Fix 1993/vanb/README.md

1993/vanb/README.md

@@ -20,8 +20,7 @@ right parameters. Thank you Cody for your assistance!
 
 	./vanb 'exp'
 	
-	where:
-	    'exp'	an octal expression
+where 'exp' is an octal expression
 
 
 ## Try:
@@ -56,44 +55,42 @@ expression, otherwise it might become hard to understand.  :-)
 
 What Makes This Program Special:
 
-    0) Of course, the fact that the program takes input and produces output
-       in octal, rather than a more useful system like decimal, makes
-       it "special".
+0. Of course, the fact that the program takes input and produces output in
+octal, rather than a more useful system like decimal, makes it "special".
 
-    1) The entire program consists of a single expression, returned from main.
+1. The entire program consists of a single expression, returned from main.
 
-    2) There are no global or local variables other than the parameters
-       to main
+2. There are no global or local variables other than the parameters to main.
 
-    3) All of the constants in the program are expressed in octal - including
-       the printf string.
+3. All of the constants in the program are expressed in octal - including the
+printf string.
 
-    4) The variables are given names which cause visual confusion with the
-       octal constants.
+4. The variables are given names which cause visual confusion with the octal
+constants.
 
-    5) The program is formatted like a roman numeral 8. The significance of
-       8 is obvious. I chose roman numerals because it is a numbering
-       system even more arcane than octal. It's not very 'deceptive',
-       but it's difficult to deceptively format a program which has no
-       explicit control structures (other than ?:'s).
+5. The program is formatted like a roman numeral 8. The significance of 8 is
+obvious. I chose roman numerals because it is a numbering system even more
+arcane than octal. It's not very 'deceptive', but it's difficult to deceptively
+format a program which has no explicit control structures (other than ?:'s).
 
-    6) Many simple tasks are done many times. I tried to do these differently
-       each time, to make the program harder to understand.
+6. Many simple tasks are done many times. I tried to do these differently each
+time, to make the program harder to understand.
 
 How It Works:
 
 It's a recursive descent parser, calling main
 for all of its recursion. O2 is a state variable, telling main
-    what grammar non-terminal to parse. **O7 is the next character.
-    O3 is an intermediate result. Whenever you see a construct like:
+what grammar non-terminal to parse. `**O7` is the next character.
+`O3` is an intermediate result. Whenever you see a construct like:
 
       !(expression with O2)?
 
 it's decrementing & testing O2 to see what state it's in. Comparisons
-    involving **O7 and octal constants are looking for certain characters.
+involving `**O7` and octal constants are looking for certain characters.
 
 Here's the grammar ('e' denotes the empty string) :
 
+
     <E>  ::=  <T><E'>
     <E'> ::=  +<T><E'> | -<T><E'> | e
     <T>  ::=  <F><T'>
@@ -104,6 +101,7 @@ expression, otherwise it might become hard to understand.  :-)
     <X>  ::=  [0-9]<X> | [A-F]<X> | [a-f]<X> | e
     <O>  ::=  [0-7]<O> | e
 
+    
 Here's how the grammar nonterminals map to octal state numbers:
 
     <E>  is  012
@@ -120,6 +118,7 @@ expression, otherwise it might become hard to understand.  :-)
 augmented with comments showing where each state begins. N1 and N2 are
 notes:
 
+
 	#define O5 main
 	O5(O2,O7,O3)char**O7;
 	{
@@ -161,14 +160,16 @@ expression, otherwise it might become hard to understand.  :-)
 
 Note N2: Since the program is properly invoked with 1 parameter, this is
 the first state it will go into. This state just invokes
-	     printf, and sends the parser to state 012 (which is <E>).
+printf, and sends the parser to state 012 (which is `<E>`).
 
 The E and T states work like this:
 
 	int e(){ return eprime( t() ); }
 
+
 The E' and T' states work like this:
 
+
 	int eprime( int intermediate )
 	{
 	    if( ch == '+' )