one-way random walks

A rather puzzling riddle from The Riddler on an 3×3 directed grid and the probability to get from the North-West to the South-East nodes following the arrows. Puzzling because while the solution could be reasonably computed with an R code like

sucz=0
for(i in 1:2^12){
  path=intToBits(i)[1:12]
  sol=0
  for(j in 1:12)sol=max(sol,
        prod(path[paz[[j]][paz[[j]]>0]]==01)*
        prod(path[-paz[[j]][paz[[j]]<0]]==00))
  sucz=sucz+sol

where paz is the list of the 12 possible paths from North-West to South-East (excluding loops!), leading to a probability of 1135/2¹², I could not find a logical reasoning to reach this number. The paths of length 4, 6, 8 are valid in 2⁸, 2⁶, 2⁴ of the cases, respectively and logically!, but this does not help as they are dependent.

Le Monde puzzle [#1011]

An combinatoric Le Monde mathematical puzzle (with two independent parts):

Given the following grid,

1. What is the longest path from A to B that does not use the same edge twice?
2.  What is the probability that two minimal length paths from A to B [of length 13] share the same middle [7th] edge?

The first question can be solved by brute force simulation. I ran a very simple minded self-avoiding random walk starting from A and restarting each time a dead-end was reached. (The details are not of capital interest: I entered the above grid as an 8×7 matrix for the nodes and associated with each node a four bit number indicating which edge had been visited. Picking at random among those not yet visited.) The longest path I found along 10⁷ simulations is 51 edges long, confirmed by an additional exploration of the paths on both square grids, separately. The associated path is as follows, the irregular shape being obtained by jittering the node locations towards a better visualisation of the order of the visits.

The second puzzle can be solved directly by looking at the number of paths sharing the seventh edge, which is ¼ (as checked by a further simulation of minimal length random walks).

a grim knight [cont’d]

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?”

as 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. The first maximal length I found this way is 32, although I also came by hand to a spiralling solution with a length of 33.

Running 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]

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”…]

board=rep(1,64)
curr=sample(1:64,1)
board[curr]=0
cont=0
stop=TRUE
while (stop){
  mov=nexx(curr,board)
  curr=mov$mov;board=mov$boa
  stop=(curr>0);cont=cont+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!

Le Monde puzzle [#882]

A terrific Le Monde mathematical puzzle:

All integers between 1 and n² are written in an (n,n)  matrix under the constraint that two consecutive integers are adjacent (i.e. 15 and 13 are two of the four neighbours of 14). What is the maximal value for the sum of the diagonal of this matrix?

Indeed, when considering a simulation resolution (for small values of m), it constitutes an example of self-avoiding random walk: when inserting the integers one by one at random, one produces a random walk over the (n,n) grid.

While the solution is trying to stick as much as possible to the diagonal vicinity for the descending sequence n²,n²-1, &tc., leaving space away from the diagonal for the terminal values, as in this example for n=5,

25 22 21 14 13
24 23 20 15 12
01 02 19 16 11
04 03 18 17 10
05 06 07 08 09

simulating such a random walk is a bit challenging as the brute force solution does not work extremely well: