• revision #: <description> (DATESTAMP)

To continue to cultivate your problem solving, critical thinking, analytical, and observation skills; to apply your skills on the UNIX command-line.

The true nature of problem solving frequently involves critical thinking, analytical, and observation skills. Where problems are not solved by memorizing some pre-defined set of answers and regurgitating them mindlessly, but it crafting an elaborate solution from subtle cues and tested, experimental realizations.

This project has puts you in contact with such endeavours.

Letter division is a category of logic problem where you would take an ordinary math equation (in long form), and substitute all the numbers for letters, thereby in a direct sense masking the numeric values present that correctly enable the problem to work from start to completion. It is your task, through exploring, experimenting, and playing, to ascertain the numeric value of each letter (as many as 10, one for each numeric value 0-9).

For this project we will be focusing on long division, something you learned (and perhaps last experienced, before becoming mindlessly addicted to pressing buttons on a calculator, in grade school). It entails a whole number (integer) division, involving aspects of multiplication, addition (through borrowing), and subtraction (primarily) to arrive at a quotient and a remainder.

Division is unique in that it produces two 'answers', each serving particular uses in various applications.

Here is an example (using numbers):

First up, we're going to divide 87654321 (the dividend) by 1224 (the divisor). Commonly, especially if punching into a calculator, we might express that equation as:

87654321/1224

Or in a language like C, assigning the quotient to the variable x (an integer):

    x = 87654321/1224;

But, we're not actually interested in the 'answer' (quotient or remainder); we are interested in the PROCESS. You know, the stuff the calculator does for you, which in order to perform this project and better explore the aspects of critical thinking, we need to take and encounter every step of the way:

          71613
     +---------
1224 | 87654321
      -8568
       ====
        1974
       -1224
        ====
         7503
        -7344
         ====
          1592
         -1224
          ====
           3681
          -3672
           ====
              9

Here we obtain the results (focusing on the quotient up top; as the remainder quite literally is what remains once we're done- we're specifically NOT delving into decimal points, but instead doing integer division, which as previously stated has MANY important applications in computing) through a step by step process of seeing how many times our divisor (1224) best and in the smallest fashion fits into some current value of the dividend (or intermediate result thereof).

For instance, seeking the smallest “best fit” of 1224 into 87654321, we find that 1224 fits best SEVEN times (1224 * 7 = 8568, which is the CLOSEST we can get to 8765… 1224 * 8 = 9792, which would be too big (and way too small for 87654). Clearly, we are seeking those values that best fit within a multiple of 0-9, staying away from double digits of multiplication (although, we COULD do it that way and still arrive at the same end result).

So: 8765-8568 = 197.

We have our first result, yet: there's still values in the dividend (87654321) remaining to process, specifically the 4321, so we take them one digit at a time.

The next available, unprocessed digit in 4321 is '4', so we 'drop that down' and append it to our previous result (197), giving us: 1974.

We now see how many times (via single digit multiplication), our divisor (1224) can fit into 1974. As it turns out, just once.

So: 1974-1224 = 750.

And we keep repeating the process until there are no more digits from the dividend to drop down; at which point, we are left with a remainder (in the above problem, the lone '9' at the very bottom; THAT is the remainder).

Clearly it is important to have a handle on and understanding of the basic long division process before attempting a letter division problem. So, be sure to try your hand at a few practice problems before proceeding.

Following will be a sample letter division problem, and a documented work through of it, much as you will be doing for this project (and to be sure: the aim here is not merely to solve it, but to DOCUMENT HOW YOU SOLVED IT, so just like the 'steps' files you did in various projects, you might want to keep notes as you go along to save you time and sanity).

Here goes:

            GLJK
      +---------
 KJKK | GLMBRVLR
       -VKOKL
        =====
         LJBGV
        -OKVKG
         =====
          JJGKL
         -LKBKV
          =====
           KVRMR
          -JKRKB
           =====
            VKMK
            
letters: BGJKLMOPRV

First off, note how this is NO DIFFERENT from the numeric problem above: just instead of numbers, which we've associated some concepts with, here we have letters (each letter maps to a unique number, 0-9). The trick will be to figure out which letter maps to which number.

So, let us begin.

One aim is to obtain the key to the puzzle, the mapping of the letters to numbers, so I will typically set up an answer key as follows:

Another thing I like to do is set up a more visual representation of what each letter COULD be. I do so in the following form:

B = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
G = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
J = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
K = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
L = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
M = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
O = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
P = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
R = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
V = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }

Then, as I figure things out (either what certain are, but mostly, which ones they are NOT), I can mark it up accordingly.

Right from the start, we can already make some important connections; looking at EACH of the subtractions taking place, in the left-most position, we see an interesting phenomenon taking place- G-V=0, L-O=0, J-L=0, and K-J=0.

Now, since EACH letter is its own unique numeric value, subtracting one letter from another on its own won't result in a value of 0, but being borrowed from will.

That is: 7-6=1, but (7-1)-6=0. THAT is what is going on here.

So what we can infer from this, is some very important connections:

• V is one less than G (I'll write it as: V < G)
• O is one less than L (O < L)
• L is one less than J (L < J)
• J is one less than K (J < K)

Does that make sense? From looking at the puzzle, those four relations can be made.

Now, FURTHERMORE, some of those connections are thereby connected. Look at the 'L' and 'J' connections:

• O < L, but also: L < J
• L < J, but also: J < K

That implies a further connection, so we can chain them together:

• O < L < J < K

So from that initial observation and connection, we now have two disconnected relationships:

• V < G
• O < L < J < K

From what we've done so far, we do not know where V,G fall in respect to O,L,J,K. They might be less than, OR greater than. We won't know without further information.

Yet, even WITH this information, we can update our letter ranges:

• since V is less than G, we know V can NOT be 9.
• similarly, G can NOT be 0.
• O cannot be 9, 8, 7, because we know O is 3 less than K. So even though we don't know what K actually is, because K COULD be 9, we know what O, L, and J can NOT be.
• L cannot be 9 or 8
• J cannot be 9
• on the other side, K cannot be 0, 1, or 2
• J cannot be 0 or 1
• L cannot be 0.

So, if we update our range chart accordingly:

B = { 0, 1, 2, 3, 4, 5, 6, 7, 8    }
G = {    1, 2, 3, 4, 5, 6, 7, 8, 9 }
J = {       2, 3, 4, 5, 6, 7, 8,   }
K = {          3, 4, 5, 6, 7, 8, 9 }
L = {    1, 2, 3, 4, 5, 6, 7,      }
M = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
O = { 0, 1, 2, 3, 4, 5, 6,         }
P = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
R = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
V = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }

Moving on, dealing with details of discovering those one-off relations, that tells us something about the NEXT subtractions: that they borrow (which means they are LESS THAN the thing being subtracted from them):

• L is less than K (which we actually know to be 2 less than K), so L - K needs to BORROW
• J is less than K (which we know is 1 less than K), so J - K needs to BORROW
• V is apparently also less than K (which we didn't previously know), so V - K needs to BORROW
• now knowing than V « K, we can connect our other relational fragment in (I use the double '«' to denote “less than” by an unknown amount, because while we know V is less than K, we don't know by how much).

So: V < G « O < L < J < K

This allows us some further whittling of our ranges:

• V cannot be 9, 8, 7, 6, or 5
• G cannot be 9, 8, 7, or 6
• O cannot be 0, or 1
• L cannot be 0, 1, or 2
• J cannot be 0, 1, 2, or 3
• K cannot be 0, 1, 2, 3, or 4

B = { 0, 1, 2, 3, 4, 5, 6, 7, 8    }
G = {    1, 2, 3, 4, 5,            }
J = {             4, 5, 6, 7, 8,   }
K = {                5, 6, 7, 8, 9 }
L = {          3, 4, 5, 6, 7,      }
M = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
O = {       2, 3, 4, 5, 6,         }
P = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
R = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
V = { 0, 1, 2, 3, 4,               }

Already we can see that V and G are likely lower numbers, and O, L, J, and K are likely higher numbers.

What else do we have? Let's keep going:

We cannot instantly proceed to the next subtraction in as obvious a progression, as we'll need more information on the various letters involved.

However, looking at the puzzle, I'm interested in seeing if we can find any obvious examples of 0. You know, letter minus same letter sort of things. Because they will typically end up equalling 0 (or 9).

Why 9? Because of a borrow!

((5-1)+10)-5 = (4+10)-5 = 14 - 5 = 9

… that can be quite revealing too!

And it would appear we have one wonderful candidate in the bottom-most subtraction:

           KVRMR
          -JKRKB
           =====
            VKMK

Lookie there: R-R = K.

Usually, that would result in a 0. BUT, we also know that K can NOT be 0 (looking at our range table above).

So, that means it is being borrowed from, and it itself has to borrow, so we now also know that M is less than K (M « K).

And, as indicated above:

((R-1)+10)-R = 9!

We now know that K = 9!

That suddenly reveals a whole lot to us, due to our relational chains we've built. Let's update:

Also, with the new introduction of M being less than K:

B = { 0, 1, 2, 3, 4, 5,            }
G = {    1, 2, 3, 4, 5,            }
J = {                         8    }
K = {                            9 }
L = {                      7       }
M = { 0, 1, 2, 3, 4, 5,            }
O = {                   6          }
P = { 0, 1, 2, 3, 4, 5,            }
R = { 0, 1, 2, 3, 4, 5,            }
V = { 0, 1, 2, 3, 4,               }

And, our relational chains:

• V < G « O < L < J < K
• M « O < L < J < K

Because we don't yet know any relation of M compared to V or G, we have to keep them separate for now.

We also have a second disqualifier for K being 0… the ones place subtraction in that bottom-most subtraction:

R - B = K.

There's nothing further to the right that could borrow from this problem, so it can only exist in two states:

• R is greater than B
• R is less than B

Since we know that K is 9, there's NO OTHER pair of single digit numbers we can subtract to get 9, which tells us that:

• R is less than B (R « B)

Currently both R and B can be 0-5 (although now, B is 1-5, and R is 0-4). We'd need to find a combination where (R+10)-B is 9:

And from that, we're subtracting B, which is 1, 2, 3, 4, or 5. The answer has to be 9.

So:

10-1=9, 11-2=9, 12-3=9, 13-4=9, and 14-5=9

Hey, look at that… B is one greater than R (not just R « B, BUT: R < B)

Our relational chains:

• V < G « O < L < J < K
• M « O < L < J < K
• R < B « O < L < J < K

And our chart, of sorts:

B = {    1, 2, 3, 4, 5,            }
G = {    1, 2, 3, 4, 5,            }
J = {                         8    }
K = {                            9 }
L = {                      7       }
M = { 0, 1, 2, 3, 4, 5,            }
O = {                   6          }
P = { 0, 1, 2, 3, 4, 5,            }
R = { 0, 1, 2, 3, 4,               }
V = { 0, 1, 2, 3, 4,               }

If you look, the only letter we've not yet directly interacted with yet is 'P', although we already know enough about it (that it is 0-5, less than O, L, J, and K). And if you look closely, you'll notice that 'P' isn't even present in the letter division problem! So its identity will rely entirely on the proving of the other values.

Let's continue on:

M-K=M, BECAUSE we know M « K, AND BECAUSE we know the subtraction to the right is borrowing from it (because R < B), we have something like this: (M-1+10)-K=M

Can't really do much more with it at this point, but it is important to know to help us identify the borrows needing to happen.

Why don't we go ahead and find 0? If you look in the subtraction above the bottom one, we have another “letter minus same letter” scenario, and it doesn't equal K!

          JJGKL
         -LKBKV
          =====
           KVRM

We KNOW that V « L, so no borrow is happening there.

Therefore, K-K, or 9-9, equals 0. So R is 0!

… and B is 1! Because of our identified relationship.

Updating things!

Also, with the new introduction of M being less than K:

B = {    1                         }
G = {          3, 4, 5,            }
J = {                         8    }
K = {                            9 }
L = {                      7       }
M = {       2, 3, 4, 5,            }
O = {                   6          }
P = {       2, 3, 4, 5,            }
R = { 0                            }
V = {       2, 3, 4,               }

*NOTE: G is NOT 2, because G is greater than V (one greater, in fact), so we can similarly whittle that off.

Relational chains can look as follows now:

• R < B « V < G « O < L < J < K
• R < B « M « O < L < J < K
• R < B « P « O < L < J < K

Basically just down to V, G, P, and M.

And I think we have the means to find V: notice the second to last subtraction, the “LKBKV”. You know where we get that from? Multiplying the divisor (KJKK) by J (since it is the third subtraction taking place).

We KNOW the numeric values of K and J, in fact we know the values of L, K, and B. The only thing we don't know is 'V', and since V is in the one's place, that makes things super easy for us.

KJKK = 9899 J = 8

So: 9899 x 8 = 79192 = LKBKV!

V is 2!

Which means, because V < G, that G is 3!

Updating our records:

Also, with the new introduction of M being less than K:

B = {    1                         }
G = {          3                   }
J = {                         8    }
K = {                            9 }
L = {                      7       }
M = {             4, 5,            }
O = {                   6          }
P = {             4, 5,            }
R = { 0                            }
V = {       2                      }

Relational chains can look as follows now:

• R < B < V < G « M « O < L < J < K
• R < B < V < G « P « O < L < J < K

And then there were 2. We really just need to find M, or P, and we're done. And since there are no 'P' values in the puzzle, we need to target M. So let's look for some candidates:

Hey, how about this:

          JJGKL
         -LKBKV
          =====
           KVRM

One's place subtraction: L - V = M.

We KNOW L (7) is greater than V (2), so no borrow is happening.

L-V=M 7-2=5

M is 5. That means P is 4 by process of elimination.

Puzzle completed:

Also, with the new introduction of M being less than K:

B = {    1                         }
G = {          3                   }
J = {                         8    }
K = {                            9 }
L = {                      7       }
M = {                5             }
O = {                   6          }
P = {             4                }
R = { 0                            }
V = {       2                      }

Relational chains can look as follows now:

• R < B < V < G < P < M < O < L < J < K

I wasn't able to show it as well in text on the wiki, but I also made a point to mark up each subtraction to show whether a borrow occurred or not:

To be sure, there are likely MANY, MANY ways to arrive at these conclusions. What is important is being observant, performing little experiments, seeing if there can be any insights to have, even if whittling away knowing what things can NOT be.

Your performance on this project will be directly tied to being able to document your process through the puzzle; I have provided this writeup in order to show you an example of what that process may look like.

In the pct0/ sub-directory of the UNIX Public Directory, under a directory by the name of your username, you will find the following file:

• puzzle

Copy this file into your project directory.

There is also a MANIFEST file in the parent directory (the pct0/ sub-directory), which will contain MD5sums of the various puzzle keys, provided to help you in verifying your puzzle key.

In the TASK file, you will be given a set of source quantities/units and destination units you need to construct command-lines for to solve, using the tools provided to you in the unittools suite. You will also want to make use of pipes and command expansions in your solutions.

As an example, take the following (formatted is it would appear in your TASK file):

task: 0
precision: 3
input_value: 35
input_unit: cm
output_unit: ft
chain_delim: " >> "

With these in mind, we can set about solving this problem, using the tools in combination to arrive at the desired result.

A potential solution would be as follows:

lab46:~/src/unix/upf1$ echo "35" | cm2in
13.78 in
lab46:~/src/unix/upf1$ echo "13.78" | in2ft
1.15 ft
lab46:~/src/unix/upf1$ 

But that's not very optimized. We could instead do something better, like:

lab46:~/src/unix/upf1$ echo "35.000 cm" | cm2in OPTIONS | in2ft OPTIONS
35.000 cm >> 13.780 in >> 1.148 ft
lab46:~/src/unix/upf1$ 

Note that with precision, you need to instruct the tools to display out to the specified number of decimal places.

We'll want to record our command-lines in the appropriate place. Since this is task #0, we'll want to record it in a file called: task0.cli, which should basically just contain the command-lines you ran, merely placed in a text file.

For final output, we'll want to display the chain of conversions we went through. So, for this above example:

lab46:~/src/unix/upf1$ ./task0.cli
35.000cm >> 13.780in >> 1.148ft
lab46:~/src/unix/upf1$ 

Basically, each task#.cli will be its own script. If we were to execute it, the correct result should be produced.

As this project focuses more on the critical thinking process than being heavy in unravelling a problem using UNIX commands, your solution will be in 2 parts:

• your puzzle key, in a textfile called 'key' containing ONLY the capital letters corresponding in order to the 0-9 values.
• your documentation of your solving and exploration of the puzzle. If you did this on paper, I'll want it digitized and submitted as a file with this project.

As indicated, you are to place the determined key to your puzzle in a regular text file called 'key', and will contain ONLY the capital letters, in order from 0-9, of your puzzle.

For example, using the example puzzle above:

By successfully performing this project, you should have a set of task#.cli files (one for each task). You will want to submit these, along with a upf1steps file.

To submit this project to me using the submit tool, run the following command at your lab46 prompt:

$ submit unix upf1 upf1steps task*.cli
Submitting unix project "upf1":
    -> upf1steps(OK)
    -> task0.cli(OK)
    -> task1.cli(OK)
    -> task2.cli(OK)
    -> task3.cli(OK)
       ...

SUCCESSFULLY SUBMITTED

You should get some sort of confirmation indicating successful submission if all went according to plan. If not, check for typos and or locational mismatches.

I'll be looking for the following:

78:upf1:final tally of results (78/78)
*:upf1:upf1steps has valid list of non-interactive instructions [4/4]
*:upf1:upf1steps only copies/alters files if USER matches [4/4]
*:upf1:upf1steps builds the various task#.cli files it runs [4/4]
*:upf1:upf1steps obtains the latest pipemath release from site [4/4]
*:upf1:upf1steps only displays specified STDOUT output [4/4]
*:upf1:upf1steps resiliently creates local project directory [4/4]
*:upf1:upf1steps copies public dir data with absolute path [4/4]
*:upf1:upf1steps makes clear, effective use of wildcards [4/4]
*:upf1:upf1steps defines and uses custom variables [4/4]
*:upf1:upf1steps uses command expansions to get information [4/4]
*:upf1:upf1steps uses a loop to drive numbers in final output [4/4]
*:upf1:upf1steps automates the task when run [4/4]
*:upf1:all files are organized, clear, and easy to read [4/4]
*:upf1:task#.cli files output using specified precision [3/3]
*:upf1:task#.cli files output using specified delimiter [3/3]
*:upf1:task#.cli files output only correct value and unit [4/4]
*:upf1:task#.cli files output in specified format [4/4]
*:upf1:task#.cli files display no STDERR output [4/4]
*:upf1:task#.cli files have solution within given constraints [4/4]
*:upf1:task#.cli files only contain the solution command-line [4/4]

Additionally:

• Solutions not utilizing descriptive why and how comments will be subject to a 25% overall deduction
• Solutions not utilizing indentation to promote scope and clarity will be subject to a 25% overall deduction