Discuss Let's Eat here

[MatthewV]

link to puzzle

[dieded*]

Ahh yes! Solved it already. It's quite easy but very tedious because you have to go through the clues over and over again

[guest*]

The picture of the calendar has the wrong dates for the preceeding week. Considering it's on GL, one can't help but wonder if it's secretly part of some obscure hidden second puzzle within the first one...

[MatthewV]

Yes, there is a very subtle joke inside the picture...
...but it wasn't too well thought out and won't be explained.

[Love Me1*]

Okay im confused someone help PLEEASSSEEE

[Nsof]

I have thought on what general ways are there to solve this puzzle.
One that came to mind was writing a program that iterates all restaurant, street, day lunch/dinner combinations and then verifies each combination against hard coded constraints. If a combination passes all constraints then it’s a solution.
Another was Integer Programming. Bellow is my rather long try at it (not invised because its a model of a solution not a solution).
Last was Graph theory but I couldn’t think of a problem to reduce to. Any ideas? Someone, Anyone?

IP solution
Variables
XijTk1 Binary variable. Intended to mean: XijTk1=1<=>Eliza lunched at Restaurant i (which is on street j) on day k
XijTk2 Binary variable. Intended to mean: XijTk0=1<=>Eliza dined at Restaurant i (which is on street j) on day k
Auxiliary variables
RiTk1 Integer variable > 0. RiTk1>0<=>Eliza lunched at Restaurant i on day k
RiTk2 Integer variable > 0. RiTk2>0<=>Eliza dined at Restaurant i on day k
RiTk Integer variable > 0. RiTk>0<=>Eliza ate at Restaurant i on day k
Where 1=Chick-Empty-A, 2=Long John Bronze's, 3=Waffle Habitat, 4=Dice's Pizza, 5=Taco Marimba, 6=Questionnos, 7=Idoaho Fried Chicken

SjTk1 Integer variable > 0. SjTk1>0<=>Eliza lunched at a restaurant located on j-th street on day k
SjTk2 Integer variable > 0. SjTk2>0<=>Eliza dined at a restaurant located on j-th street on day k
SjTk Integer variable > 0. SjTk1>0<=>Eliza ate at a restaurant located on j-th street on day k

Target function
Maximize X11T11 (doesn't matter are trying to find any solution.)

Constraints
Auxiliary variables constraints
For each i,k
AC1: RiTk1 = Sig(j){XijTk1}
AC2: RiTk2 = Sig(j){XijTk2}
AC3: RiTk = RiTk1 + RiTk2

For each j,k
AC4: SjTk1 = Sig(i){XijTk1}
AC5: SjTk2 = Sig(i){XijTk2}
AC6: SjTk = SjTk1 + SjTk2

Lets check that "RiTk1>0<=>Eliza lunched at Restaurant i on day k" is true:
if RiTk1>0 then by AC1 there exists - at least one - j such that RijTk1=1. But RijTk1=1=>"Eliza lunched at Restaurant i (which is on street j) on day k" so definitely we can conclude the less specific "Eliza lunched at Restaurant i on day k".
if "Eliza lunched at Restaurant i on street j on day k" then RijTk1=1 and therefore by AC1 RiTk1>0

Same for all auxiliary constraints

Problem constraints
P1: Eliza Pseudonym has dined at seven different local restaurants (including Idoaho Fried Chicken) twice each: once for lunch, and once on a different day for dinner.
I'll break this into three - hopefully logically equivalent (needs a proof) - sentences.

P1.1: Seven lunches at seven different restaurants
For each i
Sig(k){RiTk1}=1. (prety simple to show that: Sig(k){RiTk1}=1<=>"Seven lunches at seven different restaurants")

P1.2: Seven dinners at seven different restaurants
For each i
Sig(k){RiTk2}=1. (Same explanation)

P1.3: She didn't eat lunch and dinner at the same restaurant on the same day.
For each i,k
RiTk<=1. (from A3: RiTk1+RiTk2<=1 which can be read as, she either lunched or dined or didn’t eat at restaurant i at day k, but she certainly didn’t do both)

P2: Each restaurant is located on a different street (from 1st through 7th).
if restaurant i is on street j then Eliza will eat there twice. If its not then she cant eat there.
So summing her visits on each combination of i,j over all days can be either 0 or 2.
For each i,j
Sig(k){XijTk1+XijTk2) is in {0,2}
This is not yet a linear equation. In order to make it so we'll define yet another integer - binary in this case - variable RSij and say that the sum should equal to 2*RSij. So
RSij Binary variable. RSij=1<=>restaurant i is on street j
and accompanying constraints
For each i,j
2*RSij = Sig(k){XijK1+XijK2}

Clues constraints
C1: Wednesday was the only day on which Eliza ate at two restaurants on consecutively numbered streets.
This translates into two sentences
C1.1: On Wednesday Eliza ate at two restaurants on consecutively numbered streets
C1.2: Every day other then Wednesday Eliza did not eat at two restaurants on consecutively numbered streets.
Instead of C1.1 i'll use
C1.1': On Wednesday Eliza did not eat at two restaurants on a non consecutively numbered streets.
Which together with the fact that she must each twice on Wednesday (P1.1 and P1.2) is the same as C1.1

C1.1'
For each j,j' such that 1<=j<j+1<j'<=7 (j and j' are non consecutive)
SjT4 + Sj'T4 < 2

C1.2
For each j, k such that 1<=j<=6 and k!=4
SjTk + Sj+1Tk < 2

C2: Eliza had lunch at the restaurant located on the 6th Street on Monday.
S6T21 = 1

C3: Eliza ate at Chick-Empty-A (i=1) on Tuesday and Friday.
R1T3 + R1T6 = 1

C4: Eliza ate at Long John Bronze's (i=2) on two consecutive days
For each k, k' such that 1<=k<k+1<k'<=7 (k and k' are non consecutive)
R2Tk + R2Tk' < 2
Again the above equation together with the fact she has to eat at Long John Bronze's twice ensures that these two times will be at consecutive days.

C5: Eliza ate her lunch at the 1st Street restaurant precisely three days after she ate her dinner at the 5th Street restaurant.
This is equivalent to C5': there are k,k' such that k=k'+3 and S1Tk1+S5Tk'2 = 2.
C5.1:
For each k, k' such that k!=k'+3 (k and k' are not three days apart)
S1Tk1 + S5Tk'2 < 2
C5.2:
Sig(k=1to4){S1Tk+31+S5Tk2} = 2

I'll proof that C5.1+C5.2 <=> C5'.
C5.1 & C5.2 <= C5' is easy
C5.1 & C5.2 => C5'

Lets look at the case where S1T41=1
C5.1 (for the case k=4)=> S1T41 + S5T22 < 2 and since S1T41=1 is must be that S5T22 = 0. Same for S5T32 = S5T42 = S5T52 = S5T62 = S5T72 = 0
She cant have lunched at the same restaurant twice and therefore S1T41 + S1T51 + S1T61 + S1T71 = 1, but since S1T41 = 1 then S1T51 = S1T61 = S1T71 = 0.
C5.2 => S1T41 + S1T51 + S1T61 + S1T71 + S5T12 + S5T22 + S5T32 + S5T42 = 2
But since S1T41 = 1, S1T51 = 0, S1T61 = 0, S1T71 = 0, S5T22 = 0, S5T32 = 0, S5T42 = 0 it must be that S5T12 = 1.
Same for cases S1T51=1, S1T61=1, S1T71=1
Last case is S1T41=S1T51=S1T61=S1T71=0. This cant be because together with C5.2 it will imply that Eliza must have dined at the same restaurant twice.

C6: On Sunday, Eliza did not have the restaurant on 4th Street for lunch, nor did she eat at Waffle Habitat for dinner
C6.1: On Sunday, Eliza did not have the restaurant on 4th Street for lunch
S4T11 = 0
C6.2: On Sunday, Eliza did not eat at Waffle Habitat for dinner (i=3)
R3T12 = 0

C7: On one particular day, Eliza ate at Dice's Pizza (located on an odd numbered street) and Taco Marimba, in some order
C7.1: Dice's Pizza is located on an odd numbered street
Sig(k){X42Tk1+X42Tk2+X44Tk1+X44Tk2+X46Tk1+X46Tk2) = 0
C7.2: On one particular day, Eliza ate at Dice's Pizza and Taco Marimba, in some order
For each k,k',k'',k''' such that k!=k'!=k''!=k'''
R4Tk + R4Tk' + R5Tk'' + R5Tk''' < 4

C8: During the three-day period from Sunday through Tuesday, Eliza had eaten something at all seven restaurants excluding the one located on 7th Street; during the three-day period from Thursday through Saturday, Eliza had eaten something at all seven restaurants excluding Questionnos
C8.1: During the period from Sunday through Tuesday, Eliza had eaten something at all seven restaurants excluding the one located on 7th Street
C8.1.1: Sunday through Tuesday, Eliza had not eaten at the restaurant located on 7th Street
S7T1 + S7T2 + S7T3 = 0
C8.1.2: Sunday through Tuesday, Eliza ate each restaurant no more than once
For each i
RiT1 + RiT2 + RiT3 <= 1
C8.1.3: Sunday through Tuesday, Eliza ate at six restaurants
Sig(i){RiT1 + RiT2 + RiT3} = 6
C8.2: During the period from Thursday through Saturday, Eliza had eaten something at all seven restaurants excluding Questionnos
Similar constraints as C8.1
_________________
Will sell this place for beer

[Samadhi]

_________________
And he lived happily ever after. Except for the dieing at the end and the heartbreak in between.

[Courk]

I just want to see if that yields an answer. I won't pretend to know what it all means.

[Middle Aged Guy]

[Nsof]

[guestt+*]

i dont understand it. I found some of the answers I believe but I just cant seem to figure it out. Can more than one mean go in consecutive days?

[mrs.guest*]

I think solved it but I'm not sure if it is correct. when will the answer be posted?

[MatthewV]

That is hard to say. You could try contacting mathgrant to verify your answer. Or I could try to solve it tonight...

I do have access to an answer. You can email me at pseudonym125 [at] yahoo [dot] com and I'll verify

[AMWhy*]

Well I've solved it.
Where can I send in or post a solution?
Can I just put the solution on here under a spoiler?

[MatthewV]

I won't encourage you to post a solution....but hints aren't bad.

Also, I found out that some teacher somewhere is using this puzzle as an extra credit problem. Putting down a solution might spoil that!

[AMWhy*]

Okay.

Well, the way I solved it was to first make a chart in Excel. Put the 7 days of the week in a row along the top. Below each day, write the name of each of the 7 restaurants. Then below that, the 7 street numbers.
This will be all of the lunch possibilities.

Then below that, do the same again - the 7 restaurant names followed by the 7 street numbers.
This is for all of the dinner possibilities.

You should have something that looks a bit like this for each day:

Sunday
Lunch
Chick-Empty-A
Dices Pizza
Idoaho Fried Chicken
Long John Bronze
Questionnos
Taco Marimba
Waffle Habitat
1
2
3
4
5
6
7
Dinner
Chick-Empty-A
Dices Pizza
Idoaho Fried Chicken
Long John Bronze
Questionnos
Taco Marimba
Waffle Habitat
1
2
3
4
5
6
7


Now, work through each clue 1 by 1 elimintaing all the impossible options (just erase the name of restaurants and streets that can't be there on each day).

For example, to get started, you know from Clue 2 that lunch on Monday is at street 6. Therefore, you can eliminate street 6 from lunch on Sun, Tues, Weds, etc. You can also eliminate street 6 from Monday dinner.

Extra Hint: (spoiler): Using clue 8, we know that every restaurant name is used once between Thurs and Sat, except for Questionnos. We also know that every street is used once between Sun and Tues, except for 7. That mean that both street 7 and Questionnos must appear somewhere on Wednesday.

Further than this, from clue 4, we know that Long John Bronze is eaten at on 2 consecutive days. For this to be possible, one of those days must be Wednesday.:End of spoiler.

[kundan*]

From clue 1 and clue 8 one can easily deduce the name and street of resturant .

then clue 4 tells the name of days

then proceed on clue 2,3 and 5
then use clue 6 and 7.

if it is not still clear then i can provide the answer

[mathgrant*]

[MaartenRoode*]

There is exactly one solution to the Let's Eat Problem.
You can download it (and C# code that verifies it) from here:
http://www.verticle.co.za/Maarten/LetsEatSolution.zip

The meaning of "consecutive" gives some problems. As it turns out, it means what one would hope it does: on Wednesday, either
1) Eliza eats lunch at the restaurant on street N and dinner at the restaurant on street N+1, or
2) she eats lunch at the restaurant on street N, and dinner at the restaurant on street N-1

If one allows only (1) for Wednesday, no solutions are possible.
If one allows (1) and (2) - i.e. effectively allowing only (2) for Wednesday - then exactly one solution is possible should one also forbid the same conditions for all other days. If one applies only condition (1) to other days (i.e. forbidding it), 51 solutions exist.

The included code reveals these other solutions as well.

[Tahnan]

Very clever little logic puzzle, I thought.

[anant*]

the puzzle itself is easy...I dint get the joke though!!

[kundan*]

1st street- Waffle Habitat
2nd street - Taco Marimba
3rd street - Chick-Empty-A
4th street - Idoaho Fried Chicken
5th street - Dice's Pizza
6th street - Questionnos
7th street - Long John Bronze's

day lunch dinner
sunday Taco Marimba(2) Idoaho Fried Chicken(4)
monday Questionnos(6) Waffle Habitat(1)
tuesday Chick-Empty-A(3) Dice's Pizza(5)
wednesday Long John Bronze's(7) Questionnos(6)
thrusday Idoaho Fried Chicken(4) Long John Bronze's(7)
friday Waffle Habitat(1) Chick-Empty-A(3)
saturday Dice's Pizza(5) Taco Marimba(2)

[nick's mum*]

i think it was VERY dissapoitning