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$\begin{array}{r|lll} &\text{Name}&\text{Surname}&\text{Dessert}\\ 1&\text{Agatha}&\text{Greed}&\text{cream puffs}\\ 2&\text{Bugsy}&\text{Forager}&\text{trifle}\\ 3&\text{Delilah}&\text{Eatalot}&\text{cheesecake}\\ 4&\text{Chuck}&\text{Hunk}&\text{ice cream} \end{array}$

Four friends held a competition to see who could eat the greatest amount of dessert in weight. So they went to the Pigout Diner, ordered their preferred dessert in abundance, and the battle commenced. The results are recorded above; however, the friends ended up insensible after having crammed themselves with the richness. So, although each item is in the correct column above, the friends only managed to write down ONE item in each column in the correct position. The following facts are true about the correct order.

  1. Chuck is one place above ice cream.
  2. Trifle is not above Delilah.
  3. Greed is two places below Delilah.
  4. Trifle is one place above Forager.

This is an insanely scarce amount of information, and I believe that this requires logic at a higher level than I understand, but I don't know what that logic is. Either that, or I'm missing something very important.

Here's what I have so far (the logic grid):

$$\begin{align} &\begin{array}{r|cccc|cccc|cccc|} &1&2&3&4&A&B&C&D&E&F&G&H\\ \hline Cp&\quad&\quad&\quad&\quad&\quad&\quad&\quad&\quad&\quad&\quad&\quad&\quad\\ Tf&&&&\times&&&&&&\times\\ Cc\\ Ic&\times&&&&&&\times\\ \hline \end{array}\\[-4pt] &\begin{array}{r|cccc|cccc|} \phantom\_E&\quad&\quad&\quad&\quad&\quad&\quad&\quad&\quad\,\\ F&\times&&&&&&&\times\\ G&\times&\times&&&&&&\times\\ H\\ \hline \end{array}\\[-4pt] &\begin{array}{r|cccc|} \phantom\_A&\quad&\quad&\quad&\quad\,\\ B\\ C&&&&\times\\ D&&&\times&\times\\ \hline \end{array} \end{align}$$

The following list of position comparisons assumes position 1 to be 4, position 2 to be 3, etc. for simplicity's sake, so that someone in a lower position is truly "less than".

$4 \geq D \geq 3\\ T \leq D\\ C = Ic + 1\\ G = D - 2\\ T = F + 1\\ 2 \geq G \geq 1\\ 3 \geq F \geq 1\\ 4 \geq T \geq 2\\ F + 1 \leq D\\ G + 2 = D\\ F \leq G + 1$

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There is important information in the text outside the clues: the friends only managed to write down ONE item in each column in the correct position. A possible next step:

Neither Delilah nor Chuck can be in their correct positions, so either Agatha is 1 or Bugsy is 2. From hint 3, Delilah is in either position 1 or position 2. Therefore, positions 1 and 2 are occupied by two of those three, leaving Chuck's only option to be position 3.

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  • 1
    $\begingroup$ (I'll refrain from completely solving the puzzle since I got the impression OP wanted to get unstuck, not the full solution.) $\endgroup$ – ffao Jun 4 '17 at 17:15

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