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 −98 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −988 bit
U =
2.Number systemsHard
10-bit two's complement, N = ??sign?????????

What is the most negative integer N that can be stored in 10-bit two's complement?

10 bit
N =
3.Propositional logic & proofsMedium
¬q ∧ ¬p¬q ∧ ¬¬p¬q → ¬¬pq ∧ ¬¬p

You want to prove q → ¬p by contradiction. Which statement do you assume at the start? Click it.

q → ¬p
  1. 1¬q ∧ ¬p
  2. 2¬q ∧ ¬¬p
  3. 3¬q → ¬¬p
  4. 4q ∧ ¬¬p
4.Propositional logic & proofsHard
c¬c¬aa ∧ ¬c

The premises a → c; a are given. Which conclusion follows from them? Click it.

a → ca
  1. 1c
  2. 2¬c
  3. 3¬a
  4. 4a ∧ ¬c
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010011302141507160121130151141819011010

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 3, 4, 5, 6, 8, 9, 13, 14). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(0, 3, 4, 5, 6, 8, 9, 13, 14)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​CLKDOWN

A 3-bit binary down counter starts at 6 (decimal, Q = 110). Which count Q (in decimal) does it show after 18 clock pulses?

Q0 = 618n = 3
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 9 modulo 79: the number x with 0 < x < 79 and 9·x ≡ 1 (mod 79).

a = 9n = 79
9−1 mod 79 =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (119, 19)d = ?

An RSA key uses the primes p = 7 and q = 17 and the public exponent e = 19. Find the private exponent d (0 < d < φ(n)).

p = 7q = 17e = 19
d =
9.Algorithms & complexityMedium

What does the recurrence T(n) = T(n − 1) + O(n) solve to?

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

Which statement about little-o notation is correct?

  1. An² = o(n)
  2. Bn = o(n²), but n is not o(n)
  3. Cn = o(n), since every function is little-o of itself
  4. D2n = o(n), because 2n grows slowly
11.Data structuresMedium

The keys 50, 30, 70, 20, 40 are inserted in this order into an empty binary search tree. What is the left child of 30?

  1. A70
  2. B40
  3. C50
  4. D20
12.Data structuresHard

Which pair of traversals always determines a unique binary tree with distinct keys?

  1. ALevel-order alone
  2. BPreorder and inorder
  3. CPreorder and postorder
  4. DPreorder alone
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
1114126881413ABCDEFG

Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.

start: Aend: F
d(F) =
15.Databases & SQLMedium

A LEFT JOIN returns:

  1. ARows that match in neither table
  2. BOnly matching rows from both tables
  3. CAll rows from the left table and matching rows from the right, with NULLs where there is no match
  4. DAll rows from the right table only
16.Databases & SQLHard

Transaction T1 locks row A and waits for row B, while T2 locks row B and waits for row A. What does a typical DBMS do?

  1. ACommits both transactions at once
  2. BDetects the deadlock and aborts one transaction as the victim
  3. CMerges the two transactions into one
  4. DLets both transactions wait until one of them commits normally
17.Computer networksMedium

Which metric does RIP use to choose routes?

  1. ABusiness policy
  2. BLink bandwidth
  3. CMeasured delay
  4. DHop count
18.Computer networksHard

By Shannon's theorem, what is the capacity of a 1 MHz channel with a signal-to-noise ratio of 255?

  1. A8 Mbit/s
  2. B255 Mbit/s
  3. C1 Mbit/s
  4. D2 Mbit/s
19.IPv4 subnettingMedium
/24 → /2511111111111111111111111110000000network bitshost bits

The network 192.168.70.0/24 is divided into /25 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 2?

192.168.70.0/24/25subnet 2 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
third octet = 38012806413201608141201networkhost

A host has the address 172.20.38.231/22. What is the third octet x of its subnet's broadcast address?

172.20.38.231/22
x =

3Solutions

1.Number systems

Answer: U = 158

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

98 = 0110 0010
inverted: 1001 1101
+1 → 1001 1110

Read it as unsigned

1001 1110₂ = 256 − 98 = 158

Answer

U = 158
2.Number systems

Answer: N = −512

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

−29 = −512

Answer

N = −512
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).

Parts of the implication

a = q, b = ¬p

Negation of the implication

¬(a → b) ≡ a ∧ ¬b = q ∧ ¬¬p

Answer

q ∧ ¬¬p

If the assumption leads to a contradiction, the implication must be true.

4.Propositional logic & proofs

Answer: c

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

a = 1, c = 1: c = 1, ¬c = 0, ¬a = 0, a ∧ ¬c = 0

Conclusion

c (modus ponens)

Answer

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

Answer: N = 5

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.

ABCD000111100001111010011302141507160121130151141819011010

Group the 0s

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

(B + C′ + D)
(B′ + C′ + D′)
(A′ + C′ + D′)
(A + B + C + D′)
(A′ + B′ + C + D)

Minimal POS form

F = (B + C′ + D)(B′ + C′ + D′)(A′ + C′ + D′)(A + B + C + D′)(A′ + B′ + C + D)
N = 5

Answer

N = 5

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: Q = 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.

Full cycles

Every 8 pulses bring the counter back to the same state.

18 = 2·8 + 2

Count after the remaining pulses

Q = (6 − 2) mod 8 = 4
Q = 100₂

Answer

Q = 4
7.Number theory & modular arithmetic

Answer: 9−1 mod 79 = 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·9 (mod 79).

r = 79, t = 0
r = 9, q = 8, t = 1
r = 7, q = 1, t = −8
r = 2, q = 3, t = 9
r = 1, q = 2, t = −35

Inverse

1 ≡ −35·9 (mod 79)
x = 44
check: 9·44 = 396 = 5·79 + 1

Answer

9−1 mod 79 = 44
8.Number theory & modular arithmetic

Answer: d = 91

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.

Modulus and totient

n = 7·17 = 119
φ(n) = 6·16 = 96

Extended Euclidean algorithm

r = 96, t = 0
r = 19, q = 5, t = 1
r = 1, q = 19, t = −5

Private exponent

d = 91
check: 19·91 = 1729 = 18·96 + 1

Answer

d = 91
9.Algorithms & complexity

Answer: C · O(n²)

Why this is the answer

The work sums to about n² / 2, which is the recurrence of selection sort or of quicksort with the worst pivot.

Answer

C · O(n²)
10.Algorithms & complexity

Answer: B · n = o(n²), but n is not o(n)

Why this is the answer

f = o(g) means f(n) / g(n) tends to 0; n / n² tends to 0, but n / n = 1 does not.

Answer

B · n = o(n²), but n is not o(n)
11.Data structures

Answer: D · 20

Why this is the answer

20 is smaller than 50 and smaller than 30, so it becomes the left child of 30; 40 becomes its right child.

Answer

D · 20
12.Data structures

Answer: B · Preorder and inorder

Why this is the answer

Preorder gives the root, and its position in the inorder sequence splits the left and right subtrees; preorder with postorder is ambiguous when a node has one child.

Answer

B · Preorder and inorder
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: d(F) = 28

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

Each line: the vertex settled, its final distance, and the neighbours whose distance improves.

A (0): C: 1
C (1): G: 9
G (9): B: 17, D: 23, E: 22
B (17)
E (22): F: 28
D (23)
F (28)

Shortest path

A → C → G → E → F
d(F) = 28

Answer

d(F) = 28
15.Databases & SQL

Answer: C · All rows from the left table and matching rows from the right, with NULLs where there is no match

Why this is the answer

A left outer join keeps every left row; unmatched right columns are filled with NULL.

Answer

C · All rows from the left table and matching rows from the right, with NULLs where there is no match
16.Databases & SQL

Answer: B · Detects the deadlock and aborts one transaction as the victim

Why this is the answer

A cycle in the wait-for graph is a deadlock; the DBMS rolls back one transaction so the other can continue, and the victim may be retried.

Answer

B · Detects the deadlock and aborts one transaction as the victim
17.Computer networks

Answer: D · Hop count

Why this is the answer

RIP is a distance-vector protocol that counts hops, with 16 meaning unreachable. This limits it to small networks.

Answer

D · Hop count
18.Computer networks

Answer: A · 8 Mbit/s

Why this is the answer

C = 10⁶ × log₂(256) = 10⁶ × 8 = 8 Mbit/s. This is the upper limit for error-free data rate on a noisy channel.

Answer

A · 8 Mbit/s
19.IPv4 subnetting

Answer: x = 128

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 number 2

network: (2 − 1)·128 = 128
broadcast: 128 + 128 − 1 = 255

Answer

x = 128
20.IPv4 subnetting

Answer: x = 39

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: 172.20.36.0
broadcast: 172.20.39.255

Answer

x = 39

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