Approximate computing techniques: Worked examples — Hinglish deep dive

6.5.16 · D3 · Hardware › Advanced & Emerging Architectures › Approximate computing techniques

Scenario matrix

Is topic ka har problem inhi cells mein se ek hai. Is page ka goal hai: har cell ke liye ek worked example.

#	Cell (case class)	Kya cheez ise alag banati hai	Covered by
C1	Unbiased error, badi aggregation	errors mean-zero → relative error $1/ N$ ki tarah shrink hota hai	Example 1
C2	Biased error, badi aggregation	errors sab ek hi sign ke → error $N$ ki tarah grow karta hai (trap)	Example 2
C3	Positive product ka truncation	LSBs drop karna → hamesha under-estimate (one-sided sign)	Example 3
C4	Rounding vs truncation	same bits rakhe, par symmetric error → aadhi magnitude	Example 4
C5	Precision-scaling energy ( $n^{2}$ law)	quadratic multiplier cost, degenerate $n \to n /2$	Example 5
C6	Voltage overscaling ( $V^{2}$ vs delay)	power win vs timing-error onset; limiting $V \to V_{t h}$	Example 6
C7	Loop perforation, real image	word problem: speedup vs PSNR bound	Example 7
C8	Feedback / accumulation	tiny per-step error, par iterations ke saath amplify hota hai	Example 8
C9	Exam twist: exponent vs mantissa	GALAT field approximate karna → catastrophe	Example 9
C10	Zero / degenerate inputs	$N = 1$ , error $= 0$ , $n = 0$ bits — kya limits sahi behave karti hain?	Example 10

Compute karne se pehle, ek symbol ko pakka samajh lo, kyunki yeh neeche har jagah aata hai.

Example 1 — C1: badi sum par unbiased error

Steps.

Total error $E = \sum_{i = 1}^{N} ε_{i}$ . Iska variance hai $Var (E) = N σ^{2}$ . Yeh step kyun? Independent zero-mean errors variance mein add hote hain, magnitude mein nahi — parent note ka classic $N$ rule.
Typical error size $= N σ^{2} = N σ = 10000 \times 0.5 = 100 \times 0.5 = 50$ . Kyun? Standard deviation $= variance$ ; yahi "typical" spread hai.
Signal (true sum) $= N μ = 10000 \times 50 = 500, 000$ . Kyun? Error compare karne ke liye kuch chahiye — sum khud.
Relative error $= \frac{50}{500000} = 0.0001 = 0.01%$ .

Recall Verify

Agar $N$ ko quadruple karke $40000$ karein: error $= 40000 \cdot 0.5 = 100$ , signal $= 2, 000, 000$ , relative $= 5 \times 1 0^{- 5}$ — aadha ho gaya. Relative error $1/ N$ ki tarah girta hai. ✅ Units: error aur signal dono "reading units" mein, ratio dimensionless hai. ✅

Example 2 — C2: biased trap

Steps.

Bias linearly accumulate hota hai: $E [E] = \sum E [ε_{i}] = N \times (- 0.2) = - 2000$ . Yeh step kyun? Means directly add hote hain — jab sab ek sign share karte hain toh cancellation nahi hoti.
Zero-mean part se compare karo, jo ( $σ = 0.5$ ke saath) sirf $\pm N σ \approx \pm 50$ hai. Kyun? Dekho kaun sa term dominate karta hai. $- 2000$ , $\pm 50$ ko daba deta hai.
Relative error $= \frac{2000}{500000} = 0.004 = 0.4%$ — 40× worse than Example 1, aur yeh $N$ ke saath badhta hi rehta hai, shrink nahi hota.

Recall Verify

Bias term $∣ - 2000∣ = 40 \times 50$ (random term with $σ = 0.5$ ). ✅ $N = 40000$ par: bias $= - 8000$ ( $N$ ke saath $4 \times$ bada), jabki random term sirf $2 \times$ badhkar $100$ hua — confirm karta hai bias $\propto N$ , random $\propto N$ . ✅

Example 3 — C3: positive product truncate karna (one-sided sign)

Steps.

Exact: $1.375 \times 1.625 = 2.234375$ . Kyun? Ground truth. $A = 1.011 0_{2} = 1 + \frac{1}{4} + \frac{1}{8}$ , $B = 1.101 0_{2} = 1 + \frac{1}{2} + \frac{1}{8}$ .
Binary mein, $2.234375 = 10.00111 1_{2}$ (6 nonzero fractional bits). 4 fractional bits rakho: $10.001 1_{2} = 2.1875$ . Kyun? Truncation tail kaatta hai — hardware simply un partial-product columns ko omit karta hai.
Error $= 2.234375 - 2.1875 = 0.046875$ . Yeh hamesha $\geq 0$ hai: tumne positive weight remove kiya, isliye truncated result kabhi bhi true se bada nahi hota. Yeh kyun matter karta hai? Truncation ek biased approximation hai → yeh kaafi baar add hone par cell C2 ke danger family mein aata hai.
Relative error $= 0.046875/2.234375 \approx 0.0210 = 2.1%$ .

Recall Verify

Dropped bits hain $0.0000110 0_{2} = \frac{12}{256} = 0.046875$ exactly. ✅ Error positive hai. ✅

Example 4 — C4: iske badle rounding karna (symmetric error)

Steps.

Joh tail drop ho rahi hai woh hai $0.0000110 0_{2} = 0.046875$ , jo ek 4th-bit unit ke aadhe se zyada hai ( $2^{- 4} = 0.0625$ ; aadha hai $0.03125$ ). Yeh step kyun? Rounding tail dekhta hai: agar $\geq$ aadha unit, toh upar round karo.
Kyunki $0.046875 > 0.03125$ , upar round karo: $2.1875 + 0.0625 = 2.25$ . Kyun? Rounding nearest representable value ki taraf push karta hai, hamesha neeche nahi.
Rounding error $= 2.234375 - 2.25 = - 0.015625$ (ab negative — sign dono taraf ja sakta hai). Magnitude $0.015625 < 0.046875$ : truncation se 3× chhota.

Recall Verify

$∣ - 0.015625∣ = 0.046875/3$ exactly. ✅ Sign negative hai (true value se upar round ho gaya). ✅

Example 5 — C5: precision-scaling energy, $n^{2}$ law

Steps.

Ratio $= \frac{E _{24}}{E _{8}} = \frac{2 4 ^{2}}{8 ^{2}} = \frac{576}{64} = 9$ . $n^{2}$ kyun? Ek array multiplier $\approx n^{2}$ full-adder cells tile karta hai; har cell energy burn karta hai → total area $\propto n^{2}$ ke saath scale karta hai.
Toh ek INT8 multiply FP32 mantissa multiply se lagbhag $9 \times$ kam energy use karta hai. $24$ kyun, $32$ nahi? FP32 stored 32 bits hai, par 1 sign + 8 exponent bits mantissa multiplier mein enter nahi karte — sirf 24-bit significand multiply hota hai (dekho Precision and Number Formats (FP32, FP16, INT8)).
Degenerate check. $n \to n /2$ set karo: ratio $= (n / (n /2))^{2} = 2^{2} = 4$ . Yeh parent note ka "halving = 4×" reproduce karta hai. Kyun? $n^{2}$ law scale-free hai — sirf bit-widths ka ratio matter karta hai.

Recall Verify

$2 4^{2} / 8^{2} = 9$ , aur halving $4$ deta hai. Dono neeche check kiye. ✅

Example 6 — C6: voltage overscaling, $V^{2}$ win vs $V_{t h}$ wall

Figure — Approximate computing techniques

Steps.

Power ratio $= \frac{P ^{'}}{P} = \frac{V ^{'2}}{V ^{2}} = \frac{0. 8 ^{2}}{1. 0 ^{2}} = 0.64$ . $V^{2}$ kyun, $V$ nahi? Har switching node ek capacitor ko voltage $V$ tak charge karta hai, energy store karta hai $\frac{1}{2} C V^{2}$ — square physics hai, choice nahi.
Savings $= 1 - 0.64 = 0.36 = 36%$ . Care kyun? 20% voltage drop 36% power cut deta hai — " $V^{2}$ jackpot."
Par risk ( $V$ term): gate delay $\propto V / (V - V_{t h})^{2}$ badhta hai jaise $V \to V_{t h}$ . $V_{t h} = 0.3$ V par: delay factor $\propto V / (V - 0.3)^{2}$ . $V = 1.0$ par: $1/0.49 \approx 2.04$ . $V = 0.8$ par: $0.8/0.25 = 3.20$ → $1.57 \times$ slower. Kuch paths ab clock miss karte hain → timing errors. Yeh kyun matter karta hai? Figure par, green power curve smoothly girta hai jabki red delay curve $V_{t h}$ ke paas shoot up karta hai — yahi error cliff hai. VOS safe point se thoda left par, par cliff se right par rehta hai.

Recall Verify

Power save $= 1 - 0. 8^{2} = 0.36$ . Delay ratio $= (0.8/0.25) / (1.0/0.49) = 3.2/2.041 = 1.568$ . ✅

Example 7 — C7: loop perforation, real image (word problem)

Steps.

Speedup: hum aadhi rows skip karte hain → $\approx 2 \times$ kam iterations ⇒ ~2× faster. Acceptable kyun? Natural images mein adjacent rows highly correlated hoti hain (redundancy → C1-style resilience).
PSNR $= 10 lo g_{10} (\frac{25 5 ^{2}}{6.5}) = 10 lo g_{10} (\frac{65025}{6.5}) = 10 lo g_{10} (10004) \approx 40.0$ dB. Yeh formula kyun? Yeh page ke upar se quality ruler hai — MAX $^{2}$ over MSE.
$40.0 \geq 40$ ✅ — perforation-by-2 just pass karta hai. Perforate-by-4 push karo aur MSE roughly quadruple hoti hai, PSNR $\approx 6$ dB drop karta hai → fail. $4 \times$ par $6$ dB kyun? $10 lo g_{10} (4) \approx 6.02$ ; MSE quadruple hone par $6$ dB subtract hota hai.

Recall Verify

$10 lo g_{10} (25 5^{2} /6.5) = 40.00$ dB (2 dp tak). ✅ $4 \times$ MSE pe $10 lo g_{10} 4 = 6.02$ dB lagti hai. ✅

Example 8 — C8: feedback loop mein error accumulation

Steps.

Recursion solve karo: $x_{0} = 0$ ke saath, $x_{k} = ε \cdot \frac{g ^{k} - 1}{g - 1}$ (geometric series). Yeh step kyun? Feedback past errors ko har round mein $g$ se multiply karta hai — closed form growth expose karta hai.
Plug in karo: $x_{500} = 0.001 \cdot \frac{1.0 1 ^{500} - 1}{0.01}$ . Kyunki $1.0 1^{500} \approx 144.77$ , $x_{500} \approx 0.001 \times \frac{143.77}{0.01} = 0.001 \times 14377 \approx 14.38$ . Alarming kyun? Ek $0.001$ per-step error $\approx 14.4$ ban gaya — $\sim 14000 \times$ amplify hua.
Contrast $g = 1$ ke saath (koi amplification nahi): $x_{500} = 500 \times 0.001 = 0.5$ . Same per-step error, 28× chhota. Kyun? Jab $g > 1$ hota hai loop unstable hota hai aur errors compound hote hain; $N$ /linear intuitions APPLY NAHI HOTI.

Recall Verify

$x_{500} = 0.001 (1.0 1^{500} - 1) /0.01 \approx 14.377$ , vs $g = 1$ giving $0.5$ . Ratio $\approx 28.75$ . ✅

Example 9 — C9: exam twist — SAHI field approximate karo

Steps.

Mantissa LSB flip. FP16 mantissa mein 10 bits hain; LSB mantissa ka $2^{- 10}$ worth hai. Naya value $\approx (1.5 \pm 2^{- 10}) \times 2^{4} = 24 \pm 0.015625$ . Relative error $\approx 0.065%$ . Chhota kyun? Mantissa sirf fine detail set karta hai — yahi woh LSB hai jise hum approximate kar sakte hain.
Exponent flip (maan lo $2^{4} \to 2^{4 + 8} = 2^{12}$ high exponent bit flip karke): value $\to 1.5 \times 2^{12} = 6144$ . Yeh $256 \times$ zyada bada hai — magnitude destroy ho gayi. Catastrophic kyun? Exponent scale control karta hai; ek galat bit poore number ko power of two se multiply karta hai.
Rule. Mantissa LSBs aur pixel data approximate karo; kabhi nahi exponents, pointers, loop bounds, ya control flow. Un critical bits ko Error-Correcting Codes se protect karo. Kyun? Mantissa slip ek $0.065%$ nuisance hai; exponent slip ek $256 \times$ (ya worse) disaster hai — dono completely alag scales par rehte hain, isliye tum kaun sa field approximate karte ho yeh matter karta hai zyada, na ki kitne bits touch karte ho.

Recall Verify

Mantissa flip: $2^{- 10} \times 2^{4} = 0.015625$ , relative $= 0.015625/24 \approx 6.5 \times 1 0^{- 4}$ . ✅ Exponent flip $2^{4} \to 2^{12}$ : ratio $= 2^{8} = 256$ . ✅

Example 10 — C10: zero & degenerate limits (sanity boundary)

Steps.

$N = 1$ (koi aggregation nahi). Relative error $= \frac{σ}{μ 1} = \frac{σ}{μ}$ — poora single-op error, koi rescue nahi. $σ = 0.5, μ = 50$ ke saath: $0.01 = 1%$ . Kyun? Aggregation ko bahut saare terms chahiye; ek term kuch cancel nahi karta. $1/ N$ shrink sirf $N > 1$ ke liye help karna shuru karta hai.
Error $= 0$ (exact op). Tab $σ = 0$ → relative error $= 0$ sab $N$ ke liye. Exact machine approximate machine ka $σ \to 0$ limit hai — consistent, koi discontinuity nahi. Check kyun? Ek approximate model ko exact model reduce ho jaana chahiye jab tum error zero par dial karo.
$n = 0$ bits (degenerate precision). Multiplier energy $\propto n^{2} = 0$ : zero bits ⇒ zero hardware ⇒ zero energy aur zero information. $n^{2}$ law cleanly $n = 0$ par trivial machine par limit karta hai. Check kyun? Ek achha formula apni boundary par explode ya contradict nahi karna chahiye.

Recall Verify

$N = 1$ : $σ / (μ 1) = 0.5/50 = 0.01$ . $σ = 0$ : relative $= 0$ . $n = 0$ : $n^{2} = 0$ . Sab finite/consistent hain. ✅

Recall Pure matrix ki one-line summary

Unbiased+aggregated error shrink hota hai ( $N$ ); biased error drift karta hai ( $N$ ); truncation biased hai, rounding (almost) unbiased aur aadhi size; multiplier energy $n^{2}$ scale karta hai; voltage $V^{2}$ bachata hai par $V_{t h}$ ke paas delay wall se takraata hai; $g > 1$ wala feedback amplify karta hai; kabhi bhi exponents/control mat chhuona.

Active-recall

Kaun si technique error ko

N

ki jagah

N

ki tarah grow karati hai?

Ek biased scheme jaise pure truncation (hamesha neeche round karta hai); iska mean error linearly accumulate hota hai.

INT8 multiply FP32 mantissa multiply se ~9× sasta kyun hai?

Array multiplier energy

\propto n^{2}

; FP32 significand 24 bits hai, isliye

2 4^{2} / 8^{2} = 576/64 = 9

PSNR

\geq 40

dB kaunsi quality correspond karta hai?

Error itna chhota ki human eye essentially detect nahi kar sakti.

Gain

g > 1

wala feedback loop approximation ke liye dangerous kyun hai?

Per-step errors ko har round mein

g

se multiply kiya jaata hai aur exponentially compound hote hain, isliye chhote errors blow up karte hain.