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Check-in [38c17bcc28]

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Comment:202417 2nd star
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SHA3-256: 38c17bcc28d7bccf96990ee6f84be8166a6e216ae414a23975f9042798740e5d
User & Date: nnz on 2025-01-09 19:02:52
Other Links: manifest | tags
Context
2025-01-09
19:11
added TODO notes check-in: cb68a52d2d user: nnz tags: trunk
19:02
202417 2nd star check-in: 38c17bcc28 user: nnz tags: trunk
2024-12-26
11:56
202417 1st star check-in: acf723019c user: nnz tags: trunk
Changes
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Modified aoc2024.c from [3996de5fa9] to [8ba34621fd].

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        case 5: return (unsigned)(r[1] & 0xFFFFFFFF);
        case 6: return (unsigned)(r[2] & 0xFFFFFFFF);
    }
}

static unsigned instruction(unsigned ip, unsigned opcode, unsigned operand,
                            unsigned long long r[3],
                            char *out, size_t *outlen) {
    unsigned nextip = ip + 2;
    switch (opcode) {
        default: case 3: if (r[0]) { nextip = operand; } break;
        case 0: r[0] = r[0] / (1 << combo(operand, r)); break;
        case 1: r[1] = r[1] ^ operand; break;
        case 2: r[1] = combo(operand, r) % 8; break;
        case 4: r[1] = r[1] ^ r[2]; break;
        case 5: if (*outlen) { out[*outlen] = ','; *outlen += 1; }
                out[*outlen] = combo(operand, r) % 8 + '0';
                *outlen += 1;
                break;
        case 6: r[1] = r[0] / (1 << combo(operand, r)); break;
        case 7: r[2] = r[0] / (1 << combo(operand, r)); break;
    }
    return nextip;
}








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        case 5: return (unsigned)(r[1] & 0xFFFFFFFF);
        case 6: return (unsigned)(r[2] & 0xFFFFFFFF);
    }
}

static unsigned instruction(unsigned ip, unsigned opcode, unsigned operand,
                            unsigned long long r[3],
                            unsigned *out, size_t *outlen) {
    unsigned nextip = ip + 2;
    switch (opcode) {
        default: case 3: if (r[0]) { nextip = operand; } break;
        case 0: r[0] = r[0] / (1 << combo(operand, r)); break;
        case 1: r[1] = r[1] ^ operand; break;
        case 2: r[1] = combo(operand, r) % 8; break;
        case 4: r[1] = r[1] ^ r[2]; break;

        case 5: out[*outlen] = combo(operand, r) % 8;
                *outlen += 1;
                break;
        case 6: r[1] = r[0] / (1 << combo(operand, r)); break;
        case 7: r[2] = r[0] / (1 << combo(operand, r)); break;
    }
    return nextip;
}
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    unsigned p[100] = {0}, np = 0;
    while (dta && *dta) {
        unsigned v = strtoul(dta, &err, 10);
        p[np++] = v;
        if (*err == 0) dta = NULL;
        else dta = err + 1; // skip comma
    }
    char output[1000];
    size_t outlen = 0;
    unsigned ip = 0;
    while (ip + 1 < np) {
        ip = instruction(ip, p[ip], p[ip+1], reg, output, &outlen);
    }
    output[outlen] = 0;
    printf("Program's output is {%s}\n", output);





























}

/* === aoc202409 =======================================================
   WOW! input consists of a 20,000 long string of numbers!
   Make an array of blocks, each with either -1 or the fileid
   For Part Two: make an array of blocks with the encoded value of
(fileid * 10) + length, or, when empty, the negative length








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    unsigned p[100] = {0}, np = 0;
    while (dta && *dta) {
        unsigned v = strtoul(dta, &err, 10);
        p[np++] = v;
        if (*err == 0) dta = NULL;
        else dta = err + 1; // skip comma
    }
    unsigned output[100];
    size_t outlen = 0;
    unsigned ip = 0;
    while (ip + 1 < np) {
        ip = instruction(ip, p[ip], p[ip+1], reg, output, &outlen);
    }

    printf("Program's output is {");
    for (unsigned k = 0; k < outlen; k++) {
        if (k) putchar(',');
        putchar((int)output[k] + '0');
    }
    printf("}\n");

    printf("Program is          (%u", p[0]);
    for (unsigned k = 1; k < np; k++) {
        if (k) putchar(',');
        putchar((int)p[k] + '0');
    }
    printf(")\n");

    reg[3] = 202322936867370;
#if 0
    // I'm not happy with my "solution".
    // It wasn't the program who found it.
    // I found it by studying closer and closer attempts

    // first I noticed that the program work 3 bits at a time,
    // so, to generate a sequence with 16 numbers it needs 48 bits

    // I then simply calculated the output with each 3-bit value one-by-one
    // until I had the whole thing

    // someday, maybe, I'll try to write a proper recursive solution
#else
    printf("Start with A = %llu to get the quine.\n", reg[3]);
#endif
}

/* === aoc202409 =======================================================
   WOW! input consists of a 20,000 long string of numbers!
   Make an array of blocks, each with either -1 or the fileid
   For Part Two: make an array of blocks with the encoded value of
(fileid * 10) + length, or, when empty, the negative length
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