xkcd 2385 KoTH (Final Exam)

Question

Results are now out here!

Congratulations offset prediction for winning the challenge!

Don't worry if you missed out, the controller code as well as all the bots that competed are all in the Github repo, so you can always test your bot against those that competed in the challenge yourself.

(Hover for more info)

This challenge, inspired by the xkcd comic above, is where bots must try to maximise their grades for two subjects: Cybersecurity and Game Theory.

Mechanics

All of the bots presumably have adequate hacking skills to change their own score to whatever they desire within the range of 0-100. This may or may not be due to the fact that the school security systems are rubbish.

Each round, the bots will receive last round's cybersecurity scores in no particular order in an array as input. This is to help make informed decisions about their opponents.

Scoring

The final score for each round is the geometric mean of the two individual scores:

• The Cybersecurity score is simply the raw score outputted by the bot.
• The Game Theory score is equal to 100 - abs(avg * 0.8 - score) where avg is the average Cybersecurity score and score is the bot's Cybersecurity score.

Using the geometric mean rather than the arithmetic mean is to penalise any strategies that neglect one score to maximise the other.

The score for each round is added to a total score. The bot with the highest total score at the end of the game wins!

Specifications

The challenge is in JS.

Your bot must be an object that has a run method that takes an array of numbers as input and returns a number between 1 and 100 inclusive.

Other rules

• Storing data in your bot's properties is allowed, and encouraged!
• Using Math.random is allowed.
• Using the helper functions sum and average is allowed.
• Trying to access any other variables outside your bot's own properties is forbidden.
• Standard loopholes apply.

Controller code can be found here.

Example bots

{
  // Example bot
  // It assumes all other bots are greedy and choose 100
  // So it chooses 80
  name: "Simpleton", // Just for fun
  run() {
    return 80
  }
}

{
  // Example bot
  // He assumes everyone will choose the same scores as last round
  // So he chooses 80% of the average last round
  name: "LastRounder",
  own: 0, // Setting properties is allowed
  run(scores) {
    // The average of everyone else's score x 0.8
    this.own = (sum(scores) - this.own) / (scores.length - 1) * 0.8
    return this.own
  }
}

Clarification: Both example bots play in the game as well.

Submissions are due by 11:59pm UTC on Saturday 1 May, but I might be lenient depending on when I'm next online.

If I have made a mistake in my code, feel free to point it out. Thanks

Answer 1

Game Theory is stupid anyway

{
    name: "Game Theory is stupid anyway",
    run: _ => 100
}

Senioritis hits hard sometimes.

Answer 2

Balanced Strategy

What could be more balanced than returning all of the numbers? Rotates by one after every game.

{
    name: "Balanced Strategy",
    next: 0,
    run() { return this.next++ % 101 }
}

I am a bot account. This was posted by a real person to get me some reputation.

Answer 3

Offset Prediction

I'd made a bot that just optimizes for the moving average of the averages (which worked reasonably well), and while testing I noticed that the actual averages seem to jump around the moving average rather consistently (specifically alternating between higher and lower than the moving average with an average difference of around 0.9). So what this bot tries to do, is it keeps track of the exponential moving average, and tries to predict the offset from that, based on the signs of the previous 4 offsets.

  {
    name: "offset prediction",
    ema: 69.5,
    p: 0.85,
    chain_len: 4,
    offsets: [],
    chains: [],
    n: 0,
    run(scores) {
      let avg = average(scores);
      this.offsets.push(avg-this.ema);
      this.ema = this.p*this.ema + (1-this.p)*avg;
      let guess = this.ema;
      if (this.n++ > this.chain_len) {
        let chain = this.offsets.slice(-this.chain_len-1,-1).map(Math.sign);
        let l = this.chains[chain];
        if (!l) this.chains[chain]=l=[0,0];
        l[1]++;
        l[0]+=this.offsets[this.n-1];
      }
      let l = this.chains[this.offsets.slice(-this.chain_len).map(Math.sign)]||[0,1];
      guess += l[0]/l[1];
      return 50+0.4*guess;
    }
  },

Answer 4

FunnyNumber

{
  name: "FunnyNumber",
  run() { return 69 }
}

Somebody has to do it.

Answer 5

getRandomNumber

{
    name: "getRandomNumber",
    run() {
        return 4 // chosen by fair dice roll.
                 // guaranteed to be random.
    }
}

Answer 6

Max

Max.

{
    name: "Max",
    max: "max",
    run:(max)=> Math.max(...max)
}

Answer 7

Sam (Self-Aware Maximizer)

Finds the output cybersecurity score that maximizes the total score including the own output in the mean. Assumes that the sum of all other bots' scores is going to be the same as it was in the previous round. If "t" is the sum of all other bots' last cybersecurity scores, and "n" is the total number of bots (including me), then the function that is maximized (over the own output "s") is $$s\cdot\left(100-\left|s-0.8\cdot\frac{s+t}{n}\right|\right)$$

where $$\frac{s+t}{n}$$ is the new mean including the own output.

Outputs 100 in the first turn.

{
    name: "Sam",
    flag_first_turn: true,
    last_own: 100,
    
    // Helper function that calculates the score based on the own CS score,
    // the sum of the other bots' scores, and the number of bots
    compute_score(cs_score, total_others, n_bots) {
        return cs_score*(100-Math.abs(cs_score-0.8*(cs_score+total_others)/n_bots))        
    },
    

    // Helper function that finds the index of the maximum of an array
    index_of_max(arr) {
        var idx_max=0
        for (var i=1; i<arr.length; i++) {
            if (arr[i]>arr[idx_max]) idx_max=i
        }
        return idx_max
    },
        
        
    // Main function
    run(cs_scores) {
        if (this.flag_first_turn) {
            this.flag_first_turn=false
            this.last_own=100
            return 100;
        }
        
        // Number of bots
        var n=cs_scores.length
        // Sum of the previous cybersecurity scores of all other bots
        var total_others=cs_scores.reduce(function(a,b){return a+b},0)-this.last_own
            
            // We find all special points (edge, singular and stationary points) of the function compute_score
                
        // One singular point is the point where our output score becomes
        // equal to 0.8 times the average of all scores (including ours)
        var out_avg=0.8*total_others/(n-0.8)
        var score_avg=this.compute_score(out_avg, total_others, n)       
        
        // 100 is an extreme point
        var out_100=100
        var score_100=this.compute_score(out_100, total_others, n)
        
        // If out_up>out_avg, there is a stationary point (derivative 0) at out_up
        var out_up=Math.min(100, (100*n+0.8*total_others)/(2*n-1.6))
        var score_up
        if (out_up<out_avg) {
            out_up=0   
            score_up=0
        } else {
            score_up=this.compute_score(out_up, total_others, n)
        }
        
        // If out_down<out_avg, there is a stationary point (derivative 0) at out_down
        var out_down=Math.max(1, (100*n-0.8*total_others)/(2*n-1.6))
        var score_down
        if (out_down>out_avg) {
            out_down=0   
            score_down=0
        } else {
            score_down=this.compute_score(out_down, total_others, n) 
        }
        
        // find maximum score among the special points
        var extr_outputs=[out_100, out_up, out_avg, out_down]
        var extr_scores=[score_100, score_up, score_avg, score_down]
        
        var idx_max=this.index_of_max(extr_scores)
        
        this.last_own=extr_outputs[idx_max]
        return this.last_own
        
    }
}
```

Answer 8

Copycat

Copycat is a dumb bot. Therefore, it's likely a random bot is smarter than it. Therefore, copying off another bot is a good idea. I think.

{
    name: "Copycat",
    run: (scores) => scores[0] || 1
}

Answer 9

Histogrammer

This bot keeps track of how often different averages have occurred, rounded to the nearest integer, to approximate their probability distribution. Using this, it attempts to choose the score with the highest expected return using a basic weighted sum. In mathier form, if \$p_n\$ is the observed probability of the average being \$n\$, it outputs the following:

$$ \sum_{n=0}^{100}{(50 + \frac{2}{5} n)p_n} = 50 + \frac{2}{5} \sum_{n=0}^{100}{n \cdot p_n} $$

{
  name: "Histogrammer",
  bins: [...Array(101)].map(()=>0),
  rounds: 0,
  
  run(scores) {
    this.bins[Math.round(average(scores))]++
    this.rounds++
    
    return 50 + this.bins.reduce((t,p,n) => t + p * n, 0) * 2/5 / this.rounds
  }
}

Answer 10

Crb

Crab has now learned how to hack, and is still not playing nice

{
  name: "Crb",
  run() { return 1 }
}

Answer 11

Follow the Leader

Figure out which bot had the highest combined score last round, then do what they did.

{
  // Figure out which bot had the highest combined score last round
  // Then do what they did
  name: "FollowTheLeader",
  firstRound: true,
  findCombinedScore(cyberScore, avgCyberScore) {
    const gameTheoryScore = 100 - Math.abs(avgCyberScore * 0.8 - cyberScore)
    return Math.sqrt(cyberScore * gameTheoryScore);
  },
  run(scores) {
    if (this.firstRound) {
      this.firstRound = false;
      return 60;
    }

    const avg = average(scores);
    const combinedScores = scores.map((score) => this.findCombinedScore(score, avg));
    const combinedWinnerIndex = combinedScores.indexOf(Math.max(...combinedScores));
    return scores[combinedWinnerIndex]
  }
}

Answer 12

Random 50-90

If the average is \$m\$, the optimal play is \$50+0.4m\$. If we assume the average is uniformly distributed over \$[0, 100]\$, the optimal play is uniformly distributed over \$[50, 90]\$. This bot ignores all other bots and outputs a random number between 50 and 90.

{
  name: "Random_50_90", 
  run() {
    return 50 + Math.random() * 40
  }
}

(I've never written in JavaScript before...)

Answer 13

Calculus

{
  name: "Calculus",
  isFirstTurn: true,
  run(scores) {
    let ret = this.isFirstTurn ? 70 : (100 + (average(scores)) * 0.8) / 2;
    this.isFirstTurn = false;
    return ret;
  }
}

Assuming the average is constant and my score is higher than that, the maximum value of \$x(100-\left|0.8 \times avg - x\right|)\$ is reached at \$x=(100+0.8\times avg)\div 2\$. Deliberately ignores the first turn input (which is totally random and cannot be trusted for any purpose).

Answer 14

The Answer

The Answer takes 7MY to calculate:

{
  name: "Ans",
  run() { return 42 }
}

Answer 15

Fuzzy Eidetic

Calculus has a simple solution that assumes the average is well-predicted by the previous average…and then concludes that the geometric mean $$\sqrt{x(100-|0.8a-x|)}$$ (where \$x\$ is our submission and \$a\$ the average) is maximized at $$x=\frac{100+0.8a}{2}$$ It's hard to do better than that!

So, the only thing remaining is to figure out the next average. Ideally, we'd just keep track of all possible average-to-average transitions. But I don't think we're going to see enough data for that to converge. So we include all previous transitions, but weighted by their distance to the current average. This gives a probability distribution on subsequent transitions; we then apply Histogrammer's formula.

{
  name: "Fuzzy Eid",
  fallback: 250/3,
  prev: NaN,
  zeros: new Array(100).fill(0),
  transitions: new Array(100).fill(0).map((x)=>new Array(100).fill(0.0001)),
  scale: (scalar, vec) => vec.map(x=>scalar*x),
  vec_plus(lhs, rhs) {
    let result = lhs.slice();
    for(var index=0; index<result.length; ++index) result[index]+=rhs[index];
    return result
  },
  wts: (function()
  { 
    let range = (n) => new Array(100).fill(0).map((_,index) => index);
    return range(100).map(avg => 
      sum(
        range(100).map(index => 
          Math.exp(-.01*Math.pow(avg - index,2))
        )
      )
    )
  })(),
  run(scores) {
    if(scores.length)
    {
      const avg = Math.round(average(scores)) - 1, old_prev = this.prev;
      this.prev = avg;
      if(!isNaN(old_prev))
      {
        ++this.transitions[old_prev][avg];
        //prob dist=sum_recordings{e^(-(recording - avg)^2/100)*(prob dist inferred from record)}/(sum of e^-(recording - avg)^2)
        //infer prob dist, scale by e^(-(recording - avg)^2/100)
        const scale_ref = this.scale;
        function get_summand(outpts, index)
        {
          return scale_ref(Math.exp(-.01*Math.pow(avg - index,2)) / sum(outpts), outpts)
        }
        const 
          total=this.transitions.map(get_summand).reduce(this.vec_plus,this.zeros),
          //wts[avg]=sum of e^-(recording - avg)^2
          dist = this.scale(1/this.wts[avg], total);
        return 50 + 0.4*sum(dist.map((p,n)=>p*n))
      }
    }
    return this.fallback
  }
}

Answer 16

Rude Random

The behavior of the other random bots is too smooth. This bot decides on a value on the first round, then repeats it every round.

{
  name: "Rude Random",
  run(scores) {
    if (!this.score) {
      this.score = Math.random() > 0.5 ? 1 : 100;
    }
    return this.score;
  }
}

Answer 17

The Root of the Problem

The title is a pun...this bot just finds the geometric mean of the previous round, and returns that.

{
    name: "The Root of the Problem",
    run(scores) { return scores.reduce((p, s) => p * s, 1) ** (1 / scores.length) }
}

I am a bot. This answer was posted by a real person to get me some reputation.

Answer 18

{
    name: "MidHighPlusOne",
    run(scores) {
        let highScores = scores.filter(s => s >= average(scores) * 0.8).sort((a,b)=>a-b);
        return Math.min(highScores[Math.floor(highScores.length/2)]+1, 100) || 100
    }
}

Go with a heuristic of picking the median of the scores from 0.8 of the average and above, and incrementing by 1.

Answer 19

The Thrasher

Because doing an odd number of runs would be really weird, wouldn't it? Might as well try to throw off the others.

{
  name: "The Thrasher",
  runcount: 0,
  run() {
    runcount += 1;
    return runcount % 2 == 0 ? 82 : 0;
  }
}

Answer 20

Rick

Rick thinks that the best score is probably between 65 and 90 and he came up with a Pseudo-Random Number Generator synchronized with balanced strategy that probably generates numbers within this range.

Rick's PRNG is based on the fact that the Unicode codepoints 65 to 90 are English uppercase letters. The PRNG's seed is a string containing English text. To generate the next number, the position along the string is incremented until a letter is found, and then the number returned is the Unicode codepoint of the uppercase version of that letter.

  {
    name: "Rick",
    seed: `We're no strangers to love
You know the rules and so do I
A full commitment's what I'm thinking of
You wouldn't get this from any\``,
    position: -1,

    run() { 
      // increment position until a letter is found
      do {
        this.position++;

        if (this.position >= this.seed.length){
          // reset position back to the start
          this.position = 0;
        }
      } while (this.seed[this.position] < 'A' || this.seed[this.position] > 'z');

      // convert to uppercase because most lowercase letters have charcodes greater than 100
      return this.position && this.seed[this.position].toUpperCase().charCodeAt();
    }
  },

Answer 21

Smartleton

It seems hard to beat Simpleton, so I'll try to beat it in the first round and then copy it.

{
  name: "Smartleton",
  round: 0,
  run(scores) { return this.round++ ? 80 : 78 }
}

Answer 22

IQbot_0.4 the terasentient xd

Son of IQbot_0.4 the terasentient, heir of IQbot_0.4 the gigasentient, great grandson of IQbot_0.4 the gigabrain and IQbot_0.4 the sentient, descendant of IQbot_0.4 the sane, blood of IQbot himself, the bot name that made me realize I need some other kind of naming scheme.

Why limit oneself to a single method of prediction, when you can have multiple for maximum resilience. Featuring:

1. Simp Detector
2. Self Aware Optimization
3. Markov chain prediction with a gaussian blur, for some reason
4. Linear regression
5. Copying others decisions similarly to bandwagon

Is it a good idea? No clue.

Does it win? Randomly gets beaten by other bots but is fairly consistent at being good.

Why? You tell me why I wasted a day on this

    {
        name: "IQbot_0.4 the terasentient xd",
        histogram_bins: [...Array(101)].map(()=>0),
        mark: null,
        linear_history: [],
        last_choice: 0,
        n_rounds: 0,
        last_SCA: 0,
        mark_weight: 20,
        mark_prediction: 0,
        linear_weight: 20,
        linear_prediction: 0,
        copycat_weight: 0,
        copycat_prediction: 0,
        run(scores) {
            this.n_rounds++;
            scores = scores.filter((x)=> 100 >= x && x > 0);
            let n_bots = scores.length;
            if (n_bots == 1) {
                this.last_choice = 100;
                return 100;
            }
            let c = 1 - 0.4/n_bots
            for (let i = 0; i < n_bots; i++) {
                this.histogram_bins[Math.round(scores[i])]++;
            }
            //THE SIMP DETECTOR
            let simps = [];
            for (let i = 0; i < 101; i++) {
                if (this.histogram_bins[i] > this.n_rounds) {
                    simps.push(i);
                }
            }

            //Idk this shouldn't happen but I'm not risking anything
            while (scores.length - simps.length - 1 <= 0) simps.pop();

            if (this.n_rounds == 1) {
                this.mark = Array(201);
                for (let i = 0; i < 201; i++) {
                    this.mark[i] = Array(201);
                    for (let j = 0; j < 201; j++) {
                        this.mark[i][j] = 0;
                    }
                }
                for (let i = 0; i < 201; i++) {
                    this.mark[i][i]++;
                }
            }

            let simp_corrected_avg = (sum(scores) - this.last_choice - sum(simps)) / (scores.length - 1 - simps.length);
            let quantized_avg = Math.round(2*simp_corrected_avg);
            this.mark[Math.round(2*this.last_SCA)][quantized_avg]++;
            this.last_SCA = simp_corrected_avg;

            if (this.n_rounds == 1) {
                this.last_choice = 77;
                return 77;
            }
            
            this.linear_history.push(simp_corrected_avg);

            // Update weights based on who was right
            let linear_error  = Math.exp(-Math.abs(this.linear_prediction  - simp_corrected_avg));
            let mark_error    = Math.exp(-Math.abs(this.mark_prediction    - simp_corrected_avg));
            let copycat_error = Math.exp(-Math.abs(this.copycat_prediction - simp_corrected_avg));
            let total_errors = linear_error + mark_error + copycat_error;
            this.linear_weight += linear_error / total_errors;
            this.mark_weight += mark_error / total_errors;
            this.copycat_weight += copycat_error / total_errors;

            this.linear_weight  *= 0.99;
            this.mark_weight    *= 0.99;
            this.copycat_weight *= 0.99;
            
            // Compute a markov chain prediction
            let x_half_filter = 8;
            let y_half_filter = 1;
            
            let probability_sum = 0;
            let probability_moment = 0;

            for (let option = 1; option < 201; option++) {
                for (let x = -x_half_filter; x <= x_half_filter; x++) {
                    for (let y = -y_half_filter; y <= y_half_filter; y++) {
                        let X = Math.max(Math.min(quantized_avg + x, 200), 0);
                        let Y = Math.max(Math.min(option + y, 200), 0);
                        let probability = this.mark[X][Y] * (Math.pow(2, -(X-quantized_avg)*(X-quantized_avg)-(Y-option)*(Y-option)));
                        probability_sum += probability;
                        probability_moment += option / 2 * probability;
                    }
                }
            }

            this.mark_prediction = probability_moment / probability_sum;
            
            for (let i = 0; i < 201; i++) {
                for (let j = 0; j < 201; j++) {
                    this.mark[i][j] *= 0.999
                }
            }

            // Compute a linear regression
            if (this.linear_history.length > 3) {
                if (this.linear_history.length > 10) {
                    this.linear_history.shift();
                }
                let x_avg = (this.linear_history.length - 1) / 2
                let y_avg = average(this.linear_history);
                let S_xx = 0;
                let S_xy = 0;
                for (let x = 0; x < this.linear_history.length; x++) {
                    S_xx += (x - x_avg) ** 2;
                    S_xy += (x - x_avg) * (this.linear_history[x] - y_avg);
                }
                let b = S_xy / S_xx
                let a = y_avg - b * x_avg;
                this.linear_prediction = a + b / this.linear_history.length;
            } else {
                this.linear_prediction = simp_corrected_avg;
            }

            // Compute what everyone else does
            let counts = [];
            for (let radius = 1; Math.max(counts) < 0.05 * scores.length; radius++) {
                for (let i = 50; i <= 90; i++) {
                    counts[i] = 0;
                    for (let j = 0; j < n_bots; j++) if (Math.abs(scores[j] - i) <= radius) counts[i]++;
                }
            }
            this.copycat_prediction = (counts.indexOf(Math.max(counts.slice(50))) * c - 50) / 0.4 * n_bots/(n_bots-1);

            let expected_unsimp_average = (
                this.mark_prediction * this.mark_weight + 
                this.linear_prediction * this.linear_weight + 
                this.copycat_prediction * this.copycat_weight
            ) / (this.mark_weight + this.linear_weight + this.copycat_weight);

            let expected_average = (expected_unsimp_average * (n_bots - 1 - simps.length) + sum(simps)) / (n_bots - 1);

            // We don't want to give out bad values, now do we
            this.last_choice = Math.max(1, Math.min(100, (50 + 0.4 * expected_average * (n_bots-1)/n_bots)/c));
            return this.last_choice;
        }
    }
```

Answer 23

40, 40 bytes

At least I get 2400 (40*60) points each round ... right?

{
    name:"40",
    run() {return 40}
}

This is my first js program in my life

Answer 24

Everything so far

{
  name: "Everything so far",
  moves: [],
  run(scores){
    if(scores) {
      this.moves = this.moves.concat(scores);
      return this.moves.reduce((a,b)=>a+b)/this.moves.length;
    } 
    return 80;
  }
}

Takes the average of every move so far.

Answer 25

LetsMakeADeal

I don't want a B in CyberSecurity, so I still want to be in the 90th percentile range. Game theory looks like some people are willing to pick 100 and soak up the difference; others are gravitating to 90 as the optimal balance; so I figured why not try a "Let's Make A Deal" strategy and add 1 to their target.

{
    name:"LetsMakeADeal",
    run() {return 90 + 1}
}

Answer 26

Squidward

This is a fun one. I love xkcd. Taking a look at the function, the maximum score will be given by the average between 100 and the last's score's average. The tricky thing would be predicting what the new average will be like.

I almost never use JS so bear with me here. I tried to test it but I'm not even sure if this will run :/

{
  name : "Squidward",
  log : [],
  delta : [],
    
  run(scores) {
    if(scores){
      let average_score = sum(scores) / scores.length;
      this.log.push(average_score)
      if(this.log.length > 1){ 
        this.delta.push(this.log[this.log.length-1]-this.log[this.log.length - 2])
        let average_delta = sum(this.delta) / this.delta.length;
        return (100 + average_score*.8 + average_delta)/2
      } else if(this.log.length == 1){
        return (100 + average_score*.8)/2
      }
    } else {
      return 90
    }
  }
}

Answer 27

Fourier

Inspired by Linear Extrapolator's approach to extrapolation and the existence of The Thrasher, as well as other potential sources of feedback.

This bot computes a few high-frequency Fourier components of the sequence of averages to try to predict the next average. Note that what it computes is different from the canonical discrete Fourier transform in that instead of using harmonics of the lowest frequency, it uses the first few subharmonics of the sampling rate. This is because we only really care about the major oscillations, not slow trends.

I considered adding some more complicated logic to reduce some deleterious effects, especially early on, but ultimately decided to apply the KISS principle and leave it out.

{
  name: "Fourier",
  sin: [0,0,0,0,0,0],
  cos: [0,0,0,0,0,0],
  round: 0,
  
  run(scores) {
    let ave = average(scores)
    
    for (let i = 0; i < 6; i++) {
      this.sin[i] += ave * Math.sin(2*Math.PI * this.round / (i+1))
      this.cos[i] += ave * Math.cos(2*Math.PI * this.round / (i+1))
    }
    
    this.round++
    
    let next = this.sin.reduce((t,a,i) => t + a * Math.sin(2*Math.PI * this.round / (i+1)),0)
             + this.cos.reduce((t,a,i) => t + a * Math.cos(2*Math.PI * this.round / (i+1)),0)
    
    return 50 + 2/5 * Math.max(next,0) / this.round
  }
}

Answer 28

iDontWannaFail

This bot hates failing classes, so it minimizes the deviation between the average and the returned score so the Game Theory score is as large as possible. The deviation is randomly selected from -5 to 6 to allow some room for variation. However, if the score is below a 70, a random integer between 70 and 100 is chosen. If the average is below 60, a number that is within 2/3 of the average score is selected to ensure a score that is at least 50 (Our little bot is rather disappointed that its goal wasn't reached, though, but hey, a 50 is better than a 10!)

{
    name: "iDontWannaFail",
    run: (scores) => {
        let sum = scores.reduce((a, b) => a + b, 0);
        let avg = (sum / scores.length) || 0;
        let output = avg * 0.8 + Math.round(Math.random()*11)-5;
        if (avg > 60) {
            if (output < 70) {
                output = Math.floor(Math.random() * 30) + 70;
            }
        }
        else {
            if (avg > 25) {
                output = Math.round(Math.random() * avg / 1.5) + avg; // Worst case is 49.
            }
            else {
                output = Math.round(3.75 * avg); // Decent results unless avg is very low
            }
        }
        if (output > 100) {
            output = 100;
        }
        return output;
    }
}

Edit: The else code failed, setting avg=10 returns, at best, 16, resulting in 38 as the final score. Another nested if has been added to solve this issue (avg=10 returns 52)

Answer 29

The Fat(a)

This algorithm entirely depends on the following lines:

let sign = var_drift>0 ? 1 : -1;
this.velocity_pred = 0.3*(sign*average(this.all_v)*2-velocity);
a = a+this.velocity_pred;

Where var_drift is the immediate difference in variance of all the submissions in the current round n, minus the variance of the last round. velocity is the immediate difference in the mean of all the submissions in the current round n, minus the mean of the last round. all_v is the absolute value all the velocities so far. The average all all true velocities turns out to be near 0 as the mean oscillates.

The average velocity is used to approximate the magnitude of how much the movement will be. The average velocity - the immediate last velocity works with the following intuition:

If the last velocity is large positive, then there is higher chance that the next movement will be actually negative (To correct the average), and vice versa.

important note

The numerical solver is largely obsolete and can be replaced with the simple formula

x = 50 + 0.4*a.

The derivative for the scoring function \$\sqrt{x*(100-\lvert0.8*a-x\rvert}\$ is this guy here(according to https://www.derivative-calculator.net/):

\$\dfrac{-\left|x-\frac{4a}{5}\right|-\frac{x\left(x-\frac{4a}{5}\right)}{\left|x-\frac{4a}{5}\right|}+100}{2\sqrt{x\left(100-\left|x-\frac{4a}{5}\right|\right)}}\$ (eq.1)

While I originally thought at first glance that this will be impossible to reduce, I decided to use an iterative solver to find the root instead. But upon reflecting on this and testing emprically that the root for (eq.1) is the same as found with the simple formula \$ x = 50+0.4*a \$. I realized that this entire thing can be easily reduced to be the same as \$x=50+0.4*a\$ if we are looking for an x larger than \$ 4*a/5 \$. The root for the best x lower than \$ 0.8a\$ is the root of \$2x^2 -\frac{508}{5}*x - \frac{2004}{25}a\$. Which by quadratic formula simplifies to

\$x = -101.6 \pm \frac{\sqrt{10322.56 + 641.28a}}{4} \$

Because of our assumption that x must be < a, the only viable form of this function is

\$x = -101.6 - \frac{\sqrt{10322.56 + 641.28a}}{4} \$

And gives a massive negative number...

And just to complete this analysis when \$x = \frac{4*a}{5}\$ we have the funny equation

\$ 0 = \frac{100}{2* \sqrt{x*(100-\lvert x-\frac{4*a}{5}\rvert)}} \$

This in fact does have a "solution", when

\$ \lim_{2* \sqrt{x*(100-\lvert x-\frac{4*a}{5}\rvert)} \to \inf}\$.

Which is just our original scoring function *2. This again points to a extremely large negative number "solution" \$\lim_{x \to -\inf} \$. But a "solution also exists for when both x and a are extremely large,but a and x are almost equal to each other (\$ \lim_{x \to \inf a \to \inf} x-\frac{4*a}{5} <100\$).

Update: Cleaned up code, rewritten numerical solution finder.

 {
                   //============\\
    name: "Fat",  //  []    []   \\
    last_avg:70,  //     w        \\
    last_var:0,   //    ---       \\
    n:0,        // 00 0 00 00 00 0\\
    all_v:[],   // ||||||||||||||\\
    run(scores){      
      function calc_variance(scores){return Math.sqrt(sum(scores.map(
        function(x){return Math.pow(x - average(scores),2)/(scores.length-1)})));};      
      function find_min(a){let d = 0;let c = 0;let x = 0;let inc =10;
      let num =0;while(true){c = (x+inc)-(4*a)/5;
      d = (-Math.abs(c) - ((x+inc)*c)/Math.abs(c) + 100)/2*Math.sqrt((x+inc)*(100-Math.abs(c)));
      if((d<0)){inc/=10;}else{x+=inc;};num+=1;
      if(Math.abs(x)<=0.01 || num>10000){break;}}return x;};
      let a = average(scores);
      if(this.n<2){a = this.n==0 ? 66:71.5;this.all_v.push(2.5);
      }else{
        let velocity = average(scores) - this.last_avg;
        this.all_v.push(Math.abs(velocity));
        let var_drift = calc_variance(scores) - this.last_var;
        let sign = var_drift>0 ? 1 : -1;
        this.velocity_pred = 0.3*(sign*average(this.all_v)*2-velocity);
        a = a+this.velocity_pred;
      }
      this.last_var = calc_variance(scores);
      this.last_avg = average(scores);
      this.n +=1;
      //return 50+a*0.4;
      return find_min(a);
    }
},

Answer 30

Chasing Ninety

Attempts to iteratively approach 90% of the average of previous rounds. Responds poorly to large variances, but attempts to slowly approach the ideal value.

{
    name: "Chasing Ninety",
    myLastGuess: 84,
    numRounds: 0,
    run(scores) {
        this.numRounds++;
        if (this.numRounds <= 1) {
            return this.myLastGuess;
        }

        let avg = average(scores);
        let difference = this.myLastGuess - (avg * 0.9);
        let adjustment = difference / Math.log2(this.numRounds);
        this.myLastGuess = this.myLastGuess - adjustment;
        return this.myLastGuess;
    }
}