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
Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
The 8-bit pattern 1011 0101 stores a signed integer in two's complement. What is its value N in decimal?
What is the largest unsigned integer N that fits in 24 bits?
Which formula is logically equivalent to ¬(p ∨ r)? Click it.
How many of the 16 rows of the truth table make (d → (b ↔ ¬a)) ∧ ((a ↔ ¬b) ∨ (d ∧ c)) true? Call this number N.
Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(0, 5, 7, 9, 10, 14). How many product terms N does it have?
A 20 kHz clock drives a mod-60 counter. Find the frequency fout of its most significant output, which completes one cycle per counting cycle.
Find the inverse of 15 modulo 19: the number x with 0 < x < 19 and 15·x ≡ 1 (mod 19).
An RSA key uses the primes p = 5 and q = 17 and the public exponent e = 11. Find the private exponent d (0 < d < φ(n)).
What is the time complexity of: for i = 1 to n: (j = 1; while j < n: j = j × 2)?
How do n! and 2ⁿ compare for large n?
How does extract-min work in a binary min-heap?
In a circular buffer, front == rear can mean either empty or full. What is a common way to tell them apart?
Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.
Find the total weight W of a minimum spanning tree of the weighted graph.
Second normal form (2NF) removes:
An app runs UPDATE item SET qty = 4, version = 8 WHERE id = 1 AND version = 7, and reports a conflict if no row was updated. What technique is this?
Which medium access method did classic shared (half-duplex) Ethernet use?
How does fragmentation differ in IPv6 compared with IPv4?
The network 192.168.10.0/24 must be divided into at least 3 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
A router has routes to these networks: 10.71.198.0/24, 10.71.199.0/24, 10.71.200.0/24, 10.71.201.0/24, 10.71.202.0/24. What is the longest prefix length p of a single summary route that covers all of them?
Answer: N = −75
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.
The other bits keep their normal positive values.
Answer: N = 1.678 × 10⁷
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.
Answer: ¬p ∧ ¬r
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 columns after the bar are the options in the order shown.
Answer: N = 8
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.
Answer: N = 4
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.
Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.
Answer: fout = 0.3333 kHz
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.
Answer: 15−1 mod 19 = 14
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).
Each remainder r is written as r ≡ t·15 (mod 19).
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.
Answer: C · O(n log n)
The inner loop runs about log₂ n times for each of the n outer iterations, giving O(n log n).
Answer: D · n! grows faster than 2ⁿ
For n ≥ 4, n! exceeds 2ⁿ, and the ratio grows without bound because each extra factor of n! is larger than 2.
Answer: D · Move the last element to the root and sift it down
The last element fills the hole at the root and is swapped with its smaller child until the heap property holds, taking O(log n).
Answer: D · Keep a separate element count, or always leave one slot unused
With a count, empty means count = 0 and full means count = capacity; with one empty slot, full is defined as (rear + 1) mod n == front.
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.
Answer: W = 26
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.
Answer: C · Partial dependencies of non-key columns on part of a composite key
2NF requires 1NF and that every non-key attribute depends on the whole key, not just part of it.
Answer: B · Optimistic concurrency control
Optimistic locking assumes conflicts are rare: it checks a version column when writing and fails if another transaction has changed the row meanwhile.
Answer: C · CSMA/CD
Carrier sense multiple access with collision detection: stations stop when they detect a collision and retry after a random backoff. Full-duplex switched Ethernet has no collisions.
Answer: A · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
An IPv6 router drops an oversized packet and returns ICMPv6 Packet Too Big. The source then sends smaller packets or uses a fragment extension header.
Answer: H = 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).
Answer: p = 20
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Summary route: 10.71.192.0/20 (it also covers 11 network(s) that are not in the list).
ExamProblems · examproblems.com/courses/computer-science-fundamentals