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  • how to solve this allocation problem

    Posted by chan_chenyanj on May 14, 2022 at 12:00 am

    suppose i have a table with three columns

    first one denotes the sequence of items to pick each person allowed to pick (here only two and same seq, the front ones will have higher dollar reward) whenever available, rewards is the dollar amount corresponding the sequence of items to pick and column three is whether they allow to pick

    since person 1 can pick item 0 which has 300 as he is allowed to pick. then second person cannot pick item 0 as each item for the amount of reward is unique. so he has to skip 1 and second element is 1 which he is allowed to pick then he picks item 1.

    The question is if i have a table and row 1 correspond to first person and so on, how can i output the item number each person pick?

    pickSeq: (0;1;2);(0;1;2)

    rewards:(300;200;100);(300;200;100)

    allowToPick:(true;true;true);(true;true;true)

    chan_chenyanj replied 9 months, 1 week ago 3 Members · 2 Replies
  • 2 Replies
  • davidcrossey

    Member
    May 14, 2022 at 12:00 am

    If you had the following scenario, you could sort the reward desending, sort the pickSeq ascending and use indices to solve for your problem:

    q)n:10 
    q)people:([]pickSeq:0+til n;person:`$"person",/:string 1+til n;allowedToPick:n?01b)
    q)people 
    pickSeq person allowedToPick 
    ------------------------------ 
    0 person1 1 
    1 person2 1 
    2 person3 1 
    3 person4 1 
    4 person5 0 
    5 person6 0 
    6 person7 0 
    7 person8 1 
    8 person9 0 
    9 person10 0 
    
    q)rewards:([]prize:(1+til 10)*100) 
    q)rewards 
    prize 
    ----- 
    100 
    200 
    300 
    400 
    500 
    600 
    700 
    800 
    900 
    1000 
    
    q)(update ind:i from xdesc[`prize;rewards]) lj `ind xkey update ind:i from select person from xasc[`pickSeq;people] where allowedToPick 
    prize ind person 
    ----------------- 
    1000 0 person1 
    900 1 person2 
    800 2 person3 
    700 3 person4 
    600 4 person8 
    500 5 
    400 6 
    300 7 
    200 8 
    100 9

    You could also extract the person and their prize as follows:

    q){if[1=count x;x:enlist x]; if[1=count y;y:enlist y]; y!x iasc y}[exec desc prize from rewards;] exec person from xasc[`pickSeq;people] where allowedToPick 
    person1| 1000 
    person2| 900 
    person3| 800 
    person4| 700 
    person8| 600
  • Laura

    Administrator
    May 16, 2022 at 12:00 am

     

    q)show people:([]pickSeq:neg[n]?n;person:`$"person",/:string 1+til n;allowedToPick:n?01b) 
    pickSeq person allowedToPick 
    ------------------------------ 
    1 person1 0 
    8 person2 1 
    5 person3 0 
    7 person4 1 
    0 person5 0 
    3 person6 1 
    6 person7 0 
    4 person8 0 
    2 person9 1 
    9 person10 0 
    
    q)show prize:100*1+til 8 
    100 200 300 400 500 600 700 800

     

    The winners in order:

     

    q)`pickSeq xasc select from people where allowedToPick 
    pickSeq person allowedToPick 
    ----------------------------- 
    2 person9 1 
    3 person6 1 
    7 person4 1 
    8 person2 1

     

    And their rewards:

     

    q){select person,reward:count[x]#desc prize from x}`pickSeq xasc people where people`allowedToPick 
    person reward 
    -------------- 
    person9 800 
    person6 700 
    person4 600 
    person2 500

     

    Or in vector form:

     

    q){x!count[x]#desc prize}{x iasc y}. flip people[where people`allowedToPick;`person`pickSeq] 
    person9| 800 
    person6| 700 
    person4| 600 
    person2| 500

     

    Above, where people`allowedToPick finds the winners; and {x iasc y}. sorts the winners names by pick order. The vector form has less work to do:

    q)ts:1000 {select person,reward:count[x]#desc prize from x}`pickSeq xasc people where people`allowedToPick 
    12 3184 
    
    q)ts:1000 {x!count[x]#desc prize}{x iasc y}. flip people[where people`allowedToPick;`person`pickSeq] 
    3 3312

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