How To Multiply Two N Digit Numbers In N N Digit Formula
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1 What we still don t know about addition and multiplication Carl Pomerance, Dartmouth College Hanover, New Hampshire, USA (University of Georgia, emeritus) Bay Area Mathematical Adventures Santa Clara University January 23, 2015
2 You would think that all of the issues surrounding addition and multiplication were sewed up in third grade! Well in this talk we ll learn about some things they didn t tell you... 1
3 Here s one thing they did tell you: Find
4 If instead you had a problem with two 23-digit numbers, well you always knew deep down that math teachers are cruel and sadistic. Just kidding! (Aside: evil laugh... ) In principle if you really have to, you could work out 23-digits times 23-digits on paper, provided the paper is big enough, but it s a lot of work. So here s the real question: How much work? 3
5 Of course the amount of work depends not only on how long the numbers are, but on what they are. For example, multiplying by 10 22, that s 23-digits times 23-digits, but you can do it in your head. In general, you ll take each digit of the lower number, and multiply it painstakingly into the top number. It s less work if some digit in the lower number is repeated, and there are definitely repeats, since there are only 10 possible digits. But even if it s no work at all, you still have to write it down, and that s 23 or 24 digits. At the minimum (assuming no zeroes), you have to write down 23 2 = 529 digits for the parallelogram part of the product. And then comes the final addition, where all of those 529 digits need to be processed. 4
6 23 rows of at least 23 digits each a a a a a a a a a a a a a a a a a a a a a a a b b b b b b b b b b b b b b b b b b b b b b b c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c... + c c c c c c c c c c c c c c c c c c c c c c c 5
7 So in general if you multiply two n-digit numbers, it would seem that you d be taking n 2 steps, unless there were a lot of zeroes. This ignores extra steps, like carrying and so on, but that at worst changes n 2 to maybe 2n 2 or 3n 2. We say that the complexity of school multiplication for two n-digit numbers is of order n 2. 6
8 A. A. Karatsuba ( ): Devised a faster way to multiply two n-digit numbers, in about n 1.6 elementary steps. 7
9 Karatsuba s method was later improved by Toom, Cook, Schönhage, & Strassen. After their efforts we have the Fast Fourier Transform that allows you to multiply in about n L(n) steps, where L(n) is short-hand for the number of digits of n. (So L(n) is the number of digits of the number of digits of the numbers being multiplied!) We don t know if the Fast Fourier Transform is best possible. In particular: What is the fastest way to multiply? 8
10 Let s play Jeopardy Multiplication! Here are the rules: I give you the answer to the multiplication problem, and you give me the problem phrased as a question. You must use whole numbers larger than 1. So, if I say 15, you say What is 3 5? OK, let s play. 21 9
11 Good. That was easy. Let s up the ante
12 Good. That was easy. Let s up the ante. 91 What is 7 13? 11
13 Let s do
14 Let s do Thinking, thinking.... Hmm, 13
15 Let s do Thinking, thinking.... Hmm, 8051 = = = (90 7)(90 + 7) = Got it! What is 83 97? 14
16 So, here s what we don t know: How many steps does it take to figure out the factors if you are given an n-digit number which can be factored? (A trick problem would be: 17. The only way to write it as a b is to use 1, and that was ruled out. So, prime numbers cannot be factored, and the thing we don t know is how long it takes to factor the non-primes.) The best answer we have so far is about 10 n1/3 steps, and even this is not a theorem, but our algorithm (the number field sieve) seems to work in practice. This is all crucially important for the security of Internet commerce. Or I should say that Internet commerce relies on the premise that we cannot factor much more quickly than that. 15
17 Let s look at an unsolved problem concerning addition. We all recall the addition table:
18 The array of sums has all the numbers from 2 to 20 for a total of 19 different sums. If you were to try this for the N N addition table we d see all of the numbers from 2 to 2N for a total of 2N 1 different sums. Now, what if we were to be perverse and instead of having the numbers from 1 to N, we had some arbitrary list of N different numbers. Can you arrange it so there are fewer than 2N 1 different sums? 17
19 The array of sums has all the numbers from 2 to 20 for a total of 19 different sums. If you were to try this for the N N addition table we d see all of the numbers from 2 to 2N for a total of 2N 1 different sums. Now, what if we were to be perverse and instead of having the numbers from 1 to N, we had some arbitrary list of N different numbers. Can you arrange it so there are fewer than 2N 1 different sums? If you answered No, there are always at least 2N 1 different sums, you d be right. 18
20 Now consider the reverse problem. You have a list of N different numbers, you form the addition table with themselves, and you find that there are just 2N 1 different sums. Must the list be the numbers from 1 to N? 19
21 Now consider the reverse problem. You have a list of N different numbers, you form the addition table with themselves, and you find that there are just 2N 1 different sums. Must the list be the numbers from 1 to N? There are the 9 different sums: 4, 7, 10, 13, 16, 19, 22, 25, 28, all others are repeats of these. 20
22 In fact: If there are just 2N 1 distinct sums when adding an N-element set to itself, then the N-element set must be an arithmetic progression. Here s an example where there are many different sums:
23 So, sometimes there are few distinct sums and sometimes many. What structure is forced on the set if there are few distinct sums? We know the answer when there are very few distinct sums: Gregory Freiman 22
24 Here s something with multiplication tables. Let s look at the N N multiplication table using the numbers from 1 to N. With addition, we were able to count exactly how many distinct numbers appear in the table. How many different numbers appear in the N N multiplication table? 23
25 Let M(N) be the number of distinct entries in the N N multiplication table So, M(5) =
26 M(10) =
27 It may be too difficult to expect a neat exact formula for M(N). Instead, we could ask for its order of magnitude, or even approximate order of magnitude. For example, does M(N) go to infinity like a constant times N 2, or more slowly? That is, maybe there is a positive number c with M(N)/N 2 > c for all N. Or maybe for every positive number c, M(N)/N 2 < c for infinitely many choices for N or perhaps for all large N. 26
28 Here are some values of M(N)/N 2 (Brent & Kung 1981): N M(N) M(N)/N
29 And some more values (Brent & Kung 1981, Brent 2012): N M(N) M(N)/N
30 And some statistically sampled values (Brent & P 2012): N M(N)/N 2 N M(N)/N
31 Paul Erdős studied this problem in two papers, one in 1955, the other in Paul Erdős,
32 In 1955, Erdős proved (in Hebrew) that M(N)/N 2 tends to 0 as N runs off to infinity. In 1960, at the prodding of Linnik and Vinogradov, Erdős identified (in Russian) how quickly M(N)/N 2 tends to 0. Recall that L(N) denotes the number of (decimal) digits of N. There is a constant (let s call it E for Erdős), E = , such that M(N)/N 2 is about 1/L(N) E. That is, M(N) is about N 2 /L(N) E. 31
33 A conjecture both deep and profound is whether a circle is round. In a paper of Erdős, written in Kurdish, a counterexample is found. Leo Moser 32
34 Erdős says that M(N) is about N 2 /L(N) E. What s with this weasel word about? Answer: We re saying that no other exponent on L(N) than the Erdős constant E does better. But there could be lower order factors. In work of Tenenbaum progress was made (in French) in nailing this down. In 2008, Ford showed (in English) that M(N) is of order of magnitude N 2 L(N) E L(L(N)) 3/2. No matter the language, we still don t know an asymptotic estimate for M(N), despite this just being about multiplication tables! 33
35 Here are two famous unsolved problems involving both addition and multiplication: Goldbach s conjecture: Every even number starting with 4 is the sum of two primes. The twin prime conjecture: There are infinitely many pairs of primes that differ by 2. 34
36 And here s another famous problem (in disguised form): Let A(N) be the least common multiple of 1, 2,..., N (aka: the least common denominator when 1, 2,..., N are denominators). Let B(N) = A(N). For example: B(10) = Do we always have B(N) N < NL(N) 2? (Recall: L(N) is the number of digits of N.) The Clay Mathematics Institute offers $1,000,000 for a proof! 35
37 Let me close with one unified problem about addition and multiplication tables. It s due to Erdős & Szemerédi. Look at both the addition and multiplication tables and count the number of distinct entries. We ve seen that if we take the first N numbers we get about N 2 /L(N) E entries (since the 2N 1 from the addition table is just noise compared to the huge number from the multiplication table). At the other extreme, if we take for our N numbers the powers of 2, namely 1, 2, 4,..., 2 N 1, then there are at least 1 2 N 2 distinct entries in the addition table and only 2N 1 entries in the multiplication table. 36
38 The question is: if we make one of these small, must the other always be large?
39
40 Many products Few sums {1, 2,..., N} 38
41
42 Few products Many sums {1, 2, 4,..., 2 N 1 } 40
43 Must one always be large? Put more precisely: If we have N distinct numbers, must one of the number of distinct pairwise sums, the number of distinct pairwise products, be greater than N for all large values of N? We don t know. And it s not for lack of trying. 41
44 The game players with the sum/product problem include: Erdős, Szemerédi, Nathanson, Chen, Elekes, Bourgain, Chang, Konyagin, Green, Tao, Solymosi,... The best that s been proved (Solymosi) is that there are at least N 4/3 different entries. This list of mathematicians contains two Fields Medalists, a Wolf Prize winner, an Abel Prize winner, four Salem Prize Winners, two Crafoord Prize winners, and an Aisenstadt Prize winner. And still the problem is not solved! 42
45 My message: We could use a little help with these problems!! 43
46 My message: We could use a little help with these problems!! THANK YOU 44
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