Wikipedia:Reference desk/Archives/Mathematics/2021 December 12

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December 12

Brackets in direct products of groups

I think my math assignment wants me to answer a question with the following:

${\begin{aligned}\mathbb {Z} _{3}\times \mathbb {Z} _{63}\times \mathbb {Z} _{5}\times \mathbb {Z} _{5}\cong \mathbb {Z} _{3}\times (\mathbb {Z} _{3}\times \mathbb {Z} _{3}\times \mathbb {Z} _{7})\times \mathbb {Z} _{5}\times \mathbb {Z} _{5}\\\cong (\mathbb {Z} _{3}\times \mathbb {Z} _{3}\times \mathbb {Z} _{3})\times \mathbb {Z} _{7}\times \mathbb {Z} _{5}\times \mathbb {Z} _{5}\end{aligned}}$

I seem to be expected to bracket together groups that have the same prime order (or powers of it). What I want to know is, do the brackets change what the (produced) group is, or are they decorative? It looks stupid to me. Could I just not?  Card Zero  (talk) 02:55, 12 December 2021 (UTC)

The associative law holds for Cartesian products of groups, at least up to isomorphism. The proof is generally left as an exercise so if you're having doubts I'd recommend proving it yourself. The commutative law also holds for Cartesian products, even with non-commutative groups. Given those two facts you can rearrange and rebracket the factors at will. What I'm worried about is the first isomorphism. You weren't specific and notation varies, so I'm assuming that ℤ_n is meant to be the cyclic group of order n. If so then ℤ₆₃ is not isomorphic to ℤ₃×ℤ₃×ℤ₇. Try counting the number of elements of order 63 on the left and right; if the counts do not match then the groups are not isomorphic. --RDBury (talk) 03:58, 12 December 2021 (UTC)

Good point, I have to decompose it into distinct primes or their powers, so:

${\begin{aligned}\mathbb {Z} _{3}\times \mathbb {Z} _{63}\times \mathbb {Z} _{5}\times \mathbb {Z} _{5}\cong \mathbb {Z} _{3}\times (\mathbb {Z} _{9}\times \mathbb {Z} _{7})\times \mathbb {Z} _{5}\times \mathbb {Z} _{5}\\\cong \mathbb {Z} _{3}\times \mathbb {Z} _{3^{2}}\times \mathbb {Z} _{7}\times \mathbb {Z} _{5}\times \mathbb {Z} _{5}\end{aligned}}$

Going to keep the first set of brackets, as guidance about what was decomposed, but lose the second. The textbook would use them there, probably just as a way of emphasising "this stuff is commutative, you know", but they seem misleading to me (or at least distracting).  Card Zero  (talk) 11:28, 12 December 2021 (UTC)

The pons asinorum of logic

Wiktionary gives "A method for finding the middle term of a syllogism in Aristotlean analytics." as definition 1 in the entry for pons asinorum. But our article barely mentions this meaning at all. A quick Google search turned up inventio medii as another term for it. Does anyone know enough about Aristotelian logic to explain, at least in general terms, what the method is? (Btw, is it Aristotelian or Aristotlean?) I'm asking because to German speakers an Eselsbrücke means something completely different, and I think it's related to the logical meaning instead of the geometric meaning. --RDBury (talk) 05:30, 12 December 2021 (UTC)

An example (in English) of what Germans call an Eselsbrücke is the mantra "Some Old Horses Can Always Hear Their Owner’s Approach".^[1] Aristotelian looks like it is from Latin Aristotelianus,^[2] from Aristotelus + -ianus. I have not seen the expression pons asinorum used in relation to syllogisms, but a possible meaning could be for the situation that the first premise is (expressed symbollically) $\forall x\in S:P(x)$ and that the conclusion is $P(a)$; then the mediating second premise completing this situation to form a valid syllogism is $a\in S.$ Possibly, there was some Eselsbrücke for a drill to construct the middle term from the components of the first premise and the conclusion – which would depend on the way the propositions are formulated in natural language. More research is needed.  --Lambiam 09:42, 12 December 2021 (UTC)

The Oxford Dictionary of Philosophy writes: In traditional logic, a name for a method invented by Petrus Tartaretus around 1480, by which the middle term of a syllogism can be found. It has sometimes been called the pons asinorum or bridge of asses of logic. (Note the added qualification "of logic", which casts some doubt on the conception that the unqualified term in itself carries this sense.) This clue leads to the entry "Bridge of Asses" in the online New World Encyclopedia, which has this:

The French philosopher Jean Buridan (Joannes Buridanus, c. 1297 – 1358), professor of philosophy in the University of Paris, is credited with devising a set of rules to help slow-witted students in the discovery of syllogistic middle terms, which later became known as the pons asinorum.
    In 1480, Petrus Tartaretus applied the Latin expression “pons asinorum” to a diagram illustrating these rules, whose purpose was to help the student of logic find the middle term of a syllogism and disclose its relations to the other terms.
    The “asses’ bridge” was usually presented with the predicate, or major term, of the syllogism on the left, and the subject on the right. The three possible relations of the middle term to either the subject or the predicate (consequent, antecedent and extraneous) were represented by six points arranged in two rows of three in the middle of the diagram, between the subject and the predicate. The student was then asked to identify the nineteen valid combinations of the three figures of the syllogism and evaluate the strength of each premise.

The middle paragraph cites Dagobert D. Runes, Dictionary of Philosophy, and the last paragraph cites C. L. Hamblin, "An Improved Pons Asinorum?", Journal of the History of Philosophy xiv, 1976, pp. 131–36, and this picture on the website She-philosopher.com – not a bridge the swimming asses in the picture can cross, and also not a bridge for Buridan's ass.  --Lambiam 10:39, 12 December 2021 (UTC)

That helps. I gather it's most common meaning is "mnemonic", but I've also seen it used for a computational trick or device. I'm thinking "hack" may be a possible translation. --RDBury (talk) 12:33, 12 December 2021 (UTC)

Chance to win a chess game

Is there a mathematical treatment of the question of how likely it is for a person to win a game of chess?
I imagine it could be dependent on ELO and other criteria.--2A02:908:426:D280:C1B:60B4:C20E:F18E (talk) 10:25, 12 December 2021 (UTC)

I don't know the answer, but I think the appropriate term here is "mathematical model". A simplistic model for two-person games with a binary outcome (no draws) based on player strengths is to assign each player $p$ a strength parameter $\alpha _{p}$, where the player's performance in a game is a normally distributed random variable with mean equal to $\alpha _{p}$ and variance equal to ${\tfrac {1}{2}}.$ (The letter α stands for Greek ἀλκή (alkē), "strength as displayed in action, prowess".) Then player $p$ beats player $q$ if their performance exceeds that of $q$, an event that has probability $\Phi (\alpha _{p}-\alpha _{q}),$ where $\Phi$ denotes the cumulative distribution function of the standard normal distribution. To model draws, I suppose one can use the event $|\alpha _{p}-\alpha _{q}|<\theta ,$ in which the threshold parameter $\theta$ that needs to be reached for a conclusive win/lose outcome needs to be determined experimentally from the observed fraction of drawn games between players of equal strength. Then it should be possible to estimate the strength parameters of Elo-rated players by playing virtual versions of matches with outcomes based on the model, and finding the $\alpha$ values that produce the best correspondence with the actual outcomes of these matches. A scatter plot of these players' strength parameters versus their actual Elo ratings will, hopefully, show a strong relationship that can be expressed as a smooth curve mapping Elo ratings to strengths in the model.  --Lambiam 11:11, 12 December 2021 (UTC)

The underlying mathematical model of the Elo rating system (incidentally: note correct capitalization) is described in detail at that article. This is summarized in the lead section as A player whose rating is 100 points greater than their opponent's is expected to score 64%; if the difference is 200 points, then the expected score for the stronger player is 76%. but these are just example calculations are the article has full details. --JBL (talk) 19:27, 12 December 2021 (UTC)

• WHAAOE: see First-move advantage in chess. If you play white, you have a 52-56% chance to win, and if you play black you have a 44-48% chance to win. All other things being equal. Since chess is not yet a solved game, we have not modeled every possible chess game (see Solving chess), which would be required to definitively answer the question. Which is to say we would need the outcome of every possible chess game from beginning to end to calculate a more precise percentage; all we have now is the percentages of games that have been actually played or simulated. Given that there are 10¹²⁰ possible games in chess, it is unlikely to find such a solution before the heat death of the universe. There are some shortcuts one could hypothetically take, so it's not entirely been ruled out, but we aren't really that close. --Jayron32 04:28, 13 December 2021 (UTC)
  • Well, if chess were solved, it wouldn't be a question of "percentage" in the usual sense, or rather the only possibilities would be 0% or 100%. Chess is a perfect-information game, meaning there are only three possibilities with perfect play. It could be that white always wins; it could be (though most players consider this unlikely) that black always wins; it could be that it's always a draw. It can't be that white wins 53% of the time. See Gale–Stewart theorem. --Trovatore (talk) 05:48, 13 December 2021 (UTC)

True, but one way to read the percentages would be "what percentage of the total number of possible games ends with a victory by white?" That number exists, though it is currently unknown (and likely unknowable as we don't even know all of the possible games). Given the 10¹²⁰ possible games, and that the Information capacity of the universe is only 10⁹⁰ bits (and complete information about a single chess game would require a whole shitload of bits), it's likely not even physically possible for the observable universe to store all of the information necessary for every possible game, never mind a single human mind. --Jayron32 13:23, 13 December 2021 (UTC)

I suppose it's possible to read the question that way, and it's a well-defined number in principle, but it doesn't have much to do with a "first-move advantage in chess", unless the players are playing to sample uniformly from the space of all possible games, rather than to win. --Trovatore (talk) 16:40, 13 December 2021 (UTC)

Even if the players are just playing deterministically the outcome be the same each time. This could happen with computers playing, assuming they used the same algorithm each time they played and that there was no randomness involved in their programs. Given that chess in not solved, I think there would be some advantage in not playing predictably, for example using an weak but unusual opening that your opponent hasn't has a chance to study. In other words a certain amount of randomness makes for good strategy. The following quote by Mark Twain seems apt:

The best swordsman in the world doesn't need to fear the second best swordsman in the world; no, the person for him to be afraid of is some ignorant antagonist who has never had a sword in his hand before; he doesn't do the thing he ought to do, and so the expert isn't prepared for him; he does the thing he ought not to do; and often it catches the expert out and ends him on the spot.

--RDBury (talk) 13:12, 13 December 2021 (UTC)

So "perfect play" is a bit of an abstraction in cases like this. A "perfect" move is any move that does not convert a won position into a drawn or lost position, or a drawn position into a lost position. If you're already in a lost position, congratulations, you're always playing the perfect move, no matter what you do! --Trovatore (talk) 16:40, 13 December 2021 (UTC)

"In bad positions there are no good moves" (attributed to Tartakower, but I cannot find an original source). Refined by Ralph Betza in 1982 into "In bad positions, the result of the best move is just as bad as the current position; moves that aren't the best move make things worse." Double sharp (talk) 02:42, 15 December 2021 (UTC)

Sure, in real-life chess play, even in a lost position you may have a hope of winning or drawing if your opponent makes a mistake, and the relatively good moves are the ones that keep that possibility alive. But from the standpoint of "perfect play" that possibility is not considered, so it would be better to say that "in a lost position there are no bad moves". --Trovatore (talk) 19:38, 15 December 2021 (UTC)

There are no good moves, there are no bad moves, but there are hopeful moves and hopeless moves.  --Lambiam 22:15, 15 December 2021 (UTC)

No, in "perfect play", if you're in a lost position, you will lose no matter what. --Trovatore (talk) 22:24, 15 December 2021 (UTC)

For all we know, after 1.d4 Nf6 2.c4 g6 3.Nc3 the position may be theoretically lost for Black. Assume it is. If the player in this lost position knows the other player is perfect and that their position is lost, their struggle is waged without hope, as in a fixed match. If they don't know, their moves (such as 3. ... d5 in the Grünfeld Defence), while futile in the eyes of the omniscient onlooker, are hopeful.  --Lambiam 07:12, 16 December 2021 (UTC)

I think it is natural to make distinctions even when the position is lost, because it seems natural to say that being mated in 2 is better than being mated in 1. Then we could say that the best move in a lost position is the one that loses the most slowly. Indeed, it's pretty common to talk about it that way, e.g. Tim Krabbé doing so. When interpreted this way, Betza's statement becomes almost correct. It should just read "In bad positions, the result of the best move is one step worse than the current position; moves that aren't the best move make things even worse."

(BTW, I'd be very surprised if the Grünfeld was lost; it is one of the best defences against 1.d4, together with the Nimzo/Ragozin. GM Larry Kaufman recommends it in his recent Kaufman's New Repertoire for Black and White, for which he used a lot of computer analysis by Komodo and Leela Chess Zero.) Double sharp (talk) 09:34, 16 December 2021 (UTC)

But the point is of course that a player, not being omniscient, may not know the won/drawn/lost status of the relevant positions with absolute certainty. I'd be somewhat surprised if there are no theoretically lost positions in regularly played openings, and what we perceive to be strong may be a mirage.  --Lambiam 11:25, 16 December 2021 (UTC)

Oh, you're absolutely right. Actually, that's more or less already happened. We don't have a 32-man tablebase, but engines have already been enough. For example, we've learned from them that space advantage is much more important than had previously been thought. Kaufman puts it well in Chess Board Options: The consequence of this is that many defenses formerly considered to be playable, if slightly worse for Black, are now viewed as practically, if not theoretically, losing to a well-prepared opponent. He cites the King's Indian Defence as an example; once it was seen regularly at the top level, but now it hardly is. For another case: in 1972 Spassky-Fischer you could see Alekhine's Defence on the board, which you'd never see now. Apparently the pool of acceptable openings shrinks even further the higher you go, so that in correspondence chess, the French Defence (1.e4 e6) is already noticeably inferior to 1.e4 e5, 1.e4 c5, and 1.e4 c6. (Not sure though if this is related to the space thing or another factor, because the Caro-Kann also cedes space. Perhaps it's the mobility of the c8-bishop, though Kaufman doesn't say.) Double sharp (talk) 12:25, 16 December 2021 (UTC)

It depends not only on the player Elos, but on the board position (during a game) and what winning probability they want to have / what losing probability they are willing to tolerate. Remember that (assuming the players aren't beginners) draws in chess are quite common, and at the top levels draws are more frequent than wins/losses. The 2018 world championship match (Carlsen-Caruana) had 12 classical games, all of which were drawn (Carlsen then won a speed chess playoff). A basic guideline is that if you are (due to the tournament or match standing) satisfied with a draw, you can play toward on and probably get it, but you decrease your likelihood of a win. While if you choose sharp lines, your winning and losing chances both go up. If your Elo is 100 points higher than your opponent's, that predicts that the score of a 100 game match should be 60-40 (or whatever). But it is the player choices that determine whether that's likely to be 60 wins and 40 losses for you (winning probability = 0.6), or 80 draws and 20 wins (winning probability = 0.2), or whatever.

It might be possible to mathematically model this with more precision, but as far as I know that hasn't really been done seriously anywhere. I know that Leela Chess Zero now scores positions as WDL (win-draw-loss) probabilities instead of just WL. There is a now well known Deep Mind paper[3] about chess variants (e.g. you can push a pawn 3 squares instead of 2 on the first move) where they had Alpha Zero self-train to play the variants and then studied its games. It would be interesting if they did something like that with normal chess rules, but scoring wins and draws differently than W=1, D=0.5. That could identify sharp vs drawish strategies, etc. 2602:24A:DE47:B8E0:1B43:29FD:A863:33CA (talk) 20:04, 15 December 2021 (UTC)

Very interesting paper! Thanks for the link! --Trovatore (talk) 00:26, 16 December 2021 (UTC)

Your suggestion doesn't really work, for reasons Larry Kaufman explains in his recent book Chess Board Options: Some tournaments use so-called 'football' scoring, 0 for a loss, 1 for a draw, 3 for a win, which is the same as the normal 0 for a loss, 1 for a win, but with draws counting ⅓ instead of ½. This does decrease the incentive to draw, but it doesn't reduce the actual percentage of draws very much in top level tournaments where it's been tried, because the problem is not one of incentive, but of the fact that while White usually wants to win, his advantage just isn't large enough. In my opinion this is a terrible solution, because it goes against the very nature of the game. The inferior side should be trying to draw, and to penalize Black for obtaining a good result is crazy. It makes chess like a game of 'chicken'; who will 'blink' first and play an unsound move to avoid the mutually bad result of a draw? I think such events should not even be rateable by FIDE. A much better idea, championed for decades by U.S. chess expert Ed Epp, is to award Black 0.6, White 0.4 in case of a draw, which makes the overall expected score for White and Black about equal. I'm all for this idea, but I must admit that I think the benefit would be small, most games would have the same outcome. Then there's Armageddon chess, which takes this to the extreme by giving Black 1.0 and White 0.0 for a draw, which is obviously unfair unless White gets some other advantage, either time odds or some other initial advantage, such as an extra move or Black forfeiting the right to castle short (see separate chapter on Armageddon chess). Double sharp (talk) 09:52, 16 December 2021 (UTC)

Let $p_{\rm {W}}$ and $p_{\rm {B}}$ be the fractions of won games for, respectively, White and Black, as observed over a large number of games with top-level opponents of matching strengths. The following score assignment is then, I think, fair:

White wins: ${\frac {1}{1+p_{\rm {W}}-p_{\rm {B}}}}$ points for White;

Black wins: ${\frac {1}{1+p_{\rm {B}}-p_{\rm {W}}}}$ points for Black;

It's a draw: each player receives half of what they would have got on a win.

If $p_{\rm {W}}=p_{\rm {B}},$ this defaults to the current rule. In general, the expected value of the result for White can be calculated as:

$p_{\rm {W}}\times {\frac {1}{1+p_{\rm {W}}-p_{\rm {B}}}}+(1-p_{\rm {W}}-p_{\rm {B}})\times {\frac {1}{2}}\times {\frac {1}{1+p_{\rm {W}}-p_{\rm {B}}}}.$

This simplifies to ${\tfrac {1}{2}}.$ By symmetry, this is also the expected value for Black.  --Lambiam 14:18, 16 December 2021 (UTC)