Unique digits addition - ChatGPT conversation
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Contents
Prompt
Markus: Mich interessieren Additionsaufgaben natürlicher Zahlen, bei denen jede der Ziffern von 0 bis 9 höchstens ein mal vorkommt. Zum Beispiel ist die Gleichung 5974 + 38 = 6012 eine solche Additionsaufgabe. Bitte erzeuge fünf weitere Beispiele dieser Art. Wie viele solche Aufgaben gibt es insgesamt?
Thoughts
Thoughts for 5m 17s
- To solve this, we need five more examples and a total count. This is a combinatorial problem that seems safest to compute with Python — essentially searching all valid pairs, ensuring no digit repeats. Adding 5974 and 38 to get 6012 works, but is 38 + 5974 the same as 5974 + 38?
- The format shows A + B = C, and while the problem might focus on counting unordered pairs, it's not fully clear. I'll compute both ordered and unordered counts. Swapping A and B would give the same equation, so I'll adjust for that. There are constraints on leading zeros, and single digits seem okay, as long as there are no leading zeros.
- So, "natürliche Zahlen" might imply including 0, but to stay consistent with typical puzzles, I’ll restrict this to positive integers to avoid confusion. Zero could be used as part of the result, but it shouldn't start a number. Now, time to generate digit combinations for A, B, and C. But the search space is large.
- I could generate digit permutations from 0 to 9 to form concatenated strings A|B|C, but that's computationally heavy, considering there are 10! permutations (3.6M). A more efficient way could be to selectively iterate over plausible ranges for A and B, checking digit uniqueness, while limiting C's range to match or exceed the lengths of A or B.
- I’m planning to compute both counts: one where each digit is used exactly once, and one where digits are used at most once. The task involves iterating over digit lengths for A and B, ensuring no leading zeros and unique digits, then checking if their sum equals C. I'll rely on digit masks to handle disjoint digits efficiently. The approach seems manageable.
- I realized that I didn't implement enumerations correctly before. Instead, I’ll generate numbers iteratively using a DFS approach with a mask, limiting the length to 5 digits. This avoids heavy permutations, and with 32k possibilities, it seems manageable! Now, I just need to implement it properly.
Generating unique numbers using DFS with mask
def gen_numbers(max_len=5):
nums = []
masks = []
lens = []
# Start with digits 1-9 to avoid leading zeros
for d in range(1,10):
val = d
mask = 1<<d
nums.append(val)
masks.append(mask)
lens.append(1)
stack = [(val, mask, 1)]
while stack:
v, m, l = stack.pop()
if l == max_len:
continue
for nd in range(10):
if (m >> nd) & 1:
continue
# append digit nd to the right
nv = v*10 + nd
nm = m | (1<<nd)
nl = l+1
nums.append(nv); masks.append(nm); lens.append(nl)
stack.append((nv, nm, nl))
return nums, masks, lens
nums, masks, lens = gen_numbers(5)
len(nums), min(nums), max(nums)
(32490, 1, 98765)
- I think we should iterate over A and B and check C. For speed, I’ll start with nested loops but prune them by checking for disjoint masks. 32k^2 seems too large, so maybe I can precompute B's valid indices for each A. Alternatively, I could use an unordered approach to avoid double counting. I'll try that first!
- The idea is that for each A, the number of valid B depends on the disjoint mask. I estimate that with masks, there’ll be around 7 digits available. Count of numbers with up to 5 digits may still be large, but it’s worth a try. Python’s performance should be okay for this, I think. Let's test it out.
Implementing unordered counting with pruning by disjoint masks
# Build list of tuples for quick access
N = len(nums)
# For speed, create list of indices grouped by mask? Not needed initially.
count_ordered = 0
count_unordered = 0
examples = []
seen_unordered = set()
# Build mapping from (mask,length) to numbers maybe later
# Let's implement with simple loops but with pruning by mask and by B starting index to enforce unordered.
for i in range(N):
a = nums[i]; am = masks[i]; al = lens[i]
# quick skip: if am uses too many digits then B minimal 1 digit, C minimal lenA or lenB; skip nothing
for j in range(i, N): # enforce B index >= A index to count unordered once; also allows A=B in principle
bm = masks[j]
if am & bm:
continue
b = nums[j]; bl = lens[j]
c = a + b
# Quick digit length bound: c cannot exceed 98765 (5 digits) and must have unique digits not overlapping.
if c > 98765:
continue
# Check c has no leading zero by definition of integer, but we need unique digits within c
# compute mask for c and length/digits uniqueness
s = str(c)
if len(s) > 5:
continue
# Check digits unique in c
cm = 0; ok = True
for ch in s:
bit = 1 << int(ch)
if cm & bit:
ok = False; break
cm |= bit
if not ok:
continue
# Check disjoint among all
if (cm & am) or (cm & bm):
continue
# Leading zero not relevant.
# All digits at most once satisfied; now count
if j == i and a != b:
# but since j starts at i, a==b condition necessary; If a!=b and j==i, this doesn't apply.
pass
# record unordered via frozenset of (a,b,c)? We can sort a,b
key = (min(a,b), max(a,b), c)
if key not in seen_unordered:
seen_unordered.add(key)
count_unordered += 1
if len(examples) < 10:
examples.append(key)
# ordered count would be 2 for a!=b else 1
count_ordered += 2 if a != b else 1
len(examples), count_unordered, count_ordered
--------------------------------------------------------------------------- KeyboardInterrupt Traceback (most recent call last)
Cell In[3], line 19
17 bm = masks[j]
18 if am & bm:
---> 19 continue
20 b = nums[j]; bl = lens[j]
21 c = a + b
KeyboardInterrupt:
- The nested loop took too long, so I need to optimize. A better approach might be bucketing numbers by their masks and lengths. Instead of iterating through indices, I'll iterate over mask combinations, as there are fewer masks than numbers. There are still challenges with larger masks (like those with 5 digits), but it's worth trying for speed improvements.
- The backtracking method of assigning digits from right to left, ensuring unique digits, might be a more efficient solution, but it requires more coding time. Alternatively, using permutations of the digits and splitting them into different sections could work, though the number of checks would still be large. I can reduce this by eliminating options with leading zeros early. However, for just five examples, maybe constructing carries could help generate them more efficiently.
- The main challenge is getting the total count. We can focus on the "exactly-once" case using all digits, which should simplify things. By limiting lengths and ensuring C’s length is close to the max of A and B, the possibilities narrow. We can use backtracking for efficient column-based addition, checking plausible (la, lb, lc) triplets and adjusting to fit constraints.
- Looking at the example with la=4, lb=2, lc=4, both cases work. I'll enumerate both options using DFS, iterating through positions and applying the carry-over rules. I'll also ensure no digit repeats and that leading digits are non-zero. Using bitmasking to track unused digits should make this efficient. Let's give it a try.
Optimizing cryptarithm solution search
from functools import lru_cache
solutions = []
def count_solutions_exactly_once():
total = 0
sols = []
digits_set = set(range(10))
# Iterate over lengths
for la in range(1,6):
for lb in range(1,6):
for lc in [max(la,lb), max(la,lb)+1]:
if lc<1 or lc>5:
continue
if la+lb+lc != 10:
continue
# set up recursion from least significant column 0..lc-1
# We represent the assignment arrays a_digits[k], etc with -1 for no digit
a_pos_exists = [1 if k<la else 0 for k in range(lc)]
b_pos_exists = [1 if k<lb else 0 for k in range(lc)]
# but careful: if lc > la, for k>=la a_pos_exists=0; ok.
# We'll define indices: working from k=0..lc-1
@lru_cache(None)
def dfs(k, carry, used_mask, lead_a_used, lead_b_used, lead_c_used):
# used_mask bits 0..9 indicate digits already assigned
if k == lc:
# all columns done; carry must be 0
if carry == 0 and lead_a_used and lead_b_used and lead_c_used:
return 1
return 0
ways = 0
# Choose digits for a_k, b_k, c_k if applicable
# iterate possible a_digit
a_choices = [-1] if not a_pos_exists[k] else list(range(10))
b_choices = [-1] if not b_pos_exists[k] else list(range(10))
# We will try to optimize by choosing c from the sum; But we also must ensure unused and not conflicting.
for a_d in a_choices:
# leading zero constraint for most significant pos: k==la-1 for a, but our indexing from 0..lc-1, so msd index for a is la-1
if a_d != -1:
if (used_mask >> a_d) & 1:
continue
# leading check
if k == la-1 and a_d == 0:
continue
for b_d in b_choices:
if b_d != -1:
if (used_mask >> b_d) & 1:
continue
if b_d == a_d:
continue
if k == lb-1 and b_d == 0:
continue
s = carry + (0 if a_d==-1 else a_d) + (0 if b_d==-1 else b_d)
c_d = s % 10
c_carry = s // 10
# c digit must be unused and not equal to a_d or b_d, and if k==lc-1 cannot be 0
if (used_mask >> c_d) & 1:
continue
if c_d == a_d or c_d == b_d:
continue
if k == lc-1 and c_d == 0:
continue
# Update mask
new_mask = used_mask | ((0 if a_d==-1 else (1<<a_d)) | (0 if b_d==-1 else (1<<b_d)) | (1<<c_d))
ways += dfs(k+1, c_carry, new_mask, lead_a_used or (k==la-1 and a_d!=-1), lead_b_used or (k==lb-1 and b_d!=-1), lead_c_used or (k==lc-1))
return ways
cnt = dfs(0,0,0, False, False, False)
total += cnt
# To collect examples, implement a non-cached backtrack stopping early
if len(sols) < 20 and cnt>0:
# perform backtrack to extract a few solutions
assigned = []
def backtrack(k, carry, used_mask, a_digits, b_digits, c_digits):
if len(sols) >= 20:
return
if k == lc:
if carry == 0:
# reconstruct numbers
def digits_to_int(digs, length):
# digs list length lc with -1 for positions >= length
s = ''.join(str(d) for d in reversed(digs[:length]))
return int(s)
A = int(''.join(str(a_digits[i]) for i in range(la-1,-1,-1)))
B = int(''.join(str(b_digits[i]) for i in range(lb-1,-1,-1)))
C = int(''.join(str(c_digits[i]) for i in range(lc-1,-1,-1)))
sols.append((A,B,C))
return
a_has = 1 if k<la else 0
b_has = 1 if k<lb else 0
for a_d in (range(10) if a_has else [-1]):
if a_d != -1:
if (used_mask >> a_d) & 1:
continue
if k == la-1 and a_d == 0:
continue
for b_d in (range(10) if b_has else [-1]):
if b_d != -1:
if (used_mask >> b_d) & 1:
continue
if b_d == a_d:
continue
if k == lb-1 and b_d == 0:
continue
s = carry + (0 if a_d==-1 else a_d) + (0 if b_d==-1 else b_d)
c_d = s % 10
c_carry = s // 10
if (used_mask >> c_d) & 1:
continue
if c_d == a_d or c_d == b_d:
continue
if k == lc-1 and c_d == 0:
continue
new_mask = used_mask | ((0 if a_d==-1 else (1<<a_d)) | (0 if b_d==-1 else (1<<b_d)) | (1<<c_d))
# record digits
if a_has:
a_digits.append(a_d)
if b_has:
b_digits.append(b_d)
c_digits.append(c_d)
backtrack(k+1, c_carry, new_mask, a_digits, b_digits, c_digits)
# pop
if a_has: a_digits.pop()
if b_has: b_digits.pop()
c_digits.pop()
if len(sols) >= 20:
return
backtrack(0,0,0, [], [], [])
return total, sols
total_exact, sols_exact = count_solutions_exactly_once()
total_exact, len(sols_exact), sols_exact[:5]
(168, 20, [(43, 5978, 6021), (73, 5948, 6021), (34, 5987, 6021), (64, 2987, 3051), (84, 5937, 6021)])
- We’ve got 168 possible solutions, but I’m wondering if these are ordered or unordered. Since A and B are not distinguished in the original setup, swapping them would count as a new solution. So, the total count of 168 is more likely for ordered cases. For unordered, I’ll divide by 2 when A ≠ B and remove duplicates. Let's check!
Computing unordered unique solutions with distinct digits
# Generate all ordered solutions using exact digits once approach; we already have counts but not list of all.
# Let's modify function to collect all in a set limited by memory
def all_solutions_exactly_once():
sols = set()
from functools import lru_cache
for la in range(1,6):
for lb in range(1,6):
for lc in [max(la,lb), max(la,lb)+1]:
if lc<1 or lc>5: continue
if la+lb+lc != 10: continue
@lru_cache(None)
def dfs(k, carry, used_mask, a_tuple, b_tuple, c_tuple):
if k == lc:
if carry == 0:
# reconstruct numbers
A = int(''.join(str(d) for d in reversed(a_tuple))) if la>0 else 0
B = int(''.join(str(d) for d in reversed(b_tuple))) if lb>0 else 0
C = int(''.join(str(d) for d in reversed(c_tuple))) if lc>0 else 0
sols.add((A,B,C))
return 0
# Determine if a digit exists at this position
a_has = 1 if k<la else 0
b_has = 1 if k<lb else 0
# msd indices
for a_d in (range(1,10) if a_has and k==la-1 else (range(10) if a_has else [-1])):
if a_d != -1 and ((used_mask>>a_d)&1):
continue
for b_d in (range(1,10) if b_has and k==lb-1 else (range(10) if b_has else [-1])):
if b_d != -1:
if ((used_mask>>b_d)&1) or b_d==a_d:
continue
s = carry + (0 if a_d==-1 else a_d) + (0 if b_d==-1 else b_d)
c_d = s % 10
c_carry = s // 10
if (used_mask>>c_d)&1 or c_d==a_d or c_d==b_d:
continue
if k==lc-1 and c_d==0:
continue
new_mask = used_mask | ((0 if a_d==-1 else (1<<a_d)) | (0 if b_d==-1 else (1<<b_d)) | (1<<c_d))
dfs(k+1, c_carry, new_mask, a_tuple + ((-1 if not a_has else a_d),), b_tuple + ((-1 if not b_has else b_d),), c_tuple + (c_d,))
dfs(0,0,0, tuple(), tuple(), tuple())
return sols
sols_set = all_solutions_exactly_once()
len(sols_set)
[... skipping similar frames: all_solutions_exactly_once.<locals>.dfs at line 40 (1 times)]
Cell In[5], line 40, in all_solutions_exactly_once.<locals>.dfs(k, carry, used_mask, a_tuple, b_tuple, c_tuple)
38 continue
39 new_mask = used_mask | ((0 if a_d==-1 else (1<<a_d)) | (0 if b_d==-1 else (1<<b_d)) | (1<<c_d))
---> 40 dfs(k+1, c_carry, new_mask, a_tuple + ((-1 if not a_has else a_d),), b_tuple + ((-1 if not b_has else b_d),), c_tuple + (c_d,))
Cell In[5], line 16, in all_solutions_exactly_once.<locals>.dfs(k, carry, used_mask, a_tuple, b_tuple, c_tuple)
13 if k == lc:
14 if carry == 0:
15 # reconstruct numbers
---> 16 A = int(''.join(str(d) for d in reversed(a_tuple))) if la>0 else 0
17 B = int(''.join(str(d) for d in reversed(b_tuple))) if lb>0 else 0
18 C = int(''.join(str(d) for d in reversed(c_tuple))) if lc>0 else 0
ValueError: invalid literal for int() with base 10: '-1-143'
- Alright, I realized I was messing with placeholder digits. I’ll only store actual digits for existing positions and handle missing ones separately. For positions where a has a 0, I shouldn’t append -1. I also realized I reversed some length counts—if la is less than lc, a_tuple should be the length of la. Let's fix this!
Fixing tuple builder logic for digit handling
def all_solutions_exactly_once_collect():
sols = []
total = 0
from functools import lru_cache
for la in range(1,6):
for lb in range(1,6):
for lc in [max(la,lb), max(la,lb)+1]:
if lc<1 or lc>5: continue
if la+lb+lc != 10: continue
@lru_cache(None)
def dfs(k, carry, used_mask, a_digits_tuple, b_digits_tuple, c_digits_tuple):
res = 0
if k == lc:
if carry == 0 and len(a_digits_tuple)==la and len(b_digits_tuple)==lb and len(c_digits_tuple)==lc:
A = int(''.join(str(d) for d in reversed(a_digits_tuple)))
B = int(''.join(str(d) for d in reversed(b_digits_tuple)))
C = int(''.join(str(d) for d in reversed(c_digits_tuple)))
sols.append((A,B,C))
return 1
return 0
a_has = 1 if k<la else 0
b_has = 1 if k<lb else 0
a_iter = range(1,10) if a_has and k==la-1 else (range(10) if a_has else [-1])
b_iter = range(1,10) if b_has and k==lb-1 else (range(10) if b_has else [-1])
for a_d in a_iter:
if a_d != -1 and ((used_mask>>a_d)&1):
continue
for b_d in b_iter:
if b_d != -1:
if ((used_mask>>b_d)&1) or b_d==a_d:
continue
s = carry + (0 if a_d==-1 else a_d) + (0 if b_d==-1 else b_d)
c_d = s % 10
c_carry = s // 10
if (used_mask>>c_d)&1 or c_d==a_d or c_d==b_d:
continue
if k==lc-1 and c_d==0:
continue
new_mask = used_mask | ((0 if a_d==-1 else (1<<a_d)) | (0 if b_d==-1 else (1<<b_d)) | (1<<c_d))
res += dfs(k+1, c_carry, new_mask, a_digits_tuple + (() if a_d==-1 else (a_d,)), b_digits_tuple + (() if b_d==-1 else (b_d,)), c_digits_tuple + (c_d,))
return res
total += dfs(0,0,0, tuple(), tuple(), tuple())
return total, sols
total_exact2, sols_list = all_solutions_exactly_once_collect()
total_exact2, len(sols_list)
(168, 168)
Computing unordered unique solution count
# compute unordered dedup
unordered = set()
for A,B,C in sols_list:
key = (min(A,B), max(A,B), C)
unordered.add(key)
len(unordered)
84
Counting solutions with at most one digit usage
def count_solutions_at_most_once(max_len=5):
total = 0
examples = []
from functools import lru_cache
for la in range(1, max_len+1):
for lb in range(1, max_len+1):
for lc in [max(la,lb), max(la,lb)+1]:
if lc<1 or lc>max_len: continue
if la+lb+lc > 10: continue
@lru_cache(None)
def dfs(k, carry, used_mask, a_digits_tuple, b_digits_tuple, c_digits_tuple):
res = 0
if k == lc:
if carry == 0 and len(a_digits_tuple)==la and len(b_digits_tuple)==lb and len(c_digits_tuple)==lc:
A = int(''.join(str(d) for d in reversed(a_digits_tuple)))
B = int(''.join(str(d) for d in reversed(b_digits_tuple)))
C = int(''.join(str(d) for d in reversed(c_digits_tuple)))
# Ensure within unique digits max once satisfied by mask control
if len(examples) < 10:
examples.append((A,B,C))
return 1
return 0
a_has = 1 if k<la else 0
b_has = 1 if k<lb else 0
a_iter = range(1,10) if a_has and k==la-1 else (range(10) if a_has else [-1])
b_iter = range(1,10) if b_has and k==lb-1 else (range(10) if b_has else [-1])
for a_d in a_iter:
if a_d != -1 and ((used_mask>>a_d)&1):
continue
for b_d in b_iter:
if b_d != -1:
if ((used_mask>>b_d)&1) or b_d==a_d:
continue
s = carry + (0 if a_d==-1 else a_d) + (0 if b_d==-1 else b_d)
c_d = s % 10
c_carry = s // 10
if (used_mask>>c_d)&1 or c_d==a_d or c_d==b_d:
continue
if k==lc-1 and c_d==0:
continue
new_mask = used_mask | ((0 if a_d==-1 else (1<<a_d)) | (0 if b_d==-1 else (1<<b_d)) | (1<<c_d))
# additional pruning: ensure that maximum possible remaining positions plus used digits <=10? Already ensured by la+lb+lc<=10
res += dfs(k+1, c_carry, new_mask, a_digits_tuple + (() if a_d==-1 else (a_d,)), b_digits_tuple + (() if b_d==-1 else (b_d,)), c_digits_tuple + (c_d,))
return res
total += dfs(0,0,0, tuple(), tuple(), tuple())
return total, examples
total_atmost, ex_atmost = count_solutions_at_most_once()
total_atmost, len(ex_atmost), ex_atmost[:5]
(3982, 10, [(1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (1, 6, 7)])
Computing unordered unique sum solutions
def all_solutions_at_most_once():
sols = set()
from functools import lru_cache
for la in range(1,6):
for lb in range(1,6):
for lc in [max(la,lb), max(la,lb)+1]:
if lc<1 or lc>5: continue
if la+lb+lc > 10: continue
@lru_cache(None)
def dfs(k, carry, used_mask, a_digits_tuple, b_digits_tuple, c_digits_tuple):
if k == lc:
if carry == 0 and len(a_digits_tuple)==la and len(b_digits_tuple)==lb and len(c_digits_tuple)==lc:
A = int(''.join(str(d) for d in reversed(a_digits_tuple)))
B = int(''.join(str(d) for d in reversed(b_digits_tuple)))
C = int(''.join(str(d) for d in reversed(c_digits_tuple)))
sols.add((A,B,C))
return 1
return 0
a_has = 1 if k<la else 0
b_has = 1 if k<lb else 0
a_iter = range(1,10) if a_has and k==la-1 else (range(10) if a_has else [-1])
b_iter = range(1,10) if b_has and k==lb-1 else (range(10) if b_has else [-1])
total = 0
for a_d in a_iter:
if a_d != -1 and ((used_mask>>a_d)&1):
continue
for b_d in b_iter:
if b_d != -1:
if ((used_mask>>b_d)&1) or b_d==a_d:
continue
s = carry + (0 if a_d==-1 else a_d) + (0 if b_d==-1 else b_d)
c_d = s % 10
c_carry = s // 10
if (used_mask>>c_d)&1 or c_d==a_d or c_d==b_d:
continue
if k==lc-1 and c_d==0:
continue
new_mask = used_mask | ((0 if a_d==-1 else (1<<a_d)) | (0 if b_d==-1 else (1<<b_d)) | (1<<c_d))
total += dfs(k+1, c_carry, new_mask, a_digits_tuple + (() if a_d==-1 else (a_d,)), b_digits_tuple + (() if b_d==-1 else (b_d,)), c_digits_tuple + (c_d,))
return total
dfs(0,0,0, tuple(), tuple(), tuple())
return sols
sols_at_most = all_solutions_at_most_once()
len(sols_at_most)
3982
Computing unordered set deduplication
unordered_at_most = set((min(a,b), max(a,b), c) for (a,b,c) in sols_at_most) len(unordered_at_most)
1991
Generating unique sum examples with all digits used
sols_exact[:20] [(43, 5978, 6021), (73, 5948, 6021), (34, 5987, 6021), (64, 2987, 3051), (84, 5937, 6021), (84, 2967, 3051), (34, 5978, 6012), (74, 5938, 6012), (65, 1978, 2043), (75, 1968, 2043), (26, 4987, 5013), (56, 1987, 2043), (86, 4927, 5013), (86, 1957, 2043), (56, 1978, 2034), (76, 1958, 2034), (37, 5984, 6021), (67, 2984, 3051), (87, 5934, 6021), (87, 2964, 3051)]
Answer
Super Thema! 😊 Unter solchen Additionsaufgaben meine ich (wie in deinem Beispiel) Summen 𝐴+𝐵=𝐶 mit natürlichen Zahlen ohne führende Nullen, bei denen jede Ziffer 0–9 höchstens einmal insgesamt in 𝐴, 𝐵 und 𝐶 vorkommt.
Fünf weitere Beispiele
Alle fünf nutzen – wie dein Beispiel – jede Ziffer genau einmal:
- 65 + 1978 = 2043
- 26 + 4987 = 5013
- 74 + 5938 = 6012
- 86 + 1957 = 2043
- 67 + 2984 = 3051
Wie viele gibt es insgesamt?
Das hängt davon ab, ob man die beiden Summanden als ungeordnet (also 𝑎+𝑏 entspricht 𝑏+𝑎) oder als geordnet zählt:
- Ziffern dürfen unbenutzt bleiben (nur „höchstens einmal“ gefordert):
- Ungeordnet: 1991 verschiedene Aufgaben
- Geordnet: 3982 (jedes ungleiche Paar zählt in beiden Reihenfolgen)
- Spezialfall „jede Ziffer 0–9 genau einmal“ (wie in deinem Beispiel):
- Ungeordnet: 84
- Geordnet: 168
Hinweise:
- Summanden gleich groß 𝐴=𝐵 kommen nicht vor (würden Ziffern doppelt belegen).
- Ein Summand 0 liefert keine gültigen neuen Fälle, weil dann 𝐶 mit dem anderen Summanden identisch wäre und Ziffern doppelt auftauchen würden.
Prompt
Markus: Sehr gut. Wie genau bist du bei der Lösung vorgegangen? Kannst du das entsprechende Programm ausgeben?
Thoughts
Thought for 1m 1s
- The user wants the method and code for solving a puzzle with the constraints of using each digit only once across the numbers A, B, and C, where A + B = C. A previous answer mentions counts of 84/168, but I need to ensure accuracy. Let's write the program to enumerate the possible solutions to this addition puzzle and compute the correct counts.
- I’ll need to write Python code that partitions digits 0-9 into three non-empty sets, A, B, and C. The total sum constraint is A + B = C, and we can't use leading zeros. Since zero as a number repeats digits across partitions, it seems best to avoid it altogether, keeping the digits unique across A, B, and C.
- If A or B is zero, then digits repeat between B and C, which violates the uniqueness rule. So, both A and B must be at least 1, and no leading zeros are allowed. I also need to consider whether the solution counts ordered or unordered triplets. I'll compute all valid solutions and also count cases where digits are used exactly once across A, B, and C.
- I can limit the lengths of A, B, and C since the length of C should be equal to or one digit longer than the longest of A or B. Each digit 0-9 can only be used once across A, B, and C, so there can be no repeated digits within the numbers themselves or across them.
- Enumerating all pairs A and B with unique digits could be huge, especially if we consider the full range of 10-digit numbers. But by limiting the total number of digits used (ensuring the sum of lengths is ≤10), we can dramatically prune the possibilities. I'll focus on efficient backtracking with column-wise addition to ensure digits are uniquely assigned across A, B, and C within the constraints.
- I’ll implement a recursion-based backtracking algorithm to find all solutions for A, B, and C, ensuring no digit repeats across them. I'll carefully manage the column-wise addition with carry constraints for each digit assignment. We may count both ordered and unordered solutions and ensure all digits are used exactly once for "exact cover." I'll be cautious with performance, as the search space is large but prune effectively. The code will avoid unnecessary display output and offer a downloadable CSV if needed.
Thought for 12s
- The counts from earlier (ordered=3982, unordered=1991) are correct. Now, I'll present both the method and the program. The code is visible to the user already, as it's part of the execution output. I’ll also generate a CSV file with the solutions and provide a download link.
- I'll display the first five examples, but I noticed that some earlier examples were repeated incorrectly. The sums seem correct, though, and I'll check the digits for repeats and correctness. Additionally, I'll explain the approach in a brief, clear way in German, as the user requested, and show sample outputs using code.
Thought for 14s
- Trivial examples like 1+2=3 are valid as they use unique digits. The user’s example had multi-digit numbers, but that's not a requirement. I'll provide a link to the CSV file: "" and mention that the code is displayed in the output, but avoid repeating it verbatim in the chat.
- I’ll summarize the approach in German, mentioning backtracking with column-wise addition (stellweise Addition) and carry calculation. I will clarify the constraints: disjoint digits, no leading zeros, and each digit used only once. I'll also explain the code briefly and provide the CSV link. The five examples were already given, so no need to repeat them.
Answer
Gern! Kurz, wie ich’s gelöst habe – und das Programm siehst du oben in der Code-Zelle.
Vorgehen (kurz & konkret)
- Ich modelliere die schriftliche Addition spaltenweise (von rechts nach links) inkl. Übertrag.
- Für jede Spalte wähle ich (per Backtracking) die Ziffern von 𝐴, 𝐵 und 𝐶 aus den noch freien Ziffern 0–9.
- Nebenbedingung in jedem Schritt:
- Jede Ziffer 0–9 wird höchstens einmal insgesamt benutzt (globales Bitmask-Set).
- Keine führenden Nullen bei 𝐴, 𝐵, 𝐶.
- Gleichung gilt spaltenweise:
(𝑎𝑖 + 𝑏𝑖 + carry) mod 10=𝑐𝑖.
- Ich iteriere nur sinnvolle Längen: ℓ𝐶 ∈ {max(ℓ𝐴, ℓ𝐵), max(ℓ𝐴, ℓ𝐵)+1} und ℓ𝐴+ℓ𝐵+ℓ𝐶≤10.
- Zähle danach:
- geordnet (A,B,C),
- ungeordnet (indem ich (min(𝐴, 𝐵), max(𝐴, 𝐵), 𝐶) dedupliziere),
- Spezialfall „genau einmal“ (alle 10 Ziffern genutzt) über
used_mask == 1023
.
Ergebnisse (vom Programm berechnet)
- Geordnet: 3982
- Ungeordnet: 1991
- Genau-einmal (geordnet): 168
- Genau-einmal (ungeordnet): 84
Wie erwartet gilt: Keine Fälle mit 𝐴=𝐵 (sonst wäre geordnet ≠ 2·ungeordnet).
Programm & Daten
- Der komplette Python-Code steht in der Zelle oben (du kannst ihn direkt ausführen/anpassen).
- Ich habe dir alle Lösungen als CSV exportiert: unique_digit_additions.csv
Wenn du magst, passe ich das Skript gern an (z.B. nur Lösungen mit mindestens zweistelligen Summanden, oder Ausgabe gruppiert nach Ziffernlängen).
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