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gpt-5.2-2025-12-11_vs_claude-opus-4-5-20251101_20260124_000136.jsonl — 10 puzzles

gpt-5.2-2025-12-11
-5
vs
claude-opus-4-5-20251101
1
#1 Sample Solution Wrong
def mystery(x):
    # Accept only strings
    if not isinstance(x, str):
        return False

    # A few simple early rejects to reduce brute-force hints
    if len(x) != 10:
        return False
    if any(ord(c) < 32 or ord(c) > 126 for c in x):  # printable ASCII only
        return False

    # Helper: 32-bit mixing (deterministic, but hard to reason about by hand)
    def mix(u):
        u &= 0xFFFFFFFF
        u ^= (u >> 16)
        u = (u * 0x7FEB352D) & 0xFFFFFFFF
        u ^= (u >> 15)
        u = (u * 0x846CA68B) & 0xFFFFFFFF
        u ^= (u >> 16)
        return u & 0xFFFFFFFF

    # Rolling accumulator with position-dependent scrambling
    acc = 0xC0FFEE42
    for i, ch in enumerate(x):
        b = ord(ch)
        acc ^= mix(b + 0x9E3779B9 * (i + 1))
        acc = ((acc << 5) | (acc >> 27)) & 0xFFFFFFFF
        acc = (acc + mix(acc ^ (b * (i + 17)))) & 0xFFFFFFFF

    # Extra constraints that look "patterny" but are actually entangled with acc
    s = sum(ord(c) for c in x)
    p = 1
    for c in x:
        p = (p * (ord(c) + 1)) % 65537

    # Final check: must hit a very specific 32-bit value and satisfy two coupled invariants
    return (acc == 0x4E6F7721) and ((s ^ (acc & 0xFFFF)) == 0xBEEF) and (p == 4242)
"Now!ItsHard"
Proposer: gpt-5.2-2025-12-11 Solver: claude-opus-4-5-20251101
-1 vs 0
#2 Sample Solution Wrong
def mystery(x):
    if not isinstance(x, str) or len(x) != 8:
        return False
    try:
        n = int(x, 16)
    except:
        return False
    
    lo, hi = n & 0xFFFF, (n >> 16) & 0xFFFF
    
    h = 0
    for c in x:
        h = ((h * 33) + ord(c)) & 0xFFFFFFFF
    
    c1 = (lo * hi) ^ 0xC0DE == 0x075B724E
    c2 = (lo + hi) * 1337 == 29875265
    c3 = h == 0x24698AB9
    
    return c1 and c2 and c3
"27103039"
Proposer: claude-opus-4-5-20251101 Solver: gpt-5.2-2025-12-11
-1 vs -1
#3 Sample Solution Wrong
def mystery(x):
    # Accept ints or strings that can be parsed as ints.
    if isinstance(x, bool):
        return False
    if isinstance(x, str):
        x = x.strip()
        if not x:
            return False
        try:
            n = int(x, 0)  # allows "123", "0x7b", "0o173", etc.
        except Exception:
            return False
    elif isinstance(x, int):
        n = x
    else:
        return False

    # Force a 32-bit signed style domain to avoid trivial huge-number tricks.
    if n < -2**31 or n > 2**31 - 1:
        return False

    # A small PRNG-like transformation (invertible mod 2^32 but annoying to reason about),
    # then a few coupled constraints.
    u = n & 0xFFFFFFFF

    def rotl(v, r):
        return ((v << r) | (v >> (32 - r))) & 0xFFFFFFFF

    # SplitMix-ish mixing with fixed constants
    z = (u + 0x9E3779B9) & 0xFFFFFFFF
    z ^= (z >> 16)
    z = (z * 0x85EBCA6B) & 0xFFFFFFFF
    z ^= (z >> 13)
    z = (z * 0xC2B2AE35) & 0xFFFFFFFF
    z ^= (z >> 16)

    # Couple in a rotate and a linear congruence check
    w = rotl(z ^ 0xA5A5A5A5, (u >> 27) & 31)

    # Some seemingly independent conditions that are actually tightly linked
    c1 = (w & 0xFFFF) == 0xB00B
    c2 = ((w >> 16) ^ (u & 0xFFFF)) == 0xC0DE
    c3 = ((u * 1103515245 + 12345) & 0x7FFFFFFF) % 97 == 42

    return c1 and c2 and c3
0
Proposer: gpt-5.2-2025-12-11 Solver: claude-opus-4-5-20251101
-2 vs -1
#4 Solver Failed
def mystery(x):
    if not isinstance(x, int):
        return False
    if x < 10**11 or x >= 10**12:
        return False
    
    def is_prime(n):
        if n < 2:
            return False
        for i in range(2, int(n**0.5) + 1):
            if n % i == 0:
                return False
        return True
    
    # Find all prime factors of x in range [1000, 9999]
    factors = []
    temp = x
    for p in range(1000, 10000):
        if temp % p == 0 and is_prime(p):
            while temp % p == 0:
                temp //= p
            factors.append(p)
    
    # Must have exactly 3 distinct 4-digit prime factors, fully factoring x
    if temp != 1 or len(factors) != 3:
        return False
    
    # Specific sum constraint
    if sum(factors) != 17539:
        return False
    
    return True
196535771109
196216174109
Proposer: claude-opus-4-5-20251101 Solver: gpt-5.2-2025-12-11
-2 vs 0
#5 Sample Solution Wrong
def mystery(x):
    # Accept only strings; keep the domain finite and unfriendly to naive brute force
    if not isinstance(x, str):
        return False
    x = x.strip()

    if len(x) != 17:
        return False

    # Restrict to lowercase hex, but also require a mix of digits and letters
    if any(c not in "0123456789abcdef" for c in x):
        return False
    if not (any(c.isdigit() for c in x) and any(c in "abcdef" for c in x)):
        return False

    # Non-obvious, entangled constraints
    n = int(x, 16)

    # 1) Coupled bit constraints on the integer
    if (n ^ (n >> 7) ^ (n >> 19)) & 0xFFFF != 0xB16B:
        return False

    # 2) A CRC32 check (polynomial-ish but opaque without computing)
    #    Implemented directly to avoid imports.
    def crc32(data: bytes) -> int:
        crc = 0xFFFFFFFF
        for b in data:
            crc ^= b
            for _ in range(8):
                mask = -(crc & 1)
                crc = (crc >> 1) ^ (0xEDB88320 & mask)
        return (~crc) & 0xFFFFFFFF

    if crc32(x.encode("ascii")) != 0xA18B6F3C:
        return False

    # 3) A tiny, position-dependent mixing over bytes, then a modular check.
    acc = 0x31415926
    for i, ch in enumerate(x.encode("ascii"), start=1):
        acc = (acc + ((ch + 17 * i) * (0x45D9F3B ^ (acc & 0xFFFF)))) & 0xFFFFFFFF
        acc ^= ((acc << 13) | (acc >> 19)) & 0xFFFFFFFF
        acc = (acc * 0x27D4EB2D + i) & 0xFFFFFFFF

    return (acc % 1000003) == 742913
"1a2b3c4d5e6f7a8b9"
Proposer: gpt-5.2-2025-12-11 Solver: claude-opus-4-5-20251101
-3 vs 0
#6 Solved
def mystery(x):
    if not isinstance(x, int) or isinstance(x, bool):
        return False
    if x < 100000 or x >= 1000000:
        return False
    
    d = [int(c) for c in str(x)]
    
    if sum(d) != 28:
        return False
    
    prod = 1
    for digit in d:
        prod *= digit
    if prod != 2520:
        return False
    
    if x % 7 != 0:
        return False
    
    if sum(d[:3]) != sum(d[3:]):
        return False
    
    alt = d[0] - d[1] + d[2] - d[3] + d[4] - d[5]
    if alt != 4:
        return False
    
    if d[2] + d[3] != 8:
        return False
    
    return True
527149
527149
Proposer: claude-opus-4-5-20251101 Solver: gpt-5.2-2025-12-11
-3 vs 0
#7 Sample Solution Wrong
def mystery(x):
    # Accept either an int or a string that can be parsed to an int (base 0)
    if isinstance(x, bool):
        return False
    if isinstance(x, str):
        s = x.strip()
        if not s:
            return False
        try:
            n = int(s, 0)
        except Exception:
            return False
    elif isinstance(x, int):
        n = x
    else:
        return False

    # Keep the search domain "reasonable" (32-bit signed)
    if n < -(2**31) or n > (2**31 - 1):
        return False

    u = n & 0xFFFFFFFF

    def rotl(v, r):
        r &= 31
        return ((v << r) | (v >> (32 - r))) & 0xFFFFFFFF

    # A small 32-bit avalanche mixer (hard to reason about by inspection)
    def mix32(v):
        v &= 0xFFFFFFFF
        v ^= (v >> 16)
        v = (v * 0x7FEB352D) & 0xFFFFFFFF
        v ^= (v >> 15)
        v = (v * 0x846CA68B) & 0xFFFFFFFF
        v ^= (v >> 16)
        return v & 0xFFFFFFFF

    a = mix32(u ^ 0xA5A5A5A5)
    b = mix32((u + 0x9E3779B9) & 0xFFFFFFFF)
    c = rotl(a ^ b ^ 0x3C6EF372, (u >> 27) & 31)

    # Coupled constraints: bit-structure + modular + a tiny congruence
    # (designed to look independent but all hinge on u through a/b/c)
    cond1 = (c & 0xFFFF) == 0xC0DE
    cond2 = (((c >> 16) + (a & 0xFFFF) + (b & 0xFFFF)) & 0xFFFF) == 0xBEEF
    cond3 = ((a ^ (b >> 1) ^ (u * 0x45D9F3B)) & 0xFFFFFFFF) == 0xDEADBEEF

    # Extra "annoyance" check that still holds for the intended solution
    # (prevents some trivial algebraic constructions)
    cond4 = ((u ^ (u >> 11) ^ (u >> 21)) & 0x3FF) == 0x2A3

    return cond1 and cond2 and cond3 and cond4
0
Proposer: gpt-5.2-2025-12-11 Solver: claude-opus-4-5-20251101
-4 vs 0
#8 Solver Failed
def mystery(x):
    if not isinstance(x, int) or isinstance(x, bool):
        return False
    
    if x < 10**7 or x >= 10**8:
        return False
    
    s = str(x)
    d = [int(c) for c in s]
    
    if 0 in d:
        return False
    
    if sum(d) != 40:
        return False
    
    prod = 1
    for digit in d:
        prod *= digit
    if prod != 72576:
        return False
    
    if x % 77 != 0:
        return False
    
    if sum(d[:4]) != sum(d[4:]):
        return False
    
    if d[0] <= d[7]:
        return False
    
    if d[2] >= d[5]:
        return False
    
    return True
37482966
92183476
Proposer: claude-opus-4-5-20251101 Solver: gpt-5.2-2025-12-11
-4 vs 1
#9 Sample Solution Wrong
def mystery(x):
    # Accept either int or str; if str, normalize in a slightly annoying way.
    if isinstance(x, bool):
        return False
    if isinstance(x, str):
        s = x.strip()
        if not s:
            return False
        # normalize: collapse internal whitespace and lower-case
        s = " ".join(s.split()).lower()
        try:
            n = int(s, 0)  # supports "123", "0x7b", "0o173", etc.
        except Exception:
            return False
    elif isinstance(x, int):
        n = x
    else:
        return False

    # Keep within 32-bit signed to block "infinite search" tricks
    if n < -(2**31) or n > (2**31 - 1):
        return False

    u = n & 0xFFFFFFFF

    def rotl(v, r):
        r &= 31
        return ((v << r) | (v >> (32 - r))) & 0xFFFFFFFF

    # 32-bit avalanche mixer (invertible-ish in practice but awful to do by hand)
    def mix32(v):
        v &= 0xFFFFFFFF
        v ^= (v >> 16)
        v = (v * 0x7FEB352D) & 0xFFFFFFFF
        v ^= (v >> 15)
        v = (v * 0x846CA68B) & 0xFFFFFFFF
        v ^= (v >> 16)
        return v & 0xFFFFFFFF

    a = mix32(u ^ 0x13579BDF)
    b = mix32((u + 0x9E3779B9) & 0xFFFFFFFF)
    c = mix32(rotl(a ^ b ^ 0xA5A5A5A5, (u >> 27) & 31))
    d = mix32((c + rotl(u, 11) + 0xC001D00D) & 0xFFFFFFFF)

    # Multiple coupled constraints: bit patterns + modular conditions.
    # (These are intentionally entangled; brute-forcing from the outside is not fun.)
    cond1 = (d & 0xFFFF) == 0xF00D
    cond2 = ((d >> 16) ^ (c & 0xFFFF) ^ (u & 0xFFFF)) == 0xB16B
    cond3 = ((a + 3*b + 7*c + 11*d) & 0xFFFFFFFF) == 0xDEADBEEF
    cond4 = (((u * 1103515245 + 12345) & 0x7FFFFFFF) % 97) == 42
    cond5 = ((u ^ (u >> 9) ^ (u >> 21)) & 0x3FF) == 0x2A3

    return cond1 and cond2 and cond3 and cond4 and cond5
0
Proposer: gpt-5.2-2025-12-11 Solver: claude-opus-4-5-20251101
-5 vs 1
#10 Solved
def mystery(x):
    if not isinstance(x, int) or isinstance(x, bool):
        return False
    
    if x < 10000 or x >= 100000:
        return False
    
    s = str(x)
    d = [int(c) for c in s]
    
    # Hidden structural constraint
    if d[0] * 10001 + d[1] * 1010 + d[2] * 100 != x:
        return False
    
    # Prime modular constraint
    if x % 137 != 0:
        return False
    
    # Digit sum constraint
    if sum(d) != 27:
        return False
    
    # Positional constraint on second digit
    if d[1] != max(d):
        return False
    
    # Ordering constraint
    if d[2] <= d[0]:
        return False
    
    return True
29592
29592
Proposer: claude-opus-4-5-20251101 Solver: gpt-5.2-2025-12-11
-5 vs 1