Brian Attwell
2010-06-21 02:23:32 UTC
I remain uncertain of the conditions for going alone as the computer
player. According to the spec, the Computer Player will not go alone if
none of the following are satisfied.
1) right and left bowers, plus 1 more trump and an off-suit ace
2) right and left bowers, plus an ace and a king or queen in the same
3) right bower, 3 trump and an off-suit ace
4) right bower, at least 3 trump, and double-suited
However, if you run "./our_euchre 111" with all computers
Then "Computer (CMA) Player 1 (10S,JS,QS,KD,JH) > Player 1 is going
alone" in the first round.
Note that trump is Spades, so JS is the right bower and JC is the left
bower. So, I must be misinterpreting the conditions, because it seems
that none of the conditions are fulfilled for going alone, yet Player 1
is going alone (details below):
1) right and left bowers, plus 1 more trump and an off-suit ace
2) right and left bowers, plus an ace and a king or queen in the same
off-suit
Both False: Player 1 only has the right bower.
3) right bower, 3 trump and an off-suit ace
False: No aces, therefore, no off-suit aces.
4) right bower, at least 3 trump, and double-suited
False: three suits in the hand (if the JH was JC, then it would be two
suits, but it's not).
Do you see which of the four conditions I am analyzing incorrectly?
Thanks,
Brian Attwell
P.S. Additional details below:
./our_euchre 111
Player 1: Human (H) or Computer (C)?
C
Player 2: Human (H) or Computer (C)?
C
Player 3: Human (H) or Computer (C)?
C
Player 4: Human (H) or Computer (C)?
C
Players 1 and 3 have 0 points while Players 2 and 4 have 0 points.
Player 1 deals
The top card on the kitty is: 9S
Computer (CMA) Player 2 (9D,QC,KH,10C,QD) > Player 2 passes
Computer (CMA) Player 3 (9H,10D,KC,QH,AC) > Player 3 passes
Computer (CMA) Player 4 (JD,KS,10H,AH,9C) > Player 4 passes
Computer (CMA) Player 1 (10S,JS,QS,KD,JH) > Player 1 is going alone.
Player 1 is picking up 9S and discarding JH.
...
player. According to the spec, the Computer Player will not go alone if
none of the following are satisfied.
1) right and left bowers, plus 1 more trump and an off-suit ace
2) right and left bowers, plus an ace and a king or queen in the same
3) right bower, 3 trump and an off-suit ace
4) right bower, at least 3 trump, and double-suited
However, if you run "./our_euchre 111" with all computers
Then "Computer (CMA) Player 1 (10S,JS,QS,KD,JH) > Player 1 is going
alone" in the first round.
Note that trump is Spades, so JS is the right bower and JC is the left
bower. So, I must be misinterpreting the conditions, because it seems
that none of the conditions are fulfilled for going alone, yet Player 1
is going alone (details below):
1) right and left bowers, plus 1 more trump and an off-suit ace
2) right and left bowers, plus an ace and a king or queen in the same
off-suit
Both False: Player 1 only has the right bower.
3) right bower, 3 trump and an off-suit ace
False: No aces, therefore, no off-suit aces.
4) right bower, at least 3 trump, and double-suited
False: three suits in the hand (if the JH was JC, then it would be two
suits, but it's not).
Do you see which of the four conditions I am analyzing incorrectly?
Thanks,
Brian Attwell
P.S. Additional details below:
./our_euchre 111
Player 1: Human (H) or Computer (C)?
C
Player 2: Human (H) or Computer (C)?
C
Player 3: Human (H) or Computer (C)?
C
Player 4: Human (H) or Computer (C)?
C
Players 1 and 3 have 0 points while Players 2 and 4 have 0 points.
Player 1 deals
The top card on the kitty is: 9S
Computer (CMA) Player 2 (9D,QC,KH,10C,QD) > Player 2 passes
Computer (CMA) Player 3 (9H,10D,KC,QH,AC) > Player 3 passes
Computer (CMA) Player 4 (JD,KS,10H,AH,9C) > Player 4 passes
Computer (CMA) Player 1 (10S,JS,QS,KD,JH) > Player 1 is going alone.
Player 1 is picking up 9S and discarding JH.
...