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
place values in two's complement1−12816403211608041201

The 8-bit pattern 1101 0010 stores a signed integer in two's complement. What is its value N in decimal?

1101 00108 bit
N =
2.Number systemsHard
16 bit1 px256 px256 pxM = ? KiB

An uncompressed image is 256 × 256 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

256 × 256 px16 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
p ↔ ¬rp ∧ ¬r¬p ∨ ¬r¬p ↔ ¬r

Which formula is logically equivalent to ¬(p ↔ r)? Click it.

¬(p ↔ r)
  1. 1p ↔ ¬r
  2. 2p ∧ ¬r
  3. 3¬p ∨ ¬r
  4. 4¬p ↔ ¬r
4.Propositional logic & proofsHard
pr¬r¬r ∧ p

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

r → p¬p
  1. 1p
  2. 2r
  3. 3¬r
  4. 4¬r ∧ p
5.Digital logic: gates, K-maps & flip-flopsMedium
MUX0I0​C′I1​C′I2​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 = 0, I₁ = C′, I₂ = C′, I₃ = 0. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = 0I₁ = C′I₂ = C′I₃ = 0
N =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKJK12345678

A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown at 8 rising clock edges. How many times N does Q change state?

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

Find the inverse of 9 modulo 17: the number x with 0 < x < 17 and 9·x ≡ 1 (mod 17).

a = 9n = 17
9−1 mod 17 =
8.Number theory & modular arithmeticHard

Solve 6x ≡ 18 (mod 27). Give the smallest positive solution x.

6x ≡ 18 (mod 27)
x =
9.Algorithms & complexityMedium

In a sorted array, you look for two elements that sum to a target by moving one pointer from each end inwards. What is the time complexity?

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

Strassen's algorithm multiplies matrices using 7 instead of 8 recursive multiplications of half-size blocks. What is its running time?

  1. AΘ(n³)
  2. BΘ(n2)
  3. CΘ(n² log n)
  4. DΘ(nlog₂ 7), about n2.81
11.Data structuresMedium

You have a reference to a node in the middle of a list. In which list can you remove it in O(1)?

  1. AA singly linked list
  2. BA doubly linked list
  3. CA sorted array
  4. DA dynamic array
12.Data structuresHard

Why can elements not simply be deleted from a standard Bloom filter by clearing their bits?

  1. ABits in a Bloom filter cannot be changed after creation
  2. BThose bits may also belong to other elements, which would cause false negatives
  3. CDeleting would increase the false positive rate to 100%
  4. DBloom filters store the full elements, and these cannot be located again once hashed
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| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
13968410212ABCDEF

Prim’s algorithm builds a minimum spanning tree starting from vertex C. Click the edge it adds as edge number 2.

start: Ck = 2
  1. 1edge AB
  2. 2edge AC
  3. 3edge BD
  4. 4edge CE
  5. 5edge DE
  6. 6edge DF
  7. 7edge EF
  8. 8edge BE
15.Databases & SQLMedium

Third normal form (3NF) removes:

  1. AAtomic values
  2. BComposite keys
  3. CRedundant indexes on columns that are not part of a key
  4. DTransitive dependencies between non-key attributes
16.Databases & SQLHard

What is a covering index for a query?

  1. AAn index on the primary key only
  2. BAn index that covers every column of every table
  3. CAn index that contains all columns the query needs, so the table itself need not be read
  4. DAn index that is rebuilt every night
17.Computer networksMedium

Why is UDP often preferred for live video calls?

  1. ALow delay matters more than retransmitting every lost packet
  2. BUDP guarantees every packet arrives, so the picture never freezes
  3. CUDP encrypts data automatically
  4. DUDP is required for all video
18.Computer networksHard

A PC can reach printers on its own subnet but nothing beyond it. Its IP address and mask are correct. What is the most likely fault?

  1. AA missing or wrong default gateway
  2. BThe TCP window is too small
  3. CA duplicate MAC address on the internet
  4. DA faulty DNS server setting
19.IPv4 subnettingMedium
/22 → /2711111111111111111111111111100000network bitshost bits

The network 10.247.112.0/22 is divided into equal subnets with prefix /27. How many subnets N are there?

10.247.112.0/22/27
N =
20.IPv4 subnettingHard
third octet = 82012816403211608041201networkhost

A host has the address 172.20.82.255/22. What is the third octet x of its subnet's broadcast address?

172.20.82.255/22
x =

3Solutions

1.Number systems

Answer: N = −46

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.

Add the place values

The other bits keep their normal positive values.

−128 + 64 + 16 + 2 = −46

Answer

N = −46
2.Number systems

Answer: M = 128 KiB

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.

Bits and bytes

256·256·16 = 1048576 bit = 131072 B

Convert to KiB

131072/1024 = 128 KiB

Answer

M = 128 KiB
3.Propositional logic & proofs

Answer: p ↔ ¬r

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

¬(a ↔ b) is 1 exactly when a and b differ, which is a ↔ ¬b.
¬(p ↔ r) ≡ p ↔ ¬r

Check with the truth table

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

Answer

p ↔ ¬r

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

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: p = 0, r = 0, ¬r = 1, ¬r ∧ p = 0

Conclusion

¬r (modus tollens)

Answer

¬r
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′BC′ + AB′C′

Rows with F = 1

ABC = 010, 100
N = 2

Answer

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

Answer: N = 4

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

Count the changes

0 → 1 → 1 → 1 → 0 → 1 → 1 → 0 → 0
N = 4

Answer

N = 4
7.Number theory & modular arithmetic

Answer: 9−1 mod 17 = 2

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·9 (mod 17).

r = 17, t = 0
r = 9, q = 1, t = 1
r = 8, q = 1, t = −1
r = 1, q = 8, t = 2

Inverse

1 ≡ 2·9 (mod 17)
x = 2
check: 9·2 = 18 = 1·17 + 1

Answer

9−1 mod 17 = 2
8.Number theory & modular arithmetic

Answer: x = 3

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).

Divide by the gcd

gcd(6, 27) = 3
2x ≡ 6 (mod 9)

Solve the reduced congruence

2−1 ≡ 5 (mod 9) → x ≡ 3 (mod 9)

Solutions modulo 27

x ∈ {3, 12, 21}
x = 3

Answer

x = 3
9.Algorithms & complexity

Answer: D · O(n)

Why this is the answer

If the sum is too small the left pointer moves right, if too large the right pointer moves left, so there are at most n steps.

Answer

D · O(n)
10.Algorithms & complexity

Answer: D · Θ(nlog₂ 7), about n2.81

Why this is the answer

With a = 7 and b = 2, nlog₂ 7 dominates the O(n²) addition work, giving about n2.81.

Answer

D · Θ(nlog₂ 7), about n2.81
11.Data structures

Answer: B · A doubly linked list

Why this is the answer

A doubly linked node knows its predecessor, so you can relink prev and next directly; a singly linked list must search for the predecessor.

Answer

B · A doubly linked list
12.Data structures

Answer: B · Those bits may also belong to other elements, which would cause false negatives

Why this is the answer

Clearing a shared bit would make another element look absent; counting Bloom filters keep counters instead of bits to allow deletion.

Answer

B · Those bits may also belong to other elements, which would cause false negatives
13.Graph theory

Answer: Euler 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) = 2, deg(B) = 4, deg(C) = 2, deg(D) = 2, deg(E) = 2

Odd vertices

0
Euler circuit

Answer

Euler circuit
14.Graph theory

Answer: edge EF

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.

Prim’s algorithm

1. CE (8)
2. EF (2)
3. DE (4)
4. BD (6)
5. AC (9)

edge number 2

EF

Answer

edge EF
15.Databases & SQL

Answer: D · Transitive dependencies between non-key attributes

Why this is the answer

In 3NF no non-key attribute depends on another non-key attribute, such as storing city and postcode where city depends on postcode.

Answer

D · Transitive dependencies between non-key attributes
16.Databases & SQL

Answer: C · An index that contains all columns the query needs, so the table itself need not be read

Why this is the answer

If all referenced columns are in the index, the DBMS can do an index-only scan and skip fetching table rows, which saves I/O.

Answer

C · An index that contains all columns the query needs, so the table itself need not be read
17.Computer networks

Answer: A · Low delay matters more than retransmitting every lost packet

Why this is the answer

Retransmitting lost packets would add delay; real-time media tolerates some loss, so UDP-based protocols like RTP are used.

Answer

A · Low delay matters more than retransmitting every lost packet
18.Computer networks

Answer: A · A missing or wrong default gateway

Why this is the answer

Hosts send off-subnet packets to the default gateway. If it is wrong or missing, local traffic still works but nothing outside is reachable, even by IP address.

Answer

A · A missing or wrong default gateway
19.IPv4 subnetting

Answer: N = 32

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

27 − 22 = 5

Number of subnets

N = 25 = 32

Answer

N = 32

Each /27 subnet has 25 − 2 = 30 usable hosts.

20.IPv4 subnetting

Answer: x = 83

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 in the third octet

22 = 4

Subnet of this host

network: 172.20.80.0
broadcast: 172.20.83.255

Answer

x = 83

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