project euler: problem 62

R on Guangchuang Yu

9 years ago

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The cube, 41063625 (345³), can be permuted to produce two other cubes: 56623104 (384³) and 66430125 (405³). In fact, 41063625 is the smallest cube which has exactly three permutations of its digits which are also cube.Find the smallest cube for which exactly five permutations of its digits are cube.

—

I tried to generate all the cubic number with specific length, and iterate until exactly five numbers have the same digits.

^?View Code RSPLUS

getCubicNumber <- function(ndigit) {
    lower <- floor(10^((ndigit-1)/3))
    upper <- floor(10^(ndigit/3))
    cube <- (lower:upper)^3
    return(cube)
}
 
cubicPermutation <- function(nperm) {
    ndigit <- 1
    flag <- TRUE
    while(flag) {
        ndigit <- ndigit+1
        cubeNumber <- getCubicNumber(ndigit)
        cube <- sapply(as.character(cubeNumber), function(i)
                       paste(sort(unlist(strsplit(i, split=""))), collapse=""))
 
        cnt <- table(cube)
 
        d <- names(cnt[cnt == nperm])
        if (length(d)) {
            permDigits <- lapply(d, function(i) names(cube[cube==i]))
            res <- min(sapply(permDigits, min))
            flag <- FALSE
        }
    }
    return(res)
}
 
cat("Answer of PE 62: ", cubicPermutation(5), "\n")

This code runs in half a second.

> system.time(source("problem62.R"))
Answer of PE 62:  127035954683 
   user  system elapsed 
  0.414   0.001   0.416

October 26, 2012 — permutation test (0)
November 8, 2011 — project euler – Problem 44 (1)
November 8, 2011 — project euler-Problem 43 (0)
November 8, 2011 — project euler – Problem 32 (1)
November 7, 2011 — project euler-Problem 41 (1)

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