Programming

Have you ever wondered how FORs impact your code? How they are limiting your design and more important how they are transforming your code into an amount of lines without any human meaning?

How can you not want to read an article that starts like that? I had to steal the intro from the original. Seriously, I have used FOR since I learned BASIC back in the day. I never thought about how it was limiting my design. I. Must. Learn. How.

The article I am referring to is, Avoiding FORs – Anti-If Campaign. Eager learner I, I could not not read it.

After the resplendent intro, Avoiding FOR goes on to show “how to transform a simple example of a for […], to something more readable and well designed.”

It takes this unreadable piece of code — that I now recognize as unreadable:

[java]
public class Department {

private List resources = new ArrayList();

public void addResource(Resource resource) {
    this.resources.add(resource);
}

public void printSlips() {

    for (Resource resource : resources) {
        if(resource.lastContract().deadline().after(new Date())) {

            System.out.println(resource.name());
            System.out.println(resource.salary());
        }
    }
}

}[/java]

My eyes hurt already. Thankfully the author transforms the aberration above into this clean, much more readable snippet:

[java]public class ResourceOrderedCollection {

    private Collection<Resource> resources = new ArrayList<Resource>();



public ResourceOrderedCollection() {
    super();
}

public ResourceOrderedCollection(Collection<Resource> resources) {
    this.resources = resources;
}

public void add(Resource resource) {
    this.resources.add(resource);
}

public void forEachDo(Block block) {
    Iterator<Resource> iterator = resources.iterator();

    while(iterator.hasNext()) {
        block.evaluate(iterator.next());
    }

}

public ResourceOrderedCollection select(Predicate predicate) {

    ResourceOrderedCollection resourceOrderedCollection = new ResourceOrderedCollection();

    Iterator<Resource> iterator = resources.iterator();

    while(iterator.hasNext()) {
        Resource resource = iterator.next();
        if(predicate.is(resource)) {
            resourceOrderedCollection.add(resource);
        }
    }

    return resourceOrderedCollection;
}

}

public class Department {

private List<Resource> resources = new ArrayList<Resource>();

public void addResource(Resource resource) {
    this.resources.add(resource);
}

public void printSlips() {
    new ResourceOrderedCollection(this.resources).select(new InForcePredicate()).forEachDo(new PrintSlip());
}

}[/java]

Wait, what? Is this an Onion article?

Snarky Mode Off.

I understand what the author wanted to do, but really, the example used is so off the left field that it’s not even funny.

This weekend I took the family to celebrate Father’s Day away from town. We went around getting to know parts of the province we live in and never been to.

Since English isn’t my first language, I have a limited vocabulary, which leaves me at a disadvantage against my English-speaking friends. I can handle the “hangman” part of the game where I have to guess the word my friends come up with; but when it becomes “Scrabble” and I’ve got to form words using only a given set of letters and still make them difficult enough that a native English speaker will have problems figuring them out, then it’s tough.

An itch that needed some scratching. Enter Anagramizer.

When I woke up this morning, I decided to write a little program to help me. You call it cheating, I call it having a bit of nerd fun.

Being that I’m currently in love with Go, I decided to write in that language and it was really easy and quick to do it. It took me about half an hour to write the program that did what I needed. But then…

I succumbed to the temptation and started adding bells and whistles. Admittedly it was mostly for my own amusement and trying stuff in Go, but by the time we were leaving for lunch, the program had more options than the KDE audio volume utility (see what I did just there?)

I decided to make it available to anyone who wants to play with it. It served its purpose of entertaining me for about half a day 🙂

It’s now available on Github and released under a BSD licence.

For fun I picked one of the Euler algorithms I had played with before and rewrote it in Go. Instead of carrying over the nested-loop search, this version first eliminates c using the fixed sum and solves the Pythagorean equation for b. Only a remains to search.

package main

import (
	"fmt"
	"os"
)

func main() {
	const sum = 1000

	for a := 1; a < sum/3; a++ {
		numerator := sum * (sum - 2*a)
		denominator := 2 * (sum - a)

		if numerator%denominator != 0 {
			continue
		}

		b := numerator / denominator
		c := sum - a - b
		if a < b && b < c {
			fmt.Println(a * b * c)
			return
		}
	}

	fmt.Fprintln(os.Stderr, "no solution")
	os.Exit(1)
}

The divisibility check rejects values of a that would produce a fractional b. The program uses constant space, examines fewer than sum / 3 candidates, and prints 31875000.

This one isn’t even funny…

Starting in the top left corner of a $2 \times 2$ grid, there are 6 routes (without backtracking) to the bottom right corner.

How many routes are there through a $20 \times 20$ grid?

Your first thought would be to generate the routes, but for a $20 \times 20$ grid, those amount to BILLIONS and you’d try to do it recursively too! Forget it.

But if you have some CompSci-level math background, though, you’ll remember this one—reading The Art of Computer Programming, Vol. 4 also helps. It’s a matter of combinatorics and if we take $w$ for the width and $h$ for the height, all we need to calculate is:

Every path consists of exactly 20 moves right and 20 moves down. Choosing which 20 of the 40 positions are right moves uniquely determines a path, so the answer is the binomial coefficient $\binom{40}{20}$.

from math import comb

width = height = 20
print(comb(width + height, width))

math.comb computes the integer result directly, without constructing three factorials. The answer is 137846528820.

Project Euler’s problem #11 statement goes:

In the $20 \times 20$ grid below, four numbers along a diagonal line have been marked in bold.

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

The product of these numbers is $26 \times 63 \times 78 \times 14 = 1\,788\,696$.

What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the $20 \times 20$ grid?

This one is remarkably easy but also was quite fun. I think it’s because it reminds me of the kind of work we’d do during our Algorithms classes during my first year in college. And so this one goes to my Algorithms teacher, Ricardo Vargas Dornelles—best teacher I’ve ever had too.

Only four directions are necessary: reversing any group gives the same product. Describing those directions as row and column offsets keeps one loop responsible for every case and makes the boundary check explicit.

from math import prod

grid_text = """\
08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
"""

grid = [
    [int(number) for number in line.split()]
    for line in grid_text.splitlines()
]
directions = ((0, 1), (1, 0), (1, 1), (1, -1))
length = 4
rows = len(grid)
columns = len(grid[0])
largest = 0

for row in range(rows):
    for column in range(columns):
        for row_step, column_step in directions:
            end_row = row + (length - 1) * row_step
            end_column = column + (length - 1) * column_step
            if not (0 <= end_row < rows and 0 <= end_column < columns):
                continue

            product = prod(
                grid[row + offset * row_step][column + offset * column_step]
                for offset in range(length)
            )
            largest = max(largest, product)

print(largest)

Every valid group is visited exactly once. Since there are four fixed directions and four values per product, the traversal is linear in the number of grid cells. It prints 70600674.