Source: 1. Original. 2. Jason Telgren, who says it's public domain. 3. Sam Loyd.
We have the following 4x4 board 1 2 3 4 D . R . R C * . * . B . * . * A B . B . If every player is properly playing, nobody can win. If D2-C1,black wins with A1-B2,then D4-C3 and B2xD4 If D4-C3,black wins again with A3-B4 followed by D2-C1 and B4xD2 [Philippe missed another possibility here in C3-B2. - KD] If D2-C3, there are 2 possibilities: 1) If A3-B4, red wins with C3-B2 then A1xB3 and D4xB2 2) If A3-B2, C3-B4, B2-C1,B4-A3=crown and C1-D2=crown. The game ends on a draw.KD: I have now convinced myself that Black can always play to win.
Glenn Rhoads pointed out that Martin Gardner analyzed this puzzle in one of his Scientific American columns, showing the game is a draw (as shown above). But the 5x5 grid for checkers is not a draw.
Answer: 4 weighings. There are 2 ways of doing this puzzle; My solution#1 is probably what everyone comes up with: isolate the 7weight after 2 weighing; however (all being equal) this solution means 47 2/3 pounds on the poor scale. My solution#2 (I'm sure it's been noticed before) is a bit more complicated but means only 33 1/3 pounds on the grateful scale! I therefore think this puzzle's question should be extended to include "minimum total weights to be placed on scale, assuming each weight has equal chance of being picked up. Average SOLUTION#1 Weight: ========== ======= Weighing#1 : 2 against 2; 7weight automatically on heavy side 17 Weighing#2 : compare the 2 from heavy side; isolates 7weight 10 1/3 (Let A = 7weight, B C D = the other 3 weights) Weighing#3 : weigh A against 2 others (call them B and C) possibilities: A B C D(default) A is heavy 7 2 3 (5) 5 even 7 2 5 (7) 3 heavy 7 3 2 (5) 5 light 7 3 5 (8) 2 even 7 5 2 (7) 3 light 7 5 3 (8) 2 13 2/3 Weighing#4 : compare B to C and you got it! 6 2/3 ====== Total weight handled: 47 2/3 SOLUTION#2 ========== STEP BY STEP SOLUTION: 4 weighings, for total average weight of 33 1/3 pounds; like this: STEP 1: compare any 2 balls; call A the lighter and B the heavier. STEP 2: compare the other 2 balls; call C the lighter and D the heavier. Total weight so far is 17 (each ball used once) STEP 3: compare A to C. This isolates the 2weight; make A=2, C the other (3 or 5). 3 equally likely possiblities exist: A B C D (1) 2 3 5 7 (2) 2 5 3 7 (3) 2 7 3 5 Average weight for C is (5 + 3 + 3) / 3 = 3 2/3. Average weight for this Step = A + C = 2 + 3 2/3 = 5 2/3. So average weight so far = 17 + 5 2/3 = 22 2/3. STEP 4: compare A+C to B: if (1), B is lighter, if (2), both sides are equal, if (3), B is heavier; so we are done. Average weight for this Step is A + B + C = 2 + 3 2/3 + (3 + 5 + 7) / 3 = 10 2/3. So total average weight is 22 2/3 + 10 2/3 = 33 1/3.
It is easy to generalize with any number n included between 1 and N by determining k,the fewest number of weights, such that sum(3^k)>=N.