proof

[Brunetti]

Hello everyone,
IMHO, I believe that my old heuristic approach to this problem is much less complicated because it starts from the endgame rather than the opening. I’m sharing it out of curiosity.

A Heuristic Approach to the Hypothesis that Chess is a Forced Draw

Abstract:

This paper presents a heuristic argument supporting the hypothesis that the game of chess, when played with perfect strategy by both sides, inevitably results in a draw. The argument is based on a stepwise inductive approach, beginning with the simplest position on the chessboard and gradually adding pairs of symmetrical pieces while maintaining the draw status. Although not a formal proof, this approach provides a reasoned foundation for the belief that chess is a forced draw, highlighting the challenges of proving this assertion formally given the complexity of the game.
Introduction:

The question of whether chess, under perfect play by both sides, always results in a draw has long intrigued both theorists and practitioners of the game. Despite significant advances in chess theory and computational power, a formal proof of this hypothesis remains elusive due to the game's combinatorial complexity. This paper aims to present a heuristic argument suggesting that chess is a draw, relying on the principle of symmetry and the preservation of balance as pieces are progressively added to the board.
Methodology:

Base Case:
Consider a chessboard with only the two Kings in their starting positions. It is universally acknowledged that this is a draw, as neither King can checkmate the other.

Inductive Step:
We introduce two pawns, one for White at a2 and one for Black at a7. According to existing endgame tablebases, this position is still a draw under perfect play. The symmetry of the position suggests that neither side has a decisive advantage.

Further Additions:

Add Rooks to a1 for White and a8 for Black. The position remains a draw, as confirmed by endgame databases and supported by chess theory. The addition of pieces does not disrupt the symmetrical nature of the board.

Add two more pawns, b2 for White and b7 for Black. While we do not yet have tablebases for this exact configuration (exceeding 7 pieces), it is reasonable to assume, based on known principles and experience, that the position remains a draw. The symmetry continues to enforce balance, preventing either side from achieving a decisive advantage.

Add Knights to b1 for White and b8 for Black. Similar to the previous step, the symmetry and the principles of chess suggest that the draw status persists.

Continued Induction:
The process of adding pieces symmetrically is continued, with the assumption that each symmetrical addition maintains the draw status. The underlying logic is that the symmetry of the position, combined with perfect play by both sides, should ensure that the balance is never disrupted in favor of one side over the other.

Discussion:

The argument presented is inherently heuristic and inductive. While it suggests that the game of chess, under perfect conditions, is a draw, it does not constitute a formal proof. The complexity of chess, particularly as the number of pieces on the board increases, introduces interactions that are difficult to predict or analyze exhaustively without comprehensive computational resources.

However, the consistency of the draw status in the early stages of piece addition, supported by existing endgame tablebases and chess theory, provides a strong indication that the hypothesis holds true. The inductive nature of the argument aligns with the broader understanding of chess as a balanced game where perfect play should theoretically result in neither side being able to secure a decisive victory.
Conclusion:

While a formal proof of the hypothesis that chess is a forced draw remains out of reach due to the immense complexity of the game, the heuristic approach presented here offers a reasoned argument in support of this idea. By gradually adding pieces in a symmetrical manner and relying on known principles of chess, we can infer that the draw status is likely maintained throughout. Further research, particularly in the development of more advanced tablebases and AI-driven analysis, may bring us closer to a definitive answer.
References:

Shannon, C. E. (1950). Programming a Computer for Playing Chess. Philosophical Magazine.
Thompson, K. (1986). Retrograde analysis of certain endgames. ICCA Journal, 9(3), 131-139.
Tromp, J., & Farnebäck, G. (2014). Combinatorial Game Theory for Chess Endgames.

[jefk]

Jamie Whiting I'm going to quit arguing against such continuing negativity, but ok (as last reaction),
1) the title of my original post could better have been 'a tentative proof'; ' but then in the text (preliminary
'draft'' , etc etc. made clear i didnt' claim yet to have a sound proof, only that i'm working on that
2) if the theoretical outcome of chess has been 'determined' to be a draw it *could* not turn out to be otherwise.
maybe the word determining doesnt occur in your pregrad Cambridge math text books, but for me it's quite clear
3) not going to argue about what you think about game theory but some of your claims that UWS should contain
a proof were imo overhastily comments and to me seem to indicate you (as well as some other math boys)
seem to have some over rigid ideas about game theory especially when applied to a game as chess.

NB now noticing a new and interesting post by Brunetti but if i react further than in a separate post (and i
wish him lots of mental strength in dealing with the math inquisition btw) Preliminary comment, although
it's clear you seem to be reasoning from the endgame backwards, in opening theory it's known that symmetry
can be broken, it's known that eg. in the Italian (!) opening Black cannot endlessly repeat/copy in a symmetric
way the White moves, because after a while there will be a tactical opportunity for White (gaining significant
advantage); this used to be thought as indication for a first move advantage but Black can avoid such tactics
ofcourse and thus this first move advantage is much less than originally thought.
https://en.wikipedia.org/wiki/First-mov ... ect%20play.
Nevertheless using similar (backwards) reasoning from the endgame position may yield
interesting results, i thought about this in the past but then thought it would only be a method
for computer dat crunchers rather than an abstract reasoning methods in chess. Anyway Black doesn't
even has to be able to avoid tactical tricks by White, it's basically sufficient to avoid a forced checkmate.
But if there's not forced checkmate, then it's a draw. As i already posted now several times in this thread.
So continuing with my response to Whiting and further reasoning, the fact that i admitted with
my ref to the wikipedia article about the abduction reasoning method that it's not a -complete- proof yet,
doesn't mean it's impossible to find such a proof.
As 'complete induction' in math (attention mr towforce as well) is a valid method of finding a proof:
https://math.libretexts.org/Bookshelves ... A_New_Page
https://www.utsc.utoronto.ca/~nick/cscB ... uction.pdf
https://www.mathcentre.ac.uk/resources/ ... -proof.pdf

The method i suggested in my first post (and on which i'll continue to work) will be based
on complete induction, using the -high- degrees of freedom which exist for the Black player
(eg. a horse being able to go to eight squares, if not on the rim, etc.).
If Black can avoid a forced mate for ply n, n+1, n+2, n+3 etc, then in combination
with this high degree of freedom (not only for the knight but also for many other pieces)
this then means in a similar way a forced mate can be avoided for n+4 etc ad infinitum.
A proof by complete induction, and a valid method, thus showing that your statement that there
*could* be a forced mate further down the line simply is wrong.

Maybe you should read a bit more about examples of proving by complete induction in math
or if being lazy just watching some youtube clips, acknowledging that such a method is
completely sound well established and accepted in math; it also is used in some theorems
in eg. mathematical control theory, going beyond the example of dynamic programming
which i posted. In such a way it was proven for example that a Kalman filter in stochastic
control gives optimum results. Now control theory is not game theory, but it's still math isn't it.
End of discussion.

But i'll be back later again.
(problem with chess is that it's hard to generalize the piece movements with algebra , thus
formulating a proof -of avoiding forced checkmates- in a mathematial rigorous way; but it
can be done; i'm thinking of some topological methods, with or without network theory; simple
2D topology, and then i hope i can write it down in such a way that even bachelors in math will be able
to understand it (although they at first may still condemn it ofcourse, skeptical or even critical thinking
in science is ok with me, but there's also fine line when becoming prejudice and negativity).

PS so the proof by Wiles wasn't accepted because there was a mistake in it ? lol
So who claimed then that there was a mistake in it? (yes this was discovered later,
but not immediately ; such things can happen; Nb there also may be a much simpler
proof for Fermat last conjecture, at least i sugggest the outcome now has been 'determined' (that
Fermat was right, although his tentative proof on the back of the envelope (or margin of his A4
paper) most likely was incomplete; we will never know (now almost trolling i admit so i'll quit).

[jefk]

jw wrote

your believed value **could** be wrong.

we all know that there are chess puzzles like mate in x eg. mate in 13 or whatever
where at first you don't see the mate. Sometimes a spectacular sacrifices initiates the
mating line, and then in all variations despite the best defense, Black is being mated.

But such positions are artificially created, and are not like the start position in chess.
Thus I maintain that from the start position, Black can always avoid a forced mate
and thus there can *not* be a force mate for White in 13, 113, 333 or whatever from
the starting position. Ergo my 'believed value' (theoretical outcome) cannot be wrong.
It's a logical result from pure logical reasoning, with backwards induction, a well
known method in game theory,
https://en.wikipedia.org/wiki/Backward_induction
in chess it's called retrograde analysis
https://en.wikipedia.org/wiki/Retrograde_analysis
That my reasoning with retrograde analysis avoiding checkmates doesn't count yet as a valid rigorous
mathematical proof is not really a concern for me, i must say. The earth *could* be flat (yeah just
transform the 3d model to a 2d model, Ptolemaean or not) good luck. But *could* there be a *forced*
mate for White from the start position in eg. 547 moves? Answer, is No, impossible; for reasons I
already specified (Black can always avoid such forced mates, thus they don't exist; this
really imo should not be so hard to understand, whether it counts as a rigorous proof or
not but now i'm repeating myself i admit (but not even trolling yet).

[JacquesRW]

If Black can avoid a forced mate for ply n, n+1, n+2, n+3 etc, then in combination
with this high degree of freedom (not only for the knight but also for many other pieces)
this then means in a similar way a forced mate can be avoided for n+4 etc ad infinitum.
A proof by complete induction, and a valid method, thus showing that your statement that there
*could* be a forced mate further down the line simply is wrong.

Consider the following position

[d]3K1B2/1p6/pp6/rk2N3/b1p5/1pP5/1P2P3/8 w - - 0 1

Black can avoid a forced mate for ply 1, 2, 3, etc, combined with the high degree of freedom (not only for the knight but also for many other pieces) this then means in a similar way a forced mate can be avoided for ply 4 etc ad infinitum.

Therefore by complete induction the position must be drawn!

But its a mate in 20.

Maybe you should read a bit more about examples of proving by complete induction in math
or if being lazy just watching some youtube clips, acknowledging that such a method is
completely sound well established and accepted in math; it also is used in some theorems
in eg. mathematical control theory, going beyond the example of dynamic programming
which i posted.

Complete induction is a perfectly valid form of proof. I never disputed that. My issue is with your incorrect usage of it.

I'll restate it:
If (P(k) holds for k < n implies P(n)), and P(j) holds for some j, then P(n) holds for all n >= j.

What you have said is that from startpos, with P(n) = forced mate loss in ply n for black:
P(1) holds - this is fine.
P(1), P(2), ..., P(n) holds, because of this and the exponential nature of the search tree, P(n+1) has a high probability of holding.

This second statement isn't what is required for a complete induction, I think you've already agreed with me on that.
But the real issue is there is some lost sight of the start position - as in the example position I provided above, a position can seem fine, for 5, 10, 20 ply, and then suddenly its actually a mate in 20 (full moves), and your argument doesn't really address this issue, in my opinion - in fact you sort of handwave it away initially:

Note that some skeptics could suggest that there may be specific positions or sequences of moves where Black cannot avoid losing, even with optimal play, and that these rare situations need to be accounted for to definitively prove that chess is a draw with perfect play. My reaction is that they ignore the logical correctness of the backward reasoning (almost like complete induction) that such 'rare' situations/positions cannot occur if Black avoids them in an earlier phase. With 'perfect play' or at least adequate play, Black should, of course, avoid such positions (similar to the situation in the initial position where White should avoid losing by not playing 1. g4?).

And this really highlights why rigorous proofs are needed, because with this statement, you don't need all this other stuff you detailed:
"With 'perfect play' or at least adequate play, Black should, of course, avoid such [losing] positions"
Seems to me like you have simply used that "black can and should avoid losing positions", in an argument to prove that black can avoid losing?

[jefk]

ok i have to reread your posting, but here some quick preliminary comments

the position you give is an artificial position (just i like i mentioned chess problems/puzzles
as artificial positions); it would not occur in a perfect game of chess from the start position
because it could have been avoided.

While in my verbal reasoning there still may be (and apparently are) some tiny flaws i admit,
my main point is the reasoning from the starting position, or to be more accurate ply 4 or so
when the first checkmates can occur.
My thesis, aiming at backward induction, is that if Black can avoid a forced checkmate at
ply n, then Black can also avoid a forced checkmate at ply n+1.
And indeed at ply 4,5,6 (a bit superfluous i admit) from the start position i showed that Black can (easily)
avoid such a forced checkmate. The crucial (and possibly at this stage still weakest) point in the reasoning
is to show (ideally, prove) that if Black can avoid a checkmate at ply n = x (eg) 6, it follows from logical and
general reasoning (and the rules of chess) also a checkmate at ply N +1 eg. 6+1 ply 7.
This can't be so difficult, it's a retrograde chess analysis problem and imo can be solved. The rest of
the reasoning you know by now (complete induction, thus no checkmate at ply 1387 or whatever)

PS it's a bit like a -hypothetical- case where a certain c.queen would write something like
"1...d5?? after 1.d4! is wrong ! because i played SF1 against SF2 and then SF1 won with a Catalan !
Or do we have solved chess (with 1.d4!!) ? " etc

Upon which i would write, dear mr c.queen while Catalan is a strong opening, Black could have avoided
this by playing a Slav (with c6 instead of e6). Thus i'm sorry to disappoint you but we haven't solved chess.
And Black can also play 1...Nf6! and go for a Gruenfeld with ...g6! (but i'm digressing now a bit).

[JacquesRW]

...

Okay, we're in agreement on the current state of the argument. I am of the opinion that the weakness in the argument will be insanely difficult/effectively impossible to resolve, but good luck.

[jefk]

ok,

PS fyi
https://www.perplexity.ai/search/who-we ... .DrIZmiA#0

[JacquesRW]

PS fyi
https://www.perplexity.ai/search/who-we ... .DrIZmiA#0

Am aware. As I said, it was not accepted until the proof was corrected.

[towforce]

jefk's posts have given me an idea to provide more evidence in support of chess being drawn:

1. Pick a random set of positions from a tablebase in which both player have, say, 3 pieces of equal material, but which are winning for white

2. In each position, calculate what proportion of moves maintain the position as winning

3. Pick a random set of positions from a tablebase in which both player have, say, 3 pieces of equal material, but which are drawing for white

4. In each position, work out what proportion of the possible white moves maintain the position as drawn

This way, you'll be able to statistically demonstrate that in most positions, there are more ways to draw then there are to win.

This obviously doesn't PROVE that chess is drawn, but it would show that:

1. Getting a draw is easier than getting a win

2. Per jefk's assertion, the tree of draws grows exponentially faster than the tree of wins against number of moves

[shawn]

While we do not yet have tablebases for this exact configuration (exceeding 7 pieces), it is reasonable to assume, based on known principles and experience, that the position remains a draw.

Not a proof