Analysis

The question posed gives a (big) hint to using modular arithmetic to assess this problem. A sample of the results is therefore taken and visualised in mod 3.

Sample first 100 results in modulo 3

# Sample the first 100 meetings to visualise progress
track_vals = list(tracker.values())
track_vals = track_vals[:100]

# Create a list of each colour in mod 3
r = []
b = []
g = []
for i in range(len(track_vals)):
    r.append(list(track_vals)[i]['R'] %3)
    b.append(list(track_vals)[i]['B'] %3)
    g.append(list(track_vals)[i]['G'] %3)

# Plot graph of population change in mod 3
import pylab as plt
from matplotlib.pyplot import figure
%matplotlib inline

figure(figsize=(25, 4), dpi=100)
plt.rcParams.update({'font.size': 15})

plt.plot(range(len(b)), b, 'b')
plt.plot(range(len(r)), r, 'r')
plt.plot(range(len(g)), g, 'g')
plt.title('Total Quantity of Salamanders in mod 3')

Text(0.5, 1.0, 'Total Quantity of Salamanders in mod 3')

Modulus importance

# Prepare some data for explanation
import pandas as pd
meet_s = [['-2','+1','+1']]
meet_s_mod = [['+1','+1','+1']]
meet_d = [['-1','-1','+2']]
meet_d_mod = [['+2','+2','+2']]

We observe that the red, blue, and green numbers are always different, and hold either a value of 0, 1, or 2 in mod3. This is important, as for there to be 30 Red salamanders, there need to be 0 Blue and 0 Green (total population is 30). In mod3, this would be equivalent to 0R, 0B, and 0G. In other words, for there to be all Red salamanders, there needs to be a combination of meetings such that all colours reach 0 (mod3). In this small sample, we can see that the values of each are always different in mod 3. Why is this?

The starting position of the population is 15R, 7B, and 8G. In mod3, this equates to 0R, 1B and 2G. Upon two salamanders of the same colour, x, meeting, we get a drop in 2 of that colour, and an increase of 1 for the other two colours, y and z:

pd.DataFrame(meet_s, ['xx'], ['x', 'y', 'z'])

In mod3, this is equivalent to:

pd.DataFrame(meet_s_mod, ['xx'], ['x', 'y', 'z'])

We see that for whichever colour, if the salamanders are the same colour, the same mathematical addition applies to all colours in mod3, such that there is no convergence between colours.
Two salamanders of different colour meeting results in:

pd.DataFrame(meet_d, ['xy'], ['x', 'y', 'z'])

In mod3, this is rewritten:

pd.DataFrame(meet_d_mod, ['xy'], ['x', 'y', 'z'])

Again, where salamander colours are different, there is no convergence between colours in mod3.

This exhausts all meeting possibilities, and shows there is no possibility of convergence between quantities of each colour in mod3. With this being the case, it is impossible for all to reach 0 (mod3). This means that there can never be 30 Red salamanders.

However, 29R is possible, with 0B and 1G. This maintains the count structure in mod3 as this would be 2R, 0B, 1G (mod3).

Total Reds

Max Reds

# Show how the number of reds changes over trials
r_vals = []
for trial in tracker.values():
    r_vals.append(trial['R'])

graph_len = np.min([250,len(r_vals)])
mov_max = int(np.ceil(len(r_vals)/graph_len))

red_mov_max = []
for i in range(graph_len):
    red_mov_max.append(np.max(r_vals[i*mov_max:(i+1)*mov_max]))

figure(figsize=(25, 4))
plt.plot(range(graph_len), red_mov_max, 'r')
plt.title('Max Quantity of Red Salamanders every ' + str(mov_max) + ' trials')

Text(0.5, 1.0, 'Max Quantity of Red Salamanders every 4000 trials')

We observe that even over 1 million trials, the maximum number of Red salamanders never reaches 29. This suggests that whilst 29R is a possibility, it is highly unlikely to occur through the random sampling used.

Frequency of Red count

# Count frequency of Reds quantities over the trials
import seaborn as sns

figure(figsize=(18, 7))
sns.set_style('darkgrid')
sns.histplot(r_vals, color='r')
plt.title('Histogram of Total Quantity of Red Salamanders')

Text(0.5, 1.0, 'Histogram of Total Quantity of Red Salamanders')

We observe that the histogram shows a bell-like curve of distribution. As may be expected, the modal number of Reds is 10, or 1/3 of the total population. This reflects that with more Reds present in the population, there is a higher probability of a Red being selected and therefore the number of Reds being reduced. The opposite can be observed below this level, and a similar graph would be expected for Blues and Greens.
We can see that the graph tails off drastically above 20R - if we were to assume that the number of Reds is approximately normally distributed, we could estimate the probability of getting the maximum number of Reds (29).

from scipy.stats import norm
mean = np.mean(r_vals)
std = np.std(r_vals)

norm.pdf(29, loc=mean, scale=std)

4.750575739333807e-12

This result suggests that, as a rough figure, even if we simulated 210 billion meetings, there would still be about a 37% chance (1/$e$) we would not reach the maximum of 29 Reds at any point!

NB: This is only if assuming a normal distribution, which the bounded data does not strictly fit.

Optimising the algorithm

Initially, we used a random choice of 2 salamanders at each meeting. However, it may be possible to optimise this process to reach 29R far quicker. If we only allow for meetings that increase the number of Reds, i.e. apply a gradient descent optimisation, we should reach our target in far fewer iterations.

# Reset population to the original starting point
pop = ['R']*15 + ['B']*7 + ['G']*8
pop = np.array(pop) # Convert to numpy array

# Set max number of iterations to 1000
iters = 1000

r_vals = []
for i in tqdm(range(iters)):
    # Simulate a meeting
    meeting(gd = True) # Set `gd = True` for gradient descent
    # Save resulting population state
    counter = Counter(pop)
    r_vals.append(counter['R'])
    # Stop if 29R is reached
    if counter['R'] == 29: break

  0%|          | 0/1000 [00:00

# Show how the number of reds changes over trials
figure(figsize=(18, 7))
plt.plot(range(len(r_vals)), r_vals, 'r')
plt.title('Total Quantity of Red Salamanders (Optimised Algorithm)')

Text(0.5, 1.0, 'Total Quantity of Red Salamanders (Optimised Algorithm)')