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 −97 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −978 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1001 0101 >> 1. What is the result's value R in decimal?

1001 0101>> 1
R =
3.Propositional logic & proofsMedium
¬b ∨ a¬b ∧ a¬b ∧ ¬a¬b ∨ ¬a

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

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

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

(r ↔ (s ∨ r)) ∧ ((q → ¬p) ∧ (q → s))
N =
5.Digital logic: gates, K-maps & flip-flopsMedium
MUX1I0​1I1​0I2​C′I3​ABS1​S0​F

A 4-to-1 multiplexer has its select inputs S₁ = A and S₀ = B, and its data inputs tied to I₀ = 1, I₁ = 1, I₂ = 0, I₃ = C′. For how many of the 8 combinations of A, B and C is the output F = 1? Find N.

I₀ = 1I₁ = 1I₂ = 0I₃ = C′
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110001113X204050716112113X15114X809011110

Find the minimal SOP form of F(A, B, C, D) = Σm(1, 3, 6, 10, 12, 13, 14) + d(2, 8, 15), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(1, 3, 6, 10, 12, 13, 14) + d(2, 8, 15)
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 7 modulo 33: the number x with 0 < x < 33 and 7·x ≡ 1 (mod 33).

a = 7n = 33
7−1 mod 33 =
8.Number theory & modular arithmeticHard

Use repeated squaring to find 19145 mod 93.

a = 19k = 145n = 93
19145 mod 93 =
9.Algorithms & complexityMedium

What does the recurrence T(n) = T(n − 1) + O(1) solve to?

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

To show that a new problem X is NP-hard, what must you do?

  1. AShow that X has no polynomial algorithm for small inputs
  2. BReduce X to a known NP-hard problem in polynomial time
  3. CShow that X can be verified in polynomial time
  4. DReduce a known NP-hard problem to X in polynomial time
11.Data structuresMedium

Which data structure is used by depth-first search (iterative version)?

  1. AA stack
  2. BA hash table only
  3. CA queue
  4. DA heap
12.Data structuresHard

In a hash table using open addressing, why is deletion often done with a "tombstone" marker?

  1. AIt makes the table sorted
  2. BDeleted keys must be kept forever for security, so slots can never be reused
  3. CTombstones speed up hashing
  4. DEmptying the slot could break the probe sequence for other keys
13.Graph theoryMedium

A connected graph has 8 vertices and 15 edges. How many edges k must be removed to leave a spanning tree?

|V| = 8|E| = 15
k =
14.Graph theoryHard
2131412101615157ABCDEFG

Prim’s algorithm builds a minimum spanning tree starting from vertex A. Click the edge it adds as edge number 3.

start: Ak = 3
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge CE
  5. 5edge DE
  6. 6edge DF
  7. 7edge EF
  8. 8edge GC
  9. 9edge GD
  10. 10edge GE
15.Databases & SQLMedium

An employees table has a column manager_id that refers to another employee. How do you list each employee next to their manager's name?

  1. AUse GROUP BY manager_id only
  2. BJoin the employees table with itself using two aliases
  3. CCombine the table with itself using UNION ALL and a WHERE filter
  4. DCreate a second copy of the table on disk
16.Databases & SQLHard

What is a covering index for a query?

  1. AAn index that covers every column of every table
  2. BAn index that is rebuilt every night
  3. CAn index on the primary key only
  4. DAn index that contains all columns the query needs, so the table itself need not be read
17.Computer networksMedium

Which DNS record type maps a host name to an IPv6 address?

  1. AAAAA
  2. BMX
  3. CA
  4. DCNAME
18.Computer networksHard

What does the Diffie-Hellman key exchange achieve on its own, and what is its weakness?

  1. AIt encrypts bulk data directly over a public channel, but it is far too slow for large files and video
  2. BIt agrees a shared secret over a public channel, but alone it is open to a man-in-the-middle
  3. CIt signs certificates for websites, but only works with short keys
  4. DIt authenticates both users with passwords, but needs a central server
19.IPv4 subnettingMedium
/?11111111111111111111110000000000network bitshost bits

A department needs 809 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

809 hosts
p =
20.IPv4 subnettingHard
last octet = 155112806403211618041211networkhost

A host has the address 10.20.197.155/30. What is the fourth octet x of the last usable host address in its subnet?

10.20.197.155/30
x =

3Solutions

1.Number systems

Answer: U = 159

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

97 = 0110 0001
inverted: 1001 1110
+1 → 1001 1111

Read it as unsigned

1001 1111₂ = 256 − 97 = 159

Answer

U = 159
2.Number systems

Answer: R = 74

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

Shifting right by 1 moves every bit 1 places to the right; bits that fall off the right end are lost and zeros come in on the left.

1001 0101 >> 1
= 0100 1010

Value of the result

0100 1010₂ = 74

Answer

R = 74
3.Propositional logic & proofs

Answer: ¬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

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

Check with the truth table

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

Answer

¬b ∨ ¬a

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

4.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.

Columns

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

Split on p

p = 1: 3
p = 0: 4
N = 3 + 4 = 7

Answer

N = 7
5.Digital logic: gates, K-maps & flip-flops

Answer: N = 5

Multiplexer: a 2ⁿ-to-1 multiplexer passes one of its data inputs to the output; the select lines, read as a binary number, pick which one. For a 4-to-1 MUX with selects S₁S₀ = AB, F = A′B′·I₀ + A′B·I₁ + AB′·I₂ + AB·I₃. Tying the data inputs to 0, 1, C or C′ makes the MUX produce any function of A, B and C, which is a common way to build logic with one chip.

Output expression

F = A′B′ + A′B + ABC′

Rows with F = 1

ABC = 000, 001, 010, 011, 110
N = 5

Answer

N = 5
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.

CD′, AD′, AB, A′B′D, A′B′C
ABCD0001111000011110001113X204050716112113X15114X809011110

Minimal cover

F = CD′ + AB + A′B′D
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: 7−1 mod 33 = 19

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·7 (mod 33).

r = 33, t = 0
r = 7, q = 4, t = 1
r = 5, q = 1, t = −4
r = 2, q = 2, t = 5
r = 1, q = 2, t = −14

Inverse

1 ≡ −14·7 (mod 33)
x = 19
check: 7·19 = 133 = 4·33 + 1

Answer

7−1 mod 33 = 19
8.Number theory & modular arithmetic

Answer: 19145 mod 93 = 25

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

145 = 100100012 = 128 + 16 + 1

Repeated squaring

191 ≡ 19 (mod 93)
192 ≡ 82 (mod 93)
194 ≡ 28 (mod 93)
198 ≡ 40 (mod 93)
1916 ≡ 19 (mod 93)
1932 ≡ 82 (mod 93)
1964 ≡ 28 (mod 93)
19128 ≡ 40 (mod 93)

Multiply the needed powers

19145 ≡ 19·19·40 ≡ 25 (mod 93)

Answer

19145 mod 93 = 25
9.Algorithms & complexity

Answer: B · O(n)

Why this is the answer

There are n levels with constant work each, as in a recursive sum of an array.

Answer

B · O(n)
10.Algorithms & complexity

Answer: D · Reduce a known NP-hard problem to X in polynomial time

Why this is the answer

If a known hard problem can be transformed into X efficiently, a fast algorithm for X would solve the known problem too.

Answer

D · Reduce a known NP-hard problem to X in polynomial time
11.Data structures

Answer: A · A stack

Why this is the answer

DFS explores as deep as possible, naturally using a stack (or recursion's call stack); BFS uses a queue.

Answer

A · A stack
12.Data structures

Answer: D · Emptying the slot could break the probe sequence for other keys

Why this is the answer

If a deleted slot became empty, later searches might stop early and miss keys that were placed further along the probe chain.

Answer

D · Emptying the slot could break the probe sequence for other keys
13.Graph theory

Answer: k = 8

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.

Edges of a spanning tree

8 − 1 = 7

Edges to remove

k = 15 − 7 = 8

Answer

k = 8
14.Graph theory

Answer: edge GC

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.

Prim’s algorithm

1. AB (2)
2. AC (13)
3. GC (5)
4. GE (7)
5. EF (1)
6. DE (10)

edge number 3

GC

Answer

edge GC
15.Databases & SQL

Answer: B · Join the employees table with itself using two aliases

Why this is the answer

A self-join such as FROM employees e JOIN employees m ON e.manager_id = m.id lets one alias be the employee and the other the manager.

Answer

B · Join the employees table with itself using two aliases
16.Databases & SQL

Answer: D · An index that contains all columns the query needs, so the table itself need not be read

Why this is the answer

If all referenced columns are in the index, the DBMS can do an index-only scan and skip fetching table rows, which saves I/O.

Answer

D · An index that contains all columns the query needs, so the table itself need not be read
17.Computer networks

Answer: A · AAAA

Why this is the answer

An A record holds an IPv4 address, AAAA an IPv6 address. MX names mail servers and CNAME gives an alias.

Answer

A · AAAA
18.Computer networks

Answer: B · It agrees a shared secret over a public channel, but alone it is open to a man-in-the-middle

Why this is the answer

Diffie-Hellman lets two parties derive the same secret that eavesdroppers cannot compute. An attacker in the middle can run separate exchanges with each side unless the values are signed or otherwise authenticated.

Answer

B · It agrees a shared secret over a public channel, but alone it is open to a man-in-the-middle
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 < 809
210 − 2 = 1022 ≥ 809 → h = 10

Prefix length

p = 32 − 10 = 22

Answer

p = /22
20.IPv4 subnetting

Answer: x = 154

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

22 = 4

Subnet of this host

network: ⌊155/4⌋·4 = 152
broadcast: 152 + 4 − 1 = 155
usable hosts: 153 … 154

Answer

x = 154

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