fix Makefile for 2020/carlini and test a README.md style
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@@ -1,47 +1,56 @@
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# Best of show - abuse of libc
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```
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Nicholas Carlini <nicholas@carlini.com>
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<https://nicholas.carlini.com>
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```
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## To build:
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The code for this entry can be found in prog.c
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```sh
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make
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```
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## Judges' comments:
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### To use:
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make
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./prog
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### To run:
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```sh
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./prog
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```
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### Try:
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# Play with a friend, P1 and P2, squares are numbered 1..9
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./prog
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```sh
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# Play with a friend, P1 and P2, squares are numbered 1..9
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./prog
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```
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### Selected Judges Remarks:
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## Judges' comments:
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This program consists of a single print statement wrapped in a while loop. You
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would not think that this would amount to much, but you would be very, very
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wrong. A clue to what is happening and how this works is encoded in the
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ASCII art of the program source.
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## Author's comments:
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HOW ABOUT A NICE GAME OF <strike>CHESS</strike>TIC-TAC-TOE?
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### USAGE
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cc -o prog prog.c
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./prog
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```sh
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cc -o prog prog.c
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./prog
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```
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Alternates between P1 and P2. Enter a digit [1-9] to move:
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1 | 2 | 3
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---------
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4 | 5 | 6
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---------
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7 | 8 | 9
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```
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1 | 2 | 3
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---------
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4 | 5 | 6
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---------
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7 | 8 | 9
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```
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The game ends if:
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@@ -49,7 +58,6 @@ The game ends if:
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* All squares are taken; neither player wins
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* A player makes an illegal move; their opponent wins
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### LIMITATIONS
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Bad things may happen if the entered move is outside of the range [1..9]
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@@ -69,16 +77,19 @@ compilers unhappy, and others slow. (For example, passing GCC the flags -Wall
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Under some terminals the program may exit early. The following should always work:
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echo "1 2 3 4 5 6 7" | ./prog
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```sh
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echo "1 2 3 4 5 6 7" | ./prog
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```
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### OBFUSCATION
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The entirety of the program consists of a single iterated call to printf.
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int main() {
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while(*d) printf(fmt, arg);
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}
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```c
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int main() {
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while(*d) printf(fmt, arg);
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}
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```
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Here, `fmt` is a single string, and `arg` is a series of arguments to printf.
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@@ -102,7 +113,6 @@ This program uses three printf format specifiers.
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- `%n` takes a pointer and writes (!!) the number of bytes printed so far.
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Okay, everyone probably knows this. Let's get a bit more advanced.
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Format specifiers can take extra "arguments".
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@@ -115,14 +125,17 @@ Format specifiers can take extra "arguments".
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- `"%3$.*4$d"`: print argument 3 to printf with as many zeros as in argument 4.
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For example, the following expression
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printf("%1$.*2$d%3$hhn", 5, 10, &x)
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```c
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printf("%1$.*2$d%3$hhn", 5, 10, &x)
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```
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will have the same effect as if we had written
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x = 10;
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```c
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x = 10;
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```
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because it will print out `0000000005` (5 padded to size 10) and then write the
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number of bytes written to x.
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@@ -143,27 +156,39 @@ We can use format strings to compute the OR/NOT of arbitrary "bits".
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We'll start with the simplest, OR:
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printf("%1$s%2$s%3$hhn", a, b, c)
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```c
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printf("%1$s%2$s%3$hhn", a, b, c)
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```
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will compute
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*c = strlen(a) + strlen(b)
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```c
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*c = strlen(a) + strlen(b)
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```
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but given that `strlen(x)` is `1` for a 1-bit and `0` for a 0-bit, we have
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*c = a | b
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```c
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*c = a | b
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```
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Computing the NOT of a single value is also easy:
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printf("%1$255d%1$s%hhn", a, b)
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```c
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printf("%1$255d%1$s%hhn", a, b)
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```
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will compute
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*b = (strlen(a)+255)%256 = strlen(a)-1
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```c
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*b = (strlen(a)+255)%256 = strlen(a)-1
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```
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and again, because `strlen(x)` is either `1` or `0` we have
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*c = !b
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```c
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*c = !b
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```
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From here we can compute any binary circuit. Doing something efficient,
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though, still takes work.
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@@ -177,11 +202,13 @@ To detect who has won, we implement the following logic. Let A, B, and C be
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pointers to three squares in a row to test, and D be where to save if there is a
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win or not.
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"%A$s%B$s%C$s%1$253d%11$hhn" // r11 = !(*A & *B & *C)
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ZERO
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"%11$s%1$255d%11hhn" // r11 = !r11
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ZERO
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"%11$s%D$s%D$hhn" // *D = *D | r11
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```c
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"%A$s%B$s%C$s%1$253d%11$hhn" // r11 = !(*A & *B & *C)
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ZERO
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"%11$s%1$255d%11hhn" // r11 = !r11
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ZERO
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"%11$s%D$s%D$hhn" // *D = *D | r11
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```
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That is, we set `*D` to `1` if there is a three-in-a-row. We repeat this for all
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possible three-in-a-row configurations, for both players.
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@@ -189,7 +216,9 @@ possible three-in-a-row configurations, for both players.
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The ZERO macro ensures that the number of bytes written out is 0 mod 256 with
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the following expression
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"%1$hhn%1$s" (repeated 256 times)
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```c
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"%1$hhn%1$s" (repeated 256 times)
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```
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where argument 1 is a pointer to a temporary variable followed by a null byte.
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@@ -206,11 +235,15 @@ bits to Xs and Os to print out. This is actually rather straightforward. Given
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in 1$ the pointer to player 1's square, and 2$ the pointer to player 2's, and
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in 3$ the pointer to the board string, we can compute
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"%1$s" (repeated 47 times) "%2$s" (repeated 56 times) %1$32d%3$hhn"
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```c
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"%1$s" (repeated 47 times) "%2$s" (repeated 56 times) %1$32d%3$hhn"
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```
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which will, in effect, compute
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*r3 = (*r1) * 47 + (*r2) * 56 + 32
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```c
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*r3 = (*r1) * 47 + (*r2) * 56 + 32
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```
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which will output `' '` if neither are true, `'X'` if r1 is, or `'O'` if r2 is.
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@@ -220,7 +253,9 @@ which will output `' '` if neither are true, `'X'` if r1 is, or `'O'` if r2 is.
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In order to be able to finally display the board, while still only using one
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printf statement, we finish the statement with
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"\n\033[2J\n%26$s"
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```c
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"\n\033[2J\n%26$s"
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```
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which is the escape sequence to clear the screen, and then prints argument 26.
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Argument 26 is a pointer to a char* in memory, that initially is undefined,
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@@ -229,12 +264,14 @@ tic-tac-toe board.
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After the board, we need to print one of the following strings:
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P1>_
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P2>_
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P1 WINS
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P2 WINS
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P1 TIES
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P2 TIES
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```
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P1>_
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P2>_
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P1 WINS
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P2 WINS
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P1 TIES
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P2 TIES
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```
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Depending on if it's P1 or P2's turn to move, the game is over and someone
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won, or the game is over and it is a draw.
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@@ -242,7 +279,9 @@ won, or the game is over and it is a draw.
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This turns out not to be as hard as it might look. Using the same trick as
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before, we set byte four to be
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*byte4 = is_win * 'W' + is_tie * 'T'
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```c
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*byte4 = is_win * 'W' + is_tie * 'T'
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```
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The byte `'I'` and `'S'` can always be the same, and we do the same for `'E'`/`'N'`.
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@@ -254,19 +293,23 @@ ends. It should just exit.
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One option would be to implement the logic as
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printf()
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while (*ok) {
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scanf();
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printf();
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}
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```c
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printf()
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while (*ok) {
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scanf();
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printf();
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}
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```
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but this would DOUBLE the number of calls to printf we require. So instead we
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implement it like this
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while (*ok) {
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scanf();
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printf();
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}
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```c
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while (*ok) {
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scanf();
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printf();
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}
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```
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(In reality we actually pass `scanf()` as an argument to avoid the extra
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statement, but it has the same effect.)
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@@ -276,10 +319,10 @@ block before the first `printf()`, but we initialize the `scanf()` format to the
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string so that it returns right away without blocking. The first time the `printf()`
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call runs, it writes out `"%hhd"` to create the `scanf()` format string.
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-----------------------------------------------------------------------------------------------------
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## Copyright:
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(c) Copyright 1984-2020, [Leo Broukhis, Simon Cooper, Landon Curt Noll][judges] - All rights reserved
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This work is licensed under a [Creative Commons Attribution-ShareAlike 3.0 Unported License][cc].
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[judges]: http://www.ioccc.org/judges.html
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[cc]: http://creativecommons.org/licenses/by-sa/3.0/
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-----------------------------------------------------------------------------------------------------
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