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 −38 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −388 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
¬q ∧ p¬q ∧ ¬p¬q ∨ ¬p¬q ∨ p

Which formula is logically equivalent to ¬(q ∧ p)? Click it.

¬(q ∧ p)
  1. 1¬q ∧ p
  2. 2¬q ∧ ¬p
  3. 3¬q ∨ ¬p
  4. 4¬q ∨ p
4.Propositional logic & proofsHard
pqrq → (p ∨ r)qp111110101100011010001000

The argument with premises q → (p ∨ r); q and conclusion p is invalid. Click the row of the truth table that is a counterexample.

q → (p ∨ r)q∴ p
  1. 1p = 1, q = 1, r = 1
  2. 2p = 1, q = 1, r = 0
  3. 3p = 1, q = 0, r = 1
  4. 4p = 1, q = 0, r = 0
  5. 5p = 0, q = 1, r = 1
  6. 6p = 0, q = 1, r = 0
  7. 7p = 0, q = 0, r = 1
  8. 8p = 0, q = 0, r = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
MUXCI0​CI1​0I2​0I3​ABS1​S0​F

A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = C, I₁ = C, I₂ = 0, I₃ = 0. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = CI₁ = CI₂ = 0I₃ = 0
N =
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. Click the first clock edge at which Q goes from 1 back to 0.

Q₀ = 0JK
  1. 1clock edge 1
  2. 2clock edge 2
  3. 3clock edge 3
  4. 4clock edge 4
  5. 5clock edge 5
  6. 6clock edge 6
  7. 7clock edge 7
7.Number theory & modular arithmeticMedium

Find the inverse of 47 modulo 53: the number x with 0 < x < 53 and 47·x ≡ 1 (mod 53).

a = 47n = 53
47−1 mod 53 =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 53164 mod 86.

a = 53k = 164n = 86
53164 mod 86 =
9.Algorithms & complexityMedium

Dynamic programming is useful when a problem has:

  1. ANo subproblems at all, so it can be solved in one step
  2. BA single loop
  3. COverlapping subproblems and optimal substructure
  4. DOnly random inputs
10.Algorithms & complexityHard

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

  1. AO(n √n)
  2. BO(n log n)
  3. CO(n²)
  4. DO(n)
11.Data structuresMedium

How much memory does an adjacency list use for a graph with V vertices and E edges?

  1. AO(E²)
  2. BO(V × E)
  3. CO(V²)
  4. DO(V + E)
12.Data structuresHard

Why are B-trees used in databases and file systems?

  1. AThey store data unsorted
  2. BTheir high branching factor keeps the tree shallow, reducing slow disk reads
  3. CThey need no rebalancing
  4. DThey use less memory than any array, so the entire index always fits in main memory
13.Graph theoryMedium

A tree has 3 vertices of degree 3, 4 vertices of degree 4, 2 vertices of degree 5, 3 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?

3 × deg 34 × deg 42 × deg 53 × deg 2
L =
14.Graph theoryHard
10191815171449ABCDEF

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

|V| = 6|E| = 8
W =
15.Databases & SQLMedium

Second normal form (2NF) removes:

  1. AForeign keys
  2. BTransitive dependencies between non-key columns via another non-key column
  3. CAll NULL values
  4. DPartial dependencies of non-key columns on part of a composite key
16.Databases & SQLHard

What does the write-ahead logging (WAL) rule require?

  1. AA change's log record must reach stable storage before the changed data page does
  2. BEvery query must be logged before it is parsed
  3. CThe log is written only after a crash
  4. DData pages must be written before the log
17.Computer networksMedium

What is the IPv6 loopback address?

  1. Afe80::1
  2. B::1
  3. C::
  4. Dff02::1
18.Computer networksHard

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

  1. ATo encrypt the payload of each segment
  2. BTo compress the data more efficiently
  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
/?11111111111111111111110000000000network bitshost bits

A department needs 818 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

818 hosts
p =
20.IPv4 subnettingHard
last octet = 156112806403211618140201networkhost

A host has the address 10.20.28.156/25. What is the fourth octet x of the last usable host address in its subnet?

10.20.28.156/25
x =

3Solutions

1.Number systems

Answer: U = 218

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

38 = 0010 0110
inverted: 1101 1001
+1 → 1101 1010

Read it as unsigned

1101 1010₂ = 256 − 38 = 218

Answer

U = 218
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: ¬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).

The law

De Morgan: ¬(a ∧ b) ≡ ¬a ∨ ¬b. The ∧ turns into ∨ and both parts are negated.
¬(q ∧ p) ≡ ¬q ∨ ¬p

Check with the truth table

p = 1, q = 1: 0 | 0 0 0 1
p = 1, q = 0: 1 | 1 0 1 1
p = 0, q = 1: 1 | 0 0 1 0
p = 0, q = 0: 1 | 0 1 1 1

Answer

¬q ∨ ¬p

The columns after the bar are the options in the order shown.

4.Propositional logic & proofs

Answer: p = 0, q = 1, r = 1

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.

Premises and conclusion row by row

p = 1, q = 1, r = 1: 1 1 | 1
p = 1, q = 1, r = 0: 1 1 | 1
p = 1, q = 0, r = 1: 1 0 | 1
p = 1, q = 0, r = 0: 1 0 | 1
p = 0, q = 1, r = 1: 1 1 | 0
p = 0, q = 1, r = 0: 0 1 | 0
p = 0, q = 0, r = 1: 1 0 | 0
p = 0, q = 0, r = 0: 1 0 | 0

Counterexample

p = 0, q = 1, r = 1

Answer

p = 0, q = 1, r = 1

An argument with this form looks like a valid rule, but one row is enough to show that it is not.

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

Answer: N = 2

Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.

Output expression

F = A′B′C + A′BC

Rows with F = 1

ABC = 001, 011
N = 2

Answer

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

Answer: clock edge 3

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

First change from 1 to 0

clock edge 3

Answer

clock edge 3
7.Number theory & modular arithmetic

Answer: 47−1 mod 53 = 44

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·47 (mod 53).

r = 53, t = 0
r = 47, q = 1, t = 1
r = 6, q = 7, t = −1
r = 5, q = 1, t = 8
r = 1, q = 5, t = −9

Inverse

1 ≡ −9·47 (mod 53)
x = 44
check: 47·44 = 2068 = 39·53 + 1

Answer

47−1 mod 53 = 44
8.Number theory & modular arithmetic

Answer: 53164 mod 86 = 9

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.

Binary exponent

164 = 101001002 = 128 + 32 + 4

Repeated squaring

531 ≡ 53 (mod 86)
532 ≡ 57 (mod 86)
534 ≡ 67 (mod 86)
538 ≡ 17 (mod 86)
5316 ≡ 31 (mod 86)
5332 ≡ 15 (mod 86)
5364 ≡ 53 (mod 86)
53128 ≡ 57 (mod 86)

Multiply the needed powers

53164 ≡ 67·15·57 ≡ 9 (mod 86)

Answer

53164 mod 86 = 9
9.Algorithms & complexity

Answer: C · Overlapping subproblems and optimal substructure

Why this is the answer

Storing results of overlapping subproblems (memoisation or tabulation) avoids repeated work, turning exponential algorithms into polynomial ones.

Answer

C · Overlapping subproblems and optimal substructure
10.Algorithms & complexity

Answer: B · O(n log n)

Why this is the answer

The total is n / 1 + n / 2 + ... + n / n = n × Hn, and the harmonic number Hn is Θ(log n).

Answer

B · O(n log n)
11.Data structures

Answer: D · O(V + E)

Why this is the answer

Each vertex has a list and each edge appears once (directed) or twice (undirected), giving O(V + E).

Answer

D · O(V + E)
12.Data structures

Answer: B · Their high branching factor keeps the tree shallow, reducing slow disk reads

Why this is the answer

B-trees store many keys per node, so height is small and few block reads are needed per lookup; B+ trees also link leaves for range scans.

Answer

B · Their high branching factor keeps the tree shallow, reducing slow disk reads
13.Graph theory

Answer: L = 19

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.

Handshake lemma for a tree

L + 41 = 2(L + 12) − 2
L = 19

Answer

L = 19

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

14.Graph theory

Answer: W = 54

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

EF (4): take
BE (9): take
AB (10): take
DF (14): take
BD (15): skip, closes a cycle
CE (17): take

Total weight

W = 4 + 9 + 10 + 14 + 17 = 54

Answer

W = 54
15.Databases & SQL

Answer: D · Partial dependencies of non-key columns on part of a composite key

Why this is the answer

2NF requires 1NF and that every non-key attribute depends on the whole key, not just part of it.

Answer

D · Partial dependencies of non-key columns on part of a composite key
16.Databases & SQL

Answer: A · A change's log record must reach stable storage before the changed data page does

Why this is the answer

Writing the log first lets the DBMS redo committed work and undo uncommitted work after a crash, supporting atomicity and durability.

Answer

A · A change's log record must reach stable storage before the changed data page does
17.Computer networks

Answer: B · ::1

Why this is the answer

::1 is the loopback address. :: is the unspecified address, fe80::/10 is link-local and ff02::1 is the all-nodes multicast group.

Answer

B · ::1
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 = 22

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

29 − 2 = 510 < 818
210 − 2 = 1022 ≥ 818 → h = 10

Prefix length

p = 32 − 10 = 22

Answer

p = /22
20.IPv4 subnetting

Answer: x = 254

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

27 = 128

Subnet of this host

network: ⌊156/128⌋·128 = 128
broadcast: 128 + 128 − 1 = 255
usable hosts: 129 … 254

Answer

x = 254

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