I'm having considerable trouble trying to figure out by hand the nucleolus of the following games $[N,\nu]$ with non-empty Core (unfortunately, I no longer have access to MatLab; in that case I would not be asking here). Consider the first game $[N,\nu]$ defined by $|N|=3$ and $\nu:2^N\longrightarrow \mathbb{R}$ \begin{gather*} \nu(i)=0\forall i\in N;\\ \nu(1,2)=0.75\\ \nu(1,3)=0.50\\ \nu(2,3)=0.25\\ \nu(N)=1 \end{gather*}

The second game $[N^{\prime},\nu^{\prime}]$ is given by $|N^{\prime}|=4$ and $\nu^{\prime}:2^N\longrightarrow \mathbb{R}$ \begin{gather*} \nu^{\prime}(i)=0\forall i\in N;\\ \nu^{\prime}(1,2)=0\\ \nu^{\prime}(1,3)=1\\ \nu^{\prime}(1,4)=1\\ \nu^{\prime}(2,3)=1\\ \nu^{\prime}(2,4)=1\\ \nu^{\prime}(3,4)=0\\ \nu^{\prime}(1,2,3)=1\\ \nu^{\prime}(1,2,4)=1\\ \nu^{\prime}(1,3,4)=2\\ \nu^{\prime}(2,3,4)=2\\ \nu^{\prime}(N)=3 \end{gather*}

Then, my question is: could anyone please provide the nucleolus of these two games? Thank you all very much in advanced for your time.

  • 1
    $\begingroup$ related to your request, I have updated my answer. Hope this is useful. $\endgroup$ – Holger I. Meinhardt Mar 22 '18 at 17:41
  • $\begingroup$ Thank you very much. It certainly is. $\endgroup$ – Héctor Mar 22 '18 at 19:35

The nucleolus of the first game is given by


and for the second game, it is


You mentioned that you do not have access anymore to Matlab, in that case you might be interested in my port of my Matlab toolbox to Octave (a Matlab clone). However, you should notice that these functions are not as stable as under Matlab. In that case I will open you a link to download the files for a short period of time.

Update: SM-Nucleolus and Modiclus:

Your asking for other solutions, so I provide solutions related to the simplified and modified nucleolus, the latter is also known in the literature as the modiclus.

Solution of the simplified nucleolus:

$ spn_v=(11/24,1/3,5/24) $


$ spn_{v^{\prime}}=(2/3,2/3,5/6,5/6)$

The modiclus is given by

$ mnc_v = (3/8,3/8,1/4)$


$ mnc_{v^{\prime}} = (3,3,3,3)/4$.

By the way, the core of the first game is given as the convex hull of the following vertices:

0.2500    0.5000    0.2500
0.7500         0    0.2500
0.7500    0.2500         0
0.5000    0.5000         0
  • $\begingroup$ Thank you very much for your answer. If I figure out how to use Octave and so on; I'll reach you out. But for the moment, don't worry. For some reason, I found that the Core of the first game was given by $\phi=(\frac{14}{24},\frac{1}{3},\frac{2}{24})$. Does $\phi$ correspond to some other cooperative Game Theory solution concept that I'm not aware of? $\endgroup$ – Héctor Mar 22 '18 at 9:33

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