#mathematical puzzle

[hezigler]

Watch "The Mathematical Problem with Music, and How to Solve It" on YouTube

[leibnizstan]

fruit posting

[unnonexistence]

95% of math brainteasers that go viral are just bad notation and it drives me up the wall

[arconinternet]

The Canterbury Puzzles (Book, Henry Dudeney, 1907)

You can read it here. If the Internet Archive is still down, you can also read it on Project Gutenberg here.

Image above is from here (also archived here).

[quanyails]

Hot concept for an isekai:

The protag is a math undergrad and they're dropped in the world of their logic professor's textbook

Everyone is either wearing unidentified colored hats or in prison

[positivelyprime]

I think I found some interesting fun mathematical problem, and I could not find any literature on it.

Ok so I was thinking about the two guards riddle where one always tells the truth (guard T) and the other always lies (guard F). These guards protect a crossroads where one road is correct and lets you live and the other leads to death (or at least, wasted time). The worst part: you only have one yes-no question to ask.

The (or at least a) correct answer to this riddle is no doubt familiar with several of you: Ask one guard "If I asked which road leads to freedom to the other guard, what would he say?". This works as follows:

- Asking this question to T will lead to them saying the wrong road, as that is what F would say.

- Asking this question to F will lead them to lie about the correct road T would point to, saying the wrong road.

In either case you get the wrong road, so you simply pick the other option.

Now here is where it gets interesting, how can we generalize this problem? One way to do this is to increase the number of guards as follows:

Suppose we have k guards, with n truth-guards T and m false-guards F (n,m > 0 unknown) guarding two roads. Can we find the correct road with one yes-no question?

The main trick employed in the 2-guard problem cannot be used as it relies on there being exactly one guard of each type. However, if you ask two random guards there is now a possibility they are a TT or FF pair, in which case they will point to the right road instead.

I tried finding some kind of nested question ("if I ask the guard on your right: if I ask the guard on your right what road leads to freedom") but couldn't quite put my finger a good question to ask. Another promising method would be to incorporate some kind of boolean operators (such as AND and OR) to make a perfect question, however this turned out to be hard to do for arbitrary k.

I would love the input of anyone reading this who found textwall this interesting!

[ageofshadows666]

First person to get it right gets a shoutout, put your answers in the comments

[art-of-mathematics]

DIN A2 space filled with 4 sheets of DIN A4 paper in 4 different colors.

Each DIN A4 sheet is split into pieces ranging from DIN A5 (largest) to DIN A11 (smallest). Each size from A5 to A10 exists one time per each color. A11 exists two times per each color. (I could have continued with splitting the A11 into two A12 tiles - but these would be too tiny for my clumsy hands to handle.)

I might create some mosaic patterns with this... (using some exchange algorithms... )

[communistmishka]

Smash

[xsgamesblog]

Can you answer this? 🧠

Unlock new levels of intelligence; dive into these puzzles today!

[cecilias-cool-stuff]

New puzzle, that involves some graph theory!

An "n-cycle" is a graph that looks like a polygon with n vertices. An "independent set" of a graph is a subset of the vertices of the graph such that no pair of vertices in the subset are connected by an edge. How many k-element independent sets of an n-cycle exist?

[leibnizstan]

[thecoolerbrother]

study science like it's art and study art like it's science

[arconinternet]

Amusements in Mathematics (Book, Henry Dudeney, 1917)

You can read it here. If the Internet Archive's still down, you can also read it on Project Gutenberg here.

[andmaybegayer]

Simon Tatham added the new monotile to the Loopy game and did a write-up on dynamically placing aperiodic tilings. I have never even tried to beat the Penrose Loopy variant, I stopped at hexes.

[saimaliirajput]

Math Riddles: Only 1% genius can solve these mathematics puzzels

See Answer : Click Me to see Answer