Taking balls away: Oz’ Version

This post is based on a comment by Oz to our question about balls with two colors:

“There is an interesting (and more difficult) variation I once heard but can’t recall where:

You have a box with n red balls and n blue balls. You take out each time a ball at random as before. But, if the ball was red, you put it back in the box and take out a blue ball. If the ball was blue, you put it back in the box and take out a red ball.

You keep as before until left only with balls of the same color. How many balls will be left (as a function of n)?

1) Roughly  εn for some ε>0.

2) Roughly $\sqrt n$?

3) Roughly log n?

4) Roughly a constant?

5) Some other behavior

You have a box with n red balls and n blue balls. You take out balls one by one at random until left only with balls of the same color. How many balls will be left (as a function of n)?

1) Roughly  εn for some ε>0.

2) Roughly $\sqrt n$?

3) Roughly log n?

4) Roughly a constant?

Here is the collective intuition regarding this problem

Itai Ashlagi, Yashodhan Kanoria, and Jacob Leshno: What a Difference an Additional Man makes?

We are considering the stable marriage theorem. Suppose that there are n men and n women. If the preferences are random and men are proposing, what is the likely average women’s rank of their husbands, and what is the likely average men’s rank of their wives?

Boris Pittel proved that on average a man will be matched to the woman in place log n on his list. (Place one is his most preferred woman.) A woman will be matched on average to a man ranked n/log n on her list.

We asked in the post “Test your intuition (19)”  what is the situation if there is one additional man, and men are still proposing. This question is based on a conversation with Jacob Leshno who told me about a remarkable paper Unbalanced random matching markets by Itai Ashlagi, Yash Kanoria, and Jacob Leshno. Continue reading

Test Your Intuition (19): The Advantage of the Proposers in the Stable Matching Algorithm

Stable mariage

The Gale-Shapley stable matching theorem and the algorithm.

GALE-SHAPLEY THEOREM Consider a society of n men and n women and suppose that every man [and every woman] have a preference (linear) relation on the women [men] he [she] knows. Then there is a stable marriage, namely a perfect matching between the men and the women so that there are no men and women which are not matched so that both of them prefer the other on their spouces.

Proof: Consider the following algorithm, on day 1 every man goes to the first woman on his list and every woman select the best man among those who come to her and reject the others. On the second day every rejected men go to the second woman on his list and every woman select one man from all man that comes to her (including the man she selected in the previous day if there was such a man) and rejects all others, and so on. This process will terminate after finitely many days and with a stable marriage! To see that the process terminate note that each day at least one man will come to a new women, or go back home after beeing rejected from every women (n+1 possibilities) and none of these possibilitie will ever repeat itself so after at most $n^2+n$ days things will stabilize. When it terminates we have a stable marriage because suppose women W and men M are not married at the end. If M is married to a women he prefers less then W or to no women at all it means that M visited W and she rejected him so she had a better men than M.  Sababa!
It turns out that the above algorithm where the men are proposing and being rejected is optimal for the men! If a man M is matched to a woman W then there is not a single stable marriage where M can be matched to a woman higher on his list. Similarly this algorithm is worst for the women. But by how much?

Random independent preferences

Question 1:  There are n men and n women. If the preferences are random and men are proposing, what is the likely average women’s rank of their husbands, and what is the likely average men’s rank of their wives.

You can test your intuition, or look at the answer and for a follow up question after the fold.

Test Your Intuition (17): What does it Take to Win Tic-Tac-Toe

(A few more quantum posts are coming. But let’s have a quick break for games.)

Tic Tac Toe is played since anciant times. For the common version, where the two players X and O take turns in marking the empty squares in a 3 by 3 board – each player has a strategy that can guarantee a draw.

Now suppose that the X player has a budget of Y dollars, the O playare has a budget of 100 dollars and before each round the two players bid who makes the next move. The highest bidder makes the move and his budget is reduced by his bid.

For example, if, to start with, the X player have 120 dollars, and if for the first move X bids 40 dollars and O bids 30 dollars, then X will make the first move and will be left with 80 dollars while O will be left with his 100 dollars.

What would you expect the minimal value of Y is so the X player has a winning strategy? Of course if Y>300, X can make 3 uninterrupted moves and win, but perhaps much less is enough?

(Thanks to Reshef Meir and Moshe Tennenholtz)

Update: Another variant of this game is when the player winning the turn pays his bid to the other player. (This version is called “Richman game,” while the variant above is called “Poorman game”. In this case if Y> 800 the X player can play three moves and win. And again the question is what is the infimum value of Y for which the X player can force a win…

Discrepancy, The Beck-Fiala Theorem, and the Answer to “Test Your Intuition (14)”

The Question

Suppose that you want to send a message so that it will reach all vertices of the discrete $n$-dimensional cube. At each time unit (or round) you can send the message to one vertex. When a vertex gets the message at round $i$ all its neighbors will receive it at round $i+1$.

We will denote the vertices of the discrete $n$-cube by $\pm 1$ vectors of length $n$. The Hamming distance $d(x,y)$ between two such vertices is the number of coordinates they differ and two vertices are neighbors if their Hamming distance equals 1.

The question is

how many rounds it will take to transmit the message to all vertices?

Here is a very simple way to go about it: At round 1 you send the message to vertex (-1,-1,-1,…,-1). At round 2 you send the message to vertex (1,1,…,1). Then you sit and wait. With this strategy it will take $\lceil n/2\rceil+1$ rounds.

Test your intuition: Is there a better strategy?

The answer to the question posed in test you intuition (14) is no. There is no better strategy. The question was originated in Intel. It was posed by Joe Brandenburg and David Scott, and popularized by Vance Faber. The answer was proved by Noga Alon in the paper Transmitting in the n-dimensional cube, Discrete Applied Math. 37/38 (1992), 9-11. The result is closely related to combinatorial discrepancy theory and the proof is related to the argument in the Beck-Fiala theorem and a related lemma by Beck and Spencer. This is a good opportunity to present the Beck-Fiala Theorem.

The general form of a discrepancy problem

Let $H$ be a hypergraph, i.e., a collection of subsets of a ground set $A$. $A$ is also called the set of vertices of the hypergraphs and the subsets in $H$ are also called edges.

The discrepancy of $H$, denoted by $disc(H)$  is the minimum over all functions $f:A \to \{-1,1\}$ of the maximum over all  $S \in H$ of

$\sum \{f(s):s\in S\}$.

The Erdős Discrepancy Problem (EDP) asks about the following particular hypergraph: The vertices are {1,2,…,n} and the edges are sets of vertices which form arithmetic progressions of the form (r,2r,3r,…kr}. EDP was extensively discussed and studied in polymath5. Here is the link of the first post. Here are links to all polymath5 posts.  Here is the link to polymath5 wiki.

The Beck-Fiala theorem

Theorem (Beck-Fiala): If every element in $A$ is included in at most $t$ edges  of  $H$ then $disc(H)<2t$.

Before proving this theorem we mention that it is a famous open conjecture to show that actually $disc(H)=O(\sqrt t)$

Test Your Intuition (14): A Discrete Transmission Problem

Recall that the $n$-dimensional discrete cube is the set of all binary vectors ($0-1$ vectors) of length n. We say that two binary vectors are adjacent if they differ in precisely one coordinate. (In other words, their Hamming distance is 1.) This gives the $n$-dimensional discrete cube a structure of a graph, so from now on , we will refer to binary vectors as vertices.

Suppose that you want to send a message so that it will reach all vertices of the discrete $n$-dimensional cube. At each time unit (or round) you can send the message to one vertex. When a vertex gets the message at round $i$ all its neighbors will receive it at round $i+1$.

The question is

how many rounds it will take to transmit the message to all vertices?

Here is a very simple way to go about it: At round 1 you send the message to vertex (0,0,0,…,0). At round 2 you send the message to vertex (1,1,…,1). Then you wait and sit. With this strategy it will take $\lceil n/2\rceil+1$ rounds.

Test your intuition: Is there a better strategy?