Landon Curt Noll
4ee4c731f7
While not strictly necessary, replacing the two ASCII tab characters in the markdown copyright comment makes it easier to detect problematic TAB indentation within markdown files. After this commit, only 90 markdown files have leading ASCII TAB characters. And again, while not strictly necessary, we can consider running expand on those 90 markdown files to see what effect that might have on the HTML files they produce. Replace leading ASCII TAB characters with multiples of 8 ASCII SPACE characters (the default action of the `expand(1)` tool) might have an negative effect on some HTML formatting where **hack-a-round** formatting was attempted to fix broken indentation caused by problematic ASCII TAB characters. Think of this as a baseline for evaluating leading TAB characters within markdown files. TODO: Consider either using `expand(1)` to replace leading ASCII TAB characters with multiples of 8 ASCII SPACE characters AND/OR modifying the appropriate `bin` tool to convert leading ASCII TAB characters into multiples of 8 ASCII SPACE characters.
To build:
make all
The authors provided a spoiler version, originally uuencoded but uudecoded by us in 2023. See Alternate code below.
Bugs and (Mis)features:
The current status of this entry is:
STATUS: missing file - please provide it
For more detailed information see 1995/vanschnitz in bugs.html.
To use:
./vanschnitz
Try:
./try.sh
Alternate code:
The authors provided a spoiler version of the program. Originally uuencoded we
have decoded it for the wider audience in vanschnitz.alt.c. The
uuencoded blob is still available in their remarks below. They encourage you
to first try and figure out the program without the spoiler.
Alternate build:
make alt
Alternate use:
Use vanschnitz.alt as you would vanschnitz above.
Judges' remarks:
Back in 1995, values of LEVEL>8 caused some compilers (and in one case a
system) to crash during compilation. In 2023 with modern systems this is not
really a problem.
Back in 1995 we also suggested that for a bad/slow time to try LEVEL=15 but
this is not a big deal in 2023 either.
Authors' remarks:
A classic problem in computer science is known as the Towers of Hanoi. It involves a set of different-sized disks mounted on one of three pegs. The object is to move the pile of disks to one of the other pegs. You may only move one disk at a time, and there is an additional constraint that a disk may never be placed on top of a smaller disk.
Our program solves the Towers of Hanoi problem. Well, that's not exactly true; actually, it's the compiler that solves the problem. The resulting program just prints out the correct solution.
How do you trick a compiler into actually solving the problem?
First, you must tell it how many disks you wish to solve it
for. This is done by defining the symbol n on the compile
line. For instance, to cause the compiler to solve the Towers
of Hanoi problem with four disks, you would compile the program
like this:
gcc hanoi.c -o hanoi -Dn=4
A default value of 5 will be used for n if you do not define
it on the command line. The value of n cannot be greater than
fifteen (the compiler we used to test has a limit on the #include
depth). The compiler then solves the problem using binary
arithmetic based on whether particular symbols are defined or not.
To loop, the program #includes itself. This is, of course, expensive; one
compile we did with n=14 took about fifty minutes to compile on our system
(compiling with n=15 caused our system to crash).
The resulting program that the compiler generates simply
prints out the answer. Did I say "simply"? Actually, the
whole resulting program consists of a single printf(3) statement,
consisting of a massive string constant of length 35*(2^n-1),
followed by 3*(2^n-1) integers which get formatted into the
string. For our n=14 run, this adds up to a string constant
of length 573405, followed by 49149 integers delimited by
commas. (Generating the string constant depends on the
ANSI C feature in which adjacent character strings are
concatenated; a version that does not use this feature has been
included for people who can only run K&R). A good way to see
the resulting program (on a Unix system) is to do the command
gcc hanoi.c -E -Dn=5 | grep -v \# | grep -v ^\$
For an odd number of disks, the program will provide a solution wherein the disks end up on peg 2; for an even number of disks, they will end on peg 3. This should provide some hint as to what sort of algorithm is used.
We have included a spoiler version of the program, with meaningful symbol names and comments, but we encourage you to try to decipher the program without it...
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