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 complement?−128?64?32?16?8?4?2?1

Write −22 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −228 bit
U =
2.Number systemsHard
24-bit two's complement, N = ??sign???????????????????????

What is the most negative integer N that can be stored in 24-bit two's complement?

24 bit
N =
3.Propositional logic & proofsMedium
¬r → ¬(p ∨ q)r → (p ∨ q)(p ∨ q) ∧ ¬r¬(p ∨ q) → ¬r

You want to prove (p ∨ q) → r by contrapositive. Which implication do you prove instead? Click it.

(p ∨ q) → r
  1. 1¬r → ¬(p ∨ q)
  2. 2r → (p ∨ q)
  3. 3(p ∨ q) ∧ ¬r
  4. 4¬(p ∨ q) → ¬r
4.Propositional logic & proofsHard
p ∨ rq ∨ rq ∧ r¬q

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

p ∨ q¬p ∨ r
  1. 1p ∨ r
  2. 2q ∨ r
  3. 3q ∧ r
  4. 4¬q
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000111312041517061121131151140809011010

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(1, 2, 3, 5, 7, 12, 13, 14, 15). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C, D) = Σm(1, 2, 3, 5, 7, 12, 13, 14, 15)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​TQ3​Q3​CLKUP

A 4-bit binary up counter starts at 8 (decimal, Q = 1000). Which count Q (in decimal) does it show after 19 clock pulses?

Q0 = 819n = 4
Q =
7.Number theory & modular arithmeticMedium

Solve 18x ≡ 15 (mod 23). Give the solution x with 0 < x < 23.

18x ≡ 15 (mod 23)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod n199m = cd​ mod n?(n, e) = (391, 17)d

An RSA key uses p = 17, q = 23 and e = 17. Decrypt the ciphertext c = 199: find the message m.

p = 17q = 23e = 17c = 199
m =
9.Algorithms & complexityMedium

What is the time complexity of the standard dynamic programming algorithm for the longest common subsequence of strings of lengths m and n?

  1. AO(mn log n)
  2. BO(mn)
  3. CO(m + n)
  4. DO(2m+n)
10.Algorithms & complexityHard

Insertion sort runs in O(n) on sorted input. Does this contradict the Ω(n log n) lower bound for comparison sorting?

  1. ANo, the lower bound applies to the worst case over all inputs, not to every input
  2. BNo, because insertion sort is not a comparison sort
  3. CYes, the lower bound is therefore wrong
  4. DYes, but only for arrays shorter than 100
11.Data structuresMedium

What is the cost of inserting an element in the middle of an array of n elements?

  1. AO(1), because the index is known
  2. BO(n), because later elements must shift
  3. CO(log n), because of binary search
  4. DO(n²), because each element moves n times
12.Data structuresHard

Inserting a key into an AVL tree makes a node's left subtree too tall, and the new key went into the left subtree of its left child (left-left case). Which fix is applied?

  1. AA single right rotation at the unbalanced node
  2. BA left rotation at the child, then a right rotation at the node
  3. CA single left rotation at the unbalanced node
  4. DRecolouring the node and its child
13.Graph theoryMedium

A connected graph has 16 vertices and 16 edges. How many edges k must be removed to leave a spanning tree?

|V| = 16|E| = 16
k =
14.Graph theoryHard
ABCDE

How many walks of length 3 go from A to B? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 3
N =
15.Databases & SQLMedium

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

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

Which SQL feature is best suited to finding every employee below a manager in an organisation tree of unknown depth?

  1. AA UNIQUE constraint on manager_id
  2. BA recursive common table expression (WITH RECURSIVE)
  3. CA GROUP BY on manager_id
  4. DA single self-join of the employees table on manager_id
17.Computer networksMedium

NAT (network address translation) is mainly used to:

  1. ATranslate domain names into IP addresses for devices on a local network
  2. BIncrease the speed of a network cable
  3. CEncrypt traffic between two sites
  4. DLet many devices with private addresses share one public IP address
18.Computer networksHard

Why does IPv6 not need NAT for address shortage?

  1. AIPv6 encrypts all addresses
  2. BIts 128-bit address space is large enough to give every device a global address
  3. CIPv6 does not use addresses
  4. DIPv6 only works on local networks, so addresses never need to be shared globally
19.IPv4 subnettingMedium
/22 → /2511111111111111111111111110000000network bitshost bits

The network 10.165.0.0/22 is divided into equal subnets with prefix /25. How many subnets N are there?

10.165.0.0/22/25
N =
20.IPv4 subnettingHard

The network 192.168.79.0/24 is split with VLSM for these LANs: A: 51 hosts, B: 3 hosts, C: 10 hosts, D: 27 hosts, E: 15 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN E?

A: 51B: 3C: 10D: 27E: 15
x =

3Solutions

1.Number systems

Answer: U = 234

Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.

Invert and add one

22 = 0001 0110
inverted: 1110 1001
+1 → 1110 1010

Read it as unsigned

1110 1010₂ = 256 − 22 = 234

Answer

U = 234
2.Number systems

Answer: N = −8.389 × 10⁶

Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.

Most negative value

−223 = −8388608

Answer

N = −8388608
3.Propositional logic & proofs

Answer: ¬r → ¬(p ∨ q)

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).

Parts of the implication

a = p ∨ q, b = r

Contrapositive

¬b → ¬a = ¬r → ¬(p ∨ q)

Answer

¬r → ¬(p ∨ q)

The contrapositive is equivalent to the original statement, so proving it proves the statement.

4.Propositional logic & proofs

Answer: q ∨ 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 = 1, q = 1, r = 1: p ∨ r = 1, q ∨ r = 1, q ∧ r = 1, ¬q = 0
p = 1, q = 0, r = 1: p ∨ r = 1, q ∨ r = 1, q ∧ r = 0, ¬q = 1
p = 0, q = 1, r = 1: p ∨ r = 1, q ∨ r = 1, q ∧ r = 1, ¬q = 0
p = 0, q = 1, r = 0: p ∨ r = 0, q ∨ r = 1, q ∧ r = 0, ¬q = 0

Conclusion

q ∨ r (resolution)

Answer

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

Answer: L = 7

Karnaugh maps: the cells are ordered in Gray code (00, 01, 11, 10), so neighbouring cells, including the ones across the edges, differ in exactly one variable. Group the 1s in rectangles of 1, 2, 4, 8 or 16 cells, as large as possible and as few as possible, until every 1 is covered. Each group is one product term: the variables that stay constant inside it. A group of 2ᵏ cells loses k variables. Don't-cares (X) may be used to make groups bigger but never have to be covered. Grouping the 0s instead gives the minimal POS form.

ABCD000111100001111000111312041517061121131151140809011010

Group the 1s

Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.

A′D
AB
A′B′C

Minimal SOP form

F = A′D + AB + A′B′C
L = 7

Answer

L = 7
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 11

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.

Full cycles

Every 16 pulses bring the counter back to the same state.

19 = 1·16 + 3

Count after the remaining pulses

Q = (8 + 3) mod 16 = 11
Q = 1011₂

Answer

Q = 11
7.Number theory & modular arithmetic

Answer: x = 20

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).

Inverse of 18

r = 23, t = 0
r = 18, q = 1, t = 1
r = 5, q = 3, t = −1
r = 3, q = 1, t = 4
r = 2, q = 1, t = −5
r = 1, q = 2, t = 9
18−1 ≡ 9 (mod 23)

Solve

x ≡ 9·15 = 135 ≡ 20 (mod 23)
check: 18·20 = 360 ≡ 15 (mod 23)

Answer

x = 20
8.Number theory & modular arithmetic

Answer: m = 97

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.

Private exponent

n = 391, φ(n) = 352
d = 17−1 mod 352 = 145

Decrypt by repeated squaring

145 = 128 + 16 + 1
1991 ≡ 199 (mod 391)
1992 ≡ 110 (mod 391)
1994 ≡ 370 (mod 391)
1998 ≡ 50 (mod 391)
19916 ≡ 154 (mod 391)
19932 ≡ 256 (mod 391)
19964 ≡ 239 (mod 391)
199128 ≡ 35 (mod 391)
199145 ≡ 199·154·35 ≡ 97 (mod 391)

Message

m = 97
check: 9717 mod 391 = 199

Answer

m = 97
9.Algorithms & complexity

Answer: B · O(mn)

Why this is the answer

The table has (m + 1)(n + 1) cells and each is computed in O(1) from its neighbours.

Answer

B · O(mn)
10.Algorithms & complexity

Answer: A · No, the lower bound applies to the worst case over all inputs, not to every input

Why this is the answer

The bound says every comparison sort needs about n log n comparisons on some input of size n; special inputs can be much faster.

Answer

A · No, the lower bound applies to the worst case over all inputs, not to every input
11.Data structures

Answer: B · O(n), because later elements must shift

Why this is the answer

All elements after the insertion point move one position to the right, which takes time proportional to n.

Answer

B · O(n), because later elements must shift
12.Data structures

Answer: A · A single right rotation at the unbalanced node

Why this is the answer

In the left-left case one right rotation lifts the left child up and restores balance; zig-zag cases need a double rotation.

Answer

A · A single right rotation at the unbalanced node
13.Graph theory

Answer: k = 1

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.

Edges of a spanning tree

16 − 1 = 15

Edges to remove

k = 16 − 15 = 1

Answer

k = 1
14.Graph theory

Answer: N = 6

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 1 0 0
B: 1 0 1 1 1
C: 1 1 0 0 1
D: 0 1 0 0 1
E: 0 1 1 1 0

One more step

(M³)AB = 2·1 + 1·0 + 1·1 + 1·1 + 2·1 = 6

Answer

N = 6
15.Databases & SQL

Answer: D · 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

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

Answer: B · A recursive common table expression (WITH RECURSIVE)

Why this is the answer

A recursive CTE starts with an anchor query and repeatedly joins the previous result to the table, so it handles hierarchies of any depth.

Answer

B · A recursive common table expression (WITH RECURSIVE)
17.Computer networks

Answer: D · Let many devices with private addresses share one public IP address

Why this is the answer

NAT rewrites addresses (and ports) so private networks can access the internet through a smaller number of public addresses.

Answer

D · Let many devices with private addresses share one public IP address
18.Computer networks

Answer: B · Its 128-bit address space is large enough to give every device a global address

Why this is the answer

IPv6 has about 3.4 × 10³⁸ addresses, so the scarcity that drove NAT in IPv4 disappears; firewalls still provide security.

Answer

B · Its 128-bit address space is large enough to give every device a global address
19.IPv4 subnetting

Answer: N = 8

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

25 − 22 = 3

Number of subnets

N = 23 = 8

Answer

N = 8

Each /25 subnet has 27 − 2 = 126 usable hosts.

20.IPv4 subnetting

Answer: x = 96

VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.

Block size per LAN

A: 51 hosts → 26 = 64 (/26)
D: 27 hosts → 25 = 32 (/27)
E: 15 hosts → 25 = 32 (/27)
C: 10 hosts → 24 = 16 (/28)
B: 3 hosts → 23 = 8 (/29)

Allocation, largest first

A: .0 – .63
D: .64 – .95
E: .96 – .127
C: .128 – .143
B: .144 – .151

Answer

x = 96

LAN E: 192.168.79.96/27.

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