How to resolve the algorithm State name puzzle step by step in the C programming language

Published on 7 June 2024 03:52 AM
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How to resolve the algorithm State name puzzle step by step in the C programming language

Table of Contents

Problem Statement

Background This task is inspired by Mark Nelson's DDJ Column "Wordplay" and one of the weekly puzzle challenges from Will Shortz on NPR Weekend Edition [1] and originally attributed to David Edelheit. The challenge was to take the names of two U.S. States, mix them all together, then rearrange the letters to form the names of two different U.S. States (so that all four state names differ from one another). What states are these?

The problem was reissued on the Unicon Discussion Web which includes several solutions with analysis. Several techniques may be helpful and you may wish to refer to Gödel numbering, equivalence relations, and equivalence classes. The basic merits of these were discussed in the Unicon Discussion Web. A second challenge in the form of a set of fictitious new states was also presented.

Write a program to solve the challenge using both the original list of states and the fictitious list.

Comma separated list of state names used in the original puzzle: Comma separated list of additional fictitious state names to be added to the original (Includes a duplicate):

Let's start with the solution:

Step by Step solution about How to resolve the algorithm State name puzzle step by step in the C programming language

This C program reads a list of US states and finds pairs of states with the most similar letter frequencies. Here's a detailed explanation of the code:

  1. Data Structures:

    • states: An array of character string pointers containing the names of US states.
    • letters: A custom struct used to store letter frequencies and state names.

  2. Function Declarations:

    • count_letters(): Counts the occurrences of each letter in a given string and stores them in a letters struct.
    • lcmp(): Compares two letters structs based on letter frequencies.
    • scmp(): Compares two state names lexicographically.
    • no_dup(): Removes duplicate entries from the states array.
    • find_mix(): Finds pairs of states with the most similar letter frequencies.

  3. main() Function:

    • Calls find_mix() to find pairs of states with similar letter frequencies.

  4. find_mix() Function:

    • Remove Duplicates: Calls no_dup() to remove duplicates from the states array.
    • Create Data Structures: Allocates memory for an array of letters structs.
    • Count Letter Frequencies: Iterates over state pairs and counts the occurrences of each letter in both states' names, storing these frequencies in letters structs.
    • Sort by Letter Frequency: Sorts the letters structs in ascending order based on letter frequencies using qsort() and the lcmp() comparison function.
    • Find Similar Pairs: Iterates over the sorted letters structs and identifies pairs with identical letter frequencies.
    • Print Results: Prints pairs of states with similar letter frequencies.

  5. Helper Functions:

    • count_letters(): Iterates over a string, counts the occurrences of each letter, and stores them in a letters struct.
    • lcmp(): Compares two letters structs based on letter frequencies. It returns 1 if the first struct has a higher frequency for any letter, -1 if the second struct has a higher frequency, and 0 if both structs have the same letter frequencies.
    • scmp(): Compares two state names lexicographically.
    • no_dup(): Removes duplicate entries from the states array by sorting it and then removing adjacent duplicates.

  6. Additional Notes:

    • The USE_FAKES macro is used to generate fake states with similar names to test the program's ability to handle duplicates.
    • The program assumes that all state names are unique (excluding the fake states generated when USE_FAKES is defined).

Source code in the c programming language

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define USE_FAKES 1

const char *states[] = {
#if USE_FAKES
	"New Kory", "Wen Kory", "York New", "Kory New", "New Kory",
#endif
	"Alabama", "Alaska", "Arizona", "Arkansas",
	"California", "Colorado", "Connecticut",
	"Delaware",    
	"Florida", "Georgia", "Hawaii",
	"Idaho", "Illinois", "Indiana", "Iowa",
	"Kansas", "Kentucky", "Louisiana",
	"Maine", "Maryland", "Massachusetts", "Michigan",
	"Minnesota", "Mississippi", "Missouri", "Montana",
	"Nebraska", "Nevada", "New Hampshire", "New Jersey",
	"New Mexico", "New York", "North Carolina", "North Dakota",
	"Ohio", "Oklahoma", "Oregon",
	"Pennsylvania", "Rhode Island",
	"South Carolina", "South Dakota", "Tennessee", "Texas",
	"Utah", "Vermont", "Virginia",
	"Washington", "West Virginia", "Wisconsin", "Wyoming"
};

int n_states = sizeof(states)/sizeof(*states);
typedef struct { unsigned char c[26]; const char *name[2]; } letters;

void count_letters(letters *l, const char *s)
{
	int c;
	if (!l->name[0]) l->name[0] = s;
	else l->name[1] = s;

	while ((c = *s++)) {
		if (c >= 'a' && c <= 'z') l->c[c - 'a']++;
		if (c >= 'A' && c <= 'Z') l->c[c - 'A']++;
	}
}

int lcmp(const void *aa, const void *bb)
{
	int i;
	const letters *a = aa, *b = bb;
	for (i = 0; i < 26; i++)
		if      (a->c[i] > b->c[i]) return  1;
		else if (a->c[i] < b->c[i]) return -1;
	return 0;
}

int scmp(const void *a, const void *b)
{
	return strcmp(*(const char *const *)a, *(const char *const *)b);
}

void no_dup()
{
	int i, j;

	qsort(states, n_states, sizeof(const char*), scmp);

	for (i = j = 0; i < n_states;) {
		while (++i < n_states && !strcmp(states[i], states[j]));
		if (i < n_states) states[++j] = states[i];
	}

	n_states = j + 1;
}

void find_mix()
{
	int i, j, n;
	letters *l, *p;

	no_dup();
	n = n_states * (n_states - 1) / 2;
	p = l = calloc(n, sizeof(letters));

	for (i = 0; i < n_states; i++)
		for (j = i + 1; j < n_states; j++, p++) {
			count_letters(p, states[i]);
			count_letters(p, states[j]);
		}

	qsort(l, n, sizeof(letters), lcmp);

	for (j = 0; j < n; j++) {
		for (i = j + 1; i < n && !lcmp(l + j, l + i); i++) {
			if (l[j].name[0] == l[i].name[0]
				|| l[j].name[1] == l[i].name[0]
				|| l[j].name[1] == l[i].name[1])
				continue;
			printf("%s + %s => %s + %s\n",
				l[j].name[0], l[j].name[1], l[i].name[0], l[i].name[1]);
		}
	}
	free(l);
}

int main(void)
{
	find_mix();
	return 0;
}

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