complex-iterator.c

/**
  [2018-06-22] Challenge #364 [Hard] Tiling with Pentominos
  /r/dailyprogrammer https://redd.it/8t4440

  --> NOT A FULL SOLUTION! block flip/rotation experiment. <--

  This totally shouldn't be done this way...
  It actually works but way way too messy!
  Attempted full solution based on this was ugly and failed.
**/

#include <stdio.h>
#include <stdlib.h>

typedef struct { int mark; int width; int heigth; char flip_rots[9]; char *pattern; } pattern_t;

pattern_t pats[] = {
     { 'a', 5, 1, "1...5...", "aaaaa" },
     { 'b', 3, 3, "1.......", ".b$bbb$.b" },
     { 'c', 3, 3, "12..56..", ".cc$.c$cc" },
     { 'd', 3, 3, "1234....", "d$d$ddd" },
     { 'e', 3, 3, "1.3.56..", "eee$.e$.e" },
     { 'f', 3, 3, "1234....", "f$ff$.ff" },
     { 'g', 3, 2, "1.3.56..", "ggg$g.g" },
     { 'h', 4, 2, "12345678", "h$hhhh" },
     { 'i', 4, 2, "12345678", "..ii$iii" },
     { 'j', 4, 2, "12345678", "..j$jjjj" },
     { 'k', 3, 3, "12345678", ".k$kkk$..k" },
     { 'l', 2, 3, "12345678", "ll$ll$.l" }
};

typedef struct { int done; int x; int y; int startx; int endx; int starty; int endy; int rot; char *cmd; } itr_t;

itr_t mk_itr(char r, pattern_t pat) {
    itr_t ret;
    int w = pat.width - 1;
    int h = pat.heigth - 1;
    char *c = pat.pattern;
    switch(r) {                // d  X  Y bx ex by ey  r  c
        case '1' : ret = (itr_t){ 0, 0, 0, 0, w, 0, h, 0, c }; break;
    case '2' : ret = (itr_t){ 0, w, 0, w, 0, 0, h, 0, c }; break;
    case '3' : ret = (itr_t){ 0, 0, h, 0, w, h, 0, 0, c }; break;
    case '4' : ret = (itr_t){ 0, w, h, w, 0, h, 0, 0, c }; break;

        case '5' : ret = (itr_t){ 0, 0, 0, 0, h, 0, w, 1, c }; break;
    case '6' : ret = (itr_t){ 0, 0, w, 0, h, w, 0, 1, c }; break;
    case '7' : ret = (itr_t){ 0, h, 0, h, 0, 0, w, 1, c }; break;
    case '8' : ret = (itr_t){ 0, h, w, h, 0, w, 0, 1, c }; break;
    default : fprintf(stderr, "wrong way! no flip_rots orientation '%c' (0x%02x)\n", r, r);
          exit(1);
    };
    return ret;
}

void itr_step(itr_t *it) {
    int dx = (it->endx > it->startx) ? +1 : -1;
    int dy = (it->endy > it->starty) ? +1 : -1;
    if(it->rot) {
        if(*it->cmd == '$') { it->y = it->starty; it->x += dx; }
    else { it->y += dy; }
    }
    else {
        if(*it->cmd == '$') { it->x = it->startx; it->y += dy; }
    else { it->x += dx; }
    }
    if(*it->cmd == '\0') it->done = 1;
    else it->cmd++;
    return;
}

int main() {
    int p, r, i, j;
    char a[5][5];
    itr_t it;
    for(p = 0; p < 12; p++) {
        for(r = '1'; r <= '8'; r++) {
            for(i = 0; i < 5; i++) for(j = 0; j < 5; j++) a[i][j] = '.';

        // try our complex iterator...
        it = mk_itr(r, pats[p]);
        for(itr_t it = mk_itr(r, pats[p]); !it.done; itr_step(&it)) {
            if(*it.cmd != '.' && *it.cmd != '$') a[it.x][it.y] = *it.cmd;
        }

        // print the result
        for(j = 0; j < 5; j++) {
            for(i = 0; i < 5; i++) putc(a[i][j], stdout);
        putc('\n', stdout);
        }
        putc('\n', stdout);
        }
    }
    return 0;
}