How to resolve the algorithm Tupper's self-referential formula step by step in the Nim programming language

Published on 12 May 2024 09:40 PM

How to resolve the algorithm Tupper's self-referential formula step by step in the Nim programming language

Table of Contents

Problem Statement

Jeff Tupper, in his 2001 paper "Reliable Two-Dimensional Graphing Methods for Mathematical Formulae with Two Free Variables", shows a set of methods to graph equations and inequalities with two variables in the cartesian plane. One of the examples of the paper, refers to the inequality:

1 2

<

m o d

(

y 17

2

− 17 ⌊ x ⌋ −

m o d

(

⌊ y ⌋ , 17

)

, 2

)

{\displaystyle {\frac {1}{2}}<\left\lfloor \mathrm {mod} \left(\left\lfloor {\frac {y}{17}}\right\rfloor 2^{-17\lfloor x\rfloor -\mathrm {mod} \left(\lfloor y\rfloor ,17\right)},2\right)\right\rfloor }

That inequality, once plotted in the range 0 ≤ x ≤ 106 and k ≤ y ≤ k + 17 for k = 960, 939, 379, 918, 958, 884, 971, 672, 962, 127, 852, 754, 715, 004, 339, 660, 129, 306, 651, 505, 519, 271, 702, 802, 395, 266, 424, 689, 642, 842, 174, 350, 718, 121, 267, 153, 782, 770, 623, 355, 993, 237, 280, 874, 144, 307, 891, 325, 963, 941, 337, 723, 487, 857, 735, 749, 823, 926, 629, 715, 517, 173, 716, 995, 165, 232, 890, 538, 221, 612, 403, 238, 855, 866, 184, 013, 235, 585, 136, 048, 828, 693, 337, 902, 491, 454, 229, 288, 667, 081, 096, 184, 496, 091, 705, 183, 454, 067, 827, 731, 551, 705, 405, 381, 627, 380, 967, 602, 565, 625, 016, 981, 482, 083, 418, 783, 163, 849, 115, 590, 225, 610, 003, 652, 351, 370, 343, 874, 461, 848, 378, 737, 238, 198, 224, 849, 863, 465, 033, 159, 410, 054, 974, 700, 593, 138, 339, 226, 497, 249, 461, 751, 545, 728, 366, 702, 369, 745, 461, 014, 655, 997, 933, 798, 537, 483, 143, 786, 841, 806, 593, 422, 227, 898, 388, 722, 980, 000, 748, 404, 719 produces a drawing that visually mimics the inequality itself, hence it is called self-referential. Although the inequality is intended to be drawn on the continuum of the cartesian plane, the drawing can be performed iterating over the integer values of both the horizontal and vertical ranges. Make a drawing of the Tupper's formula, either using text, a matrix or creating an image. This task requires arbitrary precision integer operations. If your language does not intrinsically support that, you can use a library. The value of k is an encoding of the bitmap of the image, therefore any 17-width bitmap can be produced, using its associated encoded value as k.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Tupper's self-referential formula step by step in the Nim programming language

Source code in the nim programming language

import std/[algorithm, sugar]
import integers

let k = newInteger("960939379918958884971672962127852754715004339660129306651505519" &
                   "271702802395266424689642842174350718121267153782770623355993237" &
                   "280874144307891325963941337723487857735749823926629715517173716" &
                   "995165232890538221612403238855866184013235585136048828693337902" &
                   "491454229288667081096184496091705183454067827731551705405381627" &
                   "380967602565625016981482083418783163849115590225610003652351370" &
                   "343874461848378737238198224849863465033159410054974700593138339" &
                   "226497249461751545728366702369745461014655997933798537483143786" &
                   "841806593422227898388722980000748404719")

proc tuppersFormula(x, y: Integer): bool =
  ## Return true if point at (x, y) (x and y both start at 0)
  ## is to be drawn black, False otherwise
  result = (k + y) div 17 div 2^(17 * x + y mod 17) mod 2 != 0

let values = collect:
               for y in 0..16:
                 collect:
                   for x in 0..105:
                     tuppersFormula(x, y)

let f = open("tupper.txt", fmWrite)
for row in values:
  for value in reversed(row):   # x = 0 starts at the left so reverse the whole row.
    f.write if value: "\u2588" else: " "
  f.write '\n'
f.close()


  

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