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 complement1−12806413201618040201

The 8-bit pattern 1010 1000 stores a signed integer in two's complement. What is its value N in decimal?

1010 10008 bit
N =
2.Number systemsHard
607 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 607 items its own binary code of the same length. What is the smallest number of bits b needed?

607
b = bit
3.Propositional logic & proofsMedium
¬q ∧ r¬q ∨ ¬r¬q ∧ ¬rq ∨ ¬r

Which formula is logically equivalent to ¬(q ∨ r)? Click it.

¬(q ∨ r)
  1. 1¬q ∧ r
  2. 2¬q ∨ ¬r
  3. 3¬q ∧ ¬r
  4. 4q ∨ ¬r
4.Propositional logic & proofsHard
abca → bc → ba → c111110101100011010001000

The argument with premises a → b; c → b and conclusion a → c is invalid. Click the row of the truth table that is a counterexample.

a → bc → b∴ a → c
  1. 1a = 1, b = 1, c = 1
  2. 2a = 1, b = 1, c = 0
  3. 3a = 1, b = 0, c = 1
  4. 4a = 1, b = 0, c = 0
  5. 5a = 0, b = 1, c = 1
  6. 6a = 0, b = 1, c = 0
  7. 7a = 0, b = 0, c = 1
  8. 8a = 0, b = 0, c = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000011312141507161120130150140819111110

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(2, 3, 4, 5, 6, 9, 10, 11, 12). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C, D) = Σm(2, 3, 4, 5, 6, 9, 10, 11, 12)
L =
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 smallest positive integer x with x ≡ 1 (mod 5) and x ≡ 0 (mod 11).

x ≡ 1 (mod 5)x ≡ 0 (mod 11)
x =
8.Number theory & modular arithmeticHard

Solve 9x ≡ 51 (mod 78). Give the smallest positive solution x.

9x ≡ 51 (mod 78)
x =
9.Algorithms & complexityMedium

With coins of 1, 3 and 4 and an amount of 6, how many coins does the greedy "largest coin first" method use, compared to the optimum?

  1. AGreedy uses 4 coins; the optimum is 3
  2. BGreedy uses 2 coins; the optimum is 2
  3. CGreedy uses 3 coins; the optimum is 2
  4. DGreedy uses 3 coins; the optimum is 3
10.Algorithms & complexityHard

The Held–Karp dynamic programming algorithm solves the travelling salesman problem exactly. What is its running time for n cities?

  1. AO(n² 2ⁿ)
  2. BO(n!)
  3. CO(2n²)
  4. DO(n³)
11.Data structuresMedium

Why is traversing an array often much faster in practice than traversing a linked list of the same length?

  1. AArray elements are contiguous, so they make good use of the CPU cache
  2. BArrays use fewer instructions to compare values
  3. CLinked lists always store elements on disk
  4. DArray traversal is O(1) in total, while linked list traversal is O(n)
12.Data structuresHard

How long does it take to search for an arbitrary key in a binary min-heap of n elements?

  1. AO(n)
  2. BO(log n)
  3. CO(log log n)
  4. DO(1)
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| = 7
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
5410221213139ABCDEF

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 3; A itself is round 1.

start: Around 3
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F
15.Databases & SQLMedium

First normal form (1NF) requires that:

  1. AThere are no foreign keys
  2. BEvery table has exactly one column
  3. CEvery non-key column depends on the whole key and on nothing but the key
  4. DEach column contains atomic values and there are no repeating groups
16.Databases & SQLHard

Boyce–Codd normal form (BCNF) requires that for every non-trivial functional dependency X → Y:

  1. AY is a primary key
  2. BX is a superkey
  3. CX contains NULL
  4. DY is atomic
17.Computer networksMedium

What does a subnet mask of 255.255.255.0 correspond to in CIDR notation?

  1. A/32
  2. B/24
  3. C/16
  4. D/8
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. BA faulty DNS server setting
  3. CThe TCP window is too small
  4. DA duplicate MAC address on the internet
19.IPv4 subnettingMedium
/24 → /2711111111111111111111111111100000network bitshost bits

The network 192.168.86.0/24 is divided into /27 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 7?

192.168.86.0/24/27subnet 7 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard

The network 192.168.12.0/24 is split with VLSM for these LANs: A: 5 hosts, B: 31 hosts, C: 9 hosts, D: 65 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 C?

A: 5B: 31C: 9D: 65
x =

3Solutions

1.Number systems

Answer: N = −88

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 + 32 + 8 = −88

Answer

N = −88
2.Number systems

Answer: b = 10 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

29 = 512 < 607
210 = 1024 ≥ 607

Answer

b = 10
3.Propositional logic & proofs

Answer: ¬q ∧ ¬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

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

Check with the truth table

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

Answer

¬q ∧ ¬r

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

4.Propositional logic & proofs

Answer: a = 1, b = 1, c = 0

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

a = 1, b = 1, c = 1: 1 1 | 1
a = 1, b = 1, c = 0: 1 1 | 0
a = 1, b = 0, c = 1: 0 0 | 1
a = 1, b = 0, c = 0: 0 1 | 0
a = 0, b = 1, c = 1: 1 1 | 1
a = 0, b = 1, c = 0: 1 1 | 1
a = 0, b = 0, c = 1: 1 0 | 1
a = 0, b = 0, c = 0: 1 1 | 1

Counterexample

a = 1, b = 1, c = 0

Answer

a = 1, b = 1, c = 0

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: L = 14

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.

ABCD000111100001111000011312141507161120130150140819111110

Group the 1s

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

B′C
BC′D′
A′CD′
AB′D
A′BC′

Minimal SOP form

F = B′C + BC′D′ + A′CD′ + AB′D + A′BC′
L = 14

Answer

L = 14

Another grouping of the same size exists; it has the same number of terms and literals.

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

Count the changes

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

Answer

N = 4
7.Number theory & modular arithmetic

Answer: x = 11

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).

First congruence

x = 1 + 5k

Second congruence

1 + 5k ≡ 0 (mod 11) → 5k ≡ 10 (mod 11)
k ≡ 9·10 ≡ 2 (mod 11)

Result

x = 1 + 5·2 = 11
x = 11
check: 11 mod 5 = 1, 11 mod 11 = 0

Answer

x = 11
8.Number theory & modular arithmetic

Answer: x = 23

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(9, 78) = 3
3x ≡ 17 (mod 26)

Solve the reduced congruence

3−1 ≡ 9 (mod 26) → x ≡ 23 (mod 26)

Solutions modulo 78

x ∈ {23, 49, 75}
x = 23

Answer

x = 23
9.Algorithms & complexity

Answer: C · Greedy uses 3 coins; the optimum is 2

Why this is the answer

Greedy takes 4 + 1 + 1 = 3 coins, but 3 + 3 uses only 2, so greedy is not optimal for this coin system; dynamic programming is.

Answer

C · Greedy uses 3 coins; the optimum is 2
10.Algorithms & complexity

Answer: A · O(n² 2ⁿ)

Why this is the answer

There are about n 2ⁿ states, each computed in O(n); this is exponential but far faster than trying all (n − 1)! tours.

Answer

A · O(n² 2ⁿ)
11.Data structures

Answer: A · Array elements are contiguous, so they make good use of the CPU cache

Why this is the answer

Contiguous memory gives good locality of reference, while linked nodes may be scattered and cause many cache misses.

Answer

A · Array elements are contiguous, so they make good use of the CPU cache
12.Data structures

Answer: A · O(n)

Why this is the answer

A heap gives no rule for which subtree contains a key, so in the worst case all nodes must be checked; only the minimum is fast to find.

Answer

A · O(n)
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) = 3, deg(C) = 2, deg(D) = 3, deg(E) = 2, deg(F) = 2, deg(G) = 1

Odd vertices

A, B, D, G → 4
neither

Answer

neither
14.Graph theory

Answer: B

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.

Dijkstra round by round

1. A (0): B: 5, C: 4
2. C (4): E: 6
3. B (5): D: 7
4. E (6): F: 19
5. D (7)
6. F (19)

round 3

B

Answer

B
15.Databases & SQL

Answer: D · Each column contains atomic values and there are no repeating groups

Why this is the answer

1NF requires atomic values per cell and a unique row identification, without repeating groups.

Answer

D · Each column contains atomic values and there are no repeating groups
16.Databases & SQL

Answer: B · X is a superkey

Why this is the answer

BCNF demands that every determinant is a candidate key (superkey); 3NF allows an exception when Y is part of a candidate key.

Answer

B · X is a superkey
17.Computer networks

Answer: B · /24

Why this is the answer

Three octets of 255 are 24 one-bits, so 255.255.255.0 = /24.

Answer

B · /24
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: x = 223

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

25 = 32

Subnet number 7

network: (7 − 1)·32 = 192
broadcast: 192 + 32 − 1 = 223

Answer

x = 223
20.IPv4 subnetting

Answer: x = 192

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

D: 65 hosts → 27 = 128 (/25)
B: 31 hosts → 26 = 64 (/26)
C: 9 hosts → 24 = 16 (/28)
A: 5 hosts → 23 = 8 (/29)

Allocation, largest first

D: .0 – .127
B: .128 – .191
C: .192 – .207
A: .208 – .215

Answer

x = 192

LAN C: 192.168.12.192/28.

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