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Re: For 2014?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg132152] Re: For 2014?
  • From: Ulrich Arndt <ulrich.arndt at data2knowledge.de>
  • Date: Fri, 3 Jan 2014 04:43:47 -0500 (EST)
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  • References: <20131224071749.C3F1A69D5@smc.vnet.net> <3688DC0E-CE14-47F1-B93B-5BFA875C59E3@data2knowledge.de> <20131226112449.4994F6A09@smc.vnet.net> <644455E9-F8F1-491C-9EFE-1198A90B8FED@wolfram.com>

Thanks for sharing. Was obviously not aware of this.
Here a tweet conform code e.g. 139 chars - output could be nicer but 140 is a strong limitation given the long function names...


l=Range[9];r=StringJoin[Riffle[ToString/@l,#]]&/@Tuples[{"+","-","*","/",""},8];Grid[{#,Extract[r,Position[ToExpression/@r,#]]}&/@(l+2010)]

Ulrich

Am 28.12.2013 um 23:45 schrieb carlson at wolfram.com:

> I wrote a blog post about this a while ago which has a downloadable notebook with code.  There's also a reference to a post by Hans Havermann on the same topic.
>
> 	http://blog.wolfram.com/2012/02/02/happy-109876-54321/
>
> Chris
>
>
> On Dec 26, 2013, at 5:24 AM, Ulrich Arndt <ulrich.arndt at data2knowledge.de> wrote:
>
>> Actually it has to be SeedRandom[1992] - was late ;-).
>>
>> But also an complete check is possible - and therefore much better...
>>
>> char = CharacterRange["1", "9"]
>> type = {"+", "-", "*", "/", ""}
>>
>> t = Tuples[type, 8];
>> r = StringJoin[Riffle[char, #]] & /@ t;
>> e = ToExpression[#] & /@ r;
>> p = Position[e, 2014]
>> Extract[r, p]
>>
>> Works also for reversed number list.
>>
>> Ulrich
>>
>> Am 24.12.2013 um 22:05 schrieb Ulrich Arndt:
>>
>>> 123 + 45*6*7 - 8 + 9
>>>
>>> Generate is maybe a bit wrong - search ;-)
>>>
>>> char = CharacterRange["1", "9"]
>>> type = {"+", "-", "*", "/", ""}
>>> RandomSeed[1992]
>>> r = Table[StringJoin[Riffle[char, RandomChoice[type, 8]]], {1000000}];
>>> e = ToExpression[#] & /@ r;
>>> p = Position[e, 2014]
>>> Union[Extract[r, p]]
>>>
>>> 2015
>>> 1*2-3+4*567*8/9, 12*3+45*6*7+89, 12*34*5-6*7+8+9
>>>
>>>
>>>
>>>
>>>
>>>
>>> Am 24.12.2013 um 08:17 schrieb Harvey P. Dale:
>>>
>>>> 	There are some nice (very simple) math puzzles using consecutive integers that produce years.  For example, 10+(9 x 8 x (7/6) x 5 x 4)+321 and 0-12+(34 x 56)+7 x (8+9) both yield 2011, and (10 x 9 x 8) + 7 + 6 - 5 + (4 x 321) yields 2012.
>>>>
>>>> 	Two questions: (1) can anyone generate a similar puzzle yielding 2014 and (2) is there a general Mathematica program that can generate these?
>>>>
>>>> 	Best,
>>>>
>>>> 	Harvey
>>>>
>>>
>>>
>>
>>
>>
>





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