IOCCC image by Matt Zucker

The International Obfuscated C Code Contest

1989/paul - Most complex algorithm

Author:

To build:

    make all

There is an alternate version with a trace routine that the author had commented out. See alternate code below for more details.

Try:

    ./paul

Alternate code:

This version was described by the author like:

I could have squeezed the program under 1024 bytes without the trace subroutine, but I felt it was important for understanding the program. Besides it is fun to watch the tape zooming back and forth as the program runs. A much better debugger or trace could easily be added.

Alternate build:

    make alt

Alternate use:

    ./paul.alt

Judges’ remarks:

The author’s comments are a detailed program explanation, if you wish to learn more about the obfuscation method used.

Historical note:

The original source contained a long line which caused many mailers to barf. The original file may be re-constructed by removing the trailing \ on line 12 and joining lines 12 and 13 together without a space.

However as this is no longer a problem the trailing \ has been restored in all versions, both the original and the fixed for modern systems version.

NOTICE to those who wish for a greater challenge:

If you want a greater challenge, don’t read any further: just try to understand the program via the source.

If you get stuck, come back and read below for additional hints and information.

Author’s remarks:

This programs computes and prints Fibonacci numbers by simulating a Turing machine with the proper program. Understanding the C program, i.e., a Turing machine simulator, is only the first and simplest step. The Turing machine program must be understood, too! (it is trivial, perhaps even natural to write incredibly obscure Turing programs.)

If the program is invoked with an operand, the operand is used as the Turing program. It includes a “trace” facility, subroutine r (commented out for obscurity), to help write and debug Turing programs. Just the thing for some fun in a Theory of Computation class.

The Turing machine tape is represented as a doubly linked list of pointers. The forward and backward links are XORed together and stored in one pointer. If we always keep one of the links on hand, we can recover the other link at any time. The variable q is the scan head (and a pointer at some tape cell), and p is a “previous” link. The state of the tape is stored in the low order bit of the pointer. Since we always allocate an even number of bytes, the low order bit carries no information (see portability below.) Memory representing a tape cell is allocated when the cell is first scanned. Thus the simulation begins with a tape effectively the size of virtual memory set to all zeros. Since a header can be added to any Turing program to write initial data and position the scan head, this is little loss of generality.

The simulated Turing machine has a single tape with either a 1 or a 0 in each cell. The Turing machine language format is a string of three bytes. The first byte is the current state. The second byte is the next state. (The last bit of states is ignored, i.e., B and C are the same state, in an attempt to be able to have interesting words in the program.) The third byte is composed of bits. Bit 1 (2&byte) is the symbol scanned, i.e. an instruction is selected for state and for a match with the symbol under the scan head. Bit 2 (4&byte) is the new symbol to be written to the cell. Bit 3 (8&byte) is the direction to move the scan head: 0 for left and 1 for right. If bit 4 (16&byte) is true, the next character is sent to stdout. (I added this feature so programs could print results.)

The Turing machine has next state j when it begins. The cycle is (1) exit if the state is x, (2) find the next instruction (given the state and the character under the scan head). (The program string is searched forward for the next matching instruction. If the end of the string is reached, the search begins again at the first of the string. Thus states can be used as local labels in different places.) (3) change to the next state, (4) print a character if indicated, (5) write the tape symbol, and (6) move the scan head. The cycle then repeats with step 1. A call to the trace routine is just before step 2, but is commented out.

Quick outline of the Turing program:

The previous and current Fibonacci numbers are kept in base 1 form with the current on the right. The first three steps set up the first two numbers, 1 and 1. Then, beginning with ‘@ (’, a marker of ‘III’ is created and the current number is copied to the right of the marker. Then, beginning with ‘OV’, the number is converted to binary by repeatedly dividing by 2 leaving I for remainder 1, and II for remainder ‘0’. Next, beginning with ‘WV’, the binary representation is printed and its symbols and the marker are erased. Finally, beginning with ‘EEn’, the two numbers are added and the current number copied to the left to become the previous. Then the cycle repeats.

Portability

The program requires that the lowest bit of a pointer to be 0.

I could have squeezed the program under 1024 bytes without the trace subroutine, but I felt it was important for understanding the program. Besides it is fun to watch the tape zooming back and forth as the program runs. A much better debugger or trace could easily be added.

Inventory for 1989/paul

Primary files

Secondary files


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