So I had an idea for a number-based door puzzle that could readjust itself every time a wrong guess is made. Here’s the basic premise:
Given two numbers, find two more numbers in such a way that:
- no number differs from any other number in the same way,
- and none of these differences are part of the series themselves,
- and no number is bigger than it has to be, but not zero.
It seems fully deterministic to me, so let me give you an example:
Given are two numbers (1, 4). The difference is 3. 2 is out because its difference from 1 is part of the series, and in itself would be the difference from 4. 3 is out because of its difference from 2 and 4 being the same, and that difference would be 1, which is additionally part of the series. 5 is not valid either, but 6 is. 7 is not valid due to its difference to 6 being part of the series, and 8 is invalid because of the difference between (6, 8) and (4, 6). However, 9 is valid, and concludes the series.
It seems simple enough for a door puzzle, and given the above set of rules, it seems absolutely deterministic. I was considering to give the following (more vague) clue to its solution:
No siblings differ alike, nor match they their differences, and none grow taller than they must.
I need it to sound like a riddle, but still be concise and accurate and its description of the puzzle. If anyone has any suggestions, I would greatly appreciate the help.
I suppose I have four questions for this community:
- Is there a unique solution for any two starting numbers below 10?
- Can you think of a more interesting variant, like perhaps by minimising the sum of the numbers (but not having any predetermined numbers given – (1, 4, 6, 9) being the best solution I found so far)?
- It seems to me there should be some way to determine the two missing numbers in a way that minimizes the sum of all four, which is not the same as choosing the lowest possible numbers one after the other. Is there such a way?
- Is there a better way to phrase the problem in a single sentence to my players, without being too obvious or sounding too scientific?
And some more example solutions for your convenience:
- (3, 4) –> (9, 11)
- (2, 5) –> (6, 13)
- (1, 9) –> (3, 13)
- (1, 3) –> (7, 12)
- (1, 5) –> (7, 13)
- (4, 6) –> (1, 13)
- (8, 1) –> (3, 17)