Ok, I rarely post but I couldnt sleep so I decided to pass on the misery by creating a programming puzzle.

you have an exam hall represented as an array of 42 entries.

you have 19 JSS1 students, 11 JSS2 students and 12 JSS3 students sitting for a midterm exam in 2 courses.

For the first paper, all the JSS1 have exactly the same paper, all the JSS2 students too but 6 JSS3 have a different paper from the rest.

For the second paper, the JSS1 students still share the same paper and everyone else share the same papers

Please note there are 4 confirmed unrepentant cheaters among the JSS1 students who must not sit with each other.

Write a program (In Any Language of Your Choosing) or just an algorithm that seats the students in the best possible way to minimize cheating.

(If you dont mind too, Do join me in pondering this at 3am :-) )

who you give you the work? chai the person wicked o

for the first paper, do you mean questions from jss1 papers= questions from jss2 papers

or

jss1 papers != jss2 papers but each class share same papers with it's members

i cant work on this at the moment.

I think using a stack like structure solves this problem...
The knowledge that 6 JS3 students does not write a certain paper is irrelevant since their seats are not taken up by other students from other classes.....
You have your container structure of type hall that should accept data types of JS1, JS2 and JS3 students...
The stack structures of the JS students should be in such a way that whenever a student is collected, it the derefrences its content by 1... In C++, something like this
class StudentStack{
int count;
student someStudents[count];
student getStudent(){return someStudents[count--];}
};
you just have to make sure that no student should seat beside his classmate and since JS1has the largest number of students, the seating arrangement shud be like this:
JS2, JS1, JS3, JS1, JS2, JS1, JS3, JS1...... JS2, JS3, JS2, JS3...
to add students to hall according to seating positions should be something like
code:
while(hall.seats < SEAT_COUNT){
if(hall.lastAppended() != JS1){
hall.append(js1stack.getStudent());
//other codes to append other students should follow
}
}
I think this works

My brain is not functioning properly at the moment...i'll be back...

although this thread is on front page, why is only a few folks veiwing it?

veiwing this topic:
lordZOUGA(m), gidiMonsta(m), slimdiggi, ikiller and 3 guest(s)

masterpiecer: although this thread is on front page, why is only a few folks veiwing it?

veiwing this topic:
lordZOUGA(m), gidiMonsta(m), slimdiggi, ikiller and 3 guest(s)

not many are interested in stuff like this...

You are on your own ooo

webdezzi: who you give you the work? chai the person wicked o

for the first paper, do you mean questions from jss1 papers= questions from jss2 papers

or

jss1 papers != jss2 papers but each class share same papers with it's members

i cant work on this at the moment.

will appreciate this clarification too...

mkmyers45:

not many are interested in stuff like this...

Not everyone on nairaland is a programmer.

To get more viewers, just add Kim Kardashian or DBanj to the topic.

afam4eva:
Not everyone on nairaland is a programmer.

true...

Yes! exactly, using the stack like structure works well...but am busy at work to write it now, just wanna have a say..catch ya later

masterpiecer: although this thread is on front page, why is only a few folks veiwing it?

veiwing this topic:
lordZOUGA(m), gidiMonsta(m), slimdiggi, ikiller and 3 guest(s)

only few interested in programming.

and where is all the programmers in the house?
u guyz should come and detonate dis bomb o

Shaunscot: To get more viewers, just add Kim Kardashian or DBanj to the topic.

and tontoh dike

1. Can I reseat students after the first exam?
2. Can you give an example of what you mean by minimize cheating? I.e., for a smaller problem (say with 6 students rather than 42), give two configurations and say which one is better (and ideally, what the "score" associated with each configuration is).
3. Do I know the identities of the cheating students? Are they the only ones who will cheat, or might others also?

masterpiecer: and where is all the programmers in the house?
u guyz should come and detonate dis bomb o

programmers by mouth or by brain

If you don't want them to cheat,here's what to do.

Tell them to get their writing materials ready and make sure they ease themselves of urine, poo or whatever nature call, because no one will be allowed to go outside the hall after being screened in, search them to make sure they don't have any micro chips(expo) on them before letting them into the exam hall, let the students sit on separate seats with a reasonable distance from each other and get like 6 very mean supervisors to supervise them while they write the exams, no biggy.

What score does the following configuration get:

ZX_ZX_ZX_ZX_X_X_X_X_X_X_X_X_X_X_X_X_X_X_X_

where the Xs contain the locations of the JSS1 students, the first 4 Zs are say the first for JSS2 students, and the blanks take the remaining?

Does it get a perfect score (i.e., no cheating possible)? Or do JSS1 students and JSS2 students possibly have the same exam, and thus for example the first ZX pair in the configuration can cheat?

afam4eva: My brain is not functioning properly at the moment...i'll be back...

if na woman thing u go sabi shey?u no dey take last nah...hahahhaha

irrespective of the paper the java code below will make sure no two student of the same class sits directly beside each other.

String sitting[] = new String[40];
int count = 0;

for (int i = 0; i < sitting.length; i++)
{
if(count <= 19)
{
if(count == 0)
{
sitting[i] = "JSS1";
}else if(count == 1)
{
sitting[i] = "JSS3";
}
else
{
if("JSS1".equals(sitting[i - 1]))
{
if("JSS2".equals(sitting[i - 1]))
{
sitting[i] = "JSS3";

}else
{

sitting[i] = "JSS2";
}

}else
{

if("JSS2".equals(sitting[i - 1]))
{
sitting[i] = "JSS3";

}else
{

sitting[i] = "JSS2";
}




}

}

}else
{

if("JSS2".equals(sitting[i - 1]))
{
sitting[i] = "JSS3";

}else
{

sitting[i] = "JSS2";
}



}

count++;
}


holla for further clarification.

lordZOUGA: I think using a stack like structure solves this problem...
The knowledge that 6 JS3 students does not write a certain paper is irrelevant since their seats are not taken up by other students from other classes.....
You have your container structure of type hall that should accept data types of JS1, JS2 and JS3 students...
The stack structures of the JS students should be in such a way that whenever a student is collected, it the derefrences its content by 1... In C++, something like this
class StudentStack{
int count;
student someStudents[count];
student getStudent(){return someStudents[count--];}
};
you just have to make sure that no student should seat beside his classmate and since JS1has the largest number of students, the seating arrangement shud be like this:
JS2, JS1, JS3, JS1, JS2, JS1, JS3, JS1...... JS2, JS3, JS2, JS3...
to add students to hall according to seating positions should be something like
code:
while(hall.seats < SEAT_COUNT){
if(hall.lastAppended() != JS1){
hall.append(js1stack.getStudent());
//other codes to append other students should follow
}
}
I think this works

this approach still makes sense to me....
If The container "hall" increments itself anytime a student is added, You can even add 3 students in a loop
code:
while(hall.count != MAX_STUDENTS){
if(JS1students.isEmpty()){
hall.append(JS2student.getStudent());
hall.append(JS3student.getStudent());
}else{
hall.append(JS2student.getStudent());
hall.append(JS1student.getStudent());
hall.append(JS3student.getStudent());
hall.append(JS1student.getStudent());
//with the above code you can add all students in less number of loops..
}
}

^Na wa o, this una stuff must be very boring sha, just looking at what you wrote is giving me serious migraine .

@lordzouga from your code there's possibilty of having two jss1 students sit directly beside each other because you will run out of jss2 and jss3 students before the hall gets filled, a good approach will be arranging the first nineteen seats in the hall with a must jss1 and either jss2 or jss3 student, and you must start with a jss1 followed by a jss3 .This method will ensure that when we run out of jss1(seat 19) we will have exactly 7 jss2 and 7 jss3 student left that can be evenly distributed; as simulated by the java code i posted earlier on.

freecocoa: ^Na wa o, this una stuff must be very boring sha, just looking at what you wrote is giving me serious migraine .

really? Okay, here is an easy fix:
drop your device..
Go to ur drug cabinet or store and get an aspirin and then take it..
Go to the refridgerator and get an ice pack and put it against your forehead...
Then pick your device back up and continue staring at the code...
Dem no born the migraine well if em come back... No, programming is not boring at all... Just frustrating atimes...

lordZOUGA:
this approach still makes sense to me....
If The container "hall" increments itself anytime a student is added, You can even add 3 students in a loop
code:
while(hall.count != MAX_STUDENTS){
if(JS1students.isEmpty()){
hall.append(JS2student.getStudent());
hall.append(JS3student.getStudent());
}else{
hall.append(JS2student.getStudent());
hall.append(JS1student.getStudent());
hall.append(JS3student.getStudent());
hall.append(JS1student.getStudent());
//with the above code you can add all students in less number of loops..
}
}

This should work..The puzzle is quite easy...just has some irrevelant constraints...

speedzojie: @lordzouga from your code there's possibilty of having two jss1 students sit directly beside each other because you will run out of jss2 and jss3 students before the hall gets filled, a good approach will be arranging the first nineteen seats in the hall with a must jss1 and either jss2 or jss3 student, and you must start with a jss1 followed by a jss3 .This method will ensure that when we run out of jss1(seat 19) we will have exactly 7 jss2 and 7 jss3 student left that can be evenly distributed; as simulated by the java code i posted earlier on.

No sir, there is no way my algorithm runs out of jss2 and JSS3 students before JS1 students...
jss1 = 19, jss2 = 11, jss3 = 12
I must run out of JSS1 students before I run out of JSS2 and JSS3 students. Infact, I should have 4 leftover Jss2 and jss3 students
11 + 12 = 23
23 - 19 = 4
that I then arranged again
Please look carefully at the algorithm..

Javanian:
This should work..The puzzle is quite easy...just has some irrevelant constraints...

EXACTLY!! Even the 4 cheating students are irrelevant

The following information are ambigous and you should remove them from your calculations.

1. The number of courses
2. The classes JS1-JS3
3. The nature of the papers written.

Just write a code to sit 42 people putting 4 particular people among them in different seat rows.

Here's my solution below, it isn't pure but you can't blame me... time and headache.

==============

<?php 
$students = array("student_1", "student_2", "student_3", "student_4", "student_5", "student_6", "student_7", "student_8", "student_9", "student_10", "student_11", "student_12", "student_13", "student_14", "student_15", "student_16", "student_17", "student_18", "student_19", "student_20", "student_21", "student_22", "student_23", "student_24", "student_25", "student_26", "student_27", "student_28", "student_29", "student_30", "student_31", "student_32", "student_33", "student_34", "student_35", "student_36", "student_37", "student_38", "student_39", "student_40", "student_41", "student_42";

//print_r($students);
$clean_students = array_slice($students, 4, 42);
$cheaters =  array_slice($students, 0, 4);


$j = 0; $k=0; $m=0;
for($i=1; $i<12; $i++){
	echo "<p>Row ".$i.": ".$clean_students[$m]."-".$clean_students[$m+1]."-".$clean_students[$m+2]."-<span style='color:red'>".$cheaters[$j]."</span></p>";
	$m+=3;
	$j+=1;
}
?>

=========
Live demo: http://sawyerrken.net/kodewrita/

P.S: This solution isn't scalable.

^ not remotely scalable...

Dual Core: The following information are ambigous and you should remove them from your calculations.

1. The number of courses
2. The classes JS1-JS3
3. The nature of the papers written.

Just write a code to sit 42 people putting 4 particular people among them in different seat rows.

Here's my solution below, it isn't pure but you can't blame me... time and headache.

==============

<?php 


$students = array("student_1", "student_2", "student_3", "student_4", "student_5", "student_6", "student_7", "student_8", "student_9", "student_10", "student_11", "student_12", "student_13", "student_14", "student_15", "student_16", "student_17", "student_18", "student_19", "student_20", "student_21", "student_22", "student_23", "student_24", "student_25", "student_26", "student_27", "student_28", "student_29", "student_30", "student_31", "student_32", "student_33", "student_34", "student_35", "student_36", "student_37", "student_38", "student_39", "student_40", "student_41", "student_42";

//print_r($students);
$clean_students = array_slice($students, 4, 42);
$cheaters =  array_slice($students, 0, 4);


$j = 0; $k=0; $m=0;
for($i=1; $i<12; $i++){
	echo "<p>Row ".$i.": ".$clean_students[$m]."-".$clean_students[$m+1]."-".$clean_students[$m+2]."-<span style='color:red'>".$cheaters[$j]."</span></p>";
	$m+=3;
	$j+=1;
}
?>

=========
Live demo: http://sawyerrken.net/kodewrita/

P.S: This solution isn't scalable.

Good simple solution.

lordZOUGA:
EXACTLY!! Even the 4 cheating students are irrelevant

I think he just wants to make sure they don't seat together i.e the cheating students.