How powerful is the Queen compared to 2 Rooks ?

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hgm
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Re: How powerful is the Queen compared to 2 Rooks ?

Post by hgm » Wed Jan 30, 2019 4:31 pm

Nordlandia wrote:
Wed Jan 30, 2019 12:43 pm
Your implying black is the favourable side here.



For the queen white has archbishop + pawn as compensation.



Capablanca Chess Pieces in staunton:

I have to say that the chancellor looks pretty much like the queen. It resembles more like an amazon if you ask me.

Image

Definitely in the second position, which I extensively tested. With the first-move advantage averaged out, this scores about 54% in favor of the Archbishop. The first one I never tried; it doesn't really look tactically quiet, because there are so many Pawns that it will be impossible to protect them all, so that is likely to turn into a race who can capture the most. It also is a very unnatural position to have all Pawns on their starting rank with almost all piece material traded off. In practice you would be dealing with Pawn chains that already have some coherence, and about half the Pawns already gone. To 5 vs 6 Pawns would be more representative.

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Re: How powerful is the Queen compared to 2 Rooks ?

Post by Chessqueen » Wed Jan 30, 2019 7:16 pm

Nordlandia wrote:
Mon Jan 28, 2019 10:26 am
While two bishops + two pawns is stronger than rook and knight.

Example position: Stockfish is capable of beating Komodo as white here.

But if you take the d3 pawn it will end up in a draw :shock:

Here it is, I decided to stopped it since it is obvious that it will be a draw :shock:


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Re: How powerful is the Queen compared to 2 Rooks ?

Post by Dicaste » Wed Jan 30, 2019 7:36 pm

If you think simply Q = R + 2B(1 bishop can only move to 1 colored square).

Image

3 pieces vs 1 gives a little edge to them so it's Q + x = R + 2B (without positional parameters + tempo)
Last edited by Dicaste on Wed Jan 30, 2019 7:53 pm, edited 1 time in total.

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Re: How powerful is the Queen compared to 2 Rooks ?

Post by Dann Corbit » Wed Jan 30, 2019 7:38 pm

Only problem is some nut case strapped all three of them together. They would work a lot better if you could let them loose.
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Re: How powerful is the Queen compared to 2 Rooks ?

Post by Chessqueen » Wed Jan 30, 2019 10:42 pm

Nordlandia wrote:
Tue Jan 29, 2019 7:37 pm
Trivial edge for white. Q is stronger than RNP but again supressed by redundancy.



Slight edge for black.



Maybe winning for white.

At least you should give an extra pawn to black, an extra f7 pawn llke this to make it even since the Black King on f7 without an extra pawn is too exposed to the attack of two Queens but with the King on f6 and an extra pawn on f7 the position is even and there is no advantage on either side.
[D]2Q5/5p2/5kp1/3rr3/8/7P/QP4PK/3q4 b - - 52 1

And here you can see that the game is heading to another draw, which is obvious, therefore I stopped it at this point:

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Re: How powerful is the Queen compared to 2 Rooks ?

Post by Sesse » Wed Jan 30, 2019 11:50 pm

Dann Corbit wrote:
Mon Jan 28, 2019 7:48 pm
Here is what Stockfish says about the value of chessmen (note that pawns are not 100) during the midgame and during the endgame:

Code: Select all

    PawnValueMg   = 136,   PawnValueEg   = 208,
    KnightValueMg = 782,   KnightValueEg = 865,
    BishopValueMg = 830,   BishopValueEg = 918,
    RookValueMg   = 1289,  RookValueEg   = 1378,
    QueenValueMg  = 2529,  QueenValueEg  = 2687,
Note that two rooks are 2578 so a bit better than the queen in midgame. But it's close enough where it doesn't matter much.
Note that two rooks are 2756 so significantly better than the queen in the endgame.
It appears that Stockfish's lips are sealed about the value in the opening.
This is somewhat inaccurate; Stockfish also has second-order evaluation, where pieces' values are influenced by what else is on the board.

Code: Select all

  constexpr int QuadraticOurs[][PIECE_TYPE_NB] = {
    //            OUR PIECES
    // pair pawn knight bishop rook queen
    {1438                               }, // Bishop pair
    {  40,   38                         }, // Pawn
    {  32,  255, -62                    }, // Knight      OUR PIECES
    {   0,  104,   4,    0              }, // Bishop
    { -26,   -2,  47,   105,  -208      }, // Rook
    {-189,   24, 117,   133,  -134, -6  }  // Queen
  };                                                                                          

  constexpr int QuadraticTheirs[][PIECE_TYPE_NB] = {
    //           THEIR PIECES
    // pair pawn knight bishop rook queen
    {   0                               }, // Bishop pair
    {  36,    0                         }, // Pawn
    {   9,   63,   0                    }, // Knight      OUR PIECES
    {  59,   65,  42,     0             }, // Bishop
    {  46,   39,  24,   -24,    0       }, // Rook
    {  97,  100, -42,   137,  268,    0 }  // Queen
  };
So a rook gets a penalty for each friendly rook (seemingly including itself, but I might be wrong here), and the queen gets a bonus for each rook it is opposing. (The bishop pair is important enough that it's assigned its own “pseudo-piece”, so that's really a limited third order evaluation.)

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Re: How powerful is the Queen compared to 2 Rooks ?

Post by lkaufman » Thu Jan 31, 2019 6:25 am

hgm wrote:
Tue Jan 29, 2019 8:28 am
Although I do agree with the final conclusion, I am rather skeptical about the 'explanation'. I have always thought this concept of 'redundancy' to be a red herring, and that what sometimes is ascribed to redundancy in reality is due to the 'leveling effect' proposed by Ralph Betza: stronger pieces lose value when up against weaker opponent pieces.

When there is two Rooks against one, and the outcome is still at stake, the side with one Rook must have some other material in compensation. And if Queens are explicitly excluded, this must be minors. But the presence of these minors then devaluates both opponent Rooks, while the lone Rook is not bothered by minors (or by fewer minors), and thus effectively worth more. So trading it is a good deal for the player with two Rooks.

The leveling effect would predict the same for two unequal pieces of Rook value. E.g. a Rook plus a piece that moved as Knight or step one square diagonally. There would be no redundancy in that case (the pieces do not share a single move!), but it would still be favorable for the player that has R + augmented N to trade Rooks against an opponent that has R + B-pair.

A similar effect is very obvious in Capablanca Chess, where it is bad (like in orthodox Chess) to trade your Q for R+B if the other two super-pieces (Archbishop and Chancellor) are no longer present, but quite favorable when all A and C are still on the board: the remaining A and C devaluate because of the opposing R an B, while they are not so much impressed by an opposing Q. (More the other way around...) So trading A and C becomes quite favorable for the player that still has Q, while A and C are not really redundant w.r.t. Q (and in any case not more than w.r.t. R + B, which share the same moves with them as Q does).
I have been interested in this idea about weaker pieces devaluating stronger ones ever since you told me about it long ago. But we were never able to show a benefit to Komodo from this idea, the redundancy method always worked better. One issue is this: White giving knight odds (remove b1 knight), the extra Black knight reduces the value of White's majors according to your theory. Therefore White should try to exchange queens and rooks! But of course this is exactly the opposite of correct strategy, Black should be the one seeking those trades. But if all pieces of one side are considered redundant to varying degrees (except for the two bishops obviously), then Black will (correctly) seek exchanges.
Komodo rules!

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Re: How powerful is the Queen compared to 2 Rooks ?

Post by hgm » Thu Jan 31, 2019 9:07 am

I think that in the Knight-odds example the leveling effect is masked by an even more general principle: when ahead in material, simplify by trading pieces (but conserve Pawns). This principle even applies in cases where all pieces are equal (so that there cannot be any leveling effect), e.g. 4 Knights vs 3. In the usual framework of opening/end-game evaluation interpolated by game phase, the latter measured as total material, this would be implemented as a hefty (but proportionl) increase of all piece values towards the end-game. Intuitively it would make sense to express material avantages as a fraction (white - black)/(white + black) (where even the King could then count for some), but the division is a bit awkward, so we usually take a linear growth like (white - black) - C*(white + black)*(white-black), (where for simplicity I assumed that the weights for determining the game phase (white + black) are the same as the piece values, and all values decrease in the same proportion C towards the opening). The second-order term then is C*(black*black - white*white), i.e. it mainly contains interactions between own pieces. But it doesn't really discriminate between piece types, other than weighting those by their 'intrinsic' value. Basically it says: if you have many pieces, every one of those is somewhat redundant (white army strength = white - C*white*white). No matter how different they move.

Such a 'type-blind' general redundance (as an approximation to expressing the advantage as a relative quantity) should be enough to explain why trading Queens in the face of Knight odds is a bad idea. Sure, the leveling effect was making your own Queen a bit less valuable than his, because she was facing that extra Knight. But that did not take account of the fact that you needed that Queen much harder, because she also had to perform the tasks that otherwise could have been done by that missing Knight. Or, equivalently, the opponent's Queen was to some extent redundant because that extra Knight could sometimes releave her.

So perhaps I should be more precise in my statement: rather than saying that I don't believe in redundancy, I should say I do not believe in type-specific, but think that redunancy should be type-blind, and just express that it is the relative material advantage that matters. But then it can be a pretty big effect. With 6:3:2 weighting of Q:R:minor, the start total of a Knight-odds game would have 38 points, and trading Queens would reduce it to 26. That would be a 46% increase of the relative piece values, so 1.5 Pawn for a Knight's advantage. This will be much larger than the leveling effect of a single Knight on a Queen.

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Re: How powerful is the Queen compared to 2 Rooks ?

Post by Chessqueen » Thu Jan 31, 2019 12:14 pm

Chessqueen wrote:
Wed Jan 30, 2019 10:42 pm
Nordlandia wrote:
Tue Jan 29, 2019 7:37 pm

Maybe winning for white.

At least you should give an extra pawn to black, an extra f7 pawn llke this to make it even since the Black King on f7 without an extra pawn is too exposed to the attack of two Queens but with the King on f6 and an extra pawn on f7 the position is even and there is no advantage on either side.
[D]2Q5/5p2/5kp1/3rr3/8/7P/QP4PK/3q4 b - - 52 1

Even if you have an even number of pawns if Black start moving first the result will be draw:

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Re: How powerful is the Queen compared to 2 Rooks ?

Post by Nordlandia » Thu Jan 31, 2019 3:49 pm

S-Chess stockfish fork.

PawnValueMg = 166, PawnValueEg = 245,
KnightValueMg = 801, KnightValueEg = 772,
BishopValueMg = 893, BishopValueEg = 856,
RookValueMg = 1313, RookValueEg = 1261,
HawkValueMg = 1954, HawkValueEg = 2172,
ElephantValueMg = 2060, ElephantValueEg = 2556,
QueenValueMg = 2198, QueenValueEg = 2617,


Average value Q= 2407,5 : 265 ~ 9
Average value C= 2308 : 265 ~ 8,70
Average value H= 2063 : 265 ~ 7.785
Average value R= 1287 : 265 ~ 4.9
Average value B 874,5 : 265 ~ 3.3
Average value N 786,5 : 265 ~ 3
Average value P 205,5 : 265 ~ 0.8

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