1. The Intel Pentium Division Flaw
Alan Edelman
Department of Mathematics
Laboratory for Computer Science
Massachusetts Institute of Technology

2. The bug itself is (mathematically) neat!
Not So Well Known
The bug itself is (mathematically) neat!
A Lesson (for me anyway)
So much incomplete
information is out there.

3. Interesting Related Topics (but my topic is the bug)
Risk to Pentium owners
Intel’s chip replacement blunder
Kahan’s SRT division tester
Moler, Coe, and Mathisen software workaround
Only the lawyers get rich
Those ubiquitous Pentium jokes

4. Outline Nicely’s Discovery Computer Science Prerequisites
Division (SRT=Sweeney,Robertson,Tocher)
Pentium Lookup Table
Division Example
Six Ones Result
Inequality Analysis
“Send More Money” Puzzle
Always nearly five good digits

5. Nicely’s Twin Prime Bug Discovery
Twin primes: (5,7) (11,13) (17,19) (29,31) ...
Nicely was summing twin prime reciprocals:
S = 1/5 + 1/ 7 + 1/11 + 1/13 + 1/17 + 1/
S is finite.
Nicely computed on many platforms.
Nicely checked his work.

6. Computer Science Prerequisites
Carry Save Addition.
One’s vs. Two’s Complement.

7. Carry-save Addition 12 21
+ 19 52

8. Carry-save Addition 12 21
+ 19 52 20
Answer (mod 32)

9. Carry-save Addition 12 01100 21 10101 + 19 10011 52 10100 + 20
Answer (mod 32)

10. Carry-save Addition 12 01100 01100 21 10101 10101 + 19 10011 + 1 5220
01100
10101
10011
10100
01100
10101 1 + +
Sum Bits (s)
Carry Bits (c)
Answer (mod 32)

11. Carry-save Addition 12 01100 01100 21 10101 10101 + 19 10011 + 1 5220
01100
10101
10011
10100
01100
10101 1 + +
Sum Bits (s)
Carry Bits (c)
Answer (mod 32)
s = xÅyÅz = x+y+z (mod 2)
c = xÙy Ú xÙz Ú yÙz = (x+y+z | 2)

12. Carry-save Addition 12 01100 01100 21 10101 10101 + 19 10011 + 01 5220
01100
10101
10011
10100
01100
10101 01 00 + +
Sum Bits (s)
Carry Bits (c)
Answer (mod 32)
s = xÅyÅz = x+y+z (mod 2)
c = xÙy Ú xÙz Ú yÙz = (x+y+z | 2)

13. Carry-save Addition 12 01100 01100 21 10101 10101 + 19 10011 + 1100152 20
01100
10101
10011
10100
01100
10101
11001
01000 + +
Sum Bits (s)
Carry Bits (c)
Answer (mod 32)
s = xÅyÅz = x+y+z (mod 2)
c = xÙy Ú xÙz Ú yÙz = (x+y+z | 2)

14. Carry-save Addition 12 01100 01100 21 10101 10101 + 19 10011 + 1100152 20
01100
10101
10011
10100
01100
10101
11001
01000 + +
Sum Bits (s)
Carry Bits (c)
+10011
Answer (mod 32)
s = xÅyÅz = x+y+z (mod 2)
c = xÙy Ú xÙz Ú yÙz = (x+y+z | 2)

15. Carry-save Addition 12 01100 01100 21 10101 10101 + 19 10011 + 1100152 20
01100
10101
10011
10100
01100
10101
11001
01000 + +
Sum Bits (s)
Carry Bits (c)
+10011
00010
10010
Sum Bits (s)
Answer (mod 32)
Carry Bits (c)
s = xÅyÅz = x+y+z (mod 2)
c = xÙy Ú xÙz Ú yÙz = (x+y+z | 2)

16. One’s vs. Two’s Complement

17. One’s vs. Two’s Complement
Two’s Complement One’s Complement

18. One’s vs. Two’s Complement
Two’s Complement One’s Complement

19. Division Algorithms:

20. Long Division Example 7 30 28 20 14 60

21. Long Division Example 1.42857 710.00000 7 30 28 20 14 60 q0 q1q2q3q4q5
Chosen to satisfy
usual inequalities

22. Long Division Example 1.42857 710.00000 p0 7 30 p1 28 20 p2 14 60 p3
q0 q1q2q3q4q5 p0 7 p1 28 p2 14
60 p3
p = 10, d = 7
10d = 70
pk+1 = 10(pk–qkd)
0 ≤ pk+1 < 70

23. Long Division Radix 10 S Compute q = p / d. p0 := p for k=0,1,...
Find the digit qk Î {0, 1, 2, …, 9} such that
pk+1 := 10(pk - qk d) satisfies pk +1 Î [0, 10)d
end
q = p / d = q i / 10 i S
i=0

24. SRT Division Radix 4 S Compute q = p / d. Such qk exists?
p0 := p
for k=0,1,...
Look up a digit qk Î {-2,-1,0,1,2} such that
pk+1 := 4(pk - qk d) satisfies|pk +1|≤ (8/3)d
end
q = p / d = q i / 4 i S
i=0
Such qk exists?
Algorithm correct?

25. Such qk exists? S Given 1£p,d<2. Compute q=p/d. p0 := p
for k=0,1,...
Look up a digit qk Î {-2,-1,0,1,2} such that
pk+1 := 4(pk - qk d) satisfies|pk +1|≤ (8/3)d
end
q = p/d = qi / 4i S
i=0
qk := 0
qk := –2
qk := +2
qk := +1
qk := –1 3 8 d 2

26. Such qk exists? S -2 / 3d ≤ pk ≤ 2 / 3d pk+1 := 4(pk - 0)
Given 1£p,d<2. Compute q=p/d.
p0 := p
for k=0,1,...
Look up a digit qk Î {-2,-1,0,1,2} such that
pk+1 := 4(pk - qk d) satisfies|pk +1|≤ (8/3)d
end
q = p/d = qi / 4i S ∞
i=0 2 3 d
qk := 0 3 2 d 8 3 d 3 8 d
4(pk - 0)
-2 / 3d ≤ pk ≤ 2 / 3d
pk+1 := 4(pk - 0)

27. Such qk exists? S 4 / 3d ≤ pk ≤ 8 / 3d pk+1 := 4(pk - 2d)
Given 1£p,d<2. Compute q=p/d.
p0 := p
for k=0,1,...
Look up a digit qk Î {-2,-1,0,1,2} such that
pk+1 := 4(pk - qk d) satisfies|pk +1|≤ (8/3)d
end
q = p/d = qi / 4i S
i=0
qk := +2 8 3 d 3 8 d 3 4 d
4 / 3d ≤ pk ≤ 8 / 3d
pk+1 := 4(pk - 2d)

28. Such qk exists? S 4/3 ≤ pk ≤ 8/3 pk+1 := 4(pk - 2d)
Given 1£p,d<2. Compute q=p/d.
p0 := p
for k=0,1,...
Look up a digit qk Î {-2,-1,0,1,2} such that
pk+1 := 4(pk - qk d) satisfies|pk +1|≤ (8/3)d
end
q = p/d = qi / 4i S
i=0 2 3 d
pk - 2d 3 2 d
qk := +2 8 3 d 3 8 d
4/3 ≤ pk ≤ 8/3
pk+1 := 4(pk - 2d)

29. Such qk exists? S 4 / 3d ≤ pk ≤ 8 / 3d pk+1 := 4(pk - 2d)
Given 1£p,d<2. Compute q=p/d.
p0 := p
for k=0,1,...
Look up a digit qk Î {-2,-1,0,1,2} such that
pk+1 := 4(pk - qk d) satisfies|pk +1|≤ (8/3)d
end
q = p/d = qi / 4i S
i=0 2 3 d
pk - 2d 3 2 d
qk := +2 8 3 d 3 8 d
4(pk - 2d)
4 / 3d ≤ pk ≤ 8 / 3d
pk+1 := 4(pk - 2d)

30. A qk For Every Point qk := –2 qk := 0 qk := +2 qk := –1 qk := +1 2 3 d8 3 d 3 8 d
qk := –1
qk := +1

31. Algorithm Correct? Claim: Proof by Induction: p q1 qk-1 pk d 4 4k-1 d
pk+1 = 4(pk - qkd) Þ
pk qk pk+1
d k d
— = q0 + — ––– + –— 4-k
— 4-k = — + —— 4-(k+1)

32. Algorithm Correct? Claim: Proof by Induction: p q1 qk-1 qk pk+1
d k k d
Proof by Induction:
pk+1 = 4(pk - qkd) Þ
pk qk pk+1
d k d
— = q0 + — ––– + — + ––— 4-(k+1)
— 4-k = — + —— 4-(k+1)

33. Algorithm Correct? Claim: Proof by Induction: Letting k®¥ proves
p q qk qk pk+1
d k k d
Proof by Induction:
pk+1 = 4(pk - qkd) Þ
pk qk pk+1
d k d
Letting k®¥ proves
p q1 q2 d
— = q0 + — ––– + — + ––— 4-(k+1)
— 4-k = — + —— 4-(k+1)
— = q0 + — + —

34. Pentium Lookup Table (P-d plot)
Blue
Green
q := 2
q := 1
q := 0
q := -1
q:= -2
Shifted Partial Remainder
1.0000
1.0001
1.0010
1.0011
1.0100
1.0101
1.0110
1.0111
1.1000
1.1001
1.1010
1.1011
1.1100
1.1101
1.1110
1.1111
1/16
1/8
Divisor

35. Pentium Lookup Table (P-d plot)
Blue
Green
q := 2
q := 1
q := 0
q := -1
q:= -2
Shifted Partial Remainder
1.0000
1.0001
1.0010
1.0011
1.0100
1.0101
1.0110
1.0111
1.1000
1.1001
1.1010
1.1011
1.1100
1.1101
1.1110
1.1111
1/16
1/8
Divisor

36. A Close-up Look at One Column (D=1.0001)
2.875
1.5
1.375
0.375
0.25
-0.5
-0.625
-1.625
-1.75 -3 2 1 -1 -2

37. A Close-up Look at One Column (D=1.0001)
2.875
1.5
1.375
0.375
0.25
-0.5
-0.625
-1.625
-1.75 -3 2 1 -1 -2 2
Off the Chart
Off the Chart
Off the Chart
Off the Chart
Buggy Entry
Foothold

38. Pentium Division Example: 1.875/1.000
1.875 = { S C
qk:=–2
qk:=0
qk:=+2
qk:=–1
qk:=+1
pk+1 := 4(pk – qkd)

39. Pentium Division Example: 1.875/1.000
1.875 = { S C S C
– 2´1 =
qk:=–2
qk:=0
qk:=+2
= {
qk:=–1
qk:=+1
pk+1 := 4(pk – qkd)

40. Pentium Division Example: 1.875/1.000
1.875 = { S C S C
– 2´1 =
qk:=0
qk:=–2
qk:=+2
qk:=+1
qk:=–1
-0.125´4 = -0.5 = {
pk+1 := 4(pk – qkd)

41. Pentium Division Example: 1.875/1.000
1.875 = { S C S C S C
–2´1 =
qk:=–2
qk:=0
qk:=+2
qk:=–1
qk:=+1
– –1´1 =
pk+1 := 4(pk – qkd)

42. Pentium Division Example: 1.875/1.000
1.875 = { S C S C S C S C S C
–2´1 =
qk:=–2
qk:=0
qk:=+2
qk:=–1
qk:=+1
– –1´1 =
–2´1 =
–0´1 =
2/1 + –1/4 + 2/16 + 0/64 = 1.875/1.000

43. ì í î Inequality Analysis Pk ≤ pk ≤ Pk + 1/4 D ≤ d ≤ D+ º D + 1/16
Pk+1 = 4(Pk – qkD+) + Rk ì í î
3/4 if qk = –2
3/4 if qk = –1
3/4 if qk = 0
1 if qk = 1
5/4 if qk = 2
Rk ≤ Rk º
Max

44. Reaching the Flaw is Not Easy!
qk Pk+1
–2 ≤ Pbad - 1/8
–1 ≤ Pbad - 1/8
0 < Pbad - 1/8
1 < Pbad - 1/8
Pk < Pbad - 1/8 ≤ Pbad - 1/8
Pk = Pbad - 1/8 ≤ Pbad ì í î 2

45. foothold
buggy entry

46. “Send More Money” Puzzle
intel
22 Mission College Blvd.
Santa Clara, CA 95052
32¢
Massachusetts Institue of Technology
77 Massachusetts Ave.
Cambridge, MA 02139
SEND
+MORE
MONEY

47. The Path to Failure ì í î Bad Divisors: d = 1.d1d2d3d4111111d11…
six ones
.d2d3d
.d2d3d40 0 0
q = –2
q = 2
bug

48. At Least Nine Steps to Failure
Step 1
Step 2
Step 3
Step 4
: 2 more steps
Step 5
Step 6

49. At Least Nine Steps to Failure
Step 1
Step 2
Step 3
Step 4
: 2 more steps
Step 5
Step 6

50. At Least Nine Steps to Failure
Step 1
Step 2
Step 3
Step 4
: 2 more steps
Step 5
Step 6

51. At Least Nine Steps to Failure
Step 1
Step 2
Step 3
Step 4
: 2 more steps
Step 5
Step 6

52. Conclusions Mathematical analysis is possible.
Bug is more subtle and more interesting than most people realize.
One should not be so quick to laugh at Intel’s Expense.

53. Thanks to Teddy Slottow for his technical assistance in preparing this presentation