[libmarpa-r2-perl] 07/32: Github issue #151

Jonas Smedegaard dr at jones.dk
Sat Nov 22 18:38:36 UTC 2014


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js pushed a commit to annotated tag Marpa-R2-2.087_000
in repository libmarpa-r2-perl.

commit bd6d1a1a49d78303e6ee10e083f193a3ff13107f
Author: Jeffrey Kegler <JKEGL at cpan.org>
Date:   Thu Jul 3 10:43:02 2014 -0700

    Github issue #151
    
    Moved wall_proof.txt to theory repository, to resolve Debian
    licensing issue.
---
 cpan/MANIFEST                |  1 -
 cpan/inc/Marpa/R2/License.pm |  2 +-
 cpan/t/etc/wall_proof.txt    | 64 --------------------------------------------
 3 files changed, 1 insertion(+), 66 deletions(-)

diff --git a/cpan/MANIFEST b/cpan/MANIFEST
index e869eae..0b78588 100644
--- a/cpan/MANIFEST
+++ b/cpan/MANIFEST
@@ -312,7 +312,6 @@ t/deprecated1.t
 t/dsl.t
 t/duplicate_parse.t
 t/equation.t
-t/etc/wall_proof.txt
 t/final_nonnullable.t
 t/gabend.t
 t/implementation.t
diff --git a/cpan/inc/Marpa/R2/License.pm b/cpan/inc/Marpa/R2/License.pm
index b93ba7b..7a336a8 100644
--- a/cpan/inc/Marpa/R2/License.pm
+++ b/cpan/inc/Marpa/R2/License.pm
@@ -306,7 +306,6 @@ my %files_by_type = (
     'libmarpa_doc_dist/version.texi'   => \&trivial,
     'libmarpa_doc_dist/version_i.texi' => \&trivial,
     'etc/my_suppressions'              => \&trivial,
-    't/etc/wall_proof.txt'             => \&cc_a_nd,
     'xs/ppport.h' => \&ignored,    # copied from CPAN, just leave it alone
     'libmarpa/tavl/README' => \&trivial,
 
@@ -678,6 +677,7 @@ sub tex_closed {
     return @problems;
 } ## end sub tex_closed
 
+# Note!!!  This license is not Debian-compatible!!!
 sub tex_cc_a_nd {
     my ( $filename, $verbose ) = @_;
     my @problems = ();
diff --git a/cpan/t/etc/wall_proof.txt b/cpan/t/etc/wall_proof.txt
deleted file mode 100644
index b97b951..0000000
--- a/cpan/t/etc/wall_proof.txt
+++ /dev/null
@@ -1,64 +0,0 @@
-Copyright 2014 Jeffrey Kegler
-This document is licensed under
-a Creative Commons Attribution-NoDerivs 3.0 United States License.
-
-The first three numbers in A052952 are defined as 1, 1, and 3. Hand
-counts will establish that they match W(1), W(2) and W(3).
-
-Length 1, Parse 0, value==(6-1)==5
-Length 2, Parse 0, value==(6-(-1))==7
-Length 3, Parse 0, value==((6--)-1)==5
-Length 3, Parse 1, value==(6-(--1))==6
-Length 3, Parse 2, value==(6-(-(-1)))==5
-
-It remains to show that W(n) = A052952(n) for n>2.
-
-Look at the far right hand side of the sequence of minus signs.
-It's one of three things: a binary minus (subtraction), a unary
-minus (negation) or the second character of a pre-decrement. We can
-get a count of each of these three categories and add them to produce
-W(n).
-
-The unary minus category can be formed from any of the parses with
-one less minus sign, by adding a negation to the right side. So
-there are W(n-1) parses from that source.
-
-Similarly the binary minus category is all the parses with two fewer
-minus signs, with a pre-decrement added to the far right end.
-Therefore, W(n-2) parses from that source.
-
-The subtraction category is only slightly harder. The only operator
-pattern allowed to the left of the binary subtraction is a series
-of post-decrements. So if n-1 is even, there is one parse from that
-source. If n-1 is odd, the pattern of post-decrements cannot be
-formed, and there are no parses from that source. In Perlish notation,
-the number of parses is (n-1) % 2 ? 0 : 1.
-
-Putting all the sources of parses together, we get
-
-W(n) = W(n-1) + W(n-2) + ((n-1) % 2 ? 0 : 1)
-
-We can use this formula to derive a count for one of its own
-components, W(n-1):
-
-W(n-1) = W(n-2) + W(n-3) + ((n-2) % 2 ? 0 : 1)
-
-And then we can substitute in the first formula from the second:
-
-W(n) = W(n-2) + W(n-3) + ((n-2) % 2 ? 0 : 1)  + W(n-2) + ((n-1) %
-2 ? 0 : 1)
-
-Simplifying:
-
-W(n) = 2*W(n-2) + W(n-3) + ((n-2) % 2 ? 0 : 1)  + ((n-1) % 2 ? 0 :
-1)
-
-Finally, note that of two consecutive numbers one will always be
-odd and the other always even. Therefore ((n-2) % 2 ? 0 : 1)  +
-((n-1) % 2 ? 0 : 1) will always be 1. Applying the final simplification:
-
-W(n) = 2*W(n-2) + W(n-3) + 1
-
-From http://www.research.att.com/~njas/sequences/A052952, the
-recurrence for A052952 is a(n) = 2*a(n-2) + a(n-3) + 1. Exactly the
-same as the recurrence I've just proved for the Wall series. QED.

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