*some questions require pretty much one-liner answers (e.g., from 13.1, even with explanations) - keep this in mind!*

also, most problems in the book offer interesting examples and you should at least try to read these problems (e.g. in chapter 13); same goes for the extra-credit problems in the end.

**5.4-1****5.4-4**(this problem might take you longer, and we'll grade it kindly - I want all of you to try your best on it)**11.2-1**

**B.5-2**

**12.2-4****12-2**(do not forget to prove your algorithm's correctness and linear performance)

**13.1-4****13.1-5****13.1-6****13.3-2***(draw corresponding 2-3-4 tree in the intermediate steps)*

**14-1-4****14.1-6**(this is a slight twist on the question we considered in the class)**14.3-5**

**Extra credits:**

**5-1**(*extra credit*; this is a very important and useful technique - try to read the problem even if you do not do it)- 11.4-4 (extra credit)
**12.1-3***-extra credit***13.2-4**-*extra credit***13.3-5**-*extra credit***13.3-6**-*extra credit***(suggest specific pseudo-code changes;**hint: instead of the*parent*__pointer__implement*parent*__procedure/method__**)****13.4-7**-*extra credit***(think of the simplest example)****14.1-8**-*extra credit*

## FAQ:

Here is my exchange with a student, which I think might be useful to others (the bottom paragraph is the most important):

> -----Original Message-----

>

> > > Hi Professor, Problem11.2-1seems deceptively simple, I think I may

> > > be reading wrong into it. My reasoning is that if given uniform hashing

> > > we have each element landing into specific spot. So if there are >= slots

> > > there shouldn't be any collision. Otherwise #of collisions is simply

> > > #elements - #slots. Am I missing something here, this one is

> > > bugging me. Thanks in advance.

> >

> > Indeed, say, you have exactly the same numbers of slots and elements. Then

> > assuming you had no collisions up to the last element, out of n possible

> > outcomes of the hash function only 1 is non-collision. So the probability of

> > that is 1/n, and with probability 1-1/n youwillhave a collision.

>> (so, if n=100, the probability of collision would be at least 99%)

> > I think you are misunderstanding what uniform means: re-read it in the book,

> > but intuitively, it means that each element can be hashed into any slot with

> > the same probability (for each element, imagine rolling a large enough dice

> > to determine which slot it will get into).

> Load factor A would indicate how "overloaded" a slot is on the average.

> So, is A = 1.3 there's 1.3-1=.3 "extra" element in the slot-chain on the

> average. So if we have M slots than there are M*.3 collisions. Or

> (n/m - 1)*m=n-m

> Am I overlooking some cases?

I think you understand the (over)load correctly. But still, try to plug in

some simpler numbers - if you say that the load factor is 1, then your

argument would mean that there are no collisions, but in the previous email

I argued that still collisions are very likely even in this case. In fact,

the so called birthday paradox says that at least one collision is very

likely even if the size of the table m is m=n^2, i.e. the load factor is

1/n.

So, what you need to do for this problem is to carefully write down the

probability of a collision for each pair of input values, and then look up

the expected value definition and try to write it for this case.