Move 1988-2004 entries into subdirs
We we move 1988-2004 entries from all of a year's winning entries in a single directory to each winner being in their own sub-directory under the given year directory.
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Landon Curt Noll committed
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O5(O2,O7,O3)char**O7;{return!(O2+=~01+01)?00:!(O2-=02>01)?printf("\045\157\012"
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,O5(012,O7+01,00)):!(O2-=02>>01)?(**O7<=067&&**O7>057?O5(03,O7,*(*O7)++-060+010
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*O3):O3 ):!(O2 -=-O3- ~O3)? (072>**
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O7&&060 <=**O7 ?O5(04 ,O7,012 *O3-060
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+*(*O7 )++):O3 ):!(O2 -=!O3+ !!O3)?(
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**O7>057 &&**O7 <=071? O5(05, O7,*(*
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O7)+++ O3*020 -060): **O7<= 0106&&
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00101<= **O7?O5 (05,O7 ,020*O3 +*(*O7)
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++-067) :0140<** O7&&** O7<0147 ?O5(05,
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O7,-0127 +*(*O7 )+++020 *O3):O3 ):!(
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O2-=02- 01)?(** O7==050 ?050** ++*O7,
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O5(013, O7,O5( 012,O7 ,00)):* *O7<056
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&&054<* *O7?055 **++* O7,-O5( 06,O7,
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00):054 >**O7&& 052<** O7?050* *(*O7)
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++,O5(06 ,O7,00 ):!(** O7^0170 )||!(
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0130^** O7)?*++ *O7,O5 (05,O7 ,00):*
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*O7==0144 ||**O7 ==0104 ?++*O7 ,O5(04,
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O7,00): O5(03 ,O7,00 )):!-- O2?(*
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*O7==052 ?O5(07 ,O7,O3* (*++*O7 ,O5(06
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,O7,00) )):!( 045-** O7)?O5( 07,O7,
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O3%(03+( *O7)++, O5(06, O7,00) )):!(**
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O7^057)?O5(07, O7,O3/( 03-*++ *O7,O5(
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06,O7,00))):O3 ):!(O2 +=01-02 )?O5(07
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,O7,O5(06,O7, 00)):!( O2+=-02/ 02)?(!(*
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*O7-053)?O5(011,O7,O3+(++*O7,O5(010,O7,00))):!(055^**O7)?O5(011,O7,O3-(03+*(*O7
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)++,O5(0010,O7,00))):O3):!(O2-=0563&0215)?O5(011,O7,O5(010,O7,00)):(++*O7,O3);}
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@@ -0,0 +1,186 @@
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Most Irregular Expression:
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David Van Brackle
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ISX Corporation
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1165 Northchase Parkway, Suite 120
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Marietta, GA 30067
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USA
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Judges' comments:
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To run:
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make vanb
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./vanb 'exp'
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where:
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'exp' an octal expression
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The octal expression may contain:
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unary operators: +, -
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binary operators +, -, *, /, %
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hex numbers: x or X follow by hex digits
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decimal numbers: d or D follow by decimal digits
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octal numbers: octal digits
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grouping: ()
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No spaces are allowed in the expression. To avoid shell expansion,
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one should surround the expression in single quotes.
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It is a good thing that this program consists of only one
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expression, otherwise it might become hard to understand. :-)
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Selected notes from the author:
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The program does no error checking - erroneous expressions will
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produce spurious results. Note that unary '-' is an operator. Thus,
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decimal -46 would be entered as -d46 and not d-46.
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What Makes This Program Special:
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0) Of course, the fact that the program takes input and produces output
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in octal, rather than a more useful system like decimal, makes
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it "special".
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1) The entire program consists of a single expression, returned from main.
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2) There are no global or local variables other than the parameters
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to main
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3) All of the constants in the program are expressed in octal - including
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the printf string.
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4) The variables are given names which cause visual confusion with the
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octal constants.
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5) The program is formatted like a roman numeral 8. The significance of
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8 is obvious. I chose roman numerals because it is a numbering
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system even more arcane than octal. It's not very 'deceptive',
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but it's difficult to deceptively format a program which has no
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explicit control structures (other than ?:'s).
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6) Many simple tasks are done many times. I tried to do these differently
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each time, to make the program harder to understand.
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How It Works:
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It's a recursive descent parser, calling main
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for all of its recursion. O2 is a state variable, telling main
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what grammar non-terminal to parse. **O7 is the next character.
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O3 is an intermediate result. Whenever you see a construct like:
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!(expression with O2)?
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it's decrementing & testing O2 to see what state it's in. Comparisons
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involving **O7 and octal constants are looking for certain characters.
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Here's the grammar ('e' denotes the empty string) :
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<E> ::= <T><E'>
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<E'> ::= +<T><E'> | -<T><E'> | e
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<T> ::= <F><T'>
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<T'> ::= *<F><T'> | /<F><T'> | %<F><T'> | e
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<F> ::= +<F> | -<F> | (<E><C> | d<D> | D<D> | x<X> | X<X> | <O>
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<C> ::= )
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<D> ::= [0-9]<D> | e
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<X> ::= [0-9]<X> | [A-F]<X> | [a-f]<X> | e
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<O> ::= [0-7]<O> | e
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Here's how the grammar nonterminals map to octal state numbers:
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<E> is 012
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<E'> is 011
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<T> is 010
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<T'> is 07
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<F> is 06
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<C> is 013
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<D> is 04
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<X> is 05
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<O> is 03
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Here's a version of the program before it got formatted into the VIII,
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augmented with comments showing where each state begins. N1 and N2 are
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notes:
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#define O5 main
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O5(O2,O7,O3)char**O7;
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{
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return
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/* N1 */ !(O2+=~01+01) ?
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00:
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/* N2 */ !(O2-=02>01) ?
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printf("\045\157\012",O5(012,O7+01,00)):
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/* O */ !(O2-=02>>01) ?
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(**O7<=067 && **O7>057 ? O5(03,O7,*(*O7)++-060+010*O3):O3):
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/* D */ !(O2-=-O3-~O3) ?
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(072>**O7 && 060<=**O7 ? O5(04,O7,012*O3-060+*(*O7)++):O3):
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/* X */ !(O2-=!O3+!!O3) ?
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(**O7>057 && **O7<=071 ? O5(05,O7,*(*O7)+++O3*020-060):
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**O7<=0106 && 0101<=**O7 ? O5(05,O7,020*O3+*(*O7)++-067):
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0140<**O7 && **O7<0147 ? O5(05,O7,-0127+*(*O7)+++020*O3):O3):
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/* F */ !(O2-=02-01) ?
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(**O7==050 ? 050**++*O7,O5(013,O7,O5(012,O7,00)):
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**O7<056 && 054<**O7 ? 055**++*O7,-O5(06,O7,00):
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054>**O7 && 052<**O7 ? 050**(*O7)++,O5(06,O7,00):
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!(**O7^0170)||!(0130^**O7) ? *++*O7,O5(05,O7,00):
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**O7==0144||**O7==0104 ? ++*O7,O5(04,O7,00):O5(03,O7,00)):
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/* T' */ !--O2 ?
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(**O7==052 ? O5(07,O7,O3*(*++*O7,O5(06,O7,00))):
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!(045-**O7) ? O5(07,O7,O3%(03+(*O7)++,O5(06,O7,00))):
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!(**O7^057) ? O5(07,O7,O3/(03-*++*O7,O5(06,O7,00))):O3):
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/* T */ !(O2+=01-02) ?
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O5(07,O7,O5(06,O7,00)):
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/* E' */ !(O2+=-02/02) ?
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(!(**O7-053)?O5(011,O7,O3+(++*O7,O5(010,O7,00))):
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!(055^**O7) ? O5(011,O7,O3-(03+*(*O7)++,O5(010,O7,00))):O3):
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/* E */ !(O2-=0563&0215) ?
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O5(011,O7,O5(010,O7,00)):
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/* C */ (++*O7,O3);
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}
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Note N1: It should never enter this state, unless the user invokes the
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program with no parameters, in which case it just returns 0.
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Note N2: Since the program is properly invoked with 1 parameter, this is
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the first state it will go into. This state just invokes
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printf, and sends the parser to state 012 (which is <E>).
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The E and T states work like this:
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int e(){ return eprime( t() ); }
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The E' and T' states work like this:
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int eprime( int intermediate )
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{
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if( ch == '+' )
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{
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ch = nextchar();
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return eprime( intermediate + t() );
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}
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else return intermediate;
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}
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The D, X and O states work like this (they assume that they're initially
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called with 0):
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int octal( int intermediate )
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{
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if( ch>='0' && ch<='7' )
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{
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intermediate = intermediate * 8 + ch - '0';
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ch = nextchar();
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return octal( intermediate );
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}
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else return intermediate;
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}
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F and C work similarly.
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Copyright (c) 1993, Landon Curt Noll & Larry Bassel.
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All Rights Reserved. Permission for personal, educational or non-profit use is
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granted provided this this copyright and notice are included in its entirety
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and remains unaltered. All other uses must receive prior permission in writing
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from both Landon Curt Noll and Larry Bassel.
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