Why there should be wards in the City of Sydney Council

June 21, 2009

Last year I argued in favour of having wards in the City of Sydney during a referendum to introduce them held in conjunction with the September 2008 City elections.* (I have to declare that I was a Labor candidate for Council – number six on the ticket of ten candidates, and that our team supported a Yes vote in the referendum.) I argued that not having wards meant that Councillors were responsible to voters through their teams, and that having “team” votes meant that it was much more difficult for voters to vote for or against particular Councillors.

I now support having wards more strongly than I did at the election. I used to be against having wards but my view changed since being involved in community groups.

While I used to think that having wards promoted parochialism, this relatively minor point is much weaker than the negatives associated with not having wards. The fact that all nine Councillors represent the entire City of Sydney means that that no particular Councillor (or group of Councillors) is directly responsible to the voters in any particular area of the City. There is no obligation for the issues in any area to be taken up by any Councillor. Read the rest of this entry »


McKinsey Quarterly article – ‘Power curves’: What natural and economic disasters have in common

June 14, 2009

McKinsey Quarterly has a nice article discussing the oft-quoted power law distribution of events in the natural and financial worlds, together with readers’ comments:

Executives, strategists, and economic forecasters, somewhat sheepish after missing the “big one”—last year’s global credit crisis—turned to the lexicon of natural disasters, describing the shock as a tsunami hitting markets and as an earthquake shaking the world economy’s foundations. Shopworn as these metaphors may be, they aptly capture the extreme and unexpected nature of the circumstances. In fact, the parallels between the dynamics and failures of man-made systems, such as the economy or the electricity grid, and similarly complex natural ones are bringing new ideas to economic forecasting, strategic planning, and risk management. This trend may have profound implications for policy makers, economists, and corporate strategists alike.


NSW electors selecting political party candidates

June 14, 2009

Via pollbludger comes news of the NSW Nationals using a US-style primary to select its candidate for one NSW state electorate for the 2011 state election. Quoting the story:

Under the trial, polling booths will operate to allow everyone enrolled in the chosen electorate to vote on who they would like to see as the Nationals candidate.

Currently, only card-carrying members of the party have a say in that process.

Individual electorate councils within the Nationals will now be asked if they would like to have their candidate chosen via community pre-selection.

Mr Stoner previously has said that if the trial works, it would be expanded to other seats for the 2015 election.

It would be great to see the Liberals and ALP adopt a similar process for selecting a candidate for at least one seat in the next state and federal elections, perhaps in independent-held seats. This seems unlikely though given the conservatism of both party organisations.


Maths problem – which \beta satisfy \beta(i+j, k)\beta(j+k, i)\beta(k+i, j) = 1?

June 7, 2009

A problem has recently arisen when attempting to define colour traces on square traces using minimal assumptions. The problem is as follows.

Let i, j, k = 0, 1, 2, …, m-1, and let + be additive addition modulo m, i.e. we identify m and 0.

Let \beta(i, k) -> C be a complex-valued function of i and k, where \beta satisfies:

  • \beta(i, k) \beta(k, i) = 1, for all i, k, and
  • \beta(i+j, k) \beta(j+k, i) \beta(k+i, j) = 1.

The problem is: what are the allowed values of \beta(i, k) ?

Note we have: \beta(n, 0) = \beta(0, n) = 1 for all n which can be seen from fixing i=n-1, j=1, k=0, which gives \beta(n, 0) \beta(1, n-1) \beta(n-1, 1) = 1. We also have \beta(1, 1)^{2}=1.