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 complement1−12806403201608141201

The 8-bit pattern 1000 0110 stores a signed integer in two's complement. What is its value N in decimal?

1000 01108 bit
N =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1011 0110 OR 0110 0000. What is the result's value R in decimal?

1011 0110OR0110 0000
R =
3.Propositional logic & proofsMedium
tautologycontradictioncontingency

Is r ∧ ¬r a tautology, a contradiction or a contingency? Click your answer.

r ∧ ¬r
  1. 1tautology
  2. 2contradiction
  3. 3contingency
4.Propositional logic & proofsHard

How many of the 16 rows of the truth table make ((¬r → p) ∨ q) → (s ∧ (¬s → ¬r)) true? Call this number N.

((¬r → p) ∨ q) → (s ∧ (¬s → ¬r))
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010010302141517061120131150140819011110

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 4, 5, 7, 9, 10, 12, 15). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(0, 4, 5, 7, 9, 10, 12, 15)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110X0110302X4X5X706112013015X140819011110

For F(A, B, C, D) = Σm(1, 9, 10, 12) + d(0, 4, 5, 7, 14), where d marks the don't-care conditions, find the number N of essential prime implicants.

F(A, B, C, D) = Σm(1, 9, 10, 12) + d(0, 4, 5, 7, 14)
N =
7.Number theory & modular arithmeticMedium

Solve 10x ≡ 16 (mod 37). Give the solution x with 0 < x < 37.

10x ≡ 16 (mod 37)
x =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 5190 mod 57.

a = 5k = 190n = 57
5190 mod 57 =
9.Algorithms & complexityMedium

What does the Floyd–Warshall algorithm compute, and how fast?

  1. AShortest paths between all pairs of vertices in O(V³)
  2. BShortest paths from one source vertex to all others in O(V + E)
  3. CA topological order in O(V²)
  4. DA minimum spanning tree in O(E log V)
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(V + E)
  2. BO(V³)
  3. CO(V × E)
  4. DO((V + E) log V)
11.Data structuresMedium

What is true of the adjacency matrix of an undirected graph?

  1. AIt is always upper triangular
  2. BIt is symmetric about the main diagonal
  3. CEach row has exactly one 1
  4. DIts main diagonal always contains ones
12.Data structuresHard

A segment tree is typically used to:

  1. AImplement a stack
  2. BSort strings alphabetically by storing each prefix in a separate tree node
  3. CStore a graph's edges
  4. DAnswer range queries (such as range sums) and updates in O(log n)
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.

K7
H =
14.Graph theoryHard
ABCDE

How many walks of length 2 go from A to C? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =
15.Databases & SQLMedium

Why are indexes created on columns?

  1. ATo prevent NULL values
  2. BTo speed up searches and joins on those columns
  3. CTo make inserts and updates faster on those columns
  4. DTo save disk space
16.Databases & SQLHard

Two doctors on call each check that the other is still on call, then both sign off at the same time. Which anomaly can snapshot isolation still allow here?

  1. ADirty write
  2. BDirty read
  3. CNon-repeatable read within one transaction
  4. DWrite skew
17.Computer networksMedium

Why does a switch sometimes send a frame out of all ports except the one it came in on?

  1. AThe switch is acting as a router
  2. BThe frame is too large for one port
  3. CThe frame has a wrong IP checksum and must be checked by every host
  4. DThe destination MAC address is not yet in its address table
18.Computer networksHard

By the Nyquist formula, what is the maximum bit rate of a noiseless 3 kHz channel using 4 signal levels?

  1. A12 kbit/s
  2. B6 kbit/s
  3. C3 kbit/s
  4. D24 kbit/s
19.IPv4 subnettingMedium
/?11111111111111111111100000000000network bitshost bits

A department needs 1552 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1552 hosts
p =
20.IPv4 subnettingHard
last octet = 238112816413201618141201networkhost

A host has the address 10.20.159.238/25. What is the fourth octet x of the last usable host address in its subnet?

10.20.159.238/25
x =

3Solutions

1.Number systems

Answer: N = −122

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 + 4 + 2 = −122

Answer

N = −122
2.Number systems

Answer: R = 246

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.

Bit by bit

OR gives 1 where at least one bit is 1.

1011 0110
OR 0110 0000
= 1111 0110

Value of the result

1111 0110₂ = 246

Answer

R = 246
3.Propositional logic & proofs

Answer: contradiction

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

r = 1: ¬r = 0, r ∧ ¬r = 0
r = 0: ¬r = 1, r ∧ ¬r = 0

Classify

r ∧ ¬r: 0, 0
0 in every row: a contradiction.

Answer

contradiction
4.Propositional logic & proofs

Answer: N = 9

Truth tables: a formula with n different variables has 2ⁿ rows, one for every combination of truth values (1 = true, 0 = false). Work from the inside out and give every subformula its own column. ¬a flips the value; a ∧ b is 1 only when both are 1; a ∨ b is 0 only when both are 0; a → b is 0 only when a = 1 and b = 0; a ↔ b is 1 when both sides are equal.

Columns

¬r; ¬r → p; (¬r → p) ∨ q; ¬s; ¬s → ¬r; s ∧ (¬s → ¬r); ((¬r → p) ∨ q) → (s ∧ (¬s → ¬r))

Split on p

p = 1: 4
p = 0: 5
N = 4 + 5 = 9

Answer

N = 9
5.Digital logic: gates, K-maps & flip-flops

Answer: N = 6

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.

ABCD000111100001111010010302141517061120131150140819011110

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

Minimal POS form

F = (B + C′ + D′)(B′ + C′ + D)(A + C′ + D)(A + B + D′)(A′ + B + C + D)(A′ + B′ + C + D′)
N = 6

Answer

N = 6

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 = 2

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.

B′C′D, BC′D′, A′C′, ACD′, ABD′
ABCD0001111000011110X0110302X4X5X706112013015X140819011110

Essential prime implicants

B′C′D, ACD′
N = 2

Answer

N = 2
7.Number theory & modular arithmetic

Answer: x = 9

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).

Inverse of 10

r = 37, t = 0
r = 10, q = 3, t = 1
r = 7, q = 1, t = −3
r = 3, q = 2, t = 4
r = 1, q = 3, t = −11
10−1 ≡ 26 (mod 37)

Solve

x ≡ 26·16 = 416 ≡ 9 (mod 37)
check: 10·9 = 90 ≡ 16 (mod 37)

Answer

x = 9
8.Number theory & modular arithmetic

Answer: 5190 mod 57 = 43

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.

Binary exponent

190 = 101111102 = 128 + 32 + 16 + 8 + 4 + 2

Repeated squaring

51 ≡ 5 (mod 57)
52 ≡ 25 (mod 57)
54 ≡ 55 (mod 57)
58 ≡ 4 (mod 57)
516 ≡ 16 (mod 57)
532 ≡ 28 (mod 57)
564 ≡ 43 (mod 57)
5128 ≡ 25 (mod 57)

Multiply the needed powers

5190 ≡ 25·55·4·16·28·25 ≡ 43 (mod 57)

Answer

5190 mod 57 = 43
9.Algorithms & complexity

Answer: A · Shortest paths between all pairs of vertices in O(V³)

Why this is the answer

Floyd–Warshall is a dynamic programming algorithm that allows each vertex in turn as an intermediate point, using O(V³) time.

Answer

A · Shortest paths between all pairs of vertices in O(V³)
10.Algorithms & complexity

Answer: D · 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

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

Answer: B · It is symmetric about the main diagonal

Why this is the answer

In an undirected graph edge {u, v} means both M[u][v] and M[v][u] are set, so the matrix equals its transpose.

Answer

B · It is symmetric about the main diagonal
12.Data structures

Answer: D · Answer range queries (such as range sums) and updates in O(log n)

Why this is the answer

A segment tree stores aggregates for intervals, supporting range queries and point updates in O(log n) time.

Answer

D · Answer range queries (such as range sums) and updates in O(log n)
13.Graph theory

Answer: H = 360

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 = (7 − 1)!/2 = 720/2 = 360

Answer

H = 360
14.Graph theory

Answer: N = 1

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 1 0 0
B: 1 0 1 1 1
C: 1 1 0 0 1
D: 0 1 0 0 0
E: 0 1 1 0 0

Common neighbours

(M²)AC = 0·1 + 1·1 + 1·0 + 0·0 + 0·1 = 1

Answer

N = 1
15.Databases & SQL

Answer: B · To speed up searches and joins on those columns

Why this is the answer

Indexes (often B-trees) speed up lookups, but take space and slow down inserts and updates that must maintain them.

Answer

B · To speed up searches and joins on those columns
16.Databases & SQL

Answer: D · Write skew

Why this is the answer

Under snapshot isolation both transactions see the old state and change different rows, so no write conflict is detected; serializable isolation prevents this.

Answer

D · Write skew
17.Computer networks

Answer: D · The destination MAC address is not yet in its address table

Why this is the answer

For an unknown unicast address the switch floods the frame. When the destination replies, the switch learns its port from the source MAC.

Answer

D · The destination MAC address is not yet in its address table
18.Computer networks

Answer: A · 12 kbit/s

Why this is the answer

2 × 3000 × log₂ 4 = 2 × 3000 × 2 = 12,000 bit/s. More levels raise the rate, but noise limits how many can be told apart.

Answer

A · 12 kbit/s
19.IPv4 subnetting

Answer: p = 21

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).

Host bits needed

210 − 2 = 1022 < 1552
211 − 2 = 2046 ≥ 1552 → h = 11

Prefix length

p = 32 − 11 = 21

Answer

p = /21
20.IPv4 subnetting

Answer: x = 254

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 of this host

network: ⌊238/128⌋·128 = 128
broadcast: 128 + 128 − 1 = 255
usable hosts: 129 … 254

Answer

x = 254

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