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 −42 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −428 bit
U =
2.Number systemsHard
831 codes, b = ? bits?2ᵇ⁻¹…?4?2?1

A system must give every one of 831 items its own binary code of the same length. What is the smallest number of bits b needed?

831
b = bit
3.Propositional logic & proofsMedium
b ∨ (c ∧ a)(b ∧ c) ∨ (b ∧ a)(b ∨ c) ∧ (b ∨ a)(b ∧ c) ∨ a

Which formula is logically equivalent to b ∧ (c ∨ a)? Click it.

b ∧ (c ∨ a)
  1. 1b ∨ (c ∧ a)
  2. 2(b ∧ c) ∨ (b ∧ a)
  3. 3(b ∨ c) ∧ (b ∨ a)
  4. 4(b ∧ c) ∨ a
4.Propositional logic & proofsHard
¬bb¬b ∧ cc

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

b → c¬c
  1. 1¬b
  2. 2b
  3. 3¬b ∧ c
  4. 4c
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000110312141507061121131151140809111110

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(1, 2, 4, 5, 10, 11, 12, 13, 14, 15). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(1, 2, 4, 5, 10, 11, 12, 13, 14, 15)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD00011110000111101011131214150706012113X151141819111X10

For F(A, B, C, D) = Σm(0, 1, 2, 3, 4, 5, 8, 9, 11, 13, 14) + d(10, 15), where d marks the don't-care conditions, find the number N of essential prime implicants.

F(A, B, C, D) = Σm(0, 1, 2, 3, 4, 5, 8, 9, 11, 13, 14) + d(10, 15)
N =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 3 (mod 5) and x ≡ 6 (mod 9).

x ≡ 3 (mod 5)x ≡ 6 (mod 9)
x =
8.Number theory & modular arithmeticHard
101c = me​ mod n?m = cd​ mod nm(n, e) = (209, 11)d

Encrypt the message m = 101 with the RSA public key (n, e) = (209, 11). Find the ciphertext c.

n = 209e = 11m = 101
c =
9.Algorithms & complexityMedium

The time complexity of naive recursive Fibonacci (fib(n) = fib(n−1) + fib(n−2)) is:

  1. ALogarithmic
  2. BExponential
  3. CLinear
  4. DConstant
10.Algorithms & complexityHard

What is the running time of Dijkstra's algorithm with a binary heap on a graph with V vertices and E edges?

  1. AO(V³)
  2. BO((V + E) log V)
  3. CO(V × E)
  4. DO(V + E)
11.Data structuresMedium

What happens when a hash table rehashes after its load factor exceeds a threshold?

  1. AOnly the colliding keys are moved to a new overflow list at the end of the table
  2. BThe oldest keys are deleted to make room
  3. CA larger table is allocated and every key is inserted again
  4. DThe existing keys are sorted in place
12.Data structuresHard

The time complexity of decrease-key in a binary heap (given the element's position) is:

  1. AO(log n)
  2. BO(1)
  3. CO(n log n)
  4. DO(n)
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K8 have? Circuits that differ only in the starting vertex or the direction count as the same.

K8
H =
14.Graph theoryHard
14514125121411ABCDEFG

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 table has rows where col is NULL. Which rows does WHERE col = NULL return?

  1. ANo rows, because the comparison is unknown
  2. BOnly the rows where col is NULL, as with IS NULL
  3. CAn error in every database
  4. DAll rows of the table
16.Databases & SQLHard

An app runs UPDATE item SET qty = 4, version = 8 WHERE id = 1 AND version = 7, and reports a conflict if no row was updated. What technique is this?

  1. AStrict two-phase locking with row locks
  2. BDatabase sharding
  3. COptimistic concurrency control
  4. DWrite-ahead logging
17.Computer networksMedium

Which range of TCP/UDP port numbers is reserved as "well-known" ports?

  1. A0–1023
  2. B49152–65535
  3. C0–65535
  4. D1024–49151
18.Computer networksHard

What does TCP fast retransmit do?

  1. ADoubles the window after every loss
  2. BResends every segment twice to be safe
  3. CResends a segment after three duplicate ACKs, without waiting for the timeout
  4. DSkips the three-way handshake when reconnecting to known hosts, so data is resent sooner
19.IPv4 subnettingMedium
255.255.254.011111111111111111111111000000000network bitshost bits

A subnet uses the mask 255.255.254.0. How many usable host addresses H does it have?

255.255.254.0
H =
20.IPv4 subnettingHard
third octet 76 … 80012816403201618140201commonvaries

A router has routes to these networks: 10.162.76.0/24, 10.162.77.0/24, 10.162.78.0/24, 10.162.79.0/24, 10.162.80.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.162.76.0/24 … 10.162.80.0/24
p =

3Solutions

1.Number systems

Answer: U = 214

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

42 = 0010 1010
inverted: 1101 0101
+1 → 1101 0110

Read it as unsigned

1101 0110₂ = 256 − 42 = 214

Answer

U = 214
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 < 831
210 = 1024 ≥ 831

Answer

b = 10
3.Propositional logic & proofs

Answer: (b ∧ c) ∨ (b ∧ a)

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

∧ distributes over ∨: a ∧ (b ∨ c) ≡ (a ∧ b) ∨ (a ∧ c).
b ∧ (c ∨ a) ≡ (b ∧ c) ∨ (b ∧ a)

Check with the truth table

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

Answer

(b ∧ c) ∨ (b ∧ a)

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

4.Propositional logic & proofs

Answer: ¬b

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

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

Conclusion

¬b (modus tollens)

Answer

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

Answer: N = 4

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.

ABCD000111100001111000110312141507061121131151140809111110

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

Minimal POS form

F = (B + C + D)(A + C′ + D′)(A + B′ + C′)(A′ + B + C)
N = 4

Answer

N = 4
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 3

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.

Prime implicants

The largest groups that can be formed from 1s and Xs.

C′D, B′, AD, A′C′, AC
ABCD00011110000111101011131214150706012113X151141819111X10

Essential prime implicants

B′, A′C′, AC
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: x = 33

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 = 3 + 5k

Second congruence

3 + 5k ≡ 6 (mod 9) → 5k ≡ 3 (mod 9)
k ≡ 2·3 ≡ 6 (mod 9)

Result

x = 3 + 5·6 = 33
x = 33
check: 33 mod 5 = 3, 33 mod 9 = 6

Answer

x = 33
8.Number theory & modular arithmetic

Answer: c = 112

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.

Repeated squaring

11 = 8 + 2 + 1
1011 ≡ 101 (mod 209)
1012 ≡ 169 (mod 209)
1014 ≡ 137 (mod 209)
1018 ≡ 168 (mod 209)
10111 ≡ 101·169·168 ≡ 112 (mod 209)

Ciphertext

c = 112

Answer

c = 112
9.Algorithms & complexity

Answer: B · Exponential

Why this is the answer

The call tree grows roughly like φⁿ (about 1.618ⁿ). Memoisation reduces it to O(n).

Answer

B · Exponential
10.Algorithms & complexity

Answer: B · O((V + E) log V)

Why this is the answer

There are V extract-min and up to E decrease-key operations, each O(log V); a Fibonacci heap improves this to O(E + V log V).

Answer

B · O((V + E) log V)
11.Data structures

Answer: C · A larger table is allocated and every key is inserted again

Why this is the answer

Since the index is computed modulo the table size, all keys must be rehashed into the new, larger table; with doubling this stays O(1) amortised.

Answer

C · A larger table is allocated and every key is inserted again
12.Data structures

Answer: A · O(log n)

Why this is the answer

After decreasing the key, the element sifts up at most the height of the heap. Fibonacci heaps achieve O(1) amortised decrease-key.

Answer

A · O(log n)
13.Graph theory

Answer: H = 2520

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.

Count

H = (8 − 1)!/2 = 5040/2 = 2520

Answer

H = 2520
14.Graph theory

Answer: d(F) = 29

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): B: 14, C: 5
C (5): E: 17, G: 19
B (14): D: 28
E (17): D: 22, F: 29, G: 18
G (18): D: 19
D (19)
F (29)

Shortest path

A → C → E → F
d(F) = 29

Answer

d(F) = 29
15.Databases & SQL

Answer: A · No rows, because the comparison is unknown

Why this is the answer

Any comparison with NULL using = gives UNKNOWN, and WHERE keeps only TRUE rows; use col IS NULL instead.

Answer

A · No rows, because the comparison is unknown
16.Databases & SQL

Answer: C · Optimistic concurrency control

Why this is the answer

Optimistic locking assumes conflicts are rare: it checks a version column when writing and fails if another transaction has changed the row meanwhile.

Answer

C · Optimistic concurrency control
17.Computer networks

Answer: A · 0–1023

Why this is the answer

Ports 0–1023 are well-known ports for standard services. 1024–49151 are registered ports and 49152–65535 are dynamic or private ports.

Answer

A · 0–1023
18.Computer networks

Answer: C · Resends a segment after three duplicate ACKs, without waiting for the timeout

Why this is the answer

Three duplicate ACKs strongly suggest a single lost segment, so TCP retransmits it at once. This recovers much faster than waiting for the retransmission timer.

Answer

C · Resends a segment after three duplicate ACKs, without waiting for the timeout
19.IPv4 subnetting

Answer: H = 510

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).

Prefix length

255.255.254.0 = /23

Usable hosts

29 − 2 = 510

Answer

H = 510
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

76 = 01001100
80 = 01010000
common leading bits: 3

Prefix length

p = 16 + 3 = 19

Answer

p = /19

Summary route: 10.162.64.0/19 (it also covers 27 network(s) that are not in the list).

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