Standard sudoku rules applies. Fill the grid with numbers from 1 to 9 so each digit occurs exactly once in every row, column and 3x3 box.
Number outside the grid is the sum of the weight of the first X digits from the side, where X is the value of first digit from the side. Digits of the same kind have a same weight. Each kind of digit has a weight equals to its value, except one kind of digit is overweight, who has a weight that is higher than its value.
I found a lot of old puzzles that bury deep inside my memory. There might be a back story, but i didn’t record it, so, yeah. It’s a simple idea, but a little hard to write the rules. Old miles was going to interpret it as one digit got possessed by another digit, which is fine, but i may do a possess digit at some point in the future. Overweight can be a lot flexible. One digit can be a joker, meaning, it could be infinite. If possess, it could be same digit possess different digit,which needs some thinking for another day.
As for this puzzle, it’s too, normal. It could be set when my bar for my puzzles wasn’t that high. But again, in this case, outside clues are harder to design. Sure, i could leave 19 clue blank. But i guess at that time, i expect this just be a simple x-sum variant, so i didn’t try that hard, which is also a possibility.
Solution code: Row 5 and column 5(18 digits)