May 2, 2014
An n-bit Gray sequence is a sequence of 2n values, starting from 0, in which each differs from its predecessor by a single bit. There are always 2n−1 n-bit Gray sequences; for instance, the two 2-bit Gray sequences are 0, 1, 3, 2 and 0, 2, 3, 1. It is easier to see the Gray sequences when they are written in binary; the two 2-bit Gray sequences written in binary are 00, 01, 11, 10 and 0,0 10, 11, 01, where it is clear that each element of the sequence differs from the previous one by only one bit.
Although there are many possible Gray sequences for any given number of bits, there is one Gray sequence, known as the binary reflected gray code BRGC, that is almost always the Gray sequence that is being discussed. Such sequences are generated recursively from the next-smaller sequence by reversing the sequence, prefixing the entries of the original sequence with a 0-bit, prefixing the entries of the reversed sequence with a 1-bit, then concatenating the two sequences. For instance, given the 2-bit Gray sequence 00, 01, 11, 10, its reversal is 10, 11, 01, 00, adding a 0-bit to the original gives 000, 001, 011, 010, adding a 1-bit to the reversal gives 110, 111, 101, 100, and concatenating the two gives 000, 001, 011, 010, 110, 111, 101, 100, which is 0, 1, 3, 2, 6, 7, 5, 4.
Your task is to write a function that generates an n-bit binary reflected Gray sequence. 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.
April 29, 2014
We have seen the Miller-Rabin and Baillie-Wagstaff primality tests in previous exercises. Today we look at a test developed by Robert M. Solovay and Volker Strassen in 1977. The test is no longer used, having been superseded by the other two tests, but was historically of great importance in the development of the RSA cryptosystem.
The test is based on Euler’s Criterion, which states that a(p−1)/2 ≡ (a / p) (mod p) for any odd prime number p and any integer a on the range 2 to p − 1 with gcd(a, p) = 1, where (a / p) is the Legendre symbol. If a number n being tested is prime, the test will indicate that n is prime; if a number n being tested is composite, the test will indicate that it is either prime or composite with equal likelihood, a coin flip. Thus, the Solovay-Strassen test chooses k different witnesses a; if any of them indicate n is composite, then it must be composite, but if none of them indicate n is composite, it is presumed prime with odds 2−k.
If you are willing to assume the truth of the Extended Riemann Hypothesis, the Solovay-Strassen test can be made an absolute proof of primality by testing all prime a in the range 2 to min(n − 1, log2 n); if none of the a indicate the compositeness of n, then n is prime on the ERH.
The Solovay-Strassen test is no longer used, for three reasons: First, it takes more work to implement, as it involves computing the Jacobi symbol as well as modular exponentiation, whereas the Miller-Rabin test involves only modular exponentiation. Second, the Miller-Rabin test has a better error bound, k−4 compared to k−2. Third, all Euler witnesses are also Miller-Rabin strong witnesses to the compositeness of n. The presence of the absolute proof of primality on the ERH is also not a bar to the use of the Miller-Rabin test, as there is a similar proof of primality based on Miller-Rabin strong witnesses.
Your task is to write two versions of the Solovay-Strassen primality test, one probabilistic and one on the ERH. 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.
April 23, 2014
We studied a simple assembler for a hypothetical computer in the two previous exercises. Today we finish our look at the assembler by writing a program that produces neat listings, like this one for our sample program:
# print sum of input numbers (terminated by zero) 000: 03010 ld zero # initialize sum to zero 001: 04011 st sum 002: 01000 loop get # read a number 003: 08007 jz done # no more input if number is zero 004: 05011 add sum # add input to accumulated sum 005: 04011 st sum # store new value back in sum 006: 09002 j loop # go back and read another number 007: 03011 done ld sum # print sum 008: 02000 put 009: 10000 halt 010: 00000 zero const 0 011: 00000 sum const
The listing shows the memory address in the first column, the contents of memory at the beginning of the program in the second column, the mnemonic label in the third column, the opcode in the fourth column, the object of the operation in the fifth column, and the comment in the sixth column.
Your task is to write a program that produces assembler listings. 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.
April 18, 2014
In the previous exercise we built an assembler for a hypothetical computer. Today we write a simulator for that hypothetical computer. The basic idea is simple: Start with a program counter at location 0, either perform the action at the location and advance the program counter to the next location, or set the program counter to the object of a jump instruction.
April 15, 2014
In this exercise and the next one we will write a simple assembler for a hypothetical computer; we are following the assembler in the book The Awk Programming Langauge by Alfred V. Aho, Brian W. Kernighan and Peter J. Weinberger. Our computer has a memory of a thousand five-digit words, a single accumulator, and eleven opcodes:
||assembler pseudo-operator to define a constant
||read a number from the input to the accumulator|
||write the number in the accumulator to the output|
||load accumulator with contents of memory location
||store contents of accumulator in memory location
||add contents of memory location
||subtract contents of memory location
||jump to memory location
||jump to memory location
||jump to memory location
||stop program execution|
An assembly-language program is a series of blank lines and statements consisting of up to four fields: The first field, if it exists, is a label; it must start at the first position on the line and may not start with a digit. The second field, which is mandatory, is the opcode; it follows the optional label and mandatory spaces. The third field, which is used only with some opcodes, is the object; if it is present, it follows the opcode and mandatory spaces. The fourth field, which is optional, is a comment; everything on the line following a hash-sign is ignored. Here is a sample assembly-language program that prints the sum of a series of integers entered by the user, with the end of input marked by a 0:
# print sum of input numbers (terminated by zero) ld zero # initialize sum to zero st sum loop get # read a number jz done # no more input if number is zero add sum # add input to accumulated sum st sum # store new value back in sum j loop # go back and read another number done ld sum # print sum put halt zero const 0 sum const
The contents of memory after loading the sample program are shown on the next page.
Your task is to write a program that assembles a program written in our simple assembly language and loads the program into memory; we’ll write a simulator for our hypothetical computer in the next exercise. 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.
April 11, 2014
Programs that create graphs and charts are an important tool in many commercial and technical endeavors; for instance, a programmer might look at a graph of memory usage during a debugging run of his program to identify hot spots that need to be trimmed.
A simple type of graph is a scatter graph, sometimes called an x/y plot. The input is a list of points on an x,y-grid; the output is a picture with the points plotted on a grid. A simple input file appears on the next page.
Your task is to write a program to plot scatter graphs. When you are finished, you are welcome to read or run a suggested solution, or to post your own program or discuss the exercise in the comments below.
April 8, 2014
The only output functions provided by standard Scheme are
display, for plain output, and
write, for system-formatted output. That’s rather limiting. At the other extreme, Lisp provides the
format function, which has more options than you can dream about (before I discarded it in favor of the HyperSpec, the spine of my printed copy of CLtL2 was broken in two places,
loop). Many languages provide some kind of formatted output — does anyone remember PIC in COBOL? In today’s exercise we will implement the
printf function popularized by C and used in several languages.
There are three functions in the
(sprintf fmt expr …
) returns a string formatted according to the specification given by format and containing the values expr …; (printf fmt expr …
) displays a string formatted similarly to
sprintf to the current output port, and
(fprintf port fmt expr …
) displays a string formatted similarly to
sprintf to the indicated port. The fmt and port arguments are always required.
The fmt argument is a string that contains literal text, escape sequences, and specifications of how the expressions should be formatted. A specification consists of a literal percent-sign
%, zero or more modifiers, and a single-character specifier. The single-character specifiers that we will support are:
%literal percent sign
There should be as many expressions as there are format specifiers in the fmt string, except that a
% literal percent sign specifier does not consume an expression. As many as four modifiers may appear between the literal percent sign
% that starts a specifier and the single-character specifier that ends it:
-left-justify the expression in its field; if not given, the expression is right-justified
0left-pad with leading zeros instead of spaces
width pad field to this width using spaces (or zeros)
.prec digits to right of decimal point, or maximum string length
The modifiers must appear in the order shown above. The width and prec arguments are unsigned decimal integers.
Escape sequences are introduced by a backslash; the following escape sequences are supported:
\ddd character with octal value ddd, where ddd is 1 to 3 digits between 0 and 7 inclusive
\c any other character c literally, for instance
\\for backslash or
\"for quotation mark
Depending on the environment, a newline may (or may not) imply a carriage return, or vice-versa. Several examples are given on the next page.
Your task is to write a function that provides formatted output for your language, using the definition of
printf given above or some other specification that is suitable for your language and your aspirations. 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.
April 4, 2014
As penance for teasing you on April Fool’s Day, I have another exercise on factoring. Today we will simulate a machine invented by Derrick H Lehmer (the son of the famous father-son-daughter-in-law trio of mathematicians named Lehmer) that uses bicycle chains to factor integers. Unlike the previous exercise, this factoring method is real; in fact, a variant of this method was the best available method to factor integers from the mid-1930s to the mid-1960s, and Lehmer’s machine found many valuable factorizations for the Cunningham Project. We’ll discuss a simplified version of the machine today, as described by Uli Meyer, and the full version of the machine in a future exercise. I am unable to find a copyright-free picture of Lehmer’s machine, but if you’re interested you could visit the Computer History Museum, or see Lehmer himself describe the machine.
The machine is based on Fermat’s method of factoring by a difference of squares: if n = x2 − y2 then the factors of n are (x − y) and (x + y). Fermat’s method starts with y = ⌈ √n ⌉, computes y2 − n, and stops when the result is a square.
Lehmer’s machine identifies likely squares by looking at quadratic residues. An integer a is a quadratic residue modulo n if there exists some x such that x2 ≡ a (mod n). For a number y2 − n to be a square in Fermat’s algorithm, it must be a quadratic residue to many small bases. Lehmer’s machine represents quadratic residues as “open” points on a bicycle chain, then spins many bicycle chains of different lengths along a common axis and stops when all are quadratic residues; in many but not all cases, such a number will will indicate a factor of n.
Uli Meyer gives a better description, shows a useful diagram, and provides photos of his Lego sieve machine.
Your task is to write a program that simulates Lehmer’s machine to factor integers. 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.
April 1, 2014
A few days ago, an announcement was made in /r/crypto of a new factoring algorithm that runs in time O(log^4 n). A paper describes the algorithm. And even better, coding for the new algorithm is simple:
1)Set m ← n.
2) Express m as a list of its decimal digits.
3) Express each decimal digit as a list of binary bits.
4) Append the individual lists of binary bits.
5) Convert the resulting list of binary bits into a decimal number. Call it d.
6) If the greatest common divisor of n and d is between 1 and n, it is a factor of n. Report the factor and halt.
7) Set m ← d and go to Step 1.
Steps 1 through 4 form the digital code of a number. For example, 88837 forms the bit-list 1 0 0 0, 1 0 0 0, 1 0 0 0, 1 1, 1 1 1, where we used commas to delineate the decimal digits. That bit-list is 69919 in decimal, which is the digital code of 88837. To find a factor of 88837, compute successive digital codes until a gcd is greater than 1; the chain runs 88837, 69919, 54073, 2847, 2599, 5529, 2921, 333, at which point gcd(88837, 333) = 37 and we have found a factor: 88837 = 74 × 37. Factoring 96577 is even faster: the digital code of 96577 is 40319, and gcd(96577, 40319) = 23, which is a factor of 96577 = 13 × 17 × 19 × 23. But trying to factor 65059 = 17 × 43 × 89 causes an infinite loop. The paper gives a heuristic solution to that problem, but we’ll stop there and accept that sometimes a factorization fails.
Your task is to write a function that factors by digital coding. 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.
March 28, 2014
[ Today's exercise was written by guest author Paul Hofstra, who holds a PhD in Theoretical Nuclear Physics from the Free University in Amsterdam and is now retired from a major oil company. Guest authors are always welcome; contact me at the link in the menu bar if you are interested in writing for Programming Praxis. ]
We discussed various algorithms for finding the roots of a function in two previous exercises. In the first of those exercises, we looked at the bisection method, which is “slow and steady” as it always finds a solution, and the secant method, which is sometimes very fast but also sometimes very slow. In the second of those two exercises, we looked at an algorithm of Theodorus Dekker that combines the bisection and secant methods in a way that retains the speed of the secant method where possible but retains the steadiness of the bisection method where necessary. Later, Richard Brent (the same Brent that improved Pollard’s rho algorithm for finding factors), observed that there are still some functions for which Dekker’s algorithm performs poorly, and he developed the method that is most commonly used today.
Brent augmented the linear interpolation of the secant method with a quadratic interpolation that tends to get closer to the root more quickly than the secant method, and he made conditions that force a bisection step at least every three steps, so his method can never be worse than three times the bisection method, and usually much better. Brent calculates a new point using both the secant and bisection methods, as does Dekker, but instead of simply accepting the secant method when it is better, he calculates another potential point using the quadratic method and takes that instead if it is better than the secant point. You can read Brent’s description of his algorithm at http://maths-people.anu.edu.au/~brent/pd/rpb005.pdf.