## Goldbach’s Other Conjecture

### June 7, 2016

Christian Goldbach (1690-1764) was a Prussian mathematician and contemporary of Euler. One of the most famous unproven conjectures in number theory is known as Goldbach’s Conjecture, which states that every even number greater than two is the sum of two prime numbers; for example, 28 = 5 + 23. We studied Goldbach’s Conjecture in a previous exercise.

Although it is not as well known, Goldbach made another conjecture as follows: Every odd number greater than two is the sum of a prime number and twice a square; for instance, 27 = 19 + 2 * (2 ** 2). (The conjecture is sometimes stated as every odd composite number is the sum of a prime number and twice a square, since it is trivially true with 0 as the square root for all prime numbers; alternately, it is sometimes limited so that the number being squared must be positive, in which case there are some odd primes that can be so expressed.) Sadly, it is easy to find a counter-example that disproves Goldbach’s other conjecture.

Your task is to write a program that finds the smallest number that disproves Goldbach’s other conjecture. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

## A Dozen Lines Of Code

### June 3, 2016

Today’s exercise demonstrates that it is sometimes possible to do a lot with a little.

Your task is to write some interesting and useful program in no more than a dozen lines of code. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

## Learn A New Language

### May 31, 2016

It’s fun to learn new programming languages. It’s also useful, even if you never use the new language, because it forces you to think differently about how you do things.

Your task is to write a familiar program in an unfamiliar language. When you are finished, you are welcome to read or run ([1], [2]) a suggested solution, or to post your own solution or discuss the exercise in the comments below.

## Pollard’s Rho Algorithm For Discrete Logarithms

### May 27, 2016

We studied discrete logarithms in two previous exercises. Today we look at a third algorithm for computing discrete algorithms, invented by John Pollard in the mid 1970s. Our presentation follows that in the book *Prime Numbers: A Computational Perspective* by Richard Crandall and Carl Pomerance, which differs somewhat from other sources.

Our goal is to compute *l* (some browsers mess that up; it’s a lower-case ell, for “logarithm”) in the expression *g ^{l}* ≡

*t*(mod

*p*); here

*p*is a prime greater than 3,

*g*is an integer generator on the range 1 ≤

*g*<

*p*, and

*t*is an integer target on the range 1 ≤

*g*<

*p*. Pollard takes a sequence of integer pairs (

*a*,

_{i}*b*) modulo (

_{i}*p*− 1) and a sequence of integers

*x*modulo

_{i}*p*such that

*x*=

_{i}*t*g

^{ai}^{bi}(mod

*p*), beginning with

*a*

_{0}=

*b*

_{0}= 0 and

*x*

_{0}= 1. Then the rule for deriving the terms of the various sequences is:

- If 0 <
*x*<_{i}*p*/3, then*a*_{i+1}= (*a*+ 1) mod (_{i}*p*− 1),*b*_{i+1}=*b*, and_{i}*x*_{i+1}=*t x*(mod_{i}*p*). - If
*p*/3 <*x*< 2_{i}*p*/3, then*a*_{i+1}= 2*a*mod (_{i}*p*− 1),*b*_{i+1}= 2*b*mod (_{i}*p*− 1), and*x*_{i+1}=*x*_{i}^{2}mod*p*. - If 2
*p*/3 <*x*<_{i}*p*, then*a*_{i+1}=*a*,_{i}*b*_{i+1}= (*b*+ 1) mod (_{i}*p*− 1), and*x*_{i+1}=*g x*mod_{i}*p*.

Splitting the computation into three pieces “randomizes” the calculation, since the interval in which *x _{i}* is found has nothing to do with the logarithm. The sequences are computed until some

*x*=

_{j}*x*, at which point we have

_{k}*t*

^{aj}*g*=

^{bj}*t*

^{ak}*g*. Then, if

^{bk}*a*−

_{j}*a*is coprime to

_{j}*p*− 1, we compute the discrete logarithm

*l*as (

*a*−

_{j}*a*)

_{k}*l*≡

*b*−

_{k}*b*(mod (

_{j}*p*− 1)). However, if the greatest common divisor of

*a*−

_{j}*a*with

_{j}*p*− 1 is

*d*> 1, then we compute (

*a*−

_{j}*a*)

_{k}*l*

_{0}≡

*b*−

_{k}*b*(mod (

_{j}*p*− 1) /

*d*), and

*l*=

*l*

_{0}+

*m*(

*p*− 1) /

*d*for some

*m*= 0, 1, …,

*d*− 1, which must all be checked until the discrete logarithm is found.

Thus, Pollard’s rho algorithm consists of iterating the sequences until a match is found, for which we use Floyd’s cycle-finding algorithm, just as in Pollard’s rho algorithm for factoring integers. Here are outlines of the two algorithms, shown side-by-side to highlight the similarities:

# find d such that d | n # find l such that g**l = t (mod p) function factor(n) function dlog(g, t, p) func f(x) := (x*x+c) % n func f(x,a,b) := ... as above ... t, h, d := 1, 1, 1 j := (1,0,0); k := f(1,0,0) while d == 1 while j.x <> k.x t = f(t) j(x,a,b) := f(j.x, j.a, j.b) h = f(f(h)) k(x,a,b) := f(f(k.x, k.a, k.b)) d = gcd(t-h, n) d := gcd(j.a-k.a, p-1) return d return l ... as above ...

Please pardon some abuse of notation; I hope the intent is clear. In the factoring algorithm, it is possible that *d* is the trivial factor *n*, in which case you must try again with a different constant in the *f* function; the logarithm function has no such possibility. Most of the time consumed in the computation is the modular multiplications in the calculations of the *x* sequence; the algorithm itself is O(sqrt *p*), the same as the baby-steps, giant-steps algorithm of a previous exercise, but the space requirement is only a small constant, rather than the O(sqrt *p*) space required of the previous algorithm. In practice, the random split is made into more than 3 pieces, which complicates the code but speeds the computation, as much as a 25% improvement on average.

Your task is to write a program that computes discrete logarithms using Pollard’s rho algorithm. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

## Test Scores

### May 24, 2016

The high school two blocks from me just had their annual picnic, my youngest daughter just graduated from college, and my primarily academic readership suddenly dropped in half (history suggest it will stay low until mid-August), so it seems to be the right season to have a simple data-processing task involving student test scores.

Given a list of student names and test scores, compute the average of the top five scores for each student. You may assume each student has as least five scores.

Your task is to compute student scores as described above. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

## No Exercise Today

### May 20, 2016

I’ve been busy at work and haven’t had time to prepare an exercise for today. I apologize.

Your task is to solve a previous exercise that you haven’t yet solved. Have fun!

## Conditional Heap Insertion

### May 17, 2016

This is an Amazon interview question:

Given a heap (priority queue), insert an element into the heap if the element is not already present in the heap. Your solution must work in O(

n) time, wherenis the number of items in the heap.

Your task is to write a program to insert an element into a heap if the element is not already present in the heap, in logarithmic time. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

## Interleaved Increasing-Decreasing Sort

### May 13, 2016

This must be somebody’s homework:

Given an array of integers, rearrange the elements of the array so that elements in even-indexed positions are in ascending order and elements in odd-indexed positions are in descending order. For instance, given the input 0123456789, the desired output is 0927456381, with the even-indexed positions in ascending order 02468 and the odd-indexed positions in descending order 97531.

Your task is to write a program that performs the indicated rearrangement of its input. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

## Concatenate N N Times

### May 10, 2016

A number like 7777777 consists of the number 7 concatenated to itself 7 times. A number like 121212121212121212121212 consists of the number 12 concatenated to itself 12 times.

Your task is to write a program that calculates the number that is concatenated to itself the number of times as the number is (that’s hard to say). When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

## Baby Steps, Giant Steps

### May 6, 2016

In a previous exercise we discussed the discrete logarithm problem, which is to compute the exponent *y* in the expression *x ^{y}* ≡

*n*(mod

*m*), given

*x*,

*n*, and

*m*; the modulus

*m*is usually taken as prime. Today we look at an algorithm, known as baby steps, giant steps, that was developed by Daniel Shanks in 1971:

1. Compute limits:

*b* = ⌈ √*m* ⌉

*h* = (*x*^{−1})^{b}

2. Construct lists:

*A* = { *x ^{i}* :

*i*= 0, 1, …,

*b*− 1 } // giant steps

*B* = { *n h ^{j}* :

*j*= 0, 1, …,

*b*− 1 } // baby steps

3. Sort and find intersection:

Sort the lists *A* and *B*

Find an intersection, say *x ^{i}* =

*n h*

^{j}Return *y* = *i* + *j b*

Since *m* is prime, there must be some *y* ∈ [0, *m*) for which *x*^{y} ≡ *n* (mod *m*). Write *y* = *i* + *j b*, where *b* = ⌈ √*m* ⌉. Since *y* must exist, so too *i* (which counts the giant steps) and *j* (which counts the baby steps) must exist, and there must be an intersection between the baby steps and the giant steps.

Time complexity is obviously O(sqrt *m*), which beats the O(*m*) time complexity of the brute-force algorithm of the previous exercise. There are better algorithms for computing discrete logarithms, which we will study in future exercises.

Your task is to write a program that calculates discrete logarithms using the baby steps, giant steps algorithm. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.