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 complement1−12806403201608041211

The 8-bit pattern 1000 0011 stores a signed integer in two's complement. What is its value N in decimal?

1000 00118 bit
N =
2.Number systemsHard
16-bit two's complement, N = ??sign???????????????

What is the most negative integer N that can be stored in 16-bit two's complement?

16 bit
N =
3.Propositional logic & proofsMedium
¬q ∧ p¬q ∨ p¬q ∧ ¬p¬q ∨ ¬p

Which formula is logically equivalent to ¬(q ∧ p)? Click it.

¬(q ∧ p)
  1. 1¬q ∧ p
  2. 2¬q ∨ p
  3. 3¬q ∧ ¬p
  4. 4¬q ∨ ¬p
4.Propositional logic & proofsHard
prp → rrp11100100

The argument with premises p → r; r and conclusion p is invalid. Click the row of the truth table that is a counterexample.

p → rr∴ p
  1. 1p = 1, r = 1
  2. 2p = 1, r = 0
  3. 3p = 0, r = 1
  4. 4p = 0, r = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD00011110000111101111111

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(4, 5, 6, 7, 11, 12, 13)
  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
CLKT1234567

A positive-edge-triggered T flip-flop starts with Q = 0 and receives the inputs shown. Click the first clock edge at which Q goes from 1 back to 0.

Q₀ = 0T
  1. 1clock edge 1
  2. 2clock edge 2
  3. 3clock edge 3
  4. 4clock edge 4
  5. 5clock edge 5
  6. 6clock edge 6
  7. 7clock edge 7
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 2 (mod 3) and x ≡ 0 (mod 13).

x ≡ 2 (mod 3)x ≡ 0 (mod 13)
x =
8.Number theory & modular arithmeticHard

How many solutions N does 16x ≡ 52 (mod 58) have in {0, 1, …, 57}?

16x ≡ 52 (mod 58)
N =
9.Algorithms & complexityMedium

What is the time complexity of computing the n-th Fibonacci number with a bottom-up table?

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

What would follow if someone found a polynomial-time algorithm for one NP-complete problem?

  1. AAll exponential-time algorithms would immediately run in linear time
  2. BThe halting problem would become decidable
  3. CEvery problem in NP could be solved in polynomial time, so P = NP
  4. DOnly that one problem would move into P
11.Data structuresMedium

A hash table of size 7 uses h(k) = k mod 7 and linear probing. After inserting 10, where is 17 stored?

  1. AIndex 3
  2. BIndex 0
  3. CIndex 6
  4. DIndex 4
12.Data structuresHard

Which property keeps every B-tree perfectly balanced?

  1. AEvery node has exactly two children
  2. BAll leaves are at the same depth
  3. CEach node stores one key
  4. DThe keys are stored only in the root
13.Graph theoryMedium

A tree has 4 vertices of degree 3, 2 vertices of degree 4, 3 vertices of degree 5, and all its other vertices are leaves (degree 1). How many leaves L does it have?

4 × deg 32 × deg 43 × deg 5
L =
14.Graph theoryHard
9201013216457ABCDEFG

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

|V| = 7|E| = 9
W =
15.Databases & SQLMedium

What does INTERSECT return when used between two SELECT queries?

  1. ARows that appear in both results
  2. BAll rows from both results
  3. CThe Cartesian product of the results
  4. DRows in the first result but not the second
16.Databases & SQLHard

What does SELECT day, SUM(amount) OVER (ORDER BY day) FROM sales compute?

  1. AA running total of amount ordered by day
  2. BThe average amount per day
  3. COne row per day with that day's total only
  4. DThe total of all amounts on every row
17.Computer networksMedium

Which metric does RIP use to choose routes?

  1. ALink bandwidth
  2. BMeasured delay
  3. CBusiness policy
  4. DHop count
18.Computer networksHard

You must divide 192.168.1.0/24 into at least 5 subnets with up to 25 hosts each. Which prefix length fits best?

  1. A/27
  2. B/26
  3. C/28
  4. D/25
19.IPv4 subnettingMedium
/24 → /2711111111111111111111111111100000network bitshost bits

The network 192.168.90.0/24 is divided into /27 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?

192.168.90.0/24/27subnet 3 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
last octet = 78012816403201618141201networkhost

A host has the address 10.20.108.78/29. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.108.78/29
x =

3Solutions

1.Number systems

Answer: N = −125

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 + 2 + 1 = −125

Answer

N = −125
2.Number systems

Answer: N = −32768

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

−215 = −32768

Answer

N = −32768
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).

The law

De Morgan: ¬(a ∧ b) ≡ ¬a ∨ ¬b. The ∧ turns into ∨ and both parts are negated.
¬(q ∧ p) ≡ ¬q ∨ ¬p

Check with the truth table

p = 1, q = 1: 0 | 0 1 0 0
p = 1, q = 0: 1 | 1 1 0 1
p = 0, q = 1: 1 | 0 0 0 1
p = 0, q = 0: 1 | 0 1 1 1

Answer

¬q ∨ ¬p

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

4.Propositional logic & proofs

Answer: p = 0, r = 1

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.

Premises and conclusion row by row

p = 1, r = 1: 1 1 | 1
p = 1, r = 0: 0 0 | 1
p = 0, r = 1: 1 1 | 0
p = 0, r = 0: 1 0 | 0

Counterexample

p = 0, r = 1

Answer

p = 0, r = 1

An argument with this form looks like a valid rule, but one row is enough to show that it is not.

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

Answer: AB = 10, CD = 11

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.

m11: m10, m9, m15, m3 → 0
ABCD00011110000111101111111

The isolated 1

m11 = AB′CD → AB = 10, CD = 11

Answer

AB = 10, CD = 11
6.Digital logic: gates, K-maps & flip-flops

Answer: clock edge 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 T flip-flop toggles at each rising edge with T = 1 and holds with T = 0.

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

First change from 1 to 0

clock edge 3

Answer

clock edge 3
7.Number theory & modular arithmetic

Answer: x = 26

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 = 2 + 3k

Second congruence

2 + 3k ≡ 0 (mod 13) → 3k ≡ 11 (mod 13)
k ≡ 9·11 ≡ 8 (mod 13)

Result

x = 2 + 3·8 = 26
x = 26
check: 26 mod 3 = 2, 26 mod 13 = 0

Answer

x = 26
8.Number theory & modular arithmetic

Answer: N = 2

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).

Divide by the gcd

gcd(16, 58) = 2
8x ≡ 26 (mod 29)

Solve the reduced congruence

8−1 ≡ 11 (mod 29) → x ≡ 25 (mod 29)

Solutions modulo 58

x ∈ {25, 54}
N = 2

Answer

N = 2
9.Algorithms & complexity

Answer: A · O(n)

Why this is the answer

Filling the table from 0 to n takes one addition per entry, so the work is linear instead of exponential.

Answer

A · O(n)
10.Algorithms & complexity

Answer: C · Every problem in NP could be solved in polynomial time, so P = NP

Why this is the answer

Since all of NP reduces to any NP-complete problem in polynomial time, a fast algorithm for one gives a fast algorithm for all.

Answer

C · Every problem in NP could be solved in polynomial time, so P = NP
11.Data structures

Answer: D · Index 4

Why this is the answer

10 mod 7 = 3 and 17 mod 7 = 3; index 3 is taken, so linear probing tries the next slot, index 4.

Answer

D · Index 4
12.Data structures

Answer: B · All leaves are at the same depth

Why this is the answer

Nodes split upwards when full and the tree only gets taller when the root splits, so all leaves always stay on the same level.

Answer

B · All leaves are at the same depth
13.Graph theory

Answer: L = 19

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.

Handshake lemma for a tree

L + 35 = 2(L + 9) − 2
L = 19

Answer

L = 19

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

14.Graph theory

Answer: W = 46

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

DE (2): take
GC (4): take
GD (5): take
GE (7): skip, closes a cycle
AB (9): take
BC (10): take
CE (13): skip, closes a cycle
DF (16): take

Total weight

W = 2 + 4 + 5 + 9 + 10 + 16 = 46

Answer

W = 46
15.Databases & SQL

Answer: A · Rows that appear in both results

Why this is the answer

INTERSECT keeps only rows present in both queries, and like UNION it removes duplicates by default.

Answer

A · Rows that appear in both results
16.Databases & SQL

Answer: A · A running total of amount ordered by day

Why this is the answer

With ORDER BY and no explicit frame, the window runs from the first row up to the current row (and its ties), which gives a cumulative sum.

Answer

A · A running total of amount ordered by day
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 · /27

Why this is the answer

/27 gives 8 subnets with 30 usable hosts each. /26 gives only 4 subnets, and /28 allows only 14 hosts per subnet.

Answer

A · /27
19.IPv4 subnetting

Answer: x = 64

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

25 = 32

Subnet number 3

network: (3 − 1)·32 = 64
broadcast: 64 + 32 − 1 = 95

Answer

x = 64
20.IPv4 subnetting

Answer: x = 79

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

23 = 8

Subnet of this host

network: ⌊78/8⌋·8 = 72
broadcast: 72 + 8 − 1 = 79
usable hosts: 73 … 78

Answer

x = 79

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