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orientfan

25th November 2021, 18:28
I’m really struggling with the nets. I’m using the 11 2s and the 12 5s to orient myself - is that how others cracked it? If so, could some kind soul possibly tell me which 5 isn’t used? Thanks…
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orientfan

25th November 2021, 19:22
I’m really struggling with the nets. I’m using the 11 2s and the 12 5s to orient myself - is that how others cracked it? If so, could some kind soul possibly tell me which 5 isn’t used? Thanks…
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muraria

25th November 2021, 20:28
Hi orientfan

Wonder if you have some numbers wrong?
You should have 11 2s, as you say, and 11 4s, so all those cells should be used.
I have 6 spare 1s, 3 spare 3s, 2 spare 5s and 3 spare 6s, all around the perimeter of the grid.
The 11 nets interlock, the two-column one is in the NE (top cell is row1, column7) abutting the T net which is side-on

Hope this helps
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orientfan

25th November 2021, 20:37
Thanks a million muraria. Before I saw your post I had indeed just spotted that I had one digit wrong. Your summation of the spare digits confirmed my solution. I’m no fan of numericals but I have to admire the cleverness of this construction. Kept me at it till Thursday…
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muraria

25th November 2021, 21:19
No worries, OF, glad to help and agree completely about the construction of the puzzle, amazing!
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essay

26th November 2021, 01:07
Hi Moffat, I’ve ‘proved’ it can’t be done in fewer than four colours. I converted each dice net into an ‘island’ and had lines representing bridges between islands. There was, unfortunately, an island surrounded by, and connected to, an odd number (5) of other islands, all of which were connected in a circle around the central island. The result was like a spider’s web with 5 spokes. Clearly the central island would have to be a different colour from the surrounding islands. The surrounding islands would have to alternate two colours around the outside circle in order to attempt to limit the total required colours to three. Alas because five is an odd number, we would get (e.g.) red, green, red, green, red, which if connected in circular fashion would result in two reds next to each other (I’m assuming the central island is (e.g.) blue). Hence the need for a fourth colour such that the outside ring is red, green, red, green, yellow. Then yellow is adjacent to the first red. Sorry it’s a bit long winded. Hope it makes sense!
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simplesimon

26th November 2021, 09:52
Essay, sorry to have to tell you but the 4 colour problem was solved over 40 years ago. However, it is satisfying to prove it for oneself. Top tip - don’t try it for Fermat’s Last Theorem !
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essay

26th November 2021, 11:58
Thx simplesimon - very amusing, but if you are able to follow the argument (maybe with a spider web diagram with odd and even number of spokes) you’ll see how intuitively obvious it is - some forum members may also go nd it interesting.
Also the 4-colour problem was the far more complicated proof that NO MORE THAN 4 colours are ever needed. The question asked was whether ANY MORE THAN THREE ARE NEEDED IN THIS PARTICULAR CASE. The ‘proof’ has probably been around far longer than the proof of the 4-colour theorem itself.
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jif73

29th November 2021, 10:56
Desperately trying to complete this in time to submit an entry, after trying to be clever and find dice nets and thus the correct possibility to solve clues. Big mistake!
No-one seems to have had a problem with 16ac/31ac so maybe my logic is wrong. I assumed that 16ac's triangular number, 31ac's square and the sum of 16ac and 31ac were all the same "magic" number. Even if this is correct, I have too many combinations of 16 and 31…
Can some kind soul let me know if my logic is correct and or give me hint/answer to 16 and/or 32?
Also is the first digit of 32dn the higher or lower of the two possible answers and is digit 4 of 35ac twice digit 2?
Any help gratefully received!
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numbskull

29th November 2021, 14:15
jif73, you are on the the right track. I have the square in 31a as the square of 32d.
The 4th digit of 35a is twice the second
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