Support, Resistance & Price Action

Question 1

(a) Pivot computation (8 marks)

Standard formulas: $P = \frac{H+L+C}{3} = \frac{428.60+415.20+424.80}{3} = \frac{1268.60}{3} = 422.8\overline{6} \approx 422.87$ (2 marks)

$R_1 = 2P - L = 2(422.8667) - 415.20 = 845.7333 - 415.20 = 430.53$ (1.5) $S_1 = 2P - H = 845.7333 - 428.60 = 417.13$ (1.5) $R_2 = P + (H-L) = 422.8667 + 13.40 = 436.27$ (1.5) $S_2 = P - (H-L) = 422.8667 - 13.40 = 409.47$ (1.5)

(b) Round-number confluence (6 marks)

Round numbers (multiples of 5) near each level: (3 marks for identification)

• $P = 422.87$ → nearest 425 (1.13 away) / 420 (2.87) → not within ±1.
• $R_1 = 430.53$ → 430 is 0.53 away → confluence ✓
• $S_1 = 417.13$ → 415 (2.13) — no.
• $R_2 = 436.27$ → 435 (1.27) — no.
• $S_2 = 409.47$ → 410 is 0.53 away → confluence ✓

Levels with confluence: $R_1$ (≈430) and $S_2$ (≈410). (1)

Explanation (2 marks): Round numbers attract clustered orders (mental stops/limits, algorithmic reference points). When a pivot level — already a computed reference many traders watch — coincides with a psychological round number, two independent order-flow magnets overlap, increasing the density of resting orders. This raises the probability that price reacts there, making it a higher-conviction zone.

(c) Proof $P=\tfrac{R_1+S_1}{2}$ (6 marks)

$R_1 + S_1 = (2P - L) + (2P - H) = 4P - (H+L)$ (2) Since $P=\frac{H+L+C}{3}$, we have $H+L = 3P - C$. (2) $R_1+S_1 = 4P - (3P - C) = P + C$ Hmm — this gives $P+C$, not $2P$, unless we note the intended identity uses $R_1+S_1$ symmetry. Correct midpoint proof: since $R_1 = 2P-L$ and $S_1=2P-H$, $\frac{R_1+S_1}{2} = \frac{4P-(H+L)}{2} = 2P - \frac{H+L}{2}.$ This equals $P$ iff $P = \frac{H+L}{2}$, which is only true when $C=\frac{H+L}{2}$.

Correct answer: the exact midpoint identity that always holds is $\frac{R_1+S_1}{2} = 2P - \frac{H+L}{2},$ and $P$ is the midpoint of $R_1,S_1$ only when $C = \tfrac{H+L}{2}$. Award full marks to candidates who disprove the universal claim and state the precise condition, since the general statement is false. (6 marks for rigorous derivation + correct conclusion; 3 if they merely assert it without testing.)

Question 2

(a) swing_highs (8 marks)

def swing_highs(highs, k=2):
    idx = []
    n = len(highs)
    for i in range(k, n - k):
        h = highs[i]
        if all(h > highs[i-j] for j in range(1, k+1)) and \
           all(h > highs[i+j] for j in range(1, k+1)):
            idx.append(i)
    return idx

(6 marks: correct boundary range, strict comparison both sides, returns indices.) Complexity: outer loop $O(n)$, inner check $O(k)$ → $O(nk)$, which is $O(n)$ for fixed $k$. (2 marks)

(b) classify_breakout (8 marks)

def classify_breakout(closes, R):
    for t in range(len(closes) - 1):
        if closes[t] > R:                     # first break
            if closes[t+1] > R:
                return "confirmed"
            else:
                return "false"
    # handle last-bar break with no next bar
    if closes and closes[-1] > R:
        return "none"      # break but no confirmation bar yet
    return "none"

(6 marks for correct first-break detection + confirmation logic; 2 for edge handling of last bar / no break.)

(c) Trace (6 marks)

closes $=[96,98,101,99,103,105]$, $R=100$.

• $t=0$: 96 ≤ 100.
• $t=1$: 98 ≤ 100.
• $t=2$: $101>100$ → first break. Next $closes[3]=99 \le 100$ → returns "false". (3 marks)

Retest explanation (3 marks): A retest occurs when, after breaking $R$, price returns to touch the broken level (now expected to act as support via role reversal) and holds — bouncing without closing back below. It adds reliability by demonstrating that former resistance has genuinely flipped to support (fresh demand appears), filtering out the false break seen here where price failed to hold above 100.

Question 3

(a) Role reversal (6 marks) Role reversal (polarity): once a resistance level is decisively broken, the sellers who defended it are removed / trapped, and that price now tends to act as support on a pullback (and vice-versa). (3 marks) Numerical illustration (3 marks): price rejects at 100 twice (resistance), then closes 102, pulls back to 100 and bounces to 105 — the old 100 resistance became support. e.g. sequence $[97,100,96,101,102,100,104]$.

(b) Zone selection (6 marks)

• Proximity: $Z_2=[418,421]$ is nearer to price 425 (≈4 pts) than $Z_1=[410,414]$ (≈11 pts). On a decline, price meets $Z_2$ first, so $Z_2$ is the first line of defence. (2)
• Freshness: an untested (fresh) zone has unfilled resting demand and typically produces a stronger reaction; a tested zone has had orders partially consumed and is weaker. (2)
• Conclusion: $Z_2$ is the stronger first line of defence by proximity provided it is still fresh; if $Z_2$ has already been tested and $Z_1$ is fresh, $Z_1$ may give the more reliable (if deeper) bounce. A complete answer states the proximity-vs-freshness trade-off. (2)

(c) Support-strength score (6 marks)

Proposed score (Gaussian/inverse-distance weighting within tolerance $\tau$): $\Sigma(x) = \sum_{i:\,|p_i - x|\le \tau} \left(1 - \frac{|p_i - x|}{\tau}\right)$ Justification (2 marks): the sum counts touches (term (i) — more touches → higher score); the weight $(1-|p_i-x|/\tau)$ decays linearly to 0 at the band edge so closer touches contribute more (term (ii)); touches outside $\pm\tau$ contribute nothing.

Compute $\Sigma(412)$, $\tau=2.0$, lows $[411.5,412.4,419.0,412.0]$: (4 marks)

• $411.5$: $|{-}0.5|=0.5\le2$ → $1-0.25=0.75$
• $412.4$: $0.4$ → $1-0.20=0.80$
• $419.0$: $7.0>2$ → excluded
• $412.0$: $0.0$ → $1-0=1.00$

$\Sigma(412) = 0.75+0.80+1.00 = 2.55$

[
  {"claim":"Pivot P = 422.8667 for H=428.6,L=415.2,C=424.8","code":"H,L,C=428.60,415.20,424.80\nP=(H+L+C)/3\nresult = abs(P-422.86666666)<1e-4"},
  {"claim":"R1=430.53 and S1=417.13 (2dp)","code":"H,L,C=428.60,415.20,424.80\nP=(H+L+C)/3\nR1=2*P-L\nS1=2*P-H\nresult = (round(R1,2)==430.53) and (round(S1,2)==417.13)"},
  {"claim":"General midpoint of R1,S1 equals 2P-(H+L)/2, not P in general","code":"H,L,C=symbols('H L C')\nP=(H+L+C)/3\nR1=2*P-L\nS1=2*P-H\nmid=simplify((R1+S1)/2)\nresult = simplify(mid-(2*P-(H+L)/2))==0 and simplify(mid-P)!=0"},
  {"claim":"Support strength Sigma(412)=2.55","code":"tau=2.0\nx=412.0\nlows=[411.5,412.4,419.0,412.0]\nS=sum(1-abs(p-x)/tau for p in lows if abs(p-x)<=tau)\nresult = abs(S-2.55)<1e-9"}
]