Iterated Prisoner's Trilemma

19

5

CHALLENGE STATUS: OPEN

Comment, open a PR, or otherwise yell at me if I'm missing your bot.


Prisoner's dilemma ... with three choices. Crazy, huh?

Here's our payoff matrix. Player A on the left, B on the top

A,B| C | N | D
---|---|---|---
 C |3,3|4,1|0,5
 N |1,4|2,2|3,2
 D |5,0|2,3|1,1

The payoff matrix is engineered so that it's best for both players to always cooperate, but you can gain (usually) by choosing Neutral or Defection.

Here's some (competing) example bots.

# turns out if you don't actually have to implement __init__(). TIL!

class AllC:
    def round(self, _): return "C"
class AllN:
    def round(self, _): return "N"
class AllD:
    def round(self, _): return "D"
class RandomBot:
    def round(self, _): return random.choice(["C", "N", "D"])

# Actually using an identically-behaving "FastGrudger".
class Grudger:
    def __init__(self):
        self.history = []
    def round(self, last):
        if(last):
            self.history.append(last)
            if(self.history.count("D") > 0):
                return "D"
        return "C"

class TitForTat:
    def round(self, last):
        if(last == "D"):
            return "D"
        return "C"

Your bot is a Python3 class. A new instance is created for every game, and round() is called each round, with your opponent's choice from last round (or None, if it's the first round)

There's a 50 rep bounty for the winner in like a month.

Specifics

  • Every bot plays every other bot (1v1), including itself, in [REDACTED] rounds.
  • Standard loopholes disallowed.
  • No messing with anything outside your class or other underhanded shenanigains.
  • You may submit up to five bots.
  • Yes, you can implement handshake.
  • Any response other than C, N, or D will be silently taken as N.
  • Each bot's points from every game they play will be totaled up and compared.

Controller

Check!

Other Languages

I'll throw together an API if anyone needs it.

Scores: 2018-11-27

27 bots, 729 games.

name            | avg. score/round
----------------|-------------------
PatternFinder   | 3.152
DirichletDice2  | 3.019
EvaluaterBot    | 2.971
Ensemble        | 2.800
DirichletDice   | 2.763
Shifting        | 2.737
FastGrudger     | 2.632
Nash2           | 2.574
HistoricAverage | 2.552
LastOptimalBot  | 2.532
Number6         | 2.531
HandshakeBot    | 2.458
OldTitForTat    | 2.411
WeightedAverage | 2.403
TitForTat       | 2.328
AllD            | 2.272
Tetragram       | 2.256
Nash            | 2.193
Jade            | 2.186
Useless         | 2.140
RandomBot       | 2.018
CopyCat         | 1.902
TatForTit       | 1.891
NeverCOOP       | 1.710
AllC            | 1.565
AllN            | 1.446
Kevin           | 1.322

SIGSTACKFAULT

Posted 2018-11-12T17:43:03.287

Reputation: 585

1How are bots put against each other? I get from the Grudger that there are always two bots against/with each other and the enemy's last choice is passed to the bot. How many rounds are played? And for a game: Does only the result count (i.e. who won) or also the points? – Black Owl Kai – 2018-11-12T17:52:33.483

1You would get more entries if you made this language-agnostic, or at least broader. You could have a wrapper python class that spawns a process and sends it text commands to get back text responses. – Sparr – 2018-11-12T17:54:26.463

1Done. This was on the sandbox for like a month! – SIGSTACKFAULT – 2018-11-12T17:56:25.770

proposal: normalize scores so that adding more bots or more rounds doesn't inflate the numbers – Sparr – 2018-11-12T23:18:38.603

Depending on how popular this gets, you might want to throw in a "Must run under x ms" to fend off some of the nuttier sim bots. Probably not needed considering this isn't grid-scanning, though. – Veskah – 2018-11-13T00:21:29.270

@Veskah i had to limit the O(n^2) piece of PatternFinder to n=100 to avoid exploding runtime for longer matches – Sparr – 2018-11-13T00:40:30.177

@Black Owl Kai: Knowing how many rounds breaks the game. You just don't know it yet. – Joshua – 2018-11-13T01:59:44.793

We can just do trial and error at home to match the leaderboard and figure out the number of rounds being run. Normalized scores would help, but still not solve the problem. – Sparr – 2018-11-13T07:22:53.517

The game is symmetric, why run every match twice with swapped bots? Seems like a waste of time. – Sparr – 2018-11-13T07:32:40.597

2If you wrap most of main.py in while len(botlist) > 1: with botlist.remove(lowest_scoring_bot) at the bottom of the loop, you get an elimination tournament with interesting results. – Sparr – 2018-11-13T07:39:57.947

1Another version of this someday might pass the entire interaction history rather than just the last move. It doesn't change much although it simplifies user code slightly. But it would allow for extensions, such as noisy communication channels that clarify over time: "Really, a D, even though I've said C four times in a row? No, I didn't say D; what do you take me for? Oh, sorry, can we just forget that round?" – Scott Sauyet – 2018-11-13T13:31:51.150

@ScottSauyet Have fun; I was going for simplicity in the API. – SIGSTACKFAULT – 2018-11-13T13:32:56.860

@Sparr too lazy. – SIGSTACKFAULT – 2018-11-13T13:34:47.157

Should we give a second submission in an edit or new post? Would you find the new one in an edit? – Bridgeburners – 2018-11-15T18:41:38.340

@Bridgeburners if it's fundamentally different and you want both versions to compete separately, you need to submit a second answer. – Sparr – 2018-11-15T20:33:31.997

Answers

10

EvaluaterBot

class EvaluaterBot:
    def __init__(self):
        self.c2i = {"C":0, "N":1, "D":2}
        self.i2c = {0:"C", 1:"N", 2:"D"}
        self.history = [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
        self.last = [None, None]

    def round(self, last):
        if self.last[0] == None:
            ret = 2
        else:
            # Input the latest enemy action (the reaction to my action 2 rounds ago)
            # into the history
            self.history[self.last[0]][self.c2i[last]] += 1
            # The enemy will react to the last action I did
            prediction,_ = max(enumerate(self.history[self.last[1]]), key=lambda l:l[1])
            ret = (prediction - 1) % 3
        self.last = [self.last[1], ret]
        return self.i2c[ret]

Wins against all previously submitted bots except (maybe) the random bot (but it could have an advantage, because it picks D in a draw and D should optimal) and plays a constant draw against themself.

Black Owl Kai

Posted 2018-11-12T17:43:03.287

Reputation: 980

Yep, beats everything. – SIGSTACKFAULT – 2018-11-12T19:43:44.160

Scratch that, PatternFinder beats it by a bit. – SIGSTACKFAULT – 2018-11-13T14:25:48.647

7

TatForTit

class TatForTit:
    def round(self, last):
        if(last == "C"):
            return "N"
        return "D"

This bot will alternate picking D N D N while TitForTat alternates C D C D, for an average net gain of 3 points per round if I have read the payout matrix correctly. I think this might be optimal against TitForTat. Obviously it could be improved to detect a non-TFT opponent and adopt other strategies, but I was just aiming for the original bounty.

Sparr

Posted 2018-11-12T17:43:03.287

Reputation: 5 758

7

NashEquilibrium

This bot has taken a game theory class in college but was lazy and didn't go to the class where they covered iterated games. So he only plays single game mixed nash equilibrium. Turns out 1/5 2/5 2/5 is the mixed NE for the payoffs.

class NashEquilibrium:
    def round(self, _):
        a = random.random()
        if a <= 0.2:
            return "C"
        elif a <= 0.6:
            return "N"
        else:
            return "D" 

Constant Abusing Nash Equilibrium

This bot picked up a lesson or two from his lazy brother. His lazy brother's problem was that he didn't take advantage of fixed strategies. This version checks if the opponent is a constant player or titfortat and plays accordingly, else it plays the regular nash equilibrium.

It's only downside is that it averages 2.2 points per turn playing against itself.

class NashEquilibrium2:

    def __init__(self):
        self.opphistory = [None, None, None]
        self.titfortatcounter = 0
        self.titfortatflag = 0
        self.mylast = "C"
        self.constantflag = 0
        self.myret = "C"

    def round(self, last):
        self.opphistory.pop(0)
        self.opphistory.append(last)

        # check if its a constant bot, if so exploit
        if self.opphistory.count(self.opphistory[0]) == 3:
            self.constantflag = 1
            if last == "C":
                 self.myret = "D"
            elif last == "N":
                 self.myret = "C"
            elif last == "D":
                 self.myret = "N"

        # check if its a titfortat bot, if so exploit
        # give it 2 chances to see if its titfortat as it might happen randomly
        if self.mylast == "D" and last == "D":
            self.titfortatcounter = self.titfortatcounter + 1

        if self.mylast == "D" and last!= "D":
            self.titfortatcounter = 0

        if self.titfortatcounter >= 3:
            self.titfortatflag = 1

        if self.titfortatflag == 1:
            if last == "C":
                 self.myret = "D"
            elif last == "D":
                 self.myret = "N"    
            elif last == "N":
                # tit for tat doesn't return N, we made a mistake somewhere
                 self.titfortatflag = 0
                 self.titfortatcounter = 0

        # else play the single game nash equilibrium
        if self.constantflag == 0 and self.titfortatflag == 0:
            a = random.random()
            if a <= 0.2:
                self.myret = "C"
            elif a <= 0.6:
                self.myret = "N"
            else:
                self.myret = "D"


        self.mylast = self.myret
        return self.myret

Ofya

Posted 2018-11-12T17:43:03.287

Reputation: 181

1NashEquilibrium.round needs to take arguments even if it doesn't use them, so as to fit the expected function prototype. – Ray – 2018-11-13T03:18:53.333

Thank you fixed it – Ofya – 2018-11-13T03:45:10.123

Little shorter: class NashEquilibrium: def round(self, _): a = random.random() for k, v in [(0.2, "C"), (0.6, "N"), (1, "D")]: if a <= k: return v – Robert Grant – 2018-11-14T01:01:08.377

6

PatternFinder

class PatternFinder:
    def __init__(self):
        import collections
        self.size = 10
        self.moves = [None]
        self.other = []
        self.patterns = collections.defaultdict(list)
        self.counter_moves = {"C":"D", "N":"C", "D":"N"}
        self.initial_move = "D"
        self.pattern_length_exponent = 1
        self.pattern_age_exponent = 1
        self.debug = False
    def round(self, last):
        self.other.append(last)
        best_pattern_match = None
        best_pattern_score = None
        best_pattern_response = None
        self.debug and print("match so far:",tuple(zip(self.moves,self.other)))
        for turn in range(max(0,len(self.moves)-self.size),len(self.moves)):
            # record patterns ending with the move that just happened
            pattern_full = tuple(zip(self.moves[turn:],self.other[turn:]))
            if len(pattern_full) > 1:
                pattern_trunc = pattern_full[:-1]
                pattern_trunc_result = pattern_full[-1][1]
                self.patterns[pattern_trunc].append([pattern_trunc_result,len(self.moves)-1])
            if pattern_full in self.patterns:
                # we've seen this pattern at least once before
                self.debug and print("I've seen",pattern_full,"before:",self.patterns[pattern_full])
                for [response,turn_num] in self.patterns[pattern_full]:
                    score = len(pattern_full) ** self.pattern_length_exponent / (len(self.moves) - turn_num) ** self.pattern_age_exponent
                    if best_pattern_score == None or score > best_pattern_score:
                        best_pattern_match = pattern_full
                        best_pattern_score = score
                        best_pattern_response = response
                    # this could be much smarter about aggregating previous responses
        if best_pattern_response:
            move = self.counter_moves[best_pattern_response]
        else:
            # fall back to playing nice
            move = "C"
        self.moves.append(move)
        self.debug and print("I choose",move)
        return move

This bot looks for previous occurrences of the recent game state to see how the opponent responded to those occurrences, with a preference for longer pattern matches and more recent matches, then plays the move that will "beat" the opponent's predicted move. There's a lot of room for it to be smarter with all the data it keeps track of, but I ran out of time to work on it.

Sparr

Posted 2018-11-12T17:43:03.287

Reputation: 5 758

When you get the time, mind giving her an optimization pass? It's easily the largest time-sink. – SIGSTACKFAULT – 2018-11-13T03:46:52.593

2@Blacksilver I just reduced the maximum pattern length from 100 to 10. It should run almost instantly now if you're running <200 rounds – Sparr – 2018-11-13T05:55:21.520

1Maybe using a highly composite number (i.e., 12) would score better? – SIGSTACKFAULT – 2018-11-13T16:09:28.237

5

Jade

class Jade:
    def __init__(self):
        self.dRate = 0.001
        self.nRate = 0.003

    def round(self, last):
        if last == 'D':
            self.dRate *= 1.1
            self.nRate *= 1.2
        elif last == 'N':
            self.dRate *= 1.03
            self.nRate *= 1.05
        self.dRate = min(self.dRate, 1)
        self.nRate = min(self.nRate, 1)

        x = random.random()
        if x > (1 - self.dRate):
            return 'D'
        elif x > (1 - self.nRate):
            return 'N'
        else:
            return 'C'

Starts out optimistic, but gets progressively more bitter as the opponent refuses to cooperate. Lots of magic constants that could probably be tweaked, but this probably isn't going to do well enough to justify the time.

user48543

Posted 2018-11-12T17:43:03.287

Reputation:

5

Ensemble

This runs an ensemble of related models. The individual models consider different amounts of history, and have the option of either always choosing the move that will optimize the expected payout difference, or will randomly select a move in proportion to expected payout difference.

Each member of the ensemble then votes on their preferred move. They get a number of votes equal to how much more they've won than the opponent (which means that terrible models will get negative votes). Whichever move wins the vote is then selected.

(They should probably split their votes among the moves in proportion to how much they favor each, but I don't care enough to do that right now.)

It beats everything posted so far except EvaluaterBot and PatternFinder. (One-on-one, it beats EvaluaterBot and loses to PatternFinder).

from collections import defaultdict
import random
class Number6:
    class Choices:
        def __init__(self, C = 0, N = 0, D = 0):
            self.C = C
            self.N = N
            self.D = D

    def __init__(self, strategy = "maxExpected", markov_order = 3):
        self.MARKOV_ORDER = markov_order;
        self.my_choices = "" 
        self.opponent = defaultdict(lambda: self.Choices())
        self.choice = None # previous choice
        self.payoff = {
            "C": { "C": 3-3, "N": 4-1, "D": 0-5 },
            "N": { "C": 1-4, "N": 2-2, "D": 3-2 },
            "D": { "C": 5-0, "N": 2-3, "D": 1-1 },
        }
        self.total_payoff = 0

        # if random, will choose in proportion to payoff.
        # otherwise, will always choose argmax
        self.strategy = strategy
        # maxExpected: maximize expected relative payoff
        # random: like maxExpected, but it chooses in proportion to E[payoff]
        # argmax: always choose the option that is optimal for expected opponent choice

    def update_opponent_model(self, last):
        for i in range(0, self.MARKOV_ORDER):
            hist = self.my_choices[i:]
            self.opponent[hist].C += ("C" == last)
            self.opponent[hist].N += ("N" == last)
            self.opponent[hist].D += ("D" == last)

    def normalize(self, counts):
        sum = float(counts.C + counts.N + counts.D)
        if 0 == sum:
            return self.Choices(1.0 / 3.0, 1.0 / 3.0, 1.0 / 3.0)
        return self.Choices(
            counts.C / sum, counts.N / sum, counts.D / sum)

    def get_distribution(self):
        for i in range(0, self.MARKOV_ORDER):
            hist = self.my_choices[i:]
            #print "check hist = " + hist
            if hist in self.opponent:
                return self.normalize(self.opponent[hist])

        return self.Choices(1.0 / 3.0, 1.0 / 3.0, 1.0 / 3.0)

    def choose(self, dist):
        payoff = self.Choices()
        # We're interested in *beating the opponent*, not
        # maximizing our score, so we optimize the difference
        payoff.C = (3-3) * dist.C + (4-1) * dist.N + (0-5) * dist.D
        payoff.N = (1-4) * dist.C + (2-2) * dist.N + (3-2) * dist.D
        payoff.D = (5-0) * dist.C + (2-3) * dist.N + (1-1) * dist.D

        # D has slightly better payoff on uniform opponent,
        # so we select it on ties
        if self.strategy == "maxExpected":
            if payoff.C > payoff.N:
                return "C" if payoff.C > payoff.D else "D"
            return "N" if payoff.N > payoff.D else "D"
        elif self.strategy == "randomize":
            payoff = self.normalize(payoff)
            r = random.uniform(0.0, 1.0)
            if (r < payoff.C): return "C"
            return "N" if (r < payoff.N) else "D"
        elif self.strategy == "argMax":
            if dist.C > dist.N:
                return "D" if dist.C > dist.D else "N"
            return "C" if dist.N > dist.D else "N"

        assert(0) #, "I am not a number! I am a free man!")

    def update_history(self):
        self.my_choices += self.choice
        if len(self.my_choices) > self.MARKOV_ORDER:
            assert(len(self.my_choices) == self.MARKOV_ORDER + 1)
            self.my_choices = self.my_choices[1:]

    def round(self, last):
        if last: self.update_opponent_model(last)

        dist = self.get_distribution()
        self.choice = self.choose(dist)
        self.update_history()
        return self.choice

class Ensemble:
    def __init__(self):
        self.models = []
        self.votes = []
        self.prev_choice = []
        for order in range(0, 6):
            self.models.append(Number6("maxExpected", order))
            self.models.append(Number6("randomize", order))
            #self.models.append(Number6("argMax", order))
        for i in range(0, len(self.models)):
            self.votes.append(0)
            self.prev_choice.append("D")

        self.payoff = {
            "C": { "C": 3-3, "N": 4-1, "D": 0-5 },
            "N": { "C": 1-4, "N": 2-2, "D": 3-2 },
            "D": { "C": 5-0, "N": 2-3, "D": 1-1 },
        }

    def round(self, last):
        if last:
            for i in range(0, len(self.models)):
                self.votes[i] += self.payoff[self.prev_choice[i]][last]

        # vote. Sufficiently terrible models get negative votes
        C = 0
        N = 0
        D = 0
        for i in range(0, len(self.models)):
            choice = self.models[i].round(last)
            if "C" == choice: C += self.votes[i]
            if "N" == choice: N += self.votes[i]
            if "D" == choice: D += self.votes[i]
            self.prev_choice[i] = choice

        if C > D and C > N: return "C"
        elif N > D: return "N"
        else: return "D"

Test Framework

In case anyone else finds it useful, here's a test framework for looking at individual matchups. Python2. Just put all the opponents you're interested in in opponents.py, and change the references to Ensemble to your own.

import sys, inspect
import opponents
from ensemble import Ensemble

def count_payoff(label, them):
    if None == them: return
    me = choices[label]
    payoff = {
        "C": { "C": 3-3, "N": 4-1, "D": 0-5 },
        "N": { "C": 1-4, "N": 2-2, "D": 3-2 },
        "D": { "C": 5-0, "N": 2-3, "D": 1-1 },
    }
    if label not in total_payoff: total_payoff[label] = 0
    total_payoff[label] += payoff[me][them]

def update_hist(label, choice):
    choices[label] = choice

opponents = [ x[1] for x 
    in inspect.getmembers(sys.modules['opponents'], inspect.isclass)]

for k in opponents:
    total_payoff = {}

    for j in range(0, 100):
        A = Ensemble()
        B = k()
        choices = {}

        aChoice = None
        bChoice = None
        for i in range(0, 100):
            count_payoff(A.__class__.__name__, bChoice)
            a = A.round(bChoice)
            update_hist(A.__class__.__name__, a)

            count_payoff(B.__class__.__name__, aChoice)
            b = B.round(aChoice)
            update_hist(B.__class__.__name__, b)

            aChoice = a
            bChoice = b
    print total_payoff

Ray

Posted 2018-11-12T17:43:03.287

Reputation: 1 488

The controller is ready, you didn't have to do all that... – SIGSTACKFAULT – 2018-11-13T03:40:46.343

1@Blacksilver I realized that just as I was about to submit. But this one works in versions before 3.6, and gives information about individual matchups that can help identify weak spots, so it wasn't a complete waste of time. – Ray – 2018-11-13T03:42:31.447

Fair enough; running now. I'll probably add options to my controller to do similar things. – SIGSTACKFAULT – 2018-11-13T03:45:07.167

"It beats everything posted so far except Ensemble and PatternFinder" I'm honored :) – Sparr – 2018-11-13T07:19:11.283

@Sparr Oops. That was supposed to say EvaluaterBot and PatternFinder. But that's when comparing total score against the entire field. PatternFinder remains the only one that beats this in a direct match up. – Ray – 2018-11-13T08:04:21.003

4

OldTitForTat

Old school player is too lazy to update for the new rules.

class OldTitForTat:
    def round(self, last):
        if(last == None)
            return "C"
        if(last == "C"):
            return "C"
        return "D"

Joshua

Posted 2018-11-12T17:43:03.287

Reputation: 3 043

3

LastOptimalBot

class LastOptimalBot:
    def round(self, last):
        return "N" if last == "D" else ("D" if last == "C" else "C")

Assumes that the opposing bot will always play the same move again, and chooses the one that has the best payoff against it.

Averages:

Me   Opp
2.6  2    vs TitForTat
5    0    vs AllC
4    1    vs AllN
3    2    vs AllD
3.5  3.5  vs Random
3    2    vs Grudger
2    2    vs LastOptimalBot
1    3.5  vs TatForTit
4    1    vs NeverCOOP
1    4    vs EvaluaterBot
2.28 2.24 vs NashEquilibrium

2.91 average overall

Spitemaster

Posted 2018-11-12T17:43:03.287

Reputation: 695

oof. Maybe T4T would do better as return last. – SIGSTACKFAULT – 2018-11-12T18:11:00.103

I'd like that! If TitForTat were return last, LOB would go 18-9 over 6 rounds rather than the 13-10 over 5 rounds it's currently getting. I think it's fine as is - don't worry about optimizing the example bots. – Spitemaster – 2018-11-12T18:30:08.183

return last would be a better T4T for this challenge, I think – Sparr – 2018-11-12T18:43:20.877

Just tried -- the if(last): return last; else: return "C" is worse. – SIGSTACKFAULT – 2018-11-12T18:59:37.487

Right, but as @Sparr was saying, it might be more appropriate. Up to you, I suppose. – Spitemaster – 2018-11-12T19:00:54.687

3

NeverCOOP

class NeverCOOP:
    def round(self, last):
        try:
            if last in "ND":
                return "D"
            else:
                return "N"
        except:
            return "N"

If the opposing bot defects or is neutral, choose defect. Otherwise if this is the first turn or the opposing bot cooperates, choose neutral. I'm not sure how good this will work...

glietz

Posted 2018-11-12T17:43:03.287

Reputation: 101

What's the try/except for? – SIGSTACKFAULT – 2018-11-12T19:09:35.083

1@Blacksilver I'd assume it serves the same purpose as the if(last): in your Grudger bot, detecting whether there was a previous round. – ETHproductions – 2018-11-12T19:31:24.717

ahh, I see. None in "ND" errors. – SIGSTACKFAULT – 2018-11-12T19:38:07.283

Because if last and last in "ND": was too complicated? – user253751 – 2018-11-12T21:29:58.670

3

CopyCat

class CopyCat:
    def round(self, last):
        if last:
            return last
        return "C"

Copies the opponent's last move.
I don't expect this to do well, but no one had implemented this classic yet.

MannerPots

Posted 2018-11-12T17:43:03.287

Reputation: 31

2

Historic Average

class HistoricAverage:
    PAYOFFS = {
        "C":{"C":3,"N":1,"D":5},
        "N":{"C":4,"N":2,"D":2},
        "D":{"C":0,"N":3,"D":1}}
    def __init__(self):
        self.payoffsum = {"C":0, "N":0, "D":0}
    def round(this, last):
        if(last != None):
            for x in this.payoffsum:
               this.payoffsum[x] += HistoricAverage.PAYOFFS[last][x]
        return max(this.payoffsum, key=this.payoffsum.get)

Looks at history and finds the action that would have been best on average. Starts cooperative.

MegaTom

Posted 2018-11-12T17:43:03.287

Reputation: 3 787

This could run faster if it didn't re-calculate the averages every round. – Sparr – 2018-11-13T19:57:49.933

@Sparr true. I edited it so it does now. – MegaTom – 2018-11-14T20:19:52.597

2

Improved Dirichlet Dice

import random

class DirichletDice2:
    def __init__(self):

        self.alpha = dict(
                C = {'C' : 1, 'N' : 1, 'D' : 1},
                N = {'C' : 1, 'N' : 1, 'D' : 1},
                D = {'C' : 1, 'N' : 1, 'D' : 1}
        )
        self.myLast = [None, None]
        self.payoff = dict(
                C = { "C": 0, "N": 3, "D": -5 },
                N = { "C": -3, "N": 0, "D": 1 },
                D = { "C": 5, "N": -1, "D": 0 }
        )

    def DirichletDraw(self, key):
        alpha = self.alpha[key].values()
        mu = [random.gammavariate(a,1) for a in alpha]
        mu = [m / sum(mu) for m in mu]
        return mu

    def ExpectedPayoff(self, probs):
        expectedPayoff = {}
        for val in ['C','N','D']:
            payoff = sum([p * v for p,v in zip(probs, self.payoff[val].values())])
            expectedPayoff[val] = payoff
        return expectedPayoff

    def round(self, last):
        if last is None:
            self.myLast[0] = 'D'
            return 'D'

        #update dice corresponding to opponent's last response to my
        #outcome two turns ago
        if self.myLast[1] is not None:
            self.alpha[self.myLast[1]][last] += 1

        #draw probs for my opponent's roll from Dirichlet distribution and then return the optimal response
        mu = self.DirichletDraw(self.myLast[0])
        expectedPayoff = self.ExpectedPayoff(mu)
        res = max(expectedPayoff, key=expectedPayoff.get)

        #update myLast
        self.myLast[1] = self.myLast[0]
        self.myLast[0] = res

        return res    

This is an improved version of Dirichlet Dice. Instead of taking the expected multinomial distribution from the Dirichlet distribution, it draws a Multinomial distribution randomly from the Dirichlet distribution. Then, instead of drawing from the Multinomial and giving the optimal response to that, it gives the optimal expected response to the given Multinomial using the points. So the randomness has essentially been shifted from the Multinomial draw to the Dirichlet draw. Also, the priors are more flat now, to encourage exploration.

It's "improved" because it now accounts for the points system by giving the best expected value against the probabilities, while maintaining its randomness by drawing the probabilities themselves. Before, I tried simply doing the best expected payoff from the expected probabilities, but that did badly because it just got stuck, and didn't explore enough to update its dice. Also it was more predictable and exploitable.


Original submission:

Dirichlet Dice

import random

class DirichletDice:
    def __init__(self):

        self.alpha = dict(
                C = {'C' : 2, 'N' : 3, 'D' : 1},
                N = {'C' : 1, 'N' : 2, 'D' : 3},
                D = {'C' : 3, 'N' : 1, 'D' : 2}
        )

        self.Response = {'C' : 'D', 'N' : 'C', 'D' : 'N'}
        self.myLast = [None, None]

    #expected value of the dirichlet distribution given by Alpha
    def MultinomialDraw(self, key):
        alpha = list(self.alpha[key].values())
        probs = [x / sum(alpha) for x in alpha]
        outcome = random.choices(['C','N','D'], weights=probs)[0]
        return outcome

    def round(self, last):
        if last is None:
            self.myLast[0] = 'D'
            return 'D'

        #update dice corresponding to opponent's last response to my
        #outcome two turns ago
        if self.myLast[1] is not None:
            self.alpha[self.myLast[1]][last] += 1

        #predict opponent's move based on my last move
        predict = self.MultinomialDraw(self.myLast[0])
        res = self.Response[predict]

        #update myLast
        self.myLast[1] = self.myLast[0]
        self.myLast[0] = res

        return res

Basically I'm assuming that the opponent's response to my last output is a multinomial variable (weighted dice), one for each of my outputs, so there's a dice for "C", one for "N", and one for "D". So if my last roll was, for example, a "N" then I roll the "N-dice" to guess what their response would be to my "N". I begin with a Dirichlet prior that assumes that my opponent is somewhat "smart" (more likely to play the one with the best payoff against my last roll, least likely to play the one with the worst payoff). I generate the "expected" Multinomial distribution from the appropriate Dirichlet prior (this is the expected value of the probability distribution over their dice weights). I roll the weighted dice of my last output, and respond with the one with the best payoff against that dice outcome.

Starting in the third round, I do a Bayesian update of the appropriate Dirichlet prior of my opponent's last response to the thing I played two rounds ago. I'm trying to iteratively learn their dice weightings.

I could have also simply picked the response with the best "expected" outcome once generating the dice, instead of simply rolling the dice and responding to the outcome. However, I wanted to keep the randomness in, so that my bot is less vulnerable to the ones that try to predict a pattern.

Bridgeburners

Posted 2018-11-12T17:43:03.287

Reputation: 121

2

Kevin

class Kevin:
    def round(self, last):      
        return {"C":"N","N":"D","D":"C",None:"N"} [last]

Picks the worst choice. The worst bot made.

Useless

import random

class Useless:
    def __init__(self):
        self.lastLast = None

    def round(self, last):
        tempLastLast = self.lastLast
        self.lastLast = last

        if(last == "D" and tempLastLast == "N"):
            return "C"
        if(last == "D" and tempLastLast == "C"):
            return "N"

        if(last == "N" and tempLastLast == "D"):
            return "C"
        if(last == "N" and tempLastLast == "C"):
            return "D"

        if(last == "C" and tempLastLast == "D"):
            return "N"
        if(last == "C" and tempLastLast == "N"):
            return "D"

        return random.choice("CND")

It looks at the last two moves done by the opponent and picks the most not done else it picks something random. There is probably a better way of doing this.

Link1J

Posted 2018-11-12T17:43:03.287

Reputation: 19

1

Weighted Average

class WeightedAverageBot:
  def __init__(self):
    self.C_bias = 1/4
    self.N = self.C_bias
    self.D = self.C_bias
    self.prev_weight = 1/2
  def round(self, last):
    if last:
      if last == "C" or last == "N":
        self.D *= self.prev_weight
      if last == "C" or last == "D":
        self.N *= self.prev_weight
      if last == "N":
        self.N = 1 - ((1 - self.N) * self.prev_weight)
      if last == "D":
        self.D = 1 - ((1 - self.D) * self.prev_weight)
    if self.N <= self.C_bias and self.D <= self.C_bias:
      return "D"
    if self.N > self.D:
      return "C"
    return "N"

The opponent's behavior is modeled as a right triangle with corners for C N D at 0,0 0,1 1,0 respectively. Each opponent move shifts the point within that triangle toward that corner, and we play to beat the move indicated by the point (with C being given an adjustably small slice of the triangle). In theory I wanted this to have a longer memory with more weight to previous moves, but in practice the current meta favors bots that change quickly, so this devolves into an approximation of LastOptimalBot against most enemies. Posting for posterity; maybe someone will be inspired.

Sparr

Posted 2018-11-12T17:43:03.287

Reputation: 5 758

1

Tetragram

import itertools

class Tetragram:
    def __init__(self):
        self.history = {x: ['C'] for x in itertools.product('CND', repeat=4)}
        self.theirs = []
        self.previous = None

    def round(self, last):
        if self.previous is not None and len(self.previous) == 4:
            self.history[self.previous].append(last)
        if last is not None:
            self.theirs = (self.theirs + [last])[-3:]

        if self.previous is not None and len(self.previous) == 4:
            expected = random.choice(self.history[self.previous])
            if expected == 'C':
                choice = 'C'
            elif expected == 'N':
                choice = 'C'
            else:
                choice = 'N'
        else:
            choice = 'C'

        self.previous = tuple(self.theirs + [choice])
        return choice

Try to find a pattern in the opponent's moves, assuming they're also watching our last move.

user48543

Posted 2018-11-12T17:43:03.287

Reputation:

1

Handshake

class HandshakeBot:
  def __init__(self):
    self.handshake_length = 4
    self.handshake = ["N","N","C","D"]
    while len(self.handshake) < self.handshake_length:
      self.handshake *= 2
    self.handshake = self.handshake[:self.handshake_length]
    self.opp_hand = []
    self.friendly = None
  def round(self, last):
    if last:
      if self.friendly == None:
        # still trying to handshake
        self.opp_hand.append(last)
        if self.opp_hand[-1] != self.handshake[len(self.opp_hand)-1]:
          self.friendly = False
          return "D"
        if len(self.opp_hand) == len(self.handshake):
          self.friendly = True
          return "C"
        return self.handshake[len(self.opp_hand)]
      elif self.friendly == True:
        # successful handshake and continued cooperation
        if last == "C":
          return "C"
        self.friendly = False
        return "D"
      else:
        # failed handshake or abandoned cooperation
        return "N" if last == "D" else ("D" if last == "C" else "C")
    return self.handshake[0]

Recognizes when it's playing against itself, then cooperates. Otherwise mimics LastOptimalBot which seems like the best one-line strategy. Performs worse than LastOptimalBot, by an amount inversely proportional to the number of rounds. Obviously would do better if there were more copies of it in the field *cough**wink*.

Sparr

Posted 2018-11-12T17:43:03.287

Reputation: 5 758

Just submit a few clones that have different non-handshake behavior. – SIGSTACKFAULT – 2018-11-13T22:52:23.567

That seems exploit-y. I could submit one such clone for every simple behavior represented here. – Sparr – 2018-11-14T00:12:36.873

I've added an extra clause saying that you can only submit max five bots. – SIGSTACKFAULT – 2018-11-14T02:14:12.557

1

ShiftingOptimalBot

class ShiftingOptimalBot:
    def __init__(self):
        # wins, draws, losses
        self.history = [0,0,0]
        self.lastMove = None
        self.state = 0
    def round(self, last):
        if last == None:
            self.lastMove = "C"
            return self.lastMove
        if last == self.lastMove:
            self.history[1] += 1
        elif (last == "C" and self.lastMove == "D") or (last == "D" and self.lastMove == "N") or (last == "N" and self.lastMove == "C"):
            self.history[0] += 1
        else:
            self.history[2] += 1

        if self.history[0] + 1 < self.history[2] or self.history[2] > 5:
            self.state = (self.state + 1) % 3
            self.history = [0,0,0]
        if self.history[1] > self.history[0] + self.history[2] + 2:
            self.state = (self.state + 2) % 3
            self.history = [0,0,0]

        if self.state == 0:
            self.lastMove = "N" if last == "D" else ("D" if last == "C" else "C")
        elif self.state == 1:
            self.lastMove = last
        else:
            self.lastMove = "C" if last == "D" else ("N" if last == "C" else "D")
        return self.lastMove

This bot uses LastOptimalBot's algorithm as long as it's winning. If the other bot starts predicting it, however, it will start playing whichever move its opponent played last (which is the move that beats the move that would beat LastOptimalBot). It cycles through simple transpositions of those algorithms as long as it continues to lose (or when it gets bored by drawing a lot).

Honestly, I'm surprised that LastOptimalBot is sitting in 5th as I post this. I'm fairly certain this will do better, assuming that I wrote this python correctly.

Spitemaster

Posted 2018-11-12T17:43:03.287

Reputation: 695

0

HandshakePatternMatch

from .patternfinder import PatternFinder
import collections

class HandshakePatternMatch:
    def __init__(self):
        self.moves = [None]
        self.other = []
        self.handshake = [None,"N","C","C","D","N"]
        self.friendly = None
        self.pattern = PatternFinder()
    def round(self, last):
        self.other.append(last)
        if last:
            if len(self.other) < len(self.handshake):
                # still trying to handshake
                if self.friendly == False or self.other[-1] != self.handshake[-1]:
                    self.friendly = False
                else:
                    self.friendly = True
                move = self.handshake[len(self.other)]
                self.pattern.round(last)
            elif self.friendly == True:
                # successful handshake and continued cooperation
                move = self.pattern.round(last)
                if last == "C":
                    move = "C"
                elif last == self.handshake[-1] and self.moves[-1] == self.handshake[-1]:
                    move = "C"
                else:
                    self.friendly = False
            else:
                # failed handshake or abandoned cooperation
                move = self.pattern.round(last)
        else:
            move = self.handshake[1]
            self.pattern.round(last)
        self.moves.append(move)
        return move

Why pattern match yourself? Handshake and cooperate away.

Draco18s no longer trusts SE

Posted 2018-11-12T17:43:03.287

Reputation: 3 053

import PatternFinder is cheating in my books. – SIGSTACKFAULT – 2018-11-17T02:30:34.983

@Blacksilver It gets done all the time in KOTH. It's no different than copying the code in an existing answer and using it. Robot Roulette: High stakes robot gambling had it happening all over the place to the point that bots would detect if their code was being called by an opponent and sabotage the return.

– Draco18s no longer trusts SE – 2018-11-17T02:36:57.580

Alright then. TIL. – SIGSTACKFAULT – 2018-11-17T02:37:58.993

I'll do the crunching tomorrow. – SIGSTACKFAULT – 2018-11-17T02:38:23.093

Here's a perfect example of using other bots' code. Usually it comes down to "that guy worked out some tricky math, I want his results under these conditions." (My own entry did that to pretty good effect; UpYours was more scatter-shot in its approach). – Draco18s no longer trusts SE – 2018-11-17T02:41:30.433

@Blacksilver Still crunching? ;) – Draco18s no longer trusts SE – 2018-11-23T01:17:45.057

0

Hardcoded

class Hardcoded:
    sequence = "DNCNNDDCNDDDCCDNNNNDDCNNDDCDCNNNDNDDCNNDDNDDCDNCCNNDNNDDCNNDDCDCNNNDNCDNDNDDNCNDDCDNNDCNNDDCDCNNDNNDDCDNDDCCNNNDNNDDCNNDDNDCDNCNDDCDNNDDCCNDNNDDCNNNDCDNDDCNNNNDNDDCDNCDCNNDNNDDCDNDDCCNNNDNDDCNNNDNDCDCDNNDCNNDNDDCDNCNNDDCNDNNDDCDNNDCDNDNCDDCNNNDNDNCNDDCDNDDCCNNNNDNDDCNNDDCNNDDCDCNNDNNDDCDNDDCCNDNNDDCNNNDCDNNDNDDCCNNNDNDDNCDCDNNDCNNDNDDCNNDDCDNCNNDDCDNNDCDNDNCDDCNDNNDDCNNNDDCDNCNNDNNDDCNNDDNNDCDNCNDDCNNDCDNNDDCNNDDNCDCNNDNDNDDCDNCDCNNNDNDDCDCNNDNNDDCDNDDCCNNNDNNDDCNDNDNCDDCDCNNNNDNDDCDNCNDDCDNNDDCNNNDNDDCDNCNNDCNNDNDDNCDCDNNNDDCNNDDCNNDDNNDCDNCNDDCNNDDNDCDNNDNDDCCNCDNNDCNNDDNDDCNCDNNDCDNNNDDCNNDDCDCDNNDDCNDNCNNDNNDNDNDDCDNCDCNNNDNDDCDNCNNDDCDNNDCNNDDCNNDDCDCDNNDDCNDNCNNNDDCDNNDCDNDNCNNDNDDNNDNDCDDCCNNNDDCNDNDNCDDCDCNNNDNNDDCNDCDNDDCNNNNDNDDCCNDNNDDCDCNNNDNDDNDDCDNCCNNDNNDDCNNDDCDCNNDNNDDCNNDDNCNDDNNDCDNCNDDCNNDDNDCDNNDNDDCCNCDNNDCNNDNDDCNNDDNCDCDNNDCNNDNDDCDCDNNNNDDCNNDDNDCCNNDDNDDCNCDNNDCNNDDNDDCDNCNDDCNNNNDCDNNDDCNDNDDCDNCNNDCDNNDCNNDNDDNCDCNNDNDDCDNDDCCNNNNDNDDCNNDDCDCNNDNNDDCDCDNNDDC"
    def __init__(self):
        self.round_num = -1
    def round(self,_):
        self.round_num += 1
        return Hardcoded.sequence[self.round_num % 1000]

Just plays a hardcoded sequence of moves optimized to beat some of the top deterministic bots.

MegaTom

Posted 2018-11-12T17:43:03.287

Reputation: 3 787