Computer Science Fundamentals: cram pack

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

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 −102 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −1028 bit
U =
2.Number systemsHard
2130 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 2130 items its own binary code of the same length. What is the smallest number of bits b needed?

2130
b = bit
3.Propositional logic & proofsMedium
(q ∧ p) ∧ ¬rr → (q ∧ p)¬r → ¬(q ∧ p)¬(q ∧ p) → ¬r

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

(q ∧ p) → r
  1. 1(q ∧ p) ∧ ¬r
  2. 2r → (q ∧ p)
  3. 3¬r → ¬(q ∧ p)
  4. 4¬(q ∧ p) → ¬r
4.Propositional logic & proofsHard
∃x ∃y (R(x, y) → ¬R(y, x))∃x ∃y (R(x, y) ∧ ¬R(y, x))∃x ∃y (¬R(x, y) ∧ ¬R(y, x))∀x ∀y (R(x, y) ∧ ¬R(y, x))

Which statement is equivalent to ¬∀x ∀y (R(x, y) → R(y, x)), with the negation moved all the way inside? Click it.

¬∀x ∀y (R(x, y) → R(y, x))
  1. 1∃x ∃y (R(x, y) → ¬R(y, x))
  2. 2∃x ∃y (R(x, y) ∧ ¬R(y, x))
  3. 3∃x ∃y (¬R(x, y) ∧ ¬R(y, x))
  4. 4∀x ∀y (R(x, y) ∧ ¬R(y, x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

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

F = A′B′C′ + A′B + B′C′
L =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKJK1234567

A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown at 7 rising clock edges. How many times N does Q change state?

Q₀ = 0JK
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 53 modulo 64: the number x with 0 < x < 64 and 53·x ≡ 1 (mod 64).

a = 53n = 64
53−1 mod 64 =
8.Number theory & modular arithmeticHard
406c = me​ mod n?m = cd​ mod nm(n, e) = (1147, 11)d

Encrypt the message m = 406 with the RSA public key (n, e) = (1147, 11). Find the ciphertext c.

n = 1147e = 11m = 406
c =
9.Algorithms & complexityMedium

What is the time complexity of: for i = 1 to n: for j = 1 to i: do O(1) work?

  1. AO(n³)
  2. BO(n²)
  3. CO(n log n)
  4. DO(n)
10.Algorithms & complexityHard

Shortest simple paths can be found in polynomial time. What is known about finding the longest simple path between two vertices in a general graph?

  1. AIt is undecidable
  2. BIt is NP-hard
  3. CIt can be solved by Dijkstra's algorithm with negated weights
  4. DIt is always solvable in O(V + E)
11.Data structuresMedium

AVL trees and red-black trees guarantee:

  1. AConstant-time search in every case
  2. BThat keys are stored in insertion order
  3. CThat no node has two children
  4. DO(log n) height through rebalancing
12.Data structuresHard

A Bloom filter can answer "is this element in the set?" with:

  1. AExact answers always
  2. BPossible false negatives but no false positives
  3. CAnswers only for numbers
  4. DPossible false positives but no false negatives
13.Graph theoryMedium
ABCDEFGEuler 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| = 7|E| = 8
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
ABCDE

How many closed walks of length 2 start and end at C? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =
15.Databases & SQLMedium

Which ACID property means concurrent transactions do not see each other's unfinished changes?

  1. ADurability
  2. BAtomicity
  3. CAvailability
  4. DIsolation
16.Databases & SQLHard

When is a schedule of concurrent transactions called serializable?

  1. AWhen its effect is equivalent to some serial execution of the same transactions
  2. BWhen every transaction is given a serial number
  3. CWhen no transaction ever reads data
  4. DWhen the transactions run on one CPU only
17.Computer networksMedium

How many usable host addresses does a /24 IPv4 network have?

  1. A256
  2. B252
  3. C255
  4. D254
18.Computer networksHard

Why does TCP choose an unpredictable initial sequence number for each connection?

  1. ATo compress the data more efficiently
  2. BTo encrypt the payload of each segment
  3. CTo select the best route to the destination for the segments of that connection
  4. DTo avoid confusion with old segments and make spoofed segments harder to inject
19.IPv4 subnettingMedium
/?11111111111111111111111100000000network bitshost bits

A department needs 181 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

181 hosts
p =
20.IPv4 subnettingHard
third octet 242 … 246112816413211608041201commonvaries

A router has routes to these networks: 10.176.242.0/24, 10.176.243.0/24, 10.176.244.0/24, 10.176.245.0/24, 10.176.246.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.176.242.0/24 … 10.176.246.0/24
p =

3Solutions

1.Number systems

Answer: U = 154

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

102 = 0110 0110
inverted: 1001 1001
+1 → 1001 1010

Read it as unsigned

1001 1010₂ = 256 − 102 = 154

Answer

U = 154
2.Number systems

Answer: b = 12 bit

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.

Powers of two

211 = 2048 < 2130
212 = 4096 ≥ 2130

Answer

b = 12
3.Propositional logic & proofs

Answer: ¬r → ¬(q ∧ p)

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 = q ∧ p, b = r

Contrapositive

¬b → ¬a = ¬r → ¬(q ∧ p)

Answer

¬r → ¬(q ∧ p)

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

4.Propositional logic & proofs

Answer: ∃x ∃y (R(x, y) ∧ ¬R(y, x))

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.

Move the negation inwards

¬∀x ∀y (R(x, y) → R(y, x)) ≡ ∃x ∃y (R(x, y) ∧ ¬R(y, x))

Answer

∃x ∃y (R(x, y) ∧ ¬R(y, x))

The other options either keep a quantifier unchanged, change the order of the quantifiers, or negate the inner formula wrongly.

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

Answer: L = 4

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′B′C′ → m(0)
A′B → m(2, 3)
B′C′ → m(0, 4)
ABC00011110011001131214050706

Group and read off

F = B′C′ + A′B

Count the literals

L = 4

Answer

L = 4
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 5

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.

Q after every edge

A JK flip-flop: 00 holds, 01 resets (Q = 0), 10 sets (Q = 1), 11 toggles.

1: J = 1, K = 1 → Q = 1
2: J = 0, K = 0 → Q = 1
3: J = 1, K = 1 → Q = 0
4: J = 1, K = 1 → Q = 1
5: J = 1, K = 0 → Q = 1
6: J = 0, K = 1 → Q = 0
7: J = 1, K = 1 → Q = 1

Count the changes

0 → 1 → 1 → 0 → 1 → 1 → 0 → 1
N = 5

Answer

N = 5
7.Number theory & modular arithmetic

Answer: 53−1 mod 64 = 29

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·53 (mod 64).

r = 64, t = 0
r = 53, q = 1, t = 1
r = 11, q = 4, t = −1
r = 9, q = 1, t = 5
r = 2, q = 4, t = −6
r = 1, q = 2, t = 29

Inverse

1 ≡ 29·53 (mod 64)
x = 29
check: 53·29 = 1537 = 24·64 + 1

Answer

53−1 mod 64 = 29
8.Number theory & modular arithmetic

Answer: c = 850

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.

Repeated squaring

11 = 8 + 2 + 1
4061 ≡ 406 (mod 1147)
4062 ≡ 815 (mod 1147)
4064 ≡ 112 (mod 1147)
4068 ≡ 1074 (mod 1147)
40611 ≡ 406·815·1074 ≡ 850 (mod 1147)

Ciphertext

c = 850

Answer

c = 850
9.Algorithms & complexity

Answer: B · O(n²)

Why this is the answer

The inner body runs n(n + 1) / 2 times in total, which is still quadratic.

Answer

B · O(n²)
10.Algorithms & complexity

Answer: B · It is NP-hard

Why this is the answer

A longest simple path of length V − 1 is a Hamiltonian path, an NP-complete problem; on directed acyclic graphs, however, it is solvable in linear time.

Answer

B · It is NP-hard
11.Data structures

Answer: D · O(log n) height through rebalancing

Why this is the answer

Self-balancing trees keep height logarithmic, so search, insert and delete are O(log n) in the worst case.

Answer

D · O(log n) height through rebalancing
12.Data structures

Answer: D · Possible false positives but no false negatives

Why this is the answer

If any bit for an item is 0, it is definitely absent; if all are 1 it is probably present. Bloom filters trade accuracy for very little memory.

Answer

D · Possible false positives but no false negatives
13.Graph theory

Answer: neither

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) = 4, deg(C) = 3, deg(D) = 2, deg(E) = 2, deg(F) = 1, deg(G) = 3

Odd vertices

A, C, F, G → 4
neither

Answer

neither
14.Graph theory

Answer: N = 3

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

Common neighbours

(M²)CC = 1·1 + 1·1 + 0·0 + 0·0 + 1·1 = 3

Answer

N = 3
15.Databases & SQL

Answer: D · Isolation

Why this is the answer

Isolation keeps concurrent transactions apart; isolation levels decide how strictly this is enforced.

Answer

D · Isolation
16.Databases & SQL

Answer: A · When its effect is equivalent to some serial execution of the same transactions

Why this is the answer

Serializability is the main correctness criterion for concurrency: the interleaved result must equal the result of some one-at-a-time order.

Answer

A · When its effect is equivalent to some serial execution of the same transactions
17.Computer networks

Answer: D · 254

Why this is the answer

A /24 has 8 host bits, so 2⁸ = 256 addresses. The network and broadcast addresses are reserved, leaving 254 for hosts.

Answer

D · 254
18.Computer networks

Answer: D · To avoid confusion with old segments and make spoofed segments harder to inject

Why this is the answer

Random initial sequence numbers stop delayed segments from old connections being accepted and make it hard for off-path attackers to forge valid segments.

Answer

D · To avoid confusion with old segments and make spoofed segments harder to inject
19.IPv4 subnetting

Answer: p = 24

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).

Host bits needed

27 − 2 = 126 < 181
28 − 2 = 254 ≥ 181 → h = 8

Prefix length

p = 32 − 8 = 24

Answer

p = /24
20.IPv4 subnetting

Answer: p = 21

Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.

Compare the bits

242 = 11110010
246 = 11110110
common leading bits: 5

Prefix length

p = 16 + 5 = 21

Answer

p = /21

Summary route: 10.176.240.0/21 (it also covers 3 network(s) that are not in the list).

ExamProblems · examproblems.com/courses/computer-science-fundamentals