## Ternary sorting

Posted in R with tags , , , , , on July 25, 2011 by xi'an

The last Le Monde puzzle made me wonder about the ternary version of the sorting algorithms, which all seem to be binary (compare x and y, then…). The problem is, given (only) a blackbox procedure that returns the relative order of three arbitrary numbers, how many steps are necessary to sort a series of n nnumbers? The heapsort entry in Wikipedia mentions a ternary sorting version, but does not get into details. Robert Sedgewick (author of a fabulous book on algorithmic I enjoyed very much when I started programming) has a talk about the optimality of quicksort where he mentions ternary sorting, but this seems to be more related with ties than with my problem… It is of course highly formal in that I do not know of a physical device that would justify moving from binary to ternary comparisons.

## Le Monde puzzle [#28]

Posted in R with tags , , , , on July 22, 2011 by xi'an

The puzzle of last weekend in Le Monde was about finding the absolute rank of x9 when given the relative ranks of x1,….,x8 and the possibility to ask for relative ranks of three numbers at a time. In R terms, this means being able to use

```> rank(x[-9])
 1 7 4 6 8 3 2 5
> rank(x[1:3])
 1 3 2
```

or yet being able to sort the first 8 components of x

```> x[-9]=sort(x[-9])
> rank(x[c(1,8,9)])
 1 3 2
```

and then use rank() over triplets to position x9 in the list. If x9 is an extreme value, calling for its rank relative to x1 and x8  is sufficient to find its position. Else repeating with the rank relative to x2 and x7, then relative to x3 and x6  , etc., produces the absolute rank of x9 in at most 4 steps. However, if we first get the rank relative to x1 and x8,  then the rank relative to x4 and x5, we only need at most an extra step to find the absolute rank of x9 in thus at most 3 steps. Yet however, if we start with the rank relative to x3 and x6, we are left with a single rank call to determine the absolute rank of x9 since, whatever the position of x9 relative to x3 and x6, there are only two remaining numbers to compare x9 with. So we can find the absolute rank of x9 in exactly two calls to rank.

The second part of the puzzle is to determine for an unknown series x of 80 numbers the maximum number of calls to rank(c(x1,x2,x3)) to obtain rank(x). Of course, the dumb solution is to check all 82160 triplets, but I presume a sort of bubblesort would do better, even though Wikipedia tells me that bubble sort has worst-case and average complexity both О(n2), while quicksort has an average (if not worse-case) O(n log(n)) complexity…. If we follow the one-step procedure obtained in the first part to insert a new number, starting with three numbers whose relative rank is obtained with one  single call, we need in toto

1 + (9-3)*2 + ((1+3*8+2)-9)*3+(80-27)*4 =279

calls to rank since 80<1+3*(3*8+2)+2=81. (This is also the solution proposed in Le Monde, however there is no proof inserting one number at a time is the optimal solution.)