Trick 0: Warm Up Handshakes
Story. If each of us (\(n+1\) ) were to shake hands w everyone else, how many handshakes would there be?
Problem.
\(S(n) = \sum\limits_{k=1}^{n} k\)
We’ll need this, and we’ll also need:
\(S(n)=\sum\limits_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\)
Trick 1: Linearity
Story. Processing n data rows:
each row has a constant overhead of 7 units,
plus 3 units per column, with column count = row index.
Problem.
\(S(n)=\sum\limits_{i=1}^{n} (7 + 3i)\)
Trick 2: Telescoping
Story. Model accuracy at epoch \(i\) :
improvement = \(\ln(i+1) - \ln(i)\) .
Problem. \(S(n)=\sum_{i=1}^{n} \big(\ln(i+1)-\ln(i)\big)\)
Trick 3: Shifting
Story. Each day your batch size doubles (like a viral share),annotation time per item grows linearly with the day index. What’s the total work after \(n\) days?
Problem. \(S_n =\sum_{k=1}^{n} k\,2^{k}\)
Trick 4: Variable Substitution
Story. Each item \(j\) has effort that depends both on its index and how many remain.
Problem. \(S = \sum_{j=1}^{n} \big(7 + 3(n-j+1)\big)\)
Hint for later reveal: \(i = n-j+1\) . The sum becomesLinearity example.
Trick 5: Reindexing
Story. You only log every 3rd data point (rows 3, 6, 9, …).\(i\) is \(2i+1\) .
Question. What is the total cost up to row \(n\) ?
Trick 6: Double Sum
Story. Each item \(i\) must be compared with all later items \(j\ge i\) . The pairwise cost is \(c_j = 4j+1\) (depends only on the later index).
Problem. \(T = \displaystyle \sum_{i=1}^{n}\;\sum_{j=i}^{n}\,(4j+1)\)
