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 −113 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −1138 bit
U =
2.Number systemsHard
4613 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 4613 items its own binary code of the same length. What is the smallest number of bits b needed?

4613
b = bit
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is (q ∧ (q → p)) → p a tautology, a contradiction or a contingency? Click your answer.

(q ∧ (q → p)) → p
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard
¬r → ¬qr → qq → rr ∧ q

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

r → pp → q
  1. 1¬r → ¬q
  2. 2r → q
  3. 3q → r
  4. 4r ∧ q
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111011111

In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.

F = Σm(3, 6, 7, 8, 11)
  1. 1AB = 00, CD = 00
  2. 2AB = 00, CD = 01
  3. 3AB = 00, CD = 11
  4. 4AB = 00, CD = 10
  5. 5AB = 01, CD = 00
  6. 6AB = 01, CD = 01
  7. 7AB = 01, CD = 11
  8. 8AB = 01, CD = 10
  9. 9AB = 11, CD = 00
  10. 10AB = 11, CD = 01
  11. 11AB = 11, CD = 11
  12. 12AB = 11, CD = 10
  13. 13AB = 10, CD = 00
  14. 14AB = 10, CD = 01
  15. 15AB = 10, CD = 11
  16. 16AB = 10, CD = 10
6.Digital logic: gates, K-maps & flip-flopsHard
CLKD1234567

A positive-edge-triggered D 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 = 0D
N =
7.Number theory & modular arithmeticMedium

Find Euler’s totient φ(225), the number of integers from 1 to 225 that are coprime to 225.

n = 225 = 32·52
φ(225) =
8.Number theory & modular arithmeticHard
mc = me​ mod n55m = cd​ mod n?(n, e) = (1147, 23)d

An RSA key uses p = 31, q = 37 and e = 23. Decrypt the ciphertext c = 55: find the message m.

p = 31q = 37e = 23c = 55
m =
9.Algorithms & complexityMedium

Is it correct to say that an O(n²) algorithm is also O(n³)?

  1. ANo, because Big O must always state the exact growth rate of the function
  2. BNo, because n³ is smaller than n² for large n
  3. CYes, but only if n is smaller than 100
  4. DYes, because Big O is an upper bound and n² grows no faster than n³
10.Algorithms & complexityHard

What did the Cook–Levin theorem establish?

  1. AEvery NP problem can be solved in exponential time
  2. BThe halting problem is undecidable
  3. CP is not equal to NP
  4. DThe Boolean satisfiability problem (SAT) is NP-complete
11.Data structuresMedium

A binary tree has root 1 with left child 2 and right child 3; node 2 has children 4 (left) and 5 (right). What is the postorder traversal?

  1. A1 2 4 5 3
  2. B4 5 2 3 1
  3. C4 2 5 1 3
  4. D5 4 3 2 1
12.Data structuresHard

Which structure lets you compute the maximum of every window of size k in an array in O(n) total?

  1. AA min-heap rebuilt for every window
  2. BA deque holding indices in decreasing order of value
  3. CA singly linked list of the window's elements
  4. DA sorted array updated by shifting
13.Graph theoryMedium
ABCDEFEuler 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| = 6|E| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
762881561076ABCDEFG

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
  7. 7G
15.Databases & SQLMedium

What does SELECT name, CASE WHEN grade >= 5.5 THEN 'pass' ELSE 'fail' END FROM results return?

  1. AAn error because CASE is not allowed in SELECT
  2. BOnly the students who passed
  3. CThe number of students who passed and failed
  4. DEach name with the label 'pass' or 'fail' depending on the grade
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. AMerges the two transactions into one
  2. BLets both transactions wait until one of them commits normally
  3. CCommits both transactions at once
  4. DDetects the deadlock and aborts one transaction as the victim
17.Computer networksMedium

What does an MX record in DNS specify?

  1. AThe mail servers that accept email for a domain
  2. BThe name servers for a zone
  3. CThe IPv4 address of a web server
  4. DAn alias name for another host
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. A255 Mbit/s
  2. B8 Mbit/s
  3. C2 Mbit/s
  4. D1 Mbit/s
19.IPv4 subnettingMedium
/20 → /2311111111111111111111111000000000network bitshost bits

The network 10.202.64.0/20 is divided into equal subnets with prefix /23. How many subnets N are there?

10.202.64.0/20/23
N =
20.IPv4 subnettingHard
third octet 138 … 145112806403201618041201commonvaries

A router has routes to these networks: 10.155.138.0/24, 10.155.139.0/24, 10.155.140.0/24, 10.155.141.0/24, 10.155.142.0/24, 10.155.143.0/24, 10.155.144.0/24, 10.155.145.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.155.138.0/24 … 10.155.145.0/24
p =

3Solutions

1.Number systems

Answer: U = 143

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

113 = 0111 0001
inverted: 1000 1110
+1 → 1000 1111

Read it as unsigned

1000 1111₂ = 256 − 113 = 143

Answer

U = 143
2.Number systems

Answer: b = 13 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

212 = 4096 < 4613
213 = 8192 ≥ 4613

Answer

b = 13
3.Propositional logic & proofs

Answer: tautology

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).

Truth table row by row

p = 1, q = 1: q → p = 1, q ∧ (q → p) = 1, (q ∧ (q → p)) → p = 1
p = 1, q = 0: q → p = 1, q ∧ (q → p) = 0, (q ∧ (q → p)) → p = 1
p = 0, q = 1: q → p = 0, q ∧ (q → p) = 0, (q ∧ (q → p)) → p = 1
p = 0, q = 0: q → p = 1, q ∧ (q → p) = 0, (q ∧ (q → p)) → p = 1

Classify

(q ∧ (q → p)) → p: 1, 1, 1, 1
1 in every row: a tautology.

Answer

tautology
4.Propositional logic & proofs

Answer: r → q

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 = 1, q = 1, r = 1: ¬r → ¬q = 1, r → q = 1, q → r = 1, r ∧ q = 1
p = 1, q = 1, r = 0: ¬r → ¬q = 0, r → q = 1, q → r = 0, r ∧ q = 0
p = 0, q = 1, r = 0: ¬r → ¬q = 0, r → q = 1, q → r = 0, r ∧ q = 0
p = 0, q = 0, r = 0: ¬r → ¬q = 1, r → q = 1, q → r = 1, r ∧ q = 0

Conclusion

r → q (hypothetical syllogism)

Answer

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

Answer: AB = 10, CD = 00

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.

Neighbours of each 1

A 1 with a neighbouring 1 can join a group of two or more.

m8: m9, m10, m12, m0 → 0
ABCD000111100001111011111

The isolated 1

m8 = AB′C′D′ → AB = 10, CD = 00

Answer

AB = 10, CD = 00
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 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 D flip-flop copies D at each rising edge: Q⁺ = D.

1: D = 0 → Q = 0
2: D = 1 → Q = 1
3: D = 1 → Q = 1
4: D = 1 → Q = 1
5: D = 0 → Q = 0
6: D = 1 → Q = 1
7: D = 1 → Q = 1

Count the changes

0 → 0 → 1 → 1 → 1 → 0 → 1 → 1
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: φ(225) = 120

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).

Prime factors

225 = 32·52

Totient

φ(225) = φ(9)·φ(25) = 6·20 = 120

Answer

φ(225) = 120
8.Number theory & modular arithmetic

Answer: m = 757

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.

Private exponent

n = 1147, φ(n) = 1080
d = 23−1 mod 1080 = 47

Decrypt by repeated squaring

47 = 32 + 8 + 4 + 2 + 1
551 ≡ 55 (mod 1147)
552 ≡ 731 (mod 1147)
554 ≡ 1006 (mod 1147)
558 ≡ 382 (mod 1147)
5516 ≡ 255 (mod 1147)
5532 ≡ 793 (mod 1147)
5547 ≡ 55·731·1006·382·793 ≡ 757 (mod 1147)

Message

m = 757
check: 75723 mod 1147 = 55

Answer

m = 757
9.Algorithms & complexity

Answer: D · Yes, because Big O is an upper bound and n² grows no faster than n³

Why this is the answer

Big O is an upper bound, so the statement is true but not tight; Θ(n²) states the exact growth rate.

Answer

D · Yes, because Big O is an upper bound and n² grows no faster than n³
10.Algorithms & complexity

Answer: D · The Boolean satisfiability problem (SAT) is NP-complete

Why this is the answer

Cook and Levin showed independently that every problem in NP reduces to SAT in polynomial time, so SAT was the first known NP-complete problem.

Answer

D · The Boolean satisfiability problem (SAT) is NP-complete
11.Data structures

Answer: B · 4 5 2 3 1

Why this is the answer

Postorder processes 4 and 5 before 2, then 3, and finally the root 1.

Answer

B · 4 5 2 3 1
12.Data structures

Answer: B · A deque holding indices in decreasing order of value

Why this is the answer

The front of the deque is always the window's maximum; each index enters and leaves the deque once, so the work is linear.

Answer

B · A deque holding indices in decreasing order of value
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) = 2, deg(D) = 1, deg(E) = 3, deg(F) = 1

Odd vertices

A, D, E, F → 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: 7, C: 6
2. C (6): E: 14, G: 13
3. B (7): D: 15
4. G (13)
5. E (14): F: 20
6. D (15)
7. F (20)

round 3

B

Answer

B
15.Databases & SQL

Answer: D · Each name with the label 'pass' or 'fail' depending on the grade

Why this is the answer

A CASE expression evaluates its conditions per row and returns the matching value, so every row gets a label.

Answer

D · Each name with the label 'pass' or 'fail' depending on the grade
16.Databases & SQL

Answer: D · 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

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

Answer: A · The mail servers that accept email for a domain

Why this is the answer

MX (mail exchange) records list mail servers with priorities, so sending servers know where to deliver mail for that domain.

Answer

A · The mail servers that accept email for a domain
18.Computer networks

Answer: B · 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

B · 8 Mbit/s
19.IPv4 subnetting

Answer: N = 8

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

23 − 20 = 3

Number of subnets

N = 23 = 8

Answer

N = 8

Each /23 subnet has 29 − 2 = 510 usable hosts.

20.IPv4 subnetting

Answer: p = 19

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

138 = 10001010
145 = 10010001
common leading bits: 3

Prefix length

p = 16 + 3 = 19

Answer

p = /19

Summary route: 10.155.128.0/19 (it also covers 24 network(s) that are not in the list).

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