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 −84 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −848 bit
U =
2.Number systemsHard
20-bit two's complement, N = ??sign???????????????????

What is the most negative integer N that can be stored in 20-bit two's complement?

20 bit
N =
3.Propositional logic & proofsMedium
r ∧ (p ∨ q)(r ∨ p) ∧ q(r ∧ p) ∨ (r ∧ q)(r ∨ p) ∧ (r ∨ q)

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

r ∨ (p ∧ q)
  1. 1r ∧ (p ∨ q)
  2. 2(r ∨ p) ∧ q
  3. 3(r ∧ p) ∨ (r ∧ q)
  4. 4(r ∨ p) ∧ (r ∨ q)
4.Propositional logic & proofsHard
pqq → p¬q¬p11100100

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

q → p¬q∴ ¬p
  1. 1p = 1, q = 1
  2. 2p = 1, q = 0
  3. 3p = 0, q = 1
  4. 4p = 0, q = 0
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111000011312141507060121130150141809011010

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(2, 3, 4, 5, 8, 13). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(2, 3, 4, 5, 8, 13)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKJK123456

A positive-edge-triggered JK 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 = 0JK
  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.Number theory & modular arithmeticMedium

Find Euler’s totient φ(361), the number of integers from 1 to 361 that are coprime to 361.

n = 361 = 192
φ(361) =
8.Number theory & modular arithmeticHard

19 is prime. Use Fermat’s little theorem to find 2876109 mod 19.

p = 19k = 76109
2876109 mod 19 =
9.Algorithms & complexityMedium

Which sorting algorithm is stable?

  1. AMerge sort
  2. BHeap sort
  3. CSelection sort (typical version)
  4. DQuicksort (typical in-place version)
10.Algorithms & complexityHard

Which statement about P and NP is correct?

  1. ANP problems cannot be verified quickly
  2. BEvery problem in P is also in NP
  3. CIt has been proven that P ≠ NP
  4. DNP means "not polynomial"
11.Data structuresMedium

How is a new element inserted into a binary heap?

  1. ARebuild the whole heap from scratch
  2. BInsert it at its sorted position by shifting all of the larger elements one place right
  3. CPlace it at the root and sift it down
  4. DPlace it at the end of the array and sift it up while it violates the heap property
12.Data structuresHard

A Bloom filter can answer "is this element in the set?" with:

  1. AAnswers only for numbers
  2. BPossible false negatives but no false positives
  3. CExact answers always
  4. DPossible false positives but no false negatives
13.Graph theoryMedium

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

3 × deg 34 × deg 5
L =
14.Graph theoryHard
92651312111214ABCDEFG

Use Dijkstra’s algorithm to find the length d(D) of the shortest path from A to D in the weighted graph.

start: Aend: D
d(D) =
15.Databases & SQLMedium

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

  1. AAn error in every database
  2. BAll rows of the table
  3. COnly the rows where col is NULL, as with IS NULL
  4. DNo rows, because the comparison is unknown
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. ANon-repeatable read within one transaction
  2. BWrite skew
  3. CDirty read
  4. DDirty write
17.Computer networksMedium

How many usable host addresses does a /24 IPv4 network have?

  1. A254
  2. B252
  3. C255
  4. D256
18.Computer networksHard

Why does the side that closes a TCP connection first stay in the TIME_WAIT state for twice the maximum segment lifetime?

  1. ATo wait for the DNS cache to expire
  2. BSo delayed segments die out and the final ACK can be resent if it was lost
  3. CTo let the routers along the path update their routing tables for the next connection
  4. DTo keep the encryption keys for later reuse
19.IPv4 subnettingMedium
/?11111111111111111111110000000000network bitshost bits

A department needs 1016 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

1016 hosts
p =
20.IPv4 subnettingHard
last octet = 76012816403201618140201networkhost

A host has the address 10.20.56.76/25. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.56.76/25
x =

3Solutions

1.Number systems

Answer: U = 172

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

84 = 0101 0100
inverted: 1010 1011
+1 → 1010 1100

Read it as unsigned

1010 1100₂ = 256 − 84 = 172

Answer

U = 172
2.Number systems

Answer: N = −524288

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

−219 = −524288

Answer

N = −524288
3.Propositional logic & proofs

Answer: (r ∨ p) ∧ (r ∨ q)

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).
r ∨ (p ∧ q) ≡ (r ∨ p) ∧ (r ∨ q)

Check with the truth table

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

Answer

(r ∨ p) ∧ (r ∨ q)

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

4.Propositional logic & proofs

Answer: p = 1, q = 0

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

Counterexample

p = 1, q = 0

Answer

p = 1, q = 0

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: N = 5

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.

ABCD000111100001111000011312141507060121130150141809011010

Group the 0s

Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.

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

Minimal POS form

F = (B′ + C′)(B + C + D′)(A′ + C′)(A′ + B′ + D)(A + B + C)
N = 5

Answer

N = 5

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: clock edge 4

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 JK flip-flop: 00 holds, 01 resets (Q = 0), 10 sets (Q = 1), 11 toggles.

1: J = 0, K = 1 → Q = 0
2: J = 0, K = 0 → Q = 0
3: J = 1, K = 1 → Q = 1
4: J = 0, K = 1 → Q = 0
5: J = 0, K = 0 → Q = 0
6: J = 1, K = 0 → Q = 1

First change from 1 to 0

clock edge 4

Answer

clock edge 4
7.Number theory & modular arithmetic

Answer: φ(361) = 342

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

361 = 192

Totient

φ(361) = 361 − 19 = 342

Answer

φ(361) = 342
8.Number theory & modular arithmetic

Answer: 2876109 mod 19 = 16

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.

Reduce the exponent

76109 mod 18 = 5
2876109 ≡ 285 (mod 19)

Small power

285 mod 19 = 16

Answer

2876109 mod 19 = 16
9.Algorithms & complexity

Answer: A · Merge sort

Why this is the answer

Merge sort is stable when merging takes from the left half first on ties. In-place quicksort, heap sort and selection sort are usually not stable.

Answer

A · Merge sort
10.Algorithms & complexity

Answer: B · Every problem in P is also in NP

Why this is the answer

P ⊆ NP: if you can solve a problem quickly you can verify a solution quickly. Whether P = NP is still open.

Answer

B · Every problem in P is also in NP
11.Data structures

Answer: D · Place it at the end of the array and sift it up while it violates the heap property

Why this is the answer

Adding at the end keeps the tree complete; swapping with the parent restores the order in O(log n) steps.

Answer

D · Place it at the end of the array and sift it up while it violates the heap property
12.Data structures

Answer: D · Possible false positives but no false negatives

Why this is the answer

If any bit for an item is 0, it is definitely absent; if all are 1 it is probably present. Bloom filters trade accuracy for very little memory.

Answer

D · Possible false positives but no false negatives
13.Graph theory

Answer: L = 17

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 + 29 = 2(L + 7) − 2
L = 17

Answer

L = 17

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

14.Graph theory

Answer: d(D) = 13

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: 9, C: 2
C (2): B: 8, E: 15, G: 16
B (8): D: 13
D (13): F: 24
E (15)
G (16)
F (24)

Shortest path

A → C → B → D
d(D) = 13

Answer

d(D) = 13
15.Databases & SQL

Answer: D · 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

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

Answer: B · 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

B · Write skew
17.Computer networks

Answer: A · 254

Why this is the answer

A /24 has 8 host bits, so 2⁸ = 256 addresses. The network and broadcast addresses are reserved, leaving 254 for hosts.

Answer

A · 254
18.Computer networks

Answer: B · So delayed segments die out and the final ACK can be resent if it was lost

Why this is the answer

TIME_WAIT ensures a lost final ACK can be repeated and old duplicate segments expire before the same address and port pair is reused.

Answer

B · So delayed segments die out and the final ACK can be resent if it was lost
19.IPv4 subnetting

Answer: p = 22

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

29 − 2 = 510 < 1016
210 − 2 = 1022 ≥ 1016 → h = 10

Prefix length

p = 32 − 10 = 22

Answer

p = /22
20.IPv4 subnetting

Answer: x = 127

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: ⌊76/128⌋·128 = 0
broadcast: 0 + 128 − 1 = 127
usable hosts: 1 … 126

Answer

x = 127

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