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--- haas:spring2014:cprog:projects:dayofweek [2014/01/17 07:55] – [Background] wedge
+++ haas:spring2014:cprog:projects:dayofweek [2014/02/06 17:39] (current) – [Reflection] wedge
@@ Line 1: / Line 1: @@
+<WRAP centeralign round box>
+<WRAP><color red><fs 200%>Corning Community College</fs></color></WRAP>
+<WRAP><fs 150%>CSCS1320 C/C++ Programming</fs></WRAP>
+</WRAP>
+~~TOC~~
 ======Project: MENTAL MATH (DAY OF WEEK)======
-A project for C/C++ Programming.
 =====Objective=====
 To implement a programmatic solution (ie simulation) of a real life process- the mental math trick of determining what day of the week January 1 of any given year (in the 21st century) falls on.
-=====Prerequisites=====
+======Assumptions======
+To assist you in completing this project, you may make the following assumptions:
+  * all requests will be for days in the 21st century (2000-2099)
+=====Prerequisites/Corequisites=====
 In addition to the new skills required on previous projects, to successfully accomplish/perform this project, the listed resources/experiences need to be consulted/achieved:
@@ Line 15: / Line 26: @@
 The process in this case is one of simple (reduced) multiplication and mapping against a table. To wit:
-====Squaring a value====
+====Day values====
-Squaring is essentially multiplying a number by itself-
+For this trick to work, we need to be familiar with the following table (a map of days to numeric values):
-  * 5 squared (5<nowiki>^</nowiki>2) is 5*5 or 25
+^  Monday  ^  Tuesday  ^  Wednesday  ^  Thursday  ^  Friday  ^  Saturday  ^  Sunday  |
-  * 8 squared (8<nowiki>^</nowiki>2) is 8*8 or 64
+|  1  |  2  |  3  |  4  |  5  |  6  |  7 or 0  |
-While not outwardly a difficult procedure, the nature of multiplying multiple digit numbers in your head can quickly result in more steps (and more steps means more time).
+====Calculating day of the week based on year====
-Finding a shortcut through this process enables it to remain solidly within the realm of mental math, and makes for a good algorithm to practice implementing on the computer.
+Okay, time for the magic.
-This particular trick relies on a subset of the squares of double digit values (and appears to work on some triple digit values): those ending with a 5 (a five in the ones place).
+Let us try it on January 1st, 2014.
-The operational scope of this trick will be just those two-digit values ending with 5:
+===Step 1: Obtain last two digits of the year===
+In our example, we're working with 2014, the last two digits are therefore: 14
-  * 15
+You should be able to come up with a means of extracting this information in your program.
-  * 25
-  * 35
-  * 45
-  * 55
-  * 65
-  * 75
-  * 85
-  * 95
-====Squaring double digit values ending with 5====
+===Step 2: Compute 25% (drop the decimal)===
-The trick here is two-fold.
+Even this is something we can do in our heads. I can think of two approaches right off the bat:
-One, we square the 5 to get 25. That is how our result will end (so bam! we now have our tens and ones place already solved)
+==Approach 1: 10 + 10 + 5==
+% percent of anything is merely dropping the last digit. 10% of 54 is 5.
-Next, we take the value in the tens place, and multiply it by its increment:
+For our 2014 example, 10% of 14 is therefore 1.
-  * 1's increment (1+1) is 2, so 1*2
+So we need two 10 percents... 1.4 + 1.4 = 2.8
-  * 2's increment (2+1) is 3, so 2*3
-  * 3's increment (3+1) is 4, so 3*4
-  * 4's increment (4+1) is 5, so 4*5
-  * ...
-  * 9's increment (9+1) is 10, so 9*10
-We take this result and append the 25 after it.
+Finally, 5% is half of 10% (half of 1 is 0.5), so 1.4 + 1.4 + 0.5 = 3.3
-For example:
+But, since we do not care about the decimal, we drop it and are left with just 3.
-<code>
+==Approach 2: half of half==
-* 15 = 1*(1+1) 5*5
+% is a convenient value for us with respect to 100, allowing this optimized approach to work.
-        = 1*2     5*5
-        = 2       25
-        = 225
-</code>
-and:
+  * Half of 100 is 50 (50%)
+  * Half of 50 is 25 (25%) -- hence the "half of half"
-<code>
+So, 14 cut in half is 7.
-* 75 = 7*(7+1) 5*5
-        = 7*8     5*5
-        = 56      25
-        = 5625
-</code>
+cut in half is 3.5.
+Once again, dropping the decimal yields 3.
+===Step 3: Add 25% to year value===
+Once we have our 25% value, go and add it back to our two-digit year value:
++ 3 = 17
+===Step 4: Subtract the largest fitting multiple of 7===
+Some multiples of 7:
+|  0  |  7  |  14  |  21  |  28  |  35  |  42  |  49  |
+So, with a value of 17, what is the largest multiple of 7 that is still less than (or equal to) 17?
+Hopefully you identified the 14 as the likely candidate.
+- 14 = 3
+===Step 5: Look up day in table===
+We ended up with a 3 as the result for January 1st, 2014.
+Go and reference the 3 from that table... what day do we get? Does it match the actual day of the week for January 1st, 2014?
+<cli>
+lab46:~$ cal 01 2014
+    January 2014
+Su Mo Tu We Th Fr Sa
+  2  3  4
+  6  7  8  9 10 11
+13 14 15 16 17 18
+20 21 22 23 24 25
+27 28 29 30 31
+lab46:~$
+</cli>
+Pretty neat, eh?
+====Exception: Leap Years====
+In the event of a leap year, we simply subtract 1 from the 25% value, and continue on as usual.
+Makes sense, right? Leap years add a day, so something ends up being "off by one".
 =====Program=====
-It is your task to write the program that will use the above method to compute the square of the requested value ending with a 5.
+It is your task to write the program that will use the above method to determine the day of the week any given January 1st in the 21st century falls on.
 Your program should:
-  * prompt the user for the number (input)
+  * prompt the user for the four digit year (input)
   * perform the task (process)
   * display the final value (output)
@@ Line 84: / Line 121: @@
 <cli>
-lab46:~/src/cprog/squares$ ./squares
+lab46:~/src/cprog/dayofweek$ ./dayofweek
-Enter value: 75
+Which year: 2014
-x 75 = 5625
+January 1st, 2014 falls on: Wednesday
-lab46:~/src/cprog/squares$
+lab46:~/src/cprog/dayofweek$
 </cli>
@@ Line 95: / Line 132: @@
 Be sure to provide any commentary on your opus regarding realizations had and discoveries made during your pursuit of this project.
-  * Does this trick work (or can it be adapted to work) for three digit values ending with 5?
+  * 2000 by definition isn't a leap year, but how does this algorithm work for 2000 in general?
-  * How about 4 digits?
+  * Try this algorithm on some years in the 20th century and see how it fares.
+    * Is it correct? Is it correct for any of them?
+    * If it isn't correct, is it consistently off by a value?
+This isn't just about implementing a particular algorithm, it is about understanding an algorithm- its domain of correctness, and its limitations.
 =====Submission=====
 To successfully complete this project, the following criteria must be met:
@@ Line 115: / Line 155: @@
 <cli>
-$ submit cprog squares squares.c
+$ submit cprog dayofweek dayofweek.c
-Submitting cprog project "squares":
+Submitting cprog project "dayofweek":
-    -> squares.c(OK)
+    -> dayofweek.c(OK)
 SUCCESSFULLY SUBMITTED
@@ Line 124: / Line 164: @@
 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.
-====Verify submission====
-To verify you submitted successfully, you may run the following (from anywhere on lab46):
-<cli>
-lab46:~$ verify cprog squares
-squares: submitted successfully
-</cli>
-Note if automated assessment is available for the project, you may actually see results in the output as well.

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