recursion for zipcoders
It's turtles all the way down. Recursion is a key idea in computer software, and is generally despised in production software. Nonetheless, you need to know it when you see it, and potentially talk your way around it in interviews.
It is very, very cool though.
First, read much of the contents of this page: Recursion in Java
Skip over the Gray code stuff, but read down to where the Exercises (6 and up) start.
Then:
Build out a factorial class and some tests.
Build out a GCD class and some tests.
Compare GCD and GCD2 to each other using timing statements.
Remember how to do timing in Java?
Use the Instant Class
The Instant class represents an instant on the timeline. Basically, it is a numeric timestamp since the standard Java epoch of 1970-01-01T00:00:00Z.
In order to get the current timestamp, we can use the Instant.now() static method. This method allows passing in an optional Clock parameter. If omitted, it uses the system clock in the default time zone.
We can store start and finish times in two variables, as in previous examples. Next, we can calculate time elapsed between both instants.
We can additionally use the Duration class and it's between() method to obtain the duration between two Instant objects. Finally, we need to convert Duration to milliseconds:
Instant start = Instant.now();
// CODE to be timed goes HERE
Instant finish = Instant.now();
long timeElapsed = Duration.between(start, finish).toMillis();If you're up to it, Do a Longest Common subsequence algorithm. Implement the one in the princeton text.
Longest common subsequence problem.
Given two strings x and y, we wish to compute their (LCS). If we delete some characters from x and some characters from y, and the resulting two strings are equal, we call the resulting string a common subsequence. The LCS problem is to find a common subsequence of two strings that is as long as possible. For example, the LCS of GGCACCACG and ACGGCGGATACG is GGCAACG, a string of length 7.
- - G G C - - A - C C A C G A C G G C G G A T - - A C G
Longest common subsequence recurrence.
Now we describe a recursive formulation that enables us to find the LCS of two given strings s and t. Let m and n be the lengths of s and t, respectively. We use the notation s[i..m) to denote the suffix of s starting at index i, and t[j..n) to denote the suffix of t starting at index j.
- If s and t begin with the same character, then the LCS of s and t contains that first character. Thus, our problem to reduces to finding the LCS of the suffixes s[1..m) and t[1..n).
- If s and t begin with different characters, both characters cannot be part of a common subsequence, so can safely discard one or the other. In either case, the problem reduces to finding the LCS of two strings—either s[0..m) and t[1..n) or s[1..m) and t[0..n).
In general, if we let opt[i][j] denote the length of the LCS of the suffixes s[i..m) and t[j..n), then the following recurrence holds:
opt[i][j] = 0 if i = m or j = n
= opt[i+1][j+1] + 1 if s[i] = t[j]
= max(opt[i][j+1], opt[i+1][j]) otherwise
Test with
S1 = GGCACCACG S2 = ACGGCGGATACG
should be GGCAACG
and
S1 = GCCCTAGCG S2 = GCGCAATG