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Format and typo fixes in 1992/buzzard.1/README.md

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Cody Boone Ferguson committed 2023-03-29 08:27:47 -07:00
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@@ -19,159 +19,164 @@ Most Obfuscated Algorithm:
## Judges' comments ## Judges' comments
To see good C Preprocessor buzzard.1, try: To see good C Preprocessor buzzard.1, try:
make buzzard.1.cppcb make babble.cppcb
cat buzzard.1.cppcb cat babble.cppcb
Notice how many statements it takes to do anything useful? Notice how many statements it takes to do anything useful?
## Author's comments ## Author's comments;
What you shouldn't think about buzzard.1: What you shouldn't think about buzzard.1:
buzzard.1 doesn't just do a mass of #defines as its obfuscation. `buzzard.1` doesn't just do a mass of #defines as its obfuscation.
What you should be glad about: ### What you should be glad about:
I didn't just submit the post-C-preprocessed output as an entry. I didn't just submit the post-C-preprocessed output as an entry.
#### Cute trick warning:
`buzzard.1` contains only a numeric printf, yet sometimes it prints out
short strings.
Cute trick warning: ### What it does:
buzzard.1 contains only a numeric printf, yet sometimes it prints out `buzzard.1` is a simple mathematical program. It expects two numeric
short strings. arguments.
```sh
% ./buzzard.1 0 <num>
```
will print out the factorial of `<num>`.
```
% ./buzzard.1 <num1> <num2>
```
will print out the largest common factor of `num1` and `num2`.
```sh
% ./buzzard.1 1 <num2>
```
will print out a factor of `num2` if it's composite, or else
it will print a string informing you that `num2` is prime.
If the first argument is less than 0, it prints out an
error message.
What it does: ### What it is:
buzzard.1 is a simple mathematical program. It expects two numeric `buzzard.1` is a translator from a pseudo-assembly language into a subset of C
arguments. suitable for execution on Charles Babbage's Analytical Engine. Or rather, the
`#define`s in `buzzard.1` are that translator. The rest of buzzard.1 is a
% buzzard.1 0 <num> buzzard.1 program. If you run the whole mess through CPP and a beautifier, you
will print out the factorial of <num>. will see that all you have is a loop that runs until a variable is not 0, and a
sequence of assignment statements. The assignment statements are all of the
% buzzard.1 <num1> <num2> form `a op= b` or `a = b`. The Engine actually allowed 'a = b op c'. Only `+`,
will print out the largest common factor of num1 and num2. `-`, `/`, and `*` are used--no boolean or bitwise operators. The infinite loop
could have been simulated on the AE by connecting the already-processed card
% buzzard.1 1 <num2> stack to the input card stack.
will print out a factor of num2 if it's composite, or else
it will print a string informing you that num2 is prime.
If the first argument is less than 0, it prints out an
error message.
What it is: ### How to try to understand it:
buzzard.1 is a translator from a pseudo-assembly language into a *Don't* expand the `#define`s! Rather, decipher what they do.
subset of C suitable for execution on Charles Babbage's Some of them are mere obfuscation fodder, put in to encourage
Analytical Engine. Or rather, the #defines in buzzard.1 are that you to run it through CPP.
translator. The rest of buzzard.1 is a buzzard.1 program. If you run the
whole mess through CPP and a beautifier, you will see that
all you have is a loop that runs until a variable is not 0, and
a sequence of assignment statements. The assignment statements
are all of the form 'a op= b' or 'a = b'. The Engine actually
allowed 'a = b op c'. Only +, -, /, and * are used--no boolean
or bitwise operators. The infinite loop could have been
simulated on the AE by connecting the already-processed
card stack to the input card stack.
How to try to understand it: ### What it doesn't quite do right as an emulator:
*Don't* expand the #defines! Rather, decipher what they do. Because of the definitional constraints, calls to `atoi(argv[#])` also appear
Some of them are mere obfuscational fodder, put in to encourage inside the loop. These could be put outside-- the initial values of the
you to run it through CPP. "registers" on the AE--but the macro conventions being used didn't lend
themselves to it.
What it doesn't quite do right as an emulator: ### Theoretical observations:
Because of the definitional constraints, calls to atoi(argv[#]) The simulated comparison operations only work with a certain
also appear inside the loop. These could be put outside-- range of numbers; numbers that are too large will cause the
the initial values of the "registers" on the AE--but the output code to fail to simulate the input code. This means
macro conventions being used didn't lend themselves to it. that this implementation of buzzard.1 would not be Turing-complete,
even if the AE could process indefinitely-sized numbers.
However, this is actually a constraint that no actual computer
can meet, either, so we can conclude that if an AE, with
card bins connected as suggested above, were hooked up to
a memory unit of sufficient size, it would be as Turing
complete as any existing machine (or, for those who
interpret that excessively critically, i.e. "not at all" for
both, we can simply say that the AE could simulate any
existing machine--although not quickly). This would be a
good place to site some references on the AE, but I don't
have any.
Theoretical observations: ### Definition of the input language to buzzard.1 (OR What Exactly Do All The Macros Do?):
The simulated comparison operations only work with a certain [Don't read this unless you're stumped!]
range of numbers; numbers that are too large will cause the
output code to fail to simulate the input code. This means
that this implementation of buzzard.1 would not be Turing-complete,
even if the AE could process indefinitely-sized numbers.
However, this is actually a constraint that no actual computer
can meet, either, so we can conclude that if an AE, with
card bins connected as suggested above, were hooked up to
a memory unit of sufficient size, it would be as Turing
complete as any existing machine (or, for those who
interpret that excessively critically, i.e. "not at all" for
both, we can simply say that the AE could simulate any
existing machine--although not quickly). This would be a
good place to site some references on the AE, but I don't
have any.
Definition of the input language to buzzard.1:
(OR What Exactly Do All The Macros Do?)
[Don't read this unless you're stumped!]
MACRO FUNCTION MNEMONIC MACRO FUNCTION MNEMONIC
V begin variable declarations Variable
C begin program Code
Q(b,a) let b = a eQuals V begin variable declarations Variable
A(c,a,b) let c = a + b Add C begin program Code
S(c,a,b) let c = a - b Subtract
D(c,a,b) let c = a / b, where b is a *constant* Divide
U(c,a,b) let c = a / b, where b is anything Unknown
M(c,a,b) let c = a * b Multiply
O(c,a,b) let c = a boolean-or b (a,b are 0 or 1) Or Q(b,a) let b = a eQuals
B(b, a) let b = boolean value of a Boolean A(c,a,b) let c = a + b Add
P(b, a) let b = 1 if a>0, else 0. Positive S(c,a,b) let c = a - b Subtract
D(c,a,b) let c = a / b, where b is a *constant* Divide
U(c,a,b) let c = a / b, where b is anything Unknown
M(c,a,b) let c = a * b Multiply
l emit next sequential label Label O(c,a,b) let c = a boolean-or b (a,b are 0 or 1) Or
J(x) goto label #x Jump B(b, a) let b = boolean value of a Boolean
P(b, a) let b = 1 if a>0, else 0. Positive
Z(a,d) if a is 0 goto d (a is 0 or 1) Zero l emit next sequential label Label
E(a,d) if a is 1 goto d (a is 0 or 1) Else J(x) goto label #x Jump
H halt Haly Z(a,d) if a is 0 goto d (a is 0 or 1) Zero
E(a,d) if a is 1 goto d (a is 0 or 1) Else
K(x) let x = number of cmd line arguments Kount H halt Haly
G(x,y) let x = the value if the yth argument Get argument
T end of code; begin output section Terminate K(x) let x = number of cmd line arguments Kount
X(y) print out hexadecimal value of y heXadecimal G(x,y) let x = the value if the yth argument Get argument
T end of output section and program Terminate
You can figure out the other macros yourself. In the sample T end of code; begin output section Terminate
program, I've actually implemented subroutines by saving X(y) print out hexadecimal value of y heXadecimal
a return address in a variable and then jumping to a routine-- T end of output section and program Terminate
specifically, a routine that converts a number into bcd, so
it can be output by the hexadecimal output statement. You can figure out the other macros yourself. In the sample program, I've
actually implemented subroutines by saving a return address in a variable and
then jumping to a routine-- specifically, a routine that converts a number into
bcd, so it can be output by the hexadecimal output statement.
How it works, i.e., how to get useful programs from a op= b ### How it works, i.e., how to get useful programs from a op= b
The essential statement to be able to do to do interesting The essential statement to be able to do to do interesting
programs is the conditional (given that we have loops). programs is the conditional (given that we have loops).
Since all you can do in the given operation output set is Since all you can do in the given operation output set is
assignment, we implement "conditional assignment": assignment, we implement "conditional assignment":
if (x) y = z; if (x) y = z;
To implement this, we constrain x to be either 0 or 1, and To implement this, we constrain x to be either 0 or 1, and
simply compute: simply compute:
y = (z * x) + (y * (1-x)); y = (z * x) + (y * (1-x));
This is more obscured by factoring out common terms and restricting This is more obscured by factoring out common terms and restricting
ourselves to two operand operations: ourselves to two operand operations:
temp = z; (temp == z) temp = z; (temp == z)
temp -= y; (temp == z-y) temp -= y; (temp == z-y)
@@ -180,41 +185,40 @@ Most Obfuscated Algorithm:
( == z*x - y*x + y) ( == z*x - y*x + y)
( == z*x + y*(1-x)) ( == z*x + y*(1-x))
Next we imagine we have a pc. "I'm supposed to execute statement Next we imagine we have a pc. "I'm supposed to execute statement
pc next", says our emulator. But suppose we're not currently pc next", says our emulator. But suppose we're not currently
coming up on statement pc. To handle this, we simply make every coming up on statement pc. To handle this, we simply make every
operation conditional on the pc having the correct value: operation conditional on the pc having the correct value:
if (pc == some_constant) y = z; if (pc == some_constant) y = z;
To combine this with other operations, we simply multiply by our To combine this with other operations, we simply multiply by our
(pc == some_constant) flag right after the above 'temp *= x'. (`pc == some_constant`) flag right after the above 'temp `*= x'`.
To evaluate pc == some_constant requires one more trick. We assume To evaluate `pc == some_constant` requires one more trick. We assume that
that numbers can only be in some limited range. Then we use successive numbers can only be in some limited range. Then we use successive additions and
additions and divisions to reduce that number down to -1, 0, or 1. divisions to reduce that number down to `-1`, `0`, or `1`. You could, for
You could, for instance, do this by using mod by 2 and div by 2 to instance, do this by using mod by `2` and div by `2` to count bits, stopping
count bits, stopping after, say, 32 iterations. The number you get after, say, `32` iterations. The number you get is between `0..32`, so another
is between 0..32, so another 6 iterations on it reduces it to 0..6. `6` iterations on it reduces it to 0..6. Three iterations on this produces
Three iterations on this produces 0..2 (3 would be '7'), and then `0..2` (`3` would be `7`), and then two iterations on this produces 0 or 1.
two iterations on this produces 0 or 1. Instead I use a shorter Instead I use a shorter two-divide approach that assumes I'm allowed to use
two-divide approach that assumes I'm allowed to use numbers slightly numbers slightly larger than the numbers I'm operating on.
larger than the numbers I'm operating on.
A similar approach is used to detect positive numbers. A similar approach is used to detect positive numbers.
Note that the obvious code to compute y/x will not work correctly, Note that the obvious code to compute `y/x` will not work correctly,
because every value is computed, even if the pc is not set correctly; because every value is computed, even if the pc is not set correctly;
if x is ever 0 when the real C code reaches this computation the if `x` is ever `0` when the real C code reaches this computation the
program will die. (This was discovered the hard way.) The solution program will die. (This was discovered the hard way.) The solution
is to stick x in a temporary, and if the current code is not about is to stick `x` in a temporary, and if the current code is not about
to be executed, set the temporary to 1. to be executed, set the temporary to `1`.
Thanks to Bill Pugh for bringing the loopability of the AE to
my attention.
buzzard.1 was originally named 'cb', for obvious reasons, and is Thanks to Bill Pugh for bringing the loop-ability of the AE to
dedicated to the memory of Charles Barrett. my attention.
`buzzard.1` was originally named 'cb', for obvious reasons, and is
dedicated to the memory of Charles Barrett.
## Copyright and CC BY-SA 4.0 License: ## Copyright and CC BY-SA 4.0 License: