Posted on December 26, 2016
In this post we will revisit an example from the previous post Rearranging Digits with a brute force coding solution, and expand with an another example (analytically) that is trivial with with a brute force solution. The two examples are:
In the previous post we walked through a somewhat informed, yet clumsy and incomplete analytic solution that end with a less than satisfying guess and check (not to say there aren’t elegant solutions out there, I didn’t quite get there though).
This can be remedied quite easily by brute forcing our way through each of the $9^3$ possibilities for $A$, $B$, and $C$. It would look something like this in MATLAB
function [] = three_digit_with_fors()
for a=1:9
for b=1:9
for c=1:9
three_digit_geometric([a, b, c])
end
end
end
end %function
This calls the function three_digit_geometric:
function [] = three_digit_geometric(final)
tens = [100 10 1];
N1=dot(final([1 2 3]), tens);
N2=dot(final([2 3 1]), tens);
N3=dot(final([3 1 2]), tens);
if N2/N1==N3/N2 && N3~=N2
fprintf('%d, %d, %d -> ', final);
fprintf('%d, %d, %d, r=%g \n', N1, N2, N3, N3/N2);
end
end
If we wanted to generalize to higher digit numbers, the for loops would get out of hand pretty quickly, especially in with Python with forced indentation (good luck with PEP 8). To alleviate this (at least for our eyes), let’s generate a recursive function call to put these for loops in.
Now, if you’ve been around the block a bit, you know that python has itertools.permutations and MATLAB has perms – and you could use those. If you do, be wary of edge cases! The next problem is a rearrangement where some of the digits can be 0, but others are leading digits and therefore cannot. Tread lightly.
Also, perms in MATLAB appears to return the permutations in a large array (as opposed to the memory friendly generator for itertools.permutations in Python). Also, perms is recursively implemented so it’s quite easy to reach max recursive depth. Just try perms([1:n] for a large enough value of n (11 does it for me on 32-bit MATLAB 2011).
Back to our recursion. Let’s call the function rfor (short for recursive for). When given an arbitrary number of arrays, it will recursively loop through them. This will have the same shortcomings that come with any recursive algorithm, without even writing it one can posit that it will be slower and will risk reaching max recursive depth for large problems.
All of that aside, here’s what rfor would look like in MATLAB
function [] = rfor (current, final_func, level, varargin)
a=varargin;
for i=1:length(a{1})
current(level) = a{1}(i);
if length(a)>1
rfor (current, final_func, level+1, a{2:length(a)})
else
feval(final_func, current);
end
end
end
If you’re wondering whether or not time is saved by tracking the level of recursion instead of dynamically changing the size of the array, a quick benchmark showed a $>10\%$ speedup over the assignment current = [current a{1}(i)];
An example use would be
a=1:9; b=1:9; c=1:9; rfor([], 'three_digit_geometric', 1, a, b, c);
The output is
4, 3, 2 -> 432, 324, 243, r=0.75 8, 6, 4 -> 864, 648, 486, r=0.75
In case you aren't familiar with it, varargin is a way to pass a variable number of parameters to a MATLAB function. If the length is 1, we are at the lowest level of recursion required.
Python has an simple implementation as well.
def three_digit_geometric(final):
tens = [100, 10, 1]
n1 = sum([tens[i]*final[i] for i in range(len(final))])
n2 = 10 * (n1 % 100) + n1 // 100
n3 = 10 * (n2 % 100) + n2 // 100
if n2 / float(n1) == n3 / float(n2) and n3 != n2:
print n1, n2, n3, '->', final, 'r=%g' % (n3 / float(n2))
def rfor(current, final_func, level, args):
for i in args[0]:
current[level] = i
if len(args) == 1:
globals()[final_func](current)
else:
rfor(current, final_func, level+1, args[1:])
def main():
a, b, c = [range(1, 10)] * 3
rfor([0] * 3, 'three_digit_geometric', 0, [a, b, c])
As a last task, let’s solve one more fun challenge: The four six-digit numbers $ABCDEF$, $CDEFAB$, $FABCDE$, and $DEFABC$ for an arithmetic sequence. Find all possible values of ($A$, $B$, $C$, $D$, $E$, $F$).
This could probably be solved analytically. To avoid 3 equations and 6 unknowns, you would want to treat $AB$ and $DE$ as individual variables, reducing the problem to 4 unknowns (and knowledge of which should be 1-digit a 2-digit). This will be left to the reader as an exercise ;)
Here’s the code for the brute force solution (note how some digits are allowed to be zero but not others).
a=1:9; b=0:9; c=1:9; d=1:9; e=0:9; f=1:9;
rfor([], 'six_digit', a, b, c, d, e, f);
function [] = six_digit(final)
tens = [100000 10000 1000 100 10 1];
N1 = dot(final(1:6), tens);
N2 = dot(final([3:6 1:2]), tens);
N3 = dot(final([6 1:5]), tens);
N4 = dot(final([4:6 1:3]), tens);
if N2-N1==N3-N2 && N4-N3==N3-N2 && N3>N2
fprintf('%d, %d, %d, %d, %d, %d -> ', final);
fprintf('%d, %d, %d, %d, d=%d \n', N1, N2, N3, N4, N3-N2);
end
end
Here’s the output
1, 0, 2, 5, 6, 4 -> 102564, 256410, 410256, 564102, d=153846 1, 5, 3, 8, 4, 6 -> 153846, 384615, 615384, 846153, d=230769 1, 6, 2, 3, 9, 3 -> 162393, 239316, 316239, 393162, d=76923 2, 1, 3, 6, 7, 5 -> 213675, 367521, 521367, 675213, d=153846 2, 6, 4, 9, 5, 7 -> 264957, 495726, 726495, 957264, d=230769 2, 7, 3, 5, 0, 4 -> 273504, 350427, 427350, 504273, d=76923 3, 2, 4, 7, 8, 6 -> 324786, 478632, 632478, 786324, d=153846 3, 8, 4, 6, 1, 5 -> 384615, 461538, 538461, 615384, d=76923 4, 3, 5, 8, 9, 7 -> 435897, 589743, 743589, 897435, d=153846 4, 9, 5, 7, 2, 6 -> 495726, 572649, 649572, 726495, d=76923 6, 0, 6, 8, 3, 7 -> 606837, 683760, 760683, 837606, d=76923 7, 1, 7, 9, 4, 8 -> 717948, 794871, 871794, 948717, d=76923
So there are $12$ solutions.
Again if you are wondering where these came from, look no further than repeating decimals for fractions with $117$ (one of the many divisors of $999, 999$) in the denominator.
Posted on September 22, 2016
This post will include analytic solutions, a future post will show some code-based solutions.
Let's attack the first problem, repeated for clarity:
The 6-digit number $ABCDEF$ is exactly $23/76$ times the number $BCDEFA$ (where $A$, $B$, $C$, $D$, $E$, and $F$ represent single-digit non-negative integers and $A,B>0$). What is $ABCDEF?$
We will let $x=BCDEF$ be a five-digit number. Then
\begin{eqnarray*} ABCDEF &=& \frac{23}{76}\cdot BCDEFA \\ 100000\cdot A+x &=& \frac{23}{76} (10\cdot x+A) \\ 7600000\cdot A+76\cdot x &=& 230\cdot x+23\cdot A \\ 7600000\cdot A+76\cdot x &=& 230\cdot x+23\cdot A \\ 7599977\cdot A &=& 154\cdot x \\ 98701\cdot A &=& 2\cdot x \end{eqnarray*}Remember, $x$ is a five-digit integer and $A$ is a one-digit number, so one such solution is $A=2$ and $x=98701$. Are there other solutions though? Well, $A$ must be even for $x$ to be an integer, but if $A>2$ then $x$ would be a six-digit number. So there is just the one solution. As a check, this makes our numbers $ABCDEF=298701$ and $BCDEFA=987012$.
(Side note: if you are wondering where these seemingly random six-digit numbers came from, check out the repeating decimals $\frac{23}{77}$ and $\frac{76}{77}$. Fractions whose denominators are factors of $999,999$ will create repeating decimals with cycles of six digits or less and quite often use the same digits in different order.
Now let's look at the second problem, repeated for clarity: Find all ordered pairs of distinct single-digit positive integers $(A,B,C)$ such that the three 3-digit positive integers $ABC$, $BCA$, and $CAB$ form an arithmetic progression.
We will approach this the same way, let's define the 3 numbers \begin{eqnarray*} N_1&=&100A+10B+C \\ N_2&=&100B+10C+A \\ N_3&=&100C+10A+B \end{eqnarray*}
These terms are in arithmetic progression, which can be exploited. Let the common difference of this sequence be $d$, then \begin{eqnarray*} N_2-N_1&=& 100B+10C+A-(100A+10B+C)=d\\ N_3-N_2&=& 100C+10A+B-(100B+10C+A)= d\\ N_3-N_1&=& 100C+10A+B -(100A+10B+C)= 2d \end{eqnarray*}
After simplifying, we have \begin{eqnarray*} -99A +90B +~9C - d=0\\ ~9A -99B +90C - d =0\\ -90A ~-9B +99C - 2d =0 \end{eqnarray*}
We have a system with three equations and four unknowns, which should make you a little anxious (and the third equation is just the sum of the first two). However, if we simplify a little bit (one could do this by hand, but rref in MATLAB is our friend)
Using rref on the coefficient matrix gives us
1 0 -1 0.021021 0 1 -1 0.012012 0 0 0 0
We need integer solutions, so where do we go from here? Take a look at the two decimals, $0.\overline{021}$ and $0.\overline{012}$. The fraction equivalents are $\displaystyle\frac{7}{333}$ and $\displaystyle\frac{4}{333}$ respectively.
The equations are then \begin{eqnarray*} a-c+\frac{7}{333}d=0 \\ b-c+\frac{4}{333}d=0 \end{eqnarray*}
So if we need integers, let's get rid of the fraction by letting $d=333$. Then we have $a-c=-7$, and $b-c=-4$. This gives us the solutions $(a,b,c)=(1,4,8)$ and $(2,5,9)$. Did we get them all though? We can also let $d=-333$, which yields $a-c=7$, and $b-c=4$. This gives us the solutions $(a,b,c)=(8,5,1)$ and $(9,6,2)$.
We could try multiples of $333$ as well, but will quickly run into trouble. If $d=666$, then $a-c=14$, and $b-c=8$ but $a$ and $c$ must be single digit numbers and larger values of $d$ will violate this. We can also let $d=0$ but that just gives us the trivial solutions of an arithmetic progression with common difference $0$.
Therefore our four solutions are \begin{eqnarray*} 148,~481,~814 \\ 185,~518,~851 \\ 259,~592,~925 \\ 296,~629,~962 \\ \end{eqnarray*}
If you are wondering if these numbers came out of nowhere, check out the decimal equivalents of the $27$ths. It is no coincidence that $\displaystyle\frac{4}{27}=0.\overline{148}$, $\displaystyle\frac{13}{27}=0.\overline{481}$, and $\displaystyle\frac{22}{27}=0.\overline{814}$.
Now let's look at the last problem, again repeated for clarity: Find all ordered pairs of distinct single-digit positive integers $(A,B,C)$ such that the three 3-digit positive integers $ABC$, $BCA$, and $CAB$ form a geometric progression.
For the geomtric sequence \begin{eqnarray*} N_2-r\cdot N_1&=& 100B+10C+A-r(100A+10B+C)= 0\\ N_3-r\cdot N_2&=& 100C+10A+B-r(100B+10C+A)= 0 \end{eqnarray*}
There are other manipulations, but I haven't found any that are linearly independent of the two equations shown.
Not only do we have two equations and four unknowns, the equations are non-linear. With some simplification, we have \begin{eqnarray*} 100B+10C+A-r(100A+10B+C)= 0\\ 100C+10A+B-r(100B+10C+A)= 0 \end{eqnarray*}
Using the symbolic toolbox and rref we can simplify
% a b c [ 1, 0, (r - 10)/(10*r^2 - 1)] [ 0, 1, (r*(r - 10))/(10*r^2 - 1)]
Not incredibly helpful. However, we can reverse the columns of the array (now instead of A,B,C we have C,B,A) and rref gives us
% c b a [ 1, 0, (10*r^2 - 1)/(r - 10)] [ 0, 1, -r]
Which is a little cleaner. So where do we go from here? Well, from the last equation we have $b-ra=0 \rightarrow r=b/a$ from the second equation. So we could just try every $(a,b)$ combination ($72$ since $a\neq b$) and see if $$c=\displaystyle\frac{10r^2 - 1}{10 - r}a$$ is an integer. Still some brute force required though. You could also substitute $r=b/a$ into $c=\displaystyle\frac{10r^2 - 1}{10 - r}a$ yielding $$ c= \displaystyle\frac{10b^2-a^2}{10a - b} $$ which we just have to ensure is an integer. But again, it looks like guessing and checking is required. You could even simplify further with some polynomial division $$ c = \displaystyle\frac{10b^2-a^2}{10a - b} = \displaystyle\frac{999a^2}{10a - b} - 10b - 100a$$ which takes you down a path of finding $a$ and $b$ such that $10a - b$ is a divisor of $999a^2$, but that may even be more dangerous than guess and checking since it's so easy to leave out a case and every case must be checked to ensure $c$ is a positive single digit integer.
You can though, and it will lead you to two ordered triples, $(a,b,c)=(4,3,2)$ and $(8,6,4)$ (the second is just a less exciting multiple of the first, both have the common ratio $r=\frac{3}{4}$. The geometric series are $432,~324,~243$ and $864,~648,~486$.
This last solution left me feeling less than satisfied. In the next post we will do a code-based solution to ensure with some certainty that those are the only values.