I am trying to implement in python Donald Knuth's algorithm for codebreaking mastermind in not more than 5 moves. I have checked my code several times, and it seems to follow the algorithm, as its stated here: http://en.wikipedia.org/wiki/Mastermind_(board_game)#Five-guess_algorithm
However, I get that some of the secrets take 7 or even 8 moves to accomplish. Here is the code:
#returns how many bulls and cows there are
def HowManyBc(guess,secret):
invalid=max(guess)+1
bulls=0
cows=0
r=0
while r<4:
if guess[r]==secret[r]:
bulls=bulls+1
secret[r]=invalid
guess[r]=invalid
r=r+1
r=0
while r<4:
p=0
while p<4:
if guess[r]==secret[p] and guess[r]!=invalid:
cows=cows+1
secret[p]=invalid
break
p=p+1
r=r+1
return [bulls,cows]
# sends every BC to its index in HMList
def Adjustment(BC1):
if BC1==[0,0]:
return 0
elif BC1==[0,1]:
return 1
elif BC1==[0,2]:
return 2
elif BC1==[0,3]:
return 3
elif BC1==[0,4]:
return 4
elif BC1==[1,0]:
return 5
elif BC1==[1,1]:
return 6
elif BC1==[1,2]:
return 7
elif BC1==[1,3]:
return 8
elif BC1==[2,0]:
return 9
elif BC1==[2,1]:
return 10
elif BC1==[2,2]:
return 11
elif BC1==[3,0]:
return 12
elif BC1==[4,0]:
return 13
# sends every index in HMList to its BC
def AdjustmentInverse(place):
if place==0:
return [0,0]
elif place==1:
return [0,1]
elif place==2:
return [0,2]
elif place==3:
return [0,3]
elif place==4:
return [0,4]
elif place==5:
return [1,0]
elif place==6:
return [1,1]
elif place==7:
return [1,2]
elif place==8:
return [1,3]
elif place==9:
return [2,0]
elif place==10:
return [2,1]
elif place==11:
return [2,2]
elif place==12:
return [3,0]
elif place==13:
return [4,0]
# gives minimum of positive list without including its zeros
def MinimumNozeros(List1):
minimum=max(List1)+1
for item in List1:
if item!=0 and item<minimum:
minimum=item
return minimum
#list with all possible guesses
allList=[]
for i0 in range(0,6):
for i1 in range(0,6):
for i2 in range(0,6):
for i3 in range(0,6):
allList.append([i0,i1,i2,i3])
TempList=[[0,0,5,4]]
for secret in TempList:
guess=[0,0,1,1]
BC=HowManyBc(guess[:],secret[:])
counter=1
optionList=[]
for i0 in range(0,6):
for i1 in range(0,6):
for i2 in range(0,6):
for i3 in range(0,6):
optionList.append([i0,i1,i2,i3])
while BC!=[4,0]:
dummyList=[] #list with possible secrets for this guess
for i0 in range(0,6):
for i1 in range(0,6):
for i2 in range(0,6):
for i3 in range(0,6):
opSecret=[i0,i1,i2,i3]
if HowManyBc(guess[:],opSecret[:])==BC:
dummyList.append(opSecret)
List1=[item for item in optionList if item in dummyList]
optionList=List1[:] # intersection of optionList and dummyList
item1Max=0
for item1 in allList:
ListBC=[] # [list of all [bulls,cows] in optionList
for item2 in optionList:
ListBC.append(HowManyBc(item1[:],item2[:]))
HMList=[0]*14 # counts how many B/C there are for item2 in optionList
for BC1 in ListBC:
index=Adjustment(BC1)
HMList[index]=HMList[index]+1
m=max(HMList)#maximum [B,C], more left - less eliminated (min in minimax)
maxList=[i for i, j in enumerate(HMList) if j == m]
maxElement=maxList[0] #index with max
possibleGuess=[]
if m>item1Max: #max of the guesses, the max in minimax
item1Max=m
possibleGuess=[i[:] for i in optionList if AdjustmentInverse(maxElement)==HowManyBc(item1[:],i[:])]
nextGuess=possibleGuess[0][:]
guess=nextGuess[:]
BC=HowManyBc(guess[:],secret[:])
counter=counter+1
I get:
for [5, 3, 3, 4] counter is 7
for [5, 4, 4, 5] counter is 8
If someone could help I would appreciate it very much!
Thanks,mike
For Master Mind games (4 columns, 6 colors, 64 = 1296 codes, optimal logical strategy applied), the optimal code to play at first attempt is 1233 (or any other equivalent code with "1 double + 2 different colors") with an average number of attempts to find secret codes of 5660⁄64 ≈ 4.367 and a maximal number of ...
Mastermind is a code-breaking game for two players. This peg game was in- vented in 1970 by Mordecai Meirowitz. One of the players – the code-maker – chooses a secret code of four pegs of six possible repeatable colors, thus one code among 1296 possibilities.
A colored or black key peg is placed for each code peg from the guess which is correct in both color and position. A white key peg indicates the existence of a correct color code peg placed in the wrong position.
There are four mistakes.
The comment is wrong on this line:
m=max(HMList)#maximum [B,C], more left - less eliminated (min in minimax)
This is actually the "max" in the minimax (which should have been clear from the call to max
). You are trying to find the guess that minimizes the maximum size of the groups of possible secrets that yield the same evaluation. Here we are finding the maximum size of the groups, so that's the "max".
That mistake caused you to make this one:
if m>item1Max: #max of the guesses, the max in minimax
Here you need to take the min, not the max.
On the following lines you pick the first item among optionList
that would generate the same evaluation as item1
:
possibleGuess=[i[:] for i in optionList if AdjustmentInverse(maxElement)==HowManyBc(item1[:],i[:])]
nextGuess=possibleGuess[0][:]
But that's not right: the guess we want here is item1
, not some other guess that would generate the same evaluation!
Finally, you don't properly handle the case where optionList
has only one remaining item. In this case all possible guesses are equally good at distinguishing among this item, so the minimax procedure doesn't differentiate between the guesses. In this case you should just guess optionList[0]
.
The variable names are poorly chosen. For example, what is item1
? This is the guess that you are evaluating, so surely it should be called something like possible_guess
? I suspect that your mistake §1.3 above was partly caused by this poor choice of variable name.
There's vast amounts of needless copying. All of your [:]
are pointless and can be removed. The variable List1
is also pointless (why not just assign to optionList
?), as is nextGuess
(which not just assign to guess
?)
You build dummyList
consisting of all possible secrets that would match the last guess, but then you throw away all the entries in dummyList
that aren't also in optionList
. So why not just loop over optionList
and keep the entries that match? Like this:
optionList = [item for item in optionList if HowManyBc(guess, item)==BC]
You build up a table HMList
which counts the number of occurrences of each pattern of bulls and cows. You have noted the fact that there are 14 possible (bull, cow) pairs and so you've written the functions Adjustment
and AdjustmentInverse
to map back and forth between (bull, cow) pairs and their indices in the list.
These functions could have much simpler implementations if you took a data-driven approach and used the built-in list.index
method:
# Note that (3, 1) is not possible.
EVALUATIONS = [(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (1, 0), (1, 1),
(1, 2), (1, 3), (2, 0), (2, 1), (2, 2), (3, 0), (4, 0)]
def Adjustment(evaluation):
return EVALUATIONS.index(evaluation)
def AdjustmentInverse(index):
return EVALUATIONS[index]
But after fixing mistake §1.3 above, you don't need AdjustmentInverse
any more. And Adjustment
could be avoided if you kept the counts in a collections.Counter
instead of a list. So instead of:
HMList=[0]*14 # counts how many B/C there are for item2 in optionList
for BC1 in ListBC:
index=Adjustment(BC1)
HMList[index]=HMList[index]+1
m=max(HMList)
you could simply write:
m = max(Counter(ListBC).values())
Evaluating a guess (your function HowManyBc
) can be reduced to just three lines using the class collections.Counter
from the standard library.
from collections import Counter
def evaluate(guess, secret):
"""Return the pair (bulls, cows) where `bulls` is a count of the
characters in `guess` that appear at the same position in `secret`
and `cows` is a count of the characters in `guess` that appear at
a different position in `secret`.
>>> evaluate('ABCD', 'ACAD')
(2, 1)
>>> evaluate('ABAB', 'AABB')
(2, 2)
>>> evaluate('ABCD', 'DCBA')
(0, 4)
"""
matches = sum((Counter(secret) & Counter(guess)).values())
bulls = sum(c == g for c, g in zip(secret, guess))
return bulls, matches - bulls
I happen to prefer using letters for the codes in Mastermind. ACDB
is so much nicer to read and type than [0, 2, 3, 1]
. But my evaluate
function is flexible as to how you represent the codes and guesses, as long as you represent them as sequences of comparable items, so you can use lists of numbers if you prefer.
Notice also that I've written some doctests: these are a quick way to simultaneously provide examples in the documentation and to test the function.
The function itertools.product
provides a convenient way to build the list of codes without having to write four nested loops:
from itertools import product
ALL_CODES = [''.join(c) for c in product('ABCDEF', repeat=4)]
Knuth's five-guess algorithm uses the minimax principle. So why not implement it by taking the min
of a sequence of calls to max
?
def knuth(secret):
"""Run Knuth's 5-guess algorithm on the given secret."""
assert(secret in ALL_CODES)
codes = ALL_CODES
key = lambda g: max(Counter(evaluate(g, c) for c in codes).values())
guess = 'AABB'
while True:
feedback = evaluate(guess, secret)
print("Guess {}: feedback {}".format(guess, feedback))
if guess == secret:
break
codes = [c for c in codes if evaluate(guess, c) == feedback]
if len(codes) == 1:
guess = codes[0]
else:
guess = min(ALL_CODES, key=key)
Here's an example run:
>>> knuth('FEDA')
Guess AABB: feedback (0, 1)
Guess BCDD: feedback (1, 0)
Guess AEAC: feedback (1, 1)
Guess AFCC: feedback (0, 2)
Guess FEDA: feedback (4, 0)
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With