Well, it is time for a cool and easy riddle. 100 men are each assigned a number between 1-100 with repetitions (e.g. all of them may be assigned the number 17). Each of the men sees all the numbers assigned to all the other 99 men, but none of them sees the number assigned to himself.

Each of them needs to guess his own number (of course no information is exchanged between them - no one hears what others have guessed etc.). What strategy can they employ in order to make sure at least one of them makes a correct guess?

Some followup thoughts:

• Can they make sure more than one person succeeds?
• How many different solutions to the riddle are there (i.e. how many strategies can the men employ?).

A short introduction to Graph Theory is needed for this one. If you already are familiar with Graph Theoretic constructs feel free to skip it.

A graph G, is a pair (V,E) where V is a set of vertices and E is a set of edges. Each edge is an unordered pair of the form (u, v) where u and v are vertices (i.e. they belong to V). The degree of a vertex t (denoted deg(t)) is the number of edges containing it:

deg(t) = #{ e | e = (s, t) ^ s belongs to V }

(more…)

This one is a riddle of my own invention. I am really proud of it as it sounds impossible at first, but it can be resolved in a very satisfying fashion. I must admit that had I heard about it from someone else I would not have gone through the trouble of trying to solve it, as it sounds like a ”cheap-trick” riddle. I can only ask you to trust me that it is indeed a purely mathematical riddle. Although it did provoke several harsh disputes with some very smart people (Matan, Eyal, how are you guys?), I was able to finally convince them of the mathematical validity of my solution.

(more…)

I visited my parents yesterday, for the Jewish holiday of Shavuot. I found there a picture of myself and some friends from the time we sailed on the Tuichi river in Bolivia on a raft we built ourselves with machetes. Here is the picture and a legend:

(more…)

Make sure you have read the first part of this post here.

Now we are ready to define the rules for our second, more complex game - “Construction with a Straight-Edge and a Compass“. These are very similar to the rules of “Constructions with a Straight-Edge Only” (defined in the previous post) except that now, in addition to creating infinite straight lines, you can also create circles, as follows: given 2 points, a and b, you can create the circle with center at a that passes through b. A point is added to the set of points if it is the intersection of 2 lines (as before), of two circles or of a circle and a line.

Now, the addition of the new tool (the compass) enables us to construct many more points than before. If the initial set of points contains only one point then, as before, no new points can be constructed. But if the initial set of points contains two points then a the set of constructable points is infinite (we will shortly prove it). The subject for this second part of the topic will be analysing the set of constructable points in our new game.

(more…)

This post is about a conversation I once held with a good friend of mine, Nadav Sherman. It is not a very serious topic, so take it lightly. We were wondering how to calculate the total amount of fun a person has during his life time. We reasoned as follows:

The absolute state score (a.s.s.) of a person is a number ranking the person’s current state and possessions (it is a function of time). For example, if you have 500,000$ and a girl-friend, then your a.s.s. is higher than that of someone identical to you without a girlfriend, but lower than that of someone identical to you with an additional 1,000,000$. Note that we do not specify exactly how to calculate the a.s.s., but we believe that it can be defined such that claim 1 below will be true. The actual value of the a.s.s. is not directly correlated to the fun a person has - a person with a million dollars and a beautiful girl-friend can be sadder than a poor lonely guy.

(more…)

A few years ago a friend of mine asked me the following Perl riddle. Unfortunately, in order to solve it you must know Perl. As I like Python much better, I translated the riddle to Python. Attached are both versions.

I admit the Perl version is a bit more cryptic and if you know both Perl and Python you should try to solve the Perl version (but use Python for everything else in life ).

Oh, and try to solve the riddle without running it (run it only as a last resort).

perl -wle 'print "True" if (1 x shift) !~ /^1?$|^(11+?)\1+$/' [number]

python -c "import sys, re; print None == re.match('^1?$|^(11+?)\\1+$','1'*int(sys.argv[1]))" [number]

To alleviate all doubt - [number] denotes a numeric command line argument (e.g. 17).

This topic is about defining a weird game and exploring its properties. When I first encountered it I could not help but wonder, why is it interesting?

Well, this game (as you shall soon find out for yourself) has barely any practical implications. But it is an interesting subject none the less. There are 2 main reasons for that:

1. This game was analysed (with limited success) by many important mathematicians, since the times of the Greeks.
2. Using the right tools, the complete answer to the questions relating this matter is achievable and it turns out to be beautiful.

(more…)

I had a lot of thinking before deciding to include this riddle in my site. The argument for not including it, is that it is too complicated. I ended up deciding it deserves its place in my riddles section (which means I believe it is one of the most beautiful riddles ever) because, well, I think it is one of the most beautiful riddles ever. So enjoy! (but be ready for a hard one…):

(more…)

Hello all!

I am very excited to announce that my site is up and I really hope you enjoy it. I will always appreciate your feedback (and especially now, that its all brand-new!). Do not hesitate to contact me at yaniv@leviathanonline.com and please register and post comments!

Yaniv