Order by pair comparisons

Order by pair comparisons is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Sorting Algorithm
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Assume we have a set of items that can be sorted into an order by the user.

The user is presented with pairs of items from the set in no order, the user states which item is less than, equal to, or greater than the other (with respect to their relative positions if fully ordered).

Write a function that given items that the user can order, asks the user to give the result of comparing two items at a time and uses the comparison results to eventually return the items in order.

Try and minimise the comparisons the user is asked for.

Show on this page, the function ordering the colours of the rainbow:

   violet red green indigo blue yellow orange

The correct ordering being:

   red orange yellow green blue indigo violet

Note:

• Asking for/receiving user comparisons is a part of the task.
• Code inputs should not assume an ordering.
• The seven colours can form twenty-one different pairs.
• A routine that does not ask the user "too many" comparison questions should be used.

Factor

Asking the user for an ordering specifier inside a custom comparator:

Works with: Factor version 0.99 2021-02-05

<lang factor>USING: formatting io kernel math.order prettyprint qw sorting ;

qw{ violet red green indigo blue yellow orange } [ "Is %s > %s? (y/n) " printf readln "y" = +gt+ +lt+ ? ] sort .</lang>

Output:

Is violet > red? (y/n) y
Is green > indigo? (y/n) n
Is blue > yellow? (y/n) y
Is red > green? (y/n) n
Is violet > green? (y/n) y
Is violet > indigo? (y/n) y
Is yellow > orange? (y/n) y
Is red > orange? (y/n) n
Is green > orange? (y/n) y
Is green > yellow? (y/n) y
Is green > blue? (y/n) n
Is indigo > blue? (y/n) y
{ "red" "orange" "yellow" "green" "blue" "indigo" "violet" }

Julia

<lang julia>const nrequests = [0] const ordering = Dict("violet" => 7, "red" => 1, "green" => 4, "indigo" => 6, "blue" => 5,

                     "yellow" => 3, "orange" => 2)

function tellmeifgt(x, y)

   nrequests[1] += 1
   while true
       print("Is $x greater than $y?  (Y/N) =>  ")
       s = strip(readline())
       if length(s) > 0
           (s[1] == 'Y' || s[1] == 'y') && return true
           (s[1] == 'N' || s[1] == 'n') && return false
       end
   end

end

function orderbypair!(a::Vector)

   incr = div(length(a), 2)
   while incr > 0
       for i in incr+1:length(a)
           j = i
           tmp = a[i]
           while j > incr && tellmeifgt(a[j - incr], tmp)
               a[j] = a[j-incr]
               j -= incr
           end
           a[j] = tmp
       end
       if incr == 2
           incr = 1
       else
           incr = floor(Int, incr * 5.0 / 11)
       end
   end
   return a

end

const words = String.(split("violet red green indigo blue yellow orange", r"\s+")) println("Unsorted: $words") println("Sorted: $(orderbypair!(words)). Total requests: $(nrequests[1]).")

</lang>

Output:

Is violet greater than indigo?  (Y/N) =>  y
Is red greater than blue?  (Y/N) =>  n
Is green greater than yellow?  (Y/N) =>  y
Is violet greater than orange?  (Y/N) =>  y
Is indigo greater than orange?  (Y/N) =>  y
Is orange greater than red?  (Y/N) =>  n
Is red greater than yellow?  (Y/N) =>  n
Is yellow greater than indigo?  (Y/N) =>  n
Is indigo greater than blue?  (Y/N) =>  y
Is yellow greater than blue?  (Y/N) =>  n
Is indigo greater than green?  (Y/N) =>  n
Is green greater than violet?  (Y/N) =>  n
Sorted: ["orange", "red", "yellow", "blue", "indigo", "green", "violet"]. Total requests: 12.

Phix

The number of questions asked is entirely dependent on how the initial order marries in with the sorting algorithm.
This needs just 6 questions to handle an already in-order or only first two items swapped list.
I picked an initial ordering that requires a fairly easy to remember set of answers: 4Y then alternate.
The builtin sort(s) use an initial gap of 10%, ultimately balancing #comparisons against cache hits, which leads to a wider range of #questions, as said best case 6, worst case 21. A better match to the narrower range of Python (I think 10..14) could probably be made using a copy of custom_sort (it is only 52 lines) with an initial 50% gap.

integer qn = 0
function ask(string a, b)
    qn += 1
    printf(1,"%d: Is %s < %s (Y/N)?:",{qn,a,b})
    integer ch = upper(wait_key())
    printf(1,"%s\n",ch)
    return iff(ch='Y'?-1:1)
end function

?custom_sort(ask,split("violet orange red yellow green blue indigo"))

Output:

1: Is orange < violet (Y/N)?:Y
2: Is red < violet (Y/N)?:Y
3: Is red < orange (Y/N)?:Y
4: Is yellow < violet (Y/N)?:Y
5: Is yellow < orange (Y/N)?:N
6: Is green < violet (Y/N)?:Y
7: Is green < yellow (Y/N)?:N
8: Is blue < violet (Y/N)?:Y
9: Is blue < green (Y/N)?:N
10: Is indigo < violet (Y/N)?:Y
11: Is indigo < blue (Y/N)?:N
{"red","orange","yellow","green","blue","indigo","violet"}

Python

Uses binary search to insert successive items into a growing ordered list. Comparisons are asked for.

<lang python>def _insort_right(a, x, q):

   """
   Insert item x in list a, and keep it sorted assuming a is sorted.
   If x is already in a, insert it to the right of the rightmost x.
   """

   lo, hi = 0, len(a)
   while lo < hi:
       mid = (lo+hi)//2
       q += 1
       less = input(f"{q:2}: IS {x:>6} LESS-THAN {a[mid]:>6} ? y/n: ").strip().lower() == 'y'
       if less: hi = mid
       else: lo = mid+1
   a.insert(lo, x)
   return q

def order(items):

   ordered, q = [], 0
   for item in items:
       q = _insort_right(ordered, item, q)
   return ordered, q

if __name__ == '__main__':

   items = 'violet red green indigo blue yellow orange'.split()
   ans, questions = order(items)
   print('\n' + ' '.join(ans))</lang>

Output:

 1: IS    red LESS-THAN violet ? y/n: y

 2: IS  green LESS-THAN violet ? y/n: y

 3: IS  green LESS-THAN    red ? y/n: n

 4: IS indigo LESS-THAN  green ? y/n: n

 5: IS indigo LESS-THAN violet ? y/n: y

 6: IS   blue LESS-THAN indigo ? y/n: y

 7: IS   blue LESS-THAN  green ? y/n: n

 8: IS yellow LESS-THAN   blue ? y/n: y

 9: IS yellow LESS-THAN  green ? y/n: y

10: IS yellow LESS-THAN    red ? y/n: n

11: IS orange LESS-THAN   blue ? y/n: y

12: IS orange LESS-THAN yellow ? y/n: y

13: IS orange LESS-THAN    red ? y/n: n

red orange yellow green blue indigo violet

Wren

Translation of: Python

Library: Wren-ioutil

Library: Wren-fmt

<lang ecmascript>import "/ioutil" for Input import "/fmt" for Fmt

// Inserts item x in list a, and keeps it sorted assuming a is already sorted. // If x is already in a, inserts it to the right of the rightmost x. var insortRight = Fn.new{ |a, x, q|

   var lo = 0
   var hi = a.count
   while (lo < hi) {
       var mid = ((lo + hi)/2).floor
       q = q + 1
       var prompt = Fmt.swrite("$2d: Is $6s less than $6s ? y/n: ", q, x, a[mid])
       var less = Input.option(prompt, "yn") == "y"
       if (less) {
           hi = mid
       } else {
           lo = mid + 1
       }
    }
    a.insert(lo, x)
    return q

}

var order = Fn.new { |items|

   var ordered = []
   var q = 0
   for (item in items) {
       q = insortRight.call(ordered, item, q)
   }
   return ordered

}

var items = "violet red green indigo blue yellow orange".split(" ") var ordered = order.call(items) System.print("\nThe colors of the rainbow, in sorted order, are:") System.print(ordered)</lang>

Output:

 1: Is    red less than violet ? y/n: y
 2: Is  green less than violet ? y/n: y
 3: Is  green less than    red ? y/n: n
 4: Is indigo less than  green ? y/n: n
 5: Is indigo less than violet ? y/n: y
 6: Is   blue less than indigo ? y/n: y
 7: Is   blue less than  green ? y/n: n
 8: Is yellow less than   blue ? y/n: y
 9: Is yellow less than  green ? y/n: y
10: Is yellow less than    red ? y/n: n
11: Is orange less than   blue ? y/n: y
12: Is orange less than yellow ? y/n: y
13: Is orange less than    red ? y/n: n

The colors of the rainbow, in sorted order, are:
[red, orange, yellow, green, blue, indigo, violet]