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[pingpongsam]

I know in games on Isotropic I have been unpleasantly surprised when trashing certain potion cards to discover no credit for the $3 spent on the card. It seemed the effect was intermittent so I decide to run a solitaire game with several trash for benefit cards and Familiar and Philosopher's Stone and Scrying Pool as a test.

What I discovered is that Bishop, Forge, Remodel and Apprentice all respect that the potion card cost $X and allows me to convert to any other card on the board relative to the cost of the trashed card in $. Trader and Salvager also treat the potion card as any other normally priced card in $.

Remake, Develop and Farmland however operate with the potion cost of the card and require there to be relevantly priced potion cards on the board in order to function as expected and will not convert to non potion cards.

Why do some trash for benefit cards respect the potion cost while others do not?

[Grujah]

Remake and Develop (and Farmland and Governor) say "exactly $X more"
Exactly 1 more than $3P is $4P. $4 is not exactly 1 more than $3P.
Same with Ironworks, Workshop, Smugglers, Universtity and other gainers. Ironworks says "Up to $4". So you only have $4 to spend, and no Potions to spend.

Remodel (and Expand) say "Up to $2 more". So it can be exactly 2 more or anything cheaper than that.
If you trash $3P. $3P + $2 = $5P. So you can get anything that is $5P or less. That includes $5, $3, $3P, $0P.
If you trash $8, you cannot get a $2P. You don't have P to spend.

Forge and Biship say "in coins" I.E. it completely ignores potions.
Apprentice has a special case for potions.

[DStu]

Apprentice explicitly names the Potion on its card
Bishop and Forge say "its cost in coins".
Remodel (and e.g. Expand) say "costing up $x more than..."
Remake and Develop (and e.g. Upgrade) say "exactly $x more/less"

So the first means "look at the cost, take everything that is 'coin' and ignore the rest".
The second means "look at the cost, including the Potion, and take something that cost as least as much as ... . 0 Potion is less than 1 Potion, so if I say 'up to 5P' you of course also can gain up to 5 without Potion"
The third means "look at the cost, including the Potion, and take a card exactly as much. And when I say exactly, I mean exactly. So if I tell you to gain a card costing exactly 5 and a Potion, don't dare to chose the Lab."

[pingpongsam]

I'm not sure if I should feel smarter or dumber now.

[Kuildeous]

I suppose another way to look at this is on a grid. It might not serve as an effective aid, so feel free to ignore this.

Before Alchemy, cost was on a one-dimensional number line. You could map cost from 0 to 8. An increase or decrease in cost was pretty easy to map out.

Now comes Alchemy. Look, a new cost. We no longer can measure cost on a number line. Now we get to use a Cartesian plane (the x-y coordinates you learned in algebra). On the x-axis, you have the cost in coins. On the y-axis, you have the cost in potions. The costs of most cards will still be a line where the coin cost is 0-8 (and soon 11) while the potion cost is 0-1. No card requires multiple potions, but there could be in theory. You can draw a rectangle to illustrate the cost of potion cards.

Cards like Bishop and Salvager only care about the cost on the x-axis. The y-axis does not matter. Apprentice specifically looks at the cost on the x-axis and then assigns a number based on the cost on the y-axis. If you're familiar with complex numbers, it'd be like saying that you get $2 for each multiple of i. If you're not, then you probably shouldn't be reading my unnecessarily complicated analysis anyway.

Cards like Remodel and Expand give you a wide range. You can Remodel into something that costs up to 2 more. So, take the cost on the x-axis and add 2 to it. It doesn't change the y-axis at all. You now have a slightly larger rectangle. You can gain a card which cost lies in that rectangle. If Potion=1 then the new cost can also include Potion=1. Saboteur does the same thing. Shrink the rectangle down by 2 and look for a cost that falls in that rectangle. If that rectangle is actually a line, then you cannot take something that costs a Potion.

Cards like Upgrade and Remake are more specific. They say exactly 1. The understood rule is that the number refers to the number of coins. So, you go up or down on the x-axis, but the y-axis does not change. If the card has no Potion cost, then you have a line illustrating the cost.  Add 1 to the right of the cost and mark it with a red dot. The new cost must be that dot. If the card has a Potion cost, you do the same thing. The problem is that you don't change the y-axis at all. So, the red dot is one higher on the x-axis, but the Potion cost does not change. If there were a card that says, "Gain a card costing exactly 1 more Potion," then you could go up on the y-axis.

I would love to add actual illustrations to this, but I don't really have to time to fidget with that. Sorry. Not really sure that my explanation is really any better. It's just a different way of approaching it.

[theory]

This is the best rules explanation of this I've ever read.

[WanderingWinder]

I suppose another way to look at this is on a grid. It might not serve as an effective aid, so feel free to ignore this.

Before Alchemy, cost was on a one-dimensional number line. You could map cost from 0 to 8. An increase or decrease in cost was pretty easy to map out.

Now comes Alchemy. Look, a new cost. We no longer can measure cost on a number line. Now we get to use a Cartesian plane (the x-y coordinates you learned in algebra). On the x-axis, you have the cost in coins. On the y-axis, you have the cost in potions. The costs of most cards will still be a line where the coin cost is 0-8 (and soon 11) while the potion cost is 0-1. No card requires multiple potions, but there could be in theory. You can draw a rectangle to illustrate the cost of potion cards.

Cards like Bishop and Salvager only care about the cost on the x-axis. The y-axis does not matter. Apprentice specifically looks at the cost on the x-axis and then assigns a number based on the cost on the y-axis. If you're familiar with complex numbers, it'd be like saying that you get $2 for each multiple of i. If you're not, then you probably shouldn't be reading my unnecessarily complicated analysis anyway.

Cards like Remodel and Expand give you a wide range. You can Remodel into something that costs up to 2 more. So, take the cost on the x-axis and add 2 to it. It doesn't change the y-axis at all. You now have a slightly larger rectangle. You can gain a card which cost lies in that rectangle. If Potion=1 then the new cost can also include Potion=1. Saboteur does the same thing. Shrink the rectangle down by 2 and look for a cost that falls in that rectangle. If that rectangle is actually a line, then you cannot take something that costs a Potion.

Cards like Upgrade and Remake are more specific. They say exactly 1. The understood rule is that the number refers to the number of coins. So, you go up or down on the x-axis, but the y-axis does not change. If the card has no Potion cost, then you have a line illustrating the cost.  Add 1 to the right of the cost and mark it with a red dot. The new cost must be that dot. If the card has a Potion cost, you do the same thing. The problem is that you don't change the y-axis at all. So, the red dot is one higher on the x-axis, but the Potion cost does not change. If there were a card that says, "Gain a card costing exactly 1 more Potion," then you could go up on the y-axis.

I would love to add actual illustrations to this, but I don't really have to time to fidget with that. Sorry. Not really sure that my explanation is really any better. It's just a different way of approaching it.

[qmech]

Surely I can't be the only one who thought prices lived in the free commutative monoid with generators ○ and ◉?

[Thanar]

Rumor has it that card costs in Dominion: Frobenius will form a pseudo-symmetric numerical semigroup.

http://en.wikipedia.org/wiki/Coin_problem

[DStu]

Rumor has it that card costs in Dominion: Frobenius will form a pseudo-symmetric numerical semigroup.

http://en.wikipedia.org/wiki/Coin_problem

I guess this will increase the strength of Copper...

[timchen]

I guess this will increase the strength of Copper...

It won't unless the rules changes to that you have to play all treasures and have precise amount of the card cost in order to buy it!

[Jfrisch]

making the buy phase NP hard might have the side effect of increasing A.P. ...

[DStu]

I guess this will increase the strength of Copper...

It won't unless the rules changes to that you have to play all treasures and have precise amount of the card cost in order to buy it!

isn't that the point of the coin problem?

[Jeebus]

The potential misunderstanding, which the grid thing doesn't adress (I think), is based in grammar.

"A card costing exactly [1 coin] more"
This is the intended reading. It answers the question "How much more can the card cost?"

"A card costing [exactly 1] coin more"
This is the wrong reading. It answers the question "How many more coins can the card cost?"
A cost of $4 is after all one more coin than a cost of $3P.

It's not readily apparent that the second reading is wrong. But if you think about the consequences it might be, because it wouldn't only mean that you could upgrade a $3P into a $4: you could also upgrade a $2 into a $3P!

[Markov Chain]

making the buy phase NP hard might have the side effect of increasing A.P. ...

Forge is already NP-hard.  If you want a colony, you have to determine whether you have a set of cards in hand which sums to 11, which is the classical knapsack problem.

[HiveMindEmulator]

The potential misunderstanding, which the grid thing doesn't adress (I think), is based in grammar.

"A card costing exactly [1 coin] more"
This is the intended reading. It answers the question "How much more can the card cost?"

"A card costing [exactly 1] coin more"
This is the wrong reading. It answers the question "How many more coins can the card cost?"
A cost of $4 is after all one more coin than a cost of $3P.

It's not readily apparent that the second reading is wrong. But if you think about the consequences it might be, because it wouldn't only mean that you could upgrade a $3P into a $4: you could also upgrade a $2 into a $3P!

I don't know how you can accidentally parse in the second way. The number "1" is encircled by the coin logo, so separating "1" from "coin" seems a bit weird.

[Jeebus]

I don't know how you can accidentally parse in the second way. The number "1" is encircled by the coin logo, so separating "1" from "coin" seems a bit weird.

Separating it like that was only a way of illustrating the two different ways of parsing it. The point is you could interpret the text as only talking about the coin cost, since the card only mentions coin cost. And when talking about coin cost, a card costing $4 costs exactly $1 more than a card costing $3P. But I suppose strictly speaking, even without taking any other information or consequences into consideration, you could say that this can't be correct, since the card doesn't explicitly say that it only talks about coin cost. It just says "cost", which strictly speaking includes the whole cost. But I still don't think that this is readily apparent just from the card text.

[verikt]

You left out haggler. You can buy familiar or alchemist and haggle apoth, (or possession and haggle alchemist) and if you buy a vineyard or transmute after playing haggler you're stuck taking a copper or curse.

[eHalcyon]

You left out haggler. You can buy familiar or alchemist and haggle apoth, (or possession and haggle alchemist) and if you buy a vineyard or transmute after playing haggler you're stuck taking a copper or curse.

And you could buy Familiar/Alchemist to haggle Silver.  Or buy Possession to haggle Gold.

[blueblimp]

making the buy phase NP hard might have the side effect of increasing A.P. ...

Forge is already NP-hard.  If you want a colony, you have to determine whether you have a set of cards in hand which sums to 11, which is the classical knapsack problem.

To be pedantic, Forge isn't NP-hard because 11 is a constant, so dynamic programming solves this in O(hand size).

Edit: And to pedantic myself, it only even makes sense to say "isn't NP-hard" if I also say "assuming P != NP" somewhere...

[DStu]

Edit: And to pedantic myself, it only even makes sense to say "isn't NP-hard" if I also say "assuming P != NP" somewhere...

Also, if you talk about the order of the polynomial of the runtime, you probably should specify the computational model...