Surface Tension in Taxicab Geometry

Taxicab geometry is a wonderful world, where common truths of normal (Euclidean) geometry don’t hold. For example, The length of the diagonal of a square is twice the length of its side (not square root of 2). Another example: the figure whose every point is equidistant from its center is not a circle; it is a diamond shape.
Now it would be interesting to see how soap bubbles will behave in Taxicab geometry. It is easier to examine the continuous geometry version, where the size of zig-zags is infinitely small, because the tools of calculus can be used.

  1. What is the equilibrium shape of a single bubble? That is, which shape with a given area has the minimum parameter?
  2. What is the equilibrium shape of two equal-area bubbles? When the two bubbles share a common boundary, the total boundary length shrinks.
  3. What if the two bubbles have different areas?

The corresponding questions for discrete Taxicab geometry (all lines lie on a square lattice) are somewhat harder, but once some intuition has been developed for the continuous case, one can take the continuous solutions as a starting point and then look for variations.

Thousand Bottles, Revisited

Thousand bottles of wine puzzle can be analyzed in terms of information content. In the original puzzle, there is one bottle of poison amongst n = 2**m bottles of wine, and so the solution space is of size n, since  the answer can be any number from 0 to n-1.

Each prisoner yields only a single bit of information – by staying alive or dying. Hence at least m prisoners (or m bits) would be required to find the solution within the solution space of size 2**m.

Now, in the variation #1 posted earlier (https://dmranade.wordpress.com/2010/09/28/thousand-bottles-of-wine/), there are two unknowns: the poison and the antidote. There are n(n-1) possible solutions, so the information required to describe the answer is lg(n(n-1)) … almost 2m bits. Hence at least 2m prisoners would be required. Indeed, I know a solution using 2m prisoners that requires detecting the antidote in the first step and then removing that bottle, thus reducing the problem to the original 1000 bottle problem.

Now consider that there is an additional piece of information in variation no #1; half a vial of the pure poison has been recovered from the assassin. What if this vial was not available? That leads to another variation:

variation #2

All we know is that out of thousand bottles one bottle is poisoned and another one spiked with the antidote. We do have two days as before, i.e., result of one set of tests can be used in a second set of tests.

How many prisoners should suffice to solve this variation? The same argument as used for variation #1 should apply, hence 2m prisoners should be sufficient.

Here, however,  a solution with 2m prisoners is not known to me, although I did find an elegant solution using 3m-1 prisoners.

Thousand bottles of Wine

A twist on the well known puzzle:

A king has stocked 1000 bottles of wine in the wine cellar for a big party the day after tomorrow.
The chief of guards brings bad news at noon: an assassin belonging to a fanatic religious sect has been caught in the wine cellar with a vial half-full of poison. Interrogation reveals that she has put an undetectable, very deadly poison in a single wine bottle. The poison causes certain death in 20 to 22 hours, without prior symptoms, and that too with a just a drop of the poisoned wine.

To balance her karma, the assassin has also put an antidote to the poison in another wine bottle. The antidote is even more powerful: a single drop from the antidote-containing wine bottle will neutralize the effects of drinking even the whole poisoned wine bottle.

Mixing all the wines is not an option because they are of different types.

How many prisoners on death row does the chief of guards need to use determine which wine bottle has the poison?

Clock Puzzle

Which is the better clock: one that loses/gains 5 to 10 minutes a day, or one that is absolutely accurate 26 times a day?

Hint: there is an aphorism

Addendum: The above question is for a 24-hour clock. For a 12-hour clock, the puzzle becomes:

Which is the better (12-hour analog) clock: one that loses/gains 5 to 10 minutes a day, or one that is absolutely accurate 21 times a day?

Hint: the question was asked in September.

Two Girls Puzzle – solution

1. There is a couple with exactly two children. What are the odds that both are girls?

2. The same as #1, but now you are told that one child is a girl. What are the odds that both are girls?

3. The same as #2, but now you are told that one child is a girl named Idhitri. What are the odds that both are girls?

For ease of calculation, assume the boy:girl ratio is 50:50 in the population.

1. Consider Figure 1.  There are four kinds of couples with two children where each child may be a boy (b) or a girl (g) : bb, bg, gb, gg. The area of each section is proportional to the number of couples with that combination. Notice that there are four equal possibilities, but only one, gg, of interest. Hence the odds are 1 in 4.

2. Consider the same diagram, but now that it is known that one child is a girl, the combination with no girls (bb) is ruled out. Hence the odds are now 1 in 3.

Figure 1

3. Now comes the interesting case: one child is a girl with a rare name. How do we incorporate this new information? Now we need a three-way classification- boy (b), girl-named-idhitri (G), girl-not-named-idhitri (g’). Since there are two children, we have 3×3 = 9 combinations, as in Figure 2 below. Notice that the areas of these 9 sets are not equal since Idhitri is a rare name. However, four of these are ruled out because at least one G is required. The case of both girls named Idhitri (GG) is doubly rare, hence can be ignored. Thus we are left with only 4 combinations, and these are approximately equal in area. Of these, two are the ones we are interested in. Hence the odds are 1 in 2 (approximately).

Notice that bG + bg’ = bg of Figure 1.

The inaccuracy of the calculation is equal to the probability of GG.There is a subtle point there, but I won’t try to explain it further.

Figure 2