Simple cryptographic puzzle [closed] - cryptarithmetic-puzzle

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I'm looking for a way to solve this crypt arithmetic problem of:
ROBERT + GERALD = DONALD
and potentially others as well, where each letter represents a digit.
How would you go about solving this by hand and how does that relate to solving it programmatically?
Thank you in advance

You can actually work this out as a sum:
robert
+ gerald
------
= donald
and use basic mathematical knowledge.
For example, there's no carry possible in the right column (6) and we have T + D = D. That means T must be zero.
Similarly, for column 5, there's no carry from column 6 and R + L = L means R is zero as well, and no carry to column 4.
Same with column 4, E + A = A so E is zero.
So we now have:
0ob000
+ g00ald
------
= donald
From there, we can infer from columns 3 and 1 that b==n and g==d and they (along with o/a/l/d) can be any value since every digit is being added to zero so there is no chance of carry anywhere. So let's just make them all one:
011000
+ 100111
------
= 111111
In fact, you could make them all zero and end up with 000000 + 000000 = 000000.
But that's hardly programming related, so let's make it so:
#include <stdio.h>
int main (void) {
int robert, gerald, donald;
for (int r = 0; r < 10; r++) {
for (int o = 0; o < 10; o++) {
for (int b = 0; b < 10; b++) {
for (int e = 0; e < 10; e++) {
for (int t = 0; t < 10; t++) {
for (int g = 0; g < 10; g++) {
for (int a = 0; a < 10; a++) {
for (int l = 0; l < 10; l++) {
for (int d = 0; d < 10; d++) {
for (int n = 0; n < 10; n++) {
robert = r * 100000 + o * 10000 + b * 1000 + e * 100 + r * 10 + t;
gerald = g * 100000 + e * 10000 + r * 1000 + a * 100 + l * 10 + d;
donald = d * 100000 + o * 10000 + n * 1000 + a * 100 + l * 10 + d;
if (robert + gerald == donald) {
printf (" %06d\n", robert);
printf ("+ %06d\n", gerald);
printf (" ------\n");
printf ("= %06d\n", donald);
printf ("........\n");
}
}
}
}
}
}
}
}
}
}
}
return 0;
}
That will give you a whole host of solutions.
And, before you complain that you cannot have repeated digits, there is no solution if that's the case, since mathematically both T and R must be zero, as shown in the original reasoning above. And you can prove this empirically with:
#include <stdio.h>
int main (void) {
int robert, gerald, donald;
for (int r = 0; r < 10; r++) {
for (int o = 0; o < 10; o++) {
if (o==r) continue;
for (int b = 0; b < 10; b++) {
if ((b==r) || (b==o)) continue;
for (int e = 0; e < 10; e++) {
if ((e==r) || (e==o) || (e==b)) continue;
for (int t = 0; t < 10; t++) {
if ((t==r) || (t==o) || (t==b) || (t==e)) continue;
for (int g = 0; g < 10; g++) {
if ((g==r) || (g==o) || (g==b) || (g==e) || (g==t)) continue;
for (int a = 0; a < 10; a++) {
if ((a==r) || (a==o) || (a==b) || (a==e) || (a==t) || (a==g)) continue;
for (int l = 0; l < 10; l++) {
if ((l==r) || (l==o) || (l==b) || (l==e) || (l==t) || (l==g) || (l==a)) continue;
for (int d = 0; d < 10; d++) {
if ((d==r) || (d==o) || (d==b) || (d==e) || (d==t) || (d==g) || (d==a) || (d==l)) continue;
for (int n = 0; n < 10; n++) {
if ((n==r) || (n==o) || (n==b) || (n==e) || (n==t) || (n==g) || (n==a) || (n==l) || (n==d)) continue;
robert = r * 100000 + o * 10000 + b * 1000 + e * 100 + r * 10 + t;
gerald = g * 100000 + e * 10000 + r * 1000 + a * 100 + l * 10 + d;
donald = d * 100000 + o * 10000 + n * 1000 + a * 100 + l * 10 + d;
if (robert + gerald == donald) {
printf (" %06d\n", robert);
printf ("+ %06d\n", gerald);
printf (" ------\n");
printf ("= %06d\n", donald);
printf ("........\n");
}
}
}
}
}
}
}
}
}
}
}
return 0;
}
which outputs no solutions.
Now DONALD + GERALD = ROBERT, that's a different matter but you can solve that simply by modifying the code above slightly, making the if statement into:
if (donald + gerald == robert) {
printf (" %06d\n", donald);
printf ("+ %06d\n", gerald);
printf (" ------\n");
printf ("= %06d\n", robert);
printf ("........\n");
}
and you get the single solution:
526485
+ 197485
------
= 723970

Related

Programming on the Nintendo DS - issue with collisions

Sorry if this question has already been asked, but I have not managed to find any advice on the internet for my issue. I am currently trying to program a little game on the Nintendo DS, in which the player has to move a sprite (currently a square) until it reaches the exit. For this, I use a sprite I have included using a grit file, and also a background enabled in tiled mode. However, I am having a problem when it comes to checking if the sprite is going to collide with a wall. Here is the code I have both for the background configuration (where I declare the tiles and the map) and also for the sprite movements (I didn't add the condition for all cases yet, as it didn't work well) :
void configureMaze_Sub() {
int row, col;
for (row = 0; row < 24; row ++) {
for (col = 0; col < 32; col ++) {
BG_MAP_RAM_SUB(3)[row * 32 + col] = 1;
if (col == 15 && (row != 12 && row !=4 && row != 19)) {
BG_MAP_RAM_SUB(3)[row * 32 + col] = 0;
}
if ((row == 1 || row == 22) && (col > 2 && col < 29)) {
BG_MAP_RAM_SUB(3)[row * 32 + col] = 0;
}
if ((col == 3 || col == 28) && (row > 1 && row < 22 && row != 12)) {
BG_MAP_RAM_SUB(3)[row * 32 + col] = 0;
}
if ((row == 3 || row == 20) && (col > 4 && col < 27)) {
BG_MAP_RAM_SUB(3)[row * 32 + col] = 0;
}
if ((col == 5 || col == 26) && (row != 9 && row != 15 && row > 3 && row < 20)) {
BG_MAP_RAM_SUB(3)[row * 32 + col] = 0;
}
if (row == 8 && (col > 5 && col < 15)) {
BG_MAP_RAM_SUB(3)[row * 32 + col] = 0;
}
if (row == 16 && (col > 15 && col < 26)) {
BG_MAP_RAM_SUB(3)[row * 32 + col] = 0;
}
if ((row == 12) && (col > 5 && col < 26)) {
BG_MAP_RAM_SUB(3)[row * 32 + col] = 0;
}
}
}
}
void gameplayMaze() {
int x = 103, y = 41, keys;
int maze_success = 0;
while (maze_success == 0) {
scanKeys();
keys = keysHeld();
int xmod = x / 8;
int ymod = x / 8;
if ((keys & KEY_RIGHT) && BG_MAP_RAM_SUB(3)[xmod + 32 * ymod] == 1) {
x++;
printf("%d \n", x);
}
if ((keys & KEY_LEFT) && BG_MAP_RAM_SUB(3)[xmod + 32 * ymod] == 1) {
x--;
}
if (keys & KEY_UP) {
y--;
}
if (keys & KEY_DOWN) {
y++;
}
oamSet(&oamSub,
0,
x, y,
0,
0,
SpriteSize_8x8,
SpriteColorFormat_256Color,
gfxSub,
-1,
false,
false,
false, false,
false
);
swiWaitForVBlank();
oamUpdate(&oamSub);
}
The main problem I have is to try to change from the coordinates of the tiles (which are 8x8) to the ones of the map, as for the coordinates of the sprite (256x192). If any of you have any hint to help me, I would be very grateful! I am still new to programming on the NDS, so I am still struggling to get the hang of it.

How to print in the console that figure?

I want to print that in the console:
*
**
***
**
*
so, my code is:
Scanner input = new Scanner(System.in);
int n = input.nextInt();
char c = '*';
if (1 < n && n < 20) {
for (int row = 1; row <= n; row++) {
for (int col = 1; col <= row; col++) {
System.out.print(c);
}
System.out.println();
}
any proposals how to finish?
You want to print "* **". Write this:
System.out.print("* **");
If my answer is not that what you excpected, than give us more information. What is your actual problem. See StackOverflowFAQ
If n is your "*" length,below code:
if (1 < n && n < 20) {
for (int row = 1; row <= n; row++) {
for (int col = 1; col <= row; col++) {
System.out.print(c);
}
System.out.println();
}
for (int row =1; row <= n; row++) {
for (int col = n-row; col >0 ; col--) {
System.out.print(c);
}
System.out.println();
}
}

Why .hash section in my ELF file is not valid?

In a .hash section, for some x, if chain[x] != SHN_UNDEF,
it should hold hash(name(bucket[x])) === hash(name(bucket[chain[x]])) % nbucket
But why it's not the case for my shared object file?
For example, name(bucket[224]) == "_ZN9VADEnergyD0Ev" whose (ELF hash % nbucket) is 224,
name(bucket[8]) == "speex_bits_write_whole_bytes" whose (ELF hash % nbucket) is 8,
but chain[224] == 8.
(the file is avalible here)
Or my code for reading elf is wrong?
nbucket = ((int *)hash)[0];
nchain = ((int *)hash)[1];
memcpy(bucket, hash + 8, nbucket * 4);
memcpy(succ, hash + nbucket * 4 + 8, nchain * 4);
for (i = 0; i < nbucket; i++) {
printf("%d %d\n", bucket[i], succ[i]);
if (bucket[i] && succ[i])
pred[succ[i]] = i;
}
printf("%d %d\n", nbucket, nchain);
#define sym_name(x, symtbl, strtbl) (strtbl + symtbl[x].st_name)
for (i = 0; i < nbucket; i++) {
if (pred[i] == 0) {
printf("=======\n");
for (j = i; j; j = succ[j]) {
char *sname = sym_name(bucket[j], dynsym, dynstr);
printf("%d,succ=%d ", j, succ[j]);
printf("%d:%s\n", _dl_elf_hash(sname) % nbucket, sname);
}
}
}
It's my fault. It should be
hash(name(bucket[x])) === hash(name(chain[bucket[x]])) % nbucket
and
nbucket = ((int *)hash)[0];
nchain = ((int *)hash)[1];
memcpy(bucket, hash + 8, nbucket * 4);
memcpy(succ, hash + nbucket * 4 + 8, nchain * 4);
for (i = 0; i < nbucket; i++) {
printf("%d %d\n", bucket[i], succ[i]);
if (bucket[i] && succ[i])
pred[succ[i]] = i;
}
printf("%d %d\n", nbucket, nchain);
#define sym_name(x, symtbl, strtbl) (strtbl + symtbl[x].st_name)
for (i = 0; i < nbucket; i++) {
printf("=======\n");
for (j = bucket[i]; j; j = succ[j]) {
char *sname = sym_name(j, dynsym, dynstr);
printf("%d,succ=%d ", j, succ[j]);
printf("%d:%s\n", _dl_elf_hash(sname) % nbucket, sname);
}
}

Small differences in SHA1 hashes

A project I am working on uses Apache Shiro as a security framework. Passwords are SHA1 hashed (no salt, no iterations). Login is SSL secured. However, the remaining part of the application is not SSL secured. In this context (no SSL) there should be a form where a user can change the password.
Since it wouldn't be a good idea to transmit it plainly it should be hashed on the client and then transmitted to the server. As the client is GWT (2.3) based, I am trying this library http://code.google.com/p/gwt-crypto, which uses code from bouncycastle.
However, in many cases (not all) the hashes generated by both frameworks differ in 1-4(?) characters.
For instance "happa3" is hashed to
"fe7f3cffd8a5f0512a5f1120f1369f48cd6f47c2"
by both implementations, whereas just "happa" is hashed to
"fb3c3a741b4e07a87d9cb68f3db020d6fbfed00a"
by the Shiro implementation and to
"fb3c3a741b4e07a87d9cb63f3db020d6fbfed00a"
by the gwt-crypto implementation (23rd character differs).
I wonder whether there is a "correct"/standard SHA1 hashing and whether there is a bug in one of the libraries or maybe my usage of them is flawed.
One of my first thoughts was related to different encodings or strange conversions due to different transport mechanisms (RPC vs. Post). To my knowledge though (and what puzzles me most), SHA1 hashes should differ completely with a high probability if there is just a difference of a single bit. So different encodings shouldn't be the issue here.
I am using this code on the client (GWT) for hashing:
String hashed = toHex(createSHA1Hash("password"));
...
private String createSHA1Hash(String passwordString){
SHA1Digest sha1 = new SHA1Digest();
byte[] bytes;
byte[] result = new byte[sha1.getDigestSize()];
try {
bytes = passwordString.getBytes();
sha1.update(bytes, 0, bytes.length);
int val = sha1.doFinal(result, 0);
} catch (UnsupportedEncodingException e) {}
return new String(result);
}
public String toHex(String arg) {
return new BigInteger(1, arg.getBytes()).toString(16);
}
And this on the server (Shiro):
String hashed = new Sha1Hash("password").toHex()
which afaics does something very similar behind the scenes (had a quick view on the source code).
Did I miss something obvious here?
EDIT: Seems like the GWT code does not run natively for some reason (i.e. just in development mode) and silently fails (it does compile, though). Have to find out why...
Edit(2): "int val = sha1.doFinal(result, 0);" is the line that makes trouble, i.e. if present, the whole code does not run natively (JS) but only in dev-mode (with wrong results)
You could test this version:
public class SHA1 {
public static native String calcSHA1(String s) /*-{
//
// A JavaScript implementation of the Secure Hash Algorithm, SHA-1, as defined
// in FIPS 180-1
// Version 2.2 Copyright Paul Johnston 2000 - 2009.
// Other contributors: Greg Holt, Andrew Kepert, Ydnar, Lostinet
// Distributed under the BSD License
// See http://pajhome.org.uk/crypt/md5 for details.
//
//
// Configurable variables. You may need to tweak these to be compatible with
// the server-side, but the defaults work in most cases.
//
var hexcase = 0; // hex output format. 0 - lowercase; 1 - uppercase
var b64pad = ""; // base-64 pad character. "=" for strict RFC compliance
//
// These are the functions you'll usually want to call
// They take string arguments and return either hex or base-64 encoded strings
//
function b64_sha1(s) { return rstr2b64(rstr_sha1(str2rstr_utf8(s))); }
function any_sha1(s, e) { return rstr2any(rstr_sha1(str2rstr_utf8(s)), e); }
function hex_hmac_sha1(k, d)
{ return rstr2hex(rstr_hmac_sha1(str2rstr_utf8(k), str2rstr_utf8(d))); }
function b64_hmac_sha1(k, d)
{ return rstr2b64(rstr_hmac_sha1(str2rstr_utf8(k), str2rstr_utf8(d))); }
function any_hmac_sha1(k, d, e)
{ return rstr2any(rstr_hmac_sha1(str2rstr_utf8(k), str2rstr_utf8(d)), e); }
//
// Perform a simple self-test to see if the VM is working
//
function sha1_vm_test()
{
return hex_sha1("abc").toLowerCase() == "a9993e364706816aba3e25717850c26c9cd0d89d";
}
//
// Calculate the SHA1 of a raw string
//
function rstr_sha1(s)
{
return binb2rstr(binb_sha1(rstr2binb(s), s.length * 8));
}
//
// Calculate the HMAC-SHA1 of a key and some data (raw strings)
//
function rstr_hmac_sha1(key, data)
{
var bkey = rstr2binb(key);
if(bkey.length > 16) bkey = binb_sha1(bkey, key.length * 8);
var ipad = Array(16), opad = Array(16);
for(var i = 0; i < 16; i++)
{
ipad[i] = bkey[i] ^ 0x36363636;
opad[i] = bkey[i] ^ 0x5C5C5C5C;
}
var hash = binb_sha1(ipad.concat(rstr2binb(data)), 512 + data.length * 8);
return binb2rstr(binb_sha1(opad.concat(hash), 512 + 160));
}
//
// Convert a raw string to a hex string
//
function rstr2hex(input)
{
try { hexcase } catch(e) { hexcase=0; }
var hex_tab = hexcase ? "0123456789ABCDEF" : "0123456789abcdef";
var output = "";
var x;
for(var i = 0; i < input.length; i++)
{
x = input.charCodeAt(i);
output += hex_tab.charAt((x >>> 4) & 0x0F)
+ hex_tab.charAt( x & 0x0F);
}
return output;
}
//
// Convert a raw string to a base-64 string
//
function rstr2b64(input)
{
try { b64pad } catch(e) { b64pad=''; }
var tab = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
var output = "";
var len = input.length;
for(var i = 0; i < len; i += 3)
{
var triplet = (input.charCodeAt(i) << 16)
| (i + 1 < len ? input.charCodeAt(i+1) << 8 : 0)
| (i + 2 < len ? input.charCodeAt(i+2) : 0);
for(var j = 0; j < 4; j++)
{
if(i * 8 + j * 6 > input.length * 8) output += b64pad;
else output += tab.charAt((triplet >>> 6*(3-j)) & 0x3F);
}
}
return output;
}
//
// Convert a raw string to an arbitrary string encoding
//
function rstr2any(input, encoding)
{
var divisor = encoding.length;
var remainders = Array();
var i, q, x, quotient;
// Convert to an array of 16-bit big-endian values, forming the dividend
var dividend = Array(Math.ceil(input.length / 2));
for(i = 0; i < dividend.length; i++)
{
dividend[i] = (input.charCodeAt(i * 2) << 8) | input.charCodeAt(i * 2 + 1);
}
//
// Repeatedly perform a long division. The binary array forms the dividend,
// the length of the encoding is the divisor. Once computed, the quotient
// forms the dividend for the next step. We stop when the dividend is zero.
// All remainders are stored for later use.
//
while(dividend.length > 0)
{
quotient = Array();
x = 0;
for(i = 0; i < dividend.length; i++)
{
x = (x << 16) + dividend[i];
q = Math.floor(x / divisor);
x -= q * divisor;
if(quotient.length > 0 || q > 0)
quotient[quotient.length] = q;
}
remainders[remainders.length] = x;
dividend = quotient;
}
// Convert the remainders to the output string
var output = "";
for(i = remainders.length - 1; i >= 0; i--)
output += encoding.charAt(remainders[i]);
// Append leading zero equivalents
var full_length = Math.ceil(input.length * 8 /
(Math.log(encoding.length) / Math.log(2)))
for(i = output.length; i < full_length; i++)
output = encoding[0] + output;
return output;
}
//
// Encode a string as utf-8.
// For efficiency, this assumes the input is valid utf-16.
//
function str2rstr_utf8(input)
{
var output = "";
var i = -1;
var x, y;
while(++i < input.length)
{
// Decode utf-16 surrogate pairs
x = input.charCodeAt(i);
y = i + 1 < input.length ? input.charCodeAt(i + 1) : 0;
if(0xD800 <= x && x <= 0xDBFF && 0xDC00 <= y && y <= 0xDFFF)
{
x = 0x10000 + ((x & 0x03FF) << 10) + (y & 0x03FF);
i++;
}
// Encode output as utf-8
if(x <= 0x7F)
output += String.fromCharCode(x);
else if(x <= 0x7FF)
output += String.fromCharCode(0xC0 | ((x >>> 6 ) & 0x1F),
0x80 | ( x & 0x3F));
else if(x <= 0xFFFF)
output += String.fromCharCode(0xE0 | ((x >>> 12) & 0x0F),
0x80 | ((x >>> 6 ) & 0x3F),
0x80 | ( x & 0x3F));
else if(x <= 0x1FFFFF)
output += String.fromCharCode(0xF0 | ((x >>> 18) & 0x07),
0x80 | ((x >>> 12) & 0x3F),
0x80 | ((x >>> 6 ) & 0x3F),
0x80 | ( x & 0x3F));
}
return output;
}
//
// Encode a string as utf-16
//
function str2rstr_utf16le(input)
{
var output = "";
for(var i = 0; i < input.length; i++)
output += String.fromCharCode( input.charCodeAt(i) & 0xFF,
(input.charCodeAt(i) >>> 8) & 0xFF);
return output;
}
function str2rstr_utf16be(input)
{
var output = "";
for(var i = 0; i < input.length; i++)
output += String.fromCharCode((input.charCodeAt(i) >>> 8) & 0xFF,
input.charCodeAt(i) & 0xFF);
return output;
}
//
// Convert a raw string to an array of big-endian words
// Characters >255 have their high-byte silently ignored.
//
function rstr2binb(input)
{
var output = Array(input.length >> 2);
for(var i = 0; i < output.length; i++)
output[i] = 0;
for(var i = 0; i < input.length * 8; i += 8)
output[i>>5] |= (input.charCodeAt(i / 8) & 0xFF) << (24 - i % 32);
return output;
}
//
// Convert an array of big-endian words to a string
//
function binb2rstr(input)
{
var output = "";
for(var i = 0; i < input.length * 32; i += 8)
output += String.fromCharCode((input[i>>5] >>> (24 - i % 32)) & 0xFF);
return output;
}
//
// Calculate the SHA-1 of an array of big-endian words, and a bit length
//
function binb_sha1(x, len)
{
// append padding
x[len >> 5] |= 0x80 << (24 - len % 32);
x[((len + 64 >> 9) << 4) + 15] = len;
var w = Array(80);
var a = 1732584193;
var b = -271733879;
var c = -1732584194;
var d = 271733878;
var e = -1009589776;
for(var i = 0; i < x.length; i += 16)
{
var olda = a;
var oldb = b;
var oldc = c;
var oldd = d;
var olde = e;
for(var j = 0; j < 80; j++)
{
if(j < 16) w[j] = x[i + j];
else w[j] = bit_rol(w[j-3] ^ w[j-8] ^ w[j-14] ^ w[j-16], 1);
var t = safe_add(safe_add(bit_rol(a, 5), sha1_ft(j, b, c, d)),
safe_add(safe_add(e, w[j]), sha1_kt(j)));
e = d;
d = c;
c = bit_rol(b, 30);
b = a;
a = t;
}
a = safe_add(a, olda);
b = safe_add(b, oldb);
c = safe_add(c, oldc);
d = safe_add(d, oldd);
e = safe_add(e, olde);
}
return Array(a, b, c, d, e);
}
//
// Perform the appropriate triplet combination function for the current
// iteration
//
function sha1_ft(t, b, c, d)
{
if(t < 20) return (b & c) | ((~b) & d);
if(t < 40) return b ^ c ^ d;
if(t < 60) return (b & c) | (b & d) | (c & d);
return b ^ c ^ d;
}
//
// Determine the appropriate additive constant for the current iteration
//
function sha1_kt(t)
{
return (t < 20) ? 1518500249 : (t < 40) ? 1859775393 :
(t < 60) ? -1894007588 : -899497514;
}
//
// Add integers, wrapping at 2^32. This uses 16-bit operations internally
// to work around bugs in some JS interpreters.
//
function safe_add(x, y)
{
var lsw = (x & 0xFFFF) + (y & 0xFFFF);
var msw = (x >> 16) + (y >> 16) + (lsw >> 16);
return (msw << 16) | (lsw & 0xFFFF);
}
//
// Bitwise rotate a 32-bit number to the left.
//
function bit_rol(num, cnt)
{
return (num << cnt) | (num >>> (32 - cnt));
}
return rstr2hex(rstr_sha1(str2rstr_utf8(s)));
}-*/;
}
I'm using it in my client side sha generation and it worked well.

form a number using consecutive numbers

I was puzzled with one of the question in Microsoft interview which is as given below:
A function should accept a range( 3 - 21 ) and it should print all the consecutive numbers combinations to form each number as given below:
3 = 1+2
5 = 2+3
6 = 1+2+3
7 = 3+4
9 = 4+5
10 = 1+2+3+4
11 = 5+6
12 = 3+4+5
13 = 6+7
14 = 2+3+4+5
15 = 1+2+3+4+5
17 = 8+9
18 = 5+6+7
19 = 9+10
20 = 2+3+4+5+6
21 = 10+11
21 = 1+2+3+4+5+6
could you please help me in forming this sequence in C#?
Thanks,
Mahesh
So here is a straightforward/naive answer (in C++, and not tested; but you should be able to translate). It uses the fact that
1 + 2 + ... + n = n(n+1)/2,
which you have probably seen before. There are lots of easy optimisations that can be made here which I have omitted for clarity.
void WriteAsSums (int n)
{
for (int i = 0; i < n; i++)
{
for (int j = i; j < n; j++)
{
if (n = (j * (j+1) - i * (i+1))/2) // then n = (i+1) + (i+2) + ... + (j-1) + j
{
std::cout << n << " = ";
for (int k = i + 1; k <= j; k++)
{
std::cout << k;
if (k != j) // this is not the interesting bit
std::cout << std::endl;
else
std::cout << " + ";
}
}
}
}
}
This is some pseudo code to find all the combinations if any exists:
function consecutive_numbers(n, m)
list = [] // empty list
list.push_back(m)
while m != n
if m > n
first = list.remove_first
m -= first
else
last = list.last_element
if last <= 1
return []
end
list.push_back(last - 1)
m += last - 1
end
end
return list
end
function all_consecutive_numbers(n)
m = n / 2 + 1
a = consecutive_numbers(n, m)
while a != []
print_combination(n, a)
m = a.first - 1
a = consecutive_numbers(n, m)
end
end
function print_combination(n, a)
print(n + " = ")
print(a.remove_first)
foreach element in a
print(" + " + element)
end
print("\n")
end
A call to all_consecutive_numbers(21) would print:
21 = 11 + 10
21 = 8 + 7 + 6
21 = 6 + 5 + 4 + 3 + 2 + 1
I tested it in ruby (code here) and it seems to work. I'm sure the basic idea could easily be implemented in C# as well.
I like this problem. Here is a slick and slightly mysterious O(n) solution:
void DisplaySum (int n, int a, int b)
{
std::cout << n << " = ";
for (int i = a; i < b; i++) std::cout << i << " + ";
std::cout << b;
}
void WriteAsSums (int n)
{
N = 2*n;
for (int i = 1; i < N; i++)
{
if (~(N%i))
{
int j = N/i;
if (j+i%2)
{
int a = (j+i-1)/2;
int b = (j-i+1)/2;
if (a>0 & a<b) // exclude trivial & negative solutions
DisplaySum(n,a,b);
}
}
}
}
Here's something in Groovy, you should be able to understand what's going on. It's not the most efficient code and doesn't create the answers in the order you cite in your question (you seem to be missing some though) but it might give you a start.
def f(a,b) {
for (i in a..b) {
for (j in 1..i/2) {
def (sum, str, k) = [ 0, "", j ]
while (sum < i) {
sum += k
str += "+$k"
k++
}
if (sum == i) println "$i=${str[1..-1]}"
}
}
}
Output for f(3,21) is:
3=1+2
5=2+3
6=1+2+3
7=3+4
9=2+3+4
9=4+5
10=1+2+3+4
11=5+6
12=3+4+5
13=6+7
14=2+3+4+5
15=1+2+3+4+5
15=4+5+6
15=7+8
17=8+9
18=3+4+5+6
18=5+6+7
19=9+10
20=2+3+4+5+6
21=1+2+3+4+5+6
21=6+7+8
21=10+11
Hope this helps. It kind of conforms to the tenet of doing the simplest thing that could possibly work.
if we slice a into 2 digit, then a = b + (b+1) = 2*b + (0+1)
if we slice a into 3 digit, then a = b + (b+1) + (b+2) = 3*b + (0+1+2)
...
if we slice a into n digit, then a = b + (b+1) +...+ (b+n) = nb + (0+1+n-1)
the last result is a = nb + n*(n-1)/2, a,b,n are all ints.
so O(N) Algorithm is:
void seq_sum(int a)
{
// start from 2 digits
int n=2;
while(1)
{
int value = a-n*(n-1)/2;
if(value < 0)
break;
// meet the quotation we deduct
if( value%n == 0 )
{
int b=value/n;
// omit the print stage
print("......");
}
n++;
}
}