Countdown puzzles

sean d · Post by **sean d** » Tue Apr 23, 2013 1:33 pm

A couple of Countdown related things I've been pondering....

Is it possible to have a legitimate letters selection where the max is a 2 (or 1) letter word? Perhaps something like...
U U U U X J Q Z K

And we've all seen people going going way above 1,000 during a numbers solution, but what's the highest intermediate number you can get to and still get back to a legitimate target. (ie the classic 937.5 solution brings you up to the vicinity of 93,750, but how high can you go and still get back to a legitimate target under 1,000.)

Graeme Cole · Post by **Graeme Cole** » Tue Apr 23, 2013 2:23 pm

sean d wrote:A couple of Countdown related things I've been pondering....

Is it possible to have a legitimate letters selection where the max is a 2 (or 1) letter word? Perhaps something like...
U U U U X J Q Z K

In theory you could have something like UUUUUQKZV, a selection which contains no valid words at all. It's very, very unlikely though.

Graeme Cole · Post by **Graeme Cole** » Tue Apr 23, 2013 2:39 pm

sean d wrote:And we've all seen people going going way above 1,000 during a numbers solution, but what's the highest intermediate number you can get to and still get back to a legitimate target. (ie the classic 937.5 solution being you up to the vicinity of 93,750, but how high can you go and still get back to a legitimate target under 1,000.)

You can get up to 750,000, though maybe it's possible to get higher:

100 75 50 25 10 10 -> 600
100*75*10*10 = 750000
750000/(50*25) = 600

sean d · Post by **sean d** » Tue Apr 23, 2013 3:06 pm

Good man, I was wondering if this should just go in the Ask Grame thread! I suspected there might be a theoretical 'impossble' letters round. In that case would an indvidual letter count as 1-letter word?

750,000 was the highest I had got, i.e.
100 75 50 25 10 8 -> 750
(75*50*25*8) = 750,000
750,000/(100*10) = 750
That's the one I'd like to give on the show!

James Roper · Post by **James Roper** » Tue Apr 23, 2013 3:09 pm

sean d wrote:A couple of Countdown related things I've been pondering....

Is it possible to have a legitimate letters selection where the max is a 2 (or 1) letter word? Perhaps something like...
U U U U X J Q Z K

There are a few selections which yield a 2 max, and 1 or 2 that may yield 0 max, but a 1 max is impossible. A is the only valid 1 but because it stems so well as far as I know it'd be impossible to leave it as a darren.

Andy Platt · Post by **Andy Platt** » Tue Apr 23, 2013 3:25 pm

I have no idea how to prove it, but I'm pretty sure that the 750k examples are the maximum possible.
What I'd love to know is what is the highest possible whereby it's actually a necessity to go that high (rather than a flashy alternative).
No idea where to start with it!

Thomas Carey · Post by **Thomas Carey** » Tue Apr 23, 2013 4:14 pm

Andy Platt wrote:I have no idea how to prove it, but I'm pretty sure that the 750k examples are the maximum possible.
What I'd love to know is what is the highest possible whereby it's actually a necessity to go that high (rather than a flashy alternative).
No idea where to start with it!

http://www.c4countdown.co.uk/viewtopic.php?f=18&t=6321

Gavin Chipper · Post by **Gavin Chipper** » Tue Apr 23, 2013 8:02 pm

Andy Platt wrote:I have no idea how to prove it, but I'm pretty sure that the 750k examples are the maximum possible.
What I'd love to know is what is the highest possible whereby it's actually a necessity to go that high (rather than a flashy alternative).
No idea where to start with it!

I think with those 750k examples there you can do the multiplication/division in a different order to avoid going so high, so it might depend on what you consider to be "legitimate". I'd certainly consider it to be legitimate if you go high, but there's a completely different solution where you don't - completely different meaning not just a reordering.

Bob De Caux · Post by **Bob De Caux** » Wed Apr 24, 2013 12:15 am

Applying the Gev rule, I'm pretty sure the best you can do is 99,900 as follows:

(((50+3)*25+7)*75)/100 = 999

Anything higher can be rearranged, I think. Of course, there are other ways to get 999 from those numbers though...

sean d · Post by **sean d** » Fri Apr 26, 2013 9:32 am

Bob De Caux wrote:Applying the Gev rule, I'm pretty sure the best you can do is 99,900 as follows:

(((50+3)*25+7)*75)/100 = 999

Anything higher can be rearranged, I think. Of course, there are other ways to get 999 from those numbers though...

Wow... I'd just about got my head around getting to 937.5, but this is a whole new level of insanity. Can anyone explain the workings behind the Gev rule?

Gavin Chipper · Post by **Gavin Chipper** » Fri Apr 26, 2013 11:04 am

sean d wrote:
Bob De Caux wrote:Applying the Gev rule, I'm pretty sure the best you can do is 99,900 as follows:

(((50+3)*25+7)*75)/100 = 999

Anything higher can be rearranged, I think. Of course, there are other ways to get 999 from those numbers though...
Wow... I'd just about got my head around getting to 937.5, but this is a whole new level of insanity. Can anyone explain the workings behind the Gev rule?

The Gev rule is just about not going higher than you need to with the order of multiplication. This is the Bob de Caux rule.

Bob De Caux · Post by **Bob De Caux** » Fri Apr 26, 2013 11:49 am

Haha nice, always wanted my own rule. However I still it's the same as the Gev rule. Can you show me an example where you can go higher and the multiplication/division can't be rearranged?

Bob De Caux · Post by **Bob De Caux** » Fri Apr 26, 2013 12:11 pm

Right, after testing a few cases using the answer above, I think the answer to Andy's question is:

100,75,50,25,3,3 Target: 996

You have to go up to 99,600 to solve it (unique solution) and no rearrangement is possible.

Gavin Chipper · Post by **Gavin Chipper** » Fri Apr 26, 2013 7:58 pm

Bob De Caux wrote:Haha nice, always wanted my own rule. However I still it's the same as the Gev rule. Can you show me an example where you can go higher and the multiplication/division can't be rearranged?

I thought Sean was referring to the rule that you'd used to solve it (like the 937.5 rule) rather than the rule about not going higher than you need to. I'd forgotten that you'd used the same term a couple of posts back.

Dave Preece · Post by **Dave Preece** » Fri Jul 26, 2013 7:12 pm

What's the 937.5 rule please?

James Robinson · Post by **James Robinson** » Fri Jul 26, 2013 9:52 pm

Dave Preece wrote:What's the 937.5 rule please?

(75 x 25) / (100 / 50) = 937.5

This is where the 4 large experts come in now........

Dave Preece · Post by **Dave Preece** » Fri Jul 26, 2013 10:55 pm

Sorry to sound thick, or, more accurately, not-in-the-know, but can you explain in lay terms please?

Guy Barry · Post by **Guy Barry** » Sat Jul 27, 2013 7:23 am

Clearly it comes in useful if you ever get a target of 937.5

I'm just guessing, but I presume the idea is that if you're given a target close to that number, and you've got four large numbers, and one of the small numbers is odd, then you can adjust the 937.5 value appropriately. For example you could solve 25 50 75 100 3 4 -> 943 by doing (((75 x 25) + 3)/(100/50)) + 4.

Seems to be of limited usefulness though - what happens if both small numbers are even? Fractions aren't allowed in the calculation, so a solution like ((75 x 25) / (100 / 50)) + (6/4) = 939 wouldn't be allowed.

Graeme Cole · Post by **Graeme Cole** » Sat Jul 27, 2013 10:06 am

75*25/(100/50) = 937.5 is just one of the three ways you can use the rule.

How I think of it is this:

(75*25*50)/100 = 937.5

The basic principle is that any change to the numerator of that division causes 1/100 of that change in the result. For example, if you can add 50 to the numerator, the result increases by 0.5. When you do the 75*25*50 bit, you multiply two of those numbers together, add or subtract a small number, then multiply by the remaining large number. Which large number you choose depends on the target and what numbers you have available.

For example, suppose you wanted to make 939. That's 1.5 more than 937.5, which means you somehow have to add 150 before you divide by 100.

If you have a 2, you can use the 75:
((50*25+2)*75)/100 = 939

If you have a 3, you can use the 50:
((75*25+3)*50)/100 = 939

If you have a 6, you can use the 25:
((75*50+6)*25)/100 = 939

That's all you really need to know to understand how it works, the rest is practice in spotting quickly which numbers it's worth using.

The 25 and 50 cases above can be further simplified. In the 50 case, instead of doing 75*25 = 1875, 1875 + 3 = 1878, 1878 * 50 = 93900, 93900 / 100 = 939, you can get to 1878 and divide it by (100/50). For 25 you can divide by (100/25) rather than multiplying by 25 then dividing by 100. With 75 there's no such simplification as you can't make 100/75.

Dave Preece · Post by **Dave Preece** » Sat Jul 27, 2013 5:09 pm

My head hurts!

Guy Barry · Post by **Guy Barry** » Sat Jul 27, 2013 5:29 pm

Interesting. It still looks to me as though you're limited to adding or subtracting certain numbers - only odd numbers work with 50, and only 2, 6 and 10 work with 25 and 75. What about 4 and 8?

Hugh Binnie · Post by **Hugh Binnie** » Sun Jul 28, 2013 5:08 pm

Guy Barry wrote:Interesting. It still looks to me as though you're limited to adding or subtracting certain numbers - only odd numbers work with 50, and only 2, 6 and 10 work with 25 and 75. What about 4 and 8?

4 and 8 don't work with anything -- x25 = 100 or 200; x50 = 200 or 400; x75 = 300 or 600. So you'd always be adding on a whole number after dividing by 100, leaving you with 9xx.5, unfortunately.

Andy Platt · Post by **Andy Platt** » Wed Jul 31, 2013 12:03 pm

Even numbers that are not divisible by 4 are the useful ones for that trick.

I was gutted not to get a 937.5 solve on TV. If that target in R14 of the final was 969 instead, I could have had (25 x 75 + 7 x 9) x 50 / 100 or some shit. How good would that have been

David Barnard · Post by **David Barnard** » Sat Aug 17, 2013 9:33 pm

I actually messed up a 937.5 on tv which is even more gutting

sean d · Post by **sean d** » Mon Aug 19, 2013 10:14 am

One for the 4L specialists....
I believe there is only one 4L target under 130 which cannot be solved if a certain 2 small numbers are drawn. What are the 2 small numbers and the unobtainable target?

Andy Platt · Post by **Andy Platt** » Mon Aug 19, 2013 2:36 pm

Awesome question mate.

I've just tried a quick run-through with 100 75 50 25 1 1 and I can't get a target of 109 or 118, but I can get the rest.
I'm frustrated here, will keep working on it

Edit: OK I cheated and now know the answer is 109, it was a bit dumb that I saw the 116 solution, 100 + (25-1)*50/75, but couldn't see the 118 solution, 100 + (25+1+1)*50/75

sean d · Post by **sean d** » Tue Aug 20, 2013 11:49 pm

Gg wp AP. 1, 1 ... 109 is what I calculated. I programmed it in a idle hour in work a couple of years ago. I think any target under 200 is obtainable as long as you avoid two 1s, and its amazing just how many targets you can hit from 4L, 1, 1.

Graeme Cole · Post by **Graeme Cole** » Wed Aug 21, 2013 10:37 am

sean d wrote:Gg wp AP. 1, 1 ... 109 is what I calculated. I programmed it in a idle hour in work a couple of years ago. I think any target under 200 is obtainable as long as you avoid two 1s, and its amazing just how many targets you can hit from 4L, 1, 1.

Of the 899 possible targets, 324 are solvable with that selection. For 434 targets the best is 1-5 away, for 115 targets the best is 6-10 away, and for 26 targets it's not possible to get within 10.

There is only one multiple of 25 (out of the valid targets) that is not solvable exactly from 4L 1 1. Which one?

Andy Platt · Post by **Andy Platt** » Wed Aug 21, 2013 11:57 am

850 or 875, I can't remember which, and am struggling to work it out.

Gavin Chipper · Post by **Gavin Chipper** » Wed Aug 21, 2013 5:08 pm

What's the best 4-large selection in terms of the most solvable targets? (I don't know the answer)

Graeme Cole · Post by **Graeme Cole** » Wed Aug 21, 2013 5:54 pm

Andy Platt wrote:850 or 875, I can't remember which, and am struggling to work it out.

Yep, one of those. The other one has only one solution, and its difficulty is 89% according to crosswordtools.

Andy Platt · Post by **Andy Platt** » Wed Aug 21, 2013 6:30 pm

Gavin Chipper wrote:What's the best 4-large selection in terms of the most solvable targets? (I don't know the answer)

Another awesome question. Probably something like 7 and 8 as your smalls, Sean/Graeme can you work it out?

sean d · Post by **sean d** » Thu Aug 22, 2013 11:07 am

850 = (25-1) * 75/100 = 18; (18-1)*50
I've no idea now how I did the calculations before! I'd guess 3,4 would give most flexibility as the 'best' selection, possibly even 2,3.

Matthew Brockwell · Post by **Matthew Brockwell** » Fri Aug 23, 2013 4:01 pm

Gavin Chipper wrote:What's the best 4-large selection in terms of the most solvable targets? (I don't know the answer)

Charlie carried out some useful analysis of 4 large a while back here, which shows that 8 and 9 are the best pair, and answers many other questions you may have.

sean d · Post by **sean d** » Fri Aug 23, 2013 4:45 pm

Cool, thanks Matthew. May ressurect that thread some time soon

Gavin Chipper · Post by **Gavin Chipper** » Fri Aug 23, 2013 5:14 pm

Matthew Brockwell wrote:
Gavin Chipper wrote:What's the best 4-large selection in terms of the most solvable targets? (I don't know the answer)
Charlie carried out some useful analysis of 4 large a while back here, which shows that 8 and 9 are the best pair, and answers many other questions you may have.

Excellent, I can't believe I forgot about that thread.

Charlie Reams · Post by **Charlie Reams** » Fri Aug 23, 2013 7:12 pm

Gavin Chipper wrote:
Matthew Brockwell wrote:
Gavin Chipper wrote:What's the best 4-large selection in terms of the most solvable targets? (I don't know the answer)
Charlie carried out some useful analysis of 4 large a while back here, which shows that 8 and 9 are the best pair, and answers many other questions you may have.
Excellent, I can't believe I forgot about that thread.

I had also completely forgotten about it! Good remembering. Lots of nice facts in that thread.

Martin Long · Post by **Martin Long** » Fri Aug 23, 2013 8:19 pm

Matthew Brockwell wrote:
Gavin Chipper wrote:What's the best 4-large selection in terms of the most solvable targets? (I don't know the answer)
Charlie carried out some useful analysis of 4 large a while back here, which shows that 8 and 9 are the best pair, and answers many other questions you may have.

That's an awesome thread right there.

Andy Platt · Post by **Andy Platt** » Fri Aug 23, 2013 8:55 pm

Interesting that my guess of 7 and 8 as the "easiest to solve" pair doesn't even make the top 10, but 8 and 9 is number 1.

Also I nearly mentioned 8 and 3, I've genuinely noticed that there was often some really easy numbers games when those two are the small numbers.

If you remember round 14 of the S68 final, the target was 967 (one of the highest results for most likely to be impossible) and the smalls were 7 and 9 (most likely to be solvable/soluble/what is the best thing to write?) so that is an interesting little stat.