How to resolve the algorithm Set consolidation step by step in the Quackery programming language

Published on 12 May 2024 09:40 PM
#Quackery

How to resolve the algorithm Set consolidation step by step in the Quackery programming language

Table of Contents

Problem Statement

Given two sets of items then if any item is common to any set then the result of applying consolidation to those sets is a set of sets whose contents is: Given N sets of items where N>2 then the result is the same as repeatedly replacing all combinations of two sets by their consolidation until no further consolidation between set pairs is possible. If N<2 then consolidation has no strict meaning and the input can be returned.

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Let's start with the solution:

Step by Step solution about How to resolve the algorithm Set consolidation step by step in the Quackery programming language

Source code in the quackery programming language

  [ 0 swap witheach [ bit | ] ] is ->set        ( $ --> { )

  [ say "{" 0 swap
    [ dup 0 != while
      dup 1 & if [ over emit ]
      1 >> dip 1+ again ]
    2drop say "} " ]             is echoset     ( { -->   )

  [ [] swap dup size 1 - times
      [ behead over witheach
          [ 2dup & iff
              [ | swap i^ poke
                [] conclude ]
            else drop ]
           swap dip join ]
    join ]                       is consolidate ( [ --> [ )

  [ dup witheach echoset
    say "--> "
    consolidate witheach echoset
    cr ]                         is task        ( [ -->   )

  $ "AB" ->set
  $ "CD" ->set join
  task
  $ "AB" ->set
  $ "BD" ->set join
  task
  $ "AB" ->set
  $ "CD" ->set join
  $ "DB" ->set join
  task
  $ "HIK" ->set
  $ "AB"  ->set join
  $ "CD"  ->set join
  $ "DB"  ->set join
  $ "FGH" ->set join
  task

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