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Replace type parameters by their erasures, normally Object or the first declared bound, and insert checked casts where an erased result is used as its former specific type. A simple generics in Java example becomes:

// Generic version
Vector<String> names = new Vector<String>();
names.add("Ada");
String s = names.get(0);

// Non-generic version
Vector names = new Vector();
names.add("Ada");
String s = (String) names.get(0);

The two fragments are alternatives, not consecutive declarations in one scope. For a user-defined Box<T>, replace fields, parameters and returns of type T by Object, and cast the retrieved result. Bounded parameters instead erase to their bound. A full translation must also preserve method dispatch where erasure changes overridden signatures, using suitable forwarding methods; this is the principle of type erasure, not simply deleting every pair of angle brackets.

The raw collection permits accidental insertion of an unrelated object, which may fail much later at a cast. The generic collection rejects such insertion during compilation, documents its contract at the declaration, and eliminates repetitive casts at use sites. The generic version catches more errors earlier and expresses the programmer's intended element type directly. Generic types are largely erased at runtime, so their advantage is static checking rather than a new runtime vector representation. The translation mechanism is specified by the Java language specification.

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Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 5 / 3 / a / Solution by

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A generic class, interface or method in Java is parameterized by one or more types. Generics in Java allow one implementation of a collection or algorithm to be reused with different reference types while checking the allowed values and results statically. For example:

Vector<String> names = new Vector<String>();
names.add("Ada");
String first = names.get(0);

class Box<T> {
    private T value;
    public void put(T x) { value = x; }
    public T get() { return value; }
}

The type parameter is declared inside angle brackets, as in Box<T>, and supplied as a type argument, as in Box<String>. A generic method places its parameter list before its return type, for example static <T> T identity(T x). Bounds such as <T extends Number> restrict type arguments, and wildcards such as ? extends Number or ? super Integer describe restricted views. Primitive values require wrapper types such as Integer rather than a type argument int. Generics provide reusable code with compile-time checking of element types.

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Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 5 / 2 / d / Solution by

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Set-user-ID execution of a suitable executable makes its effective user identity that of the owner. When that owner is root, running it can give root privileges even to an otherwise unprivileged caller. World-write permission destroys the trust boundary: any user can alter the privileged executable, potentially substituting code which performs arbitrary privileged operations. World-execute permission then makes the compromised program available to everyone.

World-read access can reveal implementation details or embedded secrets, but reading alone is not the decisive privilege-escalation flaw; properly designed public set-user-ID utilities need not keep their code secret. Some Unix kernels clear the set-user-ID bit after an unprivileged modification. That is a mitigation, not a reason to maintain a world-writable privileged program: if the bit persists or is restored without validating the changed code, the privilege escalation remains. Root-owned privileged executables must not be writable by untrusted users.

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Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 5 / 2 / c / Solution by

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First, suppressing kernel preemption simplifies protection of shared kernel data on a uniprocessor: another ordinary kernel execution cannot be scheduled in the middle of a short update which temporarily breaks an invariant. Fewer paths require preemption-specific locking and reentrancy handling, making the operating-system kernel easier to implement correctly.

Second, it avoids involuntary kernel context switches and their scheduler, register-save and cache-disruption costs, and can make the execution of short kernel operations more predictable. The tradeoff is greater scheduling latency if a kernel operation runs too long. Simpler synchronization and lower switching overhead are the two reasons. Non-preemptive kernel code can still receive interrupts or block voluntarily, and multiprocessor execution still requires synchronization; non-preemption is not a substitute for all locking.

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Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 5 / 2 / b / Solution by

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Both kinds of context switch save and restore registers, a program counter, stack pointer and scheduler state. Switching between operating-system processes additionally changes the address-space and protection context, including the page-table root. Cached translations must be invalidated or distinguished by address-space identifiers, and the new process may have a different memory and cache locality.

Software threads of one process share its address space and most resources, so a thread switch normally preserves the page-table root and valid cached translations; only the execution context, stack and thread-specific state need change. Thus a process switch requires address-space management which a same-process thread switch usually avoids. Tagged translation caches can reduce the extra cost, so a complete translation lookaside buffer flush is not an inevitable feature of every process switch.

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Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 5 / 2 / a / Solution by

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The working set

W (t, Δ)

is the set of virtual pages referenced by an operating-system process within a recent window of

Δ

execution time or memory references. Its size estimates the number of frames needed for the process's current locality of execution. It is a set of recently used pages, not the entirety of the process's address space.

An operating system can estimate it from reference bits or fault history and try to keep those pages resident. Process scheduling and admission control should avoid simultaneously running processes whose combined working sets exceed physical memory; suspending some processes or reducing multiprogramming then allows the rest to make useful progress. If too many active pages compete for too few frames, repeated page faults produce memory thrashing. Working sets connect memory allocation to the choice of processes allowed to run.

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Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 5 / 12 / ii / Solution by

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The Fermat-Euler theorem states that for integers

m \geq 1

and

a

with

g cd (a, m) = 1

a^{φ (m)} \equiv 1 (mod m),

(1)

where Euler's totient function

φ (m)

counts the residue classes coprime to

m

. The case

m = 1

is trivial. For

m \geq 2

, list a reduced residue system

r_{1}, \dots, r_{φ (m)}

. Multiplication by

a

permutes these classes: each

a r_{i}

is coprime to

m

, and

a r_{i} \equiv a r_{j}

implies

r_{i} \equiv r_{j}

since

a

has a multiplicative inverse modulo

m

. Multiplying the permuted residues gives

a^{φ (m)} \prod_{i} r_{i} \equiv \prod_{i} r_{i} (mod m) .

(2)

The product is itself coprime to

m

and can be cancelled using its inverse, proving the theorem.

The integer 359 is prime: none of the primes

2, 3, 5, 7, 11, 13, 17

, all the primes not exceeding

359 < 19

, divides it. Therefore

φ (359) = 358

and 60 is coprime to 359. Using the supplied congruence and the Fermat-Euler theorem,

1 0^{179} \equiv (6 0^{2})^{179} = 6 0^{358} \equiv 1 (mod 359) .

(3)

Now express the decimal expansion by long division. Put

r_{0} = 1

and for

n \geq 1

define

10 r_{n - 1} = 359 a_{n} + r_{n}, 0 \leq a_{n} \leq 9, 0 \leq r_{n} < 359.

(4)

Thus

r_{n} \equiv 1 0^{n} (mod 359)

, and none of these remainders is zero. The congruence

1 0^{179} \equiv 1

implies

r_{n + 179} \equiv r_{n}

; since both are the unique representatives between zero and 358, they are equal. A digit is determined by the preceding remainder, namely

a_{n} = ⌊ 10 r_{n - 1} /359 ⌋

, so

a_{n + 179} = a_{n} for every n \geq 1.

(5)

This proves the required pure periodicity from the first digit. In fact its least period is 179: the multiplicative order of 10 modulo 359 divides the prime 179 and is not one, since

10 \neq \equiv 1 (mod 359)

. The connection between the remainder recurrence and the period is the decimal period from multiplicative order.

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Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 5 / 12 / i / Solution by

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Suppose

p

is a prime number,

p ∣ x y

and

p ∤ x

. Since the only positive divisors of

p

are one and

p

, the greatest common divisor of

p

and

x

is one. By Bézout's identity, integers

u, v

exist with

u p + vx = 1

. Multiplying by

y

gives

u p y + vx y = y

. Both terms on the left are divisible by

p

, so

p ∣ y

. Thus Euclid lemma gives

p ∣ x y ⟹ p ∣ x or p ∣ y .

(1)

The fundamental theorem of arithmetic says that every positive integer greater than one is a product of prime numbers, uniquely up to their order. Existence follows by strong mathematical induction: if

n

is prime it already has a factorization; otherwise

n = ab

with

1 < a, b < n

, and the factorizations of

a, b

combine to factor

n

For uniqueness, suppose

p_{1} \dots p_{r} = q_{1} \dots q_{s}

are two prime factorizations. Repeated application of Euclid lemma makes

p_{1}

divide one of the

q_{j}

, so primality forces

p_{1} = q_{j}

. Reorder that factor to the front, cancel it and repeat. Neither side can retain extra primes when the other has become one, since every prime exceeds one. The two multisets are therefore identical. For a negative integer there is additionally the uniquely determined sign; one has the empty positive factorization, and zero is excluded. The prime-product divisibility lemma is the step which makes factorization unique.

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Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 5 / 10 / c / Solution by

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Add a public override of toString in each concrete subclass. In TypeVar:

public String toString() {
    return "t" + id.toString();
}

In Arrow:

public String toString() {
    return "(" + from.toString() + " -> " + to.toString() + ")";
}

Inherited dynamic dispatch selects the proper method at every node of the abstract syntax tree. Full parentheses distinguish the domain and codomain nesting without relying on a reader's precedence convention. Repeated occurrences of one type variable use the same identifier, and distinct factory variables have distinct identifiers. For example, with the first three factory variables, the type from part (b) prints as ((t1 -> t2) -> (t2 -> t3)). The recursive textual form preserves both arrow structure and variable identity.

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Introduce a checked exception class, for example in its own source file:

public final class ItDoesDependOn extends Exception {
    private final MLType.TypeVar variable;
    public ItDoesDependOn(MLType.TypeVar variable) {
        this.variable = variable;
    }
    public MLType.TypeVar getVariable() { return variable; }
}

Declare the required abstract method in MLType:

public abstract void mustNotDependOn(TypeVar a)
    throws ItDoesDependOn;

Within the nested TypeVar class implement it by an identity check:

public void mustNotDependOn(TypeVar a) throws ItDoesDependOn {
    if (this == a) throw new ItDoesDependOn(a);
}

Within the nested Arrow class implement it recursively:

public void mustNotDependOn(TypeVar a) throws ItDoesDependOn {
    from.mustNotDependOn(a);
    to.mustNotDependOn(a);
}

The parameter denotes an actual non-null type variable returned by the factory; checking object identity distinguishes different variables without confusing their printed names. Structural induction on the abstract syntax tree proves the occurs check: a variable node contains the queried variable exactly when the identities match, and an arrow contains it exactly when either component does. The first detected occurrence throws and propagates the checked exception; normal return proves absence. Both subclass methods must declare the exception, and callers must catch or declare it.

For

(α \to β) \to (β \to γ)

, either recursive branch detects each of

α, β, γ

, whereas a distinct

δ

is absent and the call returns. A recursive identity-based occurs check throws exactly when the queried variable occurs.

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Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 5 / 10 / a / Solution by

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Represent the two constructors of the type grammar by subclasses of MLType: a TypeVar node and an Arrow node containing two types. Use immutable fields so a completed type expression cannot acquire cycles through later mutation. The static factory allocates a fresh object and gives it a fresh identifier shared across the whole hierarchy:

import java.math.BigInteger;

public abstract class MLType {
    private MLType() { }
    private static BigInteger next = BigInteger.ONE;

    public static synchronized TypeVar createNewTypeVar() {
        TypeVar answer = new TypeVar(next);
        next = next.add(BigInteger.ONE);
        return answer;
    }
    public static final class TypeVar extends MLType {
        private final BigInteger id;
        private TypeVar(BigInteger id) { this.id = id; }
    }
    public static final class Arrow extends MLType {
        private final MLType from, to;
        public Arrow(MLType from, MLType to) {
            if (from == null || to == null)
                throw new IllegalArgumentException();
            this.from = from; this.to = to;
        }
    }
}

The nested classes are public for use by clients, while their fields are private; TypeVar can be created only through the factory. Client code uses MLType.TypeVar and MLType.Arrow. A type variable is identified by its unique node, and a function type by its domain and codomain nodes; repeated occurrences of one logical variable share the same TypeVar object.

The monotonically increasing BigInteger identifier avoids fixed-width counter overflow, and synchronization prevents two simultaneous calls from receiving the same identifier. Actual allocation remains subject to available memory. This is an abstract syntax tree representation, possibly sharing subexpressions. Fresh variable nodes and recursive arrow nodes represent every type in the given grammar.

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Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 5 / 1 / c / Solution by

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Assume head comparison and construction of one list cell take constant time. In the merge algorithm, every comparison removes one head from the recursive inputs, and evaluation stops as soon as either is empty. For lengths

m, n > 0

, at most

m + n - 1

comparisons and the same number of newly constructed cells are required; the untouched suffix is shared. Inputs which keep both lists nonempty until the last comparison attain this bound. Thus

T_{merge} (m, n) = O (m + n), with worst case Θ (m + n) .

(1)

An initially empty input makes this particular implementation constant-time; not every pair requires linear work.

The balanced merge sort spends

Θ (n)

time per call on list length, splitting and merging. Hence its time complexity satisfies

T (n) = T (⌊ n /2 ⌋) + T (⌈ n /2 ⌉) + Θ (n), T (0), T (1) = Θ (1) .

(2)

At each recursion level the lengths of its disjoint subproblems sum to

n

, so the work per level is

Θ (n)

, and balanced splitting gives

Θ (lo g n)

levels. Consequently

T_{mergesort} (n) = Θ (n lo g n) (n \geq 2) .

(3)

The length and splitting traversals change the constant factor, not this asymptotic order; repeatedly inserting elements into a sorted list would not give the same complexity.

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Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 5 / 1 / b / Solution by

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Split the list into two contiguous halves, sort each recursively, and use the merge algorithm from part (a):

fun mergesort [] = []
  | mergesort [x] = [x]
  | mergesort xs =
      let
        val k = List.length xs div 2
        val left = List.take (xs, k)
        val right = List.drop (xs, k)
      in
        merge (mergesort left, mergesort right)
      end;

List.length counts elements, List.take(xs,k) returns the first

k

elements and List.drop(xs,k) returns the remaining suffix. Since

0 < k < n

when

n \geq 2

, both recursive inputs are smaller and differ in length by at most one. Mathematical induction on length proves that each half is sorted without losing elements, and the correctness of the merge algorithm proves the same for the combined result. Contiguous splitting and left-biased equal-key merging preserve the original relative order of equal keys. This is a stable top-down merge sort.

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Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 5 / 1 / a / Solution by

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For integer linked lists, compare their heads and copy the smaller head, retaining the other list for the next step:

fun merge ([], ys) = ys
  | merge (xs, []) = xs
  | merge (xs as x :: xt, ys as y :: yt) =
      if x <= y then x :: merge (xt, ys)
      else y :: merge (xs, yt);

This merge algorithm preserves duplicates and chooses the left list on a tie, making it stable when elements have an ordering key. Each recursive step decreases the combined length. For correctness, the chosen head is no greater than either remaining head and therefore no greater than any remaining element; structural induction then proves that the result is ordered and contains exactly the elements of both inputs. For other element types, replace the integer comparison by a supplied ordering function. Merge repeatedly emits the smaller head and reuses the remaining list when the other becomes empty.

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Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 4 / 9C / ii / Solution by

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Along the holonomic constraint

z = x^{2} / (4 a)

, we have

\overset{z}{˙} = x \overset{x}{˙} / (2 a)

and hence speed squared

(1 + x^{2} / (4 a^{2})) \overset{x}{˙}^{2}

. The smooth wire's normal reaction is perpendicular to the permitted velocity and does no work. Conservation of mechanical energy, with release from rest at height

h

, gives

\frac{1}{2} m (1 + \frac{x ^{2}}{4 a ^{2}}) \overset{x}{˙}^{2} + \frac{m g x ^{2}}{4 a} = m g h .

(1)

The turning points are

x = \pm x_{0}

with

x_{0} = 2 ah

. One full oscillation has twice the travel time between them:

T = 2 \int_{- x_{0}}^{x_{0}} \frac{1 + x ^{2} / ( 4 a ^{2} )}{2 g [ h - x ^{2} / ( 4 a )]} d x .

(2)

Set

x = x_{0} u

and

β = h / a

. After cancelling the scale factors,

T = 2 \frac{2 a}{g} \int_{- 1}^{1} \frac{1 + β u ^{2}}{1 - u ^{2}} d u,

(3)

so the bead on a parabolic wire has

G (β) = 22 \int_{- 1}^{1} \frac{1 + β u ^{2}}{1 - u ^{2}} d u .

(4)

For

β ≪ 1

, expand the numerator uniformly on

[- 1, 1]

1 + β u^{2} = 1 + \frac{1}{2} β u^{2} + O (β^{2}) .

(5)

The integrable endpoint weight gives

\int_{- 1}^{1} (1 - u^{2})^{- 1/2} d u = π

and

\int_{- 1}^{1} u^{2} (1 - u^{2})^{- 1/2} d u = π /2

, for example by

u = sin θ

. Therefore the requested small-oscillation period is

T = 2 π \frac{2 a}{g} [1 + \frac{h}{4 a} + O ((h / a)^{2})] .

(6)

The leading frequency is

g / (2 a)

, and the first finite-amplitude correction increases the period. At

h = 0

the ring is stationary; the expression is the limiting period of nonzero small oscillations.

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Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 4 / 9C / i / Solution by

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The relevant parameters are

a, h, g

and the ring mass

m

, with dimensions length, length, length/time

^{2}

and mass. The wire is fixed and smooth, so there is no additional friction parameter. Because

m

is the only parameter carrying a mass dimension, it cannot enter a dimensionless group governing the period. Using

a

and

g

as the length and time scales, the independent dimensionless variables are

T g / a

and

h / a

. Dimensional analysis therefore gives

T = \frac{a}{g} G (h / a) .

(1)

The physical mass cancellation also follows from the energy equation in part (ii), since kinetic and gravitational potential energy are both proportional to

m

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Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 4 / 8E / ii / Solution by

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The Fermat-Euler theorem states that, for a positive integer

q

and an integer

a

coprime to

q

a^{φ (q)} \equiv 1 (mod q)

, where the Euler totient function

φ (q)

counts residue classes coprime to

q

. The case

q = 1

is trivial. For

q > 1

, let

r_{1}, \dots, r_{φ (q)}

be a reduced residue system. Multiplication by

a

permutes it: equality of two resulting residues can be cancelled using the modular inverse of

a

, and the products remain coprime to

q

. Multiplying the residues before and after the permutation gives

a^{φ (q)} r_{1} \dots r_{φ (q)} \equiv r_{1} \dots r_{φ (q)} (mod q) .

(1)

Every factor on the right is a unit, so its product can be cancelled. This proves

a^{φ (q)} \equiv 1 (mod q) .

(2)

The integer

359

is prime: none of the primes

2, 3, 5, 7, 11, 13, 17

up to

359 < 19

divides it. Therefore

φ (359) = 358

. Apply the theorem to

60

and use the supplied congruence:

1 0^{179} \equiv (6 0^{2})^{179} = 6 0^{358} \equiv 1 (mod 359) .

(3)

To connect this to individual digits rather than merely assert periodicity, use long-division remainders

r_{0} = 1

and

10 r_{n - 1} = 359 a_{n} + r_{n}, 0 \leq r_{n} < 359.

(4)

Induction gives

r_{n} \equiv 1 0^{n} (mod 359)

. Since

10

is coprime to

359

, no remainder is zero. The congruence above therefore implies

r_{n + 179} = r_{n}

for every

n \geq 0

, using the unique representative in

1, \dots, 358

. In particular

a_{n + 179} = ⌊ \frac{10 r _{n + 178}}{359} ⌋ = ⌊ \frac{10 r _{n - 1}}{359} ⌋ = a_{n} .

(5)

Hence

a_{n + 179} = a_{n} (n \geq 1) .

(6)

This is the mechanism of decimal period from multiplicative order: the period divides

179

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Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 4 / 8E / i / Solution by

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If the prime

p

does not divide

x

, then

g cd (p, x) = 1

. The Bezout identity gives integers

a, b

with

a p + b x = 1

. Multiplying by

y

gives

a p y + b x y = y

. If

p ∣ x y

, both terms on the left are divisible by

p

, so

p ∣ y

. Thus Euclid lemma holds:

p ∣ x y ⟹ p ∣ x or p ∣ y .

(1)

For completeness, existence in the fundamental theorem of arithmetic follows by strong induction on a positive integer

n > 1

. If

n

is prime there is nothing to prove; otherwise

n = ab

with

1 < a, b < n

, and the inductive prime factorizations of

a, b

give one for

n

. The integer

1

has the empty factorization.

For uniqueness, suppose

p_{1} \dots p_{r} = q_{1} \dots q_{s}

are two prime factorizations. Repeated application of Euclid lemma shows that

p_{1}

divides some

q_{j}

. Both are primes, so they are equal. Reorder the second list, cancel this equal factor, and repeat. All factors must be matched, including their multiplicities. Thus prime factorization exists and is unique up to the order of its factors. For nonzero negative integers one also records the sign.

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