Re: Shortest string containing all from 0000 to 9999 challenge. by sauoq (Abbot) on May 22, 2003 at 09:51 UTC 
What is the shortest string that contains all the numbers from 0000 to 9999.
Assuming the numbers must all have 4 digits, the shortest theoretical length is 10003 digits if you can get them to overlap perfectly (10000 unique numbers and 3 digits of overhead.) As it happens, you can do it perfectly. Here's one example (be careful of newlines if you copy and paste it):
The (admittedly lousy) script I used follows. It's a bit cish...
sauoq
"My two cents aren't worth a dime.";
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Re: Shortest string containing all from 0000 to 9999 challenge. by sauoq (Abbot) on May 22, 2003 at 10:23 UTC 
By the way, you can improve your analyze function quite a bit. This piece in particular stood out to me:
for ( 0 .. 9999 ) {
my $k = 0 x ( 4  length $_ ) . $_;
Even if you had to use a for loop, you could have just written it as for ( '0000' .. '9999' ) instead. Isn't Perl great? You don't even need the for though. I wrote my analyze function like this:
sub analyze {
my $number = shift;
my $length = length $number;
my $duplicates = 0;
my %seen;
while (length $number >= 4) {
$seen{substr($number, 4)}++ and $duplicates++;
substr($number, 1, 1, '');
}
my $missing = 10000  keys %seen;
return ($length, $missing, $duplicates)
}
sauoq
"My two cents aren't worth a dime.";
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Re: Shortest string containing all from 0000 to 9999 challenge. by hardburn (Abbot) on May 22, 2003 at 17:23 UTC 
Q: What is the shortest string that contains all the numbers from 0000 to 9999.
Probably not the shortest, but likely close:
perl MCompress::Bzip2 e "
print Compress::Bzip2::compress('all the numbers from 0000 to 9999')"
 I wanted to explore how Perl's closures can be manipulated, and ended up creating an object system by accident.
 Schemer
Note: All code is untested, unless otherwise stated
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Re: Shortest string containing all from 0000 to 9999 challenge. by BrowserUk (Pope) on May 22, 2003 at 11:07 UTC 
So far, it would appear that there are at least 10000 minimal solutions. 1 for every possible set of first 4 chars. There are also multiplenumber yet to be determined, my poor ol' pII is groaning:)variations of each of these starting points.
There's probably some mathematical way of calculating the total number of unique minimum solutions, but I can't see it. tye?, tilly?, Abigail? the NSA?. Anyone?
Examine what is said, not who speaks.
"Efficiency is intelligent laziness." David Dunham
"When I'm working on a problem, I never think about beauty. I think only how to solve the problem. But when I have finished, if the solution is not beautiful, I know it is wrong." Richard Buckminster Fuller
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A better lower bound is 10! Every permutation of the 10 digits will change a valid solution to another valid solution.
Wrong.
By the way it's not that sure that you can start with every 4 digit number. I think you can  but you need to prove it.
Update: Or is it right? Let's have two differend 4 digit strings  if they are different then at least on one of the 4 position they have different digits  the permutation will change those digits to different digits as well. It means the permutation induces a 11 function on the 10000 4 digit strings. There can't be any more of them so in the whole resulting string there will be exactly 10000 different 4 digit strings. QED
That was long time since I proved a theorem.
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By the way it's not that sure that you can start with every 4 digit number. I think you can  but you need to prove it.
You can.
First, it is sufficient to prove that you can do it with every possible pattern of same/differing numbers. Permutations take care of the rest. (Your thinking on that was fine, by the way.) There are only twelve such patterns:
AAAA AAAB AABA ABAA BAAA AABC ABAC ABCA BAAC BACA BCAA ABCD. All you have to do is choose A, B, C, and D as four different digits and find a solution for each one. I chose A=0, B=1, C=2, and D=3 and then, using the same code in my original solution, found that there was indeed a solution for each of the 12 patterns.
So, for a lower bound, we have at least 12 * 10! (or the number of starting patterns multiplied by the permutations.)
Some other interesting facts (proofs left as an exercise):
 The reverse of a valid solution is a valid solution.
 No valid solutions are palindromic.
sauoq
"My two cents aren't worth a dime.";
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Based on a naive statistical analysis, we might expect about 2.8e5659 unique shortest strings. I got this number by computing the odds of a string of 10003 digits being a solution and multiplying that by the number of 10003digit strings. Of course, floating point won't handle that. I got around this by reducing the formula a bit and computing the log() of the value instead and dividing by log(10) to get about 5659.45 ( all from the "perl de 0" prompt, just in case anyone thought this was offtopic q: ).
The odds of me having made a mistake in this calculation are rather large (it was a quick hack with no double checking  both the math and the arithmatic). I'll try to find time later to post my analysis and code used to do the calculations.
I'm also not convinced (having not spent much time thinking about it) that a naive statistical analysis would be very valid. My qualms here appear to be diminishing the more I think about it (in part because the answer is so *huge*), but I still think the worry is worth mentioning.
It certainly was an interesting problem. (:
Also note that sauoq has shown us the lexicographically first of the shortest strings.
I also have an inkling that we can eliminate the backtracking in sauoq's program (or maybe not...). But I've got real work™ to do... ):
 tye
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#!/usr/bin/perl w
use strict;
my $string = '9999'; # Starting with nines simplifies the loop
my %seen = ( $string => 1 );
my $seen = 1;
while ($seen < 10000) {
for (0 .. 9) {
my $candidate = substr($string, 3) . $_;
next if $seen{$candidate};
$string .= $_;
$seen{$candidate} ++;
$seen++;
last;
}
}
print $string, "\n";
sauoq
"My two cents aren't worth a dime.";
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