Combinations with repetitions/Square digit chain: Difference between revisions
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=={{header|Ruby}}== |
=={{header|Ruby}}== |
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<lang ruby> |
<lang ruby> |
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# Count how many number chains for Natural Numbers < |
# Count how many number chains for Natural Numbers < 10**K end with a value of 1. |
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# |
# |
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# Nigel_Galloway |
# Nigel_Galloway |
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# August 26th., 2014. |
# August 26th., 2014. |
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K = 17 |
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require 'benchmark' |
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D = 8 #Calculate from 1 to 10**D (8 for task) |
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g = -> n, gn=[n,0], res=0 { while gn[0]>0 |
g = -> n, gn=[n,0], res=0 { while gn[0]>0 |
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gn = gn[0].divmod(10) |
gn = gn[0].divmod(10) |
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end |
end |
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return res==89?0:res |
return res==89?0:res |
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} |
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#An array: N[n]==1 means that n translates to 1, 0 means that it does not. |
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N = (G=Array.new( |
N = (G=Array.new(K*81+1){|n| n==0? 0:(i=g.call(n))==89 ? 0:i}).collect{|n| while n>1 do n = G[n] end; n } |
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z = 0 #Running count of numbers translating to 1 |
z = 0 #Running count of numbers translating to 1 |
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t = Benchmark.measure do |
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next if N[n.inject(:+)] == 0 #Count only ones |
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nn = Hash.new{0} #Determine how many numbers this digit combination corresponds to |
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nn = |
n.each{|n| nn[n] += 1} #and |
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z += nn.values.inject(F[K]){|gn,n| gn/F[n]} #Add to the count of numbers terminating in 1 |
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z += nn.values.inject(F[D]){|gn,n| gn/F[n]}#Add to the count of numbers terminating in 1 |
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} |
} |
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end |
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puts "\n\nTiming\n#{t}" |
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</lang> |
</lang> |
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{{out}} |
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<pre> |
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#(k=7) in the range 1 to 9999999 |
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#1418853 numbers produce 1 and 8581146 numbers produce 89 |
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#(k=8) in the range 1 to 99999999 |
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#14255666 numbers produce 1 and 85744333 numbers produce 89 |
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#(k=11) in the range 1 to 99999999999 |
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#15091199356 numbers produce 1 and 84908800643 numbers produce 89 |
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#(k=14) in the range 1 to 99999999999999 |
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#13770853279684 numbers produce 1 and 86229146720315 numbers produce 89 |
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#(k=17) in the range 1 to 99999999999999999 |
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#12024696404768024 numbers produce 1 and 87975303595231975 numbers produce 89 |
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</pre> |
Revision as of 13:28, 17 September 2014
Iterated digits squaring introduces RC the Project Euler Task #92. Combinations with repetitions introduce RC to the concept of generating all the combinations with repetitions of n types of things taken k at a time.
The purpose of this task is to combine these tasks as follows:
- The collections of k items will be taken from [0,1,4,9,16,25,36,49,64,81] and must be obtained using code from Combinations with repetitions. The collection of k zeroes is excluded.
- For each collection of k items determine if it translates to 1 using the rules from Iterated digits squaring
- For each collection which translates to 1 determine the number of different ways, c say, in which the k items can be uniquely ordered.
- Keep a running total of all the values of c obtained
- Answer the Project Euler Task #92 question (k=7).
- Answer the equivalent question for k=8,11,14.
- Optionally answer the question for k=17. These numbers will be larger than the basic integer type for many languages, if it is not easy to use larger numbers it is not necessary for this task.
Ruby
<lang ruby>
- Count how many number chains for Natural Numbers < 10**K end with a value of 1.
- Nigel_Galloway
- August 26th., 2014.
K = 17 F = Array.new(K+1){|n| n==0?1:(1..n).inject(:*)} #Some small factorials g = -> n, gn=[n,0], res=0 { while gn[0]>0
gn = gn[0].divmod(10) res += gn[1]**2 end return res==89?0:res }
- An array: N[n]==1 means that n translates to 1, 0 means that it does not.
N = (G=Array.new(K*81+1){|n| n==0? 0:(i=g.call(n))==89 ? 0:i}).collect{|n| while n>1 do n = G[n] end; n } z = 0 #Running count of numbers translating to 1 (0..9).collect{|n| n**2}.repeated_combination(K).each{|n| #Iterate over unique digit combinations
next if N[n.inject(:+)] == 0 #Count only ones nn = Hash.new{0} #Determine how many numbers this digit combination corresponds to n.each{|n| nn[n] += 1} #and z += nn.values.inject(F[K]){|gn,n| gn/F[n]} #Add to the count of numbers terminating in 1
} puts "\nk=(#{K}) in the range 1 to #{10**K-1}\n#{z} numbers produce 1 and #{10**K-1-z} numbers produce 89" </lang>
- Output:
#(k=7) in the range 1 to 9999999 #1418853 numbers produce 1 and 8581146 numbers produce 89 #(k=8) in the range 1 to 99999999 #14255666 numbers produce 1 and 85744333 numbers produce 89 #(k=11) in the range 1 to 99999999999 #15091199356 numbers produce 1 and 84908800643 numbers produce 89 #(k=14) in the range 1 to 99999999999999 #13770853279684 numbers produce 1 and 86229146720315 numbers produce 89 #(k=17) in the range 1 to 99999999999999999 #12024696404768024 numbers produce 1 and 87975303595231975 numbers produce 89