Notice that in a market s.t. all agents have unit demand valuation functions- if there is an allocation (not necessarily and a matching) that yields SW of value v, there is also a

I think one should assume that there are no outgoing edges in G_I from vertix j if j was not sold under OPT.

This makes the graph well defined and makes it possible to prove the theorem.

Is this an acceptable assumption ?

]]>Maybe it was confusing in this context, but in the previous answer I was talking about unit demand functions

I hope it is clearer now.

]]>צרו קשר דרך:

mendelsohn@mail,tau,ac,il ]]>

OPT is defined as "OPT (i.e. maximum weighted matching)" - In the definition of a "matching" in graphs that I'm familiar with, no node can be matched twice. In the context of markets, it means that each player gets at most one item.

If I understand correctly, we usually assume that an allocation in a unit-demand market is a matching (i.e. each player gets at most one item, which is more restrictive), and we can assume this without loss of generality because the value of each set is determined by only one item.

Reading your answer, it seems like that you assume players can get more than one item in OPT. Is this the definition of a "matching" we should work with? ]]>

Also, in this question we assume all valuations are unit-demand, which means that adding or removing items to a set of price 0 and value smaller than the value of the set doesn't change the set's utility.

An extreme example of this will be items of value and price 0, but in the context of the question, all unsold items which are of smaller value compared to agent i's set pose this problem as well.

Because adding and removing such items doesn't change the utility of agent i's set, I don't understand how their demand set can only have a single, unique set. I can always generate another, different set by adding/removing such an item. ]]>

Consider items of price 0. Because we prove the prices are Walrasian, all unsold items will be such items.

For any set of items in the demand set, you can add / remove items of price 0 without changing its utility. This means if there are unsold items, there are multiple sets in the demand set, contradicting what we need to prove.

Can we assume the uniqueness is "up to elements of price 0"? ]]>

Please contact me at: yotam.nitzan@gmail . com

]]>Does anyone present on the 6.6 and wants to switch with us? ]]>

I managed to prove that the greedy algorithm indeed outputs a correct result for a demand query, however I'm having trouble proving poly(m) time. In each iteration the algorithm needs to compute v(j|S) for items, and we were taught that computing v(j|S) requires computing a demand query several times. Given at worse case the greedy algorithm needs to compute v(j|S) when |S| = m-1, I realized that computing v(j|S) using the greedy algorithm as demand query oracle will result in infinite recursion, since computation of v(j|S) will require computing demand query over items set S + {j} which is the full item set. How should I proceed?

Thanks,

Eldar

or assume worst case as in assume the best poosible SW in the arrival order (which is the worst case when trying to show SW is low) ]]>

Tie breaking can be arbitrary. Assume worst case option (this should help you). ]]>

"For each pair of items j, j', let the weight of the edge (j, j') in G_I be v_i({j}) − v_i({j'}),

where i is the agent that was matched by M to j."

What happens if j was not matched to any agent in M ?

This case isn't defined explictly.

Can one assume that it means j has no outgoing edges ?

Thanks.

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