3.3.8
Intuition Core idea (WHY yeh exist karta hai)Ek fixed hash function ko hamesha defeat kiya ja sakta hai. Jab ek adversary ko aapka function h O ( n ) ek hi function pe commit mat karo . h family H randomly pick karo. Tab kisi bhi fixed pair of keys ke liye, unke collide karne ki probability bahut chhoti hoti hai — aur adversary us randomness ko exploit nahi kar sakta jo usse dikhi hi nahi.
WHAT we buy: har input ke liye expected O ( 1 )
Definition Universal familyEk family H U { 0 , 1 , … , m − 1 } universal x = y Pr h ∈ H [ h ( x ) = h ( y ) ] ≤ m 1 h H uniformly at random choose kiya gaya ho.
Common mistake Steel-man: "Random hashing matlab har key uniformly land karti hai."Yeh lagta toh sahi hai — hum keys ko evenly scatter hote imagine karte hain. Lekin universal property ek key ki distribution ke baare mein nahi hai; yeh pairs ke baare mein hai. Hum kabhi assume nahi karte ki data random hai. Randomness poori tarah h rehti hai, jabki keys x , y x , y h
Maano hum n m h ek given key x kya hogi.
α = m n x 1 + α + 1 x m n − 1 x
Intuition Therefore (HOW yeh O(1) deta hai)Expected search time = O ( 1 + α ) m = Θ ( n ) α = O ( 1 ) har operation expected O ( 1 ) input keys chahe jo bhi ho . Guarantee hamare coin flips pe hai, "nice" data pe nahi.
Definition Number-theoretic familyH p , m p > a ∈ { 1 , … , p − 1 } b ∈ { 0 , … , p − 1 } h a , b ( k ) = ( ( a k + b ) mod p ) mod m H p , m = { h a , b } a , b
Common mistake Steel-man: "Finalmod m ≤ 1/ m lagta exact hai — humne p m 1/ m m p divide nahi karta , isliye buckets thodi uneven hoti hain: kuch ko ⌈ p / m ⌉ ⌊ p / m ⌋ Pr [ h ( x ) = h ( y )] ≤ p − 1 ⌈ p / m ⌉ − 1 ≤ m 1 . ≤ p ⌈ p / m ⌉ ≤ m 1 + p 1 truly universal (≤ 1/ m m ∣ p almost-universal (≤ 1/ m + 1/ p 1/ p p Fix: ≤ 1/ m m ∣ p + 1/ p
Worked example Yeh (almost) universal kyun hai? (WHY ka sketch)Fix karo x = y r = ( a x + b ) mod p s = ( a y + b ) mod p a = 0 p ( a , b ) ↦ ( r , s ) bijection hai jahan r = s p r ≡ s ( mod m ) r s = r s ≡ r ( mod m ) ⌈ p / m ⌉ − 1 p − 1 Pr [ collision ] ≤ p − 1 ⌈ p / m ⌉ − 1 ≤ m 1 .
Worked example Example 1 — Expected probe countTable mein m = 100 n = 50 h x
Kyun: x m n E [ collisions ] ≤ 100 50 = 0.5 = 1 + 0.5 = 1.5 Constant.
Worked example Example 2 —h a , b p = 17 m = 6 a = 3 , b = 4 k = 10
ak + b = 3 ⋅ 10 + 4 = 34 Kyun: ak + b 34 mod 17 = 0 Kyun: pehle mod p 0 mod 6 = 0 Kyun: table range [ 0 , m ) h ( 10 ) = 0
Worked example Example 3 — Probability sanity checkYahan p = 17 m = 6 6 ∤ 17 17 ⌈ 17/6 ⌉ = 17 3 ≈ 0.176 1/ m = 1/6 ≈ 0.167 10 15 ( a , b ) ≤ 3/17
Recall Feynman: ek 12-saal ke bachche ko explain karoSocho ek bully hamesha wahi locker choose karta hai jo already bhara hua ho taaki tum apna saman nahi rakh sako. Agar tumhara locker-picking rule fixed hai, toh woh hamesha jeetta hai. Toh uske bajaye, class se pehle secretly dice roll karo rule decide karne ke liye. Ab bully ko nahi pata ki cheezein kaunse lockers mein gayi hain, toh average mein sab ko apna locker milta hai aur koi zyada wait nahi karta. Dice (random choice of h tumhara secret hai, aur wahi use beat karta hai — yeh nahi ki bachche "random" hain.
"RANDOM the FUNCTION, never the DATA."
Aur bound: "Pair collides at most 1-in-m." (Pr ≤ 1/ m m ∣ p + 1/ p
When is a family H Har distinct x = y Pr h ∈ H [ h ( x ) = h ( y )] ≤ 1/ m h
Where does the randomness live in universal hashing? Family se hash function ke choice mein — keys mein NAHI, jo adversarial ho sakti hain.
Why use linearity of expectation in the analysis? Yeh hamare liye E [ X x y ]
Expected number of keys colliding with x ≤ n / m = α
Expected cost of a search with universal hashing? O ( 1 + α ) m = Θ ( n ) O ( 1 )
State the classic universal family h a , b h a , b ( k ) = (( ak + b ) mod p ) mod m p > a ∈ { 1.. p − 1 } b ∈ { 0.. p − 1 }
Is h a , b Sirf jab m ∣ p Pr ≤ 1/ m almost -universal hai jab Pr ≤ 1/ m + 1/ p
Why must a = 0 h a , b Taaki map k ↦ ak + b mod p p
What does universal hashing protect against that fixed hashing cannot? Ek adversary jo worst-case keys choose kare taaki saari collisions force ho sakein.
E [ X x y ] Pr [ h ( x ) = h ( y )] X x y
Adversary forces collisions
Pick h randomly from family H
Pr collision at most 1 over m
E of X at most n-1 over m
Expected O of 1 per operation
