3 ms·
I play village cricket, and the scoreboard has cards with numbers on, then hooks to hang them up depending on the score. The problem is: Given the full range o
by mdiesel 6y ago
I play village cricket, and the scoreboard has cards with numbers on, then hooks to hang them up depending on the score.
The problem is: Given the full range of possible (or at least plausible) scores, how many of the cards do we need for a full set?
So let's simplify it to just a run tally. You could be 111, so you'd need at least 3 of the 1 cards etc. Allow for scoring up to 999 (unlikely) and that's 29 cards to keep somewhere (only 2 zeros needed)
In base 3, you need 7 digits, but only 3 cards per, so we are doing better with 19 cards needed (21=3*7, but don't need all zeros, and that gets you to 1093 so for 999 you could save another)
In roman numerals, You'd need an M, a D, 3 Cs, 3 Ls, 3 Xs, 1 V and 3 Is. Total is 15 cards.
Can we do better? Good question.
- Someone 6y agoOf course we can do better. Every card that is an identical copy of another in the set loses you some flexibility in picking different sequences. 6 different cards give you - 1 zero-card sequence - 6 different 1-card sequences - 30 different 2-card sequences - 120 different 3-card sequences - 360 different 4-card sequences - 720 different 5-card sequences - 720 different 6-card sequences That’s enough to almost get you up to 2000. And I don’t think the resulting encoding is objectively weirder than Roman digits.
- mdiesel 6y agoThe answer is: It reduces to a question of permutations. 7! > 999, so we could do it with different arrangements of just 7 cards. Can we do better? Good question...
- mdiesel 6y agoNow if we arranged them on a 2d grid, could use 4 cards, then the different shapes (even discounting the similar looking shapes) would get you to >1000. You'd have to allow more than just the standard tetrominoes.