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Lawson on Fritz & Tal

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Lots of evergreens trotted out in this one, but good writing is always worth a link. Newsman Dominic Lawson, who wrote The Inner Game on the Kasparov-Short match (and Short's run-up to it), has a piece here that starts on the Kramnik-Fritz match but focuses more on Mikhail Tal.


It's a nice piece, but he doesn't mention his chess match with Lennox Lewis or his sister Nigella's 43DD chef apron. These are very serious issues we have a right to know about.

Great piece. I was fortunate enough to see Tal in action twice (in the 70s). Forgive me for hyperbole, but he did not play chess, he WAS chess.

I never heard about Tal's fingers before. Did he have three fingers and a thumb on each hand, or just the right hand? Or was it two fingers and a thumb?

Nice article, by the way. Thanks for putting it up, Mig.

Vladimir Kramnik paid Tal what must be the greatest of all tributes in this month of his 70th anniversary: "Analyzing his games is tantamount to discussing what God looks like."

That's a fantastic line! :)

His left hand was normal. His right hand had three digits, one of them in the thumb position, all of them larger than usual. (Surely a photo exists somewhere?

William Hartston wrote somewhere that this was the hand he was given to play chess (I paraphrase). Three fingers is just right to move pieces, the rest is surplus to requirements.
I am sure he meant it respectfully, I concur.

A couple of weeks ago Mig posted the following link to a photo:


Nice piece but but but however, there are definately not more possible chess positions (TFA mentions 10^128) than atoms in the universe!! Even if chess had 32 unique pieces and every way to place them on 64 squares was legal, there can be no more than 64!/32! = 482219923991114978843459072919892677776312893440000000, or less than 10^55 positions. Idententical pieces and symmetry and lowers this number somewhat, promotions raise it a little, but thats the idea.
The number of _possible games_ is a different story, because many games lead to the same positions. But knowing all possible positions and their result would 'solve' chess just fine.

the number of legal positons in standard chess
is aprox. "10^50", while in GO is "10^480"
(see wikipedia)

Indeed; he means the number of possible games of chess; a much-quoted factoid, although I have no idea whether it's true. How many of these games are at all interesting is another matter: the application of a two-fold repetition rule would clearly cut that number drastically.

But Lawson's just another crap non-chess journalist; you wouldn't expect him to get anything right. Reminds me of that fellow Schroeder (do I mean Schroeder? That chap in the seventies whose pop history was so comprehensively rubbished by the likes of Heidenfeld and Winter). Although of the two Lawson writes slightly less badly.

I find it hard to believe that the number of possible positions in go can be greater than the number of possible games of chess, given the repetition possibilities in a chess game. Maybe I don't understand the notation.

Surely in Go there are 361 nodes and each of these, crudely, may be black, white or blank? My kindergarten mathematics suggests that makes an approximation of the number of possible positions 3 to the power of 361 (some of these positions are illegal, I realise). What relation that number might bear to 10 hat 480 I've no idea. I can't even find the hat on my computer screen.

Omigawd, I thought rdh had already plumbed the depths of asininity, pomposity and self aggrandizement, but he continues to outdo himself. I dunno what's more pathetic, the utter crap he writes, or his attempts to reformulate the theory of combinatorial mathematics. I guess you dont need to be able to think to become a lawyer. About the article, its well written, and conveys a sense of the untouchable genius of Tal. I remember writing that line from Kasparov in some thread here, and rdh's brother (in spirit if not in the conjoined flesh) koster had some silly remark to make on it. Then there are some other who think that any 2600 ELO GM has the playing strength of Tal. Oh well...

Actually I have new depths to plumb, d, - I forgot to mention that Lawson can't spell aficionado.

I would find being educated more interesting than your rather dull abuse, though, and I suspect most of us would, so why not tell us how you should calculate the number of possible positions in go? I can easily believe I'm completely wrong.

Well, according to Wikipedia


you have to divide 3 to the power of 361 by 0.012. I don't understand where that comes from - anyone care to explain?

I don't see Ovidiu's 10 hat 480 there, mind. But then like I say I don't understand what the hat is supposed to convey anyway (anyone?).


agreed, Lawson is a twit- a lot of his chess related articles had a common theme of 'Nigel is my fwend' Amazing that he got so high in the Telegraph hierarchy.

As for the numbers are you confusing positions and games? There is a fixed number of possible positions, but far more ways to get to them. You can get an enormous number of e.g. 40 move games in chess which may reach the same position via different move orders. I think the Complete Chess Addict books gave the number as greater than the number of particles in the universe (10 ^79), but don't remember offhand how much.

I don't know Go, but imagine it is the same- the number of legal positions will be smaller than in chess, but the number of routes to get to them is far greater.

On my keyboard ^ is 'SHIFT + 6'

rdh, I bill in advance whether for lessons in dull abuse, kindergarten arithmetic, spelling or the art of divination, i.e. knowing rather suspecting what most others think. Send me your name and address so that I can send you the billable hours.

^ means to the power of and is used by mathematicians and scientists when they are using simple text editors such as this one - they don't have the luxury of superscript in MS Word.

Al - thank you, I was beginning to suspect as much.

I'm fairly sure that one frequently sees the number of games quoted as 10^128, although, according to this page:


the number of games is much higher than that. On the other hand, most people seem to think the number of possible positions is smaller than that, although to judge from this there is a surprising amount of academic debate ranging from 10^120 to 10^43.

You wouldn't think it was all that hard to calculate the number of positions with the aid of a powerful computer, would you, at least to some approximation? If anyone could be bothered to do it. To get an exact figure would be quite a challenge. I suppose captures and promotions confuse things. And to get an exact number - no kings next to each other, no both kings in check, legal pawn positions, and perhaps worst of all no positions which retroanalysis proves can't be reached: one can see that would be quite a task. It's a pity they didn't try programming the machines to do that instead of playing, really.

rdh, you're right, I think that the numbers will vary according to how much the academic knows chess - illegal moves and positions (pawns on the first rank) will decrease the numbers dramatically, so those numbers at the top end are probably out.
However, I guess that a lot of people will not take promotions into account either, which will push the numbers a way back up. A Deep Blue/Fritz or equivalent could be used to find it out too. Also, the repetition rule and 50 move rulw would make things difficult as well (e.g. I think the longest legal game could be either 5998 ply or moves).

It's a reasonably simple calculation when you start off, but it's the little complexities which make the game so entertaining that will add to the numbers. Even if it is eventually solved, which I doubt, it will be impossible for any human to memorise it all. Despite what the doom-mongers say, we will be able to play it for ever.

The Lawson article also appeared in the UK newspaper The Independent on Tuesday.


Maybe that mysterious # 0.012 is expressing elimination of possible symetries (you can turn goban four ways and its basically still the same position). I just can't figure out why it's 0.012. Any idea?

Yes, good point; you'd have to divide by four to account for symmetry, I guess?

0.012 is the inverse of 3^4, roughly. That any good?

rdh, not that easy. Not each symmetry is of "4-edge" kind. There are mirror ones too. But this doesn't help us cause we're looking for more symmetries to cut 3^361, not less.
1/(3^4) is actually a nice number:
0,012345679012345679012345679012346... :-)

Aha - true. So each position has three the same but rotated. One the same but reflected in the horizontal centre line. One the same but rotated in the vertical centre line. One the same but reflected in a1-h8, and one the same but reflected in h1-a8; is that right? So in all eight duplicates, and one would need to divide by eight, is that right?

That is a quaint number. Isn't the universe strange - why on earth no 8?!

rdh, you might be on a right track. But don't forget that diagonal symmetry equals 2 "90-degree-turn" ones so you can (?) discard the diagonal ones which gives you 3x2x2 symetries. So we've got 12 and we're not sure if it's right. But we need 12x 0.001... :-(

All the positions arising from the 3^361 possibilities are not legal according to the rules of Go (groups of stones must have at least one liberty left). So the 0.012 factor doesn't seem to come from some mathematical deduction, but from statistical evaluations of how many positions are legal.

More about this at : http://homepages.cwi.nl/%7Etromp/go/legal.html


Blimey. Estimated to be 4.63 x 10^170.

Estimated?? You wouldn't think it was a very complex calculation, would you? Not so as to have academic papers written about it, anyway.

Surely not, dexx? If I have a single black stone on, say 19 (along the x axis), 18 (along the y axis), and the rest white, and I reflect in 0,0-19,19, then my black stone is now at 18,19. Or if I reflect in 0,19 - 19,0 then the black stone is now at 1,0. I couldn't have got those by rotating, could I?

And also you need to add the symmetries rather than multiplying, no?

Imagine an entirely white board with a single black stone as above, on the edge one in from the corner. There are surely seven symmetrical positions; one in from the same corner, and one in either way from the other three corners. In principle this is the same for any position, is it not?

Rameau - OK, fair enough. Hence 'estimated'. My grasp of the rules of go is sketchy. I was thinking that if a group had one liberty left you killed it by placing a stone on that liberty and then picking up the other stones, so that the position was on the board with the liberty blocked for a fleeting second. Is that wrong?

But of course there would still be illegal positions anyway - ones where more than one group had its liberty blocked.

Hmm. Tricky stuff, this...

Rameau - good grief; followed your link. Thanks. I see my suggestion it's not that hard was a bit naive, then!

I also see from my link that in fact a go board is not symmetrical - physically, although not conceptually.

Forgetting legality and illegality though and just trying to see how many ways there are to place the stones on the grid, though - that shouldn't be too hard, should it?

Of course it's not the same for any position, is it? Ones which have stones only on one of the centre lines, for example. Or where when you reflect or rotate every stone lands on another stone of the same colour. For example the one with eight black stones, each one in from the corner along an edge. That's only one of the 3^361, but it doesn't need dividing at all.

So it's tricky enough even to calculate the number of possible ways to put the stones on the board.

rdh: What you need for your calculation is Burnside's Lemma. Look it up on Wikipedia. This is a typical example of a finite group acting on a set.

I thought this blog was loaded with intelligent mathematicians. If you divide by .012 then that is the same as multiplying by 83.333



1/.0125 would be 80. even.

so actually Wikipidia will increase the number of possibilities by 80 times more. Not decrease the number because of symmetry. Maybe the idea is that a games usually lasts about 80 to 83 moves so they do the combinations 83 times or once for each move.

but then I donno.

Of course anyone can say anything on Wikipidia. The content can definitely be totally wrong. You can go and post any old stuff you want on Wikipidia any time you want.

Yes, sorry, I meant to say multiply by 0.012.

Arthur Cayley; OK, thanks, I see what you mean. Can't understand a word of Wikipedia's entry, sadly.

One of my favourite bridge teachers used to say that a very valuable skill in bridge (chess too, I suppose) was the ability to tell at once that you will be no wiser if you think for an hour. I think I have reached that point where this problem is concerned.

The link to John Tromp's page given by Rameau gives the exact number of legal go positions on boards up to 17x17 in size. He gives an estimate of about 0.011957528698 * 3^361 for a 19x19 board, which is about 2.081681994 * 10^170 possible positions.
To understand the methodology, consider a much simpler case, the 2x2 board, for which they claim 57 legal positions. There are 3^4 = 81 different positions on a 2x2 board. Evidently there is no reduction for symmetries, only for illegal positions. All positions with zero, one, or two stones are legal. All 2^4=16 positions with four stones are illegal, as neither side has liberties. Of the positions with three stones, the only illegal ones are where a player has one stone on the board, with both its potential liberties blocked by the opponent's two stones. For example, White on a1, Black on a2 and b1, with b2 open. There are 8 such positions (4 corners for the white stone to be in, and 4 more with colors reversed). So there are 16+8=24 positions with one or both players left without liberties, and the remaining 57 are legal.
Unlike chess, there is no need to alternate moves in go, since passing one's turn is legal. Hence, even though Black always moves first, it is possible to have one white stone and no black ones on the board, which requires that Black passes on his first turn.
The result indicates that the factor of 0.011957528698 represents the fraction of possible positions on a 19x19 board where all groups of stones have at least one liberty. This fraction is also estimated for an 18x18 board, but is significantly larger (about 0.017322427762). These do not seem to be simple ratios like 1/3^4. And yes, these numbers indicate that there are far more possible go positions than atoms in the universe, and something like 10^120 different possible go positions for every single possible chess position.

Has anyone here taken the time to explore "Arimaa", a board game that aims to be
-- "difficult for computers",
-- "playable with a standard chess set", and
-- "fun and interesting for humans."

Maybe we should all switch to Arimaa and leave chess to Deep Fritz and Rybka?

Every body remembers that humen's life is not very cheap, however people need cash for various stuff and not every person earns enough cash. Therefore to receive some loan and college loan should be a proper solution.

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    This page contains a single entry by Mig published on November 30, 2006 8:29 PM.

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