Difficulty: Easy · Puzzle 1 of 6 · Topic: Classic Brain Teasers
Nine identical-looking balls contain exactly one heavier ball. Using a balance scale and at most two weighings, identify the heavy ball.
First weighing should split candidates into three equal groups.
1) Split into groups A, B, C with 3 balls each. Weigh A vs B.
2) If A or B is heavier, heavy ball is inside that heavier trio. If equal, heavy ball is in C.
3) Take the identified trio and weigh one ball vs one ball.
4) If one side is heavier, that ball is the answer; if equal, the unweighed ball is heavier.
5) Why two weighings are enough: each weighing has 3 outcomes (left heavy, right heavy, equal), so 3^2 = 9 distinguishable cases.Answer: Use a 3-3-3 split, then a 1-vs-1 inside the identified trio; two weighings always isolate the heavy ball.
Checked by running a model of the rule in a Python model. A computational check: an exhaustive minimax over every pan split shows 9 balls need 2 weighings (10 would need 3), the 3-3-3 then 1-v-1 plan finds the heavy ball for all 9 positions, and a state search proves 12 balls (odd one heavier or lighter) need exactly 3 weighings, not 2.
Treating it like binary search (4 and 5 balls per side) and needing 3 weighings; a balance gives ternary information, not binary.
If one ball could be either heavier or lighter among 12 balls, how many weighings are needed? Model answer: 3 weighings. 3^3 = 27 outcomes cover the 24 cases (12 balls x heavy/light). Weigh 4 v 4 first; if balanced the odd ball is among the other 4 and you have 2 weighings left (using known-good balls), if unbalanced you track the 8 'maybe heavy/maybe light' balls with 2 more weighings.
A balance gives three outcomes, left heavy, right heavy or equal, so two weighings can separate 3 squared, which is nine, cases. I split the balls into three groups of three and weigh two groups against each other. The heavier side holds the heavy ball, and if they balance it is in the third group. Then I weigh two balls from that group of three; the heavier one is the answer, or if they balance it is the third ball. The trap is splitting four against four like binary search, which needs three weighings.