how to solve this allocation problem

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https://learninghub.kx.com/forums/topic/how-to-solve-this-allocation-problem

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)

2

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
3

 

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