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Verifying boardgame mathematics with MATLAB
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verifyingmatlabwithboardgamemathematics
Problem
I recently bought this math-based boardgame for my little cousins, and am now trying to verify its design with some MATLAB code. The relevant parts of the game are:
Now, the part I had doubts about was the coloring of the tiles - were they really colored
according to the difficulty of arriving at them? That seemed easy enough to verify - find the frequency of occurrence of each number from 1 to 100 given three dice and four mathematical operations - and so that's what this MATLAB code attempts to do:
```
function freq = compute_freq
idx = 1;
for x = 1:6
for y = 1:6
two_die_combos = get_combos(x, y);
for z = 1:6
three_die_combos = get_combos(two_die_combos, z);
valid_combos = three_die_combos(three_die_combos > 0 & ...
three_die_combos <= 100 & ...
(three_die_combos == floor(three_die_combos))); % only integer values
all_combos(idx:idx + length(valid_combos) - 1) = valid_combos;
idx = idx+length(valid_combos);
end
end
end
freq = tabulate(all_combos);
end
function combos = get_combos(a, b)
num_
- A board with numbers from 1 to 100 in one tile each, and each tile colored in one of nine colors
- There are three dice, the usual kind (1-6)
- In your turn, you roll all three dice. You have three numbers x, y and z now, from the dice. Now, using these three numbers, and applying the four basic math operations on them (+, -, , /) in any order, you must arrive at one of the numbers on the board (i.e. from 1 to 100). The catch is, the resulting numbers have scores based on their color on the board, so that for eg. arriving at 13 or 26 (red tile) gives you a much higher score compared to arriving at 25 (pink). So if you die values are 5, 5, 1, you're much better off going with
(55+1) = 26 than with(551) = 25. Choosing these operations right in each turn is the crux of the game.
Now, the part I had doubts about was the coloring of the tiles - were they really colored
according to the difficulty of arriving at them? That seemed easy enough to verify - find the frequency of occurrence of each number from 1 to 100 given three dice and four mathematical operations - and so that's what this MATLAB code attempts to do:
```
function freq = compute_freq
idx = 1;
for x = 1:6
for y = 1:6
two_die_combos = get_combos(x, y);
for z = 1:6
three_die_combos = get_combos(two_die_combos, z);
valid_combos = three_die_combos(three_die_combos > 0 & ...
three_die_combos <= 100 & ...
(three_die_combos == floor(three_die_combos))); % only integer values
all_combos(idx:idx + length(valid_combos) - 1) = valid_combos;
idx = idx+length(valid_combos);
end
end
end
freq = tabulate(all_combos);
end
function combos = get_combos(a, b)
num_
Solution
Discussion
You have three dice, which you can assume to hold three places for
The total number of possible combinations would be
Looking at your code, I don't think
Again,
Code
Considering all the assumptions and avoiding the "scandalous" cases, we would have this vectorized code -
Output
Observation
To confirm, you can create another column to the "tabulated" one and mark different weights for each number from 1 to 100 based on their colors. Then, see if such a pattern continues across the entire board. Good luck and hope this would make sense!
You have three dice, which you can assume to hold three places for
1, 2, 3, 4, 5, 6 and the four operations +, -, *, / would hold the two in-between places, which are shown as square brackets and round brackets respectively as shown in this blueprint - [] () [] () [].The total number of possible combinations would be
64646 = 3456. Out of these, you need to find the valid ones based on flooring and values in the range [1 100].Looking at your code, I don't think
-x+y would be a valid one, because you can't just pre-append a operator. If that is to be allowed, why not multiplication operator too? Moreover -x essentially means -1x, for which you are already using two operators - - and and assuming one of y or z to be 1. For these reasons, I am guesstimating, -x+y to be an invalid one, unless it says in the instructions manual to be a valid one.Again,
x\y won't induce anything new, as we are already juggling through all values of 1 to 6 across all three places around the four operators, as shown in the blueprint earlier.Code
Considering all the assumptions and avoiding the "scandalous" cases, we would have this vectorized code -
%// Define possible values of x, y and z, denoting values from three dice
x = 1:6;
y = 1:6;
z = 1:6;
%// t1 would take care of x and y and the one operator in between
t1 = [bsxfun(@plus,x',y) ; bsxfun(@minus,x',y) ; bsxfun(@times,x,y') ; bsxfun(@rdivide,x,y')]
%// t2 would take care of x, y and z and the two operators in between
t2 = [bsxfun(@plus,t1(:),z) ; bsxfun(@minus,t1(:),z) ; bsxfun(@times,t1(:),z) ; bsxfun(@rdivide,t1(:),z)]
%// Get the valid ones based on flooring and limiting within the range [1 100]
t2 = t2(t2==floor(t2) & t2=1)
freq = tabulate(t2)Output
freq =
1.0000 151.0000 7.6186
2.0000 157.0000 7.9213
3.0000 146.0000 7.3663
4.0000 148.0000 7.4672
5.0000 130.0000 6.5590
6.0000 150.0000 7.5681
7.0000 94.0000 4.7427
8.0000 106.0000 5.3481
9.0000 87.0000 4.3895
10.0000 89.0000 4.4904
11.0000 53.0000 2.6741
12.0000 91.0000 4.5913
13.0000 35.0000 1.7659
14.0000 37.0000 1.8668
15.0000 46.0000 2.3209
16.0000 36.0000 1.8163
17.0000 14.0000 0.7064
18.0000 41.0000 2.0686
19.0000 10.0000 0.5045
20.0000 34.0000 1.7154
21.0000 16.0000 0.8073
22.0000 10.0000 0.5045
23.0000 7.0000 0.3532
24.0000 42.0000 2.1191
25.0000 16.0000 0.8073
26.0000 7.0000 0.3532
27.0000 10.0000 0.5045
28.0000 11.0000 0.5550
29.0000 5.0000 0.2523
30.0000 33.0000 1.6650
......Observation
freq percentage value for 26 [red colored] is 0.3532 that is slightly lower than for 25 [pink colored] which has 0.8073. Similarly, for 13 [red colored] it is 1.7659, that is again slightly lower than for 15 [looks like pink colored], which has 2.3209.To confirm, you can create another column to the "tabulated" one and mark different weights for each number from 1 to 100 based on their colors. Then, see if such a pattern continues across the entire board. Good luck and hope this would make sense!
Code Snippets
%// Define possible values of x, y and z, denoting values from three dice
x = 1:6;
y = 1:6;
z = 1:6;
%// t1 would take care of x and y and the one operator in between
t1 = [bsxfun(@plus,x',y) ; bsxfun(@minus,x',y) ; bsxfun(@times,x,y') ; bsxfun(@rdivide,x,y')]
%// t2 would take care of x, y and z and the two operators in between
t2 = [bsxfun(@plus,t1(:),z) ; bsxfun(@minus,t1(:),z) ; bsxfun(@times,t1(:),z) ; bsxfun(@rdivide,t1(:),z)]
%// Get the valid ones based on flooring and limiting within the range [1 100]
t2 = t2(t2==floor(t2) & t2<=100 & t2>=1)
freq = tabulate(t2)freq =
1.0000 151.0000 7.6186
2.0000 157.0000 7.9213
3.0000 146.0000 7.3663
4.0000 148.0000 7.4672
5.0000 130.0000 6.5590
6.0000 150.0000 7.5681
7.0000 94.0000 4.7427
8.0000 106.0000 5.3481
9.0000 87.0000 4.3895
10.0000 89.0000 4.4904
11.0000 53.0000 2.6741
12.0000 91.0000 4.5913
13.0000 35.0000 1.7659
14.0000 37.0000 1.8668
15.0000 46.0000 2.3209
16.0000 36.0000 1.8163
17.0000 14.0000 0.7064
18.0000 41.0000 2.0686
19.0000 10.0000 0.5045
20.0000 34.0000 1.7154
21.0000 16.0000 0.8073
22.0000 10.0000 0.5045
23.0000 7.0000 0.3532
24.0000 42.0000 2.1191
25.0000 16.0000 0.8073
26.0000 7.0000 0.3532
27.0000 10.0000 0.5045
28.0000 11.0000 0.5550
29.0000 5.0000 0.2523
30.0000 33.0000 1.6650
......Context
StackExchange Code Review Q#54810, answer score: 4
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