Horrifying code written when I was 14

P r o g r a m m i n g - a - M o d e - F i n d e r
     -a guide to programming a mode finder-
                 By: Brian Gordon
 ____________
/Introduction\_________________________________
|                                              |
|This guide will outline the technique I found |
|is the best way to determine a mode from a    |
|set of data. It includes most of the source   |
|code from the original program, then explains |
|it in almost-understandable English.          |
|I use -> for STO and & for the Theta letter.  |
\______________________________________________|

First thing in the program is initialization:
 ClrHome
 DelVar L1
 DelVar L3
 DelVar L6

These first 4 lines clear a few lists (the 
input, a middleman list, and the results).

 Input L1
 dim(L1)->A
 A->dim(L2)
 For(X,1,A,1)
 For(Y,1,A,1)
 If (L1(X)=L1(Y)
 1+L2(X)->L2(X)
 End:End

Takes input to L1, the dimensision of which is 
then stored to A. A then becomes the length of 
L2, the "high-score" list. AASTAT grades which 
number is used the most (ie the mode) by scoring 
it against how often other numbers are used. 
Here, a double loop is made to step through every 
element in the list and check it against every 
other element in the list. This returns a lot 
of double and triple matches, but at least it 
gets them all. If the element matches another, 
L2 is accessed. Remember, this is the scoreboard 
list. Say the user entered {1,2,7,8,3,4,0,8} so 
that L1(4) matches with L1(7). In that case, 
L2(4) (the scorekeeper element for 8 is increased. 
It is now a simple idea to see which element in L2 
has the highest score and follow that element number 
over to L1, and there's your winner. Problem is 
repeat modes. There will be at least 3 or 4 
repeats of the same mode right now. If you were 
to disp L2 right now, you would get:
{1 1 1 2 1 1 1 2}
So, tracing that back to L1, it looks like 
1,2,7,3,4,0 all have 1 point and 8 has 2. 
Now we need to make a for loop that finds 
the highest-scoring element. Unfortunately, 
since there are two 8s, both with a score 
of 2, we then need to make some For loops that 
take into account the number of 8s and match 
it with the amount of highest-scoring elements.

 L2(1)->&
 For(Z,1,A,1
 If (L2(Z)>&)
 L2(Z)->&
 End

Ooh, scary. First off, & is initialized to be 
the first element in L2. This For loop is going 
to find the highest element in L2, so & is used 
to keep track of the highest score found so far
(not the -number- of the highest-scoring element). 
Therefore, the first element is used as the initial 
highest score. The For loop steps through L2, Z the 
number of each element being checked during that 
particular pass. It checks whether the current score 
being checked is greater than the highest score 
found so far (stored in &). If it is, then it 
becomes the highest score by being stored to &. 
This loop repeats until every element in L2 has been 
checked against the high score. After the loop is 
finished, & contains the highest score found. However, 
the element numbers (yes that's a plural; remember 
that multiple elements can share that same top score... 
multiple modes) themselves ARE NOT STORED. It is 
utterly unfeasable to combine this step and the next 
step: finding the winning element numbers and storing 
them in a compact list.

 1->O
 For(P,1,dim(L2),1)
 If (L2(P)=&):Then
 L1(P)->L3(O)
 O+1->O
 End
 End
 SortA(L3)

The For loop steps through L2 again, using P now 
instead of Z. Each time through it checks whether 
the current element, L2(P), matches the top score. 
If it does, then we know that L1(P) is a winner 
by association. In other words, if L2(P) is the 
top score, then it's corresponding L1 element, 
L1(P), is the actual number that is one of the 
modes. L1(P) is then stored to L3(O), the next 
empty element in L3, for later use. O is 
incremented for the next pass and the loop starts 
over. At the end of this loop, you have a new 
list, L3. Sort it ascending in anticipation for 
later in the program (trust me here). Stop there 
and Disp L3. You see that L3 is {8 8}. Of course 
there's only one mode here, but as mentioned 
before, it's repeated because the user entered 
-two- 8s so they both scored 2.

 0->M
 For(Q,1,dim(L3),&
 M+1->M
 M->dim(L6
 L3(Q)->L6(M)
 End

The reasoning within this For loop lies within 
a brainload of conceptual math done by me. The 
lowdown: if the top score (&) is 2, there are 2 
repeats for -each correct mode- in L3... that's 
a huge Duh. Think the Handshakes problem. If & is 
8, it means that there are 8 repeats for -each 
correct mode- in L3. So the theory is that we can 
somehow splice the correct ones out. We do that 
by only looking at every &th element in L3. This 
serves to operate on only the correct elements 
(eliminating repeats) and at the same time takes 
care of the problem of multiple modes, since at 
this point the other modes obviously must have 
the same score: &. Using a similar costruct as 
the last For loop, the loop runs through L3 at 
increments of &, each time storing the current 
L3 element to the next free element of L6. Add 
this For loop to your code and run it. Then Disp 
L6. { 8 } is the output. Congratulations, you 
just computed a mode. 

One final consideration: there are 2 types of 
NO MODE when computing with this technique: 
1) dim(L1)=dim(L6) - this is the "standard"
   NO MODE where all the numbers are different.
2) dim(L1)=dim(L3) - this is a more complex 
   NO MODE when L2 is solidly one score. An 
   example would be if the user entered:
   {7,7,7,8,8,8,5,5,5}. There is no value with 
   a lower score than the highest score, 
   therefore technically this is a NO MODE, 
   however the program will display {5,7,8} 
   as the mode.
Please add these two conditions to your program.
For example, instead of writing Disp L6, write:

 If (dim(L1)!=dim(L6) and dim(L1)!=dim(L3))
 Disp L6

!= is the not-equal-to-sign found on the Test 
menu.

I hope you learned a ton in this tutorial and I 
urge you to learn to do this on your own, without 
a tutorial.


Copyright 2004 by Brian Gordon