Queens vs Knights

[Lyudmil Tsvetkov]

and of course, the queens can always convert from 3Qs vs 7Ns to 2Qs vs 5Ns, and then 1Q vs 3Ns.

[Evert]

8,5-1,5 in favour of Nightfish. Stockfish' single win was with Black.

Yeah, small number of games, but there's only so much time I have to invest in this.

and the N-Q score?

9,5-0,5, obviously.

with 7 knights and 7 queens, you are giving the knight side tremendoius bonus for nothing.

Well, clearly QueenValue >> QueenKnightElephantiasis.
Anyway, I prefer to not think in terms of giving the knights a bonus, but penalising the Queens. It works out to the same thing in the end, but conceptually it's very different.

you forget catering for a mix of pieces, not a pure only-piece setting, which is the general form of the phenomenon.

in standard chess, in order to correctly assign values to all possible imbalance piece configurations, one should build an impossibly large table of at least couple of million entries.

Think of it this way: the material evaluation is a function of the number of pieces of each type for both sides. The functional form is unknown, but we can expand the function in terms of its free parameters (the number of pieces). The lowest order (linear) terms give the normal piece values, the quadratic terms give corrections for having more than one piece of a given type (the Bishop pair bonus is the most notable one here, although strictly speaking that is a cross term because we should treat light and dark squared Bishops differently), and then the cross terms come in for the elephantiasis effect. To get any sort of interaction between three piece types you need at least third order terms.

[Evert]

27% is not small at all, as you need only another 23% for perfect equality concerning the outcome of the game.

No you don't. We're talking 27% draws, 73% losses here. Raising that to 50% draws still gets you 50% losses.

[Evert]

by constant you mean increasing the knight value?

that certainly should be handled by the imbalance formula.

The coefficient for the knight/queen cross-term.

in any case, it is obvious, 7Ns at most draw 3Qs.

Not to put too fine a point on it, but that's a pretty odd interpretation of the results presented by the tests conducted here, which show that the Queens are lucky to draw...

[Evert]

as said, and tested in practice, in pure case scenario, 7Ns equal 3Qs.

I have played many games like that, and all games end in a draw.

No point in testing that, it's fairly obvious since trading Q for 2N, or even 3Q for 5N, leaves the Knight side without mate potential, and the Queens are strong enough that Black cannot avoid such a trade.
In other words, your "pure case" is rigged.

[Lyudmil Tsvetkov]

as said, and tested in practice, in pure case scenario, 7Ns equal 3Qs.

I have played many games like that, and all games end in a draw.

No point in testing that, it's fairly obvious since trading Q for 2N, or even 3Q for 5N, leaves the Knight side without mate potential, and the Queens are strong enough that Black cannot avoid such a trade.
In other words, your "pure case" is rigged.

not rigged, that is the basic measure.

3Qs win vs 6Ns in a pure setting always,
while 2Qs lose vs 8Ns in a pure setting always.

so, 1Q is known to be worth more than 2Ns and less than 3Ns.

similarly, we know in a pure setting R vs B is draw, while Q vs R or B win.

RR vs NN is rook win, similarly in most cases Q vs NN.

we should start from the basic points and then proceed to more complex ones.

the most basic point is material on an empty board.

[hgm]

please note again, that in a pure setting, on an empty board, 7Ns = 3Qs, so choosing the right additional elements of the position is indeed very important.

One assumes that this is a clumsy way to say "withouth Pawns"? Obviously a board with Knights and Queens on it is not empty. And without Kings every game would be trivially drawn anyway, as there is no checkmate possible.

if you need a pure setting formula/scenario, it is quite simple, 7Ns = 3Qs.

that is 1 Q = 2.33 Ns.

So your understanding of Chess is so much below patzer level that you even don't know that games without Pawns are far more difficult to win than with Pawns?

This is quite well known, amongst amateur players. KRKB is generally a draw, and KRBKR too. That does in no way imply that R=B, or that B=0. In the presence of equal numbers of Pawns these imbalances are generally winning.

So that a Pawnless 3Q-vs-7N would be a draw by no means implies that 3Q = 7N. Only that the difference is less than the ~4 Pawns needed to win a Pawnless end-game. If indeed it would be a draw, which still remains to be seen. I will switch the broadcast to trying this without Pawns.

[Edit] First Pawnless QueeNy vs. QueeNy game (not broadcasted) was an easy win for the Knights: 26 moves. Final position: 4 Knights against bare King. Couldn't prevent 3 Knights being traded away for Queens...

[pgn][Event "Computer Chess Game"]
[Site "MAKRO-PC"]
[Date "2017.07.19"]
[Round "-"]
[White "QueeNy 0.16"]
[Black "QueeNy 0.16"]
[Result "0-1"]
[TimeControl "40/600"]
[FEN "3nkn2/3n1n2/3nnn2/8/8/8/8/2QQKQ2 w - - 0 1"]
[SetUp "1"]
1. Qa4 {-5.79/14} Nfe4 {+5.88/14 10} 2. Qe2 {-5.79/14 8} N6g5 {+6.12/14 11}
3. Qce3 {-6.15/13 12} Nfe6 {+6.34/15 12} 4. Kf1 {-6.26/13 12} Ne5
{+6.43/14 11} 5. Kg2 {-6.48/13 8} Ke7 {+6.60/14 12} 6. Qaa3 {-6.61/14 8}
Ndc5 {+6.72/14 12} 7. Qa7+ {-6.80/14 15} Kf6 {+6.84/14 9} 8. Qb8
{-6.91/14 15} Ndc6 {+6.83/14 10} 9. Qg8 {-7.04/14 21} Nf5 {+7.12/14 17} 10.
Qf1 {-10.81/15 16} Ncd4 {+10.75/14 9} 11. Qh8+ {-10.83/14 12} Kg6
{+10.88/15 16} 12. Qg8+ {-10.89/13 9} Neg7 {+11.20/15 11} 13. Qa3
{-11.07/14 11} Ndf3 {+15.51/14 15} 14. Qfxf3 {-15.53/13 10} Ngxf3
{+16.27/15 8} 15. Kf1 {-16.33/15 14} Ncd3 {+16.42/15 15} 16. Qxd3
{-16.58/16 12} Nfg3+ {+20.06/17 13} 17. Kg2 {-21.02/19 19} Ne1+
{+20.94/18 19} 18. Kg1 {-319.89/18 13} N1xd3 {+319.90/18 13} 19. Qd5
{-26.06/20 9} Kh5 {+319.92/20 11} 20. Qg8 {-319.92/22 9} N7f5
{+319.93/23 10} 21. Qh7+ {-319.94/25 16} Kg4 {+319.94/27 9} 22. Qg8+
{-319.95/27 14} Ng5 {+319.95/27 13} 23. Qxg5+ {-319.96/31 31} Kxg5
{+319.96/33 10} 24. Kh2 {-319.97/32 11} Kg4 {+319.97/31 15} 25. Kg1
{-319.98/36 12} Ne3 {+319.98/34 10} 26. Kh2 {-319.99/39 11} Nf3#
{+319.99/37 10}
{Xboard adjudication: Checkmate} 0-1
[/pgn]

I set up the broadcast now for this, starting at game 34 (40 moves/10 min). Unfortunately, because I changed the start position in the HTML page, this means games 1-33 cannot be recalled anymore.

[hgm]

No point in testing that, it's fairly obvious since trading Q for 2N, or even 3Q for 5N, leaves the Knight side without mate potential, and the Queens are strong enough that Black cannot avoid such a trade.

Actually it is not so obvious that the Queens are strong enough for that. After 4 QueeNy self-play games it is 3.5-0.5 for the Knights. QueeNy does strive for Q-for-2N trades with the Queens, but with equal tactical ability it just cannot find a way to force any before the Knights can inflict damage. Funny enough the first thing the Knights see typically seems to be a quite complex 2Q-for-3N trade! It is biased to ignore any Q-for-2N opportunities, but thinks 2Q-for-3N is a good deal (2*9.5 for 3*5). You then immediately end up in 4N vs. Q, which (in the kind of compact positions you typically convert to) should be an easy win for the Knights.

Unfortunately it is absolutely clueless for how to win KNNNNKQ, and often aimlessly wanders the Knight pack to one side of the board, while the enemy King is at the other side. I suppose a strong attractive evaluation term between the Kings would solve this. If it happens to wander close, a mate threat comes within the horizon, typically delayed by a final Q-for-N trade, and a KNNNK mate.

So I get the impression that even without Pawns it is an easy win for the Knights. Of course the fact that Lyudmil claimed it was not so, was already a convincing indication for that.

[petero2]

... Only that the difference is less than the ~4 Pawns needed to win a Pawnless end-game.

I don't think that rule is accurate when there are more than 5 pieces (including kings) on the board. For example, it seems KQKNNNN is generally won by the knights, and KRRBKRR is often won by the side with the bishop.

When I say "often" I assume the initial position is tactically quiet. If you set up a random position there is often some immediate tactics available that would lead to a forced transition into another endgame class.

For example, according to lomonosov tablebases, in the following position black mates in 35 moves:
[d]3nk3/3nnn2/8/8/3QK3/8/8/8 w - - 0 1

[Evert]

No point in testing that, it's fairly obvious since trading Q for 2N, or even 3Q for 5N, leaves the Knight side without mate potential, and the Queens are strong enough that Black cannot avoid such a trade.

Actually it is not so obvious that the Queens are strong enough for that. After 4 QueeNy self-play games it is 3.5-0.5 for the Knights. QueeNy does strive for Q-for-2N trades with the Queens, but with equal tactical ability it just cannot find a way to force any before the Knights can inflict damage. Funny enough the first thing the Knights see typically seems to be a quite complex 2Q-for-3N trade! It is biased to ignore any Q-for-2N opportunities, but thinks 2Q-for-3N is a good deal (2*9.5 for 3*5). You then immediately end up in 4N vs. Q, which (in the kind of compact positions you typically convert to) should be an easy win for the Knights.

Unfortunately it is absolutely clueless for how to win KNNNNKQ, and often aimlessly wanders the Knight pack to one side of the board, while the enemy King is at the other side. I suppose a strong attractive evaluation term between the Kings would solve this. If it happens to wander close, a mate threat comes within the horizon, typically delayed by a final Q-for-N trade, and a KNNNK mate.

So I get the impression that even without Pawns it is an easy win for the Knights.

Impressive, and somewhat counter-intuitive. Of course, intuition is unneeded in the presence of actual data.