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Author Topic: Easy Puzzles  (Read 808937 times)

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Awaclus

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Re: Easy Puzzles
« Reply #1975 on: September 22, 2015, 04:16:37 pm »
+3

4-player game.  You play Ambassador to return 2 cards to the supply, onto a supply pile that now has >2 cards in it.  No reactions, on-gain cards, Lighthouse, or anything like that are anywhere.  But only 2 of your 3 opponents gain a copy of the card.  How?

It's a Ruins and the third card in the pile has a different name.
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singletee

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Re: Easy Puzzles
« Reply #1976 on: September 22, 2015, 07:55:55 pm »
0

In a 2-player game, buy 8 different cards on turn 2.

ephesos

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Re: Easy Puzzles
« Reply #1977 on: September 22, 2015, 08:03:53 pm »
0

In a 2-player game, buy 8 different cards on turn 2.
For this, events are not cards, right? But we can still buy them?
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singletee

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Re: Easy Puzzles
« Reply #1978 on: September 22, 2015, 08:05:20 pm »
0

In a 2-player game, buy 8 different cards on turn 2.
For this, events are not cards, right? But we can still buy them?

Yes, that's correct.

ephesos

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Re: Easy Puzzles
« Reply #1979 on: September 22, 2015, 08:19:22 pm »
0

In a 2-player game, buy 8 different cards on turn 2.
For this, events are not cards, right? But we can still buy them?

Yes, that's correct.

P1: Scouting Party+Lost City, using Baker token+Borrow
P2: 6C, Inheritance Squire(Baker)
P1: Play Lost City, overbuy Stonemason for 2 Lost Cities using Borrow
P2: 3xSquire/Estate, 4 Copper, 5 buys with 7 coins, buy Travelling Fairx3, Copper, Curse, 5 differently named Ruins, and Poor House
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singletee

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Re: Easy Puzzles
« Reply #1980 on: September 22, 2015, 08:46:48 pm »
0

In a 2-player game, buy 8 different cards on turn 2.
For this, events are not cards, right? But we can still buy them?

Yes, that's correct.

P1: Scouting Party+Lost City, using Baker token+Borrow
P2: 6C, Inheritance Squire(Baker)
P1: Play Lost City, overbuy Stonemason for 2 Lost Cities using Borrow
P2: 3xSquire/Estate, 4 Copper, 5 buys with 7 coins, buy Travelling Fairx3, Copper, Curse, 5 differently named Ruins, and Poor House


Nice solution, similar to what I was thinking.

But on P1's turn 2, I only count $6 to buy with: CCC(LC), play LC drawing CC for CCCCC in hand, Borrowing again for $6.

My solution:

P1: CCCCC, buy Stonemason, overpaying 4 using Baker token, gaining Throne Room and Baron
P2: CCCCC, buy Borrow, buy Stonemason, overpaying 5 using Baker token, gaining 2 Lost Cities
P1: EEECC(TR)(Baron), Throne Baron, getting +$8 and +2 buys for a total of $10 and 3 buys. Buy Borrow and Travelling Fair x 5, leaving you with $1 and 8 buys. Buy 5 different Ruins, Copper, Curse, and Poor House.

ephesos

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Re: Easy Puzzles
« Reply #1981 on: September 23, 2015, 08:40:16 pm »
0

Oops, forgot I had to pay back my Borrow...


Feel like you can do something with that extra P2 turn though...
P1: Buy Scouting Party(discarding), Ferry(on Grand Market)
P2: Borrow, Baker, Inheritance Worker's Village
P1: Messenger Grand Market
P2: Play 3 Estates, Grand Market. 5 buys, 6 money, buy Borrow, Travelling Fairx3, Poor House, Curse, Copper, and 5 differently named Ruins


There we go, got it to work with Inheritance
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JacquesTheBard

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Re: Easy Puzzles
« Reply #1982 on: October 08, 2015, 07:33:59 pm »
+14

Not sure if this one is "easy" or not, but erring on the side of caution.
There is a combination of five cards I found with incredible collective draw potential. 99% of the time, it is guaranteed to draw your deck. This is the case regardless of what your deck consists of. 50 Curses? The Silver pile? A copy of every card in Dominion? You can draw them all using this specific five-card hand.

The rules for this challenge are as follows:
  • You start your turn with a hand of 5 cards.
  • There are no durations, events, reserve cards, or Prince effects.
  • This is a solitaire game.
  • Your solution may not rely on specific cards in the deck, only in your hand of five. Therefore, Scrying Pool and Counting House are somewhat limited.
  • Your solution must have the potential to draw at least 100 cards. This eliminates, for example, a hand of 5 madmen from competing.


There are two steps to succeeding here. The first is to find the hand that can do this. The second is to find the lowest deck size at which the deck cannot be drawn.

I wish you the best. Let me know if this is too hard, too easy, or somewhere in the middle. Here are some hints, from least helpful to most helpful.

Hints:
1. King's Court is not necessary.

2. A single madman is involved.

3. This test involves three copies of the same card.

Good luck!
« Last Edit: October 08, 2015, 07:37:52 pm by JacquesTheBard »
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Simon (DK)

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Re: Easy Puzzles
« Reply #1983 on: October 08, 2015, 09:15:40 pm »
+7

Solved without looking at any of the hints.

Storyteller, 3* Philosopher's Stone, Madman will always draw your full deck.
Storyteller + 3* Philosopher's Stone will draw 3/5 of your deck. Madman will draw the rest.

Storyteller, Counterfeit, 2* Philosopher's Stone, Madman can also be used.
« Last Edit: October 08, 2015, 09:17:49 pm by Simon (DK) »
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AJD

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Re: Easy Puzzles
« Reply #1984 on: October 08, 2015, 10:32:43 pm »
0

That's great, but Philosopher's Stone rounds down, right?

So if you've got 7–9 cards in your deck, each Philosopher's Stone gives you $1. So Storyteller–3PS draws 3 cards; there's 4–6 remaining and Madman can only draw 3 three of them.


So I guess that answers Jacques's second question?
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eHalcyon

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Re: Easy Puzzles
« Reply #1985 on: October 08, 2015, 10:49:05 pm »
+3

That's great, but Philosopher's Stone rounds down, right?

So if you've got 7–9 cards in your deck, each Philosopher's Stone gives you $1. So Storyteller–3PS draws 3 cards; there's 4–6 remaining and Madman can only draw 3 three of them.


So I guess that answers Jacques's second question?

Storyteller-3PS draws 4 cards because Storyteller itself gives +$1.  Then Madman draws another 4 cards.  That's enough to draw 8, but not 9.

Edit: But you can draw it all even then -- Storyteller-Counterfeit-2PS gets you an extra $1, so it draws 5 cards.  Then Madman will actually overdraw.
« Last Edit: October 08, 2015, 10:50:30 pm by eHalcyon »
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ephesos

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Re: Easy Puzzles
« Reply #1986 on: October 08, 2015, 10:56:28 pm »
+1

That's great, but Philosopher's Stone rounds down, right?

So if you've got 7–9 cards in your deck, each Philosopher's Stone gives you $1. So Storyteller–3PS draws 3 cards; there's 4–6 remaining and Madman can only draw 3 three of them.


So I guess that answers Jacques's second question?
Storyteller gives $1, so you draw 4 cards, not 3. That exactly covers the difference for 7-8, so it would have to be 9 cards.

EDIT: ninja'd...
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ephesos

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Re: Easy Puzzles
« Reply #1987 on: October 08, 2015, 11:07:21 pm »
+1


Suppose you have 5X+Y cards in your deck(including your hand). X>=1, since you have a deck to draw. Y <= 4 is the remainder when you divide by 5.
You play Storyteller, Counterfeit(doubling PS), PS. That gives you 1+1+3(X-1) cards=3X-1 cards. (X-1) since the cards in your hand and in play aren't counted.
Madman then draws you 3X-1 cards for a total of 6X-2 cards. Add in your original hand, and that's 6X+3 cards going into your hand for the turn.
X>=1,Y>=4 imply 6X+3>=5X+4>=5X+Y
So it will always draw your deck, no matter how many cards you have.

In the original, he said he used 3 of the same card. Subtract $1 from the Counterfeit, and we get 6X+2.
The condition for drawing your deck is 6X+2>=5X+Y, or X+2>=Y. This is false when X+2<Y, but since Y<=4, X+2<4 and X<2 is the only case when this is false.
So you can't draw your deck when X=1,Y=4, or when the number of cards in your deck is 9.
« Last Edit: October 08, 2015, 11:19:34 pm by ephesos »
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ephesos

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Re: Easy Puzzles
« Reply #1988 on: October 08, 2015, 11:11:35 pm »
+4


Another modification(ignoring hint 2)
Storyteller+Counterfeit+PS+Madman+Madman

If you start with 5X+Y cards, X>=1, Y<=4,
Storyteller draws you 1+1+2(X-1)=2X+1 cards, for a total of 2X+3 in hand.
Madman doubles 2X+2 cards(1 from the other Madman) to 4X+4.
Second Madman doubles 4X+3 to 8X+6. Sufficient to overdraw any deck by about half.

And, in keeping with hint 3 having 3 of the same card(but not with hint 2):
Storyteller+PS+Madman+Madman+Madman
Storyteller draws you 1+(X-1)=X cards, putting you to X+3
Madman draws you X+2 cards, putting you to 2X+4.
Madman draws you 2X+3 cards, putting you to 4X+6
Madman draws you 4X+5 cards, putting you to 8X+10 cards. You get to overdraw by 4 more cards!
« Last Edit: October 08, 2015, 11:19:51 pm by ephesos »
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Burning Skull

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Re: Easy Puzzles
« Reply #1989 on: October 09, 2015, 05:27:16 am »
+2

...

Thank you, that's the best one in a while indeed.

Asper

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Re: Easy Puzzles
« Reply #1990 on: October 09, 2015, 10:56:36 am »
0

So, using Storyteller, PS, PS, PS, Madman you can't draw your deck if it contains exactly 3  (or 4 or 9) cards, as the combination draws 2+(cardcount mod 5)*6 cards. From 10 on, this is always more than cardcount, and for 1 and 2 the Madman-doubled draw from Storyteller (2) is enough on its own.

If you have Counterfeit, there is no amount of cards you can't draw, as ST, CF, PS, PS, MM generates 4+ (cardcount mod 5)*6 cards, which is always at least cardcount. Basically, the +4 (from Storyteller and CF doubled) balances the mod 5.


A really nice puzzle. Thanks for posting it :)
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Simon (DK)

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Re: Easy Puzzles
« Reply #1991 on: October 09, 2015, 11:36:23 am »
+3

So, using Storyteller, PS, PS, PS, Madman you can't draw your deck if it contains exactly 3  (or 4 or 9) cards

For 3 or 4 cards you just play Madman first. The only problem is with 9 cards.

I thought about the solutions with more Madmen right after I posted my solution, but I had already shut down my computer and went to bed, so I didn't want to start it up again.
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Re: Easy Puzzles
« Reply #1992 on: October 09, 2015, 06:54:00 pm »
+1

1 player game. No Shelters.  You only ever buy Silvers on $3-$5, Golds on $6-$7, Province on > $7, and nothing on < $3.  What is the longest possible such game (number of turns)?  Shortest possible game?
« Last Edit: October 09, 2015, 06:58:13 pm by Dingan »
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Elanchana

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Re: Easy Puzzles
« Reply #1993 on: October 09, 2015, 07:40:43 pm »
+3

1 player game. No Shelters.  You only ever buy Silvers on $3-$5, Golds on $6-$7, Province on > $7, and nothing on < $3.  What is the longest possible such game (number of turns)?  Shortest possible game?

For shortest I did a theoretical test run and got 18 turns.

Here's how I got there - t is turn, d is deck/draw pile, and shuffles (shuf) only include stuff that isn't in hand
shuf-7C3E
t1-3C2E-buyS (d4C1E)
t2-4C1E-buyS (empty)
shuf-2S7C3E
t3-2S2C1E-buyG (d5C2E)
t4-3C2E-buyS (d2C)
t5-2C
shuf-1G3S5C3E
t5-1S1G1C-buyP (d2S4C3E)
t6-2S2C1E-buyG (d2C2E)
t7-2C2E
shuf-2G3S5C1P1E
t7-1C-buyS (d2G3S4C1P1E)
t8-2G1S1P1E-buyP (d2S4C)
t9-1S4C-buyG (d1C)
t10-1C
shuf-3G4S6C2P2E
t10-1G1S2C-buyP (d2G3S4C2P2E)
t11-2G1S2P-buyP (d2S4C2E)
t12-1S4C-buyG (d1S2E)
t13-1S2E
shuf-4G3S7C4P1E
t13-2G-buyP (d2G3S7C4P1E)
t14-2G1S2P-buyP (d2S7C2P1E)
t15-1S4C-buyG (d1S3C2P1E)
t16-3C2P-buyS (d1S1E)
t17-1S1E
shuf-5G4S7C6P2E
t17-2G1E-buyP (d3G4S7C6P1E)
t18-2G1S2P-buyP - GAME OVER


Edit: I tried for the long game and I got up to 46 turns. I don't know if that actually is the longest possible game under those conditions but this would certainly be painful. (I went for the "strategy" of not being able to buy Gold until the deck had a lot of green and could hit $7 consistently.)
shuf-7C3E
t1-2C3E-buyX (d5C)
t2-5C-buyS (empty)
shuf-1S7C3E
t3-2C3E-buyX (d1S5C)
t4-5C-buyS (d1S)
t5-1S
shuf-1S7C3E
t5-3C1E-buyS (d1S4C2E)
t6-1S3C1E-buyS (d1C1E)
t7-1C1E
shuf-4S6C2E
t7-1S2C-buyS (d3S4C2E)
t8-1S3C1E-buyS (d2S1C1E)
t9-2S1C1E-buyS (empty)
shuf-7S7C3E
t10-1S3C1E-buyS (d6S4C2E)
t11-1S3C1E-buyS (d5S1C2E)
t12-5S-buyP (d1C2E)
t13-1C2E
shuf-9S6C1P1E
t13-2S-buyS (d7S6C1P1E)
t14-5S-buyP (d4S6C1P1E)
t15-1S3C1P-buyS (d3S3C1E)
t16-1S3C1E-buyS (d2S)
t17-2S
shuf-10S7C2P3E
t17-3S-buyP (d7S7C2P3E)
t18-5S-buyP (d2S7C2P3E)
t19-1S3C1P-buyS (d1S4C1P3E)
t20-1S3C1P-buyS (d1C3E)
t21-1C3E
shuf-14S6C4P
t21-1C-buyX (d14S5C4P)
t22-2S1C2P-buyS (d12S4C2P)
t23-2S1C2P-buyS (d10S3C)
t24-5S-buyP (d5S3C)
t25-5S-buyP (d3C)
t26-3C
shuf-16S4C6P3E
t26-2S-buyG (d14S4C6P3E)
t27-3S1C1E-buyG (d11S3C6P2E)
t28-3S1C1E-buyG (d8S2C6P1E)
t29-3S1C1E-buyG (d5S1C6P)
t30-3S1C1P-buyG (d2S5P)
t31-1S4P-buyX (d1S1P)
t32-1S1P
shuf-5G15S7C5P3E
t32-1G1S1E-buyG (d4G14S7C5P2E)
t33-1G2S2E-buyG (d3G12S7C5P)
t34-1G2S2P-buyG (d2G10S7C3P)
t35-1G2S2P-buyG (d8S7C1P)
t36-3S1C1P-buyG (d5S6C)
t37-2S3C-buyG (d3S3C)
t38-2S3C-buyG (d1S)
t39-1S
shuf-12G15S7C6P3E
t39-1G1S2E-buyG (d11G14S7C6P1E)
t40-1G2S1P1E-buyG (d10G12S7C5P)
t41-1G2S2P-buyG (d9G10S7C3P)
t42-1G2S2P-buyG (d8G8S7C1P)
t43-1G1S2C1P-buyG (d7G7S5C)
t44-1G4C-buyG (d6G7S1C)
t45-5G-buyP (d1G7S1C)
t46-5S-buyP - GAME OVER
« Last Edit: October 09, 2015, 08:20:50 pm by Elanchana »
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JacquesTheBard

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Re: Easy Puzzles
« Reply #1994 on: October 09, 2015, 10:46:47 pm »
+1

Welp... hadn't thought of 2 Madmen, or Counterfeit. I have seen the masters at work. Good job to all of you.
So yeah... easy puzzles indeed.
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Asper

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Re: Easy Puzzles
« Reply #1995 on: October 20, 2015, 07:00:46 pm »
0

I own only differently named kingdom cards. How many differently named cards can i have in play at the start of my turn if i played neither on my last turn?
Hint: It's quite a lot.
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Simon (DK)

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Re: Easy Puzzles
« Reply #1996 on: October 21, 2015, 07:08:26 am »
+3

I can get 30

With Black Market you can get as many different Kingdom Cards as you want (limitted of course by the number of different cards in the game, but we wont get anywhere near that number).
In the beginning you have Caravan Guard (played on your opponent's turn), Hireling and Champion in play.
Then you call Ratcactcher, Guide, Transmogrify and Teacher.
Then you play your Princed King's Court and play Throne Room, Procession, Golem, playing 4 different action cards (including a gainer) and calling Coin of the Realm and Duplicate. With the 2nd Procession play you play Herald, playing 2 different cards. With the 2nd Throne Room play you play Graverobber gaining Golem and Herald back from the trash, drawing Herald due to +1 card token on Graverobber. With the 3rd Throne Room play you play Herald, drawing Golem and playing 2 action cards.
Then you call Royal Carriage to replay King's Court and play Golem, playing 6 action cards.
It's still the start of your turn, and you count the cards in play: Caravan Guard, Hireling, Champion, Ratcatcher, Guide, Transmogrify, Teacher, King's Court, Throne Room, Procession, Coin of the Realm, Duplicate, Graverobber, Herald, Royal Carriage, Golem and 14 other action cards gives a total of 30.
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Asper

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Re: Easy Puzzles
« Reply #1997 on: October 21, 2015, 08:31:03 am »
0

I can get 30

With Black Market you can get as many different Kingdom Cards as you want (limitted of course by the number of different cards in the game, but we wont get anywhere near that number).
In the beginning you have Caravan Guard (played on your opponent's turn), Hireling and Champion in play.
Then you call Ratcactcher, Guide, Transmogrify and Teacher.
Then you play your Princed King's Court and play Throne Room, Procession, Golem, playing 4 different action cards (including a gainer) and calling Coin of the Realm and Duplicate. With the 2nd Procession play you play Herald, playing 2 different cards. With the 2nd Throne Room play you play Graverobber gaining Golem and Herald back from the trash, drawing Herald due to +1 card token on Graverobber. With the 3rd Throne Room play you play Herald, drawing Golem and playing 2 action cards.
Then you call Royal Carriage to replay King's Court and play Golem, playing 6 action cards.
It's still the start of your turn, and you count the cards in play: Caravan Guard, Hireling, Champion, Ratcatcher, Guide, Transmogrify, Teacher, King's Court, Throne Room, Procession, Coin of the Realm, Duplicate, Graverobber, Herald, Royal Carriage, Golem and 14 other action cards gives a total of 30.


Very good. But you can get a lot more, still. ;)
Edit 2: Allthough you can't get Golem back from the trash using Graverobber. But nice idea either way.

Edit: It's not only about the optimal combination of using Throne Room variants, by the way. I know those will make a huge difference, but i suck at that and won't claim i know how to get the most out of that. Maybe i should have excluded TR variants to keep this easier... Either way, there still is a thing or two you can do besides that.
« Last Edit: October 21, 2015, 08:40:09 am by Asper »
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AdamH

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Re: Easy Puzzles
« Reply #1998 on: October 21, 2015, 09:25:18 am »
+2

Just playing a single Black Market allows you to get a ton of treasures in play
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Asper

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Re: Easy Puzzles
« Reply #1999 on: October 21, 2015, 03:08:50 pm »
0

Just playing a single Black Market allows you to get a ton of treasures in play

Great, that's what i was thinking about. There is one more thing though, which increases the count by up to 25 on its own.
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