text:
Board Games development online course
Are there any online courses like codecademy.com or teamtreehouse.com for Board Games development? Or something similar where you can learn BG development? Also if you know any other useful resources like books, videos and similar things, please share

I don't know about online courses, but there are a number of books on the topic.  If you want to broaden your question to include those, I'll suggest some.

Best way to learn; play many different board games, including ones you wouldn't normally like.

Highly recommend the book "The Art of Game Design"

I'm voting to close this question as off-topic because it is not about a specific board or card game.

@DrunkCynic I don't think that's a good metric for closing. There are a number of questions on the site that don't refer to a specific board game. "Designing board games" is on-topic for this site, and the OP is asking for objective information. This seems to fit all of the criteria I can find for acceptable questions, except that it may generate a list.

@SocioMatt Then there is some ambiguity in the Help Center topic "What topics can I ask about here?", which has a paragraph starting with "For a question to be on topic, it must relate to a game that is on topic..."  That was my guidepost for the metric.

@DrunkCynic In rereading the entire page on the Help Center, I think that "a game" is meant to refer to games that meet the four criteria under "How do we define board games?" I agree this is confusing when taken out of context. If we assumed that questions had to relate to a game that already existed, then design would be off-topic, which would contradict the first sentence of What topics can I ask about here?

Reopening - There is no requirement that a question be about only one game.

Thank you everyone :) @bwarner I boraden question, can you suggest books please? tnx

I am voting to close this question because it is essentially the same as a game recommendation question, which is off-topic for this site.

I'm voting to close this question as off-topic because it as asking for school advice

Here are a couple EdX options. The first is about game design (both board and computer games). The second focuses on games meant for education.
https://www.edx.org/course/introduction-game-design-mitx-11-126x
https://www.edx.org/course/design-development-games-learning-mitx-11-127x


text:
Can you conquer 3 cities with the flight card?
We played Civilization - A New Dawn twice now,
And I must say that it is rather 'difficult' to built a new city.
Meaning that you can only built 1 at the time with the masonery card, and that you can't build a wonder because of the 'or' statement on the card.
During these 2 games, we only build like 3-4 cities, and that again, most of them were on easy (to conquer) terrain.
What troubles me then, is that it is very easy to attack and conquer cities if you have the flight card.  My friend had a very good attack, and was able to take out 4 easy cities and 2 city states in 2 turns.
So while it takes a while to build a city (once at the time, you can conquer 3 at the time by flight), because the tokens doesn't matter anymore.
Did we forget about a rule, or is this unbalanced?

Your title says "fight" but the body refers to a card called "flight" - you might want to edit them to match for clarity.

In case someone wants to help sort this out, the rules can be found here: https://images-cdn.fantasyflightgames.com/filer_public/a1/fe/a1fec4bd-8ffc-4317-8175-f77c1675f53c/civ2_rulebookcompressed.pdf

Yes, an attacker can conquer 3 cities with the Flight card as long as all three rival cities are within 5 spaces of one or more of the attacking player's friendly spaces at the start of his turn. (That is, an attacker cannot target spaces that have come within range due to spaces just conquered that turn.)
Based on your description, it seems you are playing by the correct rules, although the terminology is being confused. "Masonry" is the first level Military cards. The Industry cards are responsible for building cities (e.g., "Pottery", the first-level Industry card).
Strategy-wise, if you're finding Flight (the fourth and most powerful Military card) to be overpowered in your games, here are some suggestions:

Cut back on building cities on low-numbered terrain as it makes them easy targets for attacks (the exact symptom you're experiencing)
Make use of reinforced control tokens as they can greatly enhance the defense of surrounding friendly spaces. (Each adjacent reinforced control token adds +1 to its neighboring friendly space's defense.)
Allocate and stockpile trade tokens on the Military card to discourage would-be attackers.

Thanks for the reaction.
Based on what you are saying, I believe that we are skipping a rule:
Like I was saying in my question: "So while it takes a while to build a city (once at the time, you can conquer 3 at the time by flight), because the tokens doesn't matter anymore.", you are saying that the reinforced tokens still contribute to a +1 for its defence.
Note the reason why we were skipping the defence tokens in the first place for flight is that you would fly over the defence tokens, rather than face them.

@GertDeWilde The control tokens won't interrupt the path to the city as you said and therefore do not need to be attacked first. However, each reinforced token adds 1 to the base defense of adjacent friendly tiles.

I think he's saying that he was "skipping" them in terms of not applying their defense value.

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text:
Does the School of Magic count as all colours, or do you choose one colour when it's played and it is that colour permanently?
It states that it has the colour of your choice (for calculating income). Do you make that choice when the card is played and it becomes permanently yellow for example, or can you change it each round?

I believe you choose each round.
In my version of Citadels (Fantasy Flight, 2009) there isn't a card called 'Mage Tower'.  There is however a very similar card called 'School of Magic' which text states.

For purposes of Income, the School is considered to be of the colour
  of your choice.  If for you are the King this round, for example, the
  school is considered to be a nobel (yellow) district.

To me that sounds like the card you are describing with extra text clarifying which sound like it may be missing from your edition?  Also if you had to decide when it was build there should be some way of marking that and the game does not come with tokens for doing so.

Sorry, I have no idea why I typed that! I've fixed the question. Thanks for your answer

Essentially the School of Magic always adds to your colour-based income.

Note that during endgame scoring "School of Magic" does NOT change color and is always purple, so it doesn't provide additional help to get bonus for all colors. Though there is another building "Ghost city" that does exactly that, if you have met its prerequisites.

@Deo There's also the Court of Miracles (? - "Hof der Wonderen" in dutch) that can also count for any desired color at the end of the game.

@steenbergh It is the card I'm talking about. Purple district for two. Apparently correct english name is "Haunted City" rather than "Ghost City". It is top left one here.
https://boardgamegeek.com/image/147403/citadels?size=original

@Deo Yes, it's the same card. I'm kind of surprised about the difference in naming. Then again, we call this game 'Machiavelli', not Citadels...

Now that we've played about 5 games at work, Machiavelli is definitely a better name for it. I wasn't sure if there were any methods to get income from two colours during the income phase and thus it would matter if the school of magic was a single colour of your choice, or acted as all colours. But it looks like there isn't a way to do that, so it's not a problem.

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text:
How to break a Tie for a 3-of-a-kind
In a particular scenario where there is a tie for same three of a kind.
Two players -
User1 with 10h and 6d
User2 with 10d and 4c
On table 10c, 10s, Kc, 3h, 2d
Who wins?
Note: h/d/s/c = Hearts/Diamonds/Spades/Clubs

When figuring out who wins in Texas Holdem you take the best 5 card hand a player can make using their 2 cards plus the 5 community cards on the table. That gives us:
Player 1: 10, 10, 10, K, 6
Player 2: 10, 10, 10, K, 4
Since they both match on the 3 tens you look to the highest kicker card, in this case they both have kings so you need to look at the next highest card. For this card Player 1 has a 6 and Player 2 has a 4 so Player 1 will win the hand.
Reference: Texas Holdem Rules

And if they have the same best 5 cards, they split the pot.

Is suit taken into account for the three of a kind cards?


text:
How does the ability of Bloodthirsty Aerialist resolve with other life link creatures?
Scenario:
I have Bloodthirsty Aerialist and a Blood Burglar on the board. I attack with both and both get a blocker assigned to them.
Does Bloodthirsty Aerialist's ability trigger before damage because of Blood Burglar's Lifelink? As in does Bloodthirsty Aerialist get a +1/+1 counter on it before the damage is done against the blocker?

The scenario unfolds as follows:

You declare bloodthirsty Aerialist and Blood Burglar as attackers.
Your opponent declares blockers for these creatures.
All damage is dealt. At the same time, Blood Burglar's lifelink causes you to gain 2 life.
Bloodthirsty Aerialist's ability is triggered and goes on the stack (but doesn't resolve yet).
State based actions are checked. If either of these creatures have enough damage on them to die, they are put into the graveyard.
Bloodthirsty Aerialist's ability resolves (assuming players pass priority). If Bloodthirsty Aerialist is not on the battlefield, the ability does nothing. If the creature is on the battlefield, it gets a single +1/+1 counter.

Bloodthirsty Aerialist will only deal 2 damage to the creature that blocks it, and if it is dealt three damage, the +1/+1 counter won't be placed on it in time to save it from dying.

Re "All damage is dealt. At the same time, Blood Burglar's lifelink causes you to gain 2 life.", Technically, this description could be understood to mean the life gain is something that happens in addition to damage being dealt, but the life gain is the effect of the damage being dealt. The following is a more accurate description: All damage is dealt simultaneously. This has the effect of marking damage on the creatures and you gaining two life (because of lifelink).

@ikegami I have never heard before an idea of "dealing damage" as a thing that happens before the effects of that damage are applied. "Dealing damage" is synonymous with "resolving the effects of damage" as far as I know.

@Gendolkari, That's my point. But that's not the what the answer says. It says two things happen in step 3: damage being dealt and life being gained.

@ikegami I see, I misread part of your comment. Although looking it up; I see 120.4; which seems to imply that resolving effects of damage is technically a separate step from dealing damage.

@Gendolkari, I'm just saying it could be clearer. And I've often come across many questions about when life link actually happens, so it's not just a nit.

Nit-pick: Half of 4 happens after 5: The ability triggers as damage is being dealt, then state-based actions are checked, then the triggered ability goes on the stack.

In Short
The Bloodthirsty Aerialist would not get any +1/+1 counters until after the damage resolved just like Aetherfox stated above.
Longer Description
You won't gain any life until all damage is dealt and then you would gain life due to the lifelink ability. After that resolves then the Aerialist's ability would trigger (Due to the lifelink gaining you life.) and be put on the stack, but if the Aerialist is dead it can't gain any +1/+1 counters (It would be dealt damage before the ability activated.)

Minor nitpick, Lifelink is a replacement ability and not a triggered ability.  Which means that you gain the life as the damage is being dealt, not as part of a triggered ability afterwards.  It used to be triggered but got updated in the 2010 rules update.

@DenisS It's not even a replacement ability, it is a defined part of what "dealing damage" means. Just as much as, say, players who receive damage losing life.

@Arthur true, although it functions closer to a replacement effect rather than triggered because the life gain happens simultaneously.

Thanks for the comment and I did know this and I just didn't how to say it so if you have any edits please add them.

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text:
Determine the solubility of peg-solitaire (solo-noble)
Are there specific patterns we can look in a single configuration of a square peg solitaire game which we can use to determine whether we can win the game from there or not?

Have you even Googled "mathematics of peg solitaire"? Try this Wikipedia link: https://en.wikipedia.org/wiki/Peg_solitaire#Studies_on_peg_solitaire

Also this gem: http://www.link.cs.cmu.edu/15859-s11/notes/peg-solitaire.pdf

I think this is an interesting question. A quick Google search brings up a few pages which look at characteristics of peg solitaire positions, I didn't find anything about how to determine whether a position has a solution or not. Obviously it's possible to exhaustively search all possible moves, but I'd be interested to know if there is a neater way to determine if a solution exists.

@Pieter Thanks, I think that wikipedia link answers my question.


text:
Can I cast Quarantine Field with Brisela in play for my opponent?
Can I cast Quarantine Field with Brisela, Voice of Nightmares in play for my opponent?
I just tried, Magic Online wouldn't let me.
But with X as 2 Quarantine Field has a CMC of 6.
Is this right?

As far as I remember you should be able to cast it.
But my knowledge of magic rules is somewhat rusty and outdated.

You can cast Quarantine Field while your opponent controls Brisela, Voice of Nightmares, as long as you choose 1 or greater for X.
First, before you even begin to cast a spell, the game checks to see if you are allowed to do so. The game ignores any effects that require information determined during the spell’s proposal (this is an exception described in 601.2). This means that we ignore Brisela's effect for now.

601.2. [...] A player must be legally allowed to cast the spell to begin this process (see rule 601.3), ignoring any effect that would prohibit that spell from being cast based on information determined during that spell’s proposal. [...]

Second, we follow the steps for casting a spell as described by 601.2a through 601.2i.

601.2a - Put the spell on the stack.
601.2b - Choose a value for X.
...
601.2e - Check if the spell can legally be cast. (This time, no exception is made for effects that require information determined during the spell’s proposal.)

When we reach 601.2e, the spell is on the stack and has a value chosen for X. According to 202.3c, the converted mana cost of a spell on the stack includes X.

202.3c When calculating the converted mana cost of an object with an {X} in its mana cost, X is treated as 0 while the object is not on the stack, and X is treated as the number chosen for it while the object is on the stack.

If you choose a value of 2 for X while casting Quarantine field, then the converted mana cost will be 6. 6 is not "3 or less", and so the spell is legal.

The Gatherer page for Brisela has this ruling:

For spells with {X} in their mana costs, use the value chosen for X to determine if the spell’s converted mana cost is 3 or less. For example, your opponent could cast Burn from Within (a spell with mana cost {X}{R}) with X equal to 3, but not with X equal to 2.

You should have been able to cast the spell, so you should report this as a bug.

Yes, you can cast Quarantine Field. Rule 202.3c states:

202.3c When calculating the converted mana cost of an object with an {X} in its mana cost, X is treated as 0 while the object is not on the
  stack, and X is treated as the number chosen for it while the object
  is on the stack.

And Rule 601.2b specifies that the XX have a value as soon as you state that you want to play that spell:

[...] If the spell has a variable cost that will be paid as it’s being cast
  (such as an {X} in its mana cost; see rule 107.3), the player
  announces the value of that variable. [...]

Can you explain the rule that allows you to even start casting it though? As 202.3c says, x is 0 while the object is not on the stack. So while it's in your hand, it has a CMC of 2, so it seems like it's an illegal option to say "I cast this". It would only become legal once it gets to the stack.

"Casting" is moving an object from any zone (usually your hand) to the stack. When you do that, you immediately specify what options you choose for that card: any "Choose one:" modes, any X in the casting cost - everything that determines the final shape of the spell and its cost. This is an atomic action: one moment you have a card with CMC 2 in your hand, the next you have a spell with CMC 6 on the stack. Brisela bars neither of those.

You should be able to cast it.
From the gatherer's ruling on Brisela:

For spells with {X} in their mana costs, use the value chosen for X to determine if the spell’s converted mana cost is 3 or less. For example, your opponent could cast Burn from Within (a spell with mana cost {X}{R}) with X equal to 3, but not with X equal to 2.

Meaning you can cast Quarantine Field for any cost greater than 0 as it will have a cmc of at least 4

The converted mana cost you refer to applies only while the spell is on the stack.  While it's in your hand, the cmc is 2, with X being zero.  Thus, Brisela prevents you from even casting it.

Welcome to the site, and thanks for your answer! For most cases you'd be right - for example if you discard Quarantine Field using Blast of Genius, its CMC is 2. But for OP's case (can you cast the spell?) the rules explicitly have an exception: see the rulings for definitive proof, and the accepted answer for the specifics.


text:
Mirrorwing Dragon with Vigor Mortis
When I cast Vigor Mortis or a similar spell (Reanimate, Resurrection, etc.) on my Mirrorwing Dragon,  does it copy the spell on other targets in my graveyard?
I suppose not, since I do not control the creature cards in my graveyard.

I nearly thought so, but it's great to have someone for that kind of questions. I suppose more will follow :-)

No, Mirrorwing Dragon's ability only works while it's on the battlefield, and creature cards in the graveyard are not creatures.
Generally, abilities of permanent cards only work while that card is a permanent on the battlefield. Exceptions to that will always be noted on the card itself.
Even if Mirrorwing Dragon's ability would work while in the graveyard, creature cards in the graveyard are not creatures (shorthand for creature permanents), and as you noted, cards in the graveyard have no controller, so a copied Reanimate spell could not target them anyway.

112.6. Abilities of an instant or sorcery spell usually function only while that object is on the stack. Abilities of all other objects usually function only while that object is on the battlefield. The exceptions are as follows: [none of the exceptions apply]
110.1. A permanent is a card or token on the battlefield.  [..]
109.2. If a spell or ability uses a description of an object that includes a card type or subtype, but doesn’t include the word “card,” “spell,” “source,” or “scheme,” it means a permanent of that card type or subtype on the battlefield.
108.4. A card doesn’t have a controller unless that card represents a permanent or spell; in those cases, its controller is determined by the rules for permanents or spells. See rules 110.2 and 111.2.

"Exceptions to that will always be noted on the card itself." No, if the ability doesn't make sense except in a particular zone (such as Dredge), then it doesn't have to be noted on the card that it applies in that zone.

And you're saying that's somehow not noted on the card then? I didn't say explicitely, if youre into splitting hairs that much


text:
Is there a correct overcall or double in this case?
Let's say I have a 4S-3H-4D-2C distribution and 12-18 points. One of the opponents has already opened and they didn't bid clubs but one of the other suits (S, H or D). Now what can I bid? I cannot overcall because that requires a strong suit. I cannot double because that promises at least a 3 carder of all the unbid suits including clubs. What do I bid in this case?
I play Standard American 5 card major or 2/1

"can't double" depends on what you've agreed on with your partner. With this had, I'd overcall with 1NT if the hand was 15+ points.

Yup, "12-18" is too wide a range to be covered by a single bid. The strong end (however your partrnership defines that) is 1NT, the weak probably Pass.

As pointed out in the comments, the range 12-18 is a double range for which both sides must be looked at separately.
With the lower half, 12 to a bad 15, just pass. Don't hesitate; just pass in your normal tempo. Trust partner to balance when necessary, or find a partner who will. Any hesitation with this hand gives partner an ethical problem that will prevent a reopening call.
With the upper half, a good 15 to 18, just bid 1NT in tempo.

A bid of 1 NT over an opening bid of 1 of a suit holding 15-18, a balanced hand, and a stopper in the bid suit seems reasonable.  If the suit bid by the opponent was one in which the overcaller holds 3 small cards, I would favor a double, despite a 2-card club suit. It is a risk, as partner might bid the short suit, but the likelihood of the NT bid winding up in a 3NT contract where the opponents lead and run the Heart suit in the case described seems much more serious.
Given the pressure to respond in a major if possible, the risk of a club mis-fit seems tolerable to em, as no bid is perfect in this situation, and a good fifteen is too strong to make pass attractive.

Would the down voter care to explain the perceived problem with this answer? I am growing more and more convinced that unexplained down-votes should be removed from the SE model. I am not an expert, but i am a fairly experienced tournament and club player, approaching life Master status in the ACBL


text:
Does Unbridled Growth provide two mana or one?
The new Aether Revolt card Unbridled Growth reads:

Enchanted land has "{T}: Add one mana of any color to your mana pool."

Does this mean the enchanted land provides two mana?

You don't get two mana. If you enchant a basic land (mountain, forest, etc.) then it now essentially has two abilities. For example, if you enchant a mountain, then it's as if that mountain now reads:

{T}: Add {R} to your mana pool.
  {T}: Add one mana of any color to your mana pool.

So you can tap it for the first ability, or the second. You don't get to do both (just like you can't tap a creature to attack and use an activated ability).
Alternatively, you might enchant a non-basic land. For example, also from Aether Revolt is the land Spire of Industry. If you enchant it with Unbridled Growth, then it's as if the card now reads:

{T}: Add ◊ to your mana pool.
  {T}, Pay 1 life: Add one mana of any color to your mana pool. Activate this ability only if you control an artifact.
  {T}: Add one mana of any color to your mana pool.


text:
What is the optimal strategy in Quantum Tic Tac Toe?
Quantum Tic Tac Toe is a variant of Tic-Tac-Toe with "entangled" states (there are even iPhone and Android apps).
In essence, a classic naught or cross isn't placed until the 'quantum state' of the board is measured and forms a closed loop.
Here the pieces X3, X7 and O8 are measured when placing the two O8 pieces are placed and forms a closed loop:

Upon measurement, the player placing the X marks, picks whether O8 collapses onto square 3 or square 5. After picking, and resolving entanglement, the board looks like this (O2 and O4 collapse in 7, and 8 respectively. X5 to square 4.)

Is there a winning strategy for either player? If so, what is it? If not, how can one guarantee a tie?

@thesunneversets - one of the reasons for posting this was wanting to spread the knowledge - I just found it yesterday and had the same thought as you.

In Solving Quantum Tic-Tac-Toe, Ishizeki and Matsuura use a computer to search the game space to find solutions, but don't specifically speak to strategy. However, there are a few strategies we can glean from their results:

Go for a half-point victory, not a full point victory. In their search they found the first player, X, cannot guarantee a win by being the only player to get three in a row. But a win can be guaranteed where X and O both have 3 in a row, but X has the lower maximal subscript.
As player X, make your first marks split into opposite corners. If X plays perfectly, a half-point win can be guaranteed by playing the first pair of marks, X1, in opposite corners.
Don't play as O, the second player. Obviously this is not a fair strategy or one you're likely be able to follow in a real game. But since X can guarantee a half-point victory, it follows O can't guarantee a victory.

Since there is a guaranteed path to victory for X, the obvious strategy is to, after playing the first pair in opposite corners, always respond to a move by O by playing such that you are still guaranteed victory (or for O, always play the move that gives you the most options for victory, and hope X makes a mistake). Unfortunately the authors didn't include all of the paths in their search, so it would be necessary to reproduce their results to apply this strategy.
However, I have tried to glean some strategies from the one game they did include in the paper (see Figure 5. I've broken this out into each move in the image below):

Avoid entanglements. By not creating a cycle you avoid giving your opponent the advantage of choosing where your marks are fixed. Of course, if you can make a cycle such that either choice is a win for you, like the placement of X9, then you should do so. See both the resolution of X9 and the alternate resolution.
Play at least one mark of a pair in an empty square. This increases the chances of having the lowest maximum subscript in a tie breaker, as on each play you add another square where you have the mark with the lowest subscript. X does this up until the last move.
Play to increase the number of probable wins. For example on Move 3, after playing X1 in opposite corners, play one X3 in the middle and the other in a corner. When one X1 collapses, the other X1 that is now fixed will be in line with either X3. The same is true for X3 in relation to X1
Don't play on your own marks. X never does this until the final move when there is no other option.
Don't let your last mark be part of your three in a row. Of course your last move has the highest subscript, and can make you lose the tie-breaker. Making this actually happen may require some additional planning.

X could force a win on move 7 by resolving O6 the other way and then playing X7 on top of X1, which can only resolve as a diagonal win for X.  This is taking advantage of the general tic-tac-toe rule that the middle square is the most important

@ChrisDodd that solution gives up X's chance of getting the full point, the solution posted does just as well with the half point forced win but also has a chance to win the full point.


text:
How can I create that sharp, intense sound when placing a stone?
No matter the material (board, stones) being used, pros are able to create a very sharp, loud, intense and beautiful sound when placing a stone (which immediately sits perfectly placed on its intersection). I've been playing for a few years and taught myself how to place stones, but I've never been able to constantly reproduce that sound.
Do you know the secret behind it? Or is it really just experience by placing millions and millions of stones?

A large part of it is probably the quality of the board.  The table boards professionals play on in major events have a pyramid shape cut out of the bottom.
There are three widely accepted reasons for this:

To make sure the inside of the board is dry so it doesn't warp.
To improve the clack sound made when playing a stone.
To hold the blood of kibitzers who interfere with the game.

Don't agree since I've seen (heard, rather) this on normal wooden boards, too.


text:
Is this a correct formal heuristic for invalid moves due to Ko?
In the game of Go, there is the ko rule, which basically states that immediately recapturing a single stone is not a valid move.
I am working on a Go computer program and am trying to implement the ko rule. For efficiency, I am satisfied with a more relaxed definition that aims at two goals:

It always prevents moves that would illegally recapture a ko
It never prevents moves that are legal

Currently, I go with:

If in the previous half-turn exactly one single stone was captured, then it is not allowed to play at the position of this stone in the immediately following half-turn (by the other player).

This is relaxed because it does not attempt to detect that a ko is actually present, it is only concerned with preventing an illegal move. For instance:
$$B
$$ -------
$$ | . . . .
$$ | X X O O
$$ | X O 1 O
$$ | X X . .

This situation is not a ko, because there is no way to recapture. However, playing at the location of the captured stone is still an invalid move, because it would be suicide. I believe that one or the other (illegal recapture or suicide) always applies.
My question is if my heuristic fulfills the two stated goals. (I think it does.)

What evidence do you have that there is or can be a simpler, yet correct, definition? The Rules for Go already are quite simple, and have had hundreds of years of use and refinement.

Perhaps you could clarify what incorrect decisions you are happy with. I would have thought one would prefer a programme that prevents all illegal and some legal moves to one that allows all legal and some illegal moves, but it sounds to me as though you are aiming at the latter.

@ForgetIwaseverhere It was just an idea that came up and I could not think of an immediate reason why it would be wrong, that's why I asked :D

@PJTraill It should allow only legal moves and all legal moves. This Q is particularly about the ko rule.

No, this logic unfortunately doesn't work: It prevents moves that are in fact legal, because they include a snapback.
$$ white to play
$$ . . . . . . .
$$ . . X O X . .
$$ . X O X . X .
$$ . X O O O X .
$$ . . X X X . .
$$ . . . . . . .

White captures:
$$W
$$ . . . . . . .
$$ . . X O X . .
$$ . X O a 1 X .
$$ . X O O O X .
$$ . . X X X . .
$$ . . . . . . .

Now with your definition:

If in the previous half-turn exactly one single stone was captured

This is the case; white captured a single stone.

then it is not allowed to play at the position of this stone in the immediately following half-turn (by the other player).

That position is marked as a. However, it's perfectly legal for black to play there, as it captures the whole white group, and thus does not repeat the board position:
$$W
$$ . . . . . . .
$$ . . X O X . .
$$ . X . 2 . X .
$$ . X . . . X .
$$ . . X X X . .
$$ . . . . . . .

As an interesting side note, the "ko rules" in Tibetan go are such that you are not allowed to play in intersections from which a stone was just captured. This does mean that to be able to play snapbacks, and many other tactical tricks, you need to have at least one ko threat.

Your rule is adequate to detect simple kos but not all forbidden repetitions. EDIT: It incorrectly forbids snapbacks, as observed in balpha’s  answer, and thus fails your condition (2). It is unclear to me from your question whether the “moves that would illegally recapture a ko” that you want it to prevent include only immediate recaptures or also later illegal recaptures; I cannot therefore tell if your condition (1) is satisfied.
The simple rule is inadequate when there is more than one ko on the board. The ko rule is therefore more complex than you suggest and exists in several variants. The most mathematically elegant forms rule out whole-board repetition, and these are probably also most convenient for programming. You could check out Sensei on Ko. Your rule fails to prevent illegal moves in situations like Triple Ko.
P.S. A handy speed up might be that a move cannot be an illegal repetition if there are more stones of either colour on the board than ever before in the game.

Yes, I was not aiming for superko, only simple ko. I forgot to say that. Thanks for the hint in the PS!


text:
Why are the boards not square?
I noticed gobans are not square, but rectangular. Why is that, and what is the correct orientation of the board when playing?

@svante : haha ! Excellent :)

The design of Go components is all about symmetry.  Because perspective shortens the vertical height of the board, the boards are made taller than they are wide.  This way they appear to be square when you're playing.  Another similar issue occurs with the stones.  Black stones appear to be smaller than white stones, so the black stones are actually made slightly larger than the white stones to counter this effect.
Related and awesome: How many squares are there on a Go Board?

That's funny, vertigo.  I've had opponents do that in person but never realized the reason.  I can assure you, GoGrinder's boards are square, so they will help to break this habit :P

@vertigo Which go programs have non square boards? KGS for example is square. As an online player, my problems with real boards were mainly imperfect stones placement and the perspective, not seeing stones from the top.

@CodesInChaos my SGF viewer app (Qipan, for iOS) has non-square board. It simply looks wrong if it is square ;-)


text:
Counting a surrounded non-alive carpenter's square
I was playing against Crazy Stone and we ended up with (essentially) this corner situation:
$$W
$$ -------------
$$ | . . 5 X O .
$$ | 4 1 3 X O .
$$ | . 2 , X O .
$$ | X X X X O .
$$ | O O O O O .
$$ | . . . . . .

This certainly taught me (black) a valuable lesson about what I learned is called the "Carpenter's Square" and how I should not let W1 happen.
What I did not understand was the counting of this situation. I would have expected this to be counted under seki rules (as far as I can tell, Crazy Stone uses Japanese scoring), which in my understanding means all stones are alive and the four empty intersections are dame, so this corner is worth zero for either player.
The program however marked the black stones as dead and counted the corner as white territory, making the above diagram worth 22 points for white.
Is my understanding incorrect? Or the computer's? How should this situation be counted?

I do not know about Crazy Stone, but many programmes mark what they think is alive or dead but let you click on groups to change their status.

During the game
$$
$$ -------------
$$ | a b O X O .
$$ | X O O X O .
$$ | . X , X O .
$$ | X X X X O .
$$ | O O O O O .
$$ | . . . . . .

The corner is not settled yet, as both players can choose to:

either start a ten thousand year ko, white by playing at a and black by playing at b
or settle for a seki, if black plays at a or white plays at b.

At scoring time
If no move has been added in the corner, I would expect this to be considered a seki under Japanese rules (see example 12).
Under Chinese rules, both players would have to agree on the seki, or keep playing (in which case a or b will eventually be played).

Thanks, interesting. But the page you link to says "In fact, the game can end with the position remaining unplayed, in which case it is considered a seki."

@balpha I tried to clarify my answer.

Ah, I see. So indeed the corner shouldn't have counted for anything. Thank you!

Also, I believe that if it is concluded a seki then Black gets one point of territory.

@TylerDurden the Japanese rule doesn't grant points in a seki, other rules do. See "Can you count points in a seki?" http://www.britgo.org/rules/compare.html

On IGS whenever I get positiions like this, 1 point is awarded to black.

@TylerDurden it seems that IGS doesn't follow the Japanese rules: http://home.snafu.de/jasiek/igsrules.html


text:
What happens in Go if you "take back" a ko before making a ko threat?
In Go, there is a potentially repetitive sequence of moves called a "ko." So if your opponent takes a ko from you, you have to make a move elsewhere on the board, called a ko threat, before "taking back" the ko, in order to break up the repetition. But suppose someone "repeats" by taking back a ko, without making a ko threat. What are the penalties?
My understanding is that in a tournament, the penalty will be a forfeit of the game. But I've never seen this rule enforced in informal play. The standard "fix" is that the offender is forced to take back his ko "recapture," and make another move.
Is my understanding of the tournament rule correct? And is there an informal "house rule" that prevents the application of this rule in casual games?

One of the situations where online go is nice. Can't make that mistake :)

@CodesInChaos Even online you still can make that mistake, you just can't click. After trying to click several times, you will realize it: "oh, why can't I play there? oups!" It's just that your opponent doesn't know about it... Actually, I would prefer an online server where you can make such a move, and if you do, you lose immediately.

@Djaian At least on KGS the point(s) blocked by ko are marked with a rectangle.

@CodesInChaos I know. But when you are thinking, you might still want to play there, and realize your mistake only when you click.

At least in European tournaments current rule is that if you play an illegal move you have to take it back, and play a legal move. You don't lose the game by default.

Illegal move
If a player makes an illegal move, and if this is noticed within three moves, then the game should be unwound to the move just before the illegal move, and continued. The referee may allow an adjustment of the time.

EGF general tournament rules

This is reasonable!

AGA rules treat it as a pass: "Illegal Moves: An illegal move is one violating the rules. If a player makes an illegal move, it shall be taken back, treated as a pass, and a pass stone exchanged."

@CodesInChaos: Since 2danplayer seems to be inactive, perhaps we should add the AGA rules and any other rules to this answer. After all, an ideal answer should cover all sets of rules. Sensei’s Library has an article HowIllegalMovesAreHandled covering AGA, Japan and China.

In all official rules that I know this means instant forfeit of the game. Of course, in casual play, mistakes like this can be allowed to take back at the will of the players.
In tournaments I would strongly advise against making such a mistake as well as trying to ask for a take back. If the other player does not seem to allow a take back by his own active will, just resign and promise yourself to not make the same mistake again. Mental strength is part of the game, too, after all (just like time management, i.e. losing on time).
The reason players sometime forget to make a threat is often that after player A took the ko, player B takes some time to contemplate the threats, the board, counts the score, and finally simply recaptures the ko - the need to play a threat slipped out of their mind in the meantime.

Apparently this isn't true for the official EGF (taken back if noticed within three moves) and AGA rules (taken back, treated as a pass).

I was not aware of that. I thought the rules we used at the EGC (and other tournaments) are official, and did not cover this - or did they? I may have misremembered.

One important point to understand is that in an amateur tournament, whenever both players agree on something, then it's ok. For example, if one player plays an illegal move (like taking back a ko immediately) and the other says "take back your move and let's pretend nothing happened", then so it is: nothing happened.
Problems only occur when the players disagree. According to the rule, playing an illegal move is an immediate loss. That means the other player can, if he wants, declare victory.
Usually, weaker players will readily let their opponent take their move back. When it comes to stronger player, they tend to claim a win if the opponent makes such a mistake.
The worst that can happen is that someone else, not playing but just watching, interferes. As a tournament organizer I hate when people interfere in other people's games. That's usually when the fuss starts. The player agreed to let his opponent take back, but someone watching tells he should claim victory, and so on.
In a professional tournament the player making the mistake will resign as soon as he sees it (such mistakes are very rare in professional tournaments, but may happen sometimes).

Roughly how strong do you think players usually are when they start resigning/expecting the opponent to resign?

@PJTraill I have seen 6 dan players acting like ***holes during tournaments just because they see an opportunity to grab a win where they should lose (pretending the opponent is cheating while adjusting the clock for byo-yomi, continue playing a lot of tricky moves because they have more time, trying to disturb the opponent in any possible way, and so on). As a former tournament organizer myself, most problems occured with dan level players. Granted, technically speaking you can claim you won if your opponent make such a mistake. Kyu players tend to enjoy playing the game more than winning.

Amateur dan players, evidently, and presumably only some bad eggs — but sad anyway. As for •••holes, perhaps one might say trumpholes?

I have seen what you're describing in tournaments, too.  A player made a mistake and took the ko when it was illegal, and the other player just made him take the move back.

This doesn’t answer the question:(


text:
How to balance influence vs territorial plays on a 13x13 board?
Most of the answers to "What is a good opening on a 13x13 board?" deal generally with the differences between large and medium boards. I'd like to ask more generally about how I should be thinking on a 13x13 board. I feel comfortable fighting on a 9x9 board, and generally on a 19x19 board I prefer to play for influence throughout the early game.
However, I get confused in the middle. With a medium-sized board the center is small, which suggests a more territorial style, yet due to that same smallness, a "moyo" including most of a side is easy to set up, which makes me want to do that. How should I think about these differences in the fuseki?

Territory is more important on a 13X13 board than on a 19x19 board, because it's smaller. Therefore, a given chunk of territory is more valuable. That is because the corner josekis give a "fixed" amount of territory, while the resulting outer walls are worth less (and overlap more) on the 13 x13 board. So you should play a "lower" (near the edge) more territorial game on the smaller board.
I don't like to play this kind of game because it "corrupts" my thinking for the larger, regular board. The one advantage I see for 13x13 is that you get to the endgame faster (and it is a disproportionately larger part of the game, thereby giving you more "practice" in this part of the game. Endgame and opening are more important with the smaller board, while the middle game shrinks.


text:
What is a good opening for Go on a 13x13 board?
I have been playing a lot of Go recently at lunch but we are playing on a 13x13 board instead of a 19x19.  I can't seem to ever win!  I am beginning to think that my opening is leaving me in a bad position.
I generally start by playing 2 of the 4,4's (my opponent mirrors) then I do a lower approach on one of his corners.
This is based off the same way I would start on a larger 19x19 board.  Any advice?

The Stack Overflow site "Game of Go" is gestating!  http://area51.stackexchange.com/proposals/11942/game-of-go?referrer=bo6CSU3-Avu37ioaqR_AEA2

@Smandoli IMHO those questions belong on this site. It just makes more sense to have a site that can say "Come here to talk about Board and Card Games!" and not "Come to here to talk about Board and Card Games that aren't Go or Magic: The Gathering". Unless that site starts getting mad traffic once it goes into beta, I'll stand fast by this recommendation. It's the same principle of having a separate site just for Java development, or a specific video-game. :D

I don't think Smandoli was saying my question was misplaced.  Just letting me know I could find more Go information over there.  I have found the community here on B&CG very supportive so far :)

@CJD -- Yes, yours is a worthy viewpoint, the debate will stay lively, and the "mad traffic" standard is appropriate.  Probably you play Go and understand its transcendent quality.  (If not, by all means get infected.)  My "Plan A" would be for everyone to refer to (and invest in) senseis.xmp.net.

13x13 is a bit of a faster game than 19x19, but the same general principles for fuseki apply. (This is the main difference between 13x13 and 9x9, where the board is small enough to preclude a real opening)
The 4-4 is a center-facing move, emphasizing influence over strength. It leaves you open to losing the corner with an opponants 3-3 and the resulting joseki.
The 3-3 is a corner-facing move, emphasizing reliable corner territory over influence.
Another option is the side-facing 3-4 move.
Opening Theory Made Easy by Otake Hideo provides reasonably good coverage of the logic behind these different opening options. At a more fundamental level, the Janice Kim introductory series does a good job of introducing the motivations for different opening moves.

What is the joseki if I open with 4,4 and the opponent continues with 3,3 (on 19x19 or 13x13 if it is that defined)

The 3-3 invasion usually touches off a tradeoff of corner territory for outward influence, a basic summary can be found here: http://senseis.xmp.net/?33PointInvasion .  The precise value of the territory versus the influence depends on the larger board situation.

The 4-4 spot in the corner is a well balanced move on the 19x19 board. It has some influence on the corner and gives you strength in the center. I don't like it as much on the 13x13 board because the center is small and strength facing outwards often turns into over concentration. The main use of a strong group is to attack weaker groups elsewhere on the board - it can of course help to do this on the smaller board as well, but it is usually not worth as much as on the bigger board. Therefore I feel like the 4-4 is too center oriented, I prefer to open with a more territory oriented move that goes for immediate profit.
My choices for opening on 13x13 are 3-3, 3-4 and 3-5. If you want some crazy fighting you can start with your first move in the center but have to know what you are doing with such a move. You should also choose more territory oriented josekis than you would in 19x19.
As a former 3dan I am by far no pro, so please take my advice with a grain of salt. 4-4 is not an unplayable opening in 13-13, but I find it harder to utilize than the other moves I suggested.

On a 13 x13 board, I'd make your third move on the center point, and try to "control the center," as in chess.
It can be done on a full 19x19 board, but that is a very sophisticated way of playing because the center stone can be more easily gotten around. It's worth more on the smaller board because it almost links your two corner stones together and makes it almost impossible for your opponent to do the same.

I am by no means an expert at Go, but I was under the impression that controlling the center is something you don't want to do on a 19x19 until late in the game, opting instead to fight for the corners, because it is so much easier to surround territory on the corners than in the center.

@BlueRaja-DannyPflughoeft Controlling the center is always desirable in a 19x19 game, but usually not possible. If you actually can control the center early in the game, you are on the path to a big victory. However, if you try to do that, in most games you'll find that you don't actually control it.


text:
Do we have Go endgame tablebases, just like Chess?
I am a relative newcomer to Go and more interested in the programming aspects of designing a Go Engine.
 Do Go engines use a systematic list of Endgame positions with known wins, the way chess engines do? And do professional-level Go players memorize a list of endgame positions as part of their training?
Some online digging turned up  this Wikipedia article  and  computer science paper  suggesting that Solving Go Endgames by computer is really hard problem, guaranteed to take up plenty of time. Perhaps that's the reason why Go endgame tablebases don't exist / aren't widely used?

On topic at game development?

As there is two question in one I migrated the second here

While I do not recall specific comments on endgame positions in the reports on Apha Go — and I found neither “endgame” nor “end game” nor “yose” in the freely available PDF of  Mastering the Game of Go with Deep Neural Networks and
Tree Search —, I imagine that its policy neural network with 13 layers containing 192 filters would, once trained, be pretty useful in the end game.

I know much more about Go than chess, so I don't know how accurate my guesses about chess endgames will be, but... I think chess tends to be at its simplest in the endgame (like Go), and there is little information in the endgame about how the rest of the game progressed (unlike Go). And, there are  algorithms for winning (in some cases) that generalize to a large number of positions. E.g., if white has a Queen and a King, and black only has a King, unless the Q is about to be captured white will win, regardless of position.
In Go, the endgame is also the simplest part of the game, in that it is the only part of the game in which the score and the value of each move can be calculated very precisely, however position is everything, and you will never see two games end with the same board position. In chess, you can generalize many positions to what pieces remain, but in Go the only generalization is the score count and the number of points still remaining to be taken.
It would be utterly impractical to rely on whole-board level tables in Go, however many small areas of the board will fit common patterns, which are memorized by Go players and programmed into computers alike. A medium-level example would be bent four in the corner is dead, but you'll see that this relatively common position is not at all obvious to a beginner and comes with several caveats (no un-removable ko-threats, surrounding group must be alive with 2 eyes), and while relatively common, it is somewhat of a special case.
Something like knowing where to play to kill (or save!) bulky five (and many many other shapes) should be automatic for a human or computer.
Programming a computer to play good yose is a very hard problem, but is one of the few aspects of Go at which computers can beat professional Go players, as it is all calculation and, while difficult, the theory is well-understood. To do it effectively the program must be able to estimate the current score, identify the moves left on the board that are worth points, and sort them in order of worth---not trivial. This article might be a good place to start.
Summary: There is a place for systematic lists of positions in Go programs, but they won't be useful for whole-board positions. The program needs to be able to evaluate the life/death status of every group on the board, and it is in these one or two group positions that lists will be most useful---and they'll be valuable throughout the game. In trying to attack opponent's groups, your program's goal won't be just to recognize shapes from the list, but rather to force the opponent into shapes that it knows are dead, while maintaining live shapes itself. See also Common Corner Shapes.
In writing a Go program, I would give you the same advice as I would a person learning Go: start on a small board, even as small as 5x5 until you have the rules down. Move up to 7x7, and then stay with 9x9 for a long time before you even think of trying 13x13.
Good luck!

And, after writing this answer last night, I woke up to this article in my inbox: [http://www.usgo.org/resources/downloads/Supercomputer%20Go.pdf]

Clearly, the Go god was listening.

Good answer, another page on SL that might be useful (the closest thing I know of to the endgame table asked about) is the miai values of common endgame positions page - http://senseis.xmp.net/?MiaiValuesList

There are no whole-board endgame databases because there are too many possible permutations in the end. Such an approach could not be handled with state-of-the-art computer systems.
The good news is that endgame databases for confined, local situations do make sense and can be applied really well, the further the endgame has progressed. After you have analyzed, what's the result of each confined, local situation, you can calculate the best way to deal with them by finding the right order of moves. Of course it is not that easy to find a perfect solution, because you have to time "free" sente moves that do destroy your own ko-threats and so on, but AI endgame in Go is pretty good compared to midgame.
To understand the concept of the endgame, common patterns, the importance of sente/gote and so on, I can recommend the book The Endgame, if you are interested in them.

There is an endgame theory for Go using Surreal numbers. The theory is layed out in the book "Mathematical Go: Chilling Gets the Last Point" by Berlekamp and Wolfe.
An AI using this theory plays optimal and usually beats a human pro player on the endgame. Reading the book is still fun for an average Go player and teaches you a lot about endgame strategy.

Comments on the Amazon page for the book, indicate that computer Go today is very different than described in that book.

@John: The mentioned book only treats endgame theory (but this in an exhaustive way); it is not about Computer Go in general.


text:
Maximum attainable points for a single player in a two player game of Carcassonne
Let's start this question with the following assumptions about a game of Carcassonne:

It's a two player game.
The game is using the current basic set of rules/tiles. edit: just to be clear, I'm using 3rd edition rules that score all city tiles the same 2 points and count the same city for different farms as described here.
The player we want to maximize has perfect luck and perfect competence both drawing the tiles and placing them.
The opponent has minimum luck and competence.

This means we can choose what tiles are drawn by each player, how they are played and how to place each players' meeples.
Is there an optimal way of placing the tiles in order to maximize one player's score? If so, can that way be determined?

I don't remember the quantity of each pieces, but we can calculate it without building a map, assuming that we can build the most of points like there were only 1 player. So probably an excel would solve most of the case.

@user1873 Meeple placement is indeed optional. We can definitely assume the opponent never places one. I never heard of different scoring systems, although I assumed there are older and newer versions of the rules. Going to look that up as soon as I can.

If the opponent has minimum luck and competence, can we also then assume they 'accidentally' always place tiles in positions that give the first player more possibility for points, or indeed finish the first player's features for them? That would effectively be like a 1 player game where you can only place a meeple every other go.

@user1873 isn't that a rule on one of the expansions? I couldn't remember which one, but I think that when you play it on brettspielwelt it use that rule.

@user1873 the rules I link to in the question body say that each city tiles is worth 2 points regardless. I think they were the last official rules to see print.

this question solve your point problems http://boardgames.stackexchange.com/questions/7/there-seem-to-be-several-different-scoring-systems-for-fields-cities-in-carcasso

342
For third edition rules
Here's one way how:

The numbers show when tiles are placed. Black is the first player to move and always places a meeple where the number is. Blue moves second and never places a meeple.
Proof of optimality
Meeple usage optimality:
The first player to move will have 36 turns and thus only 36 opportunities to place meeples for points. The best usage of those 36 is:

7 for fields
6 for cloisters
16 for cities
7 for roads

No more can be placed on fields since you don't get those back. The number on cities and cloisters are sufficient to count all possible city/cloister points (except the odd city tile, see city optimality). So the only possible useful change would be to get more roads, which adds at most 2 because all roads without endpoints are already scored. But 2 is less than the amount we would lose by having less cities (each extra city adds 3 field points to each big field), cloisters (9 each), or small field (6 each).
City optimality:
There are 44 city tiles and 10 shields. 5 of the city tiles have two separate cities on them and so if you put them on separate cities they can be counted twice. This would all be worth 59 points and completed cities are worth double making 118. However all completed cities must have an even vertical perimeter and an even horizontal perimeter because they must be complete on both left/right and up/down (a diagonal city tile has 1 horizontal and 1 vertical perimeter, not 1 total perimeter). For example say a city is 5 tiles tall, then its vertical perimeter must be 10 if completed, yes you could have concavity but the same rules apply to that. And since the total horizontal+vertical perimeter of all available tiles is odd, you cannot complete all city tiles, the best you can do is leave out one tile. This means the maximum city score is (59-1)*2=116.
Field optimality:
We need to maximize the number of completed cities since we can construct big fields that touch all cities. A completed city must have at least 4 tiles with a corner that do not have their own city touching them. Diagonal city tiles (i.e black #29) have 1 such corner and the simple side piece cities (i.e. the starting tile) have 2 such corners (all other city tiles have 0). There are 10 diagonals city tiles and 28 simple side piece cities (tiles with 2 simple side piece cities count as 2). This gives 10+28*2=66 corners which allows for a maximum of 16 cities.
Now to maximize the number of fields that touch these 16 cities. I will attempt to explain this intuitively but user1873 came to the same conclusion and proves it using graph theory (see their answer - which is the same in other ways except only has 15 cities). At most we can have 2 big fields touching all cities. A third field must be stuck inside these big fields and therefore only touching only 2 cities. In order to touch a third city it would have to cross one of the big fields, but this would split that big field into two smaller fields.
So we have 2 big fields touching 16 cities and 5 small fields touching 2 cities. 2*16*3+5*2*3=126
Road optimality:
Each completed road has 2 end points. There are 32 road tiles without end points which can fit into any completed road. So the maximum points you can get from roads are 2 * #roads + 32. Since we only have time to place meeples on 7 roads this gives 2*7+32=46.
Cloister optimality:
Each of 6 cloisters can score a maximum of 9 for 6*9=54.
Sum of optimal features 116+126+46+54 = 342
I have to say that this is a beautiful question because not only is the proof multi faceted with outside the box thinking required, but because actually finding a configuration that achieved it was really nontrivial and fun. So much so that even 6 years later I remember the joy of figuring out to put that gap in the city to allow it to complete and the tile missing to prevent the two big fields from joining. That beauty is why I've revisited this and rewritten this answer to make it more clear and also to break it down into an example with all moves and a separate proof.
An open question would be what the maximum possible score in a one player game is. You'd have 31 more turns to place meeples but the only possible points to grab would be completing more roads. There are 12 road endpoints unused in my example but how in the world would you rearrange them to connect with each other without breaking other things? Another open question is what the total collaborative score is for games of 2-5 players. This would be tricky since sharing cities, fields and even roads would need to be done as much as possible but it surely won't be possible to share everything. These questions will be an order of magnitude harder to solve and prove optimal.

"I don't think it is actually possible to close all roads while still achieving that many fields), this could at most net 12 more points." I know my theoretical score didn't correctly count that 16 cites can be closed with the city tiles, but you might want to correct the statement above. You can only score 1 point per road, and there are only 46 tiles with roads on them. You couldn't net an additional 12 points. I agree these theoretical scores are maximums, that is why I also included average points per meeple. For example in your configuration above there are 3 roads that cannot be scored

Still, 339 is an impressive score, near enough to the maximum that it would be valuable to expand this answer to include all the moves for the 1st/2nd player that make that player score 339. It should be pretty trivial to work out that answer.

You can use the Contact Us form to have your accounts merged so that you can freely edit your answer

Note: This answer below is using 1st Edition scoring rules (Cities only scored once). Although parts of the analysis are valuable to calculating the highest score possible (36 Scoring opportunities, try to score with Meeples that have a high point value per the 36 turns, the highest score of 278 is not correct. I will have to do further analysis to determine if 7 farms can be created scoring the most cities multiple times, will result in a maximal score. (Edit: The maximum theoretical score is 338, as posted in my other 3rd Edition rules analysis)
Hackworth's analysis about optimal play exists is correct, although he is right about the difficulty in evaluating all possible moves, I believe that tengfred is closer in estimation of what we need to determine the maximum score possible. We only need to evaluate ending maps, not all possible moves from the beginning to the end. Once we have found the maximum scoring map, we only need to backtrack to find a legal set of moves that will lead to that end map. That is as tengfred points out, a much smaller problem.
The Rules show 72 tiles, and scoring as follows:

Road (1 point per tile), Cities (2 points per tile, 2 Points per pennant), Cloister (1 point for each tile, cloister tile and surrounding tiles), Farm (3 points per completed city)

A similar question was asked on The Opinionated Gamer.

My challenge to you is to see how many points you can score using only the 72 tiles in the base set – following all of the rules as printed in the box — EXCEPT that you can only use one meeple!

Wie Hwa (animation) solved that problem with 277 points.

He didn't always score on odd turns, so his solution isn't the correct answer to this question. But, someone with more desire than me might want to take his end map and modify the order of the turns so that player 1 is always the one to score.
This gets you a grand total of 278 points (the board actually will have an unmatched city tile that cannot score I think, and still place a farmer). It is amazing that Wei Hwa almost achieved this using only a single meeple in single player. Feel free to figure out if his end map is an optimal solution for 2-player.
The maximum number of scoring opportunities is 36 if you go first since you can only place meeples on your turn. The optimal solution will include:

one farmer, 45 points
15 completed cities, 116 points (10 pennants = 20 pts, 48 city edges = 96)
6 cloisters, 54 points.
The remaining points divided between the most points that can be scored on roads with the remaining 14 moves (1x Quad road = 4 pts,7x Tri Road = 21 points, 32 straight/elbow non terminals, 5x Dead End)

Wonderful answer, totally worth looking into! By the way, the question was never changed :p

Hmm, with the above graphic, you can have many more than 1 farm - the one massive central one, yes, but also potentially 6 other farmers on closed 'edge' pieces of fields around the perimeter - such as at the bottom - all those tiles with mostly city but a small bit of field underneath are all separate "farms", and thus you could score that big city 7 times (using all 7 scoring-meeple as farmers by game end) = 21 points for farmers for just that city. Add the farmer score for the remaining 14 cities for the massive farm gets you to 63 points for farming. Any advances on that for the farms?

@user1873 Those are the old Carcassonne rules.  In newer versions of the ruleset, you do very much score per farm controlled, not per city farmed.

Ah yes, old farm scoring is quite different to newer farm scoring (which is MUCH simpler).  Perhaps @rahzark needs to confirm in the question whether he means old or new rules scoring (as also old scoring also has 2-tile cities only scoring 2 points, whereas new scoring makes all city tiles worth 2 points each, so a 2-tile city is 4 points).

@NickShaw People, read the question!! I have written: "The game is using the current basic set of rules/tiles." It even links to the rules I'm talking about! Either way, I'm going to edit it now.

Whoops! Thanks for confirming @rahzark, I'd clearly missed that in the question. Thanks for clarifying the question. :)

I show 10 places on here for additional farmers. More than the meeples available. 6 around big city alone. The one in the upper right corner scores two cities.

@aramis, You should read the note at the top. This map is a near optimal solutionfor 1st Edition rules 277. 3rd Edition is covered in the other answer 338.

As Hackworth notes, testing all possible layouts is obviously infeasible. However, it might be possible to get a decent upper bound on the points.
I think we can safely assume that meeples will not be a scarce resource when playing in this way (since we can choose to complete the map in any order, it should not be difficult to ensure we always have meeples available). Thus the problem is reduced to finding the highest scoring map-layout.
The upper bounds for roads, cities and cloisters are easy; 9 for each cloister, 2 for each city tile (and 2 for each pennant) and 1 for each road-tile.
The farms are the tricky part. First we would have to figure out the maximum number of cities we can make, and second, we need to figure out the maximum number of farms that can supply each city, with the limit of max 7 fields total.
This is certainly not easy, but should be a lot easier than the original problem (there are a limited number of ways in which a city could be supplied by multiple farms, etc).
You could probably get a decent approximation by just playing around with the tiles for a few hours...

I very much concur with this answer, and was in the process of trying to ascertain the potential max farm score in such a scenario before @tengfred got in there first with this answer (nice one, tengfred).  I'm not sure I agree with the "max 9 fields" though: We only have 7 meeple to score with, so that would mean a max of 7 distinct farms from which to score from. And only shared cities (cities that adjoin two farms, bordered by road or playing-area-edge) would give extra scoring, and there's a limit to how many cities can adjoin multiple farms, so it should be calculable...

Yes, that's what I meant by 7 "scoring" farms. You can score a single city more than once if the city in question borders 2 separate fields - that's where the complexity lies; positioning cities near the edge of play such that you can use the small field areas on the very edge of play as separate farms for scoring.

Yes, it should be 7 farms, for some reason I got the idea you had 9 meeples. Edited the answer.

Is there an optimal way of placing the tiles in order to maximize one player's score?

Yes
With a finite amount of tiles, a finite number of legal placements for any tile at any point in the game, and a finite amount of meeple plays after placing a tile, it should be obvious that there are a finite number of possible games. Every game has a definite result in terms of score, so clearly, there must be at least one game that produces the highest possible score.

If so, can that way be determined?

Yes
Obviously, in theory you could just run all possible plays through a computer and find the best.
In practice, my gut feeling as a programmer tells me that an exhaustive search through all plays would, on current hardware, quickly become useless due to a combinatorial explosion, simply because the search tree becomes too broad to handle incredibly fast.
Just for a rough idea, a sample calculation:
The base set has 72 tiles. After placing the starting tile, you have 71 at your disposal.

The starter tile is designed to allow every single of those 71 tiles to be placed in 1-6 distinct ways, not counting irrelevant, merely graphical rotations.
Assuming a mean of 3 ways per tile, for the first move alone, you have roughly 200 possible maps after the 2nd tile has been placed.
Next turn, you have only 1 fewer tiles, but far more possible ways to place the average tile, because the number of single edges has increased by 2, or increased by 50% in this case. You are probably looking at around 300 possible moves now.
Even if the sum of possible tile placements for all remaining tiles stagnate at 200 per move on average (which is probably way too low an estimate), after 10+1 moves of 71+1, you already have 200^10 = 10^23 different maps.
Calculating 10^12 maps per second (probably out of reach for typical desktop hardware), it would take over 3000 years just to calculate all possible maps after 11 moves.. Make that a million Desktop PCs, and you're looking at a full day to calculate the first 11 moves. Next move, multiply by 200, and so on.
And all that is assuming nobody places any meeple during those turns, which again vastly increases the search space!
In conclusion, I find it highly unlikely that anyone has ever calculated anything close to an exhaustive set of possible Carcassonne maps, let alone the definite high score anyone can theoretically achieve.

I understand that neither of these answers is particularly helpful, but I believe that's pretty much all you can say about the problem without writing a scientific paper.

I agree with what you said. What about non-exhaustive searches? Is there a placement algorithm that can maximize this problem?

@rahzark I bet you could try something with neural nets or genetic algorithms to let the AI figure out which moves are generally the most lucrative in the long run, assuming no opponent interference and free choice of all tiles. However, the problem remains that the search space is so vast, and any sort of "learning" algorithm can only sample a vanishingly small portion of that space. You could always land in a local maximum and believe you have found a pretty good solution, while missing the mountain right next to it.

Rather than attempting to solve an extremely hard problem, you could just solve a simpler one, and figure out some good upper bounds based on the number of available tiles of each type. You'd overestimate, but it'd be nearly as useful as a theoretical maximum that you'll never achieve either!

After thinking about what you said, I don't think meeples make the problem harder. I actually think they simplify it: they are the only way to score! I think the solution to this problem is making the most out of each (for example, using only one to score all of the city tiles in a single giant city). Of course, farmers do depend on the map shape and add a lot of complexity but I son't think "brute force" is the only way to solve this.

@rahzark I don't see how meeples make the problem easier. The most trivial counter-example is the first move: Without meeples, you have only 1 move for the first turn. You place the starting tile, and pass. However, if you do place a meeple, you have 4 different game states after the first move, each of which can influence the subsequent meeple moves. It's like another game literally on top of the tile placement game.

I don't think you need to trace every single move. Imagine that you can prove it is possible to make a single city and a single road with all the existing tiles. With only two meeples you can score all those points, leaving all your other meeples to maximize cloisters and farms. It's the maximized end state that's important, not every single option you get along the way.

@rahzark That's the question though, isn't it? Is one big city really optimal, since it only counts as 1 town for your farms? Is it not better to have multiple farms, each adjacent to many small towns? Also for roads: with 3 tiles, you can make 1 road of length 3, or 2 roads of length 2 each. You see, there are many different ways to do things, and without a computer, you will have a hard time deciding which matters most.

I agree with you, but I just think it is possible there is an attainable solution that doesn't need to look at all possible board states.

The maximum theoretical score is 338. I previously answered the question for 1st Edition rules (cities only scored once by farmers), so here is my second stab at this question. First, it should be rather clear to everyone that since there are only 72 tiles and one is the starting tile, that the maximum number of scoring opportunities is 36 if you go first. You can only score by placing a meeple on your turn, regardless of whether you our your opponent lay the tile that actually scores that meeple. It should also be clear that the optimal score consists of scoring the highest possible points for each meeple that you can score. Here are the maximum scoring possibilities.

1x Long Road: A long road will score 32 pts. + 2 pts. = 34 pts.
2x Big Farm: A Big Farm will score 3 pts. x 15 cities = 45 pts.
6x Cloister: A Cloister will score 9 pts.
15x Cities: Scoring all 15 cities is worth 2x (10 pts (pennant) + 48 pts (city tiles)) = 116 pts
5x Small Farm: A small farm will score 3 pts. x 2 cities = 6 pts.
7x Short Road: A short road will score 1 pts. x 2 road tiles = 2 pts.
Total Score = 34 + 2x45 + 6x9 + 116 + 5x6 + 7x2 = 338 points

Notes regarding average score per Meeple.
A Big Farm requires that you have completed 15 cities. The maximum number of Big Farms is two. Imagine the two farms as either Inside/Outside, Left/Right, Up/Down, etc. In the graph below, the Big Farms would be A and B and there would be 15 nodes in-between them. This can be shown to be true using Euler's Theorm, regarding the K(3,3) Utility Graph problem. Average points ~10.72. This calculation comes from total city points (116) + (90) 2 Big Farms + (30) 5 Small Farms = (236 pts/(15 Knight+7 Farmer=22 Meeples))

The maximum number of small farms is 5, since you only have 7 Meeples and 2 will be unavailable as Big Farmers. These small farms will sit like node C, in-between two cities only.
The Long Road requires using all non-terminating road tiles (32), and two road tiles that dead-end into a city/cloister. Constructing a single longest road is not necessary to find a highest scoring map, as long as any non-terminating road tiles are included in the Short Roads you score.
The maximum number of cities is 15 (with one unused edge). Average points ~7.73.
The maximum number of Cloisters is 6. Average points 9.
The maximum number of Short Roads is all Meeples minus those previously used with higher average scores (36 - (1+2+6+15+5)) = 7 Meeples. As noted earlier, these short roads can use non-terminating segments that were not used in the Long Road scoring. The only thing required to obtain the highest score possible is that all 32x non-terminating road tiles are used among the 8x Long/Short Road meeples. Average score for all 8 road meeples 48 / 8 = 6 pts.

Your statement about the breakdown of the farms by number of cities isn't necessarily correct.  Consider a case with four cities and four farms.  Using your method the optimum graph would involve two farmers connected to four cities each and two connected to two cities.  In fact it's possible to connect each farmer to three cities (think of a cube with cities/farmers at the vertices).

..which scores exactly the same number of points (sorry - I miscounted!)

@tttppp, I am pretty sure the K(3,3) Utility Graph problem was solved a long time ago and verified by many different mathematicians (Euler, Kuratowski, Thomassen, etc.). I sincerely doubt that those guys who are much smarter than me got this wrong.

You didn't exactly prove the reduction from this question to that theory though!

@tttppp, I think that proving that K3,4 is a subgragh of K3,3 and therefore is also nonplanar is beyond the scope of this SE. This is not Math.SE, and I do not believe that such a proof (or explanation of one) is necessary. Is it just the Utility Graph reduction you are worried about, or do you believe that my highest Average points per Meeple reduction to the problem is flawed.

I think your answer is a valid upper bound, although I can't prove it.  The question I'm interested in is "Find the maximum number of edges of a planar bipartite graph with order (7,15)".

@tttppp, That is an interesting question, but is beyond the scope of BG.SE. I encourage you to ask that question on math.SE. I am confident that the graph K(7,15) has the subgraph K(3,3) within it and that the the maximum number of edges that can be created without overlap with that many nodes is to arrange them in a line. I assumed but didn't know the name for the Utility Graph problem, so asked there to verify my assumption.

The single line cities gets you 45+45+5x6=120 points. If you create 2 lines one of 8, and the other of 7 you get three big farms, but less overall points. The inside farm is still adjacent to 15 cities (45 pts), but the outside farms only score 45 collectively (8x3+7x3=45). This leaves you with 4 farmers left (instead of 5) who are subject to the K(3,3) Utility Graph problem. They can score at maximum 4x6=24 points. This gives you a maximum total of 114 points, 6 less than the optimal graph. This only gets worse as you increase the number of lines/big farms.

@tttppp, Actually I should note that when I speak of Big Farms, just as when I speak of Long/Short Roads, what I really mean is that each city needs to be counted the most times possible. You can instead imagine a line of farmers in the middle with one City on Top and on on the Bottom with the remaining in the middle. These middle cities can only score 6pts (13x6=78) because of Utility Graph promlem. Then the top/bottom cities need 7 edges each to connect to each farmer to score 120 pts.(7 edges per city is impossible)

7 edges is not impossible. I was thiking about a 2 tile city. A 5 tile city is sufficient for 7 edges.

I'll put a question on the maths site, that's a good idea.  Your calculation of 114 points (6 points short) is not really right, as there are an extra 6 points to be scored by connecting around the sides of the lines (e.g. the seven-city farm can be connected to the two end cities in the row of eight).

The question is here: http://math.stackexchange.com/questions/149125/what-is-the-maximum-number-of-edges-in-a-planar-bipartite-graph-which-have-parti

285
For first edition (and 272 for 2nd edition which is optimal for that edition as well). This question was for the 3rd edition, but the older editions lead to different and interesting ideas so I thought I'd answer it for that as well.
A game with the most 1st edition theoretical points possible would:

Have all roads scored and all end points on separate tiles (you can't count each endpoint on the same tile as its own point if they are on the same road).
Have as many completed and scored cities as possible and no small cities (in the 1st/2nd edition completed cities with only 2 tiles are only worth 2 points). And like roads, tiles with multiple cities must end up being in separate cities.
Any incomplete city tiles scored.
All cloisters complete and scored.
All cities scored by at least one field (cities cannot be counted by multiple fields as in the 3rd edition).

On max number of cities while minimizing small cities
For conciseness, let 2C mean a tile with a city on two sides, etc.
So from the 3rd edition answer of 342 we know that 16 cities are possible and how to calculate the number of possible cities. By that calculation (66 corners), we have 5 extra 2Cs (3 inline and 2 diagonal) that can be inserted into a small city without lowering the number of total cities. That leaves us with only 9 small cities. From here we can lower the number of small cities but to do so requires also lowering the number of total cities. Each fewer small city gains 2 points, but each less city loses 4 points (each city in a field is worth 4 points in the first edition). To start we can take 4 more 2C and then we would have 5 small cities and 15 cities, a gain of 4 points. We can't do this again though because then there is no way to complete the city with the 3Cs and 4C with using many 1C pieces inefficiently. But we can take two 3Cs and surround each with three 1Cs. Then we have 2 small cities and 14 cities, an improvement of 2 more points. Finally we can take a single 3C and surround it in the same way, this brings us to 0 small cities, and 13 cities. That doesn't add any more points but is still better because it will take less turns/meeples since each city needs to be scored. There are many city configurations that can achieve this but none can do better.
To summarize our options are:

9 small cities, 16 total cities
5 small cities, 15 total cities (+4)
2 small cities, 14 total cities (+6)
0 small cities, 13 total cities (+6)

How many points would that give?

Roads: 30 endpoints and 32 non endpoints, so 62
Cities: 44 city tiles, 1 can't be completed, 5 have 2 cities, 10 shields so (44-1+5+10)*2+1 = 117
Fields: 13 cities * 4 each = 52 (39 in 2nd edition as fields go down to being worth 3)
Cloisters: 6 cloisters * 9 each = 54

Total: 62 + 117 + 52 + 54 = 285 points
How many meeples (turns) would that take?
The fewest number of roads possible is 15 since there are 30 endpoints (it would be more if you had any pure loops).
The fewest number of fields that can touch all cities is 1.
Therefore we need at least: 15 (roads) + 13 (completed cities) + 1 (incomplete cities) + 1 (fields) + 6 (cloisters) = 36 meeples.
It just so happens that we need exactly as many opportunities to place meeples as there are turns for the first player, that's a crazy coincidence! Which means it is just barely possible to nab all possible features in an optimal solution. This also means it's impossible to score any more points even if playing solitaire.
Is it possible to construct a map that meets all these criteria?
With much guided trial and error I found one possible way to build said map:

It wouldn't be hard to find an order of laying tiles and placing meeples to achieve this because you won't need to coordinate all 7 meeples being the in the fields like you would for the solution to the third edition. At the end all you need is 1 meeple in the big field and 1 on the incomplete city tile.
This was easier to construct than the 3rd edition solution in terms of fields but much harder in terms of roads.

I dont think one needs to know all possible arrangements to calculate the maximum score. There  are a finite amount of city tiles for example which, if the cities are closed, give 2 points each (not counting shields for simplicitys sake). So basically, the question on how many points one can get from the cities is to ask if one can build one massive city with the city tiles. Or how many small cities can be made. The points will be the same in both cases

The points will not be the same, because many small towns count more for a farm. Also, one town can count for several farms, so that is good for the score as well. So, which number and arrangement of towns makes the optimal number and size of farms? And what about roads, cloisters?

@Hackworth: "Many small towns count more for a farm" - huh? A town is the same score for a farm regardless of its size (at least in the basic game).  But you are right that some towns can count for several farms - but not many; there's a limit to how many farms can share a single town with the base-game's tiles. So really, work out the max number of cities you can make, and the max number of farms that can share the max number of cities, and you have the max score for a farm.

A big city will not score the same as many small cities because several tiles can count to 2 cities (so that tile scores twice in the game), but if both parts of that tile are part of 1 city will only score once. Having said that I think it's a great way of approximating a "highest possible score". Good thinking @Jake! ... I think for it to work the best way is to make as many cities as possible with the tiles, then assume all the cities belong to two farms and all roads belong to the same player, and that that player would be the KING.

Hackworth, yes of course, but I was talking about city points only. In any case, I was not completly right anyway but my point remains valid i think, that is, one does not need to check all possible arrangements to find the maximum score.


text:
Why are stones placed with index and middle finger?
I have read that stones should be picked out of the bowl and placed on the board by pinching them between the tips of the index and middle finger. This seems to be a long-standing tradition.
Is this standard around the world? Why is this grip used instead of the more obvious thumb and  index finger grip?

This is more a matter of etiquette than anything else. One, it creates a "nice" sound on the board. Two, it minimizes the possibility of displacing stones already on the board.


text:
Game from 1978: Why does this fuseki position favor black?
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Nie Weiping's book The Art of Positional Judgment begins with a chapter on fuseki. He gives this position, which did not actually occur in the game, but might have, as one that would favor black.
He says that his opponent did not play at hoshi on the left hand side, because the position favors black. But he doesn't elaborate. He simply moves to the next diagram.
But given that it's white's move, the position seems pretty balanced to me. How does it favor black?

It seems that the move to play for White is an extension on the upper side or in the lower side. Clearly, the lower side is more interesting for the following two main reasons.
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Black's shimari in the upper right is low. Black is thus not interested in playing in this area  now.
Black has a perfect extension at 2. His moyo is developing perfectly.

Consequently, White wants to play as in the following diagram.
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However, Black has now a very good erasing move at 2. White's moyo is therefore not as good as Black's position.
In conclusion, even if White plays its best move, Black keeps the upper hand which means that the fuseki favors Black by a bit.

The san-ren-sei point in the middle of the lower part of the board, and a point around white's 3-3 point stone in the lower left hand corner are "miai." That is if White takes one, black will take the other.
Let's say white closes his corner in the lower left by taking the 5-4 point, as he did in the upper left. Black then takes the san-ren-sei point in the lower middle side, which is a perfect extension from his shimari in the lower right.
If White takes the san-ren-sei point, Black has a perfect invasion point of White's moyo at the 4-4 point in the lower left, as shown in the diagram above.
Either way, Black gets the LAST "big point," thereby retaining the advantage of the first move (even after komi is factored in).

The thought process on that looks to be around the general strength of playing at the hoshi. If white were to play there, black would have an opening to take a third corner, but by playing on the third line, white made it harder for black to take the corner from white.