Solving chess? With no BS. (moderated)

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tygxc

#160
This includes a vast majority of non sensible positions with more than 2 queens, rooks, bishops, or knights per side. The number of sensible positions with maximum 2 queens, rooks, bishops, knights per side is estimated at a factor of 1 million less.

DiogenesDue
tygxc wrote:

#160
This includes a vast majority of non sensible positions with more than 2 queens, rooks, bishops, or knights per side. The number of sensible positions with maximum 2 queens, rooks, bishops, knights per side is estimated at a factor of 1 million less.

By you, but not by Mr. Tromp's results.  He deftly avoided the notion of "sensible" positions as being too fuzzy/fluffy.

Even if you were to reduce the number to 10^38 or 10^39, that's still almost 20 orders of magnitude off your 10^20 number, and it still leaves the solving of chess off the table for the foreseeable future.  Good luck on finding a million * a million * a million more positions to eliminate happy.png.

tygxc

#162
John Tromp wrote:
"The fraction of positions with at most 1 extra queen promotion is only 0.00426%."
Sensible = possible with a standard set with 2 spare queens is neither fuzzy nor fluffy.
To solve chess is is unnecessary to visit all possible legal positions.
Hence the estimate of 10^20 nodes to visit by analogy of the way how checkers was solved visiting only a small fraction of the possible legal positions.

DiogenesDue
tygxc wrote:

#162
John Tromp wrote:
"The fraction of positions with at most 1 extra queen promotion is only 0.00426%."
Sensible = possible with a standard set with 2 spare queens is neither fuzzy nor fluffy.
To solve chess is is unnecessary to visit all possible legal positions.
Hence the estimate of 10^20 nodes to visit by analogy of the way how checkers was solved visiting only a small fraction of the possible legal positions.

You're still over 18 orders of magnitude off by your own dubious numbers wink.png

You also apparently have a problem with converting 0.00426%...that's in the thousands range, not millions, so you are actually 21 orders of magnitude off.

Circumlocutions
Unnecessary bump
DiogenesDue

I found this tidbit for Reddit interesting, and I am guessing it will also be useful in future discussions:

We can try to extrapolate that based on the number of positions that are in the endgame tablebases we have so far. I said extrapolate, so past 8 this is an estimate. Calculating it precisely can't be done without actually trying to find the positions because you don't yet know how many of your possible combination will be illegal moves.

Number of pieces
Number of positions
2
462
3
368,079
4
125,246,598
5
25,912,594,054
6
3,787,154,440,416
7
423,836,835,667,331
8
38,176,306,877,748,245
9
2,748,694,095,197,870,400
10
153,899,382,390,128,763,696
11
7,460,272,561,361,491,820,163
12
309,265,599,031,240,643,404,882
13
10,954,187,517,686,543,589,400,921
14
327,311,123,028,473,922,451,299,542
15
8,289,154,190,696,102,086,079,160,925
16
176,268,863,865,152,610,860,473,357,082
17
3,119,958,890,413,201,212,230,378,420,357
18
47,173,778,423,047,602,328,923,321,715,810
19
684,019,787,134,190,233,769,388,164,879,256
20
9,644,678,998,592,082,296,148,373,124,797,509
21
134,061,038,080,429,943,916,462,386,434,685,384
22
1850,042,325,509,933,226,047,180,932,798,658,303
23
25,345,579,859,486,085,196,846,378,779,341,618,751
24
344,699,886,089,010,758,677,110,751,399,046,015,017
25
4,687,918,450,810,546,318,008,706,219,027,025,804,242
26
60,942,939,860,537,102,134,113,180,847,351,335,455,148
27
792,258,218,186,982,327,743,471,351,015,567,360,916,934
28
10,299,356,836,430,770,260,665,127,563,202,375,691,920,150
29
133,891,638,873,600,013,388,646,658,321,630,883,994,961,953
30
1,740,591,305,356,800,174,052,406,558,181,201,491,934,505,398
31
22,627,686,969,638,402,262,681,285,256,355,619,395,148,570,184
32
294,159,930,605,299,229,414,856,708,332,623,052,136,931,412,396

Now let's add all these numbers together and we get 3,186,73,258,137,120,909,454,722,754,552,490,167,098,995,479,554

Now let's add all the numbers of the first 7 and we get 427,650,028,316,940

We divide the second number by the first number and do it times 100 to get a percentage and then we can say that

Chess is 0.0000000000000000000000000000001342% solved.

tygxc

#166
You cannot extrapolate like that. Research by John Tromp shows that most positions are with 28 men.
You also cannot add positions: the number of positions in the 7 men table base also comprises 6, 5, 4, 3, and 2 men.
7 is not the end: already now part of the 8 men table base is ready.
The 7 men table bases also contain many illegal or irrelevant positions, like with 3 light square bishops.
Your number is also wrong: John Tromp came up with better estimates. The total number of legal and sensible positions: 10^38, not 10^48 as you add.

DiogenesDue
tygxc wrote:

#166
You cannot extrapolate like that. Research by John Tromp shows that most positions are with 28 men.
You also cannot add positions: the number of positions in the 7 men table base also comprises 6, 5, 4, 3, and 2 men.
7 is not the end: already now part of the 8 men table base is ready.
The 7 men table bases also contain many illegal or irrelevant positions, like with 3 light square bishops.
Your number is also wrong: John Tromp came up with better estimates. The total number of legal and sensible positions: 10^38, not 10^48 as you add.

Not really the point, now is it?  The quote is from Reddit, and I didn't even claim the numbers were confirmed to be exactly accurate, by the way (and in fact the author disclaimers that fact himself).  But they do illustrate a point.  You seem to be stuck on the tablebase thing, though...he didn't actually say those numbers represented a cumulative tablebase.  They might just be raw calculations of the number of positions specifically for that exact number pieces.  Feel free to do the homework and confirm...it's past midnight here and honestly, the point I was making stands either way, so I have no particular reason to do a deeper dive right now.

While you are at it, go ahead and calculate the percentage chess is solved with your 10^38 number (though I believe we just got done establishing that it's 10^41 with Tromp's numbers, since you "misread" thousands as millions) and an 8 man tablebase wink.png.  It will still be a ridiculously small fraction so far below 1% as to be meaningless, which puts your whole "we're on the cusp of solving chess" into its proper perspective.

Also, stop pitching (and then extrapolating from) Mr. Tromp's numbers here.  I already linked his work, and he can speak for himself if need be.   

tygxc

#168
The numbers of #166 are way off so they are neither useful nor interesting.
John Tromp gave more detailed numbers on positions per number of men and positions per number of excess promotions and also an estimate on fraction of legal positions from sampling. To solve chess positions with 5 bishops and 3 rooks are irrelevant even if legal.
Also once more: solving chess does not need to visit all positions or establish a 32 men table base. It is enough to calculate from the initial position towards a sufficiently large table base as Schaeffer did to prove checkers a draw. It is only necessary to visit about the square root of the number of legal and sensible positions.

MARattigan
tygxc wrote:

... To solve chess positions with 5 bishops and 3 rooks are irrelevant even if legal.
...

Intuitively highly plausible (I assume you mean for the same side), but I think difficult to prove.

tygxc

#170
Here is a typical position from the John Tromp sampling. This position has been proven to be legal. It is however quite clear that it is neither sensible nor relevant to solving chess. That is not an exception: all legal positions from the John Tromp sampling look that way.



MARattigan

Probably not a typical position where one side (Black) has three rooks and five bishops. White also has unusual material.

Nevertheless, when you say it's "clear" that it is neither sensible nor relevant to solving chess, I think that means exactly that it's intuitively highly plausible. I think probably still difficult to prove even for that position. 

In fact you should say "weakly solving chess" otherwise it's false. You should also probably say under current FIDE basic rules (I don't think Tromp's estimates include different ply counts under the 50 move rule or different previous positions).

tygxc

#172
I can understand white promotes to 2 queens, but why on earth would white promote to 4 rooks, 2 dark square bishops, 3 knights and black to 3 rooks, 2 dark square bishops and 3 light square bishops? This makes no sense at all. It is not necessary to prove that this position and many many similar ones play absolutely no role in solving chess, i.e. that the proof of chess being a draw will not need this position.

This means that the total number of positions calculated by John Tromp is too large when used to reason about solving chess. John Tromp also gave the number of positions with no excess promotions, i.e. the number of positions possible with 1 set of chess pieces, without need to borrow excess queens, rooks, bishops, knights from additional sets.

John Tromp has calculated the number of positions without excess promotions i.e. possible with one chess set as:
19201527561695835455154058755564594798074
an estimated 5.2% of these is legal i.e. 9.98e+38

WPyellow
StormCentre3 wrote:

I argue before any attempt at proving a forced win by White - 1st it is necessary to prove initiative = advantage. Should be a simpler task ?

Otherwise - could not Black simply mimic White forcing a draw ? Or perhaps even win as White exhausts possible moves resulting in zug ?

We know many positions whereby the side to move losses by force.

And mimicking doesn't work. I used this against a very annoying club member

MARattigan
tygxc wrote:

#172
I can understand white promotes to 2 queens, but why on earth would white promote to 4 rooks, 2 dark square bishops, 3 knights and black to 3 rooks, 2 dark square bishops and 3 light square bishops? This makes no sense at all. It is not necessary to prove that this position and many many similar ones play absolutely no role in solving chess, i.e. that the proof of chess being a draw will not need this position.

This means that the total number of positions calculated by John Tromp is too large when used to reason about solving chess. John Tromp also gave the number of positions with no excess promotions, i.e. the number of positions possible with 1 set of chess pieces, without need to borrow excess queens, rooks, bishops, knights from additional sets.

John Tromp has calculated the number of positions without excess promotions i.e. possible with one chess set as:
19201527561695835455154058755564594798074
an estimated 5.2% of these is legal i.e. 9.98e+38

The general line of reasoning that x is impossible because I can't think of any way to achieve x led to a lot of conclusions in endgame analysis that were subsequently proved wrong by EGTBs. 

It's not a valid proof.

tygxc

#175
OK then provide one counterexample from end game table bases where promotion to more than 2 rooks, more than 2 bishops or more than 2 knights is necessary and thus relevant.

MARattigan
WPyellow wrote:
StormCentre3 wrote:

I argue before any attempt at proving a forced win by White - 1st it is necessary to prove initiative = advantage. Should be a simpler task ?

Otherwise - could not Black simply mimic White forcing a draw ? Or perhaps even win as White exhausts possible moves resulting in zug ?

We know many positions whereby the side to move losses by force.

And mimicking doesn't work. I used this against a very annoying club member

If by initiative in the starting position you mean having first move then what you suggest needs to be proved first is equivalent to the starting position being a forced win for White, i.e. an ultra-weak solution (https://en.wikipedia.org/wiki/Solved_game). Any ideas where to start?

tygxc

#177
There is concensus among all experts (Steinitz, Lasker, Capablanca, Fischer, Kasparov...) that chess is a draw and world championship games, ICCF games, TCEC games corroborate that. So to solve chess is to prove chess is a draw, like was done for checkers.
Where to start?
Calculate from the initial position towards a sufficiently large table base of 8 men or more and thus prove that for every sensible move from white there exists at least one move by black that ultimately leads to a table base draw.

Iron-Toad

Appeal to authority should not be accepted as part of a rigorous proof.  And humans are not that great at chess anyway.

tygxc

#179
Checkers was also conjectured to be a draw long before it was proved. Of course there is not yet a proof that chess is a draw, as that would imply chess being solved. However, you cannot discard expert opinions from people who dedicated their lives on the game and were world champions. The engines of TCEC and ICCF corroborate that chess is a draw. TCEC and ICCF do not apply the 50 moves draw rule. There is a scientific paper with AlphaZero that shows chess even stays a draw if stalemate were a win.
Anyway, this thread is about solving chess i.e. about the process of proving the conjecture that chess is a draw.