I don't think it would make a large difference in runtime, but the bounds aren't quite right.

The largest possible value isn't 999,999,999, because that's not a legal value. The largest value is 987,654,321

Dividing that by 9, you get 109,739,369, but that's not a legal value either, you'd want to start with 109,738,365 for the gcd as it's a valid possible row. I don't know that all gcds would need to be a valid row, but any gcd larger than 98,765,432 would need to be used at multiples from 1-9; with that gcd or smaller, multiple 10 becomes viable and the gcd doesn't need to be valid with multiple of 1.

If the floating point trick worth it? When I was young, floating point operations were very slow and I still try to avoid them like the plague.

I remembered there was a trick to avoid popcount, but I didn't remember it. So I found https://stackoverflow.com/questions/51387998/count-bits-1-on...

That version counts, the 1 bit's but here we already know the number is not 0 and we want to know that the inverted number it has exactly 1 bit set, so instead of

  popcnt(n) == 1

we can use

  n & (n-1) == 0

Moreover, n is the inverted number

  n = 0b1111111111 - s

so an alternative is to run the check in the original number, and I think that this does the trick:

  s | (s+1) == 0b1111111111

Should look into Knuth’s DLX. My JS implementation solved this case in 1.8ms

For us slow kids in the back what does this mean, “[…] such that the nine 9-digit numbers formed by the rows of the grid has the highest-possible GCD over any such grid”?

I also wrote my solution in Rust. I was pleased that my approach gave me an opportunity to write a linked list and get the chance to apply some rarely used learning[0].

My solution doesn't use SIMD, but is actually takes about the same amount of time as the the solution in the article, given the same number of cores, though an glaring weakness of my approach is that it can only scale up to 7 cores as written.

Rough outline of how mine works:

- Set current best GCD to 1.

- Search through all valid board configurations.

- Bail out from the search early if the current partially generated board couldn't have a GCD greater than the current maximum found, e.g. if we've only generated two rows of the board, and the GCD of those two rows are already less than the max.

- Update the current best GCD as you find higher ones.

- Share the current best GCD value across multiple threads. That way the longer the program runs, the earlier and earlier the searches will start bailing out.

- Don't always read from the shared variable to avoid contention. Instead, each thread has its own copy of the last value it read which we compare with first before copying and caching the shared value.

- Another interesting property of this approach is that it can be used to validate the correct solution even faster than it takes to find it. Instead of initially setting the max GCD to 1, set it to `solution - 1`. That way branches in our search will bail even sooner from the beginning. This leads to the program running about 20% more quickly.

Source: https://gist.github.com/lazytype/35b45f3ea81b5c1c5555546fe6f...

[0] https://rust-unofficial.github.io/too-many-lists/

Since we're over-engineering here, one more optimization is to skip all potential GCDs that are divisible by 2 or 5.

Suppose the GCD was divisible by 2, then all rows would be even. Since the last digit of an even integer is in {0,2,4,6,8} and we need 9 unique numbers in the final column, we know that 4 or 5 of the row numbers must be odd. So the GCD can't be even.

Similarly, the GCD can't be divisible by 5. If it were, all rows numbers would need a 0 or 5 in the final digit.

> My first attempt when solving this puzzle was to encode the constraints as an SMT optimization problem, and using Z3 to find the solution.

This is why choosing the right tool for the job matters. Z3 is hilariously overkill for such a problem (not to undermine how interesting and useful Z3 is, and not to accuse the author of doing anything wrong).

Z3 is only one class of solver; a finite domain constraint solver is much more suited to these sorts of tasks.

https://gist.github.com/Qix-/3580f268e2725848971703e74f6b26b...

    $ time node sudoku.js
    
    1 2 7  6 8 5  4 3 9  
    5 9 4  1 3 7  8 2 6  
    8 3 6  4 2 9  5 7 1  
    
    3 6 1  8 5 4  2 9 7  
    7 4 5  2 9 1  3 6 8  
    2 8 9  3 7 6  1 5 4  
    
    6 5 3  9 1 8  7 4 2  
    4 1 2  7 6 3  9 8 5  
    9 7 8  5 4 2  6 1 3 
    
    0:00.08elapsed

Sigh. My friend did this problem on pen and paper and made me feel stupid. Their solution was so wildly clever! It relied on the observation that the sum of the rows of the sudoku board (given the digits in use) is a known fixed value, and went from there (I'll leave the rest as an exercise to the reader to avoid spoilers).

This looks like fun for hours and hours, special 100 digit numbers.

The last time I saw a puzzle like this it led me to https://www.kakurokokoro.com

If anything, what a testament of the massive failure Z3 is.

  max_gcd = 0
  for a1 in range(9):
    for b1 in range(9):
      ...
        *check_valid(a1, b1, ...)*
        n1 = to_number(a1, b1, ...)
        ...

        new_gcd = gcd(n1, n2, ...)
        max_gcd = max(max_gcd, new_gcd)

  for gcd in range(111111111):
  for r1 in range(?):
    for r2 in range(?):
      ...
        a1, a2,... = split(r1*gcd)
        ...
        *check_valid(a1, b1, ...)*
        max_gcd = gcd