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L6_Large_number_multi.ppt large number multiplication

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The document discusses the time complexity of addition and multiplication algorithms, highlighting that addition has a linear time complexity of θ(n) while school multiplication has a quadratic complexity of θ(n^2). It introduces a divide-and-conquer approach to multiplication, which can optimize the computation, exploring various techniques and proving that multiplication can be performed more efficiently than traditional methods. The document concludes with an analysis of the growth rate of multiplication time complexity, providing a theoretical framework for understanding algorithm efficiency.

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Revisited: How To Multiply Two Numbers 2 X 2 = 5
Time complexity of addition * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * + * * * * * * T(n) = The amount of time addition uses to add two n-bit numbers We saw that T(n) was linear. T(n) = Θ(n)
Time complexity of school multiplication T(n) = The amount of time grade school multiplication uses to add two n-bit numbers We saw that T(n) was quadratic. T(n) = Θ(n2 ) X * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * n2
addition Addition is linear time. Is there a sub-linear time method for addition?
Any addition algorithm takes Ω(n) time Claim: Any algorithm for addition must read all of the input bits Proof: Suppose there is a mystery algorithm A that does not examine each bit Give A a pair of numbers. There must be some unexamined bit position i in one of the numbers
• If A is not correct on the inputs, we found a bug • If A is correct, flip the bit at position i and give A the new pair of numbers. A gives the same answer as before, which is now wrong. Any addition algorithm takes Ω(n) time * * * * * * * * * * * * * * * * * * * * * * * * * * * * A did not read this bit at position i
Addition can’t be improved upon by more than a constant factor.
Multiplication Multiplication: Θ(n2 ) time Is there a clever algorithm to multiply two numbers in linear time?
Repetitive addition int mult( int n, int m) { if( m==0 ) return 0; return mult(n, m-1) + n; } What is the complexity of this algorithm?
Peasant multiplication x 1011 1001 ------ 1011 0000 0000 1011 ----------- 1100011
Peasant multiplication Why is it called the “Russian” peasant’s algorithm?
Peasant multiplication It’s also called the Egyptian multiplication.
But if the number does not fit the CPU? Can we do something very different than grade school multiplication?
Divide And Conquer An approach to faster algorithms: 1. DIVIDE a problem into smaller subproblems 2. CONQUER them recursively 3. GLUE the answers together so as to obtain the answer to the larger problem
Multiplication of 2 n-bit numbers X = Y = a b c d X = a 2n/2 + b Y = c 2n/2 + d n/2 bits n/2 bits n bits X × Y = ac 2n + (ad + bc) 2n/2 + bd X Y
Same thing for decimals X = Y = a b c d X = a 10n/2 + b Y = c 10n/2 + d n/2 digits n/2 digits n digits X × Y = ac 10n + (ad + bc) 10n/2 + bd
Multiplying (Divide & Conquer style) 12345678 * 21394276 *4276 5678*2139 5678*4276 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx 1*2 1*1 2*2 2*1 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx 1*2 1*1 2*2 2*1 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Hence: 12*21 = 2*102 + (1 + 4)101 + 2 = 252 2 1 4 2 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 252 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx 1*2 1*1 2*2 2*1 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Hence: 12*21 = 2*102 + (1 + 4)101 + 2 = 252 2 1 4 2 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 252 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx 1*2 1*1 2*2 2*1 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Hence: 12*21 = 2*102 + (1 + 4)101 + 2 = 252 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 252 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx 1*3 1*9 2*3 2*9 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 242 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx 3 9 6 18 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d *102 + *101 + *101 + *1
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 242 468 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx 3 9 6 18 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d *102 + *101 + *101 + *1
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx 252 468 714 1326 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx 252 468 714 1326 *104 + *102 + *102 + *1 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx = 2639526 252 468 714 1326 *104 + *102 + *102 + *1 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 2639526 1234*4276 5678*2139 5678*4276 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 2639526 5276584 12145242 24279128 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 2639526 5276584 12145242 24279128 *108 + *104 + *104 + *1 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Multiplying (Divide & Conquer style) 12345678 * 21394276 2639526 5276584 12145242 24279128 *108 + *104 + *104 + *1 = 264126842539128 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
Divide, Conquer, and Glue MULT(X,Y): X=a;b Y=c;d
Divide, Conquer, and Glue MULT(X,Y): X=a;b Y=c;d Mult(a,c) Mult(a,d) Mult(b,c) Mult(b,d)
Divide, Conquer, and Glue MULT(X,Y): X=a;b Y=c;d Mult(a,c) Mult(a,d) Mult(b,c) Mult(b,d)
Divide, Conquer, and Glue MULT(X,Y): X=a;b Y=c;d Mult(a,d) Mult(b,c) Mult(b,d) ac ac
MULT(X,Y): X=a;b Y=c;d Mult(a,d) Mult(b,c) Mult(b,d) ac ac, Divide, Conquer, and Glue
MULT(X,Y): X=a;b Y=c;d Mult(b,c) Mult(b,d) ac ad ac, ad, Divide, Conquer, and Glue
MULT(X,Y): X=a;b Y=c;d Mult(b,c) Mult(b,d) ac ad ac, ad, Divide, Conquer, and Glue
MULT(X,Y): X=a;b Y=c;d Mult(b,d) ac ad bc ac, ad,bc, Divide, Conquer, and Glue
MULT(X,Y): X=a;b Y=c;d Mult(b,d) ac ad bc ac, ad,bc, Divide, Conquer, and Glue
Divide, Conquer, and Glue MULT(X,Y): X=a;b Y=c;d ac ad bc bd ac, ad,bc, bd XY = ac2n +(ad+bc) 2n/2 + bd
Time required by MULT T(n) = time taken by MULT on two n-bit numbers What is T(n)? What is its growth rate? Big Question: Is it Θ(n2 )?
X=a;b Y=c;d ac ad bc bd XY = ac2n + (ad + bc)2n/2 + bd T(n/2) T(n/2) T(n/2) T(n/2) Time required by MULT
X=a; b Y=c; d ac ad bc bd XY = ac2n + (ad + bc)2n/2 + bd T(n/2) T(n/2) T(n/2) T(n/2) T(n) = 4 T(n/2) + (c1 n + c2) Conquering time divide and glue
Recurrence Relation T(1) = c3 T(n) = 4 T(n/2) + c1 n + c2 What is the growth rate of T(n)?
Let’s keep it simple T(1) = 1 T(n) = 4 T(n/2) + n What is the growth rate of T(n)?
Guess: T(n) = 2n2 – n Verify: T(1) = 1 and T(n) = 4 T(n/2) + n 2n2 – n = 4 [2(n/2)2 – n/2] + n = 2n2 – 2n + n = 2n2 – n Technique: Guess and Verify
Recursion Tree Representation We visualize the recursion as a tree, where each node represents a recursive call. The root is the initial call. Leaves correspond to the exit condition.
T(n) = 4 T(n/2) + n T(n/2) T(n/2) T(n/2) T(n/2) T(n) T(n/4) T(n/4) T(n/4) T(n/4)
Recursion Tree n/2 n/2 n/2 n/2 n n/4 n/4 n/4 n/4 each node reflects the combining step
Recursion Tree n/2 n/2 n/2 n/2 n n/4 n/4 n/4 n/4 n 2n 4n write down the work at each level
= 1*n = 2*n = 4*n = 2*i n = n*n T(n)=n (1+2+4+8+ . . . +2h ), where h = log2n T(n) = n(2n-1) = Θ(n2 ) Leve l i is th e sum of 4 i copies of n/2 i 1+ 1 + 1 + 1+ 1 + 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1 + 1+ 1+ 1 + 1+ 1 + 1 + 1+ 1 + 1+ 1 + 1 + 1+ 1+ 1+ 1+ 1 + 1+ 1+ 1 + 1+ 1 + 1 + 1+ 1 + 1+ 1+ 1 + 1+ 1+ 1+ 1 . . . . . . . . . . . . . . . . . . . . . . . . . . n/2 + n/2 + n/2 + n/2 n n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/ 4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 0 1 2 i log n 2
1 + X 1 + X1 1 + X + X2 2 + X + X3 3 + … + X + … + Xn-1 n-1 + X + Xn n = = X Xn+1 n+1 - 1 - 1 X X - 1 - 1 The Geometric Series T(n)=n (1+2+4+8+ . . . +2h ), where h = log2n T(n) = n(2n-1) = Θ(n2 )
Divide and Conquer MULT: Θ(n2 ) time Multiplication: Θ(n2 ) time All that work for nothing!
Karatsuba, Anatolii Alexeevich (1937-) In 1962 Karatsuba had formulated the first algorithm to break n2 barrier!
Multiplication of 2 n-bit numbers X = a 2n/2 + b Y = c 2n/2 + d X × Y = ac 2n + (ad + bc) 2n/2 + bd z1 = (a + b) (c + d) = ac + ad + bc + bd z2 = ac z3 = bd z4 = z2 – z3 = ac - bd z5 = z1 – z2 – z3 = ad + bc X × Y = z2 2n + z5 2n/2 + z3
Karatsuba (1962) T(1) = 1 T(n) = 3 T(n/2) + n What is the growth rate of T(n)?
n T(n/2) T(n/2) T(n/2) T(n) =
n T(n/2) T(n/2) T(n) = n/2 T(n/4) T(n/4) T(n/4)
n T(n) = n/2 T(n/4) T(n/4) T(n/4) n/2 T(n/4) T(n/4) T(n/4) n/2 T(n/4) T(n/4) T(n/4)
n n/2 + n/2 + n/2 Level i is the sum of 3i copies of n/2i . . . . . . . . . . . . . . . . . . . . . . . . . . 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 0 1 2 i log n 2
= 1n = 3/2n = 9/4n = (3/2)i n T(n) = n (1+3/2+(3/2)2 + . . . + (3/2)log2 n ) = ?? Leve l i is th e sum of 3 i copies of n/2 i 1+ 1 + 1 + 1+ 1 + 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1 + 1+ 1+ 1 + 1+ 1 + 1 + 1+ 1 + 1+ 1 + 1 + 1+ 1+ 1+ 1+ 1 + 1+ 1+ 1 + 1+ 1 + 1 + 1+ 1 + 1+ 1+ 1 + 1+ 1+ 1+ 1 . . . . . . . . . . . . . . . . . . . . . . . . . . n/2 + n/2 + n/2 n n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 0 1 2 i log n 2 = (3/2)log n n
The Geometric Series 1 + X 1 + X1 1 + X + X2 2 + X + X3 3 + … + X + … + Xk-1 k-1 + X + Xk k = = X Xk+1 k+1 - 1 - 1 X X - 1 - 1 We have: X = 3/2 k = log2 n (3/2) × (3/2)log2 n - 1 ½ 3 × (3log2 n /2log2 n ) - 2 = 3 × (3log2 n /n) - 2 =
Dramatic improvement for large n T(n) = 3 nlog2 3 – 2n = Θ(n log2 3 ) = Θ(n1.58… ) A huge savings over Θ(n2 ) when n gets large.
Recursion Tree Representation f(n/b) f(n/b) f(n/b) f(n/b) f(n) f(n/b2 ) f(n/b2 ) f(n/b2 ) f(n/b2 ) T(n) = a T(n/b) + f(n) a nodes
Solve the following recurrence T(n) = p T(n/3) + n where p>0. Exercise
T(n) = p T(n/3) + n Let p>3 be an arbitrary parameter. Using the tree method, we obtain T(n) = n(1 + p/3 + p2 /9 + …+ (p/3)h ) where the tree height is log3n. It follows T(n) = Θ(nlog 3 p )
Recurrence T(n) = p T(n/3) + n has the following has the following asymptotic solution asymptotic solution T(n) = T(n) = Θ (n n log log 3 3 p p ) ) when p>3. when p>3.
T(n) = p T(n/3) + n Cases p =1, 2 and 3 should Cases p =1, 2 and 3 should be considered separately be considered separately If p =3, then If p =3, then T(n) = T(n) = Θ (n log n) n log n)
3-Way Multiplication X = a1 2n/4 + a2 2n/2 + a3 Y = b1 2n/4 + b2 2n/2 + b3 T(1) = 1 T(n) = 9 T(n/3) + n
3-Way Multiplication T(n) = p T(n/3) + n T(n) = Θ(nlog 3 p ) To get an improvement over Karatsuba’s, we have to decrease the number of multiplications to at least 5.
Toom and Cook (1963, 1966) T(1) = 1 T(n) = 5 T(n/3) + n T(n) = Θ(n log3 5 ) = Θ(n1.46… )
3-Way Multiplication X = a1 2n/4 + a2 2n/2 + a3 Y = b1 2n/4 + b2 2n/2 + b3 Is it possible to reduce the number of multiplications to 5?
Multiplication Algorithms Grade School n2 Karatsuba n1.58… 3-way n1.46… FFT n log2 n Schönhage and Strassen n logn loglogn

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L6_Large_number_multi.ppt large number multiplication

  • 1. Revisited: How To Multiply Two Numbers 2 X 2 = 5
  • 2. Time complexity of addition * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * + * * * * * * T(n) = The amount of time addition uses to add two n-bit numbers We saw that T(n) was linear. T(n) = Θ(n)
  • 3. Time complexity of school multiplication T(n) = The amount of time grade school multiplication uses to add two n-bit numbers We saw that T(n) was quadratic. T(n) = Θ(n2 ) X * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * n2
  • 4. addition Addition is linear time. Is there a sub-linear time method for addition?
  • 5. Any addition algorithm takes Ω(n) time Claim: Any algorithm for addition must read all of the input bits Proof: Suppose there is a mystery algorithm A that does not examine each bit Give A a pair of numbers. There must be some unexamined bit position i in one of the numbers
  • 6. • If A is not correct on the inputs, we found a bug • If A is correct, flip the bit at position i and give A the new pair of numbers. A gives the same answer as before, which is now wrong. Any addition algorithm takes Ω(n) time * * * * * * * * * * * * * * * * * * * * * * * * * * * * A did not read this bit at position i
  • 7. Addition can’t be improved upon by more than a constant factor.
  • 8. Multiplication Multiplication: Θ(n2 ) time Is there a clever algorithm to multiply two numbers in linear time?
  • 9. Repetitive addition int mult( int n, int m) { if( m==0 ) return 0; return mult(n, m-1) + n; } What is the complexity of this algorithm?
  • 11. Peasant multiplication Why is it called the “Russian” peasant’s algorithm?
  • 12. Peasant multiplication It’s also called the Egyptian multiplication.
  • 13. But if the number does not fit the CPU? Can we do something very different than grade school multiplication?
  • 14. Divide And Conquer An approach to faster algorithms: 1. DIVIDE a problem into smaller subproblems 2. CONQUER them recursively 3. GLUE the answers together so as to obtain the answer to the larger problem
  • 15. Multiplication of 2 n-bit numbers X = Y = a b c d X = a 2n/2 + b Y = c 2n/2 + d n/2 bits n/2 bits n bits X × Y = ac 2n + (ad + bc) 2n/2 + bd X Y
  • 16. Same thing for decimals X = Y = a b c d X = a 10n/2 + b Y = c 10n/2 + d n/2 digits n/2 digits n digits X × Y = ac 10n + (ad + bc) 10n/2 + bd
  • 17. Multiplying (Divide & Conquer style) 12345678 * 21394276 *4276 5678*2139 5678*4276 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 18. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 19. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 20. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 21. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 22. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 23. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 24. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 25. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 26. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 27. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 28. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx 1*2 1*1 2*2 2*1 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 29. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx 1*2 1*1 2*2 2*1 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Hence: 12*21 = 2*102 + (1 + 4)101 + 2 = 252 2 1 4 2 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 30. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 252 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx 1*2 1*1 2*2 2*1 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Hence: 12*21 = 2*102 + (1 + 4)101 + 2 = 252 2 1 4 2 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 31. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 252 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx 1*2 1*1 2*2 2*1 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Hence: 12*21 = 2*102 + (1 + 4)101 + 2 = 252 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 32. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 252 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx 1*3 1*9 2*3 2*9 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 33. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 242 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx 3 9 6 18 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d *102 + *101 + *101 + *1
  • 34. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 242 468 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx 3 9 6 18 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d *102 + *101 + *101 + *1
  • 35. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx 252 468 714 1326 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 36. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx 252 468 714 1326 *104 + *102 + *102 + *1 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 37. Multiplying (Divide & Conquer style) 12345678 * 21394276 1234*2139 1234*4276 5678*2139 5678*4276 12*21 12*39 34*21 34*39 xxxxxxxxxxxxxxxxxxxxxxxxx = 2639526 252 468 714 1326 *104 + *102 + *102 + *1 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 38. Multiplying (Divide & Conquer style) 12345678 * 21394276 2639526 1234*4276 5678*2139 5678*4276 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 39. Multiplying (Divide & Conquer style) 12345678 * 21394276 2639526 5276584 12145242 24279128 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 40. Multiplying (Divide & Conquer style) 12345678 * 21394276 2639526 5276584 12145242 24279128 *108 + *104 + *104 + *1 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 41. Multiplying (Divide & Conquer style) 12345678 * 21394276 2639526 5276584 12145242 24279128 *108 + *104 + *104 + *1 = 264126842539128 X = Y = X × Y = ac 10n + (ad + bc) 10n/2 + bd a b c d
  • 42. Divide, Conquer, and Glue MULT(X,Y): X=a;b Y=c;d
  • 43. Divide, Conquer, and Glue MULT(X,Y): X=a;b Y=c;d Mult(a,c) Mult(a,d) Mult(b,c) Mult(b,d)
  • 44. Divide, Conquer, and Glue MULT(X,Y): X=a;b Y=c;d Mult(a,c) Mult(a,d) Mult(b,c) Mult(b,d)
  • 45. Divide, Conquer, and Glue MULT(X,Y): X=a;b Y=c;d Mult(a,d) Mult(b,c) Mult(b,d) ac ac
  • 49. MULT(X,Y): X=a;b Y=c;d Mult(b,d) ac ad bc ac, ad,bc, Divide, Conquer, and Glue
  • 50. MULT(X,Y): X=a;b Y=c;d Mult(b,d) ac ad bc ac, ad,bc, Divide, Conquer, and Glue
  • 51. Divide, Conquer, and Glue MULT(X,Y): X=a;b Y=c;d ac ad bc bd ac, ad,bc, bd XY = ac2n +(ad+bc) 2n/2 + bd
  • 52. Time required by MULT T(n) = time taken by MULT on two n-bit numbers What is T(n)? What is its growth rate? Big Question: Is it Θ(n2 )?
  • 53. X=a;b Y=c;d ac ad bc bd XY = ac2n + (ad + bc)2n/2 + bd T(n/2) T(n/2) T(n/2) T(n/2) Time required by MULT
  • 54. X=a; b Y=c; d ac ad bc bd XY = ac2n + (ad + bc)2n/2 + bd T(n/2) T(n/2) T(n/2) T(n/2) T(n) = 4 T(n/2) + (c1 n + c2) Conquering time divide and glue
  • 55. Recurrence Relation T(1) = c3 T(n) = 4 T(n/2) + c1 n + c2 What is the growth rate of T(n)?
  • 56. Let’s keep it simple T(1) = 1 T(n) = 4 T(n/2) + n What is the growth rate of T(n)?
  • 57. Guess: T(n) = 2n2 – n Verify: T(1) = 1 and T(n) = 4 T(n/2) + n 2n2 – n = 4 [2(n/2)2 – n/2] + n = 2n2 – 2n + n = 2n2 – n Technique: Guess and Verify
  • 58. Recursion Tree Representation We visualize the recursion as a tree, where each node represents a recursive call. The root is the initial call. Leaves correspond to the exit condition.
  • 59. T(n) = 4 T(n/2) + n T(n/2) T(n/2) T(n/2) T(n/2) T(n) T(n/4) T(n/4) T(n/4) T(n/4)
  • 60. Recursion Tree n/2 n/2 n/2 n/2 n n/4 n/4 n/4 n/4 each node reflects the combining step
  • 61. Recursion Tree n/2 n/2 n/2 n/2 n n/4 n/4 n/4 n/4 n 2n 4n write down the work at each level
  • 62. = 1*n = 2*n = 4*n = 2*i n = n*n T(n)=n (1+2+4+8+ . . . +2h ), where h = log2n T(n) = n(2n-1) = Θ(n2 ) Leve l i is th e sum of 4 i copies of n/2 i 1+ 1 + 1 + 1+ 1 + 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1 + 1+ 1+ 1 + 1+ 1 + 1 + 1+ 1 + 1+ 1 + 1 + 1+ 1+ 1+ 1+ 1 + 1+ 1+ 1 + 1+ 1 + 1 + 1+ 1 + 1+ 1+ 1 + 1+ 1+ 1+ 1 . . . . . . . . . . . . . . . . . . . . . . . . . . n/2 + n/2 + n/2 + n/2 n n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/ 4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 0 1 2 i log n 2
  • 63. 1 + X 1 + X1 1 + X + X2 2 + X + X3 3 + … + X + … + Xn-1 n-1 + X + Xn n = = X Xn+1 n+1 - 1 - 1 X X - 1 - 1 The Geometric Series T(n)=n (1+2+4+8+ . . . +2h ), where h = log2n T(n) = n(2n-1) = Θ(n2 )
  • 64. Divide and Conquer MULT: Θ(n2 ) time Multiplication: Θ(n2 ) time All that work for nothing!
  • 65. Karatsuba, Anatolii Alexeevich (1937-) In 1962 Karatsuba had formulated the first algorithm to break n2 barrier!
  • 66. Multiplication of 2 n-bit numbers X = a 2n/2 + b Y = c 2n/2 + d X × Y = ac 2n + (ad + bc) 2n/2 + bd z1 = (a + b) (c + d) = ac + ad + bc + bd z2 = ac z3 = bd z4 = z2 – z3 = ac - bd z5 = z1 – z2 – z3 = ad + bc X × Y = z2 2n + z5 2n/2 + z3
  • 67. Karatsuba (1962) T(1) = 1 T(n) = 3 T(n/2) + n What is the growth rate of T(n)?
  • 71. n n/2 + n/2 + n/2 Level i is the sum of 3i copies of n/2i . . . . . . . . . . . . . . . . . . . . . . . . . . 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 0 1 2 i log n 2
  • 72. = 1n = 3/2n = 9/4n = (3/2)i n T(n) = n (1+3/2+(3/2)2 + . . . + (3/2)log2 n ) = ?? Leve l i is th e sum of 3 i copies of n/2 i 1+ 1 + 1 + 1+ 1 + 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1 + 1+ 1+ 1 + 1+ 1 + 1 + 1+ 1 + 1+ 1 + 1 + 1+ 1+ 1+ 1+ 1 + 1+ 1+ 1 + 1+ 1 + 1 + 1+ 1 + 1+ 1+ 1 + 1+ 1+ 1+ 1 . . . . . . . . . . . . . . . . . . . . . . . . . . n/2 + n/2 + n/2 n n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 0 1 2 i log n 2 = (3/2)log n n
  • 73. The Geometric Series 1 + X 1 + X1 1 + X + X2 2 + X + X3 3 + … + X + … + Xk-1 k-1 + X + Xk k = = X Xk+1 k+1 - 1 - 1 X X - 1 - 1 We have: X = 3/2 k = log2 n (3/2) × (3/2)log2 n - 1 ½ 3 × (3log2 n /2log2 n ) - 2 = 3 × (3log2 n /n) - 2 =
  • 74. Dramatic improvement for large n T(n) = 3 nlog2 3 – 2n = Θ(n log2 3 ) = Θ(n1.58… ) A huge savings over Θ(n2 ) when n gets large.
  • 75. Recursion Tree Representation f(n/b) f(n/b) f(n/b) f(n/b) f(n) f(n/b2 ) f(n/b2 ) f(n/b2 ) f(n/b2 ) T(n) = a T(n/b) + f(n) a nodes
  • 76. Solve the following recurrence T(n) = p T(n/3) + n where p>0. Exercise
  • 77. T(n) = p T(n/3) + n Let p>3 be an arbitrary parameter. Using the tree method, we obtain T(n) = n(1 + p/3 + p2 /9 + …+ (p/3)h ) where the tree height is log3n. It follows T(n) = Θ(nlog 3 p )
  • 78. Recurrence T(n) = p T(n/3) + n has the following has the following asymptotic solution asymptotic solution T(n) = T(n) = Θ (n n log log 3 3 p p ) ) when p>3. when p>3.
  • 79. T(n) = p T(n/3) + n Cases p =1, 2 and 3 should Cases p =1, 2 and 3 should be considered separately be considered separately If p =3, then If p =3, then T(n) = T(n) = Θ (n log n) n log n)
  • 80. 3-Way Multiplication X = a1 2n/4 + a2 2n/2 + a3 Y = b1 2n/4 + b2 2n/2 + b3 T(1) = 1 T(n) = 9 T(n/3) + n
  • 81. 3-Way Multiplication T(n) = p T(n/3) + n T(n) = Θ(nlog 3 p ) To get an improvement over Karatsuba’s, we have to decrease the number of multiplications to at least 5.
  • 82. Toom and Cook (1963, 1966) T(1) = 1 T(n) = 5 T(n/3) + n T(n) = Θ(n log3 5 ) = Θ(n1.46… )
  • 83. 3-Way Multiplication X = a1 2n/4 + a2 2n/2 + a3 Y = b1 2n/4 + b2 2n/2 + b3 Is it possible to reduce the number of multiplications to 5?
  • 84. Multiplication Algorithms Grade School n2 Karatsuba n1.58… 3-way n1.46… FFT n log2 n Schönhage and Strassen n logn loglogn