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Can you use workshop to gain royal blacksmith? Doesn't royal blacksmith have a cost of 0 coins and therefore is "up to 4 coins"?

David desJardins
United States Burlingame California

No, Royal Blacksmith has a cost of 0 coins plus 4 debt, which is not less than or equal to 4 coins. This is explained thoroughly in the rulebook.


DaviddesJ wrote: No, Royal Blacksmith has a cost of 0 coins plus 4 debt, which is not less than or equal to 4 coins. This is explained thoroughly in the rulebook.
Thanks for the quick response. So the key to the explanation I believe (correct me if I'm wrong) after reading it for the nth time is that since royal blacksmith cost is 0 coins and 8 debt, it is larger than what workshop allows you to get ( up to 4 coins with 0 debt) because 8 debt is greater than 0 debt?
So a workshop could never get you anything with positive debt?

Chris Schumann
United States Saint Paul Minnesota

MoviinTarget wrote: So the key to the explanation I believe (correct me if I'm wrong) after reading it for the nth time is that since royal blacksmith cost is 0 coins and 8 debt, it is larger than what workshop allows you to get ( up to 4 coins with 0 debt) because 8 debt is greater than 0 debt?
I think you have it just fine... maybe this will help even more?
Cost in Coins and Cost in Debt are orthogonal. That is, a cost of 3 coins is neither greater than nor less than a cost of 4 debt.
The very first example in the rules: Quote: 4 Debt is not "up to 4 Coins". 4 debt is not more than 4 coins and 4 coins is not more than 4 debt; both have something the other lacks.
If you want to imagine a numeric plane, where the coin cost is the x axis, and the debt cost is the y axis, workshop lets you gain a card with a cost up to 4 coins. This would mean anywhere on the x axis up to 4, but nothing with any debt cost. Nothing (yet) ever costs less than 0, so we can stay in just the first quadrant.
(You could also extrapolate a z axis being a Potion cost, since it is also orthogonal to the other two.)

David desJardins
United States Burlingame California

MoviinTarget wrote: Thanks for the quick response. So the key to the explanation I believe (correct me if I'm wrong) after reading it for the nth time is that since royal blacksmith cost is 0 coins and 8 debt, it is larger than what workshop allows you to get ( up to 4 coins with 0 debt) because 8 debt is greater than 0 debt?
The cost of the Royal Blacksmith is neither greater than 4 coins nor less than 4 coins. In mathematics, this relationship is called "incomparable". In a partially ordered set, it is not always the case that either X > Y, X = Y, or X < Y. There is a fourth possibility which is that X is incomparable with Y.

Jonathan Kift
Canada Vancouver BC

Whizkid wrote: MoviinTarget wrote: So the key to the explanation I believe (correct me if I'm wrong) after reading it for the nth time is that since royal blacksmith cost is 0 coins and 8 debt, it is larger than what workshop allows you to get ( up to 4 coins with 0 debt) because 8 debt is greater than 0 debt? I think you have it just fine... maybe this will help even more? Cost in Coins and Cost in Debt are orthogonal. That is, a cost of 3 coins is neither greater than nor less than a cost of 4 debt. The very first example in the rules: Quote: 4 Debt is not "up to 4 Coins". 4 debt is not more than 4 coins and 4 coins is not more than 4 debt; both have something the other lacks. If you want to imagine a numeric plane, where the coin cost is the x axis, and the debt cost is the y axis, workshop lets you gain a card with a cost up to 4 coins. This would mean anywhere on the x axis up to 4, but nothing with any debt cost. Nothing (yet) ever costs less than 0, so we can stay in just the first quadrant. (You could also extrapolate a z axis being a Potion cost, since it is also orthogonal to the other two.)
Now someone needs to find a way to explain complex (or hypercomplex) algebra using Dominion. We might end up with a whole new batch of abstract mathematicians!

Clive Jones
Cambridgeshire, UK

As I've noted elsewhere, it's not valid to compare debt costs to imaginary numbers.
We can't tell the two square roots of 1 apart, so we label one of them +i and the other i arbitraily. If we switched the labels round, all of mathematics would continue to work exactly as before.
This means you can't say that 2i is greater than i. But you can say that a debt of 2 is a greater cost than a debt of 1.

David desJardins
United States Burlingame California

clivej wrote: This means you can't say that 2i is greater than i. But you can say that a debt of 2 is a greater cost than a debt of 1.
Sure you can. There are lots of partial orders you can impose on the Gaussian integers. One of them has a+bi >= c+di iff a >= c and b >= d. Another partial order has a+bi >= c+di iff a >= c and b <= d. There are lots of others, too.

Clive Jones
Cambridgeshire, UK

Well yes, but those are orders being imposed on the Gaussian integers. There isn't an inherent natural ordering for them in the way that there is for the integers.


As a friend of mine once said when we were discussing such things back in the lates 60's .... "Deep"


