Berkeley CS61 series equivalence?

<p>Hi guys, I'm looking for a community college that offer equivalent courses to the 61 series at Cal, but so far I found none. My college only has equivalent courses to Math 1A, 1B, linear algebra and DE, so even after I transferred I may need to spend another 3 years at Cal if I could not do any other preregs. Please give me a few suggestion, thanks.</p>

<p>Btw, I'm also looking for COMPSCI 70 and EL ENG 42's equivalence too, please let me know if any college in the Bay Area has some or all of these 5 courses. Thanks</p>

<p>Diablo Valley has 61B & C. De Anza has 2 classes that articulate as 61B but you need an addition Cal course. Berkeley CC has 2 classes that articulate as 61B. Man, 61A is hard to find.</p>

<p>The hard way: checking all the assist.org agreements.</p>

<p>I think the only Community College with CS 61A is Laney College.</p>

<p>If you are EE bound, then I’d recommend going to those colleges above. Cause taking one or two pre-req in CS61 series is a huge advatange in taking classes and saving time.
However if you are CS bound, even if you take those pre-req, id recommend (at least) just hearing the lecture. For ex. CS bound EECS txfers from DVC took cs61b/cs61c with other people and said they learned alot. (But they also said should be careful as cs61 series takes a tremendous time in general)</p>

<p>@killmyentourage and @samuelkim1990: Thanks guys, really appreciated. DBL is kind of far a way for me but De Anza is OK bcause I’m a Foothill student.</p>

<p>@sgtbretch: ugh… never heard of that school before but I will check ASAP. Thanks a lot.</p>

<p>Btw, is there any chance that these equivalent courses at DBL, Laney, etc. offered online guys?</p>

<p>Many community colleges in the area have a discrete math course that articulates to Berkeley Math 55, which is similar to Berkeley CS 70 (use assist.org for Berkeley Math or EECS instead of L&S CS to see Math 55 articulations). Check with the Berkeley EECS department about whether such a substitution is accepted.</p>

<p>Laney may have an EE 42 articulated course.</p>

<p>If you are at Foothill:</p>

<p>Foothill or De Anza Math 22 / CIS 18 = Berkeley Math 55
West Valley CIS 4A = Berkeley CS 61B</p>

<p>Laney is in Oakland.</p>

<p>Thanks for the info ucbalumnus. I guess I have to talk with Cal Admission to substitute their discrete math with my CIS 18/ Math 22. On assist it says that the course is not articulated…</p>

<p>Can’t believe the introductory course in C++ CIS 4A at West Valley articulate with Cal’s 61B O_o . Thanks again</p>

<p>Well, that’s what assist.org says for West Valley CIS 4A. You may want to check with Berkeley CS as to whether it is full or partial articulation for CS 61B (if only partial, you will have to take the self-paced CS 47B at Berkeley to complete CS 61B).</p>

<p>You can also preview Berkeley CS courses here:</p>

<p>[EECS</a> Course WEB Sites](<a href=“http://www-inst.eecs.berkeley.edu/classes-eecs.html]EECS”>CAS - Central Authentication Service)</p>

<p>nevermind.</p>

<p>We only have 61B in my cc :(. Wish we had more.</p>

<p>According to the EECS handbook, Math 55 (Discrete Math) taken at a CC will no longer satisfy the CS70 requirement if you’re an incoming transfer for Fall 2012 or later. </p>

<p>From my experience at Laney, they only have 61A in the Spring and 61B requires CIS25 which is offered year around and CIS27 which is only offered in the Summer.</p>

<p>If you live in the east bay, it’s definitely possible to satisfy the 61ABC series but you will have to do some driving or bart and bus.</p>

<p>^^^ I was going to mention that too. CS 70 is a new requirement starting Fall 2012 and can no longer be replaced by Math 55.</p>

<p>I thought CS 70 was already a requirement for most upper div courses in addition to Math 55… Even though I heard they’re pretty similar.</p>

<p>Discrete Math is only offered online at DVC (meaning there is one Discrete Math course offered almost every semester, and it is always online never on campus)…so if it is still a requirement, you can take it there.</p>