**A:** Solution meets the stated requirements and is completely
correct. Presentation is clear, confident, and concise.

**B:** The main idea of the solution is correct, and the
presentation was fairly clear. There may be a few small mistakes in
the solution, or some faltering or missteps in the explanation.

**C:** The solution is acceptable, but there are
significant flaws or differences from the stated requirements. Group
members have difficulty explaining or analyzing their proposed
solution.

**D:** Group members fail to present a solution that
correctly solves the problem. However, there is clear evidence of
significant work and progess towards a solution.

**F:** Little to no evidence of progress towards
understanding the problem or producing a correct solution.

Problem | Final assessment |

1 | |

2 | |

3 | |

4 |

**Instructions:**
Review the course honor policy: you may not use any human sources
outside your group, and must document anything you used that's not
on the course webpage.

This cover sheet must be the front page of what you hand in. Use separate paper for the your written solutions outline and make sure they are neatly done and in order. Staple the entire packet together.

(Caution: mathematical enrichment irrelevant to the assignment.)
Integers that can be written as the sum of two squares have
interested mathematicians for a long time. For example, Fermat
showed that a prime number $p$ is the sum of two perfect squares
if and only if $p-1$ is divisible by 4.

The following algorithm takes a given integer $n$ and determines
whether it can be written as the sum of two perfect squares. If so,
it returns $a$ and $b$ such that $n = a^2 + b^2$. Otherwise, it
returns "NO".
Algorithm:sumsq(n) Input : n, a non-negative integer Output: (a,b), where a and b are integers such that n = a*a + b*b, if such integers exist, 'NO' otherwise a = 0, b = n s = a*a + b*b while a <= b and s != n: if s < n a = a + 1 else b = b - 1 s = a*a + b*b if s == n return (a, b) else return 'NO'

- Use a loop invariant to show that this algorithm is correct. State the invariant, then go through the three steps from class to show correctness.
- Determine the worst-case running time of the algorithm. Give a $\Theta$ bound on the number of primitive operations, in terms of the input integer $n$, and simplify as much as possible. Show your work.
- Develop an improved algorithm that solves the same problem. Present your new algorithm, briefly explain why it is correct (you do not have to do a formal proof with a loop invariant), and state the worst-case running time.

- It takes exactly one minute for a pair of people to meet each other.
- There is plenty of space in the room, and people can move instantly from one stranger to another.
- Each individual can only meet one other person at a time, but multiple meetings can be happening simultaneously in the room.

For example, if $n=4$, if we call the people A, B, C, D, then you could have everyone meet everyone else in the following algorithm:

- A and B meet, and C and D meet
- A and C meet, and B and D meet
- A and D meet, and B and C meet

Your task is to develop an efficient algorithm if there are any number $n$ of people, not just 4. It doesn't have to match with my algorithm for $n=4$; that's just an example to see how it could be done. Answer the following questions to get you on the right track:

- When $n=4$ above, there are a total of 6 meetings that take place. How many meetings are required for any $n$? Your answer should be a formula in terms of $n$.
- Only two people can meet at once, so for example when $n=4$ only 2
meetings can take place at once, and therefore the 3-minute solution
above is the best possible for $n=4$.
Using this same logic, what would be the least possible number of minutes to meet for any value of $n$? (Again, your answer should be a formula in terms of $n$.)

- Imagine you have split everyone up into two equal-sized groups $G_1$
and $G_2$. Everyone within $G_1$ has met everyone else in $G_1$, and
the same goes for $G_2$, but no one from $G_1$ has met anyone from
$G_2$ yet.
Describe an algorithm for everyone in $G_1$ to meet everyone in $G_2$. If there are $n/2$ people in each group, your algorithm should take $n/2$ minutes.

**Hint:**look up how "Speed Dating" works. - Come up with an efficient algorithm for everyone in the whole group
of $n$ people to meet each other. You might be able to use your
algorithm from part (c) as a useful subroutine, but that's not
required. (There are many efficient ways to solve this problem.)
Try to get an exact formula in terms of $n$ for how many minutes your algorithm will take. Ideally, your formula should match your lower bound from part (b), or at least it should have the same big-oh.

**lmove()**, which moves one tape square to the left,**rmove()**, which moves one tape square to the right, and**isblank()**, which returns true if the current tape square is blank, and false otherwise.

- Present an algorithm for the FindX problem.
- Analyze the worst-case cost of your algorithm. The "measure of difficulty"
will be $d$, the distance (in tape squares) of the square
containing X from the square you start at.
So if the X is on the square you start at, $d = 0$. If it's
on one of the squares adjacent to the start square, the
distance is 1.
The "cost measure" will be the total number of moves,
i.e. the total number of calls to lmove() or
rmove(). [Remember: "lmove(k)" and "rmove(j)" are just
short-hand for multiple calls to lmove()/rmove().]
Give a big-O bound on the worst-case number of moves
in terms of $d$. The best solutions
will have worst-case cost $O(d)$.
**Note:**this difficulty measure is not typical! Since there is no "input" to this algorithm, the measure of difficulty cannot be the input size. And the cost measure is also unusual - we're not counting primitive operations in the algorithm, rather the number of moves made. -
Asymptotics are great and all, but in this case the "hidden
constants" in the big-O really matter — tape moves on
my Turing Machine are really slow!
Refine your analysis from (b) to give an exact upper bound on the number
of miles sailed (i.e.
*not*a big-O bound). There will be a*tangible prize*for the group who gets the smallest coefficient in front of $d$.

And here is a bubbleDown algorithm that you should use:Algorithm:heapSort(A,n) Input : array A of n elements Output: leaves A is sorted order for i from n-1 to 0 do ← This loop does the heapify operation bubbleDown(A,i,n) for i from n-1 to 0 do ← This loop does the heap removals. swap(A,0,i) bubbleDown(A,0,i)

Algorithm:bubbleDown(A,start,end) Input : A - an array, start - the index of the heap root, end - the index of the last/rightmost element of the heap Note: A[start],...,A[end] is a heapexceptthat the element at the root might be out of place (hence the call to bubbleDown!) Output: at exit, A[start],...,A[end] is a heap more = true while more and 2*start + 1 ≤ end do ← as long as start is not a leaf, loop! child = 2*start + 1 ← set child to start's left child if child + 1 ≤ end and A[child+1] > A[child] ← set child to start's right child if that's larger child = child + 1 if A[start] < A[child] ← if A[start] out of place, push it swap(A, start, child) down to A[child] start = child else more = false ← start not out of place, we're done!

- Tell me the worst-case running time of InsertionSort, SelectionSort, and HeapSort. You should know this already.
- Analyze the number of *swaps* performed by a call to InsertionSort,
in the worst case. To do this, you first have to describe a
worst-case family of examples, i.e., what ordering of a list,
for *any* size, makes InsertionSort do the maximum number of swaps.
Second, you need to analyze the number of swaps that InsertionSort does on your worst case example. This should be an exact function of $n$, like $3n^3 - 6$ or something like that. Try to be as precise as possible.

Finally, turn your formula into a $\Theta$-bound on the number of swaps. - Repeat (b) for SelectionSort.
- Repeat (b) for HeapSort. It will be a little more difficult to get the exact analysis of the number of swaps, so for this one you just need to give a formula for an *upper bound* on the number of swaps. But still, try to make your formula as precise as possible.
- Show that *any* sorting algorithm that only uses swaps to move elements in the input array must always perform at least $\lceil n/2 \rceil$ swaps, in the worst case. Which of the three algorithms above comes closest to this lower bound?