Babbage problem: Difference between revisions

<lang racket>;; Text from a semicolon to the end of a line is ignored

;; This lets the racket engine know it is running racket

#lang racket

;; “define” defines a function in the engine

;; we can use an English name for the function

;; a number ends in 269696 when its remainder when

;; divided by 1000000 is 269696 (we omit commas in

;; numbers... they are used for another reason).

(define (ends-in-269696? x)

(= (remainder x 1000000) 269696))

;; we now define another function square-ends-in-269696?

;; actually this is the composition of ends-in-269696? and

;; the squaring function (which is called “sqr” in racket)

(define square-ends-in-269696? (compose ends-in-269696? sqr))

;; a for loop lets us iterate (it’s a long Latin word which

;; Victorians are good at using) over a number range.

;;

;; for/first go through the range and break when it gets to

;; the first true value

;;

;; (in-range a b) produces all of the integers from a (inclusive)

;; to b (exclusive). Because we know that 99736² ends in 269696,

;; we will stop there. The add1 is to make in-range include 99736

;;

;; we define a new variable, so that we can test the verity of

;; our result

(define first-number-that-when-squared-ends-in-269696

(for/first ((i ; “i” will become the ubiquetous looping variable of the future!

(in-range 1 (add1 99736)))

; when returns when only the first one that matches

#:when (square-ends-in-269696? i))

i))

;; display prints values out; newline writes a new line (otherwise everything

;; gets stuck together)

(display first-number-that-when-squared-ends-in-269696)

(newline)

(display (sqr first-number-that-when-squared-ends-in-269696))

(newline)

(newline)

(display (ends-in-269696? (sqr first-number-that-when-squared-ends-in-269696)))

(newline)

(display (square-ends-in-269696? first-number-that-when-squared-ends-in-269696))

(newline)

;; that all seems satisfactory</lang>

<pre>25264

638269696

#t

#t</pre>