## Ryan Thompson › Perl Weekly Review: Challenge - 043

Saturday, Jan 25, 2020| Tags: Perl Continues from previous week.

Feel free to submit a merge request or open a ticket if you found any issues with this post. We highly appreciate and welcome your feedback. You can also contact me (Ryan) directly, with any feedback on this review.

For a quick overview, go through the original tasks and recap of the weekly challenge.

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# Task #1 - Olympic Rings

The task here was to fill in the numbers 1, 2, 3, 4, and 6 into the spaces within the intersecting Olympic rings, such that the numbers in each ring sum to 11. (See the diagram.)

The problem can be solved easily enough by hand, yet still presents an interesting programming challenge that can be tackled in many different ways.

Adam’s solution this week is a Perl wrapper around a Prolog clause using AI::Prolog. It’s satisfyingly elegant.

The Prolog clause is as follows:

``````member(X,[X|T]).
member(X,[H|T]) :- member(X,T).
colors(Red, Green, Black, Yellow, Blue) :-
member(Blue, [1,2,3,4,6]),
member(Yellow, [1,2,3,4,6]),
member(Green, [1,2,3,4,6]),
member(Red, [1,2,3,4,6]),
member(Black, [1,2,3,4,6]),
R = 11, R is 9 + Red,
G = 11, G is 5 + Red + Green,
B = 11, B is 8 + Blue,
Y = 11, Y is 7 + Blue + Yellow,
Bl = 11, Bl is Green + Yellow + Black.
``````

And that clause (stored in `\$prolog`) is called in Perl like so:

``````\$prolog = new AI::Prolog(\$prolog);
\$prolog->query("colors(Red, Green, Black, Yellow, Blue).");
``````

Prolog is a great choice for this type of problem, as the solution can be simply defined as a system of equations. Prolog then understands these equations as a constraint-based problem, and abstracts away all of the complexity.

## Alicia Bielsa

Alicia’s solution sets up a hash, `%hRingsComposition`, which maps each ring to the segments that must be sum to 11 inside that ring:

``````my %hRingsComposition = (
'RED' =>  [ 'RED', 'RED_GREEN'],
'GREEN' =>  [ 'GREEN', 'RED_GREEN',  'GREEN_BLACK'],
'YELLOW' => [ 'YELLOW', 'BLUE_YELLOW', 'YELLOW_BLACK'],
'BLUE' => [ 'BLUE', 'BLUE_YELLOW'],
'BLACK' =>[ 'BLACK',  'GREEN_BLACK','YELLOW_BLACK']
);
``````

Alicia’s algorithm is efficient. It takes advantage of the fact that there is only one solution, and it can be found by simply looking for a ring with only one unknown, and filling in that unknown from one of the available values, repeating until the problem is solved. Here is the main loop:

``````my \$countAvailableNumbers = scalar @aAvailableNumbers;
while ( \$countAvailableNumbers ){
foreach my \$ringColor (keys %hRingsComposition){

my \$singleEmptyValue = findSingleEmptyValue(\$ringColor);
if (\$singleEmptyValue){

my \$currentRingSum = getCurrentRingSum(\$ringColor);
my \$difference =  \$TOTALSUM - \$currentRingSum;
if (checkValueAvailability(\$difference)){
\$hColorValue{\$singleEmptyValue} =\$difference;
} else {
die "ERROR: Value  '\$difference' not available, imposible to resolve\n";
}
}
}
\$countAvailableNumbers = scalar @aAvailableNumbers;
}
``````

Her program finds the solution in just 10 total iterations of the inner `foreach` loop. Nice!

## Andrezgz

Andrezgz’s solution followed a strategy similar to Alicia’s, but managed to do so in a concise way, by writing a program highly specialized to this particular puzzle. First, there is the `@rings_components` array, which sets up the segments contained within each ring:

``````my @rings_components = (
['Red','RedGreen'],
['Green','RedGreen','GreenBlack'],
['Blue','YellowBlue'],
['Yellow','YellowBlue','BlackYellow'],
['GreenBlack','BlackYellow','Black']
);
``````

The main loop finds the solution (`%values` contains the known (given) values):

``````foreach my \$ring (@rings_components) {
my \$internal_sum = 11;
foreach my \$c (@\$ring){
\$values{\$c} = \$internal_sum unless defined \$values{\$c};
\$internal_sum -= \$values{\$c};
}
}
``````

Andrezgz defines, but does not even need, a hash of the available numbers, 1, 2, 3, 4, and 6.

## Burkhard “Chuck” Nickels

Chuck’s solution sets up a `while` loop and a sliding window, which looks at three contiguous ring compartments at a time. If there is only one unknown in that group, that unknown can be safely calculated. Here is the loop:

``````my \$nrx;                             # Number of x in array
do {
numbers();
\$nrx = () = join("",@a) =~ /x/g; # Determine Nr of x
} while(\$nrx);                       # if Nr of x
``````

The `numbers()` routine looks at the sliding window and does the math to fill in the missing value if there is only one unknown:

``````sub numbers {
for(my \$i=0; \$i<=\$#a; \$i+=2) {  # Only every second value is a Main Ring Value

my @win = @a[\$i-1 .. \$i+1]; # Create window of 3 numbers that shifts along array
\$win = 0 if \$i==0;       # At begin, set window first element to zero
\$win = 0 if \$i==\$#a;     # At end,   set window last  element to zero

my \$nr = () = join("",@win) =~ /x/g;  # How many x are in window
print "(\$i) Win(", join("/",@win), ") Nrx \$nr\n" if \$DEBUG;
if(\$nr == 1) {                     # Only if 1 value is missing, determine next value
if(\$win eq 'x') { \$a[\$i] = 11 - \$win - \$win; }
elsif(\$win eq 'x') { \$a[\$i-1] = 11 - \$win - \$win; }
elsif(\$win eq 'x') { \$a[\$i+1] = 11 - \$win - \$win; }
}
}
}
``````

I’m impressed by this solution. It’s unique, and the sliding window makes short work of this problem. Be sure to check out Chuck’s POD blog, which goes into more detail.

## Colin Crain

Colin’s solution recognizes the fact that this puzzle can be represented as a system of linear equations. Colin derives the linear algebra equations by hand, and then uses Math::MatrixReal to do the actual matrix multiplication:

``````my \$a = Math::MatrixReal->new_from_string(<<MATRIX);
[ 1 0 0 0 0 ]
[ 1 1 0 0 0 ]
[ 0 1 1 1 0 ]
[ 0 0 0 1 1 ]
[ 0 0 0 0 1 ]
MATRIX

my \$b = Math::MatrixReal->new_from_string(<<MATRIX);
[ 3  ]
[ 4  ]
[ 11 ]
[ 6  ]
[ 2  ]
MATRIX

my \$LR = \$a->decompose_LR();
my (\$dim, \$out, \$base) = \$LR->solve_LR(\$b);
``````

Using this method, the code is very simple. The real magic is in realizing that this puzzle could be reduced to a system of linear equations in the first place!

## Cristina Heredia

Cristina’s solution provides another interesting variation on the direct calculation method we’ve seen previously. She repeats the following code for each of the rings:

``````@black = \$b;
@black = \$c;
\$q = @total - @black - @black;
@black = \$q;
validate(\$q);
print "\nValors of black:\n";
for (my \$j = 0; \$j < @black; \$j++) {
print "@black[\$j],";
}
print "\n";
``````

And the `validate()` sub simply has to check whether the remaining number, `11 - \$black - \$black`, is one of the givens. If so, the solution is valid:

``````sub validate {
\$valor = shift;
my \$valide;
for (my \$j = 0; \$j < @values; \$j++) {
if (@values[\$j] == \$valor) {
\$valide = "KO";
}
else {
\$valide = "OK";
last;
}
}
if (\$validate == KO) {
print "The number \$valor is one of the given.";
}
else {
print "The number \$valor isn't one of the given.";
}
}
``````

Stylistically, I would prefer to see `use warnings` and `use strict`, and perhaps a bit of refactoring to reduce the duplication of code. However, Cristina succeeded in the more difficult task of delivering a program that works, is efficient, and is easy to understand.

## Dave Jacoby

Dave’s solution uses some new Perl features again this week, such as the postfix deref introduced in 5.20, and signatures. Dave’s code goes through every permutation of the numbers (1, 2, 3, 4, 6) and stops if it finds one where all of the rings sum to 11. The `permute_array` function is what actually generates the permutations:

``````sub permute_array ( \$array ) {
return \$array if scalar \$array->@* == 1;
my @response = map {
my \$i        = \$_;
my \$d        = \$array->[\$i];
my \$copy->@* = \$array->@*;
splice \$copy->@*, \$i, 1;
my @out = map { unshift \$_->@*, \$d; \$_ } permute_array(\$copy);
@out
} 0 .. scalar \$array->@* - 1;
return @response;
}
``````

In terms of computational complexity, Dave’s algorithm is O(n!) on the size of the number list. In this case, there are 5! = 120 ways of arranging the numbers. While some solutions are considerably more efficient, Dave’s is general, elegant, and the factorial isn’t a problem for small lists.

## Duane Powell

Duane’s solution also looks at all the permutations of (1, 2, 3, 4, 6), but the key difference is Duane uses Math::Combinatorics to do the heavy lifting:

``````my @num = (1,2,3,4,6);

my \$c = Math::Combinatorics->new(count => 1, data => [@num]);
while (my @perm = \$c->next_permutation) {
next unless red(\$perm)                     == \$eleven;
next unless green(\$perm,\$perm)          == \$eleven;
next unless black(\$perm,\$perm,\$perm) == \$eleven;
next unless yellow(\$perm,\$perm)         == \$eleven;
next unless blue(\$perm)                    == \$eleven;

# a solution found if we made it here
say join(',',@perm);
}
``````

Whether or not to use non-core modules in these challenges is a matter of personal preference, and I, for one, love seeing both approaches when I look at the solutions every week.

## E. Choroba

Choroba’s first solution again uses `Math::Combinatorics`. However, in his blog, he notes that he saw it could be solved much more easily.

The second solution is therefore another take on the constant-time difference approach we’ve seen. As is typical of Choroba’s solutions, it is beautifully concise:

``````my (\$red, \$green, \$yellow, \$blue) = (9, 5, 7, 8);

my \$red_green    = 11 - \$red;
my \$blue_yellow  = 11 - \$blue;
my \$black_green  = 11 - \$green - \$red_green;
my \$black_yellow = 11 - \$yellow - \$blue_yellow;
my \$black        = 11 - \$black_green - \$black_yellow;

say join ' ', \$red_green, \$black_green, \$black, \$black_yellow, \$blue_yellow;)
``````

Choroba’s blog explains the math very well.

## Jaldhar H. Vyas

Jaldhar’s solution has its own `permute` sub:

``````sub permute (&@) {
my \$code = shift;
my @idx = 0..\$#_;
while ( \$code->(@_[@idx]) ) {
my \$p = \$#idx;
--\$p while \$idx[\$p-1] > \$idx[\$p];
my \$q = \$p or return;
push @idx, reverse splice @idx, \$p;
++\$q while \$idx[\$p-1] > \$idx[\$q];
@idx[\$p-1,\$q]=@idx[\$q,\$p-1];
}
}
``````

This one takes a code ref as a callback, and a list of numbers. Jaldhar then uses it like this:

``````my @permutations;
permute { push @permutations, \@_; } @numbers;
for my \$permutation (@permutations) { ... }
``````

I’m not sure why Jaldhar implemented a callback, only to `push` the values into an array to iterate over on the next line of code, but it doesn’t matter much in this case, since the list of `@permutations` is small.

The internals of the loop mostly do the arithmetic to calculate the value of each ring, and then this check is performed:

``````    if (scalar (grep { \$ringValues{\$_} == 11 } keys %ringValues) == 5) {
map { say "\$_ = \$try{\$_}"; } @unknowns;
last;
}
``````

`%ringValues` contains the sums inside each ring, so he’s checking to see if all five rings sum to 11, and then outputting the values and exiting the loop early.

## Javier Luque

Javier’s solution uses Algorithm::Combinatorics for another permutation-based approach:

``````my \$iter = permutations(\@numbers);
while (my \$p = \$iter->next) {
my \$slots = {
redgreen     => \$p->,
greenblack   => \$p->,
black        => \$p->,
blackyellow  => \$p->,
yellowblue   => \$p->
};

for my \$key (keys %\$slots) {
say 'Slot: ' . \$key .
' value: ' . \$slots->{\$key};
}
}
}
``````

`validate_answer()` is essentially the following conditional:

``````    ( \$s->{redgreen} + \$r->{red} == 11 &&
\$s->{redgreen} + \$r->{green} + \$s->{greenblack} == 11 &&
\$s->{greenblack} + \$s->{black} + \$s->{blackyellow} == 11 &&
\$s->{blackyellow} + \$r->{yellow} + \$s->{yellowblue} == 11 &&
\$r->{blue} + \$s->{yellowblue} == 11 );
``````

Javier’s blog

## Laurent Rosenfeld

Laurent’s solution takes a deductive approach to the problem. First, the red and blue rings each have only one unknown, so those numbers can be calculated directly. Once those are filled in, the green and yellow rings now have one unknown, and finally, the black ring gets the final number.

The (short) nested loops that figure out the red-green and blue-yellow intersections provide a good summary of Laurent’s approach:

``````my @ring_sequences = ( [qw <red green>], [qw <blue yellow>]
my @black_vals;

for my \$seq_ref (@ring_sequences) {
my \$diff = 0;
for my \$ring (@\$seq_ref) {
\$rings{\$ring} += \$diff;
say "Added \$diff to \$ring ring, " if \$diff;
\$diff = TARGET - \$rings{\$ring};
die "No way" unless exists \$nums{\$diff};
say "Added \$diff to \$ring ring";
\$rings{\$ring} += \$diff;
}
\$rings{black} += \$diff;
push @black_vals, \$diff;
}
``````

As always, Laurent provides solid, readable, idiomatic Perl. What’s funny is that Laurent’s code is very similar to how I solved this by hand, even though I felt like using backtracking in my Perl solution instead.

Laurent’s blog goes into more detail into the numerical analysis and algorithm design.

## Roger Bell West

Roger’s solution uses a 6-nested loop, one loop for each of the rings, and a final loop over the numbers (1, 2, 3, 4, 6), trying each number in each slot. After computing the sums in each ring, he uses List::MoreUtils`minmax` to check whether all sums are 11:

``````my @l=minmax(@sums);
if (\$l==\$target && \$l==\$target) {
print join(', ',map {"\$candidate[\$_] in \$metanames[\$index[\$_]]"} (0..\$#candidate)),"\n";
}
``````

Complexity-wise, Roger’s algorithm runs in O(m(2n)^5) time, where m is the number of candidate numbers (1, 2, 3, 4, 6), and n is the number of givens. For this problem, his algorithm will require over 200,000 operations. It still runs in about a quarter of a second, and the advantage of this approach is that the inner loop almost writes itself:

``````            my \$ix=int(\$index[\$i]/2);
\$sums[\$ix]+=\$candidate[\$i];
if (\$index[\$i]%2==1) {
\$sums[\$ix+1]+=\$candidate[\$i];
}
``````

In other words, when you just want the answer to a problem, sometimes computational complexity is less important than simply doing the first thing that might work.

## Ruben Westerberg

Ruben’s solution is another example of using techniques from linear algebra (specifically, Gauss-Jordan Elimination or row reduction). Colin Crain used a similar algorithm, but Ruben also implemented all of the matrix operations himself, without the help of any non-core modules.

The `solve` sub is a dense mix of array primitives and arithmetic:

``````sub solve {
my (\$c, \$y)=@_;
my \$j=build(\$c,\$y);
my \$s=@\$c;
my \$row;
for my \$p (0..\$s-1) {
my \$v=\$j->[\$p][\$p];
my \$prow=\$j->[\$p];
for my \$r (0..\$s-1) {
\$row=\$j->[\$r];
next if \$p==\$r;
my \$d=\$row->[\$p]/\$v;
for my \$col (0..\$s) {
\$row->[\$col]-=\$d*\$prow->[\$col];
}
}
}
\$row->[-1]/=\$row->[-2];
\$row->[-2]=1;
backsub(\$j);
}
``````

I really enjoyed going through Ruben’s implementation. If you, like me, have any doubts about your ability to sit down and write a Gauss-Jordan implementation without cracking a linear algebra textbook, it’s worth looking at Ruben’s code.

## Ryan Thompson

My own solution went a little overboard. I had already solved the problem by staring at it for a minute, so I knew it would not be a computationally difficult problem. I decided on a rather generic recursive backtracking approach.

I actually focused most of my efforts on a terminal-based animated visualizer for how backtracking algorithms work, without using any non-core modules. You can see a screenshot at the top of this page. Click the link to the solution, above, if you want to see how the visualizer works. Since that code is both lengthy and tangential to the challenge, I’ll focus on the actual solver algorithm, here.

In my backtracking approach, the base case relies on `check_sol`, which returns `solved` if the puzzle is solved, `impossible` if any of the sums do not equal 11, and `possible` otherwise. I had originally used a simpler binary `is_solved`, but having the ternary return value reduced the number of iterations from 113 (not much better than checking all 5! permutations) down to just 16.

``````sub solve {
my (%sol) = @_;
my \$check = check_sol(\%sol);
return %sol if \$check eq 'solved';
return      if \$check eq 'impossible';

# Get list of numbers still available
my %solR = reverse %sol; # keys <-> values
my @rem  = grep { not exists \$solR{\$_} } @avail;

my \$spot = first { \$sol{\$_} == 0 } @order_try;
for my \$num (@rem) {
my %new = solve(%sol, \$spot => \$num);
return %new if keys %new; # Pass back solution
}

return;
}
``````

Thanks to being able to prune early, `solve` is only ever called 16 times. Although this problem’s search space is tiny, backtracking is a powerful tool that can efficiently solve a surprising family of problems. I think I might have been the only one to explicitly use backtracking, which either makes me clever or foolish.

Olympic rings blog

## Saif Ahmed (saiftynet)

Saif’s solution iterates all of the rings in a `while` loop until the puzzle is solved. Any time there is only one unknown in a ring, Saif knows the value must be 11 minus the sum of the other number(s) in that ring. A funny comment also highlights how trivial this problem can be: “this function solves in one pass if we fix in the sequence blue red green yellow black but that would be cheating”.

Saif’s, I think, is the only other Perl solution besides mine that attempts to display the results in something resembling their actual position:

``````sub displayRings{
printf  (
"     RED %s                 BLACK %s                     BLUE %s
RedGrn %s    GrnBlk %s     BlkYel %s      YelBlu %s
GREEN %s                  YELLOW %s\n",
@list[0,4,8,1,3,5,7,2,6]
);

}
``````

The output looks like this:

``````Final state:-
RED 9                 BLACK 6                     BLUE 8
RedGrn 2    GrnBlk 4     BlkYel 1      YelBlu 3
GREEN 5                  YELLOW 7
``````

## wanderdoc

wanderdoc’s solution is another that uses `Algorithm::Combinatorics` to iterate over all permutations of the available numbers 1, 2, 3, 4, and 6.

``````while (my \$i = \$iter->next())
{
@var{@col2search} = @\$i;

next unless ( all { is_valid(\$_) } @olympic );
print join(' => ', \$_, \$var{\$_}), \$/ for @col2search;
}
``````

wanderdoc’s `ring_sum` function is interesting:

``````sub ring_sum
{
my @ring = @{\$_};
my \$sum = reduce { ('SCALAR' eq ref \$a ? \$\$a : \$a) + ('SCALAR' eq ref \$b ? \$\$b : \$b) } @ring;
return \$sum;
}
``````

This usage of `reduce` is a typical way to implement `sum`. Of course, with List::Util‘s `sum`, this could be written:

``````sum map { 'SCALAR' eq ref \$_ ? \$\$_ : \$_ } @ring;   # Or even...
sum map { ref ? \$\$_ : \$_ } @ring;
``````

Still, I like seeing `reduce` getting some love. It’s extremely powerful.

# Task #2 - Self-Descriptive Numbers

This week’s second task is to generate self-descriptive numbers in arbitrary bases. A self-descriptive number, as described by Wikipedia, is an integer m that in a given base b is b digits long in which each digit d at position n (the most significant digit being at position 0 and the least significant at position b – 1) counts how many instances of digit n are in m.

## Methods

Solutions to this challenge differed mainly on how much of the math behind self-descriptive numbers people chose to take advantage of. Most solutions used the fact that bases 7 and above have an easy formula:

``````(b - 4)b^(b-1) + 2b^(b-2) + b^(b-3) + b^3
``````

This constant-time formula does not work for bases < 7, so solutions here tended to be either hard-code the answers, or search for self-descriptive numbers. Roger Bell West and Saif Ahmed both used an interesting iterative method which munges an n-digit number to rapidly converge to a self-describing number.

Another difference in solutions is whether they output the results in base 10, or the given base, or both. As the challenge description’s only example was in base 10, there was ambiguity, so both output formats would seem to be equally valid.

Adam’s solution iterates through all non-negative integers below 100,000,000, checking if each one is self-describing, using the following function:

``````sub self_describing{
my(\$i) = @_;
my @digits = split(//, \$i);
for my \$x (0 .. @digits - 1){
my \$count = 0;
for my \$j (0 .. @digits - 1){
\$count++ if(\$digits[\$j] == \$x);
return false if(\$count > \$digits[\$x]);
}
return false if(\$count != \$digits[\$x]);
}
return true;
}
``````

I appreciate Adam’s direct approach to the problem. Adam’s blog mentions his code simply follows from the definition of a self-describing number.

## Andrezgz

Andrezgz’s solution uses the base ≥ 7 formula, supporting bases up to 35. Andrezgz hard-codes in the bases which are known to have no self-descriptive numbers, and then searches the others:

``````die "No self-descriptive numbers in base \$base" if (\$base <4 || \$base == 6);

if (\$base >= 7) {
my @symbols = (0..9,'A'..'Z');
print \$symbols[\$base - 4] . '21' . '0' x (\$base - 7) . '1000' . \$/;
}
else{
my \$from = '1' . '0' x (\$base-1);
my \$to = '9' x \$base;

for my \$n (\$from .. \$to) {
my @count = (0) x \$base;
\$count[\$_]++ for split //, \$n;
print \$n.\$/ if (\$n eq join '',@count[0..\$base-1]);
}
}
``````

## Burkhard “Chuck” Nickels

Chuck’s solution uses the base ≥ 7 formula for all bases, starting at 4. The formula does give the first self-descriptive number, 1210, but misses the second, 2020, and third, 21200. This would be easy enough to fix, however, by searching (Chuck did implement a `verify` function), or by simply hard-coding those non-generic cases.

What is interesting about Chuck’s solution, though, is his handling of bases both during the computation, and when outputting the results. Chuck noticed that bases 4..14 could be handled with decimal numbers, while bases 13..17 could be handled in hexadecimal, and bases greater than 15 could be handled by a custom `0..9, 'a'..'z'` base-35 converter that he wrote. (Yes, these ranges do overlap, but that will not negatively affect the output):

``````my \$base   = join("",@r);
my \$is_sdn;
if(\$b > 3  and \$b < 15) { \$is_sdn = verify_dec(\$base); }
if(\$b > 12 and \$b < 18) { \$is_sdn = verify_hex(\$base); }
if(\$b > 15) { \$is_sdn = verify_str(\$base); }
``````

Chuck therefore has written three different `verify_*` routines. Here is one of them:

``````sub verify_str {
my (\$sdn) = @_;
my \$i = 0;
my \$is_sdn = 1;
foreach my \$v (split(//,\$sdn)) {
my \$nr;
my \$search = \$C[\$i];
\$nr = () = \$sdn =~ m/\$search/g;
\$nr = \$C[\$nr];
if( \$v ne \$nr ) {
\$is_sdn = 0;
return \$is_sdn;
}
\$i++;
}
return \$is_sdn;
}
``````

A `convert` routine was also necessary, in order to convert between bases (note this is running under `bigint`):

``````sub convert {
my (\$b,\$n,\$erg) = @_;
my \$d = int(\$n / \$b);
my \$r = \$n % \$b;
if(\$r > 15) {
# print "Value > 15: \$r \$C[\$r]\n";
\$r = \$C[\$r];
}
elsif(\$r > 9) { \$r = sprintf("%x", \$r); }
unshift(@\$erg,\$r);
convert(\$b,\$d,\$erg) if \$d;
}
``````

Chuck’s blog (POD version) provides a refreshingly candid description of the journey towards the eventual solution.

## Colin Crain

Colin’s solution only handles bases ≥ 7. I like the additional thought Colin put in to the formula’s implementation, however. While most of us just implemented the formula numerically, or adapted it to use string concatenation, Colin showcased both methods:

``````sub self_descriptive {
## formula for creating self-descriptive numbers in base 10 for a given base ( > 7 )
my \$base = shift;
my \$dec = (\$base-4)*(\$base**(\$base-1)) + 2*(\$base**(\$base-2)) + \$base**(\$base-3) + \$base**3;
my @alphanum = (0..9, 'A'..'Z');
my \$out = "";
my \$rem;
while ( \$dec > 0 ) {
(\$dec, \$rem) = (int( \$dec/\$base ), \$dec % \$base);
\$out = \$alphanum[\$rem] . \$out;
}
return \$out;
}
``````
``````sub self_descriptive_assembled {
## or we can just assemble a graphical representation of a number manually that will fit the bill
my \$base = shift;
my @alphanum = (0..9, 'A'..'Z');
my \$out = \$alphanum[\$base-4] . "21" . "0" x (\$base-7) . "1000";
return \$out;
}
``````

The string method is more concise, and gives results directly in the target base, while the pure numerical method gives its answer in base-10, which may be desirable.

## Cristina Heredia

Cristina’s solution handles bases 7..10 in both decimal and the original base, using the numeric formula:

``````use strict;
use Math::Base::Convert qw(:base);

my \$valor;
for (my \$b = 7; \$b <= 10; \$b++) {
\$valor = (\$b - 4) * \$b**(\$b - 1) + 2 * \$b ** (\$b-2) + \$b**(\$b-3) + \$b**3;

\$converted = cnv(\$valor,10 => \$b);
print "Base is \$b, and the result is \$converted\n";
}
``````

While this is very close to being correct, the code does not quite run as-is. `\$converted` needs to be declared under `strict`, and the `cnv` function from Math::Base::Convert is not exported with `:base`.

Furthermore, `Math::Base::Convert`'s `cnv` says it supports “arbitrary bases,” yet throws an error for base 7. I looked up the error in the module source, and it seems you need to supply your own character set if not using one of the six standard bases it supports. With all this in mind, I would suggest the following minimal fix to Cristina’s code (I hope you don’t mind, Cristina!):

``````# Ryan's suggested fix
use strict;
use Math::Base::Convert qw(cnv);

for (my \$b = 7; \$b <= 10; \$b++) {
my \$valor = (\$b - 4) * \$b**(\$b - 1) + 2 * \$b ** (\$b-2) + \$b**(\$b-3) + \$b**3;

my \$converted = cnv(\$valor,10 => [0..\$b-1]);
print "Base is \$b, and the result is \$converted\n";
}
``````

## Dave Jacoby

Dave’s solution generates all self-descriptive numbers from base 7..37 using the string concatenation variant of the formula. He also takes care to verify that the numbers are indeed self-descriptive.

Here is Dave’s implementation of the formula:

``````sub get_self( \$n ) {
my @output = map { 0 } 1 .. \$n;
my \$b      = \$n - 4;
\$output  = \$from_base{\$b};
\$output  = 2;
\$output  = 1;
\$output[\$b] = 1;
return join '', @output;
}
``````

The `check_self` sub validates self-descriptive numbers by going through all of the digits.

``````sub check_self ( \$s, \$n ) {
no warnings;
my @s = split //, \$s;
my \$b = \$s;
my @check;
for my \$i ( 0 .. \$n - 1 ) {
my \$eye = \$from_base{\$i};
my \$c = \$s[\$i];
my @all = grep { \$_ eq \$eye } @s;
my \$all = join ',', @all;
my \$j   = scalar @all;
my \$jay = \$from_base{\$j};
return 0 if \$c ne \$jay;
}
return 1;
}
``````

Dave’s variable names made me smile. I used to work with someone who also used `\$i, \$j, \$eye,` and `\$jay`. His name was Dave, too.

## Duane Powell

Duane’s solution works for bases 1..10, by brute forcing the search space, and checking if each number is self-descriptive, using the following function:

``````sub SDN {
my \$n    = shift;
my \$base = shift;

my @n = split(//,\$n);                       # Split \$n into separate digits
return 0 unless (scalar @n == \$base);       # A SND is the same length as its base

my %count;
\$count{\$_} = 0 foreach (0 .. scalar(@n)-1); # Init a counter to all 0's
\$count{\$_}++   foreach (@n);                # Count the occurance of each digit

# Determine if \$n "describes" itself by comparing
# the count to the digit found at index \$i
my \$i = 0;
foreach (0 .. scalar(@n)-1) {
return 0 if (\$count{\$_} != \$n[\$i]); # not a SDN, exit
\$i++;
}
return 1; # All digits matched the counts, this is an SDN
}
``````

Duane admits this is very slow for larger bases. One optimization that Duane did that some people missed is this line:

``````  return 0 unless (scalar @n == \$base);       # A SND is the same length as its base
``````

Duane is checking whether the number is a Niven number (or Harshad number), which you might remember from week 007 of the Perl Weekly Challenge. It can be proven that all self-descriptive numbers are also Niven numbers.

## E. Choroba

Choroba’s solution uses a table of values for bases < 7. For bases ≥ 7, Choroba takes a somewhat novel approach:

``````return \$convert->(join "",
map \$_ > 9 ? chr 55 + \$_ : \$_,
(\$base - 4, 2, 1, (0) x (\$base - 7), 1, (0) x 3))
``````

The use of `join` to handle the concatenation streamlines the code somewhat. That bit of ASCII math (`chr 55 + \$_`) is a cheeky way of expanding the character set for bases greater than 9. I like it.

`\$convert` is a code ref that either returns its argument unmodified, or sends it through Convert::AnyBase‘s `decode()` method. Thus, Choroba’s `self_descriptive_number()` sub can either return the result in the original base, or be `decode()`'d to a regular Perl number:

``````my \$convert = @_ == 1
? sub { \$_ }
: sub { 'Convert::AnyBase'->new(set => \$set)->decode(\$_) };
``````

Instead of returning an empty list or `undef` if there are no self-descriptive numbers for the base, Choroba opted to use an exception model instead:

``````    if (exists \$irregular{\$base}) {
die "No self descriptive number in base \$base.\n"
unless \$irregular{\$base};

return \$convert->(\$irregular{\$base})
}
``````

One easily fixed issue is that the `%irregular` hash only lists one of the two base-4 self-descriptive numbers (1210; the other is 2020):

``````my %irregular = (
1 => undef,
2 => undef,
3 => undef,
4 => 1210,
5 => 21200,
6 => undef,
);
``````

That’s easily remedied, though. Finally, Choroba provides a comprehensive set of tests, which is always nice to see. Testing exceptions is easy with Test::Exception:

``````throws_ok { self_descriptive_number(1) } qr/base 1/, 'base 1';
throws_ok { self_descriptive_number(2) } qr/base 2/, 'base 2';
throws_ok { self_descriptive_number(3) } qr/base 3/, 'base 3';
throws_ok { self_descriptive_number(6) } qr/base 6/, 'base 6';
``````

## Jaldhar H. Vyas

Jaldhar’s solution uses the numeric formula, and hard-coded list of bases to ignore. However, the formula does not work for bases smaller than 7, so the results for bases 4 and 5 are incorrect:

``````[weekly/rjt] challenge-043/⋯/perl %> ./ch-2.pl 4
1210
[weekly/rjt] challenge-043/⋯/perl %> ./ch-2.pl 5
13100
``````

The expected results would be 1210, 2020, and 21200. This would be easily fixed, however, by either searching bases 4 and 5, or hard-coding the expected results for those non-generic cases.

Jaldhar uses the numeric formula to obtain the self-descriptive numbers, but opted to output his results in the target base, so he provided a conversion function:

``````sub base {
my (\$number, \$base) = @_;
my @digits = (0 .. 9, 'A' .. 'Z');
my @result;
while (\$number > (\$base - 1)) {
my \$digit = \$number % \$base;
push @result, \$digits[\$digit];
\$number /=  \$base;
}
push @result, \$digits[\$number];

return join '', reverse @result;
}
``````

## Javier Luque

Javier’s solution uses the base ≥ 7 definition, and has hard-coded answers for lesser bases. He shows us two similar functions: one that returns the results in base 10, and another that returns the results in the target base. I will list the target base version, here:

``````sub self_descriptive_x {
my \$b = shift;

return 'no solution' if
( \$b == 1 || \$b == 2 ||
\$b == 3 || \$b == 6);

return 1210 if (\$b == 4);
return 21200 if (\$b == 5);

return
\$NUMS[(\$b - 4)] . '2' . 1 . 0 x (\$b - 7) . '1000';
}
``````

Javier’s solution is fast. Hard-coding the bases < 7 means the solution has O(1) complexity.

There is one small bug, though: In base-4, there are actually two self-descriptive numbers: 1210 and 2020 (Happy New Year!). That’s an easy fix, though.

## Laurent Rosenfeld

Note: I’m using Laurent’s solution from his blog, as it appears to be farther along than the one currently checked in to the repository. By the way, his blog—as always—is well worth the read.

Laurent’s solution hard-codes the invalid bases, uses the numeric base ≥ 7 for larger bases, and checks all cases for bases 4 and 5.

``````sub find_self_descriptive {
my \$b = shift;
return "No self-descriptive number for base \$b"
if \$b < 4 or \$b == 6;
if (\$b == 4 or \$b == 5) {
return check_all_cases (\$b);
}
my \$dec_num = (\$b - 4) * \$b ** (\$b - 1)
+ 2 * \$b ** (\$b - 2) + \$b ** (\$b - 3) + \$b ** 3;
my \$base_num = to_base_b (\$dec_num, \$b);
return "Number in base \$b: \$base_num; decimal: \$dec_num";
}
``````

The `check_all_cases()` sub handles the iteration through every possible number of a given base, skips numbers that don’t end in zero, skips non-Niven numbers, and finally loops through the digits to check whether each digit describes its count:

``````sub check_all_cases {
my \$base = shift;;
for my \$num (\$base ** (\$base -1) .. \$base ** \$base -1) {
my \$num_in_b = to_base_b (\$num, \$base);
next unless \$num_in_b =~ /0\$/;
my @digits = split //, \$num_in_b;
my \$sum = 0;
\$sum += \$_ for split //, \$num_in_b;
next if \$sum != \$base;
my \$success = 1;
for my \$rank (0..\$base - 1) {
my \$nb_digits = \$digits[\$rank];
my \$num_occ = \$num_in_b =~ s/\$rank/\$rank/g;
if (\$num_occ != \$nb_digits) {
\$success = 0;
last;
}
}
return "Number in base \$base: \$num_in_b; decimal: \$num" if \$success;
}
}
``````

Hard-coding the invalid bases (especially base 6) provides a significant performance boost.

Laurent’s blog

## Roger Bell West

Roger’s solution also hard-codes the invalid bases, and uses a different adaptation of the base ≥ 7 formula, by keeping the digits in an array:

``````my @n=(0) x \$base;

\$n=1;

if (\$base>6) {
\$n=\$base-4;
\$n=2;
\$n=1;
\$n[\$base-4]=1
}
``````

Now see if you can figure out how the following loop works, at a glance:

``````while (1) {
my @o=@n;
my %o;
map {\$o{\$_}++} @o;
foreach my \$i (0..\$#o) {
\$n[\$i]=\$o{\$i} || 0;
}
if (join('',@o) eq join('',@n)) {
last;
}
}
``````

The `map` and `foreach` lines concisely count the digits, so the `join`s can check if a number is self-descriptive. What is much less obvious, though, is how this loop actually mutates `@n` so it will try new numbers each time through.

The `\$n[\$i]=\$o{\$i} || 0` line modifies `@n`, but the `%o` hash is dependent on `@o`, which itself was a copy of `@n` at the top of the loop. It starts to make a lot more sense when you look at the actual values being produced. For base-5, this results in the following interesting sequence:

``````1 0 0 0 0
4 1 0 0 0
3 1 0 0 1
2 2 0 1 0
2 1 2 0 0
``````

Roger is thus iteratively working his way towards a self-descriptive number by replacing each digit with its count from the current iteration. This needs to be repeated, but it converges quickly on the answer. Very nice!

One problem with Roger’s method is with the base-4 numbers; it only finds the first one, because the loop exits as soon as it finds the first number. The `last` has to be there, though; otherwise the loop would get stuck on 1210 forever. This could be easily worked around in any number of ways, such as simply hard-coding the result for base-4 or having a special case to search that small base via brute force.

## Ruben Westerberg

Ruben’s solution exhaustively searches all base-n numbers and checks whether they are self-descriptive or not.

I did not think I would see signal handling when reviewing the solutions to this challenge. Ruben hooks `\$SIG{INT}` to display all self-descriptive numbers found before exiting, so you can let the program run, press `^C` and then see the results:

``````:
NO: 3545165
NO: 3545166
NO: 3545200
^CStopping Search
Found Self Describing Numbers:
3211000
[weekly/rjt] challenge-043/⋯/perl %>
``````

Here is the main loop, and the `\$SIG{INT}` handler:

``````my \$run=1;
my @found;
\$SIG{INT}=sub {print "Stopping Search\n"; \$run=undef};
while (\$run) {
my \$res=test(\$base,\$num);
if (\$res) {
print "***OK: \$num***\n";
push @found,\$num;
sleep 1;
}
else {
print "NO: \$num\n";
}
\$i++;
\$num=decToBase(\$base,\$i);
last if length \$num > \$base
}
print "Found Self Describing Numbers:\n";
print "\$_\n" for @found;
``````

## Ryan Thompson

My solution uses the string-concatenation version of the base ≥ 7 formula, and an optimized search of smaller bases. Although I was of course aware of the bases with no self-descriptive numbers, I chose to search those anyway. It would have been trivial to skip them. The `self_descriptive_base()` function returns all self-descriptive numbers of the given base, in that base:

``````sub self_descriptive_base {
my \$b = shift;

return "\$base[\$b-4]21" . '0' x (\$b-7) . '1000' if \$b >= 7;

grep { is_self_descriptive(\$_) }
map { 10 * \$_ } 10**(\$b-2) .. 10**(\$b-1) - 1;
}
``````

Note that the `map` at the bottom multiplies the entire range by 10, so although I am iterating in base-10, I am effectively searching 1/10th of the numbers, for a 90% speedup. I can do this since self-descriptive numbers must be divisible by their base.

Here is the `is_self_descriptive()` function, which simply counts its digits and ensures that count is equal to the value of each digit:

``````sub is_self_descriptive {
my @s = split '', shift;

return if @s != sum @s; # Not a Niven number

my %count;
\$count{ \$s[\$_] }++ for 0..\$#s;

all { \$count{\$_} == \$s[\$_] } 0..\$#s;
}
``````

My blog

## Saif Ahmed

Saif’s solution is very similar to Roger’s, in that it starts with an array of digits and then mutates that array with the following `countAndPlace()` function:

``````sub countAndPlace{
my (\$string)=@_;
my @split=split  //,\$string;
foreach my \$pos (0..\$#split){
\$split[\$pos]= \$decToBase[ grep { \$_ eq  \$decToBase[\$pos] } @split];
}
return join "",@split;
}
``````

As before, this works very well, but does miss the 2nd base-4 self-descriptive number, 2020, so a small modification or special case would be necessary for that.

Saif also implemented the string-based version of the base ≥ 7 formula, but uses this only as a “cheat” to validate the results of his `countAndPlace` method:

``````sub cheatGetSelfDescriptive{
my \$base=shift;
return  \$decToBase[\$base-4]."21".("0"x(\$base-7))."1000" if \$base >6;
return "oops...failed!"
}
``````

I like the fact that Saif “assumed no prior knowledge” of self-descriptive numbers, aside from knowing which bases had no self-descriptive numbers.

## wanderdoc

wanderdoc’s solution chooses to hard-code the results for all bases < 7, and then `join` together an array built up using the base ≥ 7 formula:

``````sub descr_create
{
my \$base = \$_;
if ( \$base <= 3 or \$base == 6 ) { return "Does not exist!" }
if ( \$base == 4 ) { return "1210 or 2020"; }
if ( \$base == 5 ) { return "21200"; }
if ( \$base > 64 ) { return "Is not implemented!"; }

my @number = (0) x \$base;

\$number = \$base - 4;
\$number = 2;
\$number = 1;
\$number[\$#number - 3] = 1;
my \$num_str = join('', map \$digits{\$_}, @number);
return \$num_str;
}
``````

There isn’t much else I can say about wanderdoc’s solution, except that it is a concise, efficient implementation. Well done.

### Blogs this week:

(2) Arne SommerOlympic Numbers with Raku

(3) Burkhard NickelsOlympic Rings, Self-Descriptive Numbers

(4) Dave JacobyRings and Self-Description

(5) E. ChorobaOlympic Rings and Self-Descriptive Numbers

(6) Jaldhar H. VyasPerl Weekly Challenge: Week 43

(7) Javier LuquePerl Weekly Challenge - 043

(8) Ryan ThompsonOlympic Rings | Self-Descriptive Numbers

## SO WHAT DO YOU THINK ?

If you have any suggestions or ideas then please do share with us.