Archive for chess

Le Monde puzzle [#1141]

Posted in Kids, pictures, R, University life with tags , , , , , , , , on May 4, 2020 by xi'an

The weekly puzzle from Le Monde is in honour of John Conway, who just passed away, ending up his own game of life:

On an 8×8 checker-board, Alice picks n squares as “infected”. She then propagates the disease by having each square with least two infected neighbours to become infected as well. What is the minimal value of n for the entire board to become infected? What if three infected neighbours are required?

A plain brute force R random search for proper starting points led to n=8 (with a un-code-golfed fairly ugly rendering of the neighbourhood relation, I am afraid!) with the following initial position

With three neighbours, an similar simulation failed to return anything below n=35 as for instance:

oops, n=34 when running a little longer:

which makes sense since an upper bound is found by filling one square out of two (32) and adding both empty corners (2). But this upper bound is only considering one step ahead, so is presumably way too large. (And indeed the minimal value is 28, showing that brute force does not always work! Or it may be that forcing the number of live cells to grow at each step is a coding mistake…)

value of a chess game

Posted in pictures, Statistics, University life with tags , , , , , , , , , , , , on April 15, 2020 by xi'an

In our (internal) webinar at CEREMADE today, Miguel Oliu Barton gave a talk on the recent result his student Luc Attia and himself obtained, namely a tractable way of finding the value of a game (when minimax equals maximin), result that got recently published in PNAS:

“Stochastic games were introduced by the Nobel Memorial Prize winner Lloyd Shapley in 1953 to model dynamic interactions in which the environment changes in response to the players’ behavior. The theory of stochastic games and its applications have been studied in several scientific disciplines, including economics, operations research, evolutionary biology, and computer science. In addition, mathematical tools that were used and developed in the study of stochastic games are used by mathematicians and computer scientists in other fields. This paper contributes to the theory of stochastic games by providing a tractable formula for the value of finite competitive stochastic games. This result settles a major open problem which remained unsolved for nearly 40 years.”

While I did not see a direct consequence of this result in regular statistics, I found most interesting the comment made at one point that chess (with forced nullity after repetitions) had a value, by virtue of Zermelo’s theorem. As I had never considered the question (contrary to Shannon!). This value remains unknown.

Death won the last chess game [but it took a while]

Posted in Statistics with tags , , , , , , , , on March 30, 2020 by xi'an

a grim knight [cont’d]

Posted in Books, Kids, pictures, R, Statistics with tags , , , , , , , on October 20, 2016 by xi'an

As discussed in the previous entry, there are two interpretations to this question from The Riddler:

“…how long is the longest path a knight can travel on a standard 8-by-8 board without letting the path intersect itself?”

riddlerechckas to what constitutes a path. As a (terrible) chess player, I would opt for the version on the previous post, the knight moving two steps in one direction and one in the other (or vice-versa), thus occupying three squares on the board. But one can consider instead the graph of the moves of that knight, as in the above picture and as in The Riddler. And since I could not let the problem go I also wrote an R code (as clumsy as the previous one!) to explore at random (with some minimal degree of annealing) the maximum length of a self-avoiding knight canter. riddlerechkThe first maximal length I found this way is 32, although I also came by hand to a spiralling solution with a length of 33.

riddlerechckRunning the R code longer over the weekend however led to a path of length 34, while the exact solution to the riddle is 35, as provided by the Riddler (and earlier in many forms, including Martin Gardner’s and Donald Knuth’s).

[An unexpected side-effect of this riddle was ending up watching half of Bergman’s Seventh Seal in Swedish…]

grim knight [a riddle]

Posted in Kids, pictures, R with tags , , , on October 14, 2016 by xi'an

The Riddler of this week had a riddle that is a variation of the knight tour problem, namely

“…how long is the longest path a knight can travel on a standard 8-by-8 chessboard without letting the path intersect itself?”

the riddle being then one of a self-avoiding random walk [kind]… As I could not get back to sleep last night, I spent a couple hours (!) on this riddle, programming a random walk [or more accurately, a random canter]. This is a brute-force approach in that I pick any acceptable move with the same probability and stop when there is no further move available. [The title refers to the recommendation to avoid the rim of the chessboard with a knight: “a knight on the rim is grim”…]

while (stop){

with my function nexx a rather clumsy 50 lines business of selecting one acceptable move from the current position curr. This function returns the proposed move as well as the updated board with zeros in squares already visited by the knight. Which highlights the ambiguity in the question, namely how one defines the path of a knight? For an acceptable knight move from A to B, there are two possible paths: either take two steps in one direction and one in the orthogonal direction or the opposite. I thus pick one of the two (at random) and prohibit further visits to those squares. An alternative meaning of the question could be that the line joining A to B cannot be crossed ever again, which excludes less moves (but is more cumbersome to code). Anyway, with the former interpretation of a path, repeating the self-avoiding moves led to a maximum of 19 moves, with one solution exhibited below. (Since (64-1)/3=21, it is conceivable that the true maximum is 20 or even 21. In the path representation below, it seems possible to include yet another move by going to (4,1) instead of (4,5). But this is apparently excluded by the square representation on the right. Why is why the path representation is somewhat confusing!)


Today, namely on October 15, I received a solution of length 21, hence covering the entire board without ever using the same square twice. It was sent to me by Paul-Henry Cournède (a geographical neighbour!) and is “obvious” once you see it. Which may be why the alternative interpretation of “path” was chosen in The Riddler. And why my rhs representation is clearly misleading!