[Jprogramming] A practical problem: minimum distance with penalties

[Henry Rich]

You are right, the last 2 atoms could be *s.

Penalties & distances are measured in the same units & are essentially 
the same in this context.  To move to a new atom you must pay a penalty 
and traverse a distance, both of which are rolled into the value in P.  
D holds accumulated values of P, except for the starting values of D 
which come from "elsewhere" and are your input.

Henry Rich

On 2/23/2016 2:41 PM, Brian Schott wrote:
> Henry,
>
> Why in your example answer for D aren't the rightmost 2 atoms *s, since
> they have not changed from 20 and 3? Does that suggest that the "new" 20
> and 3 are calculated but that you are not giving their calculations?
>
> Also you say, "An atom of D indicates the minimum number of penalty points
> that must be incurred to reach that position."
> This ​suggests to me that atoms of D are penalties, not distances. Is that
> true and does that matter?​
>
> On Mon, Feb 22, 2016 at 1:28 PM, Henry Rich <henryhrich at gmail.com> wrote:
>
>> This task shows J in my hands at its best and its worst.  Best, because I
>> found a good solution; worst, because it took me 4 hours to come up with
>> something that would have been trivial in a scalar language.  See what you
>> can do with it.
>>
>> The task is to write a REALLY FAST program to solve the problem.  Solving
>> the problem correctly is not so hard, but this is the innermost loop of
>> many nested loops, and it must run at close to maximum machine speed.
>>
>> The problem arises in shortest-path-finding.  You are given two lists, the
>> list of distances D and the list of penalties P.  D is to be thought of as
>> ordered left-to-right.  An atom of D indicates the minimum number of
>> penalty points that must be incurred to reach that position.
>>
>> At each step, new elements of D are created by external code, as new best
>> ways to reach positions are discovered.  After the new elements of D have
>> been installed, you come in.  You have to process D (and P) to propagate
>> the new distances through D, leaving every atom of D showing the best
>> distance.  This best distance may be either the incumbent distance at that
>> position, or a smaller value inherited from the position to its left.  If
>> you inherit the distance from the left, you must add the value of the atom
>> of P in the position being moved into.  The incumbent value in that
>> position already includes the value of P.
>>
>> There is one more fillip.  Certain positions of D are blocked off.  A
>> blocked-off atom must not be changed, and its value may not be inherited by
>> the cell to its right.
>>
>> In the representation of D, empty cells are given a distance of HIGH-VALUE
>> (=1e6), and blocked cells are given a distance of LOW-VALUE (=_1).
>>
>> Example:
>>
>> Given the setup
>> H =. 1e6
>> L =. _1
>> D =. H, H, 1, H, 5, H, 8, H, 11, L, L, H, H, 20,3
>> P =. 1, 1, 1, 2, 4, 2, 1, 1,  1, 1, 1, 1, 2, 1, 2
>>
>> you would produce
>>
>>
>>       *  *  *  3  *  7  *  9  10  *  *  *  *  20 3
>>
>> 3 = 1+2-------|
>> 7 = 5+2-------------|
>>
>> The * values indicate 'anything greater than or equal to the incumbent
>> value' and indicate atoms that are unchanged by the process.  You are free
>> to give them any value you like, as long as it is not less than the
>> incumbent value.
>>
>> The numbered values are the inherited distances: the ones you must find.
>>
>> All the values can be integer and all are small compared with the range of
>> an integer.  The length of the lists will not exceed 10,000, the values of
>> D will not exceed 50,000, and the values of P will not exceed 20.
>>
>> Your code will be executed many times.  If you want to factor out
>> computation by pre-computing any values, do so and do not count that as
>> part of the execution time.
>>
>> Henry Rich
>>