? Is my 5-men result so misleading? You seem to need some unreasonable assumptions for your hypothesis to hold, and falls short under Occam's razor.

[Uri Blass]

Want to list the BILLIONS of things we can prove. I can prove fine #70 is a forced mate, for example. But if you can't prove something, you can't state that something as a fact either. Which is what has been done here multiple times. A knight handicap is better than no handicap, how much better is, and probably always will be unknown.

One of these "BILLIONS of things we can prove" is the following proof:

I have an EPD file of 5-men TB wins.
Using this file as openings with Komodo, I have the following:
400 Games in 40/1 minute
Komodo No TB self-games: average length of the won game -- 26.0 moves
Komodo 5-men Syzygy self-games: average length of the won game -- 23.6 moves

Therefore the perfect play SHORTENS the length of the game (the path to win) in theoretically won positions, contrary to what you are stating here all the way from the first post in this thread.

Sorry, I have not said otherwise. The point YOU miss here is "how long is the longest non-perfect game?" Yes, perfect play will shorten that. But if neither side plays a blunder? You think it will STILL be shorter? I just happen to believe that the starting position is not won or drawn in 1000 perfect moves by both sides. And I have no idea what that number will be, just that it is doubtful it will be 100. And there are MANY endgames with 500+ moves to the win, and that is just endgames. How long to reach those positions with perfect play first?

Unlike you I know that the number of moves with knight handicap is smaller than 100.

No mathematical proof for it but I am sure about it in the same way that I am sure that the sun is going to rise tomorrow.

I believe every GM in chess is going to agree with me.
It may be interesting to have the following match human against chess programs.

the program start without knight b1 when the human needs to win by checkmate with less than 100 moves.

I expect Komodo without a knight to lose most games against 2600 GM's at 90+30 and even at 45+15 in less than 100 moves.

[bob]

Want to list the BILLIONS of things we can prove. I can prove fine #70 is a forced mate, for example. But if you can't prove something, you can't state that something as a fact either. Which is what has been done here multiple times. A knight handicap is better than no handicap, how much better is, and probably always will be unknown.

One of these "BILLIONS of things we can prove" is the following proof:

I have an EPD file of 5-men TB wins.
Using this file as openings with Komodo, I have the following:
400 Games in 40/1 minute
Komodo No TB self-games: average length of the won game -- 26.0 moves
Komodo 5-men Syzygy self-games: average length of the won game -- 23.6 moves

Therefore the perfect play SHORTENS the length of the game (the path to win) in theoretically won positions, contrary to what you are stating here all the way from the first post in this thread.

Sorry, I have not said otherwise. The point YOU miss here is "how long is the longest non-perfect game?" Yes, perfect play will shorten that. But if neither side plays a blunder? You think it will STILL be shorter? I just happen to believe that the starting position is not won or drawn in 1000 perfect moves by both sides. And I have no idea what that number will be, just that it is doubtful it will be 100. And there are MANY endgames with 500+ moves to the win, and that is just endgames. How long to reach those positions with perfect play first?

Let's recapitulate a bit. It all started with:

a top gm playing komodo with knight handicap

or 2 computers one with a knight handicap but both with access to a 32 piece tablebase

and how many moves is the former ?

Clearly the latter, since they would, by definition, play perfectly.

Are you saying now that a GM would beat Komodo at Knight odds faster than Komodo would beat Komodo? Say, perfect win at Knight odds is 1000 moves. My 5-men result would suggest that Komodo then on average beats Komodo in 1100 moves at Knight odds (it doesn't seem to happen, but for the sake of the argument). Are you suggesting that a GM would beat Komodo not only faster than Komodo beats Komodo (1100 moves), but faster than perfect game (1000 moves)?

Other way around: Komodo beats Komodo at Knight odds in about 65 moves on average. My result with 5-men suggests that perfect play from both sides would result in faster than 65 moves win. Is that Knight handicap such a peculiar "won endgame" that it needs 1000 moves of perfect play instead of less than 65 suggested? Is my 5-men result so misleading? You seem to need some unreasonable assumptions for your hypothesis to hold, and falls short under Occam's razor.

Step 1: Humans ALWAYS make mistakes. So no, a human will NOT beat komodo faster than a perfect player, given knight odds. Never said nor implied that.

Step 2. We have absolutely no idea how long a perfectly played game of chess will last before it is drawn or one side wins. If a human is playing the knight advantage side, AND the human can consistently win against the perfect player (which is unknown as of today) then a perfect player with knight odds will win faster than the human.

I don't think I have been inconsistent in making the above statements, and ONLY the above statements. Most specifically, "we don't KNOW what perfect play will do at all, even or knight odds games. All one can do is guess. Stating that knight odds is enough to win period is a statement not based in fact as of today since we have no perfect player other than in 7 piece and fewer endgames (and a few 8's apparently)."

How many moves Komodo takes to win doesn't mean a thing when we talk about perfect play. It only answers the question what happens when two highly imperfect machines play that handicap game. Wonder what happens when you increase the time control? Does it follow the usual either more moves or more draws???

[bob]

Want to list the BILLIONS of things we can prove. I can prove fine #70 is a forced mate, for example. But if you can't prove something, you can't state that something as a fact either. Which is what has been done here multiple times. A knight handicap is better than no handicap, how much better is, and probably always will be unknown.

One of these "BILLIONS of things we can prove" is the following proof:

I have an EPD file of 5-men TB wins.
Using this file as openings with Komodo, I have the following:
400 Games in 40/1 minute
Komodo No TB self-games: average length of the won game -- 26.0 moves
Komodo 5-men Syzygy self-games: average length of the won game -- 23.6 moves

Therefore the perfect play SHORTENS the length of the game (the path to win) in theoretically won positions, contrary to what you are stating here all the way from the first post in this thread.

Sorry, I have not said otherwise. The point YOU miss here is "how long is the longest non-perfect game?" Yes, perfect play will shorten that. But if neither side plays a blunder? You think it will STILL be shorter? I just happen to believe that the starting position is not won or drawn in 1000 perfect moves by both sides. And I have no idea what that number will be, just that it is doubtful it will be 100. And there are MANY endgames with 500+ moves to the win, and that is just endgames. How long to reach those positions with perfect play first?

Unlike you I know that the number of moves with knight handicap is smaller than 100.

No mathematical proof for it but I am sure about it in the same way that I am sure that the sun is going to rise tomorrow.

I believe every GM in chess is going to agree with me.
It may be interesting to have the following match human against chess programs.

the program start without knight b1 when the human needs to win by checkmate with less than 100 moves.

I expect Komodo without a knight to lose most games against 2600 GM's at 90+30 and even at 45+15 in less than 100 moves.

Unlike you I don't guess. So please cite a specific reference that proves that a knight handicap is a forced with within 100 moves (200 plies). If you can't, you don't "know" this. You only "believe" this.

Your experiment is so badly flawed most anyone can see the problem. No "perfect player" involved. Without a perfect player, you can't state that a perfect player can be beaten within 100 moves. You can only speculate/extrapolate/guess/flip a coin, whatever. Pure guesswork.

I will remind you about a simple KQ vs KR ending. Until 1978 or so, one rule preached by ALL GM level players was "you must keep the rook close to the king or you lose quickly." That was disproved in 1978. How? By a "perfect player", namely Belle with Ken's 4 piece endgame databases. If you search for "the great flying rook" you might find the discussion about this. Turns out that optimal play frequently separates the king and rook to stave off mate as long as possible. BTW GM Walter Browne failed to win with KQ vs KR when he first met this "perfect player". He studied the ending in light of the new defenses Belle put up and was able to win the ending in the rematch.