Computer Science Fundamentals: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

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1Formulas and methods

Number systems

Place values
binary 2k, octal 8k, hex 16k
To decimal
Σ dk·bk
From decimal
divide by the base, read the remainders bottom-up
Hex ↔ binary
one hex digit = 4 bits
Two’s complement
invert the bits and add 1

Propositional logic & proofs

Rows of a truth table
2ⁿ
Implication
p → q ≡ ¬p ∨ q ≡ ¬q → ¬p
Negated implication
¬(p → q) ≡ p ∧ ¬q
Biconditional
p ↔ q ≡ (p → q) ∧ (q → p)
De Morgan
¬(p ∧ q) ≡ ¬p ∨ ¬q, ¬(p ∨ q) ≡ ¬p ∧ ¬q
Distributive laws
p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r), p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Negated quantifiers
¬∀x P(x) ≡ ∃x ¬P(x), ¬∃x P(x) ≡ ∀x ¬P(x)
Modus ponens, modus tollens
p → q, p ⊢ q; p → q, ¬q ⊢ ¬p

Method

  1. Count the different variables: the truth table has 2ⁿ rows, starting with all of them 1.
  2. Give every subformula its own column and fill the columns from the inside out.
  3. Apply the rule of each connective: → is 0 only for 1 → 0, ↔ is 1 when both sides agree.
  4. For equivalences, compare the final columns row by row; for an argument, look only at the rows in which every premise is 1.
  5. To negate a quantified statement, move ¬ inwards, swap ∀ and ∃, and negate the inner formula.

Digital logic: gates, K-maps & flip-flops

Basic gates
A·B, A + B, A′, (A·B)′, (A + B)′, A ⊕ B
Boolean rules
X + X′ = 1, X·X′ = 0, X + XY = X, XY + XY′ = X
De Morgan
(X·Y)′ = X′ + Y′, (X + Y)′ = X′·Y′
Karnaugh map order
00, 01, 11, 10 (Gray code)
Group of 2ᵏ cells
loses k variables
4-to-1 multiplexer
F = A′B′I0 + A′BI1 + AB′I2 + ABI3
Flip-flops
D: Q⁺ = D, T: Q⁺ = T ⊕ Q, JK: Q⁺ = JQ′ + K′Q
Counters
n-bit: 2ⁿ states; mod-N: smallest n with 2ⁿ ≥ N; fout = fclk/N
Ripple counter
fmax = 1/(n·tpd)
Ring and Johnson counters
n and 2n states

Method

  1. For a circuit, work from the inputs to the output one gate at a time and write the value on every wire, or write its expression and fill in the truth table.
  2. For simplification, put the 1s (and the don’t-cares as X) into a Karnaugh map with Gray-code rows and columns: 00, 01, 11, 10.
  3. Cover the 1s with as few and as large groups of 1, 2, 4, 8 or 16 cells as possible, wrapping round the edges; each group is one product term made of the variables that stay constant. Group the 0s for the POS form.
  4. For flip-flops and counters, read the inputs at each active clock edge, write the state after every edge, and remember that an n-bit counter wraps round after 2ⁿ states.

Number theory & modular arithmetic

Division with remainder
a = q·n + r, 0 ≤ r < n
Euclid
gcd(a, b) = gcd(b, a mod b)
Least common multiple
lcm(a, b) = a·b/gcd(a, b)
Inverse
a·a⁻¹ ≡ 1 (mod n), gcd(a, n) = 1
Linear congruence
a·x ≡ b → x ≡ a⁻¹·b (mod n)
Euler’s totient
φ(n) = n·∏(1 − 1/p), φ(pᵏ) = pᵏ − pᵏ⁻¹
Fermat
aᵖ⁻¹ ≡ 1 (mod p)
RSA
n = pq, φ = (p − 1)(q − 1), d = e⁻¹ mod φ, c = mᵉ mod n, m = cᵈ mod n

Method

  1. Reduce every number modulo n first; remainders lie between 0 and n − 1.
  2. For a gcd, divide repeatedly with Euclid’s algorithm; the last nonzero remainder is the gcd.
  3. For an inverse, run the extended algorithm and keep the coefficient of a; add n if it is negative.
  4. Solve a·x ≡ b by multiplying with a⁻¹; combine two congruences by writing x = r₁ + m₁k.
  5. For powers, write the exponent in binary and square repeatedly; in RSA, d = e⁻¹ mod φ(n) with φ(n) = (p − 1)(q − 1).

Graph theory

Handshake lemma
Σ deg(v) = 2|E|
Complete graphs
|E(Kn)| = n(n − 1)/2, |E(Km,n)| = m·n
Regular graph
|E| = n·k/2
Tree and forest
|E| = n − 1, |E| = n − c
Leaves of a tree
L = 2 + Σ (deg(v) − 2) over deg(v) ≥ 3
Euler
circuit: all degrees even; path: exactly 2 odd degrees
Hamilton circuits in Kn
(n − 1)!/2
Dijkstra update
d(v) = min(d(v), d(u) + w(u, v))
Walks of length L
(Mᴸ)ij

Method

  1. Write down the degree of every vertex; the degrees add up to twice the number of edges.
  2. For Euler questions, check that the graph is connected and count the vertices with odd degree.
  3. For trees, use n − 1 edges and the handshake lemma.
  4. For Dijkstra, keep a table of tentative distances and settle the smallest unsettled vertex each round.
  5. For a minimum spanning tree, sort the edges by weight and skip every edge that closes a cycle (Kruskal), or grow one tree from a start vertex (Prim).

IPv4 subnetting

Hosts per subnet
2(32 − prefix) − 2 (/31: 2, /32: 1)
Block size
256 − (mask octet)
Network address
IP AND mask
Broadcast address
IP OR (NOT mask): all host bits set to 1
Subnets from n borrowed bits
2n

2Practice problems

1.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −88 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −888 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1000 0000 >> 3. What is the result's value R in decimal?

1000 0000>> 3
R =
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is (c ∨ a) ∧ (¬c ∧ ¬a) a tautology, a contradiction or a contingency? Click your answer.

(c ∨ a) ∧ (¬c ∧ ¬a)
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
¬rr¬r ∧ pp

The premises r → p; ¬p are given. Which conclusion follows from them? Click it.

r → p¬p
  1. 1¬r
  2. 2r
  3. 3¬r ∧ p
  4. 4p
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

Simplify F = A′BC + AC + A′C + A′C′ to its minimal sum-of-products form. How many literals L does the simplified expression contain?

F = A′BC + AC + A′C + A′C′
L =
6.Digital logic: gates, K-maps & flip-flopsHard
DQ3​Q3​DQ2​Q2​DQ1​Q1​DQ0​Q0​CLKSHIFT

A 4-bit shift register Q₃Q₂Q₁Q₀ holds 0111. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀, Q₀ is lost) and the serial input enters Q₃. The serial input is 1, 0, 1, 0, 1 at the 5 edges. Give the final state Q = Q₃Q₂Q₁Q₀ as a decimal number.

Q₃Q₂Q₁Q₀ = 01111, 0, 1, 0, 1
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 10 modulo 79: the number x with 0 < x < 79 and 10·x ≡ 1 (mod 79).

a = 10n = 79
10−1 mod 79 =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (943, 3)d = ?

An RSA key uses the primes p = 23 and q = 41 and the public exponent e = 3. Find the private exponent d (0 < d < φ(n)).

p = 23q = 41e = 3
d =
9.Algorithms & complexityMedium

A greedy algorithm:

  1. AMakes the locally best choice at each step
  2. BAlways gives the optimal solution
  3. CUses only recursion
  4. DTries every possible solution
10.Algorithms & complexityHard

Which statement about selecting the k-th smallest element of an unsorted array is correct?

  1. AQuickselect runs in O(log n) worst case, because it recurses on only one side of the pivot
  2. BSelection is only possible in O(n²)
  3. CSelection always needs Ω(n log n) because it requires sorting
  4. DQuickselect runs in expected O(n), and median-of-medians guarantees O(n) worst case
11.Data structuresMedium

A queue is implemented with two stacks: push onto an "in" stack, and pop from an "out" stack, refilling it from "in" when empty. What is the amortised cost per operation?

  1. AO(n²)
  2. BO(log n)
  3. CO(1)
  4. DO(n)
12.Data structuresHard

Under simple uniform hashing, what is the expected time of a search in a hash table with chaining, n keys and m buckets?

  1. AO(1 + α), where α = n / m
  2. BO(n)
  3. CO(log n)
  4. DO(m)
13.Graph theoryMedium
ABCDEEuler circuitEuler path, no circuitneither

Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.

|V| = 5|E| = 5
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
54129151820817ABCDEF

Find the total weight W of a minimum spanning tree of the weighted graph.

|V| = 6|E| = 9
W =
15.Databases & SQLMedium

What does SELECT name, CASE WHEN grade >= 5.5 THEN 'pass' ELSE 'fail' END FROM results return?

  1. AEach name with the label 'pass' or 'fail' depending on the grade
  2. BAn error because CASE is not allowed in SELECT
  3. COnly the students who passed
  4. DThe number of students who passed and failed
16.Databases & SQLHard

Two transactions both read stock = 10, each subtracts 1 and writes the result, giving 9 instead of 8. Which anomaly is this?

  1. ALost update
  2. BDeletion anomaly
  3. CDirty read
  4. DPhantom read
17.Computer networksMedium

What is the network address of the host 192.168.10.77/26?

  1. A192.168.10.64
  2. B192.168.10.128
  3. C192.168.10.0
  4. D192.168.10.76
18.Computer networksHard

DNS queries normally use UDP. When does DNS typically use TCP instead?

  1. AFor responses too large for UDP and for zone transfers
  2. BOnly for reverse lookups of IP addresses
  3. COnly when the domain name is longer than ten characters
  4. DFor every query sent from a web browser
19.IPv4 subnettingMedium
/22 → /3011111111111111111111111111111100network bitshost bits

The network 10.171.92.0/22 is divided into equal subnets with prefix /30. How many subnets N are there?

10.171.92.0/22/30
N =
20.IPv4 subnettingHard
last octet = 241112816413211608040211networkhost

A host has the address 10.20.227.241/29. What is the fourth octet x of the last usable host address in its subnet?

10.20.227.241/29
x =

3Solutions

1.Number systems

Answer: U = 168

Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.

Invert and add one

88 = 0101 1000
inverted: 1010 0111
+1 → 1010 1000

Read it as unsigned

1010 1000₂ = 256 − 88 = 168

Answer

U = 168
2.Number systems

Answer: R = 16

Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.

Bit by bit

Shifting right by 3 moves every bit 3 places to the right; bits that fall off the right end are lost and zeros come in on the left.

1000 0000 >> 3
= 0001 0000

Value of the result

0001 0000₂ = 16

Answer

R = 16
3.Propositional logic & proofs

Answer: contradiction

Logical equivalence: two formulas are equivalent (≡) when they have the same value in every row of the truth table. A tautology is 1 in every row, a contradiction is 0 in every row, and a contingency is neither. Useful laws: De Morgan, ¬(a ∧ b) ≡ ¬a ∨ ¬b and ¬(a ∨ b) ≡ ¬a ∧ ¬b; a → b ≡ ¬a ∨ b ≡ ¬b → ¬a (contrapositive); ¬(a → b) ≡ a ∧ ¬b; and the distributive laws, a ∨ (b ∧ c) ≡ (a ∨ b) ∧ (a ∨ c).

Truth table row by row

a = 1, c = 1: c ∨ a = 1, ¬c = 0, ¬a = 0, ¬c ∧ ¬a = 0, (c ∨ a) ∧ (¬c ∧ ¬a) = 0
a = 1, c = 0: c ∨ a = 1, ¬c = 1, ¬a = 0, ¬c ∧ ¬a = 0, (c ∨ a) ∧ (¬c ∧ ¬a) = 0
a = 0, c = 1: c ∨ a = 1, ¬c = 0, ¬a = 1, ¬c ∧ ¬a = 0, (c ∨ a) ∧ (¬c ∧ ¬a) = 0
a = 0, c = 0: c ∨ a = 0, ¬c = 1, ¬a = 1, ¬c ∧ ¬a = 1, (c ∨ a) ∧ (¬c ∧ ¬a) = 0

Classify

(c ∨ a) ∧ (¬c ∧ ¬a): 0, 0, 0, 0
0 in every row: a contradiction.

Answer

contradiction
4.Propositional logic & proofs

Answer: ¬r

Quantifiers and arguments: ¬∀x P(x) ≡ ∃x ¬P(x) and ¬∃x P(x) ≡ ∀x ¬P(x). Move a negation inwards, swap every quantifier it passes, and negate the inner formula with the propositional laws, such as ¬(a → b) ≡ a ∧ ¬b. An argument is valid when the conclusion is 1 in every row in which all premises are 1. A single row with true premises and a false conclusion is a counterexample and shows the argument is invalid.

Rows with all premises true

p = 0, r = 0: ¬r = 1, r = 0, ¬r ∧ p = 0, p = 0

Conclusion

¬r (modus tollens)

Answer

¬r
5.Digital logic: gates, K-maps & flip-flops

Answer: L = 2

Boolean algebra: the rules that simplify expressions are X + X′ = 1 and X·X′ = 0, X + X = X, absorption X + X·Y = X, the combining rule X·Y + X·Y′ = X, X + X′Y = X + Y and the consensus theorem XY + X′Z + YZ = XY + X′Z. De Morgan's laws turn (X·Y)′ into X′ + Y′ and (X + Y)′ into X′·Y′. A Karnaugh map applies these rules graphically: every group of two neighbouring cells is one use of XY + XY′ = X.

Minterms of the expression

A′BC → m(3)
AC → m(5, 7)
A′C → m(1, 3)
A′C′ → m(0, 2)
ABC00011110011011131204151706

Group and read off

F = C + A′

Count the literals

L = 2

Answer

L = 2
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 10

Flip-flops and counters: a flip-flop changes only at the active clock edge. D copies its input (Q⁺ = D); T toggles when T = 1; JK holds for 00, resets for 01, sets for 10 and toggles for 11. An n-bit binary counter has 2ⁿ states and counts modulo 2ⁿ; a mod-N counter goes 0, 1, …, N − 1 and back to 0, and needs the smallest n with 2ⁿ ≥ N flip-flops. Each stage of a ripple counter halves the frequency, so the last output of a mod-N counter runs at fclk/N. A Johnson counter of n flip-flops has 2n states, a ring counter n.

State after every edge

0111 → 1011 → 0101 → 1010 → 0101 → 1010

Decimal value

1010₂ = 10

Answer

Q = 10
7.Number theory & modular arithmetic

Answer: 10−1 mod 79 = 8

Inverses and congruences: a has an inverse modulo n exactly when gcd(a, n) = 1. The extended Euclidean algorithm writes 1 = s·n + t·a, and then a⁻¹ ≡ t (mod n). A congruence a·x ≡ b (mod n) with gcd(a, n) = 1 has the solution x ≡ a⁻¹·b. Two congruences x ≡ r₁ (mod m₁) and x ≡ r₂ (mod m₂) with coprime moduli have one solution modulo m₁·m₂ (Chinese remainder theorem). Euler’s φ(n) counts the numbers from 1 to n that are coprime to n: φ(n) = n·∏(1 − 1/p).

Extended Euclidean algorithm

Each remainder r is written as r ≡ t·10 (mod 79).

r = 79, t = 0
r = 10, q = 7, t = 1
r = 9, q = 1, t = −7
r = 1, q = 9, t = 8

Inverse

1 ≡ 8·10 (mod 79)
x = 8
check: 10·8 = 80 = 1·79 + 1

Answer

10−1 mod 79 = 8
8.Number theory & modular arithmetic

Answer: d = 587

Modular powers and RSA: compute aᵏ mod n by repeated squaring: write k in binary, square, reduce after every step and multiply the powers whose bit is 1. Fermat: aᵖ⁻¹ ≡ 1 (mod p) for a prime p not dividing a, so the exponent can be reduced modulo p − 1. RSA: choose primes p and q, n = p·q and φ(n) = (p − 1)(q − 1); the public exponent e is coprime to φ(n), and the private exponent is d = e⁻¹ mod φ(n). Encrypt with c = mᵉ mod n and decrypt with m = cᵈ mod n.

Modulus and totient

n = 23·41 = 943
φ(n) = 22·40 = 880

Extended Euclidean algorithm

r = 880, t = 0
r = 3, q = 293, t = 1
r = 1, q = 3, t = −293

Private exponent

d = 587
check: 3·587 = 1761 = 2·880 + 1

Answer

d = 587
9.Algorithms & complexity

Answer: A · Makes the locally best choice at each step

Why this is the answer

Greedy algorithms are fast and optimal for some problems (e.g. Dijkstra, Huffman coding) but not in general (e.g. coin change with some coin systems).

Answer

A · Makes the locally best choice at each step
10.Algorithms & complexity

Answer: D · Quickselect runs in expected O(n), and median-of-medians guarantees O(n) worst case

Why this is the answer

Quickselect's expected work is n + n/2 + n/4 + ... = O(n) with random pivots; choosing the pivot as a median of medians makes the worst case linear.

Answer

D · Quickselect runs in expected O(n), and median-of-medians guarantees O(n) worst case
11.Data structures

Answer: C · O(1)

Why this is the answer

Each element is pushed and popped at most twice in total, so although one refill can take O(n), the average over a sequence is O(1).

Answer

C · O(1)
12.Data structures

Answer: A · O(1 + α), where α = n / m

Why this is the answer

Computing the hash costs O(1) and the expected chain length is α, so keeping α bounded by resizing gives O(1) expected time.

Answer

A · O(1 + α), where α = n / m
13.Graph theory

Answer: Euler path, no circuit

Euler, Hamilton and trees: a connected graph has an Euler circuit (every edge once, back to the start) when every vertex has even degree, and an Euler path that is not a circuit when exactly two vertices have odd degree; the path then runs from one odd vertex to the other. A Hamilton circuit visits every vertex once; Kn has (n − 1)!/2 of them. A tree on n vertices has n − 1 edges, and a connected graph with m edges keeps a spanning tree after removing m − n + 1 edges.

Degrees

deg(A) = 1, deg(B) = 3, deg(C) = 2, deg(D) = 2, deg(E) = 2

Odd vertices

A, B → 2
Euler path, no circuit

Answer

Euler path, no circuit

The Euler path starts at one odd vertex and ends at the other.

14.Graph theory

Answer: W = 41

Shortest paths and spanning trees: Dijkstra’s algorithm settles one vertex per round: the unsettled vertex with the smallest tentative distance. Its distance is then final, and every neighbour v gets d(v) = min(d(v), d(u) + w(u, v)). Kruskal’s algorithm builds a minimum spanning tree by taking the edges from light to heavy and skipping any edge that would close a cycle; Prim’s algorithm grows one tree from a start vertex by always adding the lightest edge that leaves it. A spanning tree of n vertices has n − 1 edges.

Kruskal’s algorithm

AC (4): take
AB (5): take
EF (8): take
BD (9): take
BC (12): skip, closes a cycle
CE (15): take

Total weight

W = 4 + 5 + 8 + 9 + 15 = 41

Answer

W = 41
15.Databases & SQL

Answer: A · Each name with the label 'pass' or 'fail' depending on the grade

Why this is the answer

A CASE expression evaluates its conditions per row and returns the matching value, so every row gets a label.

Answer

A · Each name with the label 'pass' or 'fail' depending on the grade
16.Databases & SQL

Answer: A · Lost update

Why this is the answer

In a lost update, both transactions work from the same old value and the second write erases the first; locking (SELECT ... FOR UPDATE) or atomic updates prevent it.

Answer

A · Lost update
17.Computer networks

Answer: A · 192.168.10.64

Why this is the answer

/26 subnets start at 0, 64, 128 and 192. 77 lies in the block 64–127, so the network address is 192.168.10.64.

Answer

A · 192.168.10.64
18.Computer networks

Answer: A · For responses too large for UDP and for zone transfers

Why this is the answer

If a UDP response is truncated, the resolver retries over TCP. Zone transfers between name servers also use TCP because they carry lots of data.

Answer

A · For responses too large for UDP and for zone transfers
19.IPv4 subnetting

Answer: N = 256

IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).

Borrowed bits

30 − 22 = 8

Number of subnets

N = 28 = 256

Answer

N = 256

Each /30 subnet has 22 − 2 = 2 usable hosts.

20.IPv4 subnetting

Answer: x = 246

IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).

Block size

23 = 8

Subnet of this host

network: ⌊241/8⌋·8 = 240
broadcast: 240 + 8 − 1 = 247
usable hosts: 241 … 246

Answer

x = 246

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