Perl: the Markov chain saw PerlMonks

### Re: Challenge: Optimal Animals/Pangolins Strategy

by BrowserUk (Pope)
 on May 02, 2013 at 16:34 UTC ( #1031781=note: print w/replies, xml ) Need Help??

You just end up with a really lopsided tree:

```#! perl -slw
use strict;
use Data::Dump qw[ pp ];
use List::Util qw[ reduce ];

our \$N //= 1;

my %byFreq = map{
int( rand 1000 ) => \$_
} 'A'x\$N .. 'Z'x\$N;
pp \%byFreq;

my @tree;
reduce{
\$a->[0] = \$byFreq{ \$b };
\$a->[1] = [];
} \@tree, sort{ \$a <=> \$b } keys %byFreq;

pp \@tree;

Output:

With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.

Replies are listed 'Best First'.
Re^2: Challenge: Optimal Animals/Pangolins Strategy
by Limbic~Region (Chancellor) on May 02, 2013 at 17:35 UTC
BrowserUk,
You just end up with a really lopsided tree

Then I must have made a mistake in my explanation because the solution should be closer to Huffman coding as roboticus astutely concluded above. Since I can't explain why what I am trying to accomplish is slightly different, I guess I am stuck.

Cheers - L~R

Well, you mentioned proportional which I interpreted to mean that higher frequencies should take longer to reach than lower frequencies, which barring the possibility of equal frequencies, a lop-sided tree achieves.

(Albeit you said inversely proportional which would mean reversing the order of the sort from what I posted.)

The only other sense I can get from the information provided -- brought on by the mention of Huffman -- is that you are perhaps looking to minimise the depth of the tree. This does that by building a heap and then converting it to a tree rather clumbsily. Though that could be fixed if the idea is right:

```#! perl -slw
use strict;
use Data::Dump qw[ pp ];
use List::Util qw[ reduce ];
use enum qw[ NAME FREQ LEFT RIGHT ];

our \$N //= 1;

my\$n = 0;
my @heap = map{
\$_->[LEFT] = ++\$n;
\$_->[RIGHT] = ++\$n;
\$_;
} sort {
\$a->[FREQ] <=> \$b->[FREQ]
} map[
\$_ , int( rand 1000 )
], 'A'x\$N .. 'Z'x\$N;;

my @tree = map {
\$_->[LEFT] = \$heap[ \$_->[LEFT] ], \$_->[RIGHT] = \$heap[ \$_->[RIGHT] ]
} @heap;

pp \@tree;

Output:

```C:\test>1031775.pl
do {
my \$a = [
[
"Y",
1,
[
"G",
4,
[
"D",
166,
["X", 245, ["I", 516, undef, undef], ["O", 563, undef, undef
+]],
["K", 315, ["M", 628, undef, undef], ["R", 710, undef, undef
+]],
],
[
"A",
218,
["P", 324, ["T", 731, undef, undef], ["Q", 732, undef, undef
+]],
["J", 374, ["V", 735, undef, undef], ["C", 835, undef, undef
+]],
],
],
[
"E",
33,
[
"U",
220,
["L", 393, ["S", 845, undef, undef], ["F", 930, undef, undef
+]],
["W", 471, ["Z", 944, undef, undef], undef],
],
["H", 228, ["B", 507, undef, undef], ["N", 515, undef, undef]]
+,
],
],
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
'fix',
];
\$a->[1] = \$a->[0][2];
\$a->[2] = \$a->[0][3];
\$a->[3] = \$a->[0][2][2];
\$a->[4] = \$a->[0][2][3];
\$a->[5] = \$a->[0][3][2];
\$a->[6] = \$a->[0][3][3];
\$a->[7] = \$a->[0][2][2][2];
\$a->[8] = \$a->[0][2][2][3];
\$a->[9] = \$a->[0][2][3][2];
\$a->[10] = \$a->[0][2][3][3];
\$a->[11] = \$a->[0][3][2][2];
\$a->[12] = \$a->[0][3][2][3];
\$a->[13] = \$a->[0][3][3][2];
\$a->[14] = \$a->[0][3][3][3];
\$a->[15] = \$a->[0][2][2][2][2];
\$a->[16] = \$a->[0][2][2][2][3];
\$a->[17] = \$a->[0][2][2][3][2];
\$a->[18] = \$a->[0][2][2][3][3];
\$a->[19] = \$a->[0][2][3][2][2];
\$a->[20] = \$a->[0][2][3][2][3];
\$a->[21] = \$a->[0][2][3][3][2];
\$a->[22] = \$a->[0][2][3][3][3];
\$a->[23] = \$a->[0][3][2][2][2];
\$a->[24] = \$a->[0][3][2][2][3];
\$a->[25] = \$a->[0][3][2][3][2];
\$a;
}

With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.
/div
BrowserUk,
By inversely proportional, I meant that the number of questions asked to identify an animal (Q) multiplied by how many times the animal is chosen (C) should be constant. If a goat appears a hundred times more often than a unicorn then it should take a hundred times more questions to identify the unicorn than the goat.

I apologize for not seeing it before hand, but I am pretty sure the optimal strategy for my problem is in fact Huffman Coding. To get a better idea of the fuzzy problem I am dealing with in my head, see Re^4: Challenge: Optimal Animals/Pangolins Strategy

Cheers - L~R

Aren't Huffman trees generally expected to be lop-sided?

jakeease,
I think that depends heavily on the frequency distribution.

Cheers - L~R

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