• 2 Posts
  • 70 Comments
Joined 2 years ago
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Cake day: June 12th, 2023

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  • C

    Merry Christmas everyone!

    Code
    #include "common.h"
    
    int
    main(int argc, char **argv)
    {
    	static char buf[7];
    	static short h[500][5];	/* heights */
    	static short iskey[500];
    	int p1=0, nh=0, i,j,k;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    	
    	for (nh=0; !feof(stdin) && !ferror(stdin); nh++) {
    		assert(nh < (int)LEN(h));
    
    		for (i=0; i<7; i++) {
    			fgets(buf, 7, stdin);
    			if (i==0)
    				iskey[nh] = buf[0] == '#';
    			for (j=0; j<5; j++)
    				h[nh][j] += buf[j] == '#';
    		}
    
    		/* skip empty line */
    		fgets(buf, 7, stdin);
    	}
    
    	for (i=0; i<nh; i++)
    	for (j=0; j<nh; j++)
    		if (iskey[i] && !iskey[j]) {
    			for (k=0; k<5 && h[i][k] + h[j][k] <= 7; k++) ;
    			p1 += k == 5;
    		}
    
    	printf("25: %d\n", p1);
    	return 0;
    }
    

    https://codeberg.org/sjmulder/aoc/src/branch/master/2024/c/day25.c

    Made the 1 second challenge with most of it to spare! 😎

    $ time bmake bench                                                                                                      
    day01  0:00.00  1912 Kb  0+88 faults                                                                                            
    day02  0:00.00  1992 Kb  0+91 faults 
    day03  0:00.00  1920 Kb  0+93 faults
    day04  0:00.00  1912 Kb  0+90 faults 
    day05  0:00.00  2156 Kb  0+91 faults
    day06  0:00.03  1972 Kb  0+100 faults
    day07  0:00.06  1892 Kb  0+89 faults
    day08  0:00.00  1772 Kb  0+87 faults 
    day09  0:00.02  2024 Kb  0+137 faults
    day10  0:00.00  1876 Kb  0+87 faults 
    day11  0:00.00  6924 Kb  0+3412 faults
    day12  0:00.00  1952 Kb  0+103 faults
    day13  0:00.00  1908 Kb  0+88 faults
    day14  0:00.05  1944 Kb  0+92 faults                                                                                            
    day15  0:00.00  2040 Kb  0+89 faults
    day16  0:00.03  2020 Kb  0+250 faults
    day17  0:00.00  1896 Kb  0+88 faults
    day18  0:00.00  1952 Kb  0+107 faults
    day19  0:00.01  1904 Kb  0+91 faults
    day20  0:00.01  2672 Kb  0+325 faults
    day21  0:00.00  1804 Kb  0+86 faults
    day22  0:00.03  2528 Kb  0+371 faults
    day23  0:00.02  2064 Kb  0+152 faults
    day24  0:00.00  1844 Kb  0+89 faults
    day25  0:00.00  1788 Kb  0+89 faults  
                                                                    
    real    0m0,359s
    

  • Oh my Kernigan, that was stressful. Really worried about not finishing there.

    Considered several approaches, the coolest of which would have been to test individual bits, propagate ‘suspicion’, etc, but it seemed too tricky.

    Eventually I needed to go do something other than worry about not finishing so I started writing a validator for the adder structure. Just a couple of rules later I had found 4 faults already and managed to write automated fixups for them!

    This means my solver is quite specific to my input but it can potentially be made more complete and I didn’t ‘cheat’ by hardcoding manual graph analysis.

    Code
    #include "common.h"
    
    /*
     * The approach behind part 2 was to essentially write a bunch of
     * validation rules for the structure of the adder, and then writing
     * fixups for problems it would find. That means it's likely quite
     * tailored to my input, but at least it's not hardcoding manual graph
     * analysis.
     */
    
    enum { W_NULL, W_OFF, W_ON };
    
    struct wire;
    
    struct wire {
    	struct wire *in[2];
    	char name[4];
    	char op; 		/* [A]ND, [O]R, [X]OR */
    	int8_t val;		/* W_NULL, W_OFF, W_ON */
    };
    
    static struct wire wires[320];
    static struct wire *zs[46], *swapped[8];
    static int nw, nsw;
    
    static struct wire *
    get_wire(const char *name)
    {
    	int i;
    
    	for (i=0; i<nw; i++)
    		if (!strcmp(name, wires[i].name))
    			return &wires[i];
    
    	assert(nw < (int)LEN(wires));
    	assert(strlen(name) < LEN(wires[i].name));
    
    	snprintf(wires[nw].name, sizeof(wires[nw].name), "%s", name);
    	return &wires[nw++];
    }
    
    
    static int
    cmp_wires(const void *a, const void *b)
    {
    	struct wire * const *wa = a;
    	struct wire * const *wb = b;
    
    	return strcmp((*wa)->name, (*wb)->name);
    }
    
    static int
    eval(struct wire *wire)
    {
    	int in1,in2;
    
    	if (wire->val)
    		return wire->val == W_ON;
    
    	assert(wire->in[0]);
    	assert(wire->in[1]);
    
    	in1 = eval(wire->in[0]);
    	in2 = eval(wire->in[1]);
    
    	wire->val = W_OFF + (
    	    wire->op=='A' ? in1 && in2 :
    	    wire->op=='O' ? in1 || in2 :
    	    wire->op=='X' ? in1 != in2 : (assert(!"bad op"), -1));
    
    	return wire->val == W_ON;
    }
    
    static void
    swap(struct wire *a, struct wire *b)
    {
    	struct wire tmp;
    
    	//printf("swapping %s and %s\n", a->name, b->name);
    
    	tmp = *a;
    
    	a->op = b->op;
    	a->in[0] = b->in[0];
    	a->in[1] = b->in[1];
    
    	b->op = tmp.op;
    	b->in[0] = tmp.in[0];
    	b->in[1] = tmp.in[1];
    
    	assert(nsw+2 <= (int)LEN(swapped));
    	swapped[nsw++] = a;
    	swapped[nsw++] = b;
    }
    
    static struct wire *
    find_z_xor(int bit)
    {
    	struct wire *xy_xor;
    	int i;
    
    	for (i=0; i<nw; i++) {
    		 /* must be a XOR */
    		if (wires[i].op != 'X')
    			continue;
    
    		 /* with an input XOR */
    		xy_xor = wires[i].in[0]->op == 'X' ? wires[i].in[0] :
    		         wires[i].in[1]->op == 'X' ? wires[i].in[1] :
    		         NULL;
    		if (!xy_xor)
    			continue;
    
    		 /* connected to the X and Y */
    		if (xy_xor->in[0]->name[0] != 'x' &&
    		    xy_xor->in[0]->name[0] != 'y')
    			continue;
    
    		 /* with the same bit number */
    		if (atoi(xy_xor->in[0]->name+1) != bit)
    			continue;
    
    		return &wires[i];
    	}
    
    	return NULL;
    }
    
    static struct wire *
    find_xy_and(int bit)
    {
    	int i;
    
    	for (i=0; i<nw; i++) {
    		 /* must be AND */
    		if (wires[i].op != 'A')
    			continue;
    
    		 /* must have XY inputs */
    		if ((wires[i].in[0]->name[0] != 'x'  ||
    		     wires[i].in[1]->name[0] != 'y') &&
    		    (wires[i].in[0]->name[0] != 'y'  ||
    		     wires[i].in[1]->name[0] != 'x'))
    			continue;
    		
    		 /* with the right bit number */
    		if (atoi(wires[i].in[0]->name+1) != bit ||
    		    atoi(wires[i].in[0]->name+1) != bit)
    			continue;
    		
    		return &wires[i];
    	}
    
    	return NULL;
    }
    
    static void
    fsck_carry_or(struct wire *or, int bit)
    {
    	struct wire *wire;
    	int i;
    
    	 /* both inputs must be AND; no fixup if neither */
    	assert(
    	    or->in[0]->op == 'A' ||
    	    or->in[1]->op == 'A');
    
    	for (i=0; i<2; i++) {
    		if (or->in[i]->op == 'A')
    			continue;
    
    		//printf("carry OR parent %s not AND\n", or->in[i]->name);
    
    		 /* only have fixup for the XY AND */
    		assert(
    		    or->in[!i]->in[0]->name[0] != 'x' &&
    		    or->in[!i]->in[0]->name[0] != 'y');
    
    		wire = find_xy_and(bit);
    		assert(wire);
    		swap(or->in[i], wire);
    	}
    }
    
    static void
    fsck_z(struct wire *z)
    {
    	struct wire *wire, *carry_or;
    	int bit;
    
    	assert(z->name[0] == 'z');
    
    	bit = atoi(z->name+1);
    
    	 /* first bit is very different */
    	if (!bit)
    		return;
    
    	 /* for the final bit, Z is the carry OR */
    	if (!zs[bit+1]) {
    		 /* no fixup if it isn't */
    		assert(z->op == 'O');
    
    		fsck_carry_or(z, bit-1);
    		return;
    	}
    
    	 /* must be a XOR */
    	if (z->op != 'X') {
    		//printf("Z %s isn't XOR\n", z->name);
    		wire = find_z_xor(bit);
    		assert(wire);
    		swap(z, wire);
    	}
    
    	 /* bit 2 and up must have a carry OR */
    	if (bit > 1) {
    		carry_or =
    		    z->in[0]->op == 'O' ? z->in[0] :
    		    z->in[1]->op == 'O' ? z->in[1] : NULL;
    		assert(carry_or);
    		fsck_carry_or(carry_or, bit-1);
    	}
    }
    
    int
    main(int argc, char **argv)
    {
    	struct wire *wire;
    	char buf[64], *rest, *lf, *name1,*name2, *opstr;
    	uint64_t p1=0;
    	int bit, i;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    
    	while ((rest = fgets(buf, sizeof(buf), stdin))) {
    		if (!strchr(buf, ':'))
    			break;
    
    		wire = get_wire(strsep(&rest, ":"));
    		wire->val = W_OFF + atoi(rest);
    
    	}
    
    	while ((rest = fgets(buf, sizeof(buf), stdin))) {
    		if ((lf = strchr(buf, '\n')))
    			*lf = '\0';
    
    		name1 = strsep(&rest, " ");
    		opstr = strsep(&rest, " ");
    		name2 = strsep(&rest, " ");
    		strsep(&rest, " ");
    
    		wire = get_wire(rest);
    		wire->in[0] = get_wire(name1);
    		wire->in[1] = get_wire(name2);
    		wire->op = opstr[0];
    	}
    
    	for (i=0; i<nw; i++)
    		if (wires[i].name[0] == 'z') {
    			bit = atoi(&wires[i].name[1]);
    			assert(bit >= 0);
    			assert(bit < (int)LEN(zs));
    			zs[bit] = &wires[i];
    		}
    
    	for (i=0; i < (int)LEN(zs); i++)
    		if (zs[i])
    			p1 |= (uint64_t)eval(zs[i]) << i;
    
    	for (i=0; i < (int)LEN(zs); i++)
    		if (zs[i])
    			fsck_z(zs[i]);
    
    	qsort(swapped, nsw, sizeof(*swapped), cmp_wires);
    
    	printf("24: %"PRIu64, p1);
    
    	for (i=0; i<nsw; i++)
    		printf(i ? ",%s" : " %s", swapped[i]->name);
    
    	putchar('\n');
    	return 0;
    }
    

    https://codeberg.org/sjmulder/aoc/src/branch/master/2024/c/day24.c

    Btw, spending some time on getting Graphviz output right did make studying the structure much easier!

    Graph snippet


  • C

    Really proud of this one! Started with with an O(n^atoms in the universe) scan which took 44s even after adding a dedup check.

    But iterating on a trick to encode the deltas for the dedup check, using it to build a mapping table here, a lookup there etc brought it down to a very fast, fairly low memory, linear complexity solution!

    Code
    #include "common.h"
    
    #define STEPS	2000
    #define NCODES	(19*19*19*19)
    
    int
    main(int argc, char **argv)
    {
    	static int8_t prices[STEPS];
    	static int8_t by_deltas[NCODES];
    	static int sums[NCODES];
    
    	uint64_t p1=0, secret;
    	int p2=0, i;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    	
    	while (scanf(" %"SCNu64, &secret) == 1) {
    		memset(by_deltas, 0, sizeof(by_deltas));
    
    		for (i=0; i<STEPS; i++) {
    			secret = (secret ^ secret << 6)  & 0xFFFFFF;
    			secret = (secret ^ secret >> 5);
    			secret = (secret ^ secret << 11) & 0xFFFFFF;
    
    			prices[i] = secret % 10;
    		}
    
    		/*
    		 * Build a deltas->price map for the buyer. Deltas are
    		 * encoded as an integer for easy indexing. Iterating
    		 * backwards ensures the stored price is the _earliest_
    		 * occurence of that sequence.
    		 */
    		for (i=STEPS-1; i>=4; i--)
    			by_deltas[
    			    (prices[i-3] - prices[i-4] +9) *19*19*19 +
    			    (prices[i-2] - prices[i-3] +9) *19*19 +
    			    (prices[i-1] - prices[i-2] +9) *19 +
    			    (prices[i]   - prices[i-1] +9)
    			] = prices[i];
    
    		for (i=0; i<NCODES; i++)
    			sums[i] += by_deltas[i];
    
    		p1 += secret;
    	}
    
    	for (i=0; i<NCODES; i++)
    		p2 = MAX(p2, sums[i]);
    
    	printf("22: %"PRIu64" %d\n", p1, p2);
    	return 0;
    }
    

    day22 0m00.04s real

    https://codeberg.org/sjmulder/aoc/src/branch/master/2024/c/day22.c


  • C

    Graph problems are not my cake but this one I could work out: recursively iterate unique combination of adjacent nodes. Performance benefits from using a basic, int-indexed adjacency matrix.

    Fast enough on my 2015 PC:

    day23 0:00.05 1644 Kb 0+143 faults

    Code
    #include "common.h"
    
    #define VZ 520	/* max no. vertices */
    #define SZ 32	/* max set size */
    
    static char names[VZ][3];
    static char adj[VZ][VZ];
    static int nvert;
    
    static int
    to_id(const char *name)
    {
    	int i;
    
    	for (i=0; i<nvert; i++)
    		if (!strcmp(name, names[i]))
    			return i;
    
    	assert(nvert < VZ);
    	assert(strlen(name) < LEN(*names));
    
    	snprintf(names[nvert++], sizeof(*names), "%s", name);
    	return i;
    }
    
    static int
    cmp_id(const void *a, const void *b)
    {
    	return strcmp(names[*(int*)a], names[*(int*)b]);
    }
    
    /*
     * Construct all unique combinations of nodes, with every extension,
     * confirm they're all connected. Essentally this but recursive:
     *
     *   for (a=0; a<nvert; a++)
     *   for (b=a+1; b<nvert; b++)
     *   for (c=b+1; c<nvert; c++)
     *     ...
     *
     * Note the inner loops continue forward from the point of the outside
     * loop, avoiding duplicate combinations.
     *
     * 'set' and 'best' are arrays of size SZ, length 'sz' and 'bz'. 'set'
     * is the current working state; 'best' is a copy of the longest known
     * set.
     */
    static int
    max_set(int *set, int sz, int *best, int bz)
    {
    	int bz1, v,i;
    
    	assert(sz < SZ);
    
    	/* for each potential candidate */
    	for (v = sz ? set[sz-1]+1 : 0; v < nvert; v++) {
    		 /* adjacent to all in current set? */
    		for (i=0; i<sz && adj[set[i]][v]; i++) ;
    		if (i != sz) continue;
    		 /* recur and update 'best size' if needed */
    		set[sz] = v;
    		if (bz < (bz1 = max_set(set, sz+1, best, bz))) bz = bz1;
    	}
    
    	/* store longest known set in 'best' */
    	if (sz > bz)
    		memcpy(best, set, (bz = sz) * sizeof(*best));
    
    	return bz;
    }
    
    int
    main(int argc, char **argv)
    {
    	static int set[SZ], best[SZ];
    	char buf[8];
    	int p1=0, a,b,c, i, bz;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    	
    	while (fgets(buf, sizeof(buf), stdin)) {
    		assert(strlen(buf) >= 5);
    		buf[2] = buf[5] = '\0';
    		a = to_id(buf);
    		b = to_id(buf+3);
    		adj[a][b] = adj[b][a] = 1;
    	}
    
    	for (a=0; a<nvert; a++)
    	for (b=a+1; b<nvert; b++)
    	for (c=b+1; c<nvert; c++)
    		p1 += adj[a][b] && adj[a][c] && adj[b][c] && (
    		      names[a][0] == 't' || names[b][0] == 't' ||
    		      names[c][0] == 't');
    
    	printf("23: %d ", p1);
    
    	bz = max_set(set, 0, best, 0);
    	qsort(best, bz, sizeof(*best), cmp_id);
    
    	for (i=0; i<bz; i++)
    		printf(i ? ",%s" : "%s", names[best[i]]);
    
    	putchar('\n');
    	return 0;
    }
    

    https://codeberg.org/sjmulder/aoc/src/branch/master/2024/c/day23.c


  • Finally got this one done very late last night!

    I kept getting stuck reasoning about the recursion here. For some reason I got myself convinced that after a move, the state of the ‘upper’ dpads could make it more advantageous to pick one followup move over another - i.e. steps aren’t independent.

    It took a bunch of manually working through sequences to convince myself that, after every move, every dpad above it would be on A. With that, it’s ‘just’ recursive pathfinding for independent moves.

    Since there are relatively few types of moves needed on the dpad, I just sketched them out and wrote the routes in code directly (up to two options per move, e.g. left,up or up,left).

    Code
    #include "common.h"
    
    static int64_t dpmem[26][8][2];
    
    enum {NW,NN,NE,EE,SE,SS,SW,WW};
    
    /*
     * We can sufficiently describe npad/dpad movements as: "move N x, N y,
     * then press A N times" (a 'move'). It's never faster to take a detour.
     * After pressing A on one dpad, dpads on all levels above will also be
     * on A, so these moves can be considered in isolation.
     *
     * This function gives the cost of executing a move with the dpad in
     * direction 'd' at level 'l', excluding the A presses for actually
     * tapping the selected directions, but including returning the cursor
     * back to the A button.
     *
     * E.g., moving 2 up and 3 left (NW), the steps could be <AAv<AAA>>^.
     * Excluding the A presses, that's 6 steps. Pulling out the A presses
     * lets us drop the dx and dy arguments and simplify the memoization.
     * They're easily calculated: it's |dx|+|dy|.
     *
     * The gaps present a problem: picking the lowest-cost option (e.g.
     * left-then-up vs. up-then-left) may cause us to cross a gap. The
     * "mind the gap" argument 'mtg' must be set to 1 if-and-only-if there
     * is such potential, e.g. a move from < to ^ on the dpad (up,right
     * crosses the gap) or a move from A to 1 on the npad (left,up crosses
     * the gap). With the flag set, these orderings will be avoided.
     */
    static int64_t
    dp(int d, int mtg, int l)
    {
    	int64_t ret, alt=INT64_MAX;
    
    	if (l<=0)
    		return 0;
    	
    	assert(l >= 0);
    	assert(l < (int)LEN(dpmem));
    	assert(mtg==0 || mtg==1);
    
    	if ((ret = dpmem[l][d][mtg]))
    		return ret;
    
    	/* routes avoiding gaps where necessary */
    	ret = d==SE ? 1+dp(SS,0,l-1) + 1+dp(WW,0,l-1) + 2+dp(SE,0,l-1) :
    	      d==SW ? 2+dp(SW,0,l-1) + 1+dp(WW,0,l-1) + 3+dp(NE,1,l-1) :
    	      d==SS ? 2+dp(SW,0,l-1) + 2+dp(NE,0,l-1) :
    	      d==NE ? 1+dp(SS,0,l-1) + 2+dp(NW,0,l-1) + 1+dp(EE,0,l-1) :
    	      d==NW ? 1+dp(WW,0,l-1) + 2+dp(SW,1,l-1) + 3+dp(NE,1,l-1) :
    	      d==NN ? 1+dp(WW,0,l-1) + 1+dp(EE,0,l-1) :
    	      d==EE ? 1+dp(SS,0,l-1) + 1+dp(NN,0,l-1) :
    	      d==WW ? 3+dp(SW,1,l-1) + 3+dp(NE,1,l-1) :
    	      (assert(!"bad dir"), -1);
    
    	/* alternatives crossing the gaps */
    	alt = mtg ? INT64_MAX :
    	      d==SE ? 2+dp(SW,0,l-1) + 1+dp(EE,0,l-1) + 1+dp(NN,0,l-1) :
    	      d==SW ? 3+dp(SW,1,l-1) + 1+dp(EE,0,l-1) + 2+dp(NE,0,l-1) :
    	      d==NE ? 1+dp(WW,0,l-1) + 2+dp(SE,0,l-1) + 1+dp(NN,0,l-1) :
    	      d==NW ? 3+dp(SW,1,l-1) + 2+dp(NE,1,l-1) + 1+dp(EE,0,l-1) :
    	      INT64_MAX;
    
    	return dpmem[l][d][mtg] = MIN(ret, alt);
    }
    
    static int64_t
    npcost(const char *s, int lv)
    {
    	int64_t cost=0;
    	int x=2,y=3, x1,y1, mtg, dir;
    
    	for (; *s == 'A' || (*s >= '0' && *s <= '9'); s++) {
    		x1 = *s=='A' ? 2 : *s=='0' ? 1 : 2-(9-*s+'0') % 3;
    		y1 = *s=='A' ? 3 : *s=='0' ? 3 :   (9-*s+'0') / 3;
    
    		/* potentially crossing a gap? */
    		mtg = (x==0 && y1==3) || (x1==0 && y==3);
    
    		dir = y1>y ? (x1>x ? SE : x1<x ? SW : SS) :
    		      y1<y ? (x1>x ? NE : x1<x ? NW : NN) :
    		             (x1>x ? EE : x1<x ? WW : 0);
    
    		cost += dp(dir, mtg, lv) + abs(x1-x) + abs(y1-y) +1;
    
    		x = x1;
    		y = y1;
    	}
    
    	return cost;
    }
    
    int
    main(int argc, char **argv)
    {
    	char buf[8];
    	int digits;
    	int64_t p1=0,p2=0;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    	
    	while (fgets(buf, sizeof(buf), stdin)) {
    		digits = atoi(buf);
    		p1 += digits * npcost(buf, 2);
    		p2 += digits * npcost(buf, 25);
    	}
    
    	printf("21: %"PRId64" %"PRId64"\n", p1, p2);
    	return 0;
    }
    

    https://codeberg.org/sjmulder/aoc/src/branch/master/2024/c/day21.c


  • C

    Note to self: reread the puzzle after waking up properly! I initially wrote a great solution for the wrong question (pathfinding with a given number of allowed jumps).

    For the actual question, a bit boring but flood fill plus some array iteration saved the day again. Find the cost for every open tile with flood fill, then for each position and offset combination, see if that jump yields a lower cost at the destination.

    For arbitrary inputs this would require eliminating the non-optimal paths, but the one path covered all open tiles.

    Code
    #include "common.h"
    
    #define GB 2	/* safety border on X/Y plane */
    #define GZ (GB+141+2+GB)
    
    static char G[GZ][GZ];		/* grid */
    static int  C[GZ][GZ];		/* costs */
    static int sx,sy, ex,ey;	/* start, end pos */
    
    static void
    flood(int x, int y)
    {
    	int lo = INT_MAX;
    
    	if (x<1 || x>=GZ-1 ||
    	    y<1 || y>=GZ-1 || G[y][x]!='.')
    		return;
    
    	if (C[y-1][x]) lo = MIN(lo, C[y-1][x]+1);
    	if (C[y+1][x]) lo = MIN(lo, C[y+1][x]+1);
    	if (C[y][x-1]) lo = MIN(lo, C[y][x-1]+1);
    	if (C[y][x+1]) lo = MIN(lo, C[y][x+1]+1);
    
    	if (lo != INT_MAX && (!C[y][x] || lo < C[y][x])) {
    		C[y][x] = lo;
    
    		flood(x, y-1); flood(x-1, y);
    		flood(x, y+1); flood(x+1, y);
    	}
    }
    
    static int
    count_shortcuts(int lim, int min)
    {
    	int acc=0, x,y, dx,dy;
    
    	for (y=GB; y<GZ-GB; y++)
    	for (x=GB; x<GZ-GB; x++)
    	for (dy = -lim; dy <= lim; dy++)
    	for (dx = abs(dy)-lim; dx <= lim-abs(dy); dx++)
    		acc += x+dx >= 0 && x+dx < GZ &&
    		       y+dy >= 0 && y+dy < GZ && C[y][x] &&
    		       C[y][x]+abs(dx)+abs(dy) <= C[y+dy][x+dx]-min;
    
    	return acc;
    }
    
    int
    main(int argc, char **argv)
    {
    	int x,y;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    	for (y=2; fgets(G[y]+GB, GZ-GB*2, stdin); y++)
    		assert(y < GZ-3);
    
    	for (y=GB; y<GZ-GB; y++)
    	for (x=GB; x<GZ-GB; x++)
    		if (G[y][x] == 'S') { G[y][x]='.'; sx=x; sy=y; } else
    		if (G[y][x] == 'E') { G[y][x]='.'; ex=x; ey=y; }
    
    	C[sy][sx] = 1;
    
    	flood(sx, sy-1); flood(sx-1, sy);
    	flood(sx, sy+1); flood(sx+1, sy);
    
    	printf("20: %d %d\n",
    	    count_shortcuts(2, 100),
    	    count_shortcuts(20, 100));
    
    	return 0;
    }
    

    https://codeberg.org/sjmulder/aoc/src/branch/master/2024/c/day20.c



  • I think I saw the same! At first I thought it requires pathfinding to see what nodes are connected to the wall, but then someone pointed at disjoint sets and just a glance at Wikipedia made it click right away. What an ingeniously simple but useful data structure! Maybe I’ll reimplement my solution with that - mostly as an exercise for disjoint sets and finding a convenient representation for that in C.






  • Awesome! I understood the idea behind the binary search but thought it wasn’t a good fit for the flood fill. As opposed to something like A* it will give you reachability and cost for every cell (at a cost), but that’s no use when you do repeated searches that are only meant to find a single path. So I was very happy with your suggestion, it fits better with the strengths.

    “Virtually instant” btw is measured 0.00 by time. I like it when things are fast but I also prefer simper approaches (that is: loops and arrays) over the really optimized fast stuff. People do really amazing things but the really clever algorithms lean on optimized generic data structures that C lacks. It’s fun though to see how far you can drive loops and arrays! Perhaps next year I’ll pick a compiled language with a rich data structure library and really focus on effectively applying good algorithms and appropriate data structures.

    Btw how do you measure performance? I see a lot of people including timing things in their programs but I can’t be bothered. Some people also exclude parsing - which wouldn’t work for me because I try to process the input immediately, if possible.


  • C

    Interestingly part 1 already defied a naive approach. It was fun thinking of a way to memoize without hash tables.

    Code
    #include "common.h"
    
    static char *pats[480];
    static int lens[480];
    int np;
    
    /* memoized for 's' by mem[off], 0 = unknown, >0 = value+1 */
    static int64_t
    recur(char *s, int off, int64_t *mem)
    {
    	int64_t acc=0;
    	int i;
    
    	if (!s[off]) return 1;
    	if (mem[off]) return mem[off]-1;
    
    	for (i=0; i<np; i++)
    		if (!strncmp(s+off, pats[i], lens[i]))
    			acc += recur(s, off+lens[i], mem);
    
    	mem[off] = acc+1;
    	return acc;
    }
    
    int
    main(int argc, char **argv)
    {
    	static char patbuf[3200], design[64];
    	int64_t p1=0,p2=0, mem[64], n;
    	char *rest, *lf;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    
    	rest = fgets(patbuf, sizeof(patbuf), stdin);
    
    	for (; (pats[np] = strsep(&rest, ",")); np++) {
    		while (isspace(pats[np][0]))
    			pats[np]++;	/* skip spaces */
    		if ((lf = strchr(pats[np], '\n')))
    			*lf = '\0';	/* trim trailing \n */
    		lens[np] = strlen(pats[np]);
    		assert(np+1 < (int)LEN(pats));
    	}
    
    	while (scanf(" %63s", design) == 1) {
    		memset(mem, 0, sizeof(mem));
    		n = recur(design, 0, mem);
    		p1 += n >0;
    		p2 += n;
    	}
    
    	printf("19: %"PRId64" %"PRId64"\n", p1, p2);
    	return 0;
    }
    

    https://codeberg.org/sjmulder/aoc/src/branch/master/2024/c/day19.c

    Zee

    Also a port to my cursed Dutch dialect of C, Zee:

    Code
    #ingesloten "zee.kop"
    #ingesloten "algemeen.kop"
    
    besloten letterverwijzing patronen[480];
    besloten getal lengtes[480];
    getal patroonsom;
    
    besloten zeer groot getal
    afdaling(
        letterverwijzing tekst,
        getal startpositie,
        zeergrootgetalreeksverwijzing onthouden)
    {
    	zeer groot getal deelsom=0;
    	getal sortering, teller;
    
    	tenzij (tekst[startpositie])
    		lever 1;
    	mits (onthouden[startpositie])
    		lever onthouden[startpositie]-1;
    
    	voor (teller=0; teller < patroonsom; teller++) {
    		sortering = tekstdeelvergelijking(
    		    tekst + startpositie,
    		    patronen[teller],
    		    lengtes[teller]);
    
    		mits (sortering == 0) {
    			deelsom += afdaling(
    			    tekst,
    			    startpositie + lengtes[teller],
    			    onthouden);
    		}
    	}
    
    	onthouden[startpositie] = deelsom+1;
    	lever deelsom;
    }
    
    getal
    aanvang(
        getal parametersom,
        letterverwijzingsreeksverwijzing parameters)
    {
    	blijvende letter patroonruimte[3200];
    	blijvende letter ontwerp[64];
    	zeer groot getal deel1=0, aantal;
    	zeer groot getal deel2=0, onthouden[64];
    	letterverwijzing rest;
    	letterverwijzing regeleinde;
    
    	mits (parametersom > 1)
    		VERWERP(heropen(parameters[1], "r", standaardinvoer));
    
    	rest = geefregel(patroonruimte, grootte(patroonruimte),
    	    standaardinvoer);
    
    	voor (; ; patroonsom++) {
    		verzeker(patroonsom+1 < (getal)LENGTE(patronen));
    		patronen[patroonsom] = tekstsplitsing(naar rest, ",");
    		mits (patronen[patroonsom] == NIETS)
    			klaar;
    
    		zolang (iswitruimte(patronen[patroonsom][0]))
    			patronen[patroonsom]++;
    		mits ((regeleinde = zoekletter(patronen[patroonsom], '\n')))
    			volg regeleinde = '\0';
    
    		lengtes[patroonsom] = tekstlengte(patronen[patroonsom]);
    	}
    
    	zolang (invorm(" %63s", ontwerp) == 1) {
    		overschrijf(onthouden, 0, grootte(onthouden));
    		aantal = afdaling(ontwerp, 0, onthouden);
    		deel1 += aantal >0;
    		deel2 += aantal;
    	}
    
    	uitvorm("19: %"GEEFZGG" %"GEEFZGG"\n", deel1, deel2);
    	lever 0;
    }
    

    https://codeberg.org/sjmulder/aoc/src/


  • Thanks, that’s exactly the sort of insight that I was too tired to have at that point 😅

    The other thing I had to change was to make it recursive rather than iterating over the full grid - the latter is fast for large update, but very wasteful for local updates, like removing the points. Virtually instant now!

    Code
    #include "common.h"
    
    #define SAMPLE	0
    #define PTZ	3600
    #define GZ	(SAMPLE ? 9 : 73)
    #define P1STEP	(SAMPLE ? 12 : 1024)
    #define CORR	-1
    
    static int g[GZ][GZ];
    
    static void
    flood(int x, int y)
    {
    	int lo=INT_MAX;
    
    	if (x <= 0 || x >= GZ-1 ||
    	    y <= 0 || y >= GZ-1 || g[y][x] == CORR)
    		return;
    
    	if (g[y-1][x] > 0) lo = MIN(lo, g[y-1][x] +1);
    	if (g[y+1][x] > 0) lo = MIN(lo, g[y+1][x] +1);
    	if (g[y][x-1] > 0) lo = MIN(lo, g[y][x-1] +1);
    	if (g[y][x+1] > 0) lo = MIN(lo, g[y][x+1] +1);
    
    	if (lo != INT_MAX && (!g[y][x] || g[y][x] > lo)) {
    		g[y][x] = lo;
    
    		flood(x, y-1);
    		flood(x, y+1);
    		flood(x-1, y);
    		flood(x+1, y);
    	}
    }
    
    int
    main(int argc, char **argv)
    {
    	static int xs[PTZ], ys[PTZ];
    	static char p2[32];
    	int p1=0, npt=0, i;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    
    	for (i=0; i<GZ; i++)
    		g[0][i] = g[GZ-1][i] =
    		g[i][0] = g[i][GZ-1] = CORR;
    
    	for (npt=0; npt<PTZ && scanf(" %d,%d", xs+npt, ys+npt)==2; npt++) {
    		assert(xs[npt] >= 0); assert(xs[npt] < GZ-2);
    		assert(ys[npt] >= 0); assert(ys[npt] < GZ-2);
    	}
    
    	assert(npt < PTZ);
    
    	for (i=0; i<npt; i++)
    		g[ys[i]+1][xs[i]+1] = CORR;
    
    	g[1][1] = 1;
    	flood(2, 1);
    	flood(1, 2);
    
    	for (i=npt-1; i >= P1STEP; i--) {
    		g[ys[i]+1][xs[i]+1] = 0;
    		flood(xs[i]+1, ys[i]+1);
    
    		if (!p2[0] && g[GZ-2][GZ-2] > 0)
    			snprintf(p2, sizeof(p2), "%d,%d", xs[i],ys[i]);
    	}
    
    	p1 = g[GZ-2][GZ-2]-1;
    
    	printf("18: %d %s\n", p1, p2);
    	return 0;
    }
    

  • C

    Flood fill for part 1. Little tired so for part 2 I just retry the flood fill every step. Slow by C standards (2s) but I’ll let it brew and come back to it later.

    Code
    #include "common.h"
    
    #define SAMPLE	0
    #define GZ	(SAMPLE ? 9 : 73)
    #define NCORR	(SAMPLE ? 12 : 1024)
    #define CORR	-1
    
    int g[GZ][GZ];
    
    static void
    flood(void)
    {
    	int x,y, dirty=1, lo;
    
    	for (y=1; y<GZ-1; y++)
    	for (x=1; x<GZ-1; x++)
    		if (g[y][x] > 1)
    			g[y][x] = 0;
    
    	while (dirty) {
    		dirty = 0;
    
    		for (y=1; y<GZ-1; y++)
    		for (x=1; x<GZ-1; x++) {
    			if (g[y][x] == CORR) continue;
    			lo = INT_MAX;
    			if (g[y-1][x] > 0) lo = MIN(lo, g[y-1][x]);
    			if (g[y+1][x] > 0) lo = MIN(lo, g[y+1][x]);
    			if (g[y][x-1] > 0) lo = MIN(lo, g[y][x-1]);
    			if (g[y][x+1] > 0) lo = MIN(lo, g[y][x+1]);
    			if (lo != INT_MAX && (!g[y][x] || g[y][x]>lo+1))
    				{ dirty=1; g[y][x] = lo+1; }
    		}
    	}
    }
    
    int
    main(int argc, char **argv)
    {
    	int p1=0, x,y, i;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    
    	for (i=0; i<GZ; i++)
    		g[0][i] = g[GZ-1][i] =
    		g[i][0] = g[i][GZ-1] = CORR;
    	
    	g[1][1] = 1;
    
    	for (i=0; scanf(" %d,%d", &x, &y) == 2; i++) {
    		assert(x >= 0); assert(x < GZ-2);
    		assert(y >= 0); assert(y < GZ-2);
    		g[y+1][x+1] = CORR;
    
    		flood();
    
    		if (i==NCORR-1)
    			p1 = g[GZ-2][GZ-2]-1;
    		if (g[GZ-2][GZ-2] <= 0) {
    			printf("18: %d %d,%d\n", p1, x,y);
    			return 0;
    		}
    	}
    
    	assert(!"no solution");
    	return -1;
    }
    

    https://github.com/sjmulder/aoc/blob/master/2024/c/day18.c


  • C

    Was looking forward to a VM! Really took my leisure for part 1, writing a disassembler, tracing, all pretty.

    Part 2 reminded me of 2021 day 24 where we also had to reverse an input. It’s been on my mind recently and I was thinking if there would be a way to backfeed an output through a program, yielding a representation like: “the input plus 3498 is a multiple of 40, and divisible by a number that’s 5 mod 8” (considering lossy functions like modulo and integer division).

    Today’s input didn’t lend itself to that, however, but analysing it having the solution ‘click’ was very satisfying.

    Code
    #include "common.h"
    
    #define MEMZ 32
    #define OUTZ 32
    #define BUFZ 64
    
    enum { ADV, BXL, BST, JNZ, BXC, OUT, BDV, CDV };
    
    struct vm {
    	int64_t mem[MEMZ]; int nm, pc;
    	int64_t out[OUTZ]; int no;
    	int64_t a,b,c;
    } vm;
    
    /* returns 0 if halted, 1 otherwise */
    static int
    step(void)
    {
    	int64_t op, ar, ac;	/* operator, lit. arg, combo arg */
    
    	assert(vm.pc >= 0);
    	assert(vm.pc+2 <= MEMZ);
    
    	if (vm.pc >= vm.nm)
    		return 0;
    
    	op = vm.mem[vm.pc++];
    	ar = vm.mem[vm.pc++];
    	ac = ar==4 ? vm.a : ar==5 ? vm.b : ar==6 ? vm.c : ar;
    
    	switch (op) {
    	case ADV: vm.a = vm.a >> ac; break;
    	case BDV: vm.b = vm.a >> ac; break;
    	case CDV: vm.c = vm.a >> ac; break;
    	case BXL: vm.b = vm.b ^ ar; break;
    	case BXC: vm.b = vm.b ^ vm.c; break;
    	case BST: vm.b = ac % 8; break;
    	case JNZ: if (vm.a) vm.pc = ar; break;
    	case OUT: assert(vm.no < OUTZ); vm.out[vm.no++] = ac%8; break;
    	default: assert(!"invalid opcode"); return 0;
    	}
    
    	return 1;
    }
    
    static int64_t
    recur_p2(int64_t a0, int pos)
    {
    	int64_t a, i;
    
    	/*
    	 * The code in the input uses up to 7 low bits of the A register
    	 * to produce a single number, then chops off the low 3 bits and
    	 * continues.
    	 *
    	 * That means bits above the current triplet influence what
    	 * number it generates, but bits below don't. To generate a
    	 * desired sequence then, we append, not prepend,  candidate
    	 * triplets.
    	 *
    	 * We may end up in a situation where, given the prefix found
    	 * for the numbers so far, no triplet yields the desired next
    	 * number. Then we have to backtrack and find another prefix,
    	 * hence the recursion.
    	 */
    
    	if (pos >= vm.nm)
    		return a0 >> 3;
    
    	for (i=0; i<8; i++) {
    		vm.a=a= a0+i;
    		vm.b=vm.c=vm.pc=vm.no= 0;
    
    		while (step() && !vm.no)
    			;
    
    		if (vm.no && vm.out[0] == vm.mem[vm.nm-pos-1])
    			if ((a = recur_p2(a << 3, pos+1)))
    				return a;
    	}
    
    	return 0;
    }
    
    int
    main(int argc, char **argv)
    {
    	char b[BUFZ], *tok, *rest;
    	int i;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    
    	fgets(b, BUFZ, stdin); sscanf(b, "Register A: %lld", &vm.a);
    	fgets(b, BUFZ, stdin); sscanf(b, "Register B: %lld", &vm.b);
    	fgets(b, BUFZ, stdin); sscanf(b, "Register C: %lld", &vm.c);
    	fgets(b, BUFZ, stdin);
    
    	assert(vm.b == 0);	/* assumption for part 2 */
    	assert(vm.c == 0);
    
    	rest = fgets(b, sizeof(b), stdin);
    	strsep(&rest, ":");
    
    	while ((tok = strsep(&rest, ","))) {
    		assert(vm.nm < MEMZ);
    		vm.mem[vm.nm++] = atoll(tok);
    	}
    
    	while (step())
    		;
    	for (i=0; i<vm.no; i++)
    		printf(i ? ",%lld" : "17: %lld", vm.out[i]);
    
    	printf(" %lld\n", recur_p2(0, 0));
    	return 0;
    }
    

    https://github.com/sjmulder/aoc/blob/master/2024/c/day17.c


  • C

    Yay more grids! Seemed like prime Dijkstra or A* material but I went with an iterative approach instead!

    I keep an array cost[y][x][dir], which is seeded at 1 for the starting location and direction. Then I keep going over the array, seeing if any valid move (step or turn) would yield to a lower best-known-cost for this state. It ends when a pass does not yield changes.

    This leaves us with the best-known-costs for every reachable state in the array, including the end cell (bit we have to take the min() of the four directions).

    Part 2 was interesting: I just happend to have written a dead end pruning function for part 1 and part 2 is, really, dead-end pruning for the cost map: remove any suboptimal step, keep doing so, and you end up with only the optimal steps. ‘Suboptimal’ here is a move that yields a higher total cost than the best-known-cost for that state.

    It’s fast enough too on my 2015 i5:

    day16  0:00.05  1656 Kb  0+242 faults
    
    Code
    #include "common.h"
    
    #define GZ 145
    
    enum {NN, EE, SS, WW};
    
    static const int dx[]={0,1,0,-1}, dy[]={-1,0,1,0};
    
    static char g[GZ][GZ];		/* with 1 tile border */
    static int cost[GZ][GZ][4];	/* per direction, starts at 1, 0=no info */
    
    static int traversible(char c) { return c=='.' || c=='S' || c=='E'; }
    
    static int
    minat(int x, int y)
    {
    	int acc=0, d;
    
    	for (d=0; d<4; d++)
    		if (cost[y][x][d] && (!acc || cost[y][x][d] < acc))
    			acc = cost[y][x][d];
    
    	return acc;
    }
    
    
    static int
    count_exits(int x, int y)
    {
    	int acc=0, i;
    
    	assert(x>0); assert(x<GZ-1);
    	assert(y>0); assert(y<GZ-1);
    
    	for (i=0; i<4; i++)
    		acc += traversible(g[y+dy[i]][x+dx[i]]);
    	
    	return acc;
    }
    
    /* remove all dead ends */
    static void
    prune_dead(void)
    {
    	int dirty=1, x,y;
    
    	while (dirty) {
    		dirty = 0;
    
    		for (y=1; y<GZ-1; y++)
    		for (x=1; x<GZ-1; x++)
    			if (g[y][x]=='.' && count_exits(x,y) < 2)
    				{ dirty = 1; g[y][x] = '#'; }
    	}
    }
    
    /* remove all dead ends from cost[], leaves only optimal paths */
    static void
    prune_subopt(void)
    {
    	int dirty=1, x,y,d;
    
    	while (dirty) {
    		dirty = 0;
    
    		for (y=1; y<GZ-1; y++)
    		for (x=1; x<GZ-1; x++)
    		for (d=0; d<4; d++) {
    			if (!cost[y][x][d])
    				continue;
    
    			if (g[y][x]=='E') {
    				if (cost[y][x][d] != minat(x,y))
    					{ dirty = 1; cost[y][x][d] = 0; }
    				continue;
    			}
    
    			if (cost[y][x][d]+1 > cost[y+dy[d]][x+dx[d]][d] &&
    			    cost[y][x][d]+1000 > cost[y][x][(d+1)%4] &&
    			    cost[y][x][d]+1000 > cost[y][x][(d+3)%4])
    				{ dirty = 1; cost[y][x][d] = 0; }
    		}
    	}
    }
    
    static void
    propagate_costs(void)
    {
    	int dirty=1, cost1, x,y,d;
    
    	while (dirty) {
    		dirty = 0;
    
    		for (y=1; y<GZ-1; y++)
    		for (x=1; x<GZ-1; x++)
    		for (d=0; d<4; d++) {
    			if (!traversible(g[y][x]))
    				continue;
    
    			/* from back */
    			if ((cost1 = cost[y-dy[d]][x-dx[d]][d]) &&
    			    (cost1+1 < cost[y][x][d] || !cost[y][x][d]))
    				{ dirty = 1; cost[y][x][d] = cost1+1; }
    
    			/* from right */
    			if ((cost1 = cost[y][x][(d+1)%4]) &&
    			    (cost1+1000 < cost[y][x][d] || !cost[y][x][d]))
    				{ dirty = 1; cost[y][x][d] = cost1+1000; }
    
    			/* from left */
    			if ((cost1 = cost[y][x][(d+3)%4]) &&
    			    (cost1+1000 < cost[y][x][d] || !cost[y][x][d]))
    				{ dirty = 1; cost[y][x][d] = cost1+1000; }
    		}
    	}
    }
    
    int
    main(int argc, char **argv)
    {
    	int p1=0,p2=0, sx=0,sy=0, ex=0,ey=0, x,y;
    	char *p;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    
    	for (y=1; fgets(g[y]+1, GZ-1, stdin); y++) {
    		if ((p = strchr(g[y]+1, 'S'))) { sy=y; sx=p-g[y]; }
    		if ((p = strchr(g[y]+1, 'E'))) { ey=y; ex=p-g[y]; }
    		assert(y+1 < GZ-1);
    	}
    
    	cost[sy][sx][EE] = 1;
    
    	prune_dead();
    	propagate_costs();
    	prune_subopt();
    
    	p1 = minat(ex, ey) -1;	/* costs[] values start at 1! */
    
    	for (y=1; y<GZ-1; y++)
    	for (x=1; x<GZ-1; x++)
    		p2 += minat(x,y) > 0;
    
    	printf("16: %d %d\n", p1, p2);
    	return 0;
    }
    

  • C

    3h+ train ride back home from weekend trip but a little tired and not feeling it much. Finished part 1, saw that part 2 was fiddly programming, left it there.

    Finally hacked together something before bed. The part 2 twist required rewriting the push function to be recursive but also a little care and debugging to get that right. Cleaned it up over lunch, happy enough with the solution now!

    Code
    #include "common.h"
    
    #define GW 104
    #define GH 52
    
    struct world { char g[GH][GW]; int px,py; };
    
    static int
    can_clear(struct world *w, int x, int y, int dx, int dy)
    {
    	assert(x>=0); assert(x<GW);
    	assert(y>=0); assert(y<GH);
    	assert((dx && !dy) || (dy && !dx));
    
    	return
    	    (x+dx >= 0 || x+dx < GW) &&
    	    (y+dy >= 0 || y+dy < GW) &&
    	    (w->g[y][x] == '.' || (
    	     w->g[y][x] != '#' && can_clear(w, x+dx,y+dy, dx,dy) &&
    	     (!dy || w->g[y][x]!='[' || can_clear(w, x+1,y+dy, 0,dy)) &&
    	     (!dy || w->g[y][x]!=']' || can_clear(w, x-1,y,    0,dy)) &&
    	     (!dy || w->g[y][x]!=']' || can_clear(w, x-1,y+dy, 0,dy))));
    }
    
    /* check can_clear() first! */
    static void
    clear(struct world *w, int x, int y, int dx, int dy)
    {
    	assert(x>=0); assert(x<GW); assert(x+dx>=0); assert(x+dx<GW);
    	assert(y>=0); assert(y<GH); assert(y+dy>=0); assert(y+dy<GH);
    
    	if (w->g[y][x] == '.')
    		return;
    	if (dy && w->g[y][x] == ']')
    		{ clear(w, x-1,y, dx,dy); return; }
    
    	if (dy && w->g[y][x] == '[') {
    		clear(w, x+1,y+dy, dx,dy);
    		w->g[y+dy][x+dx+1] = ']';
    		w->g[y][x+1] = '.';
    	}
    
    	clear(w, x+dx,y+dy, dx,dy);
    	w->g[y+dy][x+dx] = w->g[y][x];
    	w->g[y][x] = '.';
    }
    
    static void
    move(struct world *w, int dx, int dy)
    {
    	if (can_clear(w, w->px+dx, w->py+dy, dx,dy)) {
    		clear(w, w->px+dx, w->py+dy, dx,dy);
    		w->px += dx;
    		w->py += dy;
    	}
    }
    
    static int
    score(struct world *w)
    {
    	int acc=0, x,y;
    
    	for (y=0; y<GH && w->g[y][0]; y++)
    	for (x=0; x<GW && w->g[y][x]; x++)
    		if (w->g[y][x] == 'O' || w->g[y][x] == '[')
    			acc += 100*y + x;
    
    	return acc;
    }
    
    int
    main(int argc, char **argv)
    {
    	static struct world w1,w2;
    	int x,y, c;
    	char *p;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    
    	for (y=0; fgets(w1.g[y], GW, stdin); y++) {
    		if (!w1.g[y][0] || w1.g[y][0]=='\n')
    			break;
    
    		assert(y+1 < GH);
    		assert(strlen(w1.g[y])*2+1 < GW);
    
    		for (x=0; w1.g[y][x]; x++)
    			if (w1.g[y][x] == 'O') {
    				w2.g[y][x*2]   = '[';
    				w2.g[y][x*2+1] = ']';
    			} else {
    				w2.g[y][x*2]   = w1.g[y][x];
    				w2.g[y][x*2+1] = w1.g[y][x];
    			}
    
    		if ((p = strchr(w1.g[y], '@'))) {
    			w1.py = y; w1.px = p-w1.g[y];
    			w2.py = y; w2.px = w1.px*2;
    
    			w1.g[w1.py][w1.px]   = '.';
    			w2.g[w2.py][w2.px]   = '.';
    			w2.g[w2.py][w2.px+1] = '.';
    		}
    	}
    
    	while ((c = getchar()) != EOF)
    		switch (c) {
    		case '^': move(&w1, 0,-1); move(&w2, 0,-1); break;
    		case 'v': move(&w1, 0, 1); move(&w2, 0, 1); break;
    		case '<': move(&w1,-1, 0); move(&w2,-1, 0); break;
    		case '>': move(&w1, 1, 0); move(&w2, 1, 0); break;
    		}
    
    	printf("15: %d %d\n", score(&w1), score(&w2));
    	return 0;
    }
    

    https://github.com/sjmulder/aoc/blob/master/2024/c/day15.c


  • C

    Solved part 1 without a grid, looked at part 2, almost spit out my coffee. Didn’t see that coming!

    I used my visualisation mini-library to generate video with ffmpeg, stepped through it a bit, then thought better of it - this is a programming puzzle after all!

    So I wrote a heuristic to find frames low on entropy (specifically: having many robots in the same line of column), where each record-breaking frame number was printed. That pointed right at the correct frame!

    It was pretty slow though (.2 secs or such) because it required marking spots on a grid. I noticed the Christmas tree was neatly tucked into a corner, concluded that wasn’t an accident, and rewrote the heuristic to check for a high concentration in a single quadrant. Reverted this because the tree-in-quadrant assumption proved incorrect for other inputs. Would’ve been cool though!

    Code
    #include "common.h"
    
    #define SAMPLE 0
    #define GW (SAMPLE ? 11 : 101)
    #define GH (SAMPLE ?  7 : 103)
    #define NR 501
    
    int
    main(int argc, char **argv)
    {
    	static char g[GH][GW];
    	static int px[NR],py[NR], vx[NR],vy[NR];
    
    	int p1=0, n=0, sec, i, x,y, q[4]={}, run;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    
    	for (; scanf(" p=%d,%d v=%d,%d", px+n,py+n, vx+n,vy+n)==4; n++)
    		assert(n+1 < NR);
    
    	for (sec=1; !SAMPLE || sec <= 100; sec++) {
    		memset(g, 0, sizeof(g));
    		memset(q, 0, sizeof(q));
    
    		for (i=0; i<n; i++) {
    			px[i] = (px[i] + vx[i] + GW) % GW;
    			py[i] = (py[i] + vy[i] + GH) % GH;
    
    			g[py[i]][px[i]] = 1;
    
    			if (sec == 100) {
    				if (px[i] < GW/2) {
    					if (py[i] < GH/2) q[0]++; else
    					if (py[i] > GH/2) q[1]++;
    				} else if (px[i] > GW/2) {
    					if (py[i] < GH/2) q[2]++; else
    					if (py[i] > GH/2) q[3]++;
    				}
    			}
    		}
    
    		if (sec == 100)
    			p1 = q[0]*q[1]*q[2]*q[3];
    
    		for (y=0; y<GH; y++)
    		for (x=0, run=0; x<GW; x++)
    			if (!g[y][x])
    				run = 0;
    			else if (++run >= 10)
    				goto found_p2;
    	}
    
    found_p2:
    	printf("14: %d %d\n", p1, sec);
    	return 0;
    }
    

    https://github.com/sjmulder/aoc/blob/master/2024/c/day14.c