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Hi Bobby;
Having first encountered PDs here on MIF and only done a few, I don't know all that much about them, but it seems that small differences in wording could lead to big differences in solution strategies...which might be difficult to cater for in one solve-all algorithm.
I can't think of anything that would work, and googling dug up nothing that helped, but here's something from the 2009 "Another progressive dinner" thread:
Bobby: could this be automated? "Progressive Dinner Tool" ... ?
bobbym wrote:Hi MathsisFun;
I can't say at this time. Although it took a lot of computer help, it also took a lot of human trial and error. In other words I don't yet have an algorithm. I am going to work on her other problem and maybe see if it can be automated.
Hi Bobby;
No, I don't, sorry. I just drew it up by hand in Excel.
Option A was very straightforward, of course, but B took a few stabs at it before I got a system that worked (by adjusting A)...and that system can be used as a springboard for deriving more grid combinations.
Hi Eulero;
Hi Calabria;
For a real-life scenario I can offer the following two options:
It meets the brief in post #1 (as I understand it), and suits group placement: eg, for families with young children, so that they may sit together.
This one has a constraint that no guest may sit with the same guest at more than one table, and suits maximum guest mixing.
It can also be used for group placement (though to a much lesser degree than A's) through name/number assignment: eg, the 3-guest group {1,2,131} places a different combination of two of them at one table for each course.
I've assumed that all tables serve courses 1 to 3, in that order.
With parallel lines (AB and CD in this case), a line that is perpendicular to one (eg, as a tangent radius) is perpendicular to both.
Hi dazzle1230;
How do you know that AB=LM without using geogebra?
AB & LM are opposite sides of rectangle ABML (from post #24):
Proof that H is CD's midpoint:
(i) ABML is a rectangle (tangent perpendicular to radius theorem [see 'Tangent Angle' at bottom of page here] for points A & B; AB is parallel to CD [see post #1]; AE extends to L)
Hi;
For the original puzzle:
row B = 109
row D = 107
row F = 110col 2 = 104
col 4 = 115
col 6 = 107A7 = 5
G7 = 3That should narrow it down to one solution...
Those constraints still yield multiple solutions, but I don't know how many there are in total.
Here are twelve of them:
I could've saved a lot of time if I'd thought of doing the same!
True!
A similar program in BASIC:
x=1
y=2
d=1
ppd=0
WHILE d<13
ppd=ppd+x
x=x+y
y=y+1
d=d+1
WEND
PRINT ppd
And...welcome to the forum!
Hi Bobby;
Did you deliberately misconstrue the intended meaning of my unintentionally ambiguous statement? If so, I think we're on the same team!
I meant overlook as in fail to consider.
I don't know enough M to go the way of proper structure most of the time, so when it occurs to me to use Flatten, I do.
'Flatten' is something I often overlook until I've exhausted all other ways of successfully performing an operation on nested elements.
Ah...I see your problem! You need them!!
I'm expert at initiating 'first time' events, like:
- asking a whitegoods service technician why my machine is making an xyz noise: "Sorry, sir, but we've never heard of our machines making an xyz noise before." (maybe I should practise up on making xyz noises over the phone, or send them an mp3 recording of said noise);
- asking Customer Service if there's a fix for this or that: "Oh, we've never been asked about this or that before. Hang on a minute". That's when you discover that your understanding of the length of time a minute takes is way off the mark!
You have your specs on inside out!
Oh, I see...the word in those posts, not my search word. I sent MIF all the facts that I remembered...
It certainly looks the part, and using it probably leads people into thinking that I know what I'm doing!
Just stumbled across that. The idea popped into my head that there might be a symbol for product as there is with sigma for summation, and so I started looking.
The search word was incircle, which gave 25 results.
I was interested in 6 of them, 4 of which (years 2015 & 2016) opened up fine; but 2 (or was it 1? - I don't recall now) of the older ones (year 2014) opened up the 'forbidden access' page.
I learnt the hard way. I lost my right arm after the xth click and had to ask MIF to sew it back on again.
My symbolic version of the factorial method has about 15 fewer keypresses than yours.
Yes, I knew about censored words, but this one threw me because all I did was click on a search result.
But I clicked it several times, just in case there was some hiccough. If I'd only clicked once (or twice? or thrice?) I may not have been locked out.
That's a neat way of doing it!
I'd wanted to make a table and do a Times@@ on it, but couldn't get the table working.
I'm deeply honoured to have joined your club - unintentionally!
Hi Bobby;
Did you look at the M code in my previous post? I added the 'DivisorSigma' function to the list, having come across it in M's 'Suggestions Bar' that pops up when you click somewhere on the output screen. Would've saved me the time working out how to sum divisors...but I don't really mind learning something new.
This is a popular problem!
Other posts that may help:
- thread opened by Enshrouded_, 23rd Sep 2015
- thread opened by championmathgirl, 9th June 2015
- thread opened by SPARKS_CHAN, 30th Jan 2015
I haven't checked out the solutions, but Bob Bundy posted there and he always gives good advice.
Warning! Be careful with threads on this topic earlier than the three above (there are also two threads from 2014). I got locked out from the forum overnight after clicking multiple times on a 2014 post that opened up a 'forbidden access' page. MIF knows about it now, and unlocked me. If you do get a 'forbidden access' page after clicking on a post, don't click that post again.