Enigmatic Code
Programming Enigma Puzzles
Enigma 1427: Equal lengths
Posted by on 26 May 2013
From New Scientist #2588, 27th January 2007
I challenged Harry and Tom to draw this diagram, with the angles ABD, BEC and CED each 90° and each side of the three triangles an integral number (less than 40) of centimetres long.
They each managed to do this.
In Harry’s version (AB+BC) was the same length as (AD+DC); in Tom’s version (AB+BE) was the same length as (AD+DE); CE was the same length in both versions.
What was the length of the perimeter of the quadrilateral ABCD in (a) Harry’s version, (b) Tom’s version?
[enigma1427]
Enigmatic Code 
This Python program examines all possible combinations of Pythagorean triangles with hypotenuse less than 40 to find the answer. It runs in 51ms.
from collections import defaultdict from itertools import product from enigma import irange, printf # find pythagorean triples with a hypotenuse less than 40 ts = list() for c in irange(1, 39): for b in irange(1, c - 1): for a in irange(1, b - 1): if a * a + b * b == c * c: ts.append((a, b, c)) # now find pairs of triangles with a matching non-hypotenuse side (Hs, Ts) = (list(), list()) for (X, Y) in product(ts, repeat=2): # match the sides for (x, y) in product((0, 1), repeat=2): if X[x] != Y[y]: continue # sum the non-matching non-hypotenuse sides s = X[x ^ 1] + Y[y ^ 1] # find a matching Z for Z in ts: for z in (0, 1): if Z[z] != s: continue # check the conditions (AB, BC, BE, CE, AD, DC, DE) = (Z[z ^ 1], Y[2], Y[y ^ 1], X[x], Z[2], X[2], X[x ^ 1]) p = AB + BC + DC + AD # for Harry: if (AB + BC == AD + DC): Hs.append((CE, p, (X, Y, Z))) printf("[H: {X} {Y} {Z} / {x} {y} {z} / CE={CE} p={p}]") # for Tom: AB + BE = AD + DE if (AB + BE == AD + DE): Ts.append((CE, p, (X, Y, Z))) printf("[T: {X} {Y} {Z} / {x} {y} {z} / CE={CE} p={p}]") # choose H and T for (H, T) in product(Hs, Ts): # CE is the same if H[0] != T[0]: continue printf("pH={H[1]} pT={T[1]} [H={H[2]} T={T[2]}]")Solution: (a) 90 cm; (b) 76 cm.
In my machine which has 2.6 Ghz processor, the execution time is 16ms.
import time time1=int(round(time.time()*1000)) print(time1) Hip=[5,10,15,20,25,30,35,13,26,39,17,34,29,37]# These are the biggest side of the triangles import itertools as permt permus=list(permt.permutations(Hip,3)) def FindTwoCircumferences(): c,d,counter=0,0,0 satisfier=[[0 for i in range(2)]for i in range(20)] circumference=dict() #print(satisfier) for i in range(len(permus)): e,b,f=permus[i][1],permus[i][0],permus[i][2] a=e+f-b if a in range(40):#Every side must be less than 40." root=pow(pow(e,2)-pow(a,2),0.5).real if int(root)==root: cminusd=(b-f)*(b+f)//root#this equation comes from c^2+h^2=b^2 and d^2+h^2=f^2 c=(root+cminusd)//2# root denotes for c+d and c can be found easly by solving the eq. d=root-c h1,h2=pow(pow(b,2)-pow(c,2),0.5).real,pow(pow(f,2)-pow(d,2),0.5).real if int(h1)==h1 and h1>0 and h1==h2 : #print(a+b+e+f,a,b,e,f,h1) for j in range(len(Hip)): hip=Hip[j] if h1<hip: diff=pow(hip,2)-pow(h1,2) c1=pow(diff,0.5).real if int(c1)==c1: satisfier[counter][0],satisfier[counter][1]=c1,h1 circumference.update({h1:a+b+e+f}) counter+=1 #This part is for the other circumference for i in range(counter): for j in range(counter): c,d=satisfier[i][0],satisfier[j][0] if c>d and satisfier[i][1]==satisfier[j][1]: aminuse=d-c for k in range(len(Hip)): cplusd=c+d hip=Hip[k] if cplusd<hip: a=aminuse+hip #print(a,c1,d1,hip,cplusd,satisfier[j][1]) if (pow(hip,2)-pow(cplusd,2))==pow(a,2): h=satisfier[j][1] hsquare=pow(h,2) b,f=pow(pow(c,2)+hsquare,0.5).real,pow(pow(d,2)+hsquare,0.5).real print("CIRC1:",a+b+e+f,"CIRC2:",circumference[h]) FindTwoCircumferences() time2=int(round(time.time()*1000)) print(time2) print("s",time2-time1)Triangles BEC and DEC are the same for Tom and Harry. Only triangle ABD is different between them. To get Tom’s solution I had to comment out Harry’s constraint and output line – vice versa to get Harry’s solutions, for which the code is relevant.
Only Tom’s perimeter of 76 cm and Harry’s perimeter of 90 cm have the same value of EC (or CE)
Using Google’s Python interface to their Google Optimization Tools (or Jim’s MiniZinc wrapper), it is straightforward to produce a full constraint programming solution to this puzzle.
from ortools.constraint_solver import pywrapcp from collections import defaultdict # 1. Set up the solver solver = pywrapcp.Solver('Sunday Times Teaser 2875') # 2. Create the model variables AB, BD, AD, BC, CD, BE, ED, EC = vars = [solver.IntVar(1, 40, x) for x in 'AB BD AD BC CD BE ED EC'.split()] ABCD = solver.IntVar(4, 160, 'ABCD') all_vars = vars + [ABCD] # 3. Set up the constraints solver.Add(BE + ED == BD) solver.Add(AB * AB + BD * BD == AD * AD) solver.Add(BE * BE + EC * EC == BC * BC) solver.Add(ED * ED + EC * EC == CD * CD) solver.Add(ABCD == AB + BC + CD + AD) # index solutions on the length of EC ec_to_sol = defaultdict(list) # 4. Initialise the solver solution = solver.Assignment() solution.Add(all_vars) solver.NewSearch(solver.Phase(all_vars, solver.CHOOSE_FIRST_UNBOUND, solver.ASSIGN_MIN_VALUE)) # 5. Find possible solutions while solver.NextSolution(): ab, bd, ad, bc, cd, be, ed, ec = [x.Value() for x in (AB, BD, AD, BC, CD, BE, ED, EC)] # record solutions for Harry if ab + bc == ad + cd: ec_to_sol[EC.Value()].append(('H', ABCD.Value())) # record solutions for Tom if ab + be == ad + ed: ec_to_sol[EC.Value()].append(('T', ABCD.Value())) solver.EndSearch() # look for solutions for both Harry and Tom with the same EC length for ce, sl in ec_to_sol.items(): if len(sl) > 1: h, t = ([y for x, y in sl if x == c] for c in 'HT') if len(h) == len(t) == 1: print(f'(a) {h[0]}cm (b) {t[0]}cm.') else: print(f'Multiple solutions: (a) {h}cm (b) {t}cm.')A blast from the past!
Here’s a MiniZinc model that generates possibilities for Harry’s and Tom’s triangles as output. It runs in 67ms (using the [[
mzn-gecode -a]] solver).In fact there is only one possible triangle for Tom. There are two possible triangles for Harry, but only one of them has the same CE length as Tom’s triangle.
The following Python program executes the above MiniZinc code and analyses the output to find matching possibilities for Tom and Harry. It uses the minizinc.py wrapper library (which, of course, I hadn’t written in 2013 when this puzzle was posted to the site). The overall execution time is 132ms.
from collections import defaultdict from enigma import printf from minizinc import MiniZinc; # load the minizinc model for possible diagrams p = MiniZinc("enigma1427.mzn") # record perimters by CE for Harry and Tom Hs = defaultdict(list) Ts = defaultdict(list) # record results n = 0 for (AB, AD, BC, BE, CD, CE, DE) in p.solve(result="AB AD BC BE CD CE DE"): if AB + BC == AD + CD: # Harry's diagram printf("Harry: CE={CE}, AB={AB} BC={BC}, AD={AD} CD={CD}") Hs[CE].append(AB + BC + CD + AD) if AB + BE == AD + DE: # Tom's diagram printf("Tom: CE={CE}, AB={AB} BE={BE}, AD={AD} DE={DE}") Ts[CE].append(AB + BC + CD + AD) n += 1 printf("[considered {n} diagrams]") # find common CE values for CE in set(Hs.keys()).intersection(Ts.keys()): # output perimiters for Harry and Tom printf("CE={CE}: Harry={H}, Tom={T}", H=Hs[CE], T=Ts[CE])We can remove line 21 from the MiniZinc model to generate all possible diagrams (there are 45, the same number that my original program considers). The Python program only selects those that are applicable to Tom and Harry anyway, and the overall run time is the same.