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 complement1−12806403211618140211

The 8-bit pattern 1001 1101 stores a signed integer in two's complement. What is its value N in decimal?

1001 11018 bit
N =
2.Number systemsHard
32 bit1 px256 px256 pxM = ? KiB

An uncompressed image is 256 × 256 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

256 × 256 px32 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
(r ∨ q) ∧ ¬p¬(r ∨ q) → ¬pp → (r ∨ q)¬p → ¬(r ∨ q)

You want to prove (r ∨ q) → p by contrapositive. Which implication do you prove instead? Click it.

(r ∨ q) → p
  1. 1(r ∨ q) ∧ ¬p
  2. 2¬(r ∨ q) → ¬p
  3. 3p → (r ∨ q)
  4. 4¬p → ¬(r ∨ q)
4.Propositional logic & proofsHard
b → c¬c → ¬bc ∧ bc → b

The premises c → a; a → b are given. Which conclusion follows from them? Click it.

c → aa → b
  1. 1b → c
  2. 2¬c → ¬b
  3. 3c ∧ b
  4. 4c → b
5.Digital logic: gates, K-maps & flip-flopsMedium
MUXCI0​1I1​0I2​1I3​ABS1​S0​F

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

I₀ = CI₁ = 1I₂ = 0I₃ = 1
N =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKJK12345678

A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown at 8 rising clock edges. How many times N does Q change state?

Q₀ = 0JK
N =
7.Number theory & modular arithmeticMedium

Solve 13x ≡ 56 (mod 72). Give the solution x with 0 < x < 72.

13x ≡ 56 (mod 72)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (1927, 13)d = ?

An RSA key uses the primes p = 41 and q = 47 and the public exponent e = 13. Find the private exponent d (0 < d < φ(n)).

p = 41q = 47e = 13
d =
9.Algorithms & complexityMedium

Which algorithm finds the shortest paths from one source in a graph with non-negative edge weights?

  1. ADijkstra's algorithm
  2. BKruskal's algorithm
  3. CBellman–Ford is required in all cases
  4. DDepth-first search
10.Algorithms & complexityHard

What is the difference between an undecidable problem and an intractable one?

  1. AUndecidable problems are in NP, while intractable problems are in P
  2. BBoth mean the problem has no algorithm at all
  3. CAn undecidable problem has no algorithm for all inputs; an intractable one has one, but it is too slow
  4. DIntractable problems cannot be solved by any algorithm, while undecidable ones only need exponential time
11.Data structuresMedium

Which data structure does a running program use to keep track of active function calls and their local variables?

  1. AA binary search tree of frames
  2. BA hash table of functions
  3. CThe call stack
  4. DThe heap memory queue
12.Data structuresHard

Why are B-trees used in databases and file systems?

  1. AThey store data unsorted
  2. BThey use less memory than any array, so the entire index always fits in main memory
  3. CThey need no rebalancing
  4. DTheir high branching factor keeps the tree shallow, reducing slow disk reads
13.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K5 have? Circuits that differ only in the starting vertex or the direction count as the same.

K5
H =
14.Graph theoryHard
4143197815ABCDE

Prim’s algorithm builds a minimum spanning tree starting from vertex E. Click the edge it adds as edge number 2.

start: Ek = 2
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge BE
15.Databases & SQLMedium

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

  1. AOnly the rows where col is NULL, as with IS NULL
  2. BNo rows, because the comparison is unknown
  3. CAn error in every database
  4. DAll rows of the table
16.Databases & SQLHard

Why might a query with WHERE YEAR(order_date) = 2025 not use an index on order_date?

  1. AYEAR is not valid SQL
  2. BIndexes never work with dates
  3. CApplying a function to the column prevents a simple index range lookup
  4. DAn index on order_date is only used for ORDER BY clauses, never for WHERE
17.Computer networksMedium

How does TLS typically use asymmetric and symmetric cryptography?

  1. AOnly symmetric cryptography is used, with a fixed shared password
  2. BSymmetric methods agree keys and authenticate; asymmetric ciphers encrypt all the data
  3. CAsymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
  4. DOnly asymmetric cryptography is used throughout
18.Computer networksHard

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

  1. AHow much unacknowledged data the sender may send, based on the receiver's buffer
  2. BThe number of open ports
  3. CThe size of each IP packet header
  4. DThe retransmission time-out, calculated from the measured round-trip time of each segment
19.IPv4 subnettingMedium
/24 → /2911111111111111111111111111111000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 20 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/2420 subnets
H =
20.IPv4 subnettingHard
last octet = 16012806403211608040201networkhost

A host has the address 10.20.22.16/26. What is the fourth octet x of the last usable host address in its subnet?

10.20.22.16/26
x =

3Solutions

1.Number systems

Answer: N = −99

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 + 16 + 8 + 4 + 1 = −99

Answer

N = −99
2.Number systems

Answer: M = 256 KiB

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.

Bits and bytes

256·256·32 = 2097152 bit = 262144 B

Convert to KiB

262144/1024 = 256 KiB

Answer

M = 256 KiB
3.Propositional logic & proofs

Answer: ¬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).

Parts of the implication

a = r ∨ q, b = p

Contrapositive

¬b → ¬a = ¬p → ¬(r ∨ q)

Answer

¬p → ¬(r ∨ q)

The contrapositive is equivalent to the original statement, so proving it proves the statement.

4.Propositional logic & proofs

Answer: c → b

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.

Rows with all premises true

a = 1, b = 1, c = 1: b → c = 1, ¬c → ¬b = 1, c ∧ b = 1, c → b = 1
a = 1, b = 1, c = 0: b → c = 0, ¬c → ¬b = 0, c ∧ b = 0, c → b = 1
a = 0, b = 1, c = 0: b → c = 0, ¬c → ¬b = 0, c ∧ b = 0, c → b = 1
a = 0, b = 0, c = 0: b → c = 1, ¬c → ¬b = 1, c ∧ b = 0, c → b = 1

Conclusion

c → b (hypothetical syllogism)

Answer

c → b
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′C + A′B + AB

Rows with F = 1

ABC = 001, 010, 011, 110, 111
N = 5

Answer

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

Answer: N = 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 = 1, K = 1 → Q = 1
2: J = 0, K = 0 → Q = 1
3: J = 1, K = 0 → Q = 1
4: J = 1, K = 0 → Q = 1
5: J = 0, K = 1 → Q = 0
6: J = 1, K = 0 → Q = 1
7: J = 0, K = 0 → Q = 1
8: J = 1, K = 1 → Q = 0

Count the changes

0 → 1 → 1 → 1 → 1 → 0 → 1 → 1 → 0
N = 4

Answer

N = 4
7.Number theory & modular arithmetic

Answer: x = 32

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 13

r = 72, t = 0
r = 13, q = 5, t = 1
r = 7, q = 1, t = −5
r = 6, q = 1, t = 6
r = 1, q = 6, t = −11
13−1 ≡ 61 (mod 72)

Solve

x ≡ 61·56 = 3416 ≡ 32 (mod 72)
check: 13·32 = 416 ≡ 56 (mod 72)

Answer

x = 32
8.Number theory & modular arithmetic

Answer: d = 1557

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 = 41·47 = 1927
φ(n) = 40·46 = 1840

Extended Euclidean algorithm

r = 1840, t = 0
r = 13, q = 141, t = 1
r = 7, q = 1, t = −141
r = 6, q = 1, t = 142
r = 1, q = 6, t = −283

Private exponent

d = 1557
check: 13·1557 = 20241 = 11·1840 + 1

Answer

d = 1557
9.Algorithms & complexity

Answer: A · Dijkstra's algorithm

Why this is the answer

Dijkstra's algorithm works with non-negative weights; Bellman–Ford handles negative weights; Kruskal builds minimum spanning trees.

Answer

A · Dijkstra's algorithm
10.Algorithms & complexity

Answer: C · An undecidable problem has no algorithm for all inputs; an intractable one has one, but it is too slow

Why this is the answer

The halting problem has no algorithm at all; an NP-hard problem such as TSP can be solved by brute force, just not in known polynomial time.

Answer

C · An undecidable problem has no algorithm for all inputs; an intractable one has one, but it is too slow
11.Data structures

Answer: C · The call stack

Why this is the answer

Each call pushes a stack frame and each return pops it, which is why the most recent call finishes first.

Answer

C · The call stack
12.Data structures

Answer: D · Their high branching factor keeps the tree shallow, reducing slow disk reads

Why this is the answer

B-trees store many keys per node, so height is small and few block reads are needed per lookup; B+ trees also link leaves for range scans.

Answer

D · Their high branching factor keeps the tree shallow, reducing slow disk reads
13.Graph theory

Answer: H = 12

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 = (5 − 1)!/2 = 24/2 = 12

Answer

H = 12
14.Graph theory

Answer: edge BC

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. CE (7)
2. BC (3)
3. AB (4)
4. DE (8)

edge number 2

BC

Answer

edge BC
15.Databases & SQL

Answer: B · 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

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

Answer: C · Applying a function to the column prevents a simple index range lookup

Why this is the answer

The expression is not "sargable"; using order_date >= '2025-01-01' AND order_date < '2026-01-01' allows an index range scan.

Answer

C · Applying a function to the column prevents a simple index range lookup
17.Computer networks

Answer: C · Asymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data

Why this is the answer

Public-key methods and certificates set up a shared secret and prove identity. Fast symmetric ciphers such as AES then protect the bulk data.

Answer

C · Asymmetric methods agree keys and authenticate; symmetric ciphers encrypt the data
18.Computer networks

Answer: A · 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

A · How much unacknowledged data the sender may send, based on the receiver's buffer
19.IPv4 subnetting

Answer: H = 6

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).

Bits to borrow

25 = 32 ≥ 20 → b = 5, /29

Hosts per subnet

23 − 2 = 6

Answer

H = 6
20.IPv4 subnetting

Answer: x = 62

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 of this host

network: ⌊16/64⌋·64 = 0
broadcast: 0 + 64 − 1 = 63
usable hosts: 1 … 62

Answer

x = 62

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