Get 1989 entries to compile
There were some which cannot be compiled under clang. One entry we used a patch from Yusuke Endoh to get the entry to compile; thank you Yusuke for your assistance! Another one we used a patch from Yusuke (thank you!) but I further fixed it so that it would work with clang. It had to be done: IOCCC Tetris has to work! This was done by changing the environment arg to main() to be a char * in main() itself rather than a char **. Using a char ** in main() third arg got it to compile but only one key worked with clang so I changed it to be not the third argument but a regular char * in main(). **a (third arg) evaluated correctly but *a[1] and onward was actually part of the environment. Again IOCCC Tetris has to work! I also fixed fubar to work which took a few tweaks. README.md files were also updated.
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@@ -1,4 +1,4 @@
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Most complex algorithm:
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## Most complex algorithm:
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Paul E. Black
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CIRRUS LOGIC, Inc.
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@@ -7,100 +7,109 @@ Most complex algorithm:
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95035
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USA
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Judges notes:
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## To build:
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The original source contained a long line which caused many
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mailers to barf. The original file may be re-constructed by
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removeing the trailing "\" on line 12 and joining lines 12 and
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13 together without a space.
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make all
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WHAT FOLLOWS IS A DETAILED PROGRAM EXPLINATION AND SPOILER.
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IF YOU WANT A REAL CHALLENGE, DON'T READ ANY FURTHER AND TRY
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TO UNDERSTAND THE PROGRAM VIA THE SOURCE.
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NOTE: this entry will likely dump core with arm64 chips. We could get it to work
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with modern systems with x86_64 but systems but we at least with macOS arm64
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this program dumps core. We could not test Intel Macs.
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## Judges' comments:
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The original source contained a long line which caused many
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mailers to barf. The original file may be re-constructed by
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removing the trailing "\" on line 12 and joining lines 12 and
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13 together without a space.
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WHAT FOLLOWS IS A DETAILED PROGRAM EXPLANATION AND SPOILER.
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IF YOU WANT A REAL CHALLENGE, DON'T READ ANY FURTHER AND TRY
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TO UNDERSTAND THE PROGRAM VIA THE SOURCE.
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Selected notes from the author:
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This programs computes and prints Fibonacci numbers by
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simulating a Turing machine with the proper program.
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Understanding the C program, i.e., a Turing machine simulator,
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is only the first and simplest step. The Turing machine
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program must be understood, too! (it is trivial, perhaps even
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natural to write incredibly obscure Turing programs.)
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This programs computes and prints Fibonacci numbers by
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simulating a Turing machine with the proper program.
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Understanding the C program, i.e., a Turing machine simulator,
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is only the first and simplest step. The Turing machine
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program must be understood, too! (it is trivial, perhaps even
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natural to write incredibly obscure Turing programs.)
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If the program is invoked with an operand, the operand is used
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as the Turing program. It includes a "trace" facility,
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subroutine r (commented out for obscurity), to help write and
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debug Turing programs. Just the thing for some fun in a Theory
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of Computation class.
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If the program is invoked with an operand, the operand is used
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as the Turing program. It includes a "trace" facility,
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subroutine r (commented out for obscurity), to help write and
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debug Turing programs. Just the thing for some fun in a Theory
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of Computation class.
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The Turing machine tape is represented as a doubly linked list
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of pointers. The forward and backward links are XOR'd together
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and stored in one pointer. If we always keep one of the links
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on hand, we can recover the other link at any time. The
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variable q is the scan head (and a pointer at some tape cell),
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and p is a "previous" link. The state of the tape is stored in
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the low order bit of the pointer. Since we always allocate an
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even number of bytes, the low order bit carries no information
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(see portability below.) Memory representing a tape cell is
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allocated when the cell is first scanned. Thus the simulation
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begins with a tape effectively the size of virtual memory set
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to all zeros. Since a header can be added to any Turing
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program to write initial data and position the scan head, this
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is little loss of generality.
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The Turing machine tape is represented as a doubly linked list
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of pointers. The forward and backward links are XOR'd together
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and stored in one pointer. If we always keep one of the links
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on hand, we can recover the other link at any time. The
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variable q is the scan head (and a pointer at some tape cell),
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and p is a "previous" link. The state of the tape is stored in
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the low order bit of the pointer. Since we always allocate an
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even number of bytes, the low order bit carries no information
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(see portability below.) Memory representing a tape cell is
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allocated when the cell is first scanned. Thus the simulation
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begins with a tape effectively the size of virtual memory set
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to all zeros. Since a header can be added to any Turing
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program to write initial data and position the scan head, this
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is little loss of generality.
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The simulated Turing machine has a single tape with either an 1
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or a 0 in each cell. The Turing machine language format is a
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string of three bytes. The first byte is the current state.
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The second byte is the next state. (The last bit of states is
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ignored, e.g., B and C are the same state, in an attempt to be
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able to have interesting words in the program.) The third byte
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is composed of bits. Bit 1 (2&byte) is the symbol scanned,
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i.e. an instruction is selected for state and for a match with
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the symbol under the scan head. Bit 2 (4&byte) is the new
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symbol to be written to the cell. Bit 3 (8&byte) is the
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direction to move the scan head: 0 for left and 1 for right.
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If bit 4 (16&byte) is true, the next character is sent to
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stdout. (I added this feature so programs could print
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results.)
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The simulated Turing machine has a single tape with either an 1
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or a 0 in each cell. The Turing machine language format is a
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string of three bytes. The first byte is the current state.
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The second byte is the next state. (The last bit of states is
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ignored, e.g., B and C are the same state, in an attempt to be
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able to have interesting words in the program.) The third byte
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is composed of bits. Bit 1 (2&byte) is the symbol scanned,
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i.e. an instruction is selected for state and for a match with
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the symbol under the scan head. Bit 2 (4&byte) is the new
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symbol to be written to the cell. Bit 3 (8&byte) is the
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direction to move the scan head: 0 for left and 1 for right.
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If bit 4 (16&byte) is true, the next character is sent to
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stdout. (I added this feature so programs could print
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results.)
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The Turing machine has next state 'j' when it begins. The
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cycle is 1) exit if the state is 'x', 2) find the next
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instruction (given the state and the character under the scan
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head). [The program string is searched forward for the next
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matching instruction. If the end of the string is reached, the
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search begins again at the first of the string. Thus states
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can be used as local labels in different places.] 3) change to
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the next state, 4) print a character if indicated, 5) write the
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tape symbol, and 6) move the scan head. The cycle then repeats
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with step 1. A call to the trace routine is just before step
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2, but is commented out.
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The Turing machine has next state 'j' when it begins. The
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cycle is 1) exit if the state is 'x', 2) find the next
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instruction (given the state and the character under the scan
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head). [The program string is searched forward for the next
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matching instruction. If the end of the string is reached, the
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search begins again at the first of the string. Thus states
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can be used as local labels in different places.] 3) change to
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the next state, 4) print a character if indicated, 5) write the
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tape symbol, and 6) move the scan head. The cycle then repeats
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with step 1. A call to the trace routine is just before step
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2, but is commented out.
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The following ROT13'ed text is a quick outline of the actual
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Turing program:
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The following ROT13'ed text is a quick outline of the actual
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Turing program:
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Urer vf n dhvpx bhgyvar bs gur Ghevat cebtenz: gur
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cerivbhf naq pheerag Svobanppv ahzoref ner xrcg va onfr
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1 sbez jvgu gur pheerag ba gur evtug. Gur svefg guerr
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fgrcf frg hc gur svefg gjb ahzoref, 1 naq 1. Gura
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[ortvaavat jvgu "@ ("] n znexre bs VVV vf perngrq naq
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gur pheerag ahzore vf pbcvrq gb gur evtug bs gur
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znexre. Gura [ortvaavat jvgu "BI "] gur ahzore vf
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pbairegrq gb ovanel ol ercrngrq qvivqvat ol 2 yrnivat V
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sbe erznvaqre 1, naq VV sbe erznvaqre 0. Arkg
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[ortvaavat jvgu "JI "] gur ovanel ercerfragngvba vf
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cevagrq naq vgf flzobyf naq gur znexre ner renfrq.
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Svanyyl [ortvaavat jvgu "RRa"] gur gjb ahzoref ner
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nqqrq naq gur pheerag ahzore pbcvrq gb gur yrsg gb
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orpbzr gur cerivbhf. Gura gur plpyr ercrngf.
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Urer vf n dhvpx bhgyvar bs gur Ghevat cebtenz: gur
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cerivbhf naq pheerag Svobanppv ahzoref ner xrcg va onfr
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1 sbez jvgu gur pheerag ba gur evtug. Gur svefg guerr
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fgrcf frg hc gur svefg gjb ahzoref, 1 naq 1. Gura
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[ortvaavat jvgu "@ ("] n znexre bs VVV vf perngrq naq
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gur pheerag ahzore vf pbcvrq gb gur evtug bs gur
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znexre. Gura [ortvaavat jvgu "BI "] gur ahzore vf
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pbairegrq gb ovanel ol ercrngrq qvivqvat ol 2 yrnivat V
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sbe erznvaqre 1, naq VV sbe erznvaqre 0. Arkg
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[ortvaavat jvgu "JI "] gur ovanel ercerfragngvba vf
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cevagrq naq vgf flzobyf naq gur znexre ner renfrq.
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Svanyyl [ortvaavat jvgu "RRa"] gur gjb ahzoref ner
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nqqrq naq gur pheerag ahzore pbcvrq gb gur yrsg gb
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orpbzr gur cerivbhf. Gura gur plpyr ercrngf.
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The program requires that the lowest bit of a pointer to be 0.
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The program requires that the lowest bit of a pointer to be 0.
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I could have squeezed the program under 1024 bytes without the
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trace subroutine, but I felt it was important for understanding
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the program. Besides it is fun to watch the tape zooming back
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and forth as the program runs. A much better debugger or trace
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could easily be added.
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I could have squeezed the program under 1024 bytes without the
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trace subroutine, but I felt it was important for understanding
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the program. Besides it is fun to watch the tape zooming back
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and forth as the program runs. A much better debugger or trace
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could easily be added.
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Copyright (c) 1989, 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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