Counting Fingers
June 16, 2020
A little girl counts on her fingers in a curious way. She counts 1 on her thumb, 2 on her index finger, 3 on her middle finger, 4 on her ring finger, and 5 on her pinkie finger, then works back, counting 6 on her ring finger, 7 on her middle finger, 8 on her index finger, and 9 on her thumb, when she again turns around and counts 10 on her index finger, 11 on her middle finger, and so on.
Your task is to write a program that determines which finger the little girl will be on when she reaches a thousand. 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.
Programming Praxis 
I presume that should be: “counts 10 on her index finger, 11 on her middle finger, and so on”
Correct. Fixed. Thank you.
Here is my take on Julia: https://pastebin.com/NLWPwL5C
Cool drill for getting acquainted with modulo arithmetic. Cheers!
f = lambda n: 4-abs(4-(n-1)%8) print(f(1000)) print("timrp"[f(1000)]) # see Mark's solutionMy thorouhly documented solution implemented in SML can be found at <ahref=”https://pastebin.com/HQ5WkqKx”>. It takes a different approach from modular arithmetic.
Hi, I’m beginner, can I resolve task in this way?
public class Exercises1 {
public static void main(String[] args) {
Solution solution = new Solution();
while (solution.getCounter_fingers() < 1006586890){
solution.setCounter_fingers(solution.getCounter_fingers() + 4);
solution.setCounter_iterations(solution.getCounter_iterations() + 1);
}
System.out.println(solution.getCounter_iterations());
solution.setCounter_fingers(solution.getCounter_fingers() – 1);
if(solution.getCounter_iterations() % 2 == 0){
System.out.println(“Girl stop count on index finger”);
}else {
System.out.println(“Girl stop count on ring finger”);
}
}
Using Spanish abbreviations for finger names (Haskell):
finger n = ['i', 'p', 'i', 'm', 'a', 'c', 'a', 'm'] !! (nmod8)Here’s Clojure solution with infinity lazy sequence of fingers :)
(defn counting-fingers [n] (let [fingers (cycle [:thumb :index :middle :ring :pinky :ring :middle :index])] (first (drop (dec n) fingers))))(defn counting-fingers [n] ([:thumb :index :middle :ring :pinky :ring :middle :index] (mod (dec n) 8)))Here’s a solution in C.
#include <assert.h> #include <stdio.h> #include <stdlib.h> int main(int argc, char* argv[]) { assert(argc == 2); int x = atoi(argv[1]); assert(x > 0); char finger = "timrprmi"[(x - 1) % 8]; printf("%c\n", finger); return EXIT_SUCCESS; }Usage:
I took a very different approach from you all– my original thought was to just use x%8, but I decided to instead use the absolute value of a triangle wave. At least I thought it was neat.
int finger(int x){ return (int)(fabs(8/M_PI * asin(sin(M_PI*(x-1) / 8)))+0.5); } int main(){ printf("%c\n", "TIMRP"[finger(1000)]); }Here’s a Python solution inspired by @Antonio’s solution, adapted to use integers in place of floating point.
def finger(x): return 'timrp'[abs(8 * ((x + 3) // 8) - x + 1)] for x in range(1, 11): print(x, finger(x)) print(1000, finger(1000))Output:
In my approach, I took advantage of array pointers in PHP. Every time the pointer reaches an outer finger (thumb/pinky), the array elements are reversed in order. You simply scroll down to see the finger that is declared at count number 1000.
<?php // Put your fingers in the array $fingers = [ 'thumb', 'index', 'middle', 'ring', 'pinky', ]; for ($i = 1; $i <= 1000; $i++) { // Declare both the count and finger with each iteration echo $i.". ".current($fingers)."<br>"; // Move the array pointer up one element next($fingers); // If the array pointer is at either the pinky or the thumb if (current($fingers) == 'pinky' || current($fingers) == 'thumb') { // reverse the order of the elements in the array $fingers = array_reverse($fingers); } } ?>