City Grid Logic Puzzle (Hard) Walkthrough

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1.) B4 tells you that A3 and B3 are both (r)oad, as well as another 3 units in Row 3

2a) It’s important to realize that the implication of A3 + F1 is that no (h)ousing can be any farther south than Row 3. You will use this information later.

2b.) B3 tells you that either C3 or D3 is not (r)oad, so according to B4, E3 and F3 are also both (r)oad.

3.) E3 tells you that B4 must be next to a (s)tore. But A3 tells you that neither A4 nor B5 can be a (s)tore. Therefore C4 must be (s)tore.

4.) F3 tells you that neither C5 nor D4 is (r)oad. According to F1 and step 2a above, nothing in Row 5 is (h)ousing. C4 tells you that C5 is not (e)ducational. Therefore, according to C3, C5, which is on the outer edge, must be (w)ater.

5.) According to C5, St. John’s School is not in a corner. This is because any (a)ttraction next to a corner is sure to be on the outer edge, and no (a)ttraction can be on the outer edge according to A3. But C5 also tells you that St. John’s cannot be A4 or B5 because neither of those cells can be next to a skating rink (a)ttraction, since A5 is on the outer edge. C4 tells you that neither A4 nor B5 is road since B4 is already next to a road at B3. And again, according to F1 and step 2a above, nothing in Row 4 or 5 can be (h)ousing. Therefore, both A4 and B5 are (w)ater, according to A3.

6.) St. John’s School must be on the outer edge (A3), it must be next to an (a)ttraction (C5), and that attraction must be next to a (s)tore (E3). So where can St. John’s be?
Corners are out because (a)ttractions cannot be on the outer edge.
A2 and B1 are out because the (a)ttraction would be B2 and the (s)tore would be C2, but B5 says C2 is not a (s)tore.
C1 is out because C2 cannot be an (a)ttraction, according to B5.
D1 is out because D2 would be an (a)ttraction, but then so would E2 according to B5, and if D2 and E2 are both (a)ttractions, they cannot both be next to a (s)tore in order to satisfy E3.
E1 is out for the same basic reason as D1.
F2 is out for the same basic reason ad D1 and E1.
D5 is out according to A4.
This leaves only E5 or F4, both of which would dictate the (a)ttraction at E4.

7.) According to E3 and A3, D4 must be a (s)tore.

7.) If either E5 or F4 is St. John’s School (step 6 above), then F5 is not (w)ater, according to A4. F5 is also not (h)ousing, according to F1 and step 2 above. F5 is also not St. John’s School, since that is either E5 or F4. Therefore, according to A3, F5 is (r)oad.

8.) Dillard’s and Sam’s Club cannot both be next to 0 roads, according to B4. Sam’s Club clearly cannot be next to 2 (r)oads, so according to F5, both Sam’s Club and Dillard’s are next to 1 (r)oad. Sam’s Club’s (r)oad must be C3, and according to B4, Dillard’s’ (r)oad must be D5.

9.) At this point you know that there are no more (a)ttractions. They cannot be on the outer edge (A3) and they must be next to a (s)tore (E3) and C2 is neither (a)ttraction nor (s)tore (B5) and D2 and E2 must be the same (B5) but cannot both be (a)ttractions. Since all the (a)ttractions are known and both are in Row 4, according to F1, D3 must be (h)ousing.

10.) The only 2 cells that are next to each other, both of which can be (s)tores, are D2 and E2. Therefore, according to D3, D2 and E2 are (s)tores.

11.) According to E4, Mach City High School exists. It can only go in B2. Therefore, B2 is (e)ducational.

12.) According to step 6 above, St. John’s School is E5 or F4. The other must be (w)ater – can’t be (h)ousing according to F1 and can’t be (r)oad according to C4. F2 can’t be (e)ducational, so according to B2, F2 must be the same as whichever of E5 or F4 is (w)ater. So F2 is (w)ater.

13.) F2 tells you that nothing in Column E is (h)ousing. D3 tells you that E1 is not (r)oad. Therefore, according to A3, E1 is (w)ater.

14.) All the (s)tores are placed, since the only remaining cells are either outer edge or C2, which is not a (s)tore according to B5. So E1 can only be talking about C2. It is 2 units away from C4 and E2. So C2 is (r)oad.

15.) D2 tells you that D2 is not next to (w)ater, because C4 and E2 already are. D2 is also not next to a second (r)oad according to D5, because C3 is next to 2 (r)oads, a (s)tore, and (h)ousing. Therefore D2 must be (h)ousing.

16.) D1 must be talking about A1. So A2 and B1 are (h)ousing.

17.) There are 8 (r)oad units. According to A2, there will be at least 9. There are 5 (w)ater units, sure to be at least 6, possibly 7. So (r)oad units do outnumber (w)ater units. Therefore, according to B1, E5 is (e)ducational.

18.) According to step 12 above, F4 must be (w)ater.

19.) According to A2, there will be at least 9 (r)oad units. According to F4, there can’t be 10. A5 cannot be housing according to F1, so it must be (w)ater.

20.) According to A2, either A1 or C1 is (r)oad. This means that the “if” statements in both A5 and E5 are false. So if C1 is neither housing nor water, it must be (r)oad.

21.) C1 tells you that A1 must be (h)ousing.

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