Every month New Scientist magazine publishes a little mathematical challenge. Normally these things irritate me because they rely on some cute but irrelevant mathematical trick. This month, however, they have tried to obscure the question by replacing numbers with letters, which makes it a Perlish challenge. Here we go:
A triangular number is a term from the sequence generated by this equation (in Perl): foreach my $n (1..1000) {
$x=.5*$n*($n+1);
print $x;
}
This gives 1, 3, 6, 10 and so on. This months challenge is to find a certain four numbers from this sequence. They don't give us the numbers, but they have substituted letters for digits of these numbers, and made words from them. The words are: ONE, THREE, SIX and TEN.
So for instance, we could have O=1, N=2, E=3, and the first number would be 123.
So I wrote a Perl program to have a go at it. I make a hash of the letters, assign a value (starting at zero) to each letter, then start incrementing. It's been going for the last few hours and I just realised it could take quite a long time to finish.
I am declaring that the challenge is to make it efficient, not write it with the fewest characters. This is not part of the official challenge, if you feel like writing it in one line, go right ahead.
The prize is drawn on the 28 June
Update
jeroenes asked if there were restrictions on the numbers  there are. It's one letter for one digit. The digits may be 0..9 (the article is a little vague on this, but that's my reading). But I forgot to mention that none of the numbers start with a zero. i.e. 003 is not a valid solution for this challenge
The challenge does not mention the order of the numbers, so I guess we can assume 'any'.
If you would also like to apply for the 15pound prize you can send your solution to enigma@newscientist.com (include your postal address)
#!/usr/bin/perl
use strict;
my @words=('ONE', 'THREE', 'SIX', 'TEN');
my %letters;
foreach (@words) {
my @a=split //, $_;
foreach (@a) {
$letters{$_}=0;
}
}
#Construct an array of the letters (you'll see why
#in a few lines)
my @letters=(keys %letters);
my $count;
my $oldtime;
do {
$letters{$letters[$#letters]}++;
foreach( reverse(0..$#letters)) {
if ($count++>10000) {
print time$oldtime,"\n";
$oldtime=time;
$count=0;
}
#Increment the number for the letter at the end
#of the array, and don't forget to carry
if ($letters{$letters[$_]} > 9) {
$letters{$letters[$_]}=0;
$letters{$letters[$_1]}++;
}
#Turn the letters into numbers
prep(%letters);
}
} until ($letters{$letters[0]} > 9);
sub prep {
my %letters=@_;
my $match=0;
my %match;
foreach my $a (@words) {
my $test= $a;
#Change the letters into numbers
foreach (keys %letters) {
$test=~ s/$_/$letters{$_}/g;
}
$match{"$a=$test is triangular\n"}=1 and $match ++ if is_triangular
+($test);
}
if ($match>0) {
# print "\n";
if ($match>(@words1)) {
print $_ foreach keys %match;
print "Found it!\n";
#exit;
}
}
}
#This should be memoised, but I'm afraid of the
#memory blowout
sub is_triangular {
my $num=shift;
my ($n,$x)=(1,0);
do {
$x=.5*$n*(++$n);
return 1 if $x==$num;
} until $x > $num;
return 0;
}
Update II
Since tachyon has indeed beaten me to the answer I shall enter PerlMonks in the competetion, winnings go in the plate. (More likely they will go into an envelope and off to vroom).
Update III
My bad. The challenge for the magazine is to get the answer. The answers are then drawn out of a hat, first right one gets the prize.
My challenge to you is to do it most efficiently.
____________________
Jeremy
I didn't believe in evil until I dated it.
Re: (Efficiency Golf) Triangular numbers by jeroenes (Priest) on May 30, 2001 at 09:59 UTC 
Is there any constraint on the way the letters form the
numbers? Like, did they state that the letters code
for the numbers 0..9 in an unknown order? Can it also be
that they code for 0..25 each? Or can they encode say
1000..1025, but are always in the same order?
Jeroen  [reply] 
Re: (Efficiency Golf) Triangular numbers by tachyon (Chancellor) on May 30, 2001 at 14:03 UTC 
Looks like we may need benchmark for this game. Here is the answer, I'll post the complete code in 24 hours so as not to spoil all the fun.
It's ~ 30 average lines long and runs under strict. I have included the begining and end. My woodwork teacher
used to say look after the edges and the middle will look after itself.....
BTW Please enter perlmonks for the prize and donate it to the offering plate when we win!
Hint:
There is only one solution,
brute force is not the key
With code in evolution,
consider the holy three
tachyon
one 435 three 17955 six 820 ten 153
Elapsed 0 seconds
#!/usr/bin/perl w
use strict;
my $time = time();
my %tri;
# make hash of tiangular numbers 5 digits or less
map{$tri{.5*$_*($_+1)}=1}1..446;
...
...
# prove we are right!
print "one $o$n$e " if defined $tri{"$o$n$e"};
print "three $t$h$r$e$e " if defined $tri{"$t$h$r$e$e"};
print "six $s$i$x " if defined $tri{"$s$i$x"};
print "ten $t$e$n\n" if defined $tri{"$t$e$n"};
$time = time()$time;
print "Elapsed $time seconds\n";
 [reply] [d/l] 
Re: (Efficiency Golf) Triangular numbers by tachyon (Chancellor) on May 30, 2001 at 16:03 UTC 
my$time=time();local$";map{$t{.5*$_*($_+1)}=1}1..446;map{
push@p,$1 if/(...(.)\2)/}keys%t;map{@a[0..3]=split'';map{
push@p1,"@a$_"if defined$t{"$a[0]$a[3]$_"}}0..9}@p;map{@a
=split'';map{push@p2,"@a$_"if defined$t{"$_$a[4]$a[3]"}}
0..9}@p1;map{/(.)(.)(.)(.)(.)(.)/;push @p3,$_ if/[^$2$3$4
$5$6][^$1$3$4$5$6][^$1$2$4$5$6][^$1$2$3$5$6][^$1$2$3$4$6]
[^$1$2$3$4$5]/x;}@p2;for(@p3){$r='0123456789';$r=~s/$_//
for split'';@r=split'',$r;for$a(@r){for$b(@r){next if$b==
$a;for$c(@r){next if$c==$b or$c==$a;if(defined$t{"$a$b$c"})
{@a=split'';$w="one:$a[5]$a[4]$a[3] three:$a[0]$a[1]$a[2]"
."$a[3]$a[3] six:$a$b$c ten:$a[0]$a[3]$a[4]\n"}}}}}$time=
time()$time;print$w."Elapsed time $time seconds\n";
 [reply] [d/l] 
Re: (Efficiency Golf) Triangular numbers by Abigail (Deacon) on May 30, 2001 at 17:58 UTC 
There is no need for the program to run very long. A paper
and pencil puzzler will beat your 'running for many hours'
program.
I will not post the solution here, but I will provide a
link to the program. It's about 80 lines, but more than half
of the lines are comments.  [reply] 

I'm familiar with assigning to slices of a hash, and its
use in this line:
@seen {$t, $h, $r, $e} = ();
was very clever. However, why did you use slices
in these lines?
@seen {$n,} = ();
....
@seen {$o,} = ();
Aren't they equivalent to:
$seen{$n} = undef;
$seen{$o} = undef;
Or even just:
@seen {$n} = (); # note the lack of comma after $n
....
@seen {$o} = ();
I guess its that trailing comma that looks odd to me.
Blake
p.s. props on the @HIGH and @LOW generation... very sneaky!
 [reply] [d/l] [select] 

 [reply] 
Re: (Efficiency Golf) Triangular numbers by Zaxo (Archbishop) on May 30, 2001 at 23:13 UTC 
I've pasted in the full program and its output. The strategy is to preselect the numbers to consider by populating arrays with only the triangular numbers which match the pattern of digits implied by 'THREE' and friends. Then I look for pairs of 5 and 3 digit numbers which can correspond to 'THREE' and 'TEN'. That leaves just two pairs, which speeds the search for 'ONE' and 'SIX' a lot.
This isn't very general or clever, just gets the job done.
Full Output:
There are 31 three digit triangular numbers
There are 307 five digit triangular numbers
There are 26 three digit ones which lack matching digits.
There are 9 five digit ones matching the pattern 'THREE'.
THREE: 17955
ONE: 435
SIX: 820
TEN: 153
Elapsed time: 0.02 0 0 0
Full Listing:
#!/usr/bin/perl w # *Perl*
use strict;
sub triangular {
use integer;
my $n=shift;
$n*($n+1)/2;
}
my @threedigit = map triangular($_),
(int(sqrt(2*100))..int(sqrt(2*1000)));
my @fivedigit = map triangular($_),
(int(sqrt(2*10000))..int(sqrt(2*100000)));
print "There are ".scalar @threedigit." three digit triangular numbers
+\n";
print "There are ".scalar @fivedigit." five digit triangular numbers\n
+";
my @threes;
for (@threedigit){
$_ !~ m/(\d)\d*\1/g and push @threes,$_;
}
@threedigit = @threes;
@threes = ();
print "There are ".scalar @threedigit." three digit ones which lack ma
+tching digits.\n";
my @fives;
for (@fivedigit){
$_ =~ m/^\d*(\d)\1$/ and
$_ !~ m/^\d*(\d)\d*\1\d+$/g and
push @fives,$_;}
@fivedigit = @fives;
@fives = ();
print "There are ".scalar @fivedigit." five digit ones matching the pa
+ttern 'THREE'.\n";
my @THREE;
my @TEN;
# eliminate fives which dont give a TEN
for (@fivedigit) {
my $t5 = $_;
my ($T,$H,$R,$E) = split //,$t5;
my $N = "[^$T$H$R$E]";
for (@threedigit) {
my $t3 = $_;
my @t3 = split //, $t3;
if ($t3 =~ m/$T$E$N/){
$N=$t3[2];
push @threes, $t3;
push @fives, $t5;
}
}
}
# Now Brute Force
my $size = $#fives;
my %solution = ('THREE'=>[],'TEN'=>[],'ONE'=>[],'SIX'=>[]);
my $idx;
for $idx (0..$size) {
my ($T,$H,$R,$E) = split //, $fives[$idx];
my ($XT,$XE,$N) = split //, $threes[$idx];
my $cc = "[^$T$H$R$E$N]";
for (@threedigit){
my $num = $_;
if ($num =~ m/$cc$N$E/){
my $O = (split //, $num)[0];
my $cc="[^$T$H$R$E$N$O]";
for (@threedigit){
if ($_ =~ /$cc{3}/){
my ($S,$I,$X) = split //, $_;
push @{$solution{'THREE'}}, $fives[$idx];
push @{$solution{'TEN'}}, $threes[$idx];
push @{$solution{'ONE'}}, $num;
push @{$solution{'SIX'}}, $_;
}
}
}
}
}
# write out solution and timings
for (keys %solution) {
print "$_:\t",join("\t",@{$solution{$_}}),"\n";
}
my @tim = times();
print "Elapsed time:\t",join "\t", @tim,"\n";
After Compline
Zaxo
 [reply] [d/l] [select] 
Re: (Efficiency Golf) Triangular numbers by blakem (Monsignor) on May 31, 2001 at 10:30 UTC 
This was a fun one. My solution looks quite similiar
to Abigail's in structure, but is not nearly so elegant.
I question how they are really going to judge these
based on execution time. Its basically a "find the needle in a haystack" problem, but after you've found it, its pretty easy to write a program that finds it faster the
next time around.
For instance: Which one of these wins? Are any thrown
out for "cheating"?
1: Brute force, try every concievable set of four numbers
(loooong time)
2: Using some sort of filtering at each stage, so not
every set of four is used (much quicker)
3: Cleverly ordering the stages, so they are done in
an optimum order  perhaps 'THREE>TEN>ONE>SIX'.
(slightly quicker, though is uses information gathered
in program 2)
4: Brute force, but coded so that the first set of four
you try is the correct answer.
(even faster!!!)
5: print "$answer"; (clearly the fastest)
Since there is only one haystack and one needle, how do
you determine which program "finds" it, and which has
already been told where it is....
Blake  [reply] 

Thanks for the efficiency clarification... (i.e. only
counts here, not for the magazine prize) I was about
to add some "presorting" to my arrays, allowing
the answer to magically be the first one I tried.
Blake
 [reply] 

 [reply] 
Re: (Efficiency Golf) Triangular numbers by tachyon (Chancellor) on May 31, 2001 at 14:51 UTC 
Here is the code, commented and with follow the progress prints. There are ~35 code lines, the rest are comments, prints or whitespace.
As hinted at in my short poem, the secret (such as there is one) is to analyse the problem.
THREE must be a five digit number, but also it must be a 5 digit number that
ends with the same two digits. With this we remove 90% of the possibles
for the letters THRE, in fact there are a mere 27 in all as you can see if you run the code.
As we go we just push the possibilities for our letters into an array, we then split the letters out as required.
TEN uses the same T and E as found in THREE so we only
need to search for N within the constraint of Ts and Es found initially. Once we have
our possibles that satisfy THREN we look for ONE. Once again we only have to
look for O within the constraint of the Ns and Es already generated.
At this stage there are only six possible combinations of digits for THRENO.
We get rid of 4 as they contain the same digit for two or more letters to leave just two.
We then brute force the possibilities for SIX from the remaining 4 digits (only 4*3*2 = 24 cases)
to get the answer.
Tachyon
#!/usr/bin/perl w
use strict;
my $time = time();
my (%tri,@pos,@pos1,@pos2,@pos3);
# make hash of tiangular numbers 5 digits or less
# the 447th has 6 digits so we don't map past 446
map{$tri{.5*$_*($_+1)}=1}1..446;
# find all possible matches for 'three'
# these are 5 digits long, but last two digits are the same
# this allows us to limit the search
for my $key(keys %tri){
push @pos,$1 if $key =~/(\d\d\d(\d)\2)/;
}
# let's see how many possibilities we have
print "Initially we have ".@pos." possibles for \$t\$h\$r\$e\$e\n";
print "$_\n" for @pos;
# find all possible matches for 'ten' within constraint
# of $t and $e possibilities generated above, we are looking for 'n'
for (@pos) {
my($t,$h,$r,$e)=split'',$_;
for my $n(0..9) {
push @pos1, "$t$h$r$e$n" if defined $tri{"$t$e$n"}
}
}
# let's see how many possibilities we have left
print "\nNext we have ".@pos1." possibles for \$t\$h\$r\$e\$n\n";
print "$_\n" for @pos1;
#now look at 'one' in same way, we are looking for 'o'
for (@pos1) {
my($t,$h,$r,$e,$n)=split'',$_;
for my $o(0..9) {
push @pos2, "$t$h$r$e$n$o" if defined $tri{"$o$n$e"}
}
}
# let's see how many possibilities we have left
print "\nNow we have ".@pos2." possibles for \$t\$h\$r\$e\$n\$o\n";
print "$_\n" for @pos2;
# remove dulicates where digits for $t$h$r$e$n$o are not unique
# I'm sure there is something more elegant but this works
for (@pos2) {
$_ =~ /(.)(.)(.)(.)(.)(.)/;
push @pos3,$_ if $_=~m/[^$2$3$4$5$6][^$1$3$4$5$6][^$1$2$4$5$6][^$1
+$2$3$5$6][^$1$2$3$4$6][^$1$2$3$4$5]/;
}
# let's see how many possibilities we have left
print "\nAfter removing cases where we have duplicate digits\n";
print "we have ".@pos3." possible matches for \$t\$h\$r\$e\$n\$o\n";
print "$_\n" for @pos3;
# find the solution
for my $pos(@pos3) {
# get the remaining digits available for 'six'
# we erase the 6 digits we are currently using
# for t h r e n o
my $remaining = '0123456789';
for (split'',$pos) {$remaining =~ s/$_//;}
# look at the remaining cases
print "\nBrute forcing\n";
print "If \$t\$h\$r\$e\$n\$o\ is $pos then \$s\$i\$x must come fro
+m $remaining\n\n";
# brute force possibilities for six, it's only 4 digits
my @rem = split'',$remaining;
for my $s(@rem){
i: for my $i(@rem){
next i if $i==$s;
x: for my $x(@rem) {
next x if $x==$i or $x==$s;
if (defined $tri{"$s$i$x"}){
my($t,$h,$r,$e,$n,$o)=split'',$pos;
# prove we are right!
print "\nfound solution\n";
print "###################################\n";
print "one $o$n$e " if defined $tri{"$o$n$e"};
print "three $t$h$r$e$e " if defined $tri{"$t$h$r$
+e$e"};
print "six $s$i$x " if defined $tri{"$s$i$x"};
print "ten $t$e$n\n" if defined $tri{"$t$e$n"};
print "###################################\n\n";
} else {print "No match \$s\$i\$x > $s$i$x\n"}
}
}
}
}
$time = time()$time;
print "\nElapsed $time seconds\n";
If anyone has an elegant way of deleting the cases where we have duplicate characters I would love to see it.
My 2 line regex is functional, but fairly agricultural!  [reply] [d/l] 

foreach (@array) {
push @other, $_ if some_condition;
}
Of course, that's much clearer, and more efficiently, written
as:
@other = grep {some_condition} @array;
 Abigail  [reply] [d/l] [select] 

If anyone has an elegant way of deleting the cases where we have duplicate characters I would love to see it. My 2 line regex is functional, but fairly agricultural!
I won't claim elegance, but above I constructed character classes on the fly for that. Each partial solution like ($T,$H,$R,$E,$N) is used to form "[^$T$H$R$E$N]". That is applied to 3digit candidates which are already filtered for unlike digits.
After Compline
Zaxo
Update: Rereading your statement, I saw I'd filtered that too :) I used $_ !~ m/(\d)\d*\1/g to filter out matching digits.
 [reply] 
Re: (Efficiency Golf) Triangular numbers by blakem (Monsignor) on May 31, 2001 at 22:30 UTC 
Since others are posting, here is my original code. It
uses a slightly different tactic, which I hoped would
be a bit of an optimization (though I don't think it
actually was) When I generate my triangular numbers, I
also calculate a "pattern mask", i.e.
THREE => 'abcdd'
55345 => 'aabca'
66456 => 'aabca'
98766 => 'abcdd' (aha, same as THREE's)
So: wordmask('THREE') eq wordmask(98766);
This was supposed to eliminate running repeated regexes
against the same numbers. Anyway, it complicates the code a bit and others have submitted faster solutions using
the same basic algorithm.
Anyway here is the code:
#!/usr/bin/perl
use strict;
# Calculate some "masks" outside the loops
my (%nummasks,%wordmasks);
map{my $k=.5*$_*($_+1); $nummasks{$k}=wordmask($k)}1..446; # ok, this
+line was tweaked from another poster
map{$wordmasks{$_}=wordmask($_)}('ONE','THREE','SIX','TEN');
# Loop through the THREE candidates
for my $three (keys %nummasks) {
next unless $nummasks{$three} eq $wordmasks{'THREE'};
my @three = split(//,$three);
# Loop through the TEN candidates
for my $ten (keys %nummasks) {
next unless $nummasks{$ten} eq $wordmasks{'TEN'};
my @ten = split(//,$ten);
next unless $ten[0] == $three[0] && $ten[1] == $three[3];
my %used = map{$_,1} (@three);
next if $used{$ten[2]};
# Loop through the ONE candidates
for my $one (keys %nummasks) {
next unless $nummasks{$one} eq $wordmasks{'ONE'};
my @one = split(//,$one);
next unless $one[1] == $ten[2] && $one[2] == $ten[1];
my %used = map{$_,1} (@three,@ten);
next if $used{$one[0]};
# Find a SIX and we've solved it
for my $six (keys %nummasks) {
next unless $nummasks{$six} eq $wordmasks{'SIX'};
my @six = split(//,$six);
my %used = map{$_,1} (@three,@ten,@one);
next if $used{$six[0]}  $used{$six[1]}  $used{$six[2]};
print "ONE:\t$one\nTHREE:\t$three\nSIX:\t$six\nTEN:\t$ten\n";
}
}
}
}
sub wordmask { # generate a "pattern mask", 56675 => abbca
my $mask = shift;
my $letter = 'a';
while ($mask =~ /([^az])/) {
$mask =~ s/$1/$letter/g;
$letter++;
}
return $mask;
}
$ time ./abigails.pl
ONE = 435; THREE = 17955; SIX = 820; TEN = 153.
0.06user 0.00system 0:00.07elapsed 83%CPU
$ time ./mine.pl
ONE: 435
THREE: 17955
SIX: 820
TEN: 153
0.25user 0.01system 0:00.26elapsed 98%CPU
So, Abigail's beats mine by a factor of about 4.5!
Blake
 [reply] [d/l] 

