Earlier today I set a logic problem and a word problem from Puzzlebomb, a monthly problem sheet.

I’ll start with the word problem.

*Find two words (or phrases) that each use every cell in each honeycomb below. Both words (or phrases) start and end on a gold cell, and follow a path through adjacent cells that passes through every cell exactly once. The cell with a bee hides two different letters (one for each of the solutions.)*

The answers are: on the left left, RATIONALES, and LACERATION; on the right CATTLE GRID and CLATTERING.

The logic problem was a meta-puzzle about bar graphs:

*In the diagram below are three empty bar graphs that all refer to each other. Your task is to read the axes and the titles, and then fill in the bars. (All of which fit in the space provided.) There are three positions for bars in each graph, although some of the bars may have zero height. The only information you are given is that the total height of all the bars is 23.*

The solution is presented here, with the workings below:

In the build up to the question I gave you the hint that the total heights of the bars in Graph A is 9 and in Graph C is 3. (Because A is counting all nine bars, and C is counting only 3 bars.) Thus the total height of the bars n B must be 11.

For Graph C to have a ‘uniquely tallest bar’ its tallest bar must either be 2 or 3. (If the tallest bar was 1, then all three bars would be 1, so it would not be the ‘uniquely’ tallest.)

Bar B3 by definition has the same height as C’s tallest bar, so it must be either 2 or 3. Which means that B1 + B2 = 8 or 9. Since the highest possible value for a bar is 5 (since otherwise it would not fit in the space provided), B1 and B2 must be 5 and 4, or 5 and 3. (Not 4 and 4, since then Graph B wouldn’t have a uniquely tallest bar.) Since B2 must be Graph B’s highest bar, B2 = 5, and B1 = 3 or 4.

We know that B2 is the highest bar in Graph B, so C2 is at least 1. Let’s assume that C2 is 3. If so, then B3 = 3, and thus A1 is at least 2 (from the C1 and C3), and A3 is at least 4 (from B1, B2, B3 and C2). We deduce that A2 cannot be the highest bar in Graph A, and this C2 cannot be 3.

So, using the logic from three paragraphs ago, the three heights in Graph C must be some combination of 0, 1 and 2. Thus B3 = 2, and B1 = 4.

So far, we know that the heights in B are 2, 5, and 4, and in C are 0, 1 and 2. Thus A1 is at least 2, A2 is at least 2 and A3 is at least 2.

We deduce therefore that there are no more bars o height 0 or 1, so A1 = 2, and thus A2 must be at least 3, which means A3 must be at least 3, which means A4 = 4. (This was my favourite part of the deduction.)

The tallest bar in A is on the right. The tallest bar in B is in the middle. We deduce that C3 = 0 , and the only combination which works now is C1 = 1, C2 = 2.

Lovely!

*Thanks again to Katie Steckles and Paul Taylor for these puzzles. If you want to subscribe to Puzzlebomb, please check out their Patreon page.*

*I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.*