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 −13 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −138 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 0111 0100 OR 1001 0011. What is the result's value R in decimal?

0111 0100OR1001 0011
R =
3.Propositional logic & proofsMedium
abcc → a(c → a) ∨ b(b → ¬c) → ((c → a) ∨ b)111?110?101?100?011?010?001?000?

In how many of the 8 rows of its truth table is (b → ¬c) → ((c → a) ∨ b) true? Call this number N.

(b → ¬c) → ((c → a) ∨ b)
N =
4.Propositional logic & proofsHard
∃x ∃y (R(x, y) → ¬R(y, x))∃x ∃y (R(x, y) ∧ ¬R(y, x))∃x ∃y (¬R(x, y) ∧ ¬R(y, x))∀x ∀y (R(x, y) ∧ ¬R(y, x))

Which statement is equivalent to ¬∀x ∀y (R(x, y) → R(y, x)), with the negation moved all the way inside? Click it.

¬∀x ∀y (R(x, y) → R(y, x))
  1. 1∃x ∃y (R(x, y) → ¬R(y, x))
  2. 2∃x ∃y (R(x, y) ∧ ¬R(y, x))
  3. 3∃x ∃y (¬R(x, y) ∧ ¬R(y, x))
  4. 4∀x ∀y (R(x, y) ∧ ¬R(y, x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC000111100101324576

Simplify F = A′BC + AC + ABC′ to its minimal sum-of-products form. How many literals L does the simplified expression contain?

F = A′BC + AC + ABC′
L =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD000111100001111010X1030204050716X121131151141819011X10

For F(A, B, C, D) = Σm(0, 6, 8, 9, 13, 14, 15) + d(1, 10, 12), where d marks the don't-care conditions, find the number N of essential prime implicants.

F(A, B, C, D) = Σm(0, 6, 8, 9, 13, 14, 15) + d(1, 10, 12)
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 58 modulo 79: the number x with 0 < x < 79 and 58·x ≡ 1 (mod 79).

a = 58n = 79
58−1 mod 79 =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (65, 11)d = ?

An RSA key uses the primes p = 5 and q = 13 and the public exponent e = 11. Find the private exponent d (0 < d < φ(n)).

p = 5q = 13e = 11
d =
9.Algorithms & complexityMedium

What is the running time of LSD radix sort on n integers with d digits in base k?

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

Why does binary search not give O(log n) time on a singly linked list?

  1. AReaching the middle element takes O(n) steps, since there is no direct index access
  2. BBinary search needs at least twice the memory
  3. CComparisons on linked list nodes are slower
  4. DLinked lists cannot be sorted
11.Data structuresMedium

What is the cost of inserting an element in the middle of an array of n elements?

  1. AO(n), because later elements must shift
  2. BO(log n), because of binary search
  3. CO(1), because the index is known
  4. DO(n²), because each element moves n times
12.Data structuresHard

Which data structure supports an LRU (least recently used) cache with O(1) get and put?

  1. AA binary search tree only
  2. BA hash map combined with a doubly linked list
  3. CA single sorted array
  4. DA stack only
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
89462010171431119ABCDEFG

Find the total weight W of a minimum spanning tree of the weighted graph.

|V| = 7|E| = 11
W =
15.Databases & SQLMedium

Second normal form (2NF) removes:

  1. APartial dependencies of non-key columns on part of a composite key
  2. BTransitive dependencies between non-key columns via another non-key column
  3. CAll NULL values
  4. DForeign keys
16.Databases & SQLHard

Table b contains the values 1, 2 and NULL in column y. What does SELECT * FROM a WHERE x NOT IN (SELECT y FROM b) return?

  1. ARows of a where x is not 1 or 2
  2. BOnly rows of a where x is NULL
  3. CNo rows at all
  4. DEvery row of a
17.Computer networksMedium

Which single prefix summarises both 192.168.0.0/24 and 192.168.1.0/24?

  1. A192.168.1.0/23
  2. B192.168.0.0/23
  3. C192.168.0.0/25
  4. D192.168.0.0/22
18.Computer networksHard

In a TCP connection, what does the receive window (rwnd) control?

  1. AThe retransmission time-out, calculated from the measured round-trip time of each segment
  2. BThe number of open ports
  3. CHow much unacknowledged data the sender may send, based on the receiver's buffer
  4. DThe size of each IP packet header
19.IPv4 subnettingMedium
/24 → /2611111111111111111111111111000000network bitshost bits

The network 192.168.79.0/24 is divided into /26 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 3?

192.168.79.0/24/26subnet 3 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
third octet = 17012806403211608040211networkhost

A host has the address 172.20.17.113/20. What is the third octet x of its subnet's broadcast address?

172.20.17.113/20
x =

3Solutions

1.Number systems

Answer: U = 243

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

13 = 0000 1101
inverted: 1111 0010
+1 → 1111 0011

Read it as unsigned

1111 0011₂ = 256 − 13 = 243

Answer

U = 243
2.Number systems

Answer: R = 247

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.

0111 0100
OR 1001 0011
= 1111 0111

Value of the result

1111 0111₂ = 247

Answer

R = 247
3.Propositional logic & proofs

Answer: N = 7

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.

Truth table row by row

a = 1, b = 1, c = 1: ¬c = 0, b → ¬c = 0, c → a = 1, (c → a) ∨ b = 1, (b → ¬c) → ((c → a) ∨ b) = 1
a = 1, b = 1, c = 0: ¬c = 1, b → ¬c = 1, c → a = 1, (c → a) ∨ b = 1, (b → ¬c) → ((c → a) ∨ b) = 1
a = 1, b = 0, c = 1: ¬c = 0, b → ¬c = 1, c → a = 1, (c → a) ∨ b = 1, (b → ¬c) → ((c → a) ∨ b) = 1
a = 1, b = 0, c = 0: ¬c = 1, b → ¬c = 1, c → a = 1, (c → a) ∨ b = 1, (b → ¬c) → ((c → a) ∨ b) = 1
a = 0, b = 1, c = 1: ¬c = 0, b → ¬c = 0, c → a = 0, (c → a) ∨ b = 1, (b → ¬c) → ((c → a) ∨ b) = 1
a = 0, b = 1, c = 0: ¬c = 1, b → ¬c = 1, c → a = 1, (c → a) ∨ b = 1, (b → ¬c) → ((c → a) ∨ b) = 1
a = 0, b = 0, c = 1: ¬c = 0, b → ¬c = 1, c → a = 0, (c → a) ∨ b = 0, (b → ¬c) → ((c → a) ∨ b) = 0
a = 0, b = 0, c = 0: ¬c = 1, b → ¬c = 1, c → a = 1, (c → a) ∨ b = 1, (b → ¬c) → ((c → a) ∨ b) = 1

Count the rows with value 1

(b → ¬c) → ((c → a) ∨ b): 1, 1, 1, 1, 1, 1, 0, 1
N = 7

Answer

N = 7
4.Propositional logic & proofs

Answer: ∃x ∃y (R(x, y) ∧ ¬R(y, x))

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.

Move the negation inwards

¬∀x ∀y (R(x, y) → R(y, x)) ≡ ∃x ∃y (R(x, y) ∧ ¬R(y, x))

Answer

∃x ∃y (R(x, y) ∧ ¬R(y, x))

The other options either keep a quantifier unchanged, change the order of the quantifiers, or negate the inner formula wrongly.

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

Answer: L = 6

Boolean algebra: the rules that simplify expressions are X + X′ = 1 and X·X′ = 0, X + X = X, absorption X + X·Y = X, the combining rule X·Y + X·Y′ = X, X + X′Y = X + Y and the consensus theorem XY + X′Z + YZ = XY + X′Z. De Morgan's laws turn (X·Y)′ into X′ + Y′ and (X + Y)′ into X′·Y′. A Karnaugh map applies these rules graphically: every group of two neighbouring cells is one use of XY + XY′ = X.

Minterms of the expression

A′BC → m(3)
AC → m(5, 7)
ABC′ → m(6)
ABC00011110010001130204151716

Group and read off

F = BC + AC + AB

Count the literals

L = 6

Answer

L = 6
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.

B′C′, BCD′, AD′, AC′, AB
ABCD000111100001111010X1030204050716X121131151141819011X10

Essential prime implicants

B′C′, BCD′, AB
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: 58−1 mod 79 = 15

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·58 (mod 79).

r = 79, t = 0
r = 58, q = 1, t = 1
r = 21, q = 2, t = −1
r = 16, q = 1, t = 3
r = 5, q = 3, t = −4
r = 1, q = 5, t = 15

Inverse

1 ≡ 15·58 (mod 79)
x = 15
check: 58·15 = 870 = 11·79 + 1

Answer

58−1 mod 79 = 15
8.Number theory & modular arithmetic

Answer: d = 35

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 = 5·13 = 65
φ(n) = 4·12 = 48

Extended Euclidean algorithm

r = 48, t = 0
r = 11, q = 4, t = 1
r = 4, q = 2, t = −4
r = 3, q = 1, t = 9
r = 1, q = 3, t = −13

Private exponent

d = 35
check: 11·35 = 385 = 8·48 + 1

Answer

d = 35
9.Algorithms & complexity

Answer: B · O(d(n + k))

Why this is the answer

Each of the d passes is a counting sort costing O(n + k), so for fixed d and k the total is linear in n.

Answer

B · O(d(n + k))
10.Algorithms & complexity

Answer: A · Reaching the middle element takes O(n) steps, since there is no direct index access

Why this is the answer

Binary search needs O(1) access to the middle; on a linked list walking there costs linear time, so the total is O(n).

Answer

A · Reaching the middle element takes O(n) steps, since there is no direct index access
11.Data structures

Answer: A · O(n), because later elements must shift

Why this is the answer

All elements after the insertion point move one position to the right, which takes time proportional to n.

Answer

A · O(n), because later elements must shift
12.Data structures

Answer: B · A hash map combined with a doubly linked list

Why this is the answer

The hash map finds nodes in O(1) and the doubly linked list moves accessed items to the front and evicts from the back in O(1).

Answer

B · A hash map combined with a doubly linked list
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) = 2, deg(B) = 3, deg(C) = 1, deg(D) = 2, deg(E) = 3, deg(F) = 1

Odd vertices

B, C, E, F → 4
neither

Answer

neither
14.Graph theory

Answer: W = 48

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.

Kruskal’s algorithm

GC (3): take
BC (4): take
BD (6): take
AB (8): take
AC (9): skip, closes a cycle
DE (10): take
GD (11): skip, closes a cycle
GB (14): skip, closes a cycle
EF (17): take

Total weight

W = 3 + 4 + 6 + 8 + 10 + 17 = 48

Answer

W = 48
15.Databases & SQL

Answer: A · Partial dependencies of non-key columns on part of a composite key

Why this is the answer

2NF requires 1NF and that every non-key attribute depends on the whole key, not just part of it.

Answer

A · Partial dependencies of non-key columns on part of a composite key
16.Databases & SQL

Answer: C · No rows at all

Why this is the answer

The comparison with NULL is UNKNOWN, so the whole condition is never TRUE; NOT EXISTS or filtering out NULLs avoids this trap.

Answer

C · No rows at all
17.Computer networks

Answer: B · 192.168.0.0/23

Why this is the answer

0 and 1 share all bits of the third octet except the last, so one bit less of prefix (/23) starting at 192.168.0.0 covers both exactly.

Answer

B · 192.168.0.0/23
18.Computer networks

Answer: C · How much unacknowledged data the sender may send, based on the receiver's buffer

Why this is the answer

The receiver advertises how much buffer space it has; the sender limits unacknowledged data accordingly, preventing receiver overload.

Answer

C · How much unacknowledged data the sender may send, based on the receiver's buffer
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

26 = 64

Subnet number 3

network: (3 − 1)·64 = 128
broadcast: 128 + 64 − 1 = 191

Answer

x = 128
20.IPv4 subnetting

Answer: x = 31

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

24 = 16

Subnet of this host

network: 172.20.16.0
broadcast: 172.20.31.255

Answer

x = 31

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