Seven pumpkins are placed in a circle. The seven pairs formed by adjacent pairs of pumpkins are weighed, much like in prior
pumpkin puzzles. The weights recorded, sorted in ascending order, are 126, 149, 152, 160, 162, 182, and 191 pounds.
The same seven pumpkins (still in the same order) are weighted in pairs again, this time taking pairs separated by one pumpkin. (If A, B, and C are consecutive pumpkins in the circle then A+C is weighed.) The weights recorded, sorted in ascending order, are 125, 138, 163, 169, 170, 173, and 184 pounds.
From these two sets of weights determine the weights of the seven pumpkins and the order they were placed in the circle.
Just so I can do the numbers in my head, let's start by subtracting 62 from every pumpkin. We will add them back at the end.
That makes the side by side weights 2 25 28 36 38 58 67
and the skipping one weights 1 14 39 45 46 49 60
Next consider 4 adjacent weights, abcd
(a+b) and (c+d) are two side by side weights, and
(a+c) and (b+d) are two of the skipping one weights.
So, there will be at least 7 pairs of weights in the first group that sum to 7 pairs of weights in the 2nd.
In fact, there are 9 pairs, which are two more than we need.
From the first group,
40 = 2+38
53 = 25+28
60 = 2+58
61 = 25+36
63 = 25+38
74 = 36+38
94 = 36+58
95 = 28+67
105 = 38+67
From the second group,
40 = 1+39
53 = 14+39
60 = 14+46
61 = 1+60
63 = 14+49
74 = 14+60
94 = 45+49
95 = 46+49
105 = 45+60
Focusing on the first group, we need to see each paired weight exactly twice. But we see 38 four times and 25 and 36 three times. So we can eliminate the equalities involving just 25 and 36 and 38, leaving
40 = 2+38
53 = 25+28
60 = 2+58
61 = 25+36
94 = 36+58
95 = 28+67
105 = 38+67
And now these can be linked in a chain
2 38 67 28 25 36 58 2 38
So we know that the pair that weighs 2 is next to a pair that weighs 38 is next to a pair that 67, going around the ring twice.
But the total weight of all 7 weights are = (2+25+28+36+38+58+67)/2 = 127. (This means, by the way that the weights are all whole numbers. Until now, I had considered that they might all be half-pounds).
2 + 38 + 67 = 107 are the weights of three adjacent pairs, so the weight that is not in any of these pairs is 127 - 107 = 20
And two weights away from it (let's say clockwise) are 127 - (38+67+28) = -6 (don't worry, I'll fix this up soon).
Continuing around, we get a full list of 20, -6, 7, 38, 8, 31, 29 (looking at every other weight). But because this skips one, the true order is 20,8,-6,31,7,29,38.
Adding 62 back to each one gives 82,70,56,93,69,91,100. This solution is unique, except for rotations and reflections
And because this matches Charlie's computer solution, I will not bother to check it.
Edited on February 7, 2016, 9:15 am
Analytical solution (spoiler)
