A large number can be comma formatted to read more easily into groups of three. E.g. 1050 = 1,050 and 10200 = 10,200.
The sum of each of these groups of three would be:
1050=1,050 gives: 1+50=51
10200=10,200 gives: 10+200=210
I need to search for matches in the sum of the groups of threes.
Namely, if I am searching for 1234, then I am looking for numbers whose sum of threes = 1234.
The smallest match is 235,999 since 235+999=1234. No other integer less than 235,999 gives a sum of threes equal to 1234.
The next smallest match is 236,998 since 236+998=1234.
One can add 999 each time, but this fails after reaching 999 since an extra digit of 1 is added to the number due to overflow in the 999.
More generally, I am asking for the solutions (smallest to highest) to:
a+b+c+d… = x
where a,b,c,d… is an arbitrary number of integers between 0-999 and x
is a fixed integer
Note there are infinite solutions to this for any positive integer x.
How would one get the solutions to this beginning with the smallest number solutions (for y number of solutions where y can be an arbitrarily large number)?
Is there a way to do this without brute force looping one by one? I'm dealing with potentially very large numbers, which could take years to loop through in a straight loop. Ideally, one should do this without failed attempts.
The problem is easier to think about if instead of groups of 3 digits, you just consider 1 digit at a time.
An algorithm:
Some Python code for illustration:
x = 1234
grouping = 3
max_iterations = 200
max_in_group = 10**grouping - 1
a = [x]
while max_iterations > 0:
#step 1: while a[0] is too large: redistribute to the left
i = 0
while a[i] > max_in_group:
if i == len(a) - 1:
a.append(0)
a[i + 1] += a[i] - max_in_group
a[i] = max_in_group
i += 1
num = sum(10**(grouping*i) * a[i] for i, n in enumerate(a))
print(f"{num} {num:,}")
# print("".join([str(t) for t in a[::-1]]), ",".join([str(t) for t in a[::-1]]))
# step 2: add one to the penultimate group, while group already full: set to 0 and increment the
# group left of it;
# while the surplus is too large (because a[0] is too small) repeat the incrementing
i0 = 1
surplus = 0
while True: # needs to be executed at least once, and repeated if the surplus became too large
i = i0
while True: # increment a[i] by 1, which can carry to the left
if i == len(a):
a.append(1)
surplus += 1
break
else:
if a[i] == max_in_group:
a[i] = 0
surplus -= max_in_group
i += 1
else:
a[i] += 1
surplus += 1
break
if a[0] >= surplus:
break
else:
surplus -= a[i0]
a[i0] = 0
i0 += 1
#step 3: a[0] should absorb the surplus created in step 1, although a[0] can get out of bounds
a[0] -= surplus
surplus = 0
max_iterations -= 1
Abbreviated output:
235,999 236,998 ... 998,236 999,235 ... 1,234,999 1,235,998 ... 1,998,235 1,999,234 2,233,999 2,234,998 ...
Output for grouping=3 and x=3456:
459,999,999,999 460,998,999,999 460,999,998,999 460,999,999,998 461,997,999,999
461,998,998,999 461,998,999,998 461,999,997,999 461,999,998,998 461,999,999,997
462,996,999,999 ...
Output for grouping=1 and x=16:
79 88 97 169 178 187 196 259 268 277 286 295 349 358 367 376 385 394 439 448 457 466
475 484 493 529 538 547 556 565 574 583 592 619 628 637 646 655 664 673 682 691 709
718 727 736 745 754 763 772 781 790 808 817 826 835 844 853 862 871 880 907 916 925
934 943 952 961 970 1069 1078 1087 1096 1159 1168 1177 1186 1195 1249 1258 1267 1276
1285 1294 1339 1348 1357 1366 1375 1384 1393 1429 1438 1447 1456 1465 1474 1483 1492
1519 1528 1537 1546 1555 1564 1573 1582 1591 1609 1618 1627 1636 1645 1654 1663 1672
1681 1690 1708 1717 1726 1735 1744 1753 1762 1771 1780 1807 1816 1825 1834 1843 1852
1861 1870 1906 1915 1924 1933 1942 1951 1960 2059 2068 2077 2086 2095 2149 2158 2167
2176 2185 2194 2239 2248 2257 2266 2275 2284 2293 2329 2338 2347 2356 2365 2374 2383
2392 2419 2428 2437 2446 2455 2464 2473 2482 2491 2509 2518 2527 2536 2545 2554 2563
2572 2581 2590 2608 2617 2626 2635 2644 2653 2662 2671 2680 2707 2716 2725 2734 ...
