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\li For each of the following functions, determine $f: \bb Z \times \bb
Z \to \bb Z,$ determine whether the function is:

\item{i)} onto and one-to-one

\item{ii)} onto but not one-to-one

\item{iii)} not onto but one-to-one

\item{iv)} neither onto nor one-to-one

\item{v)} not a function

{\startlist

\li $f(x,y) = 3x^2-2y.$

Neither onto nor one-to-one.
$1$ is unreachable, and $f(1,0) = f(-1,0).$

\li $f(x,y) = y! + x.$

Not a function. $y!$ is not defined on $y<0.$

\li $f(x,y) = 2x\cdot 3y.$

Neither onto nor one-to-one.
$1$ is unreachable, but $f(0,1) = f(1,0).$

\li $f(x,y) = x - y.$

Onto but not one-to-one
All integers are reachable, but $f(0,0) = f(1,1).$

\li $f(x,y) = |x| + y.$

Onto but not one-to-one
All integers are reachable, but $f(0,1) = f(1,0).$
}

\li Use the cashier's algorithm to make change using quarters, dimes,
nickels, and pennies for the following amounts of money. You do not have
to specifically show how you greedily formed change, but rather there
are many ways to make this change and only the cashier's/greedy
distribution will be accepted.

{\startlist

\li 43 cents

One quarter, a dime, a nickel, 3 pennies

\li 68 cents

Two quarters, a dime, a nickel, 3 pennies.

\li 98 cents

Three quarters, two dimes, 3 pennies
}

\li Imagine that a new coin that is worth exactly 4 cents has been
introduced to our existing currency system.  Prove or disprove the
statement: ``The cashier’s algorithm using quarters, dimes, nickels,
4-cent coins, and pennies and will always produce change using the
fewest coins possible.''

This is false. 8 cents is decomposed into 1 nickel and 3 pennies by the
greedy algorithm, but is decomposed as 2 4-coins, which is fewer coins.

\li State whether the following is True or False and explain your
reasoning for full credit:

{\startlist

\li Given two integers $x$ and $c,$ if $x+c < 10,$ then $\lfloor x/10
\rfloor = \lfloor (x+c)/10 \rfloor.$

False. For $x = 0$ and $c=-100,$ then $x+c < 10$ but $\lfloor
x/10\rfloor = 0 \neq -10 = \lfloor (x+c)/10\rfloor.$

\li Given two functions $f(x)$ and $g(x),$ if $f(g(x))$ is defined, then
$g(f(x))$ must also be defined.

False. Let $f: \bb N \to \bb R$ and $g: \bb N \to \bb N.$
$f\circ g: \bb N \to \bb R,$ but $g\circ f$ undefined.
}

\li For each of these function from $\bb R^+ \to \bb R,$ find the least
integer $n$ such that $f(x)$ is $O(x^n)$ if possible. If not, explain
why the function cannot be $O(x^n).$

{\startlist

\li $f(x) = x^2\sqrt x.$

This function is $O(x^3).$ $x \geq \sqrt x,$ and the product of
$O(f(x))$ and $O(g(x))$ is $O(f(x)g(x)).$

\li $f(x) = 2x^3 + x^2\log(3x)$

This function is $O(x^3).$ All polynomials are $O(x^n)$ where $n$ is the
highest degree in the polynomial.

\li $f(x) = {x^5 + x^4 \over x^4 + 5\log_5(5^x)}$

This function is $O(x).$ ($n=1$)
The top polynomial is $O(x^5)$ because polynomials are on the order of
their highest degree, and the bottom polynomial is $x^4 + 5x = O(x^4).$
A rational function $g/h$ is on the order of $O(g)/O(h),$ so $f = O(x).$

\li $f(x) = \log(1000)$

This function is $O(1).$ ($n=0$) Any constant is $O(1).$

\li $f(x) = {3^x\over x^3} + {x^3\over\log(x)}$

This function is not $O(x^n)$ for any $n.$
The exponential function grows faster than any polynomial function.
}

\li Consider the linear search algorithm as outlined in class. How many
values would 15 be compared to in the sequence $(1,7,13,22,44,15,7,9).$

It would be compared to 6 values and on the last comparison, 15 is
found and the algorithm exits.

\li Consider the binary search algorithm as outlined in class. (You must
use this exact version of the binary search algorithm):

{\startlist

\li List the values that 44 would be compared to in a search for 44 in
the following sequence: $(1,7,9,13,15,44,57,88).$ Make sure to include
the final value in the ``equality'' check as well.

We will examine $(13, 44, 15, 44),$ the 4th, 6th, and 5th elements until
$i=j=6,$ and we exit the loop and check we have the right element.

\li List the values that 103 would be compared to in a search for 103 in
the following sequence: $(3,8,17,21,44,73,88,101,113).$ Make sure to
include the final value in the ``equality'' check as well.

We examine $(44, 88, 101, 113),$ the 5th, 7th, and 8th elements until
$i=j=9,$ and we check $x = a_9$ and find that our number is not in the
list.
}

\li Prove or disprove the following statements.

{\startlist

\li $\lfloor 2x \rfloor = \lfloor x \rfloor + \lfloor x + 0.5 \rfloor,$
for all $x\in\bb R.$

{\bf Proof.}

Let $x\in\bb R.$
$x$ can be uniquely represented as $n+r$ where $0\leq r < 1,$ and
$n\in\bb Z.$

If $r < 1/2,$ $\lfloor x \rfloor = \lfloor x + 0.5 \rfloor = n.$
$2x = 2n + 2r,$ and $0\leq 2r < 1,$ so $\lfloor 2x \rfloor = 2n =
\lfloor x \rfloor + \lfloor x + 0.5 \rfloor.$

If $r\geq1/2,$ $\lfloor x \rfloor = n$ and $\lfloor x+0.5 \rfloor =
n+1.$ $2x = 2n + 1 + 2(r-1/2)$ and $0 \leq r-1/2 < 1,$ so $\lfloor 2x
\rfloor = 2n+1,$ so $\lfloor x \rfloor + \lfloor x + 0.5 \rfloor.$

\li $x^3 + x^2 + \log(x)$ is $O(x^3).$ (If you choose to prove this
statement you must do so using witnesses).

{\bf Proof.}

On $x \geq 1,$
$x^3 \geq x^3,$ $x^3 \geq x^2,$ and $x^3 \geq \log x,$ so
$3x^3 \geq x^3 + x^2 + \log x$ for $x \geq 1,$ so we know that $x^3 +
x^2 + \log(x) = O(x^3).$

\bye
